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a. Iron metabolism: Transferrin (Tf) is a plasma protein that transports iron. Tf binds to cell-surface receptors (transferrin receptors [TfR]) that get internalized and provide cells, such as erythroblasts, with iron. The mRNA for the TfR has several cis-acting iron-responsive elements (IRE) in its 3′-UTR. IRE have a short stem-loop structure that can be bound by trans-acting iron regulatory proteins (IRP), as shown in Figure 33.14. When the iron concentration in the cell is low, the IRP bind to the 3′-IRE and stabilize the mRNA for TfR, allowing TfR synthesis. When intracellular iron levels are high, the IRP dissociate. The lack of IRP bound to the mRNA hastens its destruction, resulting in decreased TfR synthesis. [Note: The mRNA for ferritin, an intracellular protein of iron storage, has a single IRE in its 5′-UTR. When iron levels in the cell are low, IRP bind the 5′-IRE and prevent the use of the mRNA, and less ferritin is made. When iron accumulates in the cell, the IRP dissociate, allowing synthesis of ferritin molecules to store the excess iron. Aminolevulinic acid synthase 2, the regulated enzyme of heme synthesis (see p. 278) in erythroblasts, also contains a 5′-IRE.] (See Chapter 29 for a discussion of iron metabolism.) b. RNA interference: RNAi is a mechanism of gene silencing through decreased expression of mRNA, either by repression of translation or by increased degradation. It plays a key role in such fundamental processes as cell proliferation, differentiation, and apoptosis. RNAi is mediated by short (~22 nucleotides), noncoding RNA called microRNA (miRNA). The miRNA arise from far longer, genomically encoded nuclear transcripts, primary miRNA (pri-miRNA), that are partially processed in the nucleus to pre-miRNA by an endonuclease (Drosha) then transported to the cytoplasm. There, an endonuclease (Dicer) completes the processing and generates short, double-stranded miRNA. A single strand (the guide or antisense strand) of the miRNA associates with a cytosolic protein complex known as the RNA-induced silencing complex (RISC). The guide strand hybridizes with a complementary sequence in the 3′-UTR of a full-length target mRNA, bringing RISC to the mRNA. This can result in repression of translation of the mRNA or its degradation by an endonuclease (Argonaute/Ago/Slicer) of the RISC. The extent of complementarity appears to be the determining factor (Fig. 33.15). RNAi can also be triggered by the introduction of exogenous double-stranded short interfering RNA (siRNA) into a cell, a process that has enormous therapeutic potential.
The first clinical trial of RNAi-based therapy involved the neovascular form of age-related macular degeneration (AMD), which is triggered by overproduction of vascular endothelial growth factor (VEGF), leading to the sprouting of excess blood vessels behind the retina. The vessels leak, clouding and often entirely destroying vision (therefore, neovascular AMD is also referred to as wet AMD). An siRNA was designed to target the mRNA of VEGF and promote its degradation. Although considerable effort and resources have been expended to develop RNAi-based therapeutics, especially for the treatment of cancer, no products have gone from trials to the market. The research applications of RNAi, however, have grown rapidly.
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5. Messenger RNA translation: Regulation of gene expression can also occur at the level of mRNA translation. One mechanism by which translation is regulated is through phosphorylation of the eukaryotic translation initiation factor, eIF-2 (Fig. 33.16). Phosphorylation of eIF-2 inhibits its function and so inhibits translation at the initiation step (see p. 459). [Note: Phosphorylation of eIF-2 prevents its reactivation by inhibiting GDP-GTP exchange.] Phosphorylation is catalyzed by kinases that are activated in response to environmental conditions, such as amino acid starvation, heme deficiency in erythroblasts, the presence of double-stranded RNA (signaling viral infection), and the accumulation of misfolded proteins in the rough endoplasmic reticulum (see p. 460).
phosphate; = phosphate.
C. Regulation through variations in DNA
Gene expression in eukaryotes is also influenced by the accessibility of DNA to the transcriptional apparatus, the amount of DNA, and the arrangement of DNA. [Note: Localized transitions between the B and Z forms of DNA (see p. 414) can also affect gene expression.] 1.
Access to DNA: In eukaryotes, DNA is found complexed with histone and nonhistone proteins to form chromatin (see p. 425). Transcriptionally active, decondensed chromatin (euchromatin) differs from the more condensed, inactive form (heterochromatin) in a number of ways. Active chromatin contains histone proteins that have been covalently modified at their amino terminal ends by reversible methylation, acetylation, or phosphorylation (see p. 438 for a discussion of histone acetylation/deacetylation by histone acetyltransferase and histone deacetylase). Such modifications decrease the positive charge of these basic proteins, thereby decreasing the strength of their association with negatively charged DNA. This relaxes the nucleosome (see p. 425), allowing transcription factors access to specific regions on the DNA. Nucleosomes can also be repositioned, an ATP-requiring process that is part of chromatin remodeling. Another difference between transcriptionally active and inactive chromatin is the extent of methylation of cytosine bases in CG-rich regions (CpG islands) in the promoter region of many genes. Methylation is by methyltransferases that use S-adenosylmethionine as the methyl donor (Fig. 33.17). Transcriptionally active genes are less methylated (hypomethylated) than their inactive counterparts, suggesting that DNA hypermethylation silences gene expression. Modification of histones and methylation of DNA are epigenetic in that they are heritable changes in DNA that alter gene expression without altering the base sequence.
2.
Amount of DNA: A change up or down in the number of copies of a gene can affect the amount of gene product produced. An increase in copy number (gene amplification) has contributed to increased genomic complexity and is still a normal developmental process in certain nonmammalian species. In mammals, however, gene amplification is seen with some diseases and in response to particular chemotherapeutic drugs such as methotrexate, an inhibitor of the enzyme dihydrofolate reductase (DHFR), required for the synthesis of thymidine triphosphate (TTP) in the pyrimidine biosynthetic pathway (see p. 303). TTP is essential for DNA synthesis. Gene amplification results in an increase in the number of DHFR genes and resistance to the drug, allowing TTP to be made.
3.
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Arrangement of DNA: The process by which immunoglobulins (antibodies) are produced by B lymphocytes involves permanent rearrangements of the DNA in these cells. The immunoglobulins (for example, IgG) consist of two light and two heavy chains, with each chain containing regions of variable and constant amino acid sequence. The variable region is the result of somatic recombination of segments within both the light-and the heavy-chain genes. During B-lymphocyte development, single variable (V), diversity (D), and joining (J) gene segments are brought together through gene rearrangement to form a unique variable region (Fig. 33.18). This process allows the generation of 109−1011 different immunoglobulins from a single gene, providing the diversity needed for the recognition of an enormous number of antigens. [Note: Pathologic DNA rearrangement is seen with translocation, a process by which two different chromosomes exchange DNA segments.] 4.
Mobile DNA elements: Transposons (Tn) are mobile segments of DNA that move in an essentially random manner from one site to another on the same or a different chromosome. Movement is mediated by transposase, an enzyme encoded by the Tn itself. Movement can be direct, in which transposase cuts out and then inserts the Tn at a new site, or replicative, in which the Tn is copied and the copy inserted elsewhere while the original remains in place. In eukaryotes, including humans, replicative transposition frequently involves an RNA intermediate made by a reverse transcriptase (see p. 424), in which case the Tn is called a retrotransposon. Transposition has contributed to structural variation in the genome but also has the potential to alter gene expression and even to cause disease. Tn comprise ~50% of the human genome, with retrotransposons accounting for 90% of Tn. Although the vast majority of these retrotransposons have lost the ability to move, some are still active. Their transposition is thought to be the basis for some rare cases of hemophilia A and Duchenne muscular dystrophy. [Note: The growing problem of antibiotic-resistant bacteria is a consequence, at least in part, of the exchange of plasmids among bacterial cells. If the plasmids contain Tn-carrying antibiotic resistance genes, the recipient bacteria gain resistance to one or more antimicrobial drugs.]
V. CHAPTER SUMMARY
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Gene expression results in the production of a functional gene product (either RNA or protein) through the processes of transcription and translation (Fig. 33.19). Genes can be either constitutive (always expressed, housekeeping genes) or regulated (expressed only under certain conditions in all cells or in a subset of cells). The ability to appropriately induce (positively regulate) or repress (negatively regulate) genes is essential in all organisms. Regulation of gene expression occurs primarily at transcription in both prokaryotes and eukaryotes and is mediated through trans-acting proteins binding to cis-acting regulatory DNA elements. In eukaryotes, regulation also occurs through DNA modifications and through posttranscriptional and posttranslational processing. In prokaryotes, such as Escherichia coli, the coordinate regulation of genes whose protein products are required for a particular process is achieved through operons (groups of genes sequentially arranged on the chromosome along with the regulatory elements that determine their transcription). The lac operon contains the Z, Y, and A structural genes, the protein products of which are needed for the catabolism of lactose. It is subject to negative and positive regulation. When glucose is available, the operon is repressed by the binding of the repressor protein (the product of the lacI gene) to the operator, thus preventing transcription. When only lactose is present, the operon is induced by an isomer of lactose (allolactose) that binds the repressor protein, preventing it from binding to the operator. In addition, cyclic adenosine monophosphate (cAMP) binds the catabolite activator protein (CAP), and the complex binds the DNA at the CAP site. This increases promoter efficiency and results in the expression of the structural genes through the production of a polycistronic messenger RNA (mRNA). When both glucose and lactose are present, glucose prevents formation of cAMP, and transcription of these genes is negligible. The trp operon contains genes needed for the synthesis of tryptophan (Trp), and, like the lac operon, it is regulated by negative control. Unlike the lac operon, it is also regulated by attenuation, in which mRNA synthesis that escaped repression by Trp is terminated before completion. Transcription of ribosomal RNA and transfer RNA is selectively inhibited in prokaryotes by the stringent response to amino acid starvation. Translation is also a site of prokaryotic gene regulation: Excess ribosomal proteins bind the Shine-Dalgarno sequence on their own polycistronic mRNA, preventing ribosomes from binding. Gene regulation is more complex in eukaryotes. Operons are not present, but coordinate regulation of the transcription of genes located on different chromosomes can be achieved through the binding of trans-acting proteins to cis-acting elements as seen in the galactose circuit in unicellular yeast. In multicellular organisms, hormones can cause coordinated regulation, either through the binding of the hormone receptor–hormone complex to the DNA (as with steroid hormones) or through the binding of a protein that is activated in response to a second messenger (as with glucagon). In each case, binding to DNA is mediated through structural motifs such as the zinc finger. Co-and posttranscriptional regulation is also seen in eukaryotes and includes alternative mRNA splicing and polyadenylation, mRNA editing, and variations in mRNA stability as seen with transferrin receptor synthesis and with RNA interference. Regulation at the translational level can be caused by the phosphorylation and inhibition of eukaryotic initiation factor 2. Gene expression in eukaryotes is also influenced by accessibility of DNA to the transcriptional apparatus (as seen with epigenetic changes to histone proteins), the amount of DNA, and the arrangement of the DNA.
reticulum.
Choose the ONE best answer.
3.1. Which of the following mutations is most likely to result in reduced expression of the lac operon?
A. cya− (no adenylyl cyclase made)
B. i− (no repressor protein made)
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C. Oc (operator cannot bind repressor protein)
D. One resulting in impaired glucose uptake
Correct answer = A. In the absence of glucose, adenylyl cyclase makes cyclic adenosine monophosphate (cAMP), which forms a complex with the catabolite activator protein (CAP). The cAMP–CAP complex binds the CAP site on the DNA, causing RNA polymerase to bind more efficiently to the lac operon promoter, thereby increasing expression of the operon. With cya− mutations, adenylyl cyclase is not made, and so the operon is unable to be maximally expressed even when glucose is absent and lactose is present. The absence of a repressor protein or decreased ability of the repressor to bind the operator results in constitutive (essentially constant) expression of the lac operon.
3.2. Which of the following is best described as cis-acting?
A. Cyclic adenosine monophosphate response element–binding protein
B. Operator
C. Repressor protein
D. Thyroid hormone nuclear receptor
Correct answer = B. The operator is part of the DNA itself, and so is cis-acting. The cyclic adenosine monophosphate response element–binding protein, repressor protein, and thyroid hormone nuclear receptor protein are molecules that diffuse (transit) to the DNA, bind, and affect the expression of that DNA and so are trans-acting.
3.3. Which of the following is the basis for the intestine-specific expression of apolipoprotein B-48?
A. DNA rearrangement and loss
B. DNA transposition
C. RNA alternative splicing
D. RNA editing
E. RNA interference
Correct answer = D. The production of apolipoprotein (apo) B-48 in the intestine and apo B-100 in liver is the result of RNA editing in the intestine, where a sense codon is changed to a nonsense codon by posttranscriptional deamination of cytosine to uracil. DNA rearrangement and transposition, as well as RNA interference and alternative splicing, do alter gene expression but are not the basis of apo B-48 tissue-specific production.
3.4. Which of the following is most likely to be true in hemochromatosis, a disease of iron accumulation?
A. The messenger RNA for the transferrin receptor is stabilized by the binding of iron regulatory proteins to its 3′-iron-responsive elements.
B. The messenger RNA for the transferrin receptor is not bound by iron regulatory proteins and is degraded.
C. The messenger RNA for ferritin is not bound by iron regulatory proteins at its 5′-iron-responsive element and is translated.
D. The messenger RNA for ferritin is bound by iron regulatory proteins and is not translated.
E. Both B and C are correct.
Correct answer = E. When iron levels in the body are high, as is seen with hemochromatosis, there is increased synthesis of the iron-storage molecule, ferritin, and decreased synthesis of the transferrin receptor (TfR) that mediates iron uptake by cells. These effects are the result of cis-acting iron-responsive elements not being bound by trans-acting iron regulatory proteins, resulting in degradation of the messenger RNA (mRNA) for TfR and increased translation of the mRNA for ferritin.
3.5. Patients with estrogen receptor–positive (hormone responsive) breast cancer may be treated with the drug tamoxifen, which binds the estrogen nuclear receptor without activating it. Which of the following is the most logical outcome of tamoxifen use?
A. Increased acetylation of estrogen-responsive genes
B. Increased growth of estrogen receptor–positive breast cancer cells
C. Increased production of cyclic adenosine monophosphate
D. Inhibition of the estrogen operon
E. Inhibition of transcription of estrogen-responsive genes
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Correct answer = E. Tamoxifen competes with estrogen for binding to the estrogen nuclear receptor. Tamoxifen fails to activate the receptor, preventing its binding to DNA sequences that upregulate expression of estrogen-responsive genes. Tamoxifen, then, blocks the growth-promoting effects of these genes and results in growth inhibition of estrogen-dependent breast cancer cells. Acetylation increases transcription by relaxing the nucleosome. Cyclic adenosine monophosphate is a regulatory signal mediated by cell-surface rather than nuclear receptors. Mammalian cells do not have operons.
3.6. The ZYA region of the lac operon will be maximally expressed if:
A. cyclic adenosine monophosphate levels are low.
B. glucose and lactose are both available.
C. the attenuation stem-loop is able to form.
D. the CAP site is occupied.
Correct answer = D. It is only when glucose is gone, cyclic adenosine monophosphate (cAMP) levels are increased, the cAMP–catabolite activator protein (CAP) complex is bound to the CAP site, and lactose is available that the operon is maximally expressed (induced). If glucose is present, the operon is off as a result of catabolite repression. The lac operon is not regulated by attenuation, a mechanism for stopping transcription in some operons such as the trp operon.
3.7. X chromosome inactivation is a process by which one of two X chromosomes in mammalian females is condensed and inactivated to prevent overexpression of X-linked genes. What would most likely be true about the degree of DNA methylation and histone acetylation on the inactivated X chromosome?
Cytosines in CpG islands would be hypermethylated, and histone proteins would be deacetylated. Both conditions are associated with decreased gene expression, and both are important in maintaining X inactivation.
For additional ancillary materials related to this chapter, please visit thePoint.
I. OVERVIEW
In the past, efforts to understand genes and their expression have been confounded by the immense size and complexity of human deoxyribonucleic acid (DNA). The human genome contains ~3 billion (109) base pairs (bp) that encode 20,000–25,000 protein-coding genes located on 23 chromosomes in the haploid genome. It is now possible to determine the nucleotide sequence of long stretches of DNA, and the entire human genome has been sequenced. This effort (called the Human Genome Project and completed in 2003) was made possible by several tools that have already contributed to our understanding of many genetic diseases (Fig. 34.1). These include 1) the discovery of restriction endonucleases that permit the cleavage of huge DNA molecules into defined fragments, 2) the development of cloning techniques that provide a mechanism for amplification of specific nucleotide sequences, and 3) the ability to synthesize specific probes, which has allowed the identification and manipulation of nucleotide sequences of interest. These and other experimental approaches have permitted the identification of both normal and mutant nucleotide sequences in DNA. This knowledge has led to the development of methods for the diagnosis of genetic diseases and some successes in the treatment of patients by gene therapy. [Note: The genomes of several viruses, prokaryotes, and nonhuman eukaryotes have also been sequenced.]
II. RESTRICTION ENDONUCLEASES
One of the major obstacles to molecular analysis of genomic DNA is the immense size of the molecules involved. The discovery of a special group of bacterial enzymes, called restriction endonucleases (restriction enzymes), which cleave double-stranded DNA (dsDNA) into smaller, more manageable fragments, opened the way for DNA analysis. Because each enzyme cleaves dsDNA at a specific nucleotide sequence (restriction site), restriction enzymes are used experimentally to obtain precisely defined DNA segments called restriction fragments.
A. Specificity
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Restriction endonucleases recognize short stretches of dsDNA (4–8 bp) that contain specific nucleotide sequences. These sequences, which differ for each restriction enzyme, are palindromes, that is, they exhibit twofold rotational symmetry (Fig. 34.2). This means that, within a short region of the dsDNA, the nucleotide sequence on the two strands is identical if each is read in the 5′→3′ direction. Therefore, if you turn the page upside down (that is, rotate it 180° around its axis of symmetry) the sequence remains the same.
In bacteria, restriction endonucleases limit (restrict) the expression of nonbacterial (foreign) DNA through cleavage. Bacterial DNA is protected from cleavage by methylation of adenine at the restriction site.
B. Nomenclature
A restriction enzyme is named according to the organism from which it was isolated. The first letter of the name is from the genus of the bacterium. The next two letters are from the name of the species. An additional letter indicates the type or strain (as needed), and a number (Roman numeral) is appended to indicate the order in which the enzyme was discovered in that particular organism. For example, HaeIII is the third restriction endonuclease isolated from the bacterium Haemophilus aegyptius.
C. Sticky and blunt ends
Restriction enzymes cleave dsDNA so as to produce a 3′-hydroxyl group on one end and a 5′-phosphate group on the other. Some restriction endonucleases, such as TaqI, form staggered cuts that produce sticky or cohesive ends (that is, the resulting DNA fragments have single-stranded regions that are complementary to each other), as shown in Figure 34.3. Other restriction endonucleases, such as HaeIII, produce fragments that have blunt ends that are entirely double stranded and, therefore, do not form hydrogen bonds with each other. Using the enzyme DNA ligase (see p. 418), sticky ends of a DNA fragment of interest can be covalently joined with other DNA fragments that have sticky ends produced by cleavage with the same restriction endonuclease (Fig. 34.4). [Note: A ligase encoded by bacteriophage T4 can covalently join blunt-ended fragments.]
D. Restriction sites
A DNA sequence that is recognized and cut by a restriction enzyme is called a restriction site. Restriction endonucleases cleave dsDNA into fragments of different sizes depending upon the size of the sequence recognized. For example, an enzyme that recognizes a specific 4-bp sequence produces many cuts in the DNA molecule, one every 44 bp. In contrast, an enzyme requiring a unique sequence of 6 bp produces fewer cuts (one every 46 bp) and, therefore, longer pieces. Hundreds of these enzymes, each having different cleavage specificities (varying in both nucleotide sequences and length of recognition sites), are commercially available.
III. DNA CLONING
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Introduction of a foreign DNA molecule into a replicating cell permits the cloning or, amplification (that is, the production of many identical copies) of that DNA. [Note: Human DNA for cloning can be obtained from blood, saliva, and solid tissue.] In some cases, a single DNA fragment can be isolated and purified prior to cloning. More commonly, to clone a nucleotide sequence of interest, the total cellular DNA is first cleaved with a specific restriction enzyme, creating hundreds of thousands of fragments. Each of the resulting DNA fragments is joined to a DNA vector molecule (referred to as a cloning vector) to form a hybrid, or recombinant, DNA molecule. Each recombinant molecule carries its inserted DNA fragment into a single host cell (for example, a bacterium), where it is replicated. [Note: The process of introducing foreign DNA into a cell is called transformation for bacteria and yeast and transfection for higher eukaryotes.] As the host cell multiplies, it forms a clone in which every bacterium contains copies of the same inserted DNA fragment, hence the name “cloning.” The cloned DNA can be released from its vector by cleavage (using the appropriate restriction endonuclease) and isolated. By this mechanism, many identical copies of the DNA of interest can be produced. [Note: An alternative to amplification by biologic cloning, the polymerase chain reaction (PCR), is described on p. 495.]
A. Vectors
A vector is a molecule of DNA to which the fragment of DNA to be cloned is joined. Essential properties of a vector include the 1) capacity for autonomous replication within a host cell, 2) presence of at least one specific nucleotide sequence recognized by a restriction endonuclease, and 3) presence of at least one gene (such as an antibiotic resistance gene) that confers the ability to select for the vector. Commonly used vectors include plasmids and viruses.
1.
Prokaryotic plasmids: Prokaryotic organisms typically contain single, large, circular chromosomes. In addition, most species of bacteria also normally contain small, circular, extrachromosomal DNA molecules called plasmids (Fig. 34.5). Plasmid DNA undergoes replication that may or may not be synchronized to chromosomal division. Plasmids may carry genes that convey antibiotic resistance to the host bacterium and may facilitate the transfer of genetic information from one bacterium to another. They can be readily isolated from bacterial cells, their circular DNA cleaved at specific sites by restriction endonucleases, and up to 15 kb (kilobases) of foreign DNA (cut with the same restriction enzyme) inserted. The recombinant plasmid vector can be introduced into a bacterium, producing large numbers of copies of the plasmid. The bacteria are grown in the presence of antibiotics, thus selecting for cells containing the hybrid plasmids, which provide antibiotic resistance (Fig. 34.6). Artificial plasmids are routinely constructed. An example is the classic pBR322 (see Fig. 34.5), which contains an origin of replication, two antibiotic resistance genes, and >40 unique restriction sites. Use of plasmids is limited by the size of the DNA that can be inserted.
2.
Other vectors: The development of improved vectors that can more efficiently accommodate larger DNA segments, or express the passenger genes in different cell types, has aided molecular genetics research and therapeutics. In addition to the prokaryotic plasmids described above, naturally occurring viruses that infect bacteria (bacteriophage λ, for example) or mammalian cells (retroviruses, for example), as well as artificial constructs such as cosmids and bacterial or yeast artificial chromosomes (BAC or YAC, respectively), are currently used as cloning vectors. [Note: BAC and YAC can accept DNA inserts of 100–300 kb and 250–1,000 kb, respectively.]
B. DNA libraries
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A DNA library is a collection of cloned restriction fragments of the DNA of an organism. Two kinds of libraries are commonly used: genomic libraries and complementary DNA (cDNA) libraries. Genomic libraries ideally contain a copy of every DNA nucleotide sequence in the genome. In contrast, cDNA libraries contain those DNA sequences that only appear as processed messenger RNA (mRNA) molecules, and these differ according to cell type and environmental conditions. [Note: cDNA lacks introns and the control regions of the genes, whereas these are present in genomic DNA.] 1.
Genomic DNA libraries: A genomic library is created by digestion of the total DNA of an organism with a restriction endonuclease and subsequent ligation to an appropriate vector. The recombinant DNA molecules replicate within host bacteria. Thus, the amplified DNA fragments collectively represent the entire genome of the organism and are called a genomic library. Regardless of the restriction enzyme used, the chances are good that the gene of interest contains more than one restriction site recognized by that enzyme. If this is the case, and if the digestion is allowed to go to completion, the gene of interest is fragmented (that is, it is not contained in any one clone in the library). To avoid this usually undesirable result, a partial digestion is performed in which either the amount or the time of action of the enzyme is limited. This results in cleavage occurring at only a fraction of the restriction sites on any one DNA molecule, thus producing fragments of ~20 kb. Enzymes that cut very frequently (that is, those that recognize 4-bp sequences) are generally used for this purpose so that the result is an almost random collection of fragments. This insures a high degree of probability that the gene of interest is contained, intact, in some fragment.
2.
Complementary DNA libraries: If a protein-coding gene of interest is expressed at a high level in a particular tissue, the mRNA transcribed from that gene is likely also present at high concentrations in the cells of that tissue. For example, reticulocyte mRNA is composed largely of molecules that code for the α-globin and β-globin chains of hemoglobin
A (HbA). This mRNA can be used as a template to make a cDNA molecule using the enzyme reverse transcriptase (Fig. 34.7). Therefore, the resulting cDNA is a double-stranded copy of mRNA. [Note: The template mRNA is isolated from transfer RNA and ribosomal RNA by the presence of its poly-A tail.] cDNA can be amplified by biologic cloning or by PCR. It can be used as a probe to locate the gene that encodes the original mRNA (or fragments of the gene) in mixtures containing many unrelated DNA fragments. If the mRNA used as a template is a mixture of many different size species, the resulting cDNA is heterogeneous. These mixtures can be cloned to form a cDNA library. Because cDNA has no introns, it can be cloned into an expression vector for the synthesis of eukaryotic proteins by bacteria (Fig. 34.8). These special plasmids contain a bacterial promoter for transcription of the cDNA and a Shine-Dalgarno (SD) sequence (see p. 454) that allows the bacterial ribosome to initiate translation of the resulting mRNA molecule. The cDNA is inserted downstream of the promoter and within a gene for a protein that is expressed in the bacterium (for example, lacZ; see p. 466), such that the mRNA produced contains an SD sequence, a few codons for the bacterial protein, and all the codons for the eukaryotic protein. This allows for more efficient expression and results in the production of a fusion protein. [Note: Therapeutic human insulin is made in bacteria through this technology. However, the extensive co-and posttranslational modifications required for most other human proteins (for example, blood clotting factors) necessitates the use of eukaryotic, even mammalian, hosts.]
C. Sequencing cloned DNA fragments
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The base sequence of DNA fragments that have been cloned can be determined. The original procedure for this purpose was the Sanger dideoxy chain termination method illustrated in Figure 34.9. In this method, the single-stranded DNA (ssDNA) to be sequenced is used as the template for DNA synthesis by DNA polymerase (DNA pol). A radiolabeled primer complementary to the 3′-end of the target DNA is added, along with the four deoxyribonucleoside triphosphates (dNTP). The sample is divided into four reaction tubes, and a small amount of one of the four dideoxyribonucleoside triphosphates (ddNTP) is added to each tube. Because it contains no 3′-hydroxyl group, incorporation of a ddNMP terminates elongation at that point. The products of this reaction, then, consist of a mixture of DNA strands of different lengths, each terminating at a specific base. Separation of the various DNA products by size in an electric field using polyacrylamide gel electrophoresis, followed by autoradiography, yields a pattern of bands from which the DNA base sequence can be read. [Note: The shorter the fragment, the farther it travels on the gel, with the shortest fragment representing that which was made first (that is, the 5′-end).] In place of a labeled primer, a mixture of the four ddNTP linked to different fluorescent dyes and in a single reaction tube is now commonly used. The mixture is separated by capillary electrophoresis, the fluorescent labels are detected, and a color readout of the sequence is generated (Fig. 34.10). [Note: The Human Genome Project used variations of this technique to sequence the human genome.] Advances in sequencing technology, so-called next generation, or high-throughput sequencing, now allow the rapid sequencing of an entire genome with increased fidelity and decreased cost through the simultaneous (parallel) sequencing of many DNA pieces. [Note: Sequencing of the exome, that portion of the genome that encodes proteins, is now possible.]
IV. PROBES
Cleavage of large DNA molecules by restriction enzymes produces an enormous array of fragments. How can the DNA sequence of interest be picked out of such a mixture? The answer lies in the use of a probe, a short piece of ssDNA or RNA, labeled with a radioisotope, such as 32P, or with a nonradioactive molecule, such as biotin or a fluorescent dye. The sequence of a probe is complementary to a sequence in the DNA of interest, called the target DNA. Probes are used to identify which band on a gel or which clone in a library contains the target DNA, a process called screening.
A. Hybridization to DNA
The utility of probes hinges on the process of hybridization (or annealing) in which a probe containing a complementary sequence binds a single-stranded sequence of a target DNA. ssDNA, produced by alkaline denaturation of dsDNA, is first bound to a solid support, such as a nitrocellulose membrane. The immobilized DNA strands are prevented from self-annealing but are available for hybridization to the exogenous, radiolabeled, single-stranded probe. The extent of hybridization is measured by the retention of radioactivity on the membrane. Excess probe molecules that do not hybridize are removed by washing the membrane.
B. Synthetic oligonucleotide probes
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If the sequence of all or part of the target DNA is known, short, single-stranded oligonucleotide probes can be synthesized that are complementary to a small region of the gene of interest. If the sequence of the gene is unknown, the amino acid sequence of the protein, the final gene product, may be used to construct a nucleic acid probe using the genetic code as a guide. Because of the degeneracy of the genetic code (see p. 449), it is necessary to synthesize several oligonucleotides. [Note: Oligonucleotides can be used to detect single-base changes in the sequence to which they are complementary. In contrast, cDNA probes contain many thousands of bases, and their binding to a target DNA with a single-base change is unaffected.] 1. Detecting the βS-globin mutation: A synthetic allele-specific oligonucleotide (ASO) probe can be used to detect the presence of the sickle cell mutation in the β-globin gene (Fig. 34.11). DNA, isolated from white blood cells (WBC) and amplified, is denatured and applied to a membrane. A radiolabeled oligonucleotide probe, complementary to the point mutation (GAG → GTG, glutamate → valine) at codon 6 in patients with the βS gene, is applied to the membrane. DNA isolated from a heterozygous individual (sickle cell trait) or a homozygous patient (sickle cell anemia) contains a sequence that is complementary to the probe and a double-stranded hybrid form can be detected. In contrast, DNA obtained from normal individuals is not complementary at this position and, therefore, does not form a hybrid (see Fig. 34.11). Use of a pair of such ASO probes (one specific for the normal allele and one specific for the mutant allele) allows all three possible genotypes (homozygous normal, heterozygous, and homozygous mutant) to be distinguished (Fig. 34.12). [Note: ASO probes are useful only if the mutation and its location are known.]
C. Biotinylated probes
Because the disposal of radioactive waste is becoming increasingly expensive, nonradiolabeled probes have been developed. One of the most successful is based on the vitamin biotin (see p. 385), which can be chemically linked to the nucleotides used to synthesize the probe. Biotin was chosen because it binds very tenaciously to avidin, a readily available protein contained in chicken egg whites. Avidin can be attached to a fluorescent dye detectable optically with great sensitivity. Thus, a DNA fragment (displayed, for example, by gel electrophoresis) that hybridizes with the biotinylated probe can be made visible by immersing the gel in a solution of dye-coupled avidin. After washing away the excess avidin, the DNA fragment that binds the probe is fluorescent. [Note: Labeled probes can allow detection and localization of DNA or RNA sequences in cell or tissue preparations, a process called in situ hybridization (ISH). If the probe is fluorescent (F), the technique is called FISH.]
D. Antibodies
If no amino acid sequence information is available to guide the synthesis of a probe for direct detection of the DNA of interest, a gene can be identified indirectly by cloning cDNA in an expression vector that allows the cloned cDNA to be transcribed and translated. A labeled antibody is used to identify which bacterial colony produces the protein and, therefore, contains the cDNA of interest.
V. SOUTHERN BLOTTING
Southern blotting is a technique that combines the use of restriction enzymes, electrophoresis, and DNA probes to generate, separate, and detect pieces of DNA.
A. Procedure
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This method, named after its inventor, Edward Southern, involves the following steps (Fig. 34.13). First, DNA is extracted from cells, for example, a patient’s WBC. Second, the DNA is cleaved into many fragments using a restriction enzyme. Third, the resulting fragments (all of which are negatively charged) are separated on the basis of size by electrophoresis. [Note: Because the large fragments move more slowly than the smaller ones, the lengths of the fragments, usually expressed as the number of base pairs, can be calculated from comparison of the position of the band relative to standard fragments of known size.] The DNA fragments in the gel are denatured and transferred (blotted) to a nitrocellulose membrane for analysis. If the original DNA represents the individual’s entire genome, the enzymic digest contains ≥106 fragments. The gene of interest is on only one (or a few if the gene itself was fragmented) of these pieces of DNA. If all the DNA fragments were visualized by a nonspecific technique, they would appear as an unresolved blur of overlapping bands. To avoid this, the last step in Southern blotting uses a probe to identify the DNA fragments of interest. The patterns observed on Southern blot analysis depend both on the specific restriction endonuclease and on the probe used to visualize the restriction fragments. [Note: Variants of the Southern blot have been facetiously named northern if RNA is being studied (see p. 499) and western if protein is being studied (see p. 500), neither of which relates to anyone’s name or to points of the compass.]
B. Mutation detection
Southern blotting can detect DNA mutations such as large insertions or deletions, trinucleotide repeat expansions, and rearrangements of nucleotides. It can also detect point mutations (replacement of one nucleotide by another; see p. 449) that cause the loss or gain of restriction sites. Such mutations cause the pattern of bands to differ from those seen with a normal gene. Longer fragments are generated if a restriction site is lost. For example, in Figure 34.13, person 2 lacks a restriction site present in person 1. Alternatively, the point mutation may create a new cleavage site with the production of shorter fragments. [Note: Most sequence differences at restriction sites are harmless variations in the DNA.]
VI. RESTRICTION FRAGMENT LENGTH POLYMORPHISM
It has been estimated that the genomes of any two unrelated people are 99.5% identical. With 6 billion bp in the diploid human genome, that represents variation in ~30 million bp. These genome variations are the result of mutations that lead to polymorphisms. A polymorphism is a change in genotype that can result in no change in phenotype or a change in phenotype that is harmless, causes increased susceptibility to a disease, or, rarely, causes the disease. It is traditionally defined as a sequence variation at a given locus (allele) in >1% of a population. Polymorphisms primarily occur in the 98% of the genome that does not encode proteins (that is, in introns and intergenic regions). A restriction fragment length polymorphism (RFLP) is a genetic variant that can be observed by cleaving the DNA into fragments (restriction fragments) with a restriction endonuclease. The length of the restriction fragments is altered if the variant alters the DNA so as to create or abolish a restriction site. RFLP can be used to detect human genetic variations, for example, in prospective parents or in fetal tissue.
A. DNA variations resulting in RFLP
Two types of DNA variations commonly result in RFLP: single-base changes in the DNA sequence and tandem repeats of DNA sequences.
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1. Single-base changes: About 90% of human genome variation comes in the form of single nucleotide polymorphisms (SNPs, pronounced “snips”), that is, variations that involve just one base (Fig. 34.14). The substitution of one nucleotide at a restriction site can render the site unrecognizable by a particular restriction endonuclease. A new restriction site can also be created by the same mechanism. In either case, cleavage with an endonuclease results in fragments of lengths that differ from the normal and can be detected by DNA hybridization (see Fig. 34.13). The altered restriction site can be either at the site of a disease-causing mutation (rare) or at a site some distance from the mutation. [Note: The HapMap, developed by The International Haplotype Map Project, is a catalog of common SNP in the human genome. The data are being used in genome-wide association studies (GWAS) to identify those alleles that affect health and disease.] 2. Tandem repeats: Polymorphisms in chromosomal DNA can also arise from the presence of a variable number of tandem repeats (VNTR), as shown in Figure 34.15. These are short sequences of DNA at scattered locations in the genome, repeated in tandem (one after another). The number of these repeat units varies from person to person but is unique for any given individual and, therefore, serves as a molecular “fingerprint.” Cleavage by restriction enzymes yields fragments that vary in length depending on how many repeated segments are contained in the fragment (see Fig. 34.15). Many different VNTR loci have been identified and are extremely useful for DNA fingerprint analysis, such as in forensic and paternity cases. It is important to emphasize that these polymorphisms, whether SNP or VNTR, are simply markers, which, in most cases, have no known effect on the structure, function, or rate of production of any particular protein.
B. Tracing chromosomes from parent to offspring
If the DNA of an individual has gained a restriction site by base substitution, then enzymic cleavage yields at least one additional fragment.
Conversely, if a mutation results in loss of a restriction site, fewer fragments are produced by enzymic cleavage. An individual who is heterozygous for a polymorphism has a sequence variation in the DNA of one chromosome and not in the homologous chromosome. In such individuals, each chromosome can be traced from parent to offspring by determining the presence or absence of the polymorphism.
C. Prenatal diagnosis
Families with a history of severe genetic disease, such as an affected previous child or near relative, may wish to determine the presence of the disorder in a developing fetus. Prenatal diagnosis, in association with genetic counseling, allows for an informed reproductive decision if the fetus is affected.
1.
Methods available: The available diagnostic methods vary in sensitivity and specificity. Visualization of the fetus, for example, by ultrasound or fiberoptic devices (fetoscopy), is useful only if the genetic abnormality results in gross anatomic defects (for example, neural tube defects [NTD]). The chemical composition of the amniotic fluid can also provide diagnostic clues. For example, the presence of high levels of αfetoprotein is associated with NTD. Fetal cells obtained from amniotic fluid or from biopsy of the chorionic villi can be used for karyotyping, which assesses the morphology of metaphase chromosomes. Staining and cell sorting techniques permit the rapid identification of trisomies and translocations that produce an extra chromosome or chromosomes of abnormal lengths. However, molecular analysis of fetal DNA provides the most detailed genetic picture.
2.
DNA sources: DNA may be obtained from blood cells, amniotic fluid, or chorionic villi (Fig. 34.16). For amniotic fluid, it was formerly necessary to grow cells in culture for 2–3 weeks in order to have sufficient DNA for analysis. The ability to amplify DNA by PCR has dramatically shortened the time needed for a DNA analysis.
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3. Direct diagnosis of sickle cell anemia using RFLP: The genetic disorders of Hb are the most common genetic diseases in humans. In the case of sickle cell anemia (Fig. 34.17), the point mutation that gives rise to the disease (see p. 35) is actually one and the same mutation that gives rise to the polymorphism. However, direct detection by RFLP of diseases that result from point mutations is limited to only a few genetic diseases.
a. Early diagnostic efforts: In the past, prenatal diagnosis of sickle cell anemia involved the determination of the amount and kinds of Hb synthesized in the nucleated red cells obtained from fetal blood.
However, the invasive procedures to obtain fetal blood have a high mortality rate (~5%), and analysis cannot be carried out until late in the second trimester of pregnancy when HbA (and its HbS variant) begins to be produced.
b. RFLP analysis: In sickle cell anemia, the sequence alteration caused by the point mutation abolishes the recognition site of the restriction endonuclease MstII: CCTNAGG (where N is any nucleotide; see Fig. 34.17). Thus, the A-to-T mutation in codon 6 of the βS-globin gene eliminates a cleavage site for the enzyme. Normal DNA digested with MstII yields a 1.15-kb fragment, whereas a 1.35-kb fragment is generated from the βS gene as a result of the loss of one MstII cleavage site. Diagnostic techniques that allow analysis of fetal DNA from amniotic cells or chorionic villus sampling rather than fetal blood have proved valuable because they provide safe, early detection of sickle cell anemia as well as other genetic diseases. [Note: Genetic disorders caused by insertions or deletions between two restriction sites, rather than by the creation or loss of cleavage sites, will also display RFLP.] 4. Indirect diagnosis of phenylketonuria using RFLP: The gene for phenylalanine hydroxylase (PAH), the enzyme deficient in phenylketonuria ([PKU] see p. 270), is located on chromosome 12. It spans ~90 kb of genomic DNA and contains 13 exons separated by introns (Fig. 34.18; see p. 442 for a description of exons and introns). Mutations in the PAH gene usually do not directly affect any restriction endonuclease recognition site. To establish a diagnostic protocol for PKU, DNA from family members of the affected individual must be analyzed. The goal is to identify genetic markers (RFLP) that are tightly linked to the disease trait. Once these markers are identified, RFLP analysis can be used to carry out prenatal diagnosis.
a.
Mutant gene identification: Determining the presence of the mutant gene by identifying the polymorphism marker can be done if two conditions are satisfied. First, if the polymorphism is closely linked to a disease-producing mutation, the defective gene can be traced by detection of the RFLP. For example, if DNA from a family carrying a disease-causing gene is examined by restriction enzyme cleavage and Southern blotting, it is sometimes possible to find an RFLP that is consistently associated with that gene (that is, they show close linkage and are coinherited). It is then possible to trace the inheritance of the gene within a family without knowledge of the nature of the genetic defect or its precise location in the genome. [Note: The polymorphism may be known from the study of other families with the disorder or may be discovered to be unique in the family under investigation.] Second, for autosomal-recessive disorders, such as PKU, the presence of an affected individual in the family would aid in the diagnosis. This individual would have the mutation present on both chromosomes, allowing identification of the RFLP associated with the genetic disorder.
b.
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RFLP analysis: The presence of abnormal genes for PAH can be shown using DNA polymorphisms as markers to distinguish between normal and mutant genes. For example, Figure 34.19 shows a typical pattern obtained when DNA from members of an affected family is cleaved with an appropriate restriction enzyme and subjected to electrophoresis. The vertical arrows represent the cleavage sites for the restriction enzyme used. The presence of a polymorphic site creates fragment “b” in the autoradiogram (after hybridization with a labeled PAH-cDNA probe), whereas the absence of this site yields only fragment “a.” Note that subject II-2 demonstrates that the polymorphism, as shown by the presence of fragment “b,” is associated with the mutant gene. Therefore, in this particular family, the appearance of fragment “b” corresponds to the presence of a polymorphic site that marks the abnormal gene for PAH. The absence of fragment “b” corresponds to having only the normal gene. In Figure 34.19, examination of fetal DNA shows that the fetus inherited two abnormal genes from the parents and, therefore, has PKU.
c.
Value of DNA testing: DNA-based testing is useful not only in determining if an unborn fetus is affected by PKU but also in detecting unaffected carriers of the mutated gene to aid in family planning. [Note: PKU is treatable by dietary restriction of phenylalanine. Early diagnosis and treatment are essential in preventing severe neurologic damage in affected individuals.]
VII. POLYMERASE CHAIN REACTION
PCR is an in vitro method for amplifying a selected DNA sequence that does not rely on the biologic (in vivo) cloning method described on p. 483. PCR permits the synthesis of millions of copies of a specific nucleotide sequence in a few hours. It can amplify the sequence, even when the targeted sequence makes up less than one part in a million of the total initial sample. The method can be used to amplify DNA sequences from any source, including viral, bacterial, plant, or animal. The steps in PCR are summarized in Figures 34.20 and 34.21.
A. Procedure
PCR uses DNA pol to repetitively amplify targeted portions of genomic or cDNA. Each cycle of amplification doubles the amount of DNA in the sample, leading to an exponential increase (2n, where n = cycle number) in DNA with repeated cycles of amplification. The amplified DNA products can then be separated by gel electrophoresis, detected by Southern blotting and hybridization, and sequenced.
1.
Constructing primer: It is not necessary to know the nucleotide sequence of the target DNA in the PCR method. However, it is necessary to know the nucleotide sequence of short segments on each side of the target DNA. These stretches, called flanking sequences, bracket the DNA sequence of interest. The nucleotide sequences of the flanking regions are used to construct two, single-stranded oligonucleotides, usually 20– 35 nucleotides long, which are complementary to the respective flanking sequences. The 3′-hydroxyl end of each oligonucleotide points toward the target sequence (see Fig. 34.20). These synthetic oligonucleotides function as primers in PCR.
2.
Denaturing DNA: The target DNA to be amplified is heated to ~95°C to separate the dsDNA into single strands.
3.
Annealing primers: The separated strands are cooled to ~50°C and the two primers (one for each strand) anneal to a complementary sequence on the ssDNA.
4.
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Extending primers: DNA pol and dNTP (in excess) are added to the mixture (~72°C) to initiate the synthesis of two new strands complementary to the original DNA strands. DNA pol adds nucleotides to the 3′-hydroxyl end of the primer, and strand growth extends in the 5′→3′ direction across the target DNA, making complementary copies of the target. [Note: PCR products can be several thousand base pairs long.] At the completion of one cycle of replication, the reaction mixture is heated again to separate the strands (of which there are now four). Each strand binds a complementary primer, and the step of primer extension is repeated. By using a heat-stable DNA pol (for example, Taq from the bacterium Thermus aquaticus that normally lives at high temperatures), the polymerase is not denatured and, therefore, does not have to be added at each successive cycle. However, Taq lacks proofreading activity. Typically, 20–30 cycles are run during this process, amplifying the DNA by a million-fold (220) to a billion-fold (230). [Note: Each extension product includes a sequence at its 5′-end that is complementary to the primer (see Fig. 34.20). Thus, each newly synthesized strand can act as a template for the successive cycles (see Fig. 34.21). This leads to an exponential increase in the amount of target DNA with each cycle, hence, the name “polymerase chain reaction.”] Probes can be made during PCR by adding labeled nucleotides to the last few cycles.
B. Advantages
The major advantages of PCR over biologic cloning as a mechanism for amplifying a specific DNA sequence are sensitivity and speed. DNA sequences present in only trace amounts can be amplified to become the predominant sequence. PCR is so sensitive that DNA sequences present in an individual cell can be amplified and studied. Isolating and amplifying a specific DNA sequence by PCR is faster and less technically difficult than traditional cloning methods using recombinant DNA techniques.
C. Applications
PCR has become a very common tool in research, forensics, and clinical diagnostics.
1.
Comparison of a normal gene to its mutant form: PCR allows the synthesis of mutant DNA in sufficient quantities for a sequencing protocol without laborious biologic cloning of the DNA.
2.
Forensic analysis of DNA samples: DNA fingerprinting by means of PCR has revolutionized the analysis of evidence from crime scenes. DNA isolated from a single human hair, a tiny spot of blood, or a sample of semen is sufficient to determine whether the sample comes from a specific individual. The DNA markers analyzed for such fingerprinting are most commonly a type of polymorphism known as short tandem repeats. These are very similar to the VNTR described previously (see p.
491) but are smaller in size. [Note: Paternity testing uses the same techniques.] 3.
Detection of low-abundance nucleic acid sequences: Viruses that have a long latency period, such as human immunodeficiency virus (HIV), are difficult to detect at the early stage of infection using conventional methods. PCR offers a rapid and sensitive method for detecting viral DNA sequences even when only a small proportion of cells harbors the virus. [Note: Quantitative PCR (qPCR), also known as real-time PCR, allows quantification of the amount (copy number) of the target nucleic acid after each cycle of amplification (that is, in real time) rather than at the end and is useful in determining viral load (the amount of virus).] 4.
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Prenatal diagnosis and carrier detection of cystic fibrosis: Cystic fibrosis is an autosomal-recessive genetic disease resulting from mutations in the gene for the cystic fibrosis transmembrane conductance regulator (CFTR) protein. The most common mutation is a three-base deletion that results in the loss of a phenylalanine residue from the CFTR protein (see p. 450). Because the mutant allele is three bases shorter than the normal allele, it is possible to distinguish them from each other by the size of the PCR products obtained by amplifying that portion of the DNA. Figure 34.22 illustrates how the results of such a PCR test can distinguish between homozygous normal, heterozygous (carriers), and homozygous mutant (affected) individuals.
The simultaneous amplification of multiple regions of a target DNA using multiple primer pairs is known as multiplex PCR. It allows detection of the loss of ≥1 exons in a gene with many exons such as the gene for CFTR, which has 27 exons.
VIII. GENE EXPRESSION ANALYSIS
The tools of biotechnology not only allow the study of gene structure, but also provide ways of analyzing the mRNA and protein products of gene expression.
A. Determining messenger RNA levels mRNA levels are usually determined by the hybridization of labeled probes to either mRNA itself or to cDNA produced from mRNA. [Note: Amplification by PCR of cDNA made from mRNA by retroviral reverse transcriptase (RT) is referred to as RT-PCR.] 1. Northern blots
Northern blots are similar to Southern blots (see Fig. 34.13), except that the sample contains a mixture of mRNA molecules that are separated by electrophoresis, then transferred to a membrane and hybridized with a radiolabeled probe. The bands obtained by autoradiography give a measure of the amount and size of the mRNA molecules in the sample.
2. Microarrays: DNA microarrays contain thousands of immobilized ssDNA sequences organized in an area no larger than a microscope slide. These microarrays are used to analyze a sample for the presence of gene variations or mutations (genotyping) or to determine the patterns of mRNA production (gene expression analysis), analyzing thousands of genes at the same time. For genotyping analysis, the sample is from genomic DNA. For expression analysis, the population of mRNA molecules from a particular cell type is converted to cDNA and labeled with a fluorescent tag (Fig. 34.23). This mixture is then exposed to a gene (or, DNA) chip, which is a glass slide or membrane containing thousands of tiny spots of DNA, each corresponding to a different gene. The amount of fluorescence bound to each spot is a measure of the amount of that particular mRNA in the sample. DNA microarrays are used to determine the differing patterns of gene expression in two different types of cell (for example, normal and cancer cells; see Fig. 34.23). They can also be used to subclassify cancers, such as breast cancer, to optimize treatment. [Note: Microarrays involving proteins and the antibodies or other proteins that recognize them are being used to identify biomarkers to aid in the diagnosis, prognosis, and treatment of disease based on a patient’s protein expression profile. Protein (and DNA) microarrays are important tools in the development of personalized (precision) medicine in which the treatment and/or prevention strategies consider the genetic, environmental, and lifestyle variations among individuals.]
B. Protein analysis
The kinds and amounts of proteins in cells do not always directly correspond to the amounts of mRNA present. Some mRNA are translated more efficiently than others, and some proteins undergo posttranslational modification. When analyzing the abundance and interactions of a large number of proteins, automated methods involving a variety of techniques, such as mass spectrometry and two-dimensional electrophoresis, are used. When investigating one, or a limited number of proteins, labeled antibodies (Ab) are used to detect and quantify specific proteins and to determine posttranslational modifications.
1.
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Enzyme-linked immunosorbent assays: These assays (known as ELISA) are performed in the wells of a microtiter dish. The antigen (protein) is bound to the plastic of the dish. The probe used consists of an Ab specific for the protein (such as troponin, see p. 66) to be measured. The Ab is covalently bound to an enzyme, which will produce a colored product when exposed to its substrate. The amount of color produced is proportional to the amount of Ab present and, indirectly, to the amount of protein in a test sample.
2.
Western blots: Western blots (also called immunoblots) are similar to Southern blots, except that it is protein molecules in the sample that are separated by electrophoresis and blotted (transferred) to a membrane. The probe is a labeled Ab, which produces a band at the location of its antigen.
3.
Detecting exposure to human immunodeficiency virus: ELISA and western blots are commonly used to detect exposure to HIV by measuring the amount of anti-HIV Ab present in a patient’s blood sample. ELISA are used as the primary screening tool because they are very sensitive. Because these assays sometimes give false positives, however, western blots, which are more specific, are often used as a confirmatory test (Fig. 34.24). [Note: ELISA and western blots can only detect HIV exposure after anti-HIV Ab appear in the bloodstream. PCR based testing for HIV is more useful in the first few months after exposure.]
C. Proteomics
The study of the proteome, or all the proteins expressed by a genome, including their relative abundance, distribution, posttranslational modifications, functions, and interactions with other macromolecules, is known as proteomics. The 20,000–25,000 protein-coding genes of the human genome translate into well over 100,000 proteins when posttranscriptional and posttranslational modifications are considered. Although a genome remains essentially unchanged, the amounts and types of proteins in any particular cell change dramatically as genes are turned on and off. [Note: Proteomics (and genomics) required the parallel development of bioinformatics, the computer-based organization, storage, and analysis of biologic data.] Figure 34.25 compares some of the analytic techniques discussed in this chapter.
IX. GENE THERAPY
The goal of gene therapy is to treat disease through delivery of the normal, cloned DNA for a gene into the somatic cells of a patient who has a defect in that gene as a result of a disease-causing mutation. Because somatic gene therapy changes only the targeted somatic cells, the change is not passed on to the next generation. [Note: In germline gene therapy, the germ cells are modified, and so the change is passed on. A long-standing moratorium on germline gene therapy is in effect worldwide.] There are two types of gene transfer: 1) ex vivo, in which cells from the patient are removed, transduced, and returned, and 2) in vivo, in which the cells are directly transduced. Both types require use of a viral vector to deliver the DNA. Challenges of gene therapy include development of vectors, achievement of long-lived expression, and prevention of side effects such as an immune response. The first successful gene therapy involved two patients with severe combined immunodeficiency disease (SCID) caused by mutations to the gene for adenosine deaminase (ADA, see p. 301). It utilized mature T lymphocytes transduced ex vivo with a retroviral vector (Fig. 34.26). [Note: Human ADA cDNA is now used.] Since 1990, only a small number of patients (with a variety of disorders, such as hemophilia, cancers, and certain types of blindness) have been treated with gene therapy, with varying degrees of success.
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Gene editing, as opposed to gene addition, allows a mutated gene to be repaired. Combinations of DNA-binding molecules (proteins or RNA) and endonucleases are used to identify and cleave the mutated sequence. Cleavage activates homologous recombination repair of dsDNA breaks (see p. 429) that integrates DNA containing the correct sequence into the gene. [Note: An endonuclease guided to a specific DNA sequence by a custom-designed RNA has been used in gene editing in human cell lines. The technique is based on (and named for) the prokaryotic CRISPR-Cas9 (clustered regularly interspaced short palindromic repeats [CRISPR]-associated protein) system that identifies and cleaves foreign DNA in bacterial cells. CRISPR is currently used in the laboratory but not in the clinic.]
X. TRANSGENIC ANIMALS
Transgenic animals can be produced by injecting a cloned foreign gene (a transgene) into a fertilized egg. If the gene randomly and stably integrates into a chromosome, it will be present in the germline of the resulting animal and can be passed from generation to generation. A giant mouse called “Supermouse” was produced in this way by injecting the gene for rat growth hormone into a fertilized mouse egg. [Note: Transgenic animals have been designed that produce therapeutic human proteins in their milk, a process called “pharming.” Antithrombin, an anticlotting protein (see online Chapter 35), was produced by transgenic goats and approved for clinical use in 2009.] If the functional transgene undergoes targeted (not random) insertion, a knockin (KI) mouse that expresses the gene is created. Targeted insertion of a nonfunctional version of the transgene creates a knockout (KO) mouse that does not express the gene. Such genetically engineered animals can serve as models for the study of a corresponding human disease.
XI. CHAPTER SUMMARY
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Restriction endonucleases are bacterial enzymes that cleave double-stranded DNA (dsDNA) into smaller fragments. Each enzyme cleaves at a specific 4–8 base-pair sequence (a restriction site), producing DNA segments called restriction fragments. The sequences that are recognized are palindromic. Restriction enzymes form either staggered cuts (sticky ends) or blunt-end cuts on the DNA. Bacterial DNA ligases can join two DNA fragments from different sources if they have been cut by the same restriction endonuclease. This hybrid combination of two fragments is called a recombinant DNA molecule. Introduction of a foreign DNA molecule into a replicating cell permits the amplification (production of many copies) of the DNA, a process called cloning. A vector is a molecule of DNA to which the fragment of DNA to be cloned is joined. Vectors must be capable of autonomous replication within the host cell, must contain at least one specific nucleotide sequence recognized by a restriction endonuclease, and must carry at least one gene that confers the ability to select for the vector such as an antibiotic resistance gene. Prokaryotic organisms normally contain small, circular, extrachromosomal DNA molecules called plasmids that can serve as vectors. They can be readily isolated from the bacterium (or artificially constructed), joined with the DNA of interest, and reintroduced into the bacterium, which will replicate, thus making multiple copies of the hybrid plasmid. A DNA library is a collection of cloned restriction fragments of the DNA of an organism. A genomic library is a collection of fragments of dsDNA obtained by digestion of the total DNA of the organism with a restriction endonuclease and subsequent ligation to an appropriate vector. It ideally contains a copy of every DNA nucleotide sequence in the genome. In contrast, complementary DNA (cDNA) libraries contain only those DNA sequences that are complementary to processed messenger RNA (mRNA) molecules present in a cell and differ according to cell type and environmental conditions. Because cDNA has no introns, it can be cloned into an expression vector for the synthesis of human proteins by bacteria or eukaryotes. Cloned, then purified, fragments of DNA can be sequenced, for example, using the Sanger dideoxy chain termination method. A probe is a small piece of RNA or single-stranded DNA (usually labeled with a radioisotope, such as 32P, or another identifiable compound, such as biotin or a fluorescent dye) that has a nucleotide sequence complementary to the DNA molecule of interest (target DNA). Probes can be used to identify which clone of a library or which band on a gel contains the target DNA. Southern blotting is a technique that can be used to detect specific sequences present in DNA. The DNA is cleaved using a restriction endonuclease, after which the pieces are separated by gel electrophoresis and are denatured and transferred (blotted) to a nitrocellulose membrane for analysis. The fragment of interest is detected using a probe. The human genome contains many thousands of polymorphisms (DNA sequence variations at a given locus). Polymorphisms can arise from single-base changes and from tandem repeats. A polymorphism can serve as a genetic marker that can be followed through families. A restriction fragment length polymorphism (RFLP) is a genetic variant that can be observed by cleaving the DNA into restriction fragments using a restriction enzyme. A base substitution in one or more nucleotides at a restriction site can render the site unrecognizable by a particular restriction endonuclease. A new restriction site also can be created by the same mechanism. In either case, cleavage with the endonuclease results in fragments of lengths differing from the normal that can be detected by hybridization with a probe. RFLP analysis can be used to diagnose genetic diseases early in the gestation of a fetus. The polymerase chain reaction (PCR), another method for amplifying a selected DNA sequence, does not rely on the biologic cloning method. PCR permits the synthesis of millions of copies of a specific nucleotide sequence in a few hours. It can amplify the sequence, even when the targeted sequence makes up less than one part in a million of the total initial sample. The method can be used to amplify DNA sequences from any source. Applications of the PCR technique include 1) efficient comparison of a normal gene with a mutant form of the gene, 2) forensic analysis of DNA samples, 3) detection of low-abundance nucleic acid sequences, and 4) prenatal diagnosis and carrier detection (for example, of cystic fibrosis). The products of gene expression (mRNA and proteins) can be measured by techniques such as northern blots, which are like Southern blots except that the sample contains a mixture of mRNA molecules that are separated by electrophoresis, then hybridized to a radiolabeled probe; microarrays are used to determine the differing patterns of gene expression in two different types of cells (for example, normal and cancer cells); enzyme-linked immunosorbent assays (ELISA); and western blots (immunoblots) are used to detect specific proteins. Proteomics is the study of all the proteins expressed by a genome. The goal of gene therapy is the insertion of a normal cloned gene to replace a defective gene in a somatic cell, whereas the goal of gene editing is the repair of a mutated gene. Insertion of a foreign gene (transgene) into the germline of an animal creates a transgenic animal that can produce therapeutic proteins or serve as gene knockin or knockout models for human diseases.
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Choose the ONE best answer.
4.1. HindIII is a restriction endonuclease. Which of the following is most likely to be the recognition sequence for this enzyme?
A. AAGAAG
B. AAGAGA
C. AAGCTT
D. AAGGAA
E. AAGTTC
Correct answer = C. The vast majority of restriction endonucleases recognize palindromes in double-stranded DNA, and AAGCTT is the only palindrome among the choices. Because the sequence of only one DNA strand is given, the base sequence of the complementary strand must be determined. To be a palindrome, both strands must have the same sequence when read in the 5′→3′ direction. Thus, the complement of 5′-AAGCTT-3′ is also 5′-AAGCTT-3′.
4.2. An Ashkenazi Jewish couple has their 6-month-old son evaluated for listlessness, poor head control, and a fixed gaze. Tay-Sachs disease, an autosomal-recessive disease of lipid degradation, is diagnosed. The couple also has a daughter. The family’s pedigree is shown to the right, along with Southern blots of a restriction fragment length polymorphism very closely linked to the gene for hexosaminidase A, which is defective in Tay-Sachs disease. Which of the statements below is most accurate with respect to the daughter?
A. She has a 25% chance of having Tay-Sachs disease.
B. She has a 50% chance of having Tay-Sachs disease.
C. She has Tay-Sachs disease.
D. She is a carrier for Tay-Sachs disease.
E. She is homozygous normal.
Correct answer = E. Because they have an affected son, both the biological father and mother must be carriers for this disease. The affected son must have inherited a mutant allele from each parent. Because he shows only the 3kilobase (kb) band on the Southern blot, the mutant allele for this disease must be linked to the 3-kb band. The normal allele must be linked to the 4-kb band, and because the daughter inherited only the 4-kb band, she must be homozygous normal for the hexosaminidase A gene.
4.3. A physician would like to determine the global patterns of gene expression in two different types of tumor cells in order to develop the most appropriate form of chemotherapy for each patient. Which of the following techniques would be most appropriate for this purpose?
A. Enzyme-linked immunosorbent assay
B. Microarray
C. Northern blot
D. Southern blot
E. Western blot
Correct answer = B. Microarray analysis allows the determination of messenger RNA (mRNA) production (gene expression) from thousands of genes at once. A northern blot only measures mRNA production from one gene at a time. Western blots and enzyme-linked immunosorbent assay measure protein production (also gene expression) but only from one gene at a time. Southern blots are used to analyze DNA, not the products of DNA expression.
4.4. A 2-week-old infant is diagnosed with a urea cycle defect. Enzymic analysis showed no activity for ornithine transcarbamoylase (OTC), an enzyme of the cycle. Molecular analysis revealed that the messenger RNA (mRNA) product of the gene for OTC was identical to that of a control. Which of the techniques listed below was most likely used to analyze mRNA?
A. Dideoxy chain termination
B. Northern blot
C. Polymerase chain reaction
D. Southern blot
E. Western blot
Correct answer = B. Northern blot allows analysis of the messenger RNA present (expressed) in a particular cell or tissue. Southern blot is used for DNA analysis, whereas western blot is used for protein analysis. Dideoxy chain termination is used to sequence DNA. Polymerase chain reaction is used to generate multiple, identical copies of a DNA sequence in vitro.
4.5. For the patient above, which phase of the central dogma was most likely affected?
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Correct answer = Translation. The gene is present and is able to be expressed as evidenced by normal production of messenger RNA. The lack of enzymic activity means that some aspect of protein synthesis is affected.
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I. OVERVIEW
Blood clotting (coagulation) is designed to rapidly stop bleeding from a damaged blood vessel in order to maintain a constant blood volume (hemostasis). Coagulation is accomplished through vasoconstriction and the formation of a clot (thrombus) that consists of a plug of platelets (primary hemostasis) and a meshwork of the protein fibrin (secondary hemostasis) that stabilizes the platelet plug. Clotting occurs in association with membranes on the surface of platelets and damaged blood vessels (Fig. 35.1). [Note: If clotting occurs within an intact vessel such that the lumen is occluded and blood flow is impeded, a condition known as thrombosis, serious tissue damage, and even death can occur. This is what happens, for example, during a myocardial infarction (MI).] Processes to limit clot formation to the area of damage and remove the clot once vessel repair is underway also play essential roles in hemostasis. [Note: Separate discussions of the formation of the platelet plug and the fibrin meshwork facilitate presentation of these multistep, multicomponent processes. However, the two work together to maintain hemostasis.]
II. FIBRIN MESHWORK FORMATION
The formation of the fibrin meshwork involves two unique pathways that converge to form a common pathway (Fig. 35.2). In each pathway, the major components are proteins (called factors [F]) designated by Roman numerals. The factors are glycoproteins that are synthesized and secreted by the liver, primarily. [Note: Several factors are also denoted by alternative names. For example, factor X (FX), the point of pathway convergence, is also known as Stuart factor.]
A. Proteolytic cascade
Within the pathways, a cascade is set up in which proteins are converted from an inactive form, or zymogen, to an active form by proteolytic cleavage in which the protein product of one activation reaction initiates another. The active form of a factor is denoted by a lowercase “a” after the numeral. The active proteins FIIa, FVIIa, FIXa, FXa, FXIa, and FXIIa are enzymes that function as serine proteases with trypsin-like specificity and, therefore, cleave a peptide bond on the carboxyl side of an arginine or lysine residue in a polypeptide. For example, FIX (Christmas factor) is activated through cleavage at arginine 145 and arginine 180 by FXIa (Fig. 35.3). The proteolytic cascade results in enormous rate acceleration, because one active protease can produce many molecules of active product each of which, in turn, can activate many molecules of the next protein in the cascade. In some cases, activation can be caused by a conformational change in the protein in the absence of proteolysis. [Note: Nonproteolytic proteins play a role as accessory proteins (cofactors) in the pathways. FIII, FV, and FVIII are the accessory proteins.]
B. Role of phosphatidylserine and calcium
The presence of the negatively charged phospholipid phosphatidylserine (PS) and positively charged calcium ions (Ca2+) accelerates the rate of some steps in the clotting cascade.
1.
Phosphatidylserine: PS is located primarily on the intracellular (cytosolic) face of the plasma membrane. [Note: Flippases create the asymmetry (see p. 205).] Its exposure signals injury to the endothelial cells that line blood vessels. PS is also exposed on the surface of activated platelets.
2.
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Calcium ions: Ca2+ binds the negatively charged γ-carboxyglutamate (Gla) residues present in four of the serine proteases of clotting (FII, FVII, FIX, and FX), facilitating the binding of these proteins to exposed phospholipids (Fig. 35.4). The Gla residues are good chelators of Ca2+ because of their two adjacent negatively charged carboxylate groups (Fig. 35.5). [Note: The use of chelating agents such as sodium citrate to bind Ca2+ in blood-collecting tubes or bags prevents the blood from clotting.]
C. Formation of γ-carboxyglutamate residues γ-Carboxylation is a posttranslational modification in which 9–12 glutamate residues (at the amino [N]-terminus of the target protein) get carboxylated at the γ carbon, thereby forming Gla residues. The process occurs in the rough endoplasmic reticulum (RER) of the liver.
1. γ-Carboxylation: This carboxylation reaction requires a protein substrate, oxygen (O2), carbon dioxide (CO2), γ-glutamyl carboxylase, and the hydroquinone form of vitamin K as a coenzyme (Fig. 35.6). In the reaction, the hydroquinone form of vitamin K gets oxidized to its epoxide form as O2 is reduced to water. [Note: Dietary vitamin K, a fat-soluble vitamin (see p. 393), is reduced from the quinone form to the hydroquinone coenzyme form by vitamin K reductase (Fig. 35.7).] 2. Inhibition by warfarin: The formation of Gla residues is sensitive to inhibition by warfarin, a synthetic analog of vitamin K that inhibits the enzyme vitamin K epoxide reductase (VKOR). The reductase, an integral protein of the RER membrane, is required to regenerate the functional hydroquinone form of vitamin K from the epoxide form generated in the γ-carboxylation reaction. Thus, warfarin is an anticoagulant that inhibits clotting by functioning as a vitamin K antagonist. Warfarin salts are used therapeutically to limit clot formation. [Note: Warfarin is used commercially as a pest control agent such as in rat poison. It was developed by the Wisconsin Alumni Research Foundation, hence the name.]
Genetic differences (genotypes) in the gene for catalytic subunit 1 of VKOR (VKORC1) influence patient response to warfarin. For example, a polymorphism (see p. 491) in the promoter region of the gene decreases gene expression, resulting in less VKOR being made, thereby necessitating a lower dose of warfarin to achieve a therapeutic level. Polymorphisms in the cytochrome P450 enzyme (CYP2C9) that metabolizes warfarin are also known. In 2010, the U.S. Food and Drug Administration added a genotype-based dose table to the warfarin label (package insert). The influence of genetics on an individual’s response to drugs is known as pharmacogenetics.
D. Pathways
Three distinct pathways are involved in formation of the fibrin meshwork: the extrinsic pathway, the intrinsic pathway, and the common pathway. Production of FXa by the extrinsic and intrinsic pathways initiates the common pathway (see Fig. 35.2).
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1. Extrinsic: This pathway involves a protein, tissue factor (TF), that is not in the blood but becomes exposed when blood vessels get injured. TF (or, FIII) is a transmembrane glycoprotein abundant in vascular subendothelium. It is an extravascular accessory protein and not a protease. Any injury that exposes FIII to blood rapidly (within seconds) initiates the extrinsic (or, TF) pathway. Once exposed, TF binds a circulating Gla-containing protein, FVII, activating it through conformational change. [Note: FVII can also be activated proteolytically by thrombin (see 3. below).] Binding of FVII to TF requires the presence of Ca2+ and phospholipids. The TF–FVIIa complex then binds and activates FX by proteolysis (Fig. 35.8). Therefore, activation of FX by the extrinsic pathway occurs in association with the cell membrane. The extrinsic pathway is quickly inactivated by tissue factor pathway inhibitor (TFPI) that, in a FXa-dependent process, binds to the TF–FVIIa complex and prevents further production of FXa. [Note: TF and FVII are unique to the extrinsic pathway.] 2. Intrinsic: All of the protein factors involved in the intrinsic pathway are present in the blood and are, therefore, intravascular. The intrinsic pathway involves two phases: the contact phase and the FX-activation phase, each with known deficiencies.
a.
Contact phase: This phase results in the activation of FXII (Hageman factor) by conformational change through binding to a negative surface. Deficiencies in FXII (or in the other proteins of this phase, high molecular weight kininogen and prekallikrein) do not result in bleeding, calling into question the importance of this phase in coagulation. However, the contact phase does play a role in inflammation. [Note: FXII can be activated proteolytically by thrombin (see 3. below)].
b.
Factor X–activation phase: The sequence of events leading to the activation of FX to FXa by the intrinsic pathway is initiated by FXIIa (Fig. 35.9). FXIIa activates FXI, and FXIa activates FIX, a Glacontaining serine protease. FIXa combines with FVIIIa (a bloodborne accessory protein), and the complex activates FX, a Gla-containing serine protease. [Note: The complex containing FIXa, FVIIIa, and FX forms on exposed negatively charged membrane regions, and FX gets activated to FXa. This complex is sometimes referred to as Xase. Binding of the complex to membrane phospholipids requires Ca2+.] c.
Factor XII deficiency: A deficiency in FXII does not lead to a bleeding disorder. This is because FXI, the next protein in the cascade, can be activated proteolytically by thrombin (see 3. below).
d.
Hemophilia: Hemophilia is a coagulopathy, a defect in the ability to clot. Hemophilia A, which accounts for 80% of all hemophilia, results from deficiency of FVIII, whereas deficiency of FIX results in hemophilia B. Each deficiency is characterized by decreased and delayed ability to clot and/or formation of abnormally friable (easily disrupted) clots. This can be manifested, for example, by bleeding into the joints (Fig. 35.10). The extent of the factor deficiency determines the severity of the disease. Current treatment for hemophilia is factor replacement therapy using FVIII or FIX obtained from pooled human blood or from recombinant DNA technology. However, antibodies to the factors can develop. Gene therapy is a goal. Because the genes for both proteins are on the X chromosome, hemophilia is an X-linked disorder. [Note: Deficiency of FXI results in a bleeding disorder that sometimes is referred to as hemophilia C.] with hemophilia.
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The inactivation of the extrinsic pathway by TFPI results in dependence on the intrinsic pathway for continued production of FXa. This explains why individuals with hemophilia bleed even though they have an intact extrinsic pathway.
3. Common: FXa produced by both the intrinsic and the extrinsic paths initiates the common pathway, a sequence of reactions that results in the generation of fibrin (FIa), as shown in Figure 35.11. FXa associates with FVa (a bloodborne accessory protein) and, in the presence of Ca2+ and phospholipids, forms a membrane-bound complex referred to as prothrombinase. The complex cleaves prothrombin (FII) to thrombin (FIIa). [Note: FVa potentiates the proteolytic activity of FXa.] The binding of Ca2+ to the Gla residues in FII facilitates the binding of FII to the membrane and to the prothrombinase complex, with subsequent cleavage to FIIa. Cleavage excises the Gla-containing region, releasing FIIa from the membrane and, thereby, freeing it to activate fibrinogen (FI) in the blood. [Note: This is the only example of cleavage of a Gla protein that results in the release of a Gla-containing peptide. The peptide travels to the liver where it is thought to act as a signal for increased production of clotting proteins.] Oral, direct inhibitors of FXa have been approved for clinical use as anticoagulants. In contrast to warfarin, they have a more rapid onset and shorter half-life and do not require routine monitoring.
A common point mutation (G20210A) in which an adenine (A) replaces a guanine (G) at nucleotide 20210 in the 3′ untranslated region of the gene for FII leads to increased levels of FII in the blood. This results in thrombophilia, a condition characterized by an increased tendency for blood to clot.
a. Fibrinogen cleavage to fibrin: FI is a soluble glycoprotein made by the liver. It consists of dimers of three different polypeptide chains [(αβγ)2] held together at the N termini by disulfide bonds. The N termini of the α and β chains form “tufts” on the central of three globular domains (Fig. 35.12). The tufts are negatively charged and result in repulsion between FI molecules. Thrombin (FIIa) cleaves the charged tufts (releasing fibrinopeptides A and B), and FI becomes FIa. As a result of the loss of charge, the FIa monomers are able to noncovalently associate in a staggered array, and a soft (soluble) fibrin clot is formed.
b.
Fibrin cross-linking: The associated FIa molecules get covalently cross-linked. This converts the soft clot to a hard (insoluble) clot. FXIIIa, a transglutaminase, covalently links the γ-carboxamide of a glutamine residue in one FIa molecule to the ε-amino of a lysine residue in another through formation of an isopeptide bond and release of ammonia (Fig. 35.13). [Note: FXIII is also activated by thrombin.] c.
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Importance of thrombin: The activation of FX by the extrinsic pathway provides the “spark” of FXa that results in the initial activation of thrombin. FIIa then activates factors of the common (FV, FI, FXIII), intrinsic (FXI, FVIII), and extrinsic (FVII) pathways (Fig. 35.14). It also activates FXII of the contact phase. The extrinsic pathway, then, initiates clotting by the generation of FXa, and the intrinsic pathway amplifies and sustains clotting after the extrinsic pathway has been inhibited by TFPI. [Note: Hirudin, a peptide secreted from the salivary gland of medicinal leeches, is a potent direct thrombin inhibitor (DTI).
Injectable recombinant hirudin has been approved for clinical use. Dabigatran is an oral DTI.] Additional crosstalk between the pathways of clotting is achieved by the FVIIa–TF-mediated activation of the intrinsic pathway and the FXIIa-mediated activation of the extrinsic pathway. The complete picture of physiologic blood clotting via the formation of a hard fibrin clot is shown in Figure 35.15. The factors of the clotting cascade are shown organized by function in Figure 35.16.
activated form would be denoted by an “a” after the numeral. [Note: Calcium is
IV. There is no VI. I (fibrin) is neither a protease nor an accessory protein. XIII is a transglutaminase.] Gla = γ-carboxyglutamate.
Clinical laboratory tests are available to evaluate the extrinsic through common pathways (prothrombin time [PT] using thromboplastin and expressed as the international normalized ratio [INR]) and the intrinsic through common pathways (activated partial thromboplastin time [aPTT]). Thromboplastin is a combination of phospholipids + FIII. A derivative, partial thromboplastin contains just the phospholipid portion because FIII is not needed to activate the intrinsic pathway.
III. LIMITING CLOTTING
The ability to limit clotting to areas of damage (anticoagulation) and to remove clots once repair processes are underway (fibrinolysis) are exceedingly important aspects of hemostasis. These actions are performed by proteins that inactivate clotting factors either by binding to them and removing them from the blood or by degrading them and also by proteins that degrade the fibrin meshwork.
A. Inactivating proteins
Proteins synthesized by the liver and by the blood vessels themselves balance the need to form clots at sites of vessel injury with the need to limit their formation beyond the injured area.
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1. Antithrombin: Antithrombin III (ATIII), also referred to simply as antithrombin (AT), is a hepatic protein that circulates in the blood. It inactivates free FIIa by binding to it and carrying it to the liver (Fig. 35.17). Thus, ATIII removes FIIa from the blood, preventing it from participating in coagulation. [Note: ATIII is a serine protease inhibitor, or “serpin.” A serpin contains a reactive loop to which a specific protease binds. Once bound, the protease cleaves a peptide bond in the serpin causing a conformational change that traps the enzyme in a covalent complex. α1-Antitrypsin (see p. 50) is also a serpin.] The affinity of ATIII for FIIA is greatly increased when ATIII is bound to heparin, an intracellular glycosaminogly-can (see p. 159) released in response to injury by mast cells associated with blood vessels. Heparin, an anticoagulant, is used therapeutically to limit clot formation. [Note: In contrast to the anticoagulant warfarin, which has a slow onset and a long half-life and is administered orally, heparin has a rapid onset and a short half-life and requires intravenous administration. The two drugs are commonly used in an overlapping manner in the treatment (and prevention) of thrombosis.] ATIII also inactivates FXa and the other serine proteases of clotting, FIXa, FXIa, FXIIa, and the FVIIa–TF complex. [Note: ATIII binds to a specific pentasaccharide within the oligosaccharide form of heparin. Inhibition of FIIa requires the oligosaccharide form, whereas inhibition of FXa requires only the pentasaccharide form. Fondaparinux, a synthetic version of the pentasaccharide, is used clinically to inhibit FXa.] 2. Protein C–protein S complex: Protein C, a circulating Gla-containing protein made in the liver, is activated by FIIa complexed with thrombomodulin. Thrombomodulin, an integral membrane glycoprotein of endothelial cells, binds FIIa, thereby decreasing FIIa’s affinity for fibrinogen and increasing its affinity for protein C. Protein C in complex with protein S, also a Gla-containing protein, forms the activated protein C (APC) complex that cleaves the accessory proteins FVa and FVIIIa, which are required for maximal activity of FXa (Fig. 35.18). Protein S helps anchor APC to the clot. Thrombomodulin, then, modulates the activity of thrombin, converting it from a protein of coagulation to a protein of anticoagulation, thereby limiting the extent of clotting. Factor V Leiden is a mutant form of FV (glutamine is substituted for arginine at position 506) that is resistant to APC. It is the most common inherited cause of thrombophilia in the United States, with highest frequency in the Caucasian population. Heterozygotes have a 7-fold increase in the risk for venous thrombosis, and homozygotes have up to a 50-fold increase. [Note: Women with FV Leiden are at even greater risk of thrombosis during pregnancy or when taking estrogen.]
Thrombophilia (hypercoagulability) can result from deficiencies of proteins C, S, and ATIII; from the presence of FV Leiden and antiphospholipid antibodies; and from excess production of FII (G20210A mutation). [Note: A thrombus that forms in the deep veins of the leg (deep venous thrombosis, or DVT) can cause a pulmonary embolism (PE) if the clot (or a piece of it) breaks off, travels to the lungs, and blocks circulation.]
B. Fibrinolysis
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Clots are temporary patches that must be removed once wound repair has begun. The fibrin clot is cleaved by the protein plasmin to fibrin degradation products (Fig. 35.19). [Note: Measurement of D-dimer, a fibrin degradation product containing two cross-linked D domains released by the action of plasmin, can be used to assess the extent of clotting (see Fig. 35.12).] Plasmin is a serine protease that is generated from plasminogen by plasminogen activators. Plasminogen, secreted by the liver into the circulation, binds to FIa and is incorporated into clots as they form. Tissue plasminogen activator (TPA, t-PA), made by vascular endothelial cells and secreted in an inactive form in response to FIIa, becomes active when bound to FIa–plasminogen. Bound plasmin and TPAa are protected from their inhibitors, α2-antiplasmin and plasminogen activator inhibitors, respectively. Once the fibrin clot is dissolved, plasmin and TPAa become available to their inhibitors. Therapeutic fibrinolysis in patients with an MI or an ischemic stroke can be achieved by treatment with commercially available TPA made by recombinant DNA techniques. Mechanical clot removal (thrombectomy) is also possible. [Note: Urokinase is a plasminogen activator (u-PA) made in a variety of tissues and originally isolated from urine. Streptokinase (from bacteria) activates both free and fibrin-bound plasminogen.]
Plasminogen contains structural motifs known as “kringle domains” that mediate protein–protein interactions. Because lipoprotein (a) [Lp(a)] also contains kringle domains, it competes with plasminogen for binding to FIa. The potential to inhibit fibrinolysis may be the basis for the association of elevated Lp(a) with increased risk for cardiovascular disease (see p. 236).
IV. PLATELET PLUG FORMATION
Platelets (thrombocytes) are small, anucleate fragments of megakaryocytes that adhere to exposed collagen of damaged endothelium, get activated, and aggregate to form a platelet plug (Fig. 35.20; also see Fig. 35.1). Formation of the platelet plug is referred to as primary hemostasis because it is the first response to bleeding. In a normal adult, there are 150,000–450,000 platelets per µl of blood. They have a life span of up to 10 days, after which they are taken up by the liver and spleen and destroyed. Clinical laboratory tests to measure platelet number and activity are available.
A. Adhesion
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Adhesion of platelets to exposed collagen at the site of vessel injury is mediated by the protein von Willebrand factor (VWF). VWF binds to collagen, and platelets bind to VWF via glycoprotein Ib (GPIb), a component of a membrane receptor complex (GPIb–V–IX) on the platelet surface (Fig. 35.21). Binding to VWF stops the forward movement of platelets. [Note: Deficiency in the receptor for VWF results in Bernard-Soulier syndrome, a disorder of decreased platelet adhesion.] VWF is a glycoprotein that is released from platelets. It also is made and secreted by endothelial cells. In addition to mediating the binding of platelets to collagen, VWF also binds to and stabilizes FVIII in the blood. Deficiency of VWF results in von Willebrand disease (VWD), the most common inherited coagulopathy. VWD results from decreased binding of platelets to collagen and a deficiency in FVIII (due to increased degradation). Platelets can also bind directly to collagen via the membrane receptor glycoprotein VI (GPVI). Once adhered, platelets get activated. [Note: Damage to the endothelium also exposes FIII, initiating the extrinsic pathway of blood clotting and activation of FX (see Fig. 35.8).]
B. Activation
Once adhered to areas of injury, platelets get activated. Platelet activation involves morphologic (shape) changes and degranulation, the process by which platelets secrete the contents of their α and δ (or, dense) storage granules. Activated platelets also expose PS on their surface. The externalization of PS is mediated by a Ca2+-activated enzyme known as scramblase that disrupts the membrane asymmetry created by flippases (see p. 205). Thrombin is the most potent platelet activator. FIIa binds to and activates protease-activated receptors, a type of G protein–coupled receptor (GPCR), on the surface of platelets (Fig. 35.22). FIIa is primarily associated with Gq proteins (see p. 205), resulting in activation of phospholipase C and a rise in diacylglycerol (DAG) and inositol trisphosphate (IP3). [Note: Thrombomodulin, through its binding of FIIa, decreases the availability of FIIa for platelet activation (see Fig. 35.18).] 1. Degranulation: DAG activates protein kinase C, a key event for degranulation. IP3 causes the release of Ca2+ (from dense granules). The Ca2+ activates phospholipase A2, which cleaves membrane phospholipids to release arachidonic acid, the substrate for the synthesis of thromboxane A2 (TXA2) in activated platelets by cyclooxygenase-1 (COX-1) (see p. 214). TXA2 causes vasoconstriction, augments degranulation, and binds to platelet GPCR, causing activation of additional platelets. Recall that aspirin irreversibly inhibits COX and, consequently, TXA2 synthesis and is referred to as an antiplatelet drug.
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Degranulation also results in release of serotonin and adenosine diphosphate (ADP) from dense granules. Serotonin causes vasoconstriction. ADP binds to GPCR on the surface of platelets, activating additional platelets. [Note: Some antiplatelet drugs, such as clopidogrel, are ADP-receptor antagonists.] Platelet-derived growth factor (involved in wound healing), VWF, FV, FXIII, and FI are among other proteins released from α granules. [Note: Platelet-activating factor (PAF), an ether phospholipid (see p. 202) synthesized by a variety of cell types including endothelial cells and platelets, binds PAF receptors (GPCR) on the surface of platelets and activates them.] 2. Morphologic change: The change in shape of activated platelets from discoidal to spherical with pseudopod-like processes that facilitate platelet–platelet and platelet–surface interactions (Fig. 35.23) is initiated by the release of Ca2+ from dense granules. Ca2+ bound to calmodulin (see p. 133) mediates the activation of myosin light chain kinase that phosphorylates the myosin light chain, resulting in a major reorganization of the platelet cytoskeleton.
C. Aggregation
Activation causes dramatic changes in platelets that lead to their aggregation. Structural changes in a surface receptor (GPIIb/IIIa) expose binding sites for fibrinogen. Bound FI molecules link activated platelets to one another (Fig. 35.24), with a single FI able to bind two platelets. FI is converted to FIa by FIIa and then covalently cross-linked by FXIIIa coming from both the blood and the platelets. [Note: The exposure of PS on the surface of activated platelets allows formation of the Xase complex (VIIIa, IXa, X, and Ca2+) with subsequent formation of FXa and generation of FIIa.] Fibrin formation (secondary hemostasis) strengthens the platelet plug. [Note: Rare defects in the platelet receptor for FI result in Glanzmann thrombasthenia (decreased platelet function), whereas autoantibodies to this receptor are a cause of immune thrombocytopenia (decreased platelet number).]
Unnecessary activation of platelets is prevented because 1) an intact vascular wall is separated from the blood by a monolayer of endothelial cells, preventing the contact of platelets with collagen; 2) endothelial cells synthesize prostaglandin I2 (PGI2, or prostacyclin) and nitric oxide, each of which causes vasodilation; and 3) endothelial cells have a cell surface ADPase that converts ADP to adenosine monophosphate.
V. CHAPTER SUMMARY
Blood clotting (coagulation) is designed to rapidly stop bleeding from a damaged blood vessel in order to maintain a constant blood volume (hemostasis). Coagulation is accomplished through formation of a clot (thrombus) consisting of a plug of platelets and a meshwork of the protein fibrin (Fig. 35.25).
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The formation of the fibrin meshwork by the clotting cascade involves the extrinsic and intrinsic pathways (and their associated protein factors [F]) that converge at FXa to form the common pathway. Many of the protein factors are serine proteases with trypsin-like specificity. Calcium binds the negatively charged γ-carboxyglutamate (Gla) residues present in certain of the clotting proteases (FII, FVII, FIX, and FX), facilitating the binding of these proteins to exposed negatively charged phosphatidylserine at the site of injury and on the surface of platelets. γ-Glutamyl carboxylase and its coenzyme, the hydroquinone form of vitamin K, are required for formation of Gla residues. In the reaction, vitamin K gets oxidized to the nonfunctional epoxide form. Warfarin, a synthetic analog of vitamin K used clinically to reduce clotting, inhibits the enzyme vitamin K epoxide reductase that regenerates the functional reduced form. The extrinsic pathway is initiated by exposure of FIII (tissue factor [TF]), an accessory protein, in vascular subendothelium. Exposed TF binds a circulating Glacontaining protein, FVII, activating it through conformational change. The TF–FVIIa complex then binds and activates FX by proteolysis. FXa from the extrinsic pathway allows thrombin production by the common pathway. Thrombin then activates components of the intrinsic pathway. The extrinsic pathway is rapidly inhibited by tissue factor pathway inhibitor. The intrinsic pathway is initiated by FXIIa. FXIIa activates FXI, and FXIa activates FIX. FIXa combines with FVIIIa (an accessory protein), and the complex activates FX. FVIII deficiency results in hemophilia A, whereas FIX deficiency results in the less common hemophilia B. FXa associates with FVa (an accessory protein), forming prothrombinase that cleaves prothrombin (FII) to thrombin (FIIa). Thrombin then cleaves fibrinogen to fibrin (FIa). Fibrin monomers associate, forming a soluble (soft) fibrin clot. The fibrin molecules get cross-linked by FXIIIa, a transglutaminase, forming an insoluble (hard) fibrin clot. Proteins synthesized by the liver and by blood vessels themselves balance coagulation with anticoagulation. Antithrombin III, a serine protease inhibitor, or serpin, binds to and removes thrombin from the blood. Its affinity for thrombin is increased by heparin, which is used therapeutically to limit clot formation. Protein C, a Gla-containing protein, is activated by the thrombin–thrombomodulin complex. Thrombomodulin decreases thrombin’s affinity for fibrinogen, converting it from a protein of coagulation to a protein of anticoagulation. Protein C in complex with protein S (a Gla-containing protein) forms the activated protein C (APC) complex that cleaves the accessory proteins FVa and FVIIIa. FV Leiden is resistant to APC. It is the most common inherited thrombophilic condition in the United States. The fibrin clot is cleaved (fibrinolysis) by the protein plasmin, a serine protease that is generated from plasminogen by plasminogen activators such as tissue plasminogen activator (TPA, t-PA).
Recombinant TPA is used clinically.
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Wound to a tissue damages blood vessels and exposes collagen. Platelets (thrombocytes) adhere to the exposed collagen, get activated, and aggregate to form a platelet plug. Adhesion is mediated by von Willebrand factor (VWF). VWF binds collagen, and platelets bind VWF via glycoprotein Ib (GPIb) within a receptor complex on the platelet surface. Deficiency of VWF results in von Willebrand disease, the most common inherited coagulopathy. Once adhered, platelets get activated. Platelet activation involves changes in shape (discoidal to spherical with pseudopodia) and degranulation, the process by which platelets release the contents of their storage granules. Thrombin is the most potent activator of platelets. Thrombin binds to protease-activated G protein–coupled receptors on the surface of platelets. Activated platelets release substances that cause vasoconstriction (serotonin and thromboxane A2 [TXA2]), recruit and activate other platelets (adenosine diphosphate and TXA2), and support the formation of a fibrin clot (FV, FXIII, and fibrinogen). Activation causes changes in platelets that lead to their aggregation. Structural changes in a surface receptor (GPIIb/IIIa) expose binding sites for fibrinogen. Fibrinogen molecules link activated platelets to one another. The fibrinogen is activated to fibrin by thrombin and then cross-linked by FXIIIa coming both from the blood and from platelets. The initial loose plug of platelets (primary hemostasis) is strengthened by the fibrin meshwork (secondary hemostasis). Disorders of platelets and coagulation proteins can result in deviations in the ability to clot. Prothrombin time (PT) and activated partial thromboplastin time (aPTT) are clinical laboratory tests used to evaluate the clotting cascade.
Choose the ONE best answer.
For Questions 31.1–31.5, match the most appropriate protein factors (F) of clotting to the description.
5.1. This factor activates components of the intrinsic, extrinsic, and common pathways.
5.2. This factor converts the soluble clot to an insoluble clot.
5.3. This factor initiates the common pathway.
5.4. This factor is an accessory protein that potentiates the activity of factor Xa.
5.5. This factor is a γ-carboxyglutamate–containing serine protease of the extrinsic pathway.
Correct answers = B, J, H, D, E. Thrombin (FII) is formed in the common pathway and activates components in each of the three pathways of the clotting cascade. FXIII, a transglutaminase, covalently cross-links associated fibrin monomers, thereby converting a soluble clot to an insoluble one. The generation of FXa by the intrinsic and extrinsic pathways initiates the common pathway. FV increases the activity of FXa. It is one of three accessory (nonprotease) proteins. The others are FIII (tissue factor) and FVIII (complexes with FIX to activate FX). FVII is a γ-carboxyglutamate–containing serine protease that complexes with FIII in the extrinsic pathway.
5.6. In which patient would prothrombin time be unaffected and activated partial thromboplastin time be prolonged?
A. A patient on aspirin therapy
B. A patient with end-stage liver disease
C. A patient with hemophilia
D. A patient with thrombocytopenia
Correct answer = C. Prothrombin time (PT) measures the activity of the extrinsic through the common pathways, and activated partial thromboplastin time (aPTT) measures the activity of the intrinsic through the common pathways. Patients with hemophilia are deficient in either FVIII (hemophilia
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A) or FIX (hemophilia B), components of the common pathway. They have an intact extrinsic pathway. Therefore, the PT is unaffected, and the aPTT is prolonged. Patients on aspirin therapy and those with thrombocytopenia have alterations in platelet function and number, respectively, and not in the proteins of the clotting cascade. Therefore, both the PT and the aPTT are unaffected. Patients with end-stage liver disease have decreased ability to synthesize clotting proteins. They show prolonged PT and aPTT.
5.7. Which one of the following can be ruled out in a patient with thrombophilia?
A. A deficiency of antithrombin III
B. A deficiency of FIX
C. A deficiency of protein C
D. An excess of prothrombin
E. Expression of FV Leiden
Correct answer = B. Symptomatic deficiencies in clotting factors will present with a decreased ability to clot (coagulopathy). Thrombophilia, however, is characterized by an increased tendency to clot. Choices A, C, D, and E result in thrombophilia.
5.8. Current guidelines for the treatment of patients with acute ischemic stroke (a stroke caused by a blood clot obstructing a vessel that supplies blood to the brain) include the recommendation that tissue plasminogen activator (TPA) be used shortly after the onset of symptoms. The basis of the recommendation for TPA is that it activates:
A. antithrombin III.
B. the activated protein C complex.
C. the receptor for von Willebrand factor.
D. the serine protease that degrades fibrin.
E. thrombomodulin.
Correct answer = D. TPA converts plasminogen to plasmin. Plasmin (a serine protease) degrades the fibrin meshwork, removing the obstruction to blood flow. Antithrombin III in association with heparin binds thrombin and carries it to the liver, decreasing thrombin's availability in the blood. The activated protein C complex degrades the accessory proteins FV and FVIII. The platelet receptor for von Willebrand factor is not affected by TPA. Thrombomodulin binds thrombin and converts it from a protein of coagulation to one of anticoagulation by decreasing its activation of fibrinogen and increasing its activation of protein C.
5.9. The adhesion, activation, and aggregation of platelets provide the initial plug at the site of vessel injury. Which of the following statements concerning the formation of this platelet plug is correct?
A. Activated platelets undergo a shape change that decreases their surface area.
B. Formation of a platelet plug is prevented in intact vessels by the production of thromboxane A2 by endothelial cells.
C. The activation phase requires production of cyclic adenosine monophosphate.
D. The adhesion phase is mediated by the binding of platelets to von Willebrand factor via glycoprotein Ib.
E. Thrombin activates platelets by binding to a protease-activated G protein–coupled receptor and causing activation of protein kinase A.
Correct answer = D. The adhesion phase of platelet plug formation is initiated by the binding of von Willebrand factor to a receptor (glycoprotein Ib) on the surface of platelets. Shape change from discoidal to spherical with pseudopodia increases the surface area of platelets. Thromboxane A2 is made by platelets. It causes platelet activation and vasoconstriction. Adenosine diphosphate is released from activated platelets, and it itself activates platelets. Thrombin works primarily through receptors coupled to Gq proteins causing activation of phospholipase C.
5.11. Nephrotic syndrome is a kidney disease characterized by protein loss in the urine (≥3 g/day) that is accompanied by edema. The loss of protein results in a hypercoagulable state. Excretion of which of the following proteins would explain the thrombophilia seen in the syndrome?
A. Antithrombin III
B. FV
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C. FVIII
D. Prothrombin Correct answer = A. Antithrombin III (ATIII) inhibits the action of thrombin (FIIa), a Gla-containing protein of clotting that activates the extrinsic, intrinsic, and common pathways. Excretion of ATIII in nephrotic syndrome allows the actions of FIIa to continue, resulting in a hypercoagulable state. The other choices are proteins required for clotting. Their excretion in the urine would decrease clotting.
5.12. Blocking the action of which of the following proteins would be a rational therapy for hemophilia B?
A. FIX
B. FXIII
C. Protein C
D. Tissue factor pathway inhibitor
Correct answer = D. Hemophilia B is a coagulopathy caused by decreased thrombin production by the common pathway as a result of a deficiency in FIX of the intrinsic pathway. Because the extrinsic pathway also can result in thrombin production, blocking the inhibitor of this pathway (tissue factor pathway inhibitor) should, in principle, increase thrombin production.
5.13. The parents of a newborn baby girl refuse to allow the baby to be given the injection of vitamin K that is recommended shortly after birth to prevent vitamin K deficiency bleeding, which is caused by the low levels of the vitamin in newborns. The activity of which one of the following protein factors involved in clotting would be decreased with vitamin K deficiency?
A. FV
B. FVII
C. FXI
D. FXII
E. FXIII
Correct answer = B. FVII is a γ-carboxyglutamate (Gla)-containing protein of clotting. The creation of Gla residues by γ-glutamyl carboxylase requires vitamin K as a coenzyme. FII, FIX, and FX, as well as proteins C and S that limit clotting, also contain Gla residues. The other choices do not contain Gla residues.
5.14. Thrombin, produced in the common pathway of clotting, has both procoagulant and anticoagulant activities. Which one of the following is an anticoagulant activity of thrombin?
A. Activating FXIII
B. Binding to thrombomodulin
C. Increasing nitric oxide production
D. Inhibiting FV and FVIII
E. Inhibiting platelet activation
F. Inhibiting tissue factor pathway inhibitor
Correct answer = B. Thrombin bound to thrombomodulin activates protein C that degrades the accessory proteins FV and FVIII, thereby inhibiting clotting. Activation of FXIII by thrombin strengthens the fibrin clot. Nitric oxide, a vasodilator made by endothelial cells, decreases clot formation. It is not affected by thrombin. Thrombin is a powerful activator of platelets. Inhibition of tissue factor pathway inhibitor would increase clotting.
5.15. A student is reviewing the use of prothrombin time (PT) and activated partial thromboplastin time (aPTT) in evaluating a suspected deficiency of a clotting protein. Which one of the following results would be correct for a deficiency in FXIII?
A. Both prothrombin time and activated partial thromboplastin time are decreased.
B. Both prothrombin time and activated partial thromboplastin time are increased.
C. Both prothrombin time and activated partial thromboplastin time are unchanged.
D. Only prothrombin time is affected.
E. Only activated partial thromboplastin time is affected.
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Correct answer = C. FXIII is a transglutaminase that cross-links fibrin molecules in a soft clot to form a hard clot. Its deficiency does not affect the PT or aPTT tests. [Note: It is evaluated by a clot solubility test.] 5.16. Why do individuals with Scott syndrome, a rare disorder caused by mutations to scramblase in platelets, have a tendency to bleed?
Scramblase moves phosphatidylserine (PS) from the cytosolic leaflet to the extracellular leaflet in the plasma membrane of platelets. This disrupts the asymmetrical localization of membrane phospholipids created by ATP-dependent flippases (move PS from extracellular to cytosolic leaflet) and floppases (move phosphatidylcholine [PC] in the opposite direction). Having PS on the outer face of platelet membranes provides a site for protein clotting factors to interact and activate thrombin. If scramblase is inactive, PS is not available to these factors, and bleeding results.
5.10. Several days after having had their home treated for an infestation of rats, the parents of a 3-year-old girl become concerned that she might be ingesting the poison-containing pellets. After calling the Poison Hotline, they take her to the emergency department. Blood studies reveal a prolonged prothrombin and activated partial thromboplastin time and a decreased concentration of FII, FVII, FIX, and FX. Why might administration of vitamin K be a rational approach to the treatment of this patient?
Many rodent poisons are super warfarins, drugs that have a long half-life in the body. Warfarin inhibits γ-carboxylation (production of γ-carboxyglutamate, or Gla, residues), and the clotting proteins reported as decreased are the Glacontaining proteases of the clotting cascade. [Note: Proteins C and S of anticlotting are also Gla-containing proteins.] Because warfarin functions as a vitamin K antagonist, administration of vitamin K is a rational approach to treatment.
I. I. Integrative Cases
Metabolic pathways, initially presented in isolation, are, in fact, linked to form an interconnected network. The following four integrative case studies illustrate how a perturbation in one process can result in perturbations in other processes of the network.
Case 1: Chest Pain
Patient Presentation: BJ, a 35-year-old man with severe substernal chest pain of ~2 hours’ duration, is brought by ambulance to his local hospital at 5 AM. The pain is accompanied by dyspnea (shortness of breath), diaphoresis (sweating), and nausea.
Focused History: BJ reports episodes of exertional chest pain in the last few months, but they were less severe and of short duration. He smokes (2–3 packs per day), drinks alcohol only rarely, eats a “typical” diet, and walks with his wife most weekends. His blood pressure has been normal. Family history reveals that his father and paternal aunt died of heart disease at age 45 and 39 years, respectively. His mother and younger (age 31 years) brother are said to be in good health.
Physical Examination (Pertinent Findings): BJ is pale and clammy and is in distress due to chest pain. Blood pressure and respiratory rate are elevated. Lipid deposits are noted on the periphery of his corneas (corneal arcus; see left image) and under the skin on and around his eyelids (xanthelasmas; see right image). No deposits on his tendons (xanthomas) are detected.
Pertinent Test Results: BJ’s electrocardiogram is consistent with a myocardial infarction (MI). Angiography reveals areas of severe stenosis (narrowing) of several coronary arteries. Initial results from the clinical laboratory include the following:
H = High; L = Low. [Note: BJ had not eaten for ~8 hours prior to the blood draw.]
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Diagnosis: MI, the irreversible necrosis (death) of heart muscle secondary to ischemia (decreased blood supply), is caused by the occlusion (blockage) of a blood vessel most commonly by a blood clot (thrombus). BJ subsequently is determined to have heterozygous familial hypercholesterolemia (FH), also known as type IIa hyperlipidemia.
Immediate Treatment: BJ is given O2, a vasodilator, pain medication, and drugs to dissolve blood clots (thrombolytics) and reduce clotting (antithrombotics).
Long-Term Treatment: Lipid-lowering drugs (for example, high-potency statins, bile acid [BA] sequestrants, and niacin); daily aspirin; β-blockers; and counseling on nutrition, exercise, and smoking cessation would be part of the long-term treatment plan.
Prognosis: Patients with heterozygous FH have ~50% of the normal number of functional LDL receptors and a hypercholesterolemia (two to three times normal) that puts them at high risk (>50% risk) for premature coronary heart disease (CHD). However, <5% of patients with hypercholesterolemia have FH.
Nutrition Nugget: Dietary recommendations for individuals with heterozygous FH include limiting saturated fats to <7% of total calories and cholesterol to <200 mg/day, substituting unsaturated fats for saturated fats, and adding soluble fiber (10–20 g/day) and plant sterols (2 g/day) for their hypocholesterolemic effects. Fiber increases BA excretion. This results in increased hepatic uptake of cholesterol-rich LDL to supply the substrate for BA synthesis. Plant sterols decrease cholesterol absorption in the intestine.
Genetics Gem: FH is caused by hundreds of different mutations in the gene for the LDL receptor (on chromosome 19) that affect receptor amount and/or function. FH is an autosomal-dominant disease in which homozygotes are more seriously affected than heterozygotes. Heterozygous FH has an incidence of ~1:500 in the general population. It is associated with increased risk of cardiovascular disease. Genetic screening of the first-degree relatives of BJ would identify affected individuals for treatment.
Review Questions: Choose the ONE best answer.
RQ1. Triacylglycerols are glycerol-based lipids. Which of the following is also a glycerol-based lipid? A. Ganglioside GM2
B. Phosphatidylcholine C. Prostaglandin PGI2
D. Sphingomyelin
E. Vitamin D RQ2. Statins are of benefit to patients with hypercholesterolemia because they:
A. decrease a rate-limiting and regulated step of de novo cholesterol biosynthesis by inhibiting hydroxymethylglutaryl coenzyme A (HMG CoA) reductase.
B. decrease expression of the gene for the LDL receptor by preventing the movement of the sterol regulatory element–binding protein-2 (SREBP 2) in complex with SREBP cleavage–activating protein (SCAP) from the membrane of the endoplasmic reticulum to the membrane of the Golgi.
C. increase the oxidation of cholesterol to CO2 + H2O.
D. interfere with the absorption of bile salts in the enterohepatic circulation, thereby causing the liver to take up cholesterol from the blood for use in BA synthesis.
E. reduce cholesterol by increasing steroid hormone and vitamin D synthesis.
RQ3. Statins are competitive inhibitors of HMG CoA reductase. Which of the following statements about competitive inhibitors is correct?
A. Competitive inhibitors are examples of irreversible inhibitors.
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B. Competitive inhibitors increase both the apparent Michaelis constant (Km) and the apparent maximal velocity (Vmax).
C. Competitive inhibitors increase the apparent Km and have no effect on the Vmax.
D. Competitive inhibitors decrease both the apparent Km and the apparent
Vmax.
E. Competitive inhibitors have no effect on the Km and decrease the apparent Vmax. RQ4. In an MI, a blood clot forms as a result of injury to a blood vessel that leads to production of a platelet plug and a fibrin meshwork. The clot occludes the blood vessel, preventing blood flow and, therefore, delivery of O2. Destruction of the clot (thrombolysis) restores blood flow. Which one of the following is an example of a thrombolytic agent?
A. Activated protein C complex
B. Antithrombin III
C. Aspirin
D. Factor XIII
E. Heparin
F. Tissue plasminogen activator
G. Vitamin K
H. Warfarin
RQ5. Decreased tissue perfusion results in hypoxia (decreased O2 availability). Relative to normoxia, in hypoxia the:
A. electron transport chain will be upregulated to provide protons for ATP synthesis.
B. ratio of the oxidized form of nicotinamide adenine dinucleotide (NAD+) to the reduced form (NADH) will increase.
C. pyruvate dehydrogenase complex will be active.
D. process of substrate-level phosphorylation will be increased in the cytosol.
E. tricarboxylic acid cycle will be upregulated to provide the reducing equivalents needed for oxidative phosphorylation to occur.
RQ6. Genetic screening of BJ’s first-degree relatives would be accomplished by mutation analysis via polymerase chain reaction–based amplification followed by automated sequencing of the promoter region and the 18 exons of the LDL receptor gene. This process would involve the:
A. generation and use of complementary DNA (cDNA).
B. initiation of DNA synthesis with dideoxynucleotides.
C. isolation of genomic DNA from germ cells.
D. use of fluorescently labeled nucleotides.
TQ1. Relative to an individual with familial defective LDL receptors, what would be the expected phenotype in an individual with familial defective apolipoprotein B-100? With apolipoprotein E-2, the isoform that only poorly binds its receptor?
TQ2. Why was aspirin prescribed? Hint: What pathway of lipid metabolism is affected by aspirin?
TQ3. Heart muscle normally uses aerobic metabolism to meet its energy needs. However, in hypoxia, anaerobic glycolysis is increased. What allosteric activator of glycolysis is responsible for this effect? With hypoxia, what will be the end product of glycolysis?
TQ4. One of the reasons for encouraging smoking cessation and exercise for BJ is that these changes raise the level of HDL, and elevated HDL reduces the risk for CHD. How does a rise in HDL reduce the risk for CHD?
Case 2: Severe Fasting Hypoglycemia
Patient Presentation: JS is a 4-month-old boy whose mother is concerned about the “twitching” movements he makes just before feedings. She tells the pediatrician that the movements started ~1 week ago, are most apparent in the morning, and disappear shortly after eating.
Focused History: JS is the product of a normal pregnancy and delivery. He appeared normal at birth. On his growth charts, he has been at the 30th percentile for both weight and length since birth. His immunizations are up to date. JS last ate a few hours ago.
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Physical Examination (Pertinent Findings): JS appears sleepy and feels clammy to the touch. His respiratory rate is elevated. His temperature is normal. JS has a protuberant, firm abdomen that appears to be nontender. His liver is palpable 4 cm below the right costal margin and is smooth. His kidneys are enlarged and symmetrical.
Pertinent Test Results:
H = High; L = Low.
JS is sent to the regional children’s hospital for further evaluation. Ultrasound studies confirm hepatomegaly and renomegaly and show no evidence of tumors. A liver biopsy is performed. The hepatocytes are distended. Staining reveals large amounts of lipid (primarily triacylglycerol) and carbohydrate. Liver glycogen is elevated in amount and normal in structure. Enzyme assay using liver homogenate treated with detergent reveals <10% of the normal activity of glucose 6-phosphatase, an enzyme of the endoplasmic reticular (ER) membrane in the liver and the kidneys.
Diagnosis: JS has glucose 6-phosphatase deficiency (glycogen storage disease [GSD] type Ia, von Gierke disease).
Treatment (Immediate): JS was given glucose intravenously, and his blood glucose level rose into the normal range. However, as the day progressed, it fell to well below normal. Administration of glucagon had no effect on blood glucose levels but increased blood lactate. JS’s blood glucose levels were able to be maintained by constant infusion of glucose.
Prognosis: Individuals with glucose 6-phosphatase deficiency develop hepatic adenomas starting in the second decade of life and are at increased risk for hepatic carcinoma. Kidney glomerular function is impaired and can result in kidney failure. Patients are at increased risk for developing gout, but this rarely occurs before puberty.
Nutrition Nugget: Long-term medical nutrition therapy for JS is designed to maintain his blood glucose levels in the normal range. Frequent (every 2–3 hours) daytime feedings rich in carbohydrate (provided by uncooked cornstarch that is slowly hydrolyzed) and nighttime nasogastric infusion (pump assisted) of glucose are advised. Avoidance of fructose and galactose is recommended because they are metabolized to glycolytic intermediates and lactate, which can exacerbate the metabolic problems. Calcium and vitamin D supplements are prescribed.
Genetics Gem: GSD Ia is an autosomal-recessive disorder caused by >100 known mutations to the gene for glucose 6-phosphatase located on chromosome 17. It has an incidence of 1:100,000 and accounts for ~25% of all cases of GSD in the United States. It is one of the few genetic causes of hypoglycemia in newborns. GSD Ia is not routinely screened for in newborns. [Note: Deficiency of the translocase that moves glucose 6phosphate into the ER is the cause of GSD Ib. Hypoglycemia and neutropenia are seen.]
Review Questions: Choose the ONE best answer.
RQ1. JS is hypoglycemic because:
A. free (nonphosphorylated) glucose cannot be produced from either glycogenolysis or gluconeogenesis as a result of the deficiency in glucose 6-phosphatase.
B. glycogen phosphorylase is dephosphorylated and inactive, and glycogen cannot be degraded.
C. hormone-sensitive lipase is dephosphorylated and inactive, and fatty acid substrates for gluconeogenesis cannot be generated.
D. the decrease in the insulin/glucagon ratio upregulates glucose transporters in the liver and kidneys, resulting in increased uptake of blood glucose.
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RQ2. JS was prescribed calcium supplements because chronic acidosis can cause bone demineralization, resulting in osteopenia. Vitamin D (1,25-diOH-D3) was also prescribed because vitamin D:
A. binds Gq protein–coupled membrane receptors and causes a rise in inositol trisphosphate with release of calcium from intracellular stores.
B. cannot be synthesized by humans and, therefore, must be supplied in the diet.
C. is a fat-soluble vitamin that increases intestinal absorption of calcium.
D. is the coenzyme-prosthetic group for calbindin, a calcium transporter in the intestine.
RQ3. The hepatomegaly and renomegaly seen in JS are primarily the result of an increase in the amount of glycogen stored in these organs. What is the basis for glycogen accumulation in these organs?
A. Glycolysis is downregulated, which pushes glucose to glycogenesis.
B. Increased oxidation of fatty acids spares glucose for glycogenesis.
C. Glucose 6-phosphate is an allosteric activator of glycogen synthase b.
D. The rise in the insulin/glucagon ratio favors glycogenesis.
RQ4. Glucose 6-phosphatase is an integral protein of the ER membrane. Which of the following statements about such proteins is correct?
A. If glycosylated, the carbohydrate is on the portion of the protein that extends into the cytosol.
B. They are synthesized on ribosomes that are free in the cytosol.
C. The membrane-spanning domain consists of hydrophilic amino acids.
D. The initial targeting signal is an amino terminal hydrophobic signal sequence.
TQ1. What is the likely reason for JS’s twitching movements? TQ2. Why was the liver homogenate treated with detergent? Hint: Think about where the enzyme is located. TQ3. Why is JS’s blood glucose level unaffected by glucagon? Hint: What is the role of glucagon in normal individuals who experience a drop in blood glucose?
TQ4. Why are urate and lactate elevated in a disorder of glycogen metabolism? Hint: It is the result of a decrease in inorganic phosphate (Pi), but why is Pi decreased?
TQ5. A.Why are triacylglycerols and cholesterol elevated? Hint: Glucose is the primary carbon source for their synthesis. B.Why are ketone bodies not elevated?
Case 3: Hyperglycemia and Hyperketonemia
Patient Presentation: MW, a 40-year-old woman, was brought to the hospital in a disoriented, confused state by her husband.
Focused History: As noted on her medical alert bracelet, MW has had type 1 diabetes (T1D) for the last 24 years. Her husband reports that this is her first medical emergency in 2 years.
Physical Examination (Pertinent Findings): MW displayed signs of dehydration (such as dry mucous membranes and skin, poor skin turgor, and low blood pressure) and acidosis (such as deep, rapid breathing [Kussmaul respiration]). Her breath had a faintly fruity odor. Her temperature was normal.
Pertinent Test Results: Rapid, bedside tests were strongly positive for glucose and acetoacetate and negative for protein. Results on blood tests performed by the clinical laboratory are shown below:
H=High;L=Low.
Microscopic examination of her urine revealed a urinary tract infection (UTI).
Diagnosis: MW is in diabetic ketoacidosis (DKA) that was precipitated by a UTI. [Note: Diabetes increases the risk for infections such as UTI.]
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Immediate Treatment: MW was rehydrated with normal saline given intravenously (IV). She also was given insulin IV. Blood glucose, ketone bodies, and electrolytes were measured periodically. Antibiotic treatment of her UTI was started.
Long-Term Treatment: Diabetes increases the risk for macrovascular complications (such as coronary artery disease and stroke) and microvascular complications (such as retinopathy, nephropathy, and neuropathy). Ongoing monitoring for these complications will be continued.
Prognosis: Diabetes is the seventh leading cause of death by disease in the United States. Individuals with diabetes have a reduced life expectancy relative to those without diabetes.
Nutrition Nugget: Monitoring total intake of carbohydrates is primary in blood glucose control. Carbohydrates should come from whole grains, vegetables, legumes, and fruits. Low-fat dairy products and nuts and fish rich in ω-3 fatty acids are encouraged. Intake of saturated and trans fats should be minimized.
Genetics Gem: Autoimmune destruction of pancreatic β cells is characteristic of T1D. Of the genetic loci that confer risk for T1D, the human-leukocyte antigen (HLA) region on chromosome 6 has the strongest association. The majority of genes in the HLA region are involved in the immune response.
Review Questions: Choose the ONE best answer.
RQ1. Which of the following statements concerning T1D is correct?
A. Diagnosis can be made by measuring the level of glucose or glycated hemoglobin (HbA1c) in the blood.
B. During periods of physiologic stress, the urine of an individual with T1D would likely test negative for reducing sugars.
C. T1D is associated with obesity and a sedentary lifestyle.
D. The characteristic metabolic abnormalities seen in T1D result from insensitivity to both insulin and glucagon.
E. Treatment with exogenous insulin allows normalization of blood glucose (euglycemia).
RQ2. DKA occurs when the rate of ketone body production is greater than the rate of utilization. Which of the following statements concerning ketone body metabolism is correct? Ketone bodies:
A. are made in mitochondria from acetyl coenzyme A (CoA) primarily produced by the oxidation of glucose.
B. are utilized by many tissues, particularly the liver, after conversion to acetyl CoA.
C. include acetoacetate, which can impart a fruity odor to the breath.
D. require albumin for transport through the blood.
E. utilized in energy metabolism are organic acids that can add to the proton load of the body.
RQ3. Adipose lipolysis followed by β-oxidation of the fatty acid (FA) products is required for the generation of ketone bodies. Which of the following statements concerning the generation and use of FA is correct?
A. Mitochondrial β-oxidation of FA is inhibited by malonyl CoA.
B. Production of FA from adipose lipolysis is upregulated by insulin.
C. The acetyl CoA product of FA β-oxidation favors the use of pyruvate for gluconeogenesis by activating the pyruvate dehydrogenase complex.
D. The β-oxidation of FA utilizes reducing equivalents generated by gluconeogenesis.
E. The FA produced by lipolysis are taken up by the brain and oxidized for energy.
TQ1. At admission, MW was hypoinsulinemic, and she was given insulin. Why did MW’s hypoinsulinemia result in hyperglycemia? Hint: What is the role of insulin in glucose metabolism?
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TQ2. Why is there glucose in MW’s urine (glucosuria)? How is the glucosuria related to her dehydrated state? TQ3. Why is the majority of the acetyl CoA from FA β-oxidation being used for ketogenesis rather than being oxidized in the tricarboxylic acid cycle? TQ4. Was MW in positive or negative nitrogen balance when she was brought to the hospital?
TQ5. What response to the DKA is apparent in MW? What response is likely occurring in the kidney? Hint: In addition to conversion to urea, how is toxic ammonia removed from the body?
TQ6. What would be true about the levels of ketone bodies and glucose during periods of physiologic stress in individuals with impaired FA oxidation?
Case 4: Hypoglycemia, Hyperketonemia, and Liver Dysfunction
Patient Presentation: AK, a 59-year-old male with slurred speech, ataxia (loss of skeletal muscle coordination), and abdominal pain, was dropped off at the Emergency Department (ED).
Focused History: AK is known to the ED staff from previous visits. He has a 6year history of chronic, excessive alcohol consumption. He is not known to take illicit drugs. At this ED visit, AK reports that he has been drinking heavily in the past day or so. He cannot recall having eaten anything in that time. There is evidence of recent vomiting, but no blood is apparent.
Physical Examination (Pertinent Findings): The physical examination was remarkable for AK’s emaciated appearance. (His body mass index was later determined to be 17.5, which put him in the underweight category.) His facial cheeks were erythematous (red in color) due to dilated blood vessels in the skin (telangiectasia). Eye movement was normal. Neither icterus (jaundice) nor edema (swelling due to fluid retention) was seen. The liver was slightly enlarged. Bedside tests revealed hypoglycemia and hyperketonemia (as acetoacetate). Blood was drawn and sent to the clinical laboratory.
Pertinent Test Results:
DUI = driving under the influence; H = High; L = Low.
Additional Tests: Complete blood count (CBC) and blood smear revealed a macrocytic anemia (see right image). Folate and B12 levels were ordered.
Diagnosis: AK is diagnosed with alcoholism.
Treatment (Immediate): Thiamine and glucose were given intravenously.
Prognosis: Alcoholism (alcohol dependence) is the third most common cause of preventable death in the United States. People with alcoholism are at increased risk for liver cirrhosis, pancreatitis, gastrointestinal bleeding, and some cancers.
Nutrition Nugget: Those with alcoholism are at risk for vitamin deficiencies as a result of decreased intake and absorption. Thiamine (vitamin B1) deficiency is common and can have serious consequences such as Wernicke-Korsakoff syndrome with its neurologic effects.
Thiamine pyrophosphate (TPP), the coenzyme form, is required for the dehydrogenase-mediated oxidation of α-keto acids (such as pyruvate) as well as the transfer of two-carbon ketol groups by transketolase in the reversible sugar interconversions in the pentose phosphate pathway.
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Genetics Gem: Acetaldehyde, the product of ethanol oxidation by the hepatic, cytosolic, nicotinamide adenine dinucleotide (NAD+)-requiring enzyme alcohol dehydrogenase (ADH), is oxidized to acetate by the mitochondrial, NAD+-requiring aldehyde dehydrogenase (ALDH2). The majority of individuals of East Asian (but not European or African) heritage have a single nucleotide polymorphism (SNP) that renders ALDH2 essentially inactive. This results in aldehyde-induced facial flushing and mild to moderate intoxication after consumption of small amounts of ethanol.
Review Questions: Choose the ONE best answer.
RQ1. Many of the metabolic consequences of chronic excessive alcohol consumption seen in AK are the result of an increase in the ratio of reduced nicotinamide adenine dinucleotide (NADH) to its oxidized form (NAD+) in both the cytoplasm and mitochondria. Which of the following statements concerning the effects of the rise in mitochondrial NADH is correct?
A. Fatty acid oxidation is increased.
B. Gluconeogenesis is increased.
C. Lipolysis is inhibited.
D. The tricarboxylic acid cycle is inhibited.
E. The reduction of malate to oxaloacetate in the malate–aspartate shuttle is increased.
RQ2. Ethanol can also be oxidized by cytochrome P450 (CYP) enzymes, and CYP2E1 is an important example. CYP2E1, which is ethanol inducible, generates reactive oxygen species (ROS) in its metabolism of ethanol.
Which of the following statements concerning the CYP proteins is correct?
A. CYP proteins are heme-containing dioxygenases.
B. CYP proteins of the inner mitochondrial membrane are involved in detoxification reactions.
C. CYP proteins of the smooth endoplasmic reticular membrane are involved in the synthesis of steroid hormones, bile acids, and calcitriol.
D. ROS such as hydrogen peroxide generated by CYP2E1 can be oxidized by glutathione peroxidase.
E. The pentose phosphate pathway is an important source of the nicotinamide adenine dinucleotide phosphate (NADPH) that provides the reducing equivalents needed for activity of CYP proteins and the regeneration of functional glutathione.
RQ3. Alcohol is known to modulate the levels of serotonin in the central nervous system, where the monoamine functions as a neurotransmitter. Which of the following statements about serotonin is correct? Serotonin is:
A. associated with anxiety and depression.
B. degraded via methylation by monoamine oxidase, which also degrades the catecholamines.
C. released by activated platelets.
D. synthesized from tyrosine in a two-step process that utilizes a tetrahydrobiopterin-requiring hydroxylase and a pyridoxal phosphate– requiring carboxylase.
RQ4. Chronic, excessive consumption of alcohol is a leading cause of acute pancreatitis, a painful inflammatory condition that results from autodigestion of the gland by premature activation of pancreatic enzymes. Which of the following statements concerning the pancreas is correct?
A. Autodigestion of the pancreas would be expected to result in a decrease in pancreatic proteins in the blood.
B. In individuals who progress from acute to chronic pancreatitis, with the characteristic structural changes that result in decreased pancreatic function, diabetes and steatorrhea are expected findings.
C. In response to secretin, the exocrine pancreas secretes protons to lower the pH in the intestinal lumen.
D. Pancreatitis may also be seen in individuals with hypercholesterolemia.
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TQ1. A. What effect does the rise in cytosolic NADH seen with ethanol metabolism have on glycolysis? Hint: What coenzyme is required in glycolysis?
B. How does this relate to the fatty liver (hepatic steatosis) commonly seen in alcohol-dependent individuals? TQ2. Why might individuals with a history of gouty attacks be advised to reduce their consumption of ethanol? TQ3. Why might prothrombin time be affected in alcohol-dependent individuals? TQ4. Folate and vitamin B12 deficiencies cause a macrocytic anemia that may be seen in those with alcoholism. Why is it advisable to measure vitamin B12 levels before supplementing with folate in an individual with macrocytic anemia?
Case 1: Answers to Review Questions
RQ1.Answer = B. Phosphatidylcholine is a glycerol-based phospholipid derived from diacylglycerol phosphate (phosphatidic acid) and cytidine diphosphatecholine. Gangliosides are derived from ceramides, lipids with a sphingosine backbone. Prostaglandins of the 2 series (such as PGI2) are derived from the 20 carbon polyunsaturated fatty acid arachidonic acid. Sphingomyelin is a sphingophospholipid derived from ceramide. Vitamin D is derived from an intermediate in the biosynthetic pathway for the sterol cholesterol. RQ2.Answer = A. Statins inhibit hydroxymethylglutaryl coenzyme A (HMG CoA) reductase, thereby preventing the nicotinamide adenine dinucleotide phosphate (NADPH)-dependent reduction of HMG CoA to mevalonate and decreasing cholesterol biosynthesis (see figure below). The decrease in cholesterol content caused by statins results in movement of the sterol regulatory element–binding protein-2 (SREBP-2) in complex with SREBP cleavage– activating protein (SCAP) from the endoplasmic reticular membrane to the Golgi membrane. SREBP-2 is cleaved, generating a transcription factor that moves to the nucleus and binds to the sterol regulatory element upstream of the genes for HMG CoA reductase and the low-density lipoprotein (LDL) receptor, increasing their expression. Humans are unable to degrade the steroid nucleus to CO2 +
H2O. Bile acid (BA) sequestrants, such as cholestyramine, prevent the absorption of bile salts by the liver, thereby increasing their excretion. The liver then takes up cholesterol via the LDL receptor and uses it to make BA, thereby reducing blood cholesterol levels. Steroid hormones are synthesized from cholesterol, and vitamin D is synthesized in skin from an intermediate (7dehydrocholesterol) in the cholesterol biosynthetic pathway. Therefore, inhibition of cholesterol synthesis would be expected to decrease their production as well.
RQ3.Answer = C. Competitive inhibitors bind to the same site as the substrate (S) and prevent the S from binding. This results in an increase in the apparent Km (Michaelis constant, or that S concentration that gives one half of the maximal velocity [Vmax]). However, because the inhibition can be reversed by adding additional substrate, the Vmax is unchanged (see figure at right). It is noncompetitive inhibitors that decrease the apparent Vmax and have no effect on Km.
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RQ4.Answer = F. Tissue plasminogen activator (TPA) converts plasminogen to plasmin that degrades fibrin (fibrinolysis), thereby degrading the clot (thrombolysis). Aspirin, an inhibitor of cyclooxygenase, is an antiplatelet drug. Antithrombin III (ATIII) removes thrombin from the blood, and its action is potentiated by heparin. Activated protein C (APC) complex cleaves the accessory proteins factor (F)Va and FVIIIa. ATIII and APC are involved in anticoagulation. FXIII is a transglutaminase that cross-links the fibrin meshwork. Vitamin K is a fat-soluble vitamin required for the γ-carboxylation of FII, FVII, FIX, and FX. Warfarin prevents regeneration of the functional, reduced form of vitamin K. RQ5.Answer = D. In hypoxia, substrate-level phosphorylation in glycolysis provides ATP. Oxidative phosphorylation is inhibited by the lack of O2. Because the rate of ATP synthesis by oxidative phosphorylation controls the rate of cellular respiration, electron transport is inhibited. The resulting rise in the ratio of the reduced form of nicotinamide adenine dinucleotide (NADH) to the oxidized form (NAD+) inhibits the tricarboxylic acid cycle and the pyruvate dehydrogenase complex. RQ6.Answer = D. Fluorescently labeled nucleotides allow the base sequence of the DNA of interest to be determined. Complementary DNA (cDNA) is generated from processed messenger RNA and would not contain the promoter. Dideoxynucleotides lack the 3′-OH needed to form the 3′→5′-phosphodiester bond that joins the nucleotides and, thus, will terminate DNA synthesis. Genomic DNA obtained from white cells isolated from a blood sample would be the source of the DNA.
Case 1: Answers to Thought Questions
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TQ1.The phenotype would be the same. In familial defective apolipoprotein (apo) B-100, LDL receptors are normal in number and function, but the ligand for the receptor is altered such that binding to the receptor is decreased. Decreased ligand–receptor binding results in increased levels of LDL in the blood with hypercholesterolemia. [Note: The phenotype would be the same in individuals with a gain-of-function mutation to PCSK9, the protease that decreases recycling of the LDL receptor, thereby increasing its degradation.] With the apo E-2 isoform, cholesterol-rich chylomicron remnants and intermediate-density lipoproteins would accumulate in blood. TQ2.Aspirin irreversibly inhibits cyclooxygenase (COX) and, therefore, the synthesis of prostaglandins (PG), such as PGI2 in vascular endothelial cells, and thromboxanes (TX), such as TXA2 in activated platelets. TXA2 promotes vasoconstriction and formation of a platelet plug, whereas PGI2 inhibits these events. Because platelets are anucleate, they cannot overcome this inhibition by synthesizing more COX. However, endothelial cells have a nucleus. Aspirin, then, inhibits formation of blood clots by preventing production of TXA2 for the life of the platelet. TQ3.The decrease in ATP (as the result of a decrease in O2 and, thus, a decrease in oxidative phosphorylation) causes an increase in adenosine monophosphate (AMP). AMP allosterically activates phosphofructokinase-1, the key regulated enzyme of glycolysis. The rise in glycolysis increases the production of ATP by substrate-level phosphorylation. It also increases the ratio of the reduced to oxidized forms of NAD. Under anaerobic conditions, pyruvate produced in glycolysis is reduced to lactate by lactate dehydrogenase as NADH is oxidized to NAD+. NAD+ is required for continued glycolysis. Because fewer ATP molecules are produced per molecule of substrate in substrate-level phosphorylation relative to oxidative phosphorylation, there is a compensatory increase in the rate of glycolysis under anaerobic conditions.
TQ4.High-density lipoprotein (HDL) functions in reverse cholesterol transport. It takes cholesterol from nonhepatic (peripheral) tissues (for example, the endothelial layer of arteries) and brings it to the liver (see figure on the next page). The ABCA1 transporter mediates the efflux of cholesterol to HDL. The cholesterol is esterified by extracellular lecithin–cholesterol acyltransferase (LCAT) that requires apo A-1 as a coenzyme. Some cholesteryl ester is transferred to very-low-density lipoproteins (VLDL) by cholesteryl ester transfer protein (CETP) in exchange for triacylglycerol. The remainder is taken up by a scavenger receptor (SR-B1) on the surface of hepatocytes. The liver can use the cholesterol from HDL in the synthesis of bile acids. Removal of cholesterol from endothelial cells prevents its accumulation (as cholesterol or cholesteryl ester), decreasing the risk of heart disease. [Note: In contrast, LDL carries cholesterol from the liver to peripheral tissues or back to the liver.]
Case 2: Answers to Review Questions
RQ1.Answer = A. Deficiency of glucose 6-phosphatase prevents the glucose 6phosphate generated by glycogenolysis and gluconeogenesis from being dephosphorylated and released into the blood (see figure below). Blood glucose levels fall, and a severe, fasting hypoglycemia results. [Note: JS’s symptoms appeared only recently because, at age 4 months, his feedings are less frequent.]
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Hypoglycemia stimulates release of glucagon, which leads to phosphorylation and activation of glycogen phosphorylase kinase that phosphorylates and activates glycogen phosphorylase. Epinephrine is also released and leads to phosphorylation and activation of hormone-sensitive lipase. However, typical fatty acids (FA) cannot serve as substrates for gluconeogenesis. The glucose transporters in the liver and kidneys are insulin insensitive.
RQ2.Answer = C. Vitamin D is a fat-soluble vitamin that functions as a steroid hormone. In complex with its intracellular nuclear receptor, it increases transcription of the gene for calbindin, a calcium (Ca2+) transporter protein in the intestine (see figure at right). Vitamin D does not bind to a membrane receptor and does not produce second messengers. It can be synthesized in the skin by the action of ultraviolet light on an intermediate of cholesterol synthesis, 7dehydrocholesterol. Of the fat-soluble vitamins (A, D, E, and K), only K functions as a coenzyme.
RQ3.Answer = C. Glucose 6-phosphate is a positive allosteric effector of the covalently inhibited (phosphorylated) glycogen synthase b. With the rise in glucose 6-phosphate, glycogen synthesis is activated, and glycogen stores are increased in both the liver and the kidneys. The increased availability of glucose 6-phosphate also drives glycolysis. The increase in glycolysis provides substrates for lipogenesis, thereby increasing synthesis of FA and triacylglycerols (TAG). In hypoglycemia, the insulin/glucagon ratio is low, not high.
RQ4.Answer = D. Membrane proteins are initially targeted to the endoplasmic reticulum (ER) by an amino terminal hydrophobic signal sequence. Glycosylation is the most common posttranslational modification found in proteins. The glycosylated portion of membrane proteins is found on the extracellular face of the membrane. The membrane-spanning domain consists of ~22 hydrophobic amino acids. Proteins destined for secretion or for membranes, the ER lumen, Golgi, or lysosomes are synthesized on ribosomes associated with the ER.
Case 2: Answers to Thought Questions
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TQ1.The twitching is the result of the adrenergic response to hypoglycemia and is mediated by the rise in epinephrine. The adrenergic response includes tremor and sweating. Neuroglycopenia (impaired delivery of glucose to the brain) results in impairment of brain function that can lead to seizures, coma, and death. Neuroglycopenic symptoms develop if the hyperglycemia persists. TQ2.Detergents are amphipathic molecules (that is, they have both hydrophilic [polar] and hydrophobic [nonpolar] regions). Detergents solubilize membranes, thereby disrupting membrane structure. If the problem were the translocase needed to move the glucose 6-phosphate substrate into the ER, rather than the phosphatase, disruption of the ER membrane would allow the substrate access to the phosphatase. TQ3.Glucagon, a peptide hormone released from pancreatic α cells in hypoglycemia, binds its plasma membrane G protein–coupled receptor on hepatocytes. The αs subunit of the associated trimeric G protein is activated (guanosine diphosphate is replaced by guanosine triphosphate), separates from the β and γ subunits, and activates adenylyl cyclase that generates cyclic adenosine monophosphate (cAMP) from ATP. cAMP activates protein kinase A (PKA) that phosphorylates and activates glycogen phosphorylase kinase, which phosphorylates and activates glycogen phosphorylase. The phosphorylase degrades glycogen, generating glucose 1-phosphate that is converted to glucose 6-phosphate. With glucose 6-phosphatase deficiency, the degradative process stops here (see figure below). Consequently, administration of glucagon is unable to cause a rise in blood glucose. [Note: Epinephrine would be similarly ineffective.]
TQ4.The availability of inorganic phosphate (Pi) is decreased because it is trapped as phosphorylated glycolytic intermediates as a result of the upregulation of glycolysis by the rise in glucose 6-phosphate. Urate is elevated because the trapping of Pi decreases the ability to phosphorylate adenosine diphosphate (ADP) to ATP, and the fall in ATP causes a rise in adenosine monophosphate (AMP). The AMP is degraded to urate. Additionally, the availability of glucose 6-phosphate drives the pentose phosphate pathway, resulting in a rise in ribose 5-phosphate (from ribulose 5-phosphate) and, consequently, a rise in purine synthesis. Nicotinamide adenine dinucleotide phosphate (NADPH) also rises. Purines made beyond need are degraded to urate (see figure on the next page). [Note: The decrease in Pi reduces the activity of glycogen phosphorylase, resulting in increased storage of glycogen with a normal structure.] Lactate is elevated because the decrease in phosphorylation of ADP to ATP results in a decrease in cellular respiration (respiratory control) as a result of these processes being coupled. As a consequence, reduced nicotinamide adenine dinucleotide (NADH) from glycolysis cannot be oxidized by Complex I of the electron transport chain. Instead, it is oxidized by cytosolic lactate dehydrogenase with its coenzyme NADH as pyruvate is reduced to lactate. [Note: Pyruvate is increased as a result of the increase in glycolysis.] The lactate ionizes, releasing protons (H+) and leading to a metabolic acidosis (low pH caused here by increased production of acid). Respiratory compensation causes an increased respiratory rate.
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TQ5.Increased glycolysis results in increased availability of glycerol 3phosphate for hepatic TAG synthesis. Additionally, some of the pyruvate generated in glycolysis will be oxidatively decarboxylated to acetyl coenzyme A (CoA). However, the tricarboxylic acid cycle is inhibited by the rise in NADH, and the acetyl CoA is transported to the cytosol as citrate. The rise of acetyl CoA in the cytosol results in increased fatty acid (FA) synthesis. Recall that citrate is an allosteric activator of acetyl CoA carboxylase (ACC). The malonyl product of ACC inhibits FA oxidation at the carnitine palmitoyltransferase I step. Because mitochondrial FA oxidation generates the acetyl CoA substrate for hepatic ketogenesis, ketone body levels do not rise. The FA gets esterified to the glycerol backbone, resulting in an increase in TAG that gets sent out of the liver as components of very-low-density lipoproteins (VLDL). [Note: The hypoglycemia results in release of epinephrine and the activation of TAG lipolysis with release of free FA into the blood. The FA are oxidized, with the excess used in hepatic TAG synthesis.] The acetyl CoA is also a substrate for cholesterol synthesis. Thus, the increase in glycolysis results in the hyperlipidemia seen in JS (see figure below).
Case 3: Answers to Review Questions
RQ1.Correct answer = A. Diabetes is characterized by hyperglycemia. Chronic hyperglycemia can result in the nonenzymatic glycosylation (glycation) of hemoglobin (Hb), producing HbA1c. Therefore, measurement of glucose or
HbA1c in the blood is used to diagnose diabetes. In response to physiologic stress (for example, a urinary tract infection), secretion of counterregulatory hormones (such as the catecholamines) results in a rise in blood glucose. Glucose is a reducing sugar. It is type 2 diabetes (T2D) that is associated with obesity and a sedentary lifestyle and is caused by insensitivity to insulin (insulin resistance). T1D is caused by lack of insulin as a result of the autoimmune destruction of pancreatic β cells. Even individuals on a program of tight glycemic control do not achieve euglycemia. RQ2.Correct answer = E. The ketone bodies 3-hydroxybutyrate and acetoacetate are organic acids, and their ionization contributes to the proton load of the body. Ketone bodies are made in the mitochondria of liver cells using acetyl coenzyme A (CoA) generated primarily from the β-oxidation of fatty acids ([FA]; see figure on the next page). Because they are water soluble, they do not require a transporter. The liver cannot use them because it lacks the enzyme thiophorase, which moves CoA from succinyl CoA to acetoacetate for conversion to two molecules of acetyl CoA. It is the acetone released in the breath that can impart a fruity odor.
RQ3.Correct answer = A. Malonyl CoA, an intermediate of FA synthesis, inhibits FA β-oxidation through inhibition of carnitine palmitoyltransferase I. Lipolysis occurs when the insulin/counterregulatory hormone ratio decreases. Acetyl CoA, the product of FA β-oxidation, inhibits the pyruvate dehydrogenase (PDH) complex through activation of PDH kinase and activates pyruvate carboxylase. Acetyl CoA, then, pushes pyruvate to gluconeogenesis. β-Oxidation generates reduced nicotinamide adenine dinucleotide (NADH), the reducing equivalent required for gluconeogenesis. FA are not readily catabolized for energy by the brain.
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Case 3: Answers to Thought Questions
TQ1.Hypoinsulinemia results in hyperglycemia because insulin is required for the uptake of blood glucose by muscle and adipose tissue. Their glucose transporter (GLUT-4) is insulin dependent in that insulin is required for movement of the transporter to the cell surface from intracellular storage sites. Insulin is also required to suppress hepatic gluconeogenesis. Insulin suppresses the release of glucagon from pancreatic α cells. The resulting rise in the insulin/glucagon ratio results in the dephosphorylation and activation of the kinase domain of bifunctional phosphofructokinase-2 (PFK-2). The fructose 2,6bisphosphate produced by PFK-2 activates phosphofructokinase-1 of glycolysis (see figure below). It also inhibits fructose 1,6-bisphosphatase (FBP-2), thereby inhibiting gluconeogenesis. With hypoinsulinemia, the failure to take up glucose from the blood while simultaneously sending it out into the blood results in hyperglycemia.
TQ2.The blood glucose level has exceeded the capacity of the kidney to reabsorb glucose (via a sodium-dependent glucose transporter [SGLT]). The high concentration of glucose in the urine osmotically draws water from the body. This causes increased urination (polyuria) with loss of water that results in dehydration. TQ3.The NADH generated in FA β-oxidation inhibits the tricarboxylic acid (TCA) cycle at the three NADH-producing dehydrogenase steps. This shifts acetyl CoA away from oxidation in the TCA cycle and toward use as a substrate in hepatic ketogenesis.
TQ4.MW was in negative nitrogen balance: More nitrogen was going out than coming in. This is reflected in the elevated blood urea nitrogen (BUN) level seen in the patient (see figure at top right). [Note: The BUN value also reflects dehydration.] Muscle proteolysis and amino acid catabolism are occurring as a result of the fall in insulin. (Recall that skeletal muscle does not express the glucagon receptor.) Amino acid catabolism produces ammonia (NH3), which is converted to urea by the hepatic urea cycle and sent into the blood. [Note: Urea in the urine is reported as urinary urea nitrogen.] TQ5.The Kussmaul respiration seen in MW is a respiratory response to the metabolic acidosis. Hyperventilation blows off CO2 and water, reducing the concentration of protons (H+) and bicarbonate (HCO3−) as reflected in the following equation: The renal response includes, in part, the excretion of H+ as ammonium (NH4+). Degradation of branched-chain amino acids in skeletal muscle results in the release of large amounts of glutamine (Gln) into the blood. The kidneys take up and catabolize the Gln, generating NH3 in the process. The NH3 is converted to NH4+ by secreted H+ and is excreted (see figure at middle right). [Note: When ketone bodies are plentiful, enterocytes shift to using them as a fuel instead of
Gln. This increases the amount of Gln going to the kidney.] TQ6.Because FA β-oxidation supplies the acetyl CoA substrate for ketogenesis, impaired β-oxidation decreases the ability to make ketone bodies. Ketone bodies are an alternate to the use of glucose, and, thus, dependence on glucose increases. Because FA β-oxidation supplies the NADH and the nucleoside triphosphates needed for gluconeogenesis, glucose production decreases. The result is a hypoketotic hypoglycemia. Recall that this was seen with medium-chain acyl CoA dehydrogenase (MCAD) deficiency.
Case 4: Answers to Review Questions
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RQ1.Answer = D. The rise in reduced nicotinamide adenine dinucleotide (NADH) in the mitochondria decreases the tricarboxylic acid (TCA) cycle, fatty acid (FA) oxidation, and gluconeogenesis. NADH inhibits the isocitrate dehydrogenase reaction, the key regulated step of the TCA cycle, and the αketoglutarate dehydrogenase reaction (see figure at bottom right). It also favors the reduction of oxaloacetate (OAA) to malate (not malate to OAA), decreasing the availability of OAA for condensation with acetyl coenzyme A (CoA) in the TCA cycle and for gluconeogenesis. FA oxidation requires the oxidized form of nicotinamide adenine dinucleotide (NAD+) for the 3-hydroxyacyl CoA dehydrogenase step and, thus, is inhibited by the rise in NADH. The decrease in FA oxidation decreases the production of ATP and acetyl CoA (the allosteric activator of pyruvate carboxylase) needed for gluconeogenesis. Lipolysis is activated in fasting as a consequence of the fall in insulin and the rise in catecholamines that result in activation of hormone-sensitive lipase.
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RQ2.Answer = E. The irreversible, oxidative portion of the pentose phosphate pathway provides the nicotinamide adenine dinucleotide phosphate (NADPH) that supplies the reducing equivalents needed for activity of cytochrome P450 (CYP) proteins and for the regeneration of functional (reduced) glutathione. It is also an important source of NADPH for reductive biosynthetic processes in the cytosol, such as FA and cholesterol synthesis. [Note: Malic enzyme is another source.] CYP proteins are monooxygenases (mixed-function oxidases). They incorporate one O atom from O2 into the substrate as the other is reduced to water. It is the CYP proteins of the smooth endoplasmic reticular membrane that are involved in detoxification reactions. Those of the inner mitochondrial membrane are involved in the synthesis of steroid hormones, bile acids, and vitamin D. Reactive oxygen species are reduced by glutathione peroxidase as glutathione is oxidized. RQ3.Answer = C. Serotonin is released by activated platelets and causes vasoconstriction and platelet aggregation. [Note: Platelets do not synthesize serotonin, but they take up that which was made in the intestine and secreted into the blood.] Serotonin is associated with a feeling of well-being. It is degraded to 5-hydroxyindoleacetic acid by monoamine oxidase that catalyzes oxidative deamination. It is catechol-O-methyltransferase that catalyzes the methylation step in the degradation of the catecholamines. Serotonin is synthesized from tryptophan in a two-step process that utilizes tetrahydrobiopterin (BH4) requiring tryptophan hydroxylase and a pyridoxal phosphate (PLP)-requiring decarboxylase (see figure at right). RQ4.Answer = B. The exocrine pancreas secretes enzymes required for the digestion of dietary carbohydrate, protein, and fat. The endocrine pancreas secretes the peptide hormones insulin and glucagon. Damage that affects the functions of the pancreas would lead to diabetes (decreased insulin) and steatorrhea (fatty stool), with the latter the consequence of maldigestion of dietary fat. As was seen with the rise of troponins in a myocardial infarction and transaminases in liver damage, loss of cellular integrity (as would be seen in autodigestion of the pancreas) results in proteins that normally are intracellular being found in higher-than-normal concentrations in the blood. Secretin causes the pancreas to release bicarbonate to raise the pH of the chyme coming to the intestine from the stomach. Pancreatic enzymes work best at neutral or slightly alkaline pH. Pancreatitis is seen in individuals with hypertriglyceridemia as a result of a deficiency in lipoprotein lipase or its coenzyme, apolipoprotein C-II.
Case 4: Answers to Thought Questions
TQ1.A. The rise in cytosolic NADH seen with ethanol metabolism inhibits glycolysis. The glyceraldehyde 3-phosphate dehydrogenase step requires NAD+, which gets reduced as glyceraldehyde 3-phosphate gets oxidized. With the rise in NADH, glyceraldehyde 3-phosphate accumulates.
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B. Glyceraldehyde 3-phosphate from glycolysis is converted to glycerol 3phosphate, the initial acceptor of FA in triacylglycerol (TAG) synthesis (see figure at right). FA are available because of increased synthesis (from acetyl CoA, which is increased as a result of both increased production from the acetate product of acetaldehyde oxidation and decreased use in the TCA cycle), increased availability from lipolysis in adipose tissue, and decreased degradation. The TAG produced in the liver accumulate (due, in part, to decreased production of very-low-density lipoproteins) and cause fatty liver (steatosis). Hepatic steatosis is an early (and reversible) stage in alcohol-related liver disease. Subsequent stages are alcohol-related hepatitis (sometimes reversible) and cirrhosis (irreversible).
TQ2.The rise in NADH favors the reduction of pyruvate to lactate by lactate dehydrogenase. Lactate decreases the renal excretion of uric acid, thereby causing hyperuricemia, a necessary step in an acute gouty attack. [Note: The shift from pyruvate to lactate decreases the availability of pyruvate, a substrate for gluconeogenesis. This contributes to the hypoglycemia seen in AK.] TQ3.Prothrombin time (PT) measures the time it takes for plasma to clot after the addition of tissue factor (FIII), thereby allowing evaluation of the extrinsic (and common) pathways of coagulation. In the extrinsic pathway, FIII activates FVII in a calcium (Ca2+)-and phospholipid (PL)-dependent process (see figure at bottom right). FVII, like most of the proteins of clotting, is made by the liver. Alcohol-induced liver damage can decrease its synthesis. Additionally, FVII has a short half-life, and, as a γ-carboxyglutamate (Gla)-containing protein, its synthesis requires vitamin K. Poor nutrition can result in decreased availability of vitamin K and, therefore, decreased ability to clot. [Note: Severe liver disease results in prolonged PT and activated partial thromboplastin time, or aPPt.]
TQ4.Administration of folate can mask a deficiency in vitamin B12 by reversing the hematologic manifestation (macrocytic anemia) of the deficiency. However, folate has no effect on the neurologic damage caused by B12 deficiency. Over time, then, the neurologic effects can become severe and irreversible. Thus, folate can mask a deficiency of B12 and prevent treatment until the neuropathy is apparent.
III. Focused Cases
Case 1: Microcytic Anemia
Patient Presentation: ME is a 24-year-old man who is being evaluated as a follow-up to a preplacement medical evaluation he had prior to starting his new job.
Focused History: ME has no significant medical issues. His family history is unremarkable, but he knows little of the health status of those family members who remain in Greece.
Pertinent Findings: The physical examination was normal. Routine analysis of his blood included the following results:
Based on the data, hemoglobin (Hb) electrophoresis was performed. The results are as follows:
H = High; L = Low. [Note: HbA includes HbA1c.]
Diagnosis: ME has β-thalassemia trait (β-thalassemia minor) that is causing a microcytic anemia (see image at right). Ethnicity (such as being of Mediterranean origin) influences the risk for thalassemia.
Treatment: None is required at this time. Patients are advised that iron supplements will not prevent their anemia.
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Prognosis: β-Thalassemia trait does not cause mortality or significant morbidity. Patients should be informed of the genetic nature of their autosomal-recessive condition for family planning considerations because homozygous β-thalassemia (Cooley anemia) is a serious disorder.
Case-Related Questions: Choose the ONE best answer.
Q1. Mutations to the gene for β globin that result in decreased production of the protein are the cause of β-thalassemia. The mutations primarily affect gene transcription or posttranscriptional processing of the messenger RNA (mRNA) product. Which of the following statements concerning mRNA is correct?
A. Eukaryotic mRNA is polycistronic.
B. mRNA synthesis involves trans-acting factors binding to cis-acting elements.
C. mRNA synthesis is terminated at the DNA base sequence thymine adenine guanine (TAG).
D. Polyadenylation of the 5′-end of eukaryotic mRNA requires a methyl donor.
E. Splicing of eukaryotic mRNA involves removal of exons and joining of introns by the proteasome. Q2. HbA, a tetramer of 2 α and 2 β globin chains, delivers O2 from the lungs to the tissues and protons and CO2 from the tissues to the lungs. Increased concentration of which of the following will result in decreased O2 delivery by HbA? A. 2,3-Bisphosphoglycerate
B. Carbon dioxide
C. Carbon monoxide
D. Protons Q3. What is the basis for the increase in HbA2 and HbF (fetal Hb) in the βthalassemias?
Q4. Why is the allele-specific oligonucleotide (ASO) hybridization technique useful in the diagnosis of all cases of sickle cell anemia but not all cases of β-thalassemia?
Case 2: Skin Rash
Patient Presentation: KL is a 34-year-old woman who presents with a red, nonitchy rash on her left thigh and flu-like symptoms.
Focused History: KL reports that the rash first appeared a little over 2 weeks ago. It started out small but has gotten larger. She also thinks she is getting the flu because her muscles and joints ache (myalgia and arthralgia, respectively), and she has had a headache for the last few days. Upon questioning, KL reports that she and her husband took a camping trip through New England last month.
Pertinent Findings: The physical examination is remarkable for the presence of a red, circular, flat lesion ~11 cm in size that resembles a bullseye (erythema migrans) (see image at right). KL also has a low-grade fever.
Diagnosis: KL has Lyme disease caused by the bacterium Borrelia burgdorferi, which is transmitted by the bite of a tick in the genus Ixodes. Infected ticks are endemic in the Northeast region of the United States.
Treatment: KL is prescribed doxycycline, an antibiotic in the tetracycline family. Monitoring of KL will continue until all symptoms have completely resolved. Blood is drawn for clinical laboratory tests.
Prognosis: Patients treated with the appropriate antibiotic in the early stages of Lyme disease typically recover quickly and completely.
Case-Related Questions: Choose the ONE best answer.
Q1. Antibiotics in the tetracycline class inhibit protein synthesis (translation) at the initiation step. Which of the following statements about translation is correct?
A. In eukaryotic translation, the initiating amino acid is formylated methionine.
B. Only the charged initiating transfer RNA goes directly to the ribosomal A site.
C. Peptidyltransferase is a ribozyme that forms the peptide bond between two amino acids.
D. Prokaryotic translation can be inhibited by the phosphorylation of initiation factor 2.
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E. Termination of translation is independent of guanosine triphosphate hydrolysis.
F. The Shine-Dalgarno sequence facilitates the binding of the large ribosomal subunit to eukaryotic messenger RNA (mRNA).
Q2. The Centers for Disease Control and Prevention recommends a two-tier testing procedure for Lyme disease that involves a screening enzyme-linked immunosorbent assay (ELISA) followed by a confirmatory western blot analysis on any sample with a positive or equivocal ELISA result. Which of the following statements about these testing procedures is correct?
A. Both techniques are used to detect specific mRNA.
B. Both techniques involve the use of antibodies.
C. ELISA requires the use of electrophoresis.
D. Western blots require use of the polymerase chain reaction. Q3. Why are eukaryotic cells unaffected by antibiotics in the tetracycline class?
Case 3: Blood on the Toothbrush
Patient Presentation: LT is an 84-year-old man whose gums have been bleeding for several months.
Focused History: LT is a widower and lives alone in a suburban community on the East Coast. He no longer drives. His two children live on the West Coast and come east infrequently. Since the death of his wife 11 months ago, he has been isolated and finds it hard to get out of the house. His appetite has changed, and he is content with cereal, coffee, and packaged snacks. Chewing is difficult.
Pertinent Findings: The physical examination was remarkable for the presence of swollen dark-colored gums (see image at right). Several of LT’s teeth were loose, including one that anchors his dental bridge. Several black and blue marks (ecchymoses) were noted on the legs, and an unhealed sore was present on the right wrist. Inspection of his scalp revealed tiny red spots (petechiae) around some of the hair follicles. Blood was drawn for testing.
Results of tests on LT’s blood:
The test for blood in his stool (occult blood test) was negative.
Results of follow-up tests (obtained several days after the appointment) included the following:
H = High; L = Low.
Diagnosis: LT has vitamin C deficiency with a microcytic, hypochromic anemia secondary to the deficiency.
Treatment: LT was prescribed vitamin C (as oral ascorbic acid) and iron (as oral ferrous sulfate) supplements. He will also be referred to social services.
Prognosis: The prognosis for recovery is excellent.
Case-Related Questions: Choose the ONE best answer.
Q1. Which of the following statements about vitamin C is correct? Vitamin C is:
A. a competitor of iron absorption in the intestine.
B. a fat-soluble vitamin with a 3-month supply typically stored in adipose tissue.
C. a coenzyme in several enzymic reactions such as the hydroxylation of proline.
D. required for the cross-linking of collagen.
Q2. In contrast to the microcytic anemia characteristic of iron deficiency (common in older adults), a macrocytic anemia is seen with deficiencies of vitamin B12 and/or folic acid. These vitamin deficiencies are also common in older adults. Which of the following statements concerning these vitamins is correct?
A. An inability to absorb B12 results in pernicious anemia.
B. Both vitamins cause changes in gene expression.
C. Folic acid plays a key role in energy metabolism in most cells.
D. Treatment with methotrexate can result in toxic levels of the coenzyme form of folic acid.
E. Vitamin B12 is the coenzyme for enzymes catalyzing amino acid deaminations, decarboxylations, and transaminations. Q3. How do hemolytic anemias differ from nutritional anemias?
Case 4: Rapid Heart Rate, Headache, and Sweating
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Patient Presentation: BE is a 45-year-old woman who presents with concerns about sudden (paroxysmal), intense, brief episodes of headache, sweating (diaphoresis), and a racing heart (palpitations).
Focused History: BE reports that the attacks started ~3 weeks ago. They last from 2 to 10 minutes, during which time she feels quite anxious. During the attacks, it feels as though her heart is skipping beats (arrhythmia). At first, she thought the attacks were related to recent stress at work and maybe even menopause. The last time it happened, she was in a pharmacy and had her blood pressure taken. She was told it was 165/110 mm Hg. BE notes that she has lost weight (~8 lbs) in this period even though her appetite has been good.
Pertinent Findings: The physical examination was remarkable for BE’s thin, pale appearance. Blood pressure was elevated (150/100 mm Hg), as was the heart rate (110–120 beats/minute). Based on BE’s history, blood levels of normetanephrine and metanephrine were ordered. They were found to be elevated.
Diagnosis: BE has a pheochromocytoma, a rare catecholamine-secreting tumor of the adrenal medulla.
Treatment: Imaging studies of the abdomen were done to locate the tumor. Surgical resection of the tumor was performed. The tumor was found to be nonmalignant. Follow-up measurement of plasma metanephrines was performed 2 weeks later and was in the normal range.
Prognosis: The 5-year survival rate for nonmalignant pheochromocytomas is >95%.
Case-Related Questions: Choose the ONE best answer.
Q1. Pheochromocytomas secrete norepinephrine (NE) and epinephrine. Which of the following statements concerning the synthesis and degradation of these two biogenic amines is correct?
A. The substrate for their synthesis is tryptophan, which is hydroxylated to 3,4-dihydroxyphenylalanine (DOPA) by tetrahydrobiopterin-requiring tryptophan hydroxylase.
B. The conversion of DOPA to dopamine utilizes a pyridoxal phosphate– requiring carboxylase.
C. The conversion of NE to epinephrine requires vitamin C.
D. Degradation involves methylation by catechol-O-methyltransferase and produces normetanephrine from NE and metanephrine from epinephrine.
E. Normetanephrine and metanephrine are oxidatively deaminated to homovanillic acid by monoamine oxidase.
Q2. Which of the following statements concerning the actions of epinephrine and/or NE are correct?
A. NE functions as a neurotransmitter and a hormone.
B. They are initiated by autophosphorylation of select tyrosine residues in their receptors.
C. They are mediated by binding to adrenergic receptors, which are a class of nuclear receptors.
D. They result in the activation of glycogen and triacylglycerol synthesis.
Q3. NE bound to certain receptors causes vasoconstriction and an increase in blood pressure. Why might NE be used clinically in the treatment of septic shock?
Case 5: Sun Sensitivity
Patient Presentation: AZ is a 6-year-old boy who is being evaluated for freckle-like areas of hyperpigmentation on his face, neck, forearms, and lower legs.
Focused History: AZ’s father reports that the boy has always been quite sensitive to the sun. His skin turns red (erythema) and his eyes hurt (photophobia) if he is exposed to the sun for any period of time.
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Pertinent Findings: The physical examination was remarkable for the presence of thickened, scaly areas (actinic keratosis) and hyperpigmented areas on skin exposed to ultraviolet (UV) radiation from the sun. Small dilated blood vessels (telangiectasia) were also seen. Tissue from several sites on his face was biopsied, and two were later determined to be squamous cell carcinomas.
Diagnosis: AZ has xeroderma pigmentosum, a rare defect in nucleotide excision repair of DNA.
Treatment: Protection from sunlight through use of sunscreens such as protective clothing that reflect UV radiation and chemicals that absorb it is essential. Frequent skin and eye examinations are recommended.
Prognosis: Most patients with xeroderma pigmentosum die at an early age from skin cancers. However, survival beyond middle age is possible.
Case-Related Questions: Choose the ONE best answer.
Q1. Which of the following statements about DNA repair mechanisms is correct? DNA repair:
A. is performed only by eukaryotes.
B. of double-strand breaks is error free.
C. of mismatched bases involves repair of the parental strand.
D. of UV radiation–induced pyrimidine dimers involves removal of a short oligonucleotide containing the dimer.
E. of uracil produced by the deamination of cytosine requires the actions of endo-and exonucleases to remove the uracil base.
Q2. Which one of the following statements about DNA synthesis (replication) is correct? Replication:
A. in both eukaryotes and prokaryotes requires an RNA primer.
B. in eukaryotes requires condensation of chromatin.
C. in prokaryotes is accomplished by a single DNA polymerase.
D. is initiated at random sites in the genome.
E. produces a polymer of deoxyribonucleoside monophosphates linked by 5′→3′-phosphodiester bonds. . What is the difference between DNA proofreading and repair?
Case 6: Dark Urine and Yellow Sclerae
Patient Presentation: JF is a 13-year-old boy who presents with fatigue and yellow sclerae.
Focused History: JF began treatment ~4 days ago with a sulfonamide antibiotic and a urinary analgesic for a urinary tract infection. He had been told that his urine would change color (become reddish) with the analgesic, but he reports that it has gotten darker (more brownish) over the last 2 days. Last night, his mother noticed that his eyes had a yellow tint. JF says he feels as though he has no energy.
Pertinent Findings: The physical examination was remarkable for JF’s pale appearance, mild scleral icterus (jaundice), mild splenomegaly, and increased heart rate (tachycardia). JF’s urine tested positive for hemoglobin (hemoglobinuria). A peripheral blood smear reveals a lower-than-normal number of red blood cells (RBC), with some containing precipitated hemoglobin (Heinz bodies; see image at right), and a higher-than-normal number of reticulocytes (immature RBC). Results of the complete blood count (CBC) and blood chemistry tests are pending.
Diagnosis: JF has glucose 6-phosphate dehydrogenase (G6PD) deficiency, an X-linked disorder that causes hemolysis (RBC lysis).
Treatment: G6PD deficiency can result in a hemolytic anemia in affected individuals exposed to oxidative agents. JF will be switched to a different antibiotic. He will be advised that he is susceptible to certain drugs (for example, sulfa drugs), foods (fava or broad beans), and certain chemicals (for example, naphthalene), and must avoid exposure to them.
Prognosis: In the absence of exposure to oxidative agents, G6PD deficiency does not cause mortality or significant morbidity.
Case-Related Questions: Choose the ONE best answer.
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Q1. G6PD catalyzes the regulated step in the pentose phosphate pathway. Which of the following statements concerning G6PD and the pentose phosphate pathway is correct?
A. Deficiency of G6PD occurs only in RBC.
B. Deficiency of G6PD results in an inability to keep glutathione in its functional, reduced form.
C. The pentose phosphate pathway includes one reversible reductive reaction followed by a series of phosphorylated sugar interconversions.
D. The reduced nicotinamide adenine dinucleotide phosphate (NADPH) product of the pentose phosphate pathway is utilized in processes such as fatty acid oxidation.
Q2. The results of JF's CBC were consistent with a hemolytic anemia. Blood chemistry tests revealed an elevation in the bilirubin level. Which of the following statements concerning bilirubin is correct?
A. Hyperbilirubinemia can cause deposition of bilirubin in the skin and sclerae resulting in jaundice.
B. The solubility of bilirubin is increased by conjugating it with two molecules of ascorbic acid in the liver.
C. The conjugated form of bilirubin increases in the blood with a hemolytic anemia.
D. Phototherapy can increase the solubility of the excess bilirubin generated in the porphyrias.
Q3. Why is urinary urobilinogen increased relative to normal in hemolytic jaundice and absent in obstructive jaundice?
Case 7: Joint Pain
Patient Presentation: IR is a 22-year-old male who presents for follow-up 10 days after having been treated in the Emergency Department (ED) for severe inflammation at the base of his thumb.
Focused History: This was IR’s first occurrence of severe joint pain. In the ED, he was given an anti-inflammatory medication. Fluid aspirated from the carpometacarpal joint of the thumb was negative for organisms but positive for needle-shaped monosodium urate (MSU) crystals (see image at right). The inflammatory symptoms have since resolved. IR reports he is in good health otherwise, with no significant past medical history. His body mass index (BMI) is 31. No tophi (deposits of MSU crystals under the skin) were detected in the physical examination.
Pertinent Findings: Results on a 24-hour urine specimen and blood tests requested in advance of this visit reveal that IR is not an undersecretor of uric acid. His blood urate was 8.5 mg/dl (reference = 2.5–8.0). The unusually young age of presentation is suggestive of an enzymopathy of purine metabolism, and additional blood tests are ordered.
Diagnosis: IR has gout (MSU crystal deposition disease), a type of inflammatory arthritis.
Treatment: IR was given prescriptions for allopurinol and colchicine. The treatment goals are to reduce his blood urate levels to <6.0 mg/dl and prevent additional attacks. He was advised to lose weight because being overweight or obese is a risk factor for gout. His BMI of 31 puts him in the obese category. He was also given written information on the association between diet and gout.
Prognosis: Gout increases the risk of developing renal stones. It is also associated with hypertension, diabetes, and heart disease.
Case-Related Questions: Choose the ONE best answer.
Q1. Allopurinol is converted in the body to oxypurinol, which functions as a noncompetitive inhibitor of an enzyme in purine metabolism. Which of the following statements concerning purine metabolism and its regulation is correct?
A. As a noncompetitive inhibitor, oxypurinol increases the apparent Michaelis constant (Km) of the target enzyme.
B. Colchicine inhibits xanthine oxidase, an enzyme of purine degradation.
C. Glutamate provides two of the nitrogen atoms of the purine ring.
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D. In purine nucleotide synthesis, the ring system is first constructed and then attached to ribose 5-phosphate.
E. Oxypurinol inhibits the amidotransferase that initiates degradation of the purine ring system.
F. Partial or complete enzymic deficiencies in the salvage of purine bases are characterized by hyperuricemia.
Q2. Purines are one type of nitrogenous base found in nucleotides. Pyrimidines are the other. Which of the following statements is true of the pyrimidines?
A. Carbamoyl phosphate synthetase I is the regulated enzymic activity in pyrimidine ring synthesis.
B. Methotrexate decreases synthesis of the pyrimidine nucleotide thymidine monophosphate.
C. Orotic aciduria is a pathology of pyrimidine degradation.
D. Pyrimidine nucleotide synthesis is independent of 5-phosphoribosyl-1pyrophosphate (PRPP).
Q3. IR is subsequently shown to have a form of PRPP synthetase that shows increased enzymic activity. Why does this result in hyperuricemia?
Case 8: No Bowel Movement
Patient Presentation: DW is a 48-hour-old female who has not yet had a bowel movement.
Focused History: DW is the full-term product of a normal pregnancy and delivery. She appeared normal at birth. DW is the first child of parents of Northern European ethnicity. The parents are both in good health, and their family histories are unremarkable.
Pertinent Findings: DW has a distended abdomen. She recently vomited small amounts of bilious (green-colored) material.
Diagnosis: Meconium ileus (obstruction of the ileum by meconium, the first stool produced by newborns) was confirmed by abdominal x-rays. About 98% of full-term newborns with meconium ileus have cystic fibrosis (CF). Diagnosis of CF was subsequently confirmed with a chloride sweat test.
Treatment: The ileus was successfully treated nonsurgically. For management of the CF, the family was referred to the CF center at the regional children’s hospital.
Prognosis: CF is the most common life-limiting autosomal-recessive disease in Caucasians.
Case-Related Questions: Choose the ONE best answer.
Q1. CF is the result of mutations to the gene that encodes the CF transmembrane conductance regulator (CFTR) protein that functions as a chloride channel in the apical membrane of epithelial cells on a mucosal surface. Which of the following statements concerning CF is correct?
A. Clinical manifestations of CF are the consequence of chloride retention with increased water reabsorption that causes mucus on the epithelial surface to be excessively thick and sticky.
B. Excessive pancreatic secretion of insulin in CF commonly results in hypoglycemia.
C. Genetic testing for CF may involve the use of a set of probes for the most common mutations, a technique known as restriction fragment length polymorphism analysis.
D. Some mutations result in premature degradation of the CFTR protein through tagging with ubiquinone followed by proteasome-mediated proteolysis.
E. The most common mutation, ∆F508, results in the loss of a codon for phenylalanine (F) and is classified as a frameshift mutation.
Q2. The CFTR protein is an intrinsic plasma membrane glycoprotein. Targeting of proteins destined to function as components of membranes:
A. includes transport to and through the Golgi.
B. involves an amino-terminal signal sequence that is retained in the functional protein.
C. occurs after the protein has been completely synthesized (that is, posttranslationally).
D. requires the presence of mannose 6-phosphate residues on the protein. Q3. Why might steatorrhea be seen with CF?
Case 9: Elevated Ammonia
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Patient Presentation: RL is a 40-hour-old male with signs of cerebral edema.
Focused History: RL is the full-term product of a normal pregnancy and delivery. He appeared normal at birth. At age 36 hours, he became irritable, lethargic, and hypothermic. He fed only poorly and vomited. He also displayed tachypneic (rapid) breathing and neurologic posturing. At age 38 hours, he had a seizure.
Pertinent Findings: Respiratory alkalosis (increased pH, decreased CO2 [hypocapnia]), increased ammonia, and decreased blood urea nitrogen were found. An amino acid screen revealed that argininosuccinate was increased >60fold over baseline, and citrulline was increased 4-fold. Glutamine was elevated, and arginine (Arg) was decreased relative to normal.
Diagnosis: RL has a urea cycle enzyme defect with neonatal onset.
Treatment: Hemodialysis was performed to remove ammonia. Sodium phenylacetate and sodium benzoate were administered to aid in excretion of waste nitrogen, as was Arg. Long-term treatment will include lifelong limitation of dietary protein; supplementation with essential amino acids; and administration of Arg, sodium phenylacetate, and sodium phenylbutyrate.
Prognosis: Survival into adulthood is possible. The degree of neurologic impairment is related to the degree and extent of the hyperammonemia.
Case-Related Questions: Choose the ONE best answer.
Q1. Based on the findings, which enzyme of the urea cycle is most likely to be deficient in this patient?
A. Arginase
B. Argininosuccinate lyase
C. Argininosuccinate synthetase
D. Carbamoyl phosphate synthetase I
E. Ornithine transcarbamoylase Q2. Why is Arg supplementation helpful in this case?
Q3. In individuals with partial (milder) deficiency of urea cycle enzymes, the level of which one of the following would be expected to be decreased during periods of physiologic stress?
A. Alanine
B. Ammonia
C. Glutamine
D. Insulin
E. pH
Case 10: Calf Pain
Patient Presentation: CR is a 19-year-old female who is being evaluated for pain and swelling in her right calf.
Focused History: Ten days ago, CR had her spleen removed following a bicycle accident in which she fractured her tibial eminence, necessitating immobilization of the right knee. She has had a good recovery from the surgery. CR is no longer taking pain medication but has continued her oral contraceptives (OCP).
Pertinent Findings: CR’s right calf is reddish in color (erythematous) and warm to the touch. It is visibly swollen. The left calf is normal in appearance and is without pain. An ultrasound is ordered.
Diagnosis: CR has a deep venous thrombosis (DVT). OCP are a risk factor for DVT, as are surgery and immobilization.
Treatment (Immediate): Heparin and warfarin are administered.
Prognosis: In the 10 years following a DVT, about one third of individuals have a recurrence.
Case-Related Questions: Choose the ONE best answer.
Q1. A DVT is a blood clot that occludes the lumen of a deep vein, most commonly in the leg. Which of the following statements about the clotting cascade is correct?
A. A deficiency in factor (F)IX of the intrinsic pathway results in hemophilia A.
B. FIII of the extrinsic pathway is a serine protease.
C. Formation of the fibrin meshwork is referred to as primary hemostasis.
D. Thrombin proteolytically activates components of the extrinsic, intrinsic, and common pathways.
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E. Vitamin K is required for the activation of fibrinogen. Q2. Which one of the following would increase the risk of thrombosis?
A. Excess production of antithrombin
B. Excess production of protein S
C. Expression of FV Leiden
D. Hypoprothrombinemia
E. von Willebrand disease Q3. Compare and contrast the actions of heparin and warfarin.
Focused Cases: Answers to Case-Based Questions
Case 1: Anemia with β-Thalassemia Minor
Answer = B. Transcription (synthesis of single-stranded RNA from the template strand of double-stranded DNA) requires the binding of proteins (trans-acting factors) to sequences on the DNA (cis-acting elements). Eukaryotic messenger RNA (mRNA) is monocistronic because it contains information from just one gene (cistron). The base sequence TAG (thymine adenine guanine) in the coding strand of DNA is U(uracil)AG in the mRNA. UAG is a signal that terminates translation (protein synthesis), not transcription. It is formation of the 5′-cap of eukaryotic mRNA that requires methylation (using S-adenosylmethionine), not 3′-end polyadenylation. Splicing is the spliceosome-mediated process by which introns are removed from eukaryotic mRNA and exons joined.
Answer = C. Carbon monoxide (CO) increases the affinity of hemoglobin (Hb)A for O2, thereby decreasing the ability of HbA to offload O2 in the tissues. CO stabilizes the R (relaxed) or oxygenated form and shifts the O2 dissociation curve to the left, decreasing O2 delivery (see figure at top right). The other choices decrease the affinity for O2, stabilize the T (tense) or deoxygenated form, and cause a right shift in the curve. HbA2 and fetal Hb (HbF) do not contain β globin. As β globin production decreases, synthesis of HbA2 (α2δ2) and HbF (α2γ2) increases.
Sickle cell anemia is caused by a single point mutation (A→T) in the gene for β globin that results in the replacement of glutamate by valine at the sixth amino acid position in the protein. Mutational analysis using allele-specific oligonucleotide (ASO) probes for that mutation (βS) and for the normal sequence (βA) is used in diagnosis (see figure at lower right). β-Thalassemia, in contrast, is caused by hundreds of different mutations. Mutational analysis using ASO probes can assess common mutations, including point mutations, in at-risk populations (for example, those of Greek ancestry). However, less common mutations are often not included in the panel and can be detected only by DNA sequencing.
Case 2: Skin Rash with Lyme Disease
Answer = C. Peptide-bond formation between the amino acid in the A site of the ribosome and the amino acid last added to the growing peptide in the P site is catalyzed by an RNA of the large ribosomal subunit. Any RNA with catalytic activity is referred to as a ribozyme (see figure on the next page). Formylated methionine is used to initiate prokaryotic translation. The charged initiating transfer RNA (tRNAi) is the only tRNA that goes directly to the P site, leaving the A site available for the tRNA carrying the next amino acid of the protein being made. Eukaryotic translation is inhibited by the phosphorylation of initiation factor 2 (eIF-2). The Shine-Dalgarno sequence is found in prokaryotic messenger RNA (mRNA) and facilitates the interaction of the mRNA with the small ribosomal subunit. In eukaryotes, the cap-binding proteins perform that task.
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Answer = B. The enzyme-linked immunosorbent assay (ELISA) and western blot are used to analyze proteins. Each makes use of antibodies to detect and quantify the protein of interest. It is western blots that utilize electrophoresis. The polymerase chain reaction (PCR) is used to amplify DNA. Antibiotics in the tetracycline family inhibit protein synthesis by binding to and blocking the A site of the small (30S) ribosomal subunit in prokaryotes. Tetracycline specifically interacts with the 16S ribosomal RNA (rRNA) component of the 30S subunit, inhibiting translation initiation. Eukaryotes do not contain 16S rRNA. Their small (40S) subunit contains 18S rRNA, which does not bind tetracycline.
Case 3: Blood on the Toothbrush with Vitamin C Deficiency
Answer = C. Vitamin C (ascorbic acid) functions as a coenzyme in the hydroxylation of proline and lysine in the synthesis of collagen, a fibrous protein of the extracellular matrix. Vitamin C is also the coenzyme for duodenal cytochrome b (Dcytb) that reduces dietary iron from the ferric (Fe3+) to the ferrous (Fe2+) form that is required for absorption via the divalent metal transporter (DMT) of enterocytes (see figure below). With a deficiency of vitamin C, uptake of dietary iron is impaired and results in a microcytic, hypochromic anemia. As a water-soluble vitamin, vitamin C is not stored. Cross-linking of collagen by lysyl oxidase requires copper, not vitamin C.
Answer = A. An inability to absorb vitamin B12 leads to pernicious anemia and is most commonly caused by decreased production of intrinsic factor (IF) by the parietal cells of the stomach (see figure at right). Vitamins D and A, in complex with their receptors, bind to DNA and alter gene expression. Thiamine (vitamin B1) is a coenzyme in the oxidative decarboxylation of pyruvate and αketoglutarate and, therefore, is important in energy metabolism in most cells. Methotrexate inhibits dihydrofolate reductase, the enzyme that reduces dihydrofolate to tetrahydrofolate (THF), the functional coenzyme form of folate. This results in decreased availability of THF. It is pyridoxine (vitamin B6) as pyridoxal phosphate that is the coenzyme for most reactions involving amino acids. [Note: Tetrahydrobiopterin is required by aromatic amino acid hydroxylases and nitric oxide synthases.] Nutritional anemias are characterized by either increased red blood cell (RBC) size (folate and B12 deficiencies) or decreased RBC size (iron and vitamin C deficiencies). In hemolytic anemias, such as is seen in glucose 6-phosphate dehydrogenase and pyruvate kinase deficiencies and in sickle cell anemia, RBC size typically is normal, and RBC number is decreased.
Case 4: Rapid Heart Rate, Headache, and Sweating with a Pheochromocytoma
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Answer = D. Degradation of both epinephrine and norepinephrine (NE) involves methylation by catechol-O-methyltransferase (COMT) that produces normetanephrine from NE and metanephrine from epinephrine (see figure at right). Both of these products are deaminated to vanillylmandelic acid by monoamine oxidase (MAO). The substrate for the synthesis of the catecholamines is tyrosine, which gets hydroxylated to 3,4dihydroxyphenylalanine (DOPA) by tetrahydrobiopterin-requiring tyrosine hydroxylase. DOPA is converted to dopamine by a pyridoxal phosphate– requiring decarboxylase. [Note: Most carboxylases require biotin.] NE is converted to epinephrine by methylation, and S-adenosylmethionine provides the methyl group.
Answer = A. NE released from the sympathetic nervous system functions as a neurotransmitter that acts on postsynaptic neurons and causes, for example, increased heart rate. It is also released from the adrenal medulla and, along with epinephrine, functions as a counterregulatory hormone that results in mobilization of stored fuels (for example, glucose and triacylglycerols). These actions are mediated by the binding of NE to adrenergic receptors, which are G protein–coupled receptors of the plasma membrane, and not to nuclear receptors like those of steroid hormones or membrane tyrosine kinase receptors like that of insulin. Septic shock is vasodilatory hypotension (low blood pressure caused by blood vessel dilation) resulting from the production of large amounts of nitric oxide by inducible nitric oxide synthase in response to infection. NE bound to receptors on smooth muscle cells causes vasoconstriction and, thus, raises blood pressure.
Case 5: Sun Sensitivity with Xeroderma Pigmentosum
Answer = D. Pyrimidine dimers are the characteristic DNA lesions caused by ultraviolet (UV) radiation. Their repair involves the excision of an oligonucleotide containing the dimer and replacement of that oligonucleotide, a process known as nucleotide excision repair (NER). (See figure at right for a representation of the process in prokaryotes.) DNA repair systems are found in prokaryotes and eukaryotes. Nothing is error free, but the homologous recombination (HR) method of double-strand break repair is much less prone to error than is the nonhomologous end joining (NHEJ) method because any DNA that was lost is replaced. Mismatched-base repair (MMR) involves identification and repair of the newly synthesized (daughter) strand. In prokaryotes, the extent of strand methylation is used to discriminate between the strands. Base excision repair (BER), the mechanism by which uracil is removed from DNA, utilizes a glycosylase to remove the base, creating an apyrimidinic or apurinic (AP) site. The sugar-phosphate is then removed by the actions of an endo-and exonuclease.
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Answer = A. All replication requires an RNA primer because DNA polymerases (pol) cannot initiate DNA synthesis. The chromatin of eukaryotes gets decondensed (relaxed) for replication. Relaxation can be accomplished, for example, by acetylation via histone acetyltransferases. Prokaryotes have more than one DNA pol. For example, pol III extends the RNA primer with DNA, and pol I removes the primer and replaces it with DNA. Replication is initiated at specific locations (one in prokaryotes, many in eukaryotes) that are recognized by proteins (for example, DnaA in prokaryotes). Deoxynucleoside monophosphates (dNMP) are joined by a phosphodiester bond that links the 3′hydroxyl group of the last dNMP added with the 5′-phosphate group of the incoming nucleotide, thereby forming a 3′→5′-phosphodiester bond as pyrophosphate is released. Proofreading occurs during replication in the S (synthesis of DNA) phase of the cell cycle and involves the 3′→5′ exonuclease activity possessed by some DNA pol (see figure below). Because repair can occur independently of replication, it can be performed outside of the S phase.
Case 6: Dark Urine and Yellow Sclerae with
Answer = B. Glutathione in its reduced form (G-SH) is an important antioxidant. The selenium-containing enzyme glutathione peroxidase reduces hydrogen peroxide (H2O2, a reactive oxygen species) to water as glutathionine is oxidized (G-S-S-G). Reduced nicotinamide adenine dinucleotide phosphate (NADPH)requiring glutathionine reductase regenerates G-SH from G-S-S-G (see Figure A). The NADPH is supplied by the oxidative reactions of the pentose phosphate pathway (see Figure B), which is regulated by the availability of NADPH at the glucose 6-phosphate dehydrogenase (G6PD)-catalyzed step (the first step). Deficiency of G6PD occurs in all cells, but the effects are seen in red blood cells where the pentose phosphate pathway is the only source of NADPH. The pathway involves two irreversible oxidative reactions, each of which generates NADPH. The NADPH is used in reductive processes such as fatty acid synthesis (not oxidation) as well as steroid hormone and cholesterol synthesis.
Answer = A. Jaundice (icterus) refers to the yellow color of the skin, nail beds, and sclerae that results from bilirubin deposition when the bilirubin level in the blood is elevated (hyperbilirubinemia; see Image C). Bilirubin has low solubility in aqueous solutions, and its solubility is increased by conjugation with uridine diphosphate–glucuronic acid in the liver, forming bilirubin diglucuronide or conjugated bilirubin (CB). In hemolytic conditions, such as G6PD deficiency, both CB and unconjugated bilirubin (UCB) are increased, but it is UCB that is found in the blood. CB is sent into the intestine. Phototherapy is used to treat unconjugated hyperbilirubinemia because it converts bilirubin to isomeric forms that are more water soluble. Bilirubin is the product of heme degradation in cells of the mononuclear phagocyte system, particularly in the liver and the spleen. The porphyrias are pathologies of heme synthesis and, therefore, are not characterized by hyperbilirubinemia.
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With hemolysis, more bilirubin is produced and conjugated. CB is sent to the intestine where it is converted to urobilinogen, some of which is reabsorbed, enters the portal blood, and travels to the kidney. Because the source of urinary urobilinogen is intestinal urobilinogen, urinary urobilinogen will be low in obstructive jaundice because intestinal urobilinogen will be low as a result of the obstruction of the common bile duct (see Figure D).
Case 7: Joint Pain with Gout
Answer = F. Salvage of the purine bases hypoxanthine and guanine to the purine nucleotides inosine monophosphate (IMP) and guanosine monophosphate (GMP) by hypoxanthine-guanine phosphoribosyltransferase (HGPRT) requires 5-phosphoribosyl-1-pyrophosphate (PRPP) as the source of the ribose 1phosphate. Salvage decreases the amount of substrate available for degradation to uric acid. Therefore, a deficiency in salvage results in hyperuricemia (see figure at right). Noncompetitive inhibitors such as oxypurinol have no effect on the Michaelis constant (Km) but decrease the apparent maximal velocity (Vmax).
Colchicine is an anti-inflammatory drug. It has no effect on the enzymes of purine synthesis or degradation. Glutamine (not glutamate) is a nitrogen source for purine ring synthesis. In purine nucleotide synthesis, the purine ring system is constructed on the ribose 5-phosphate provided by PRPP. Allopurinol and its metabolite, oxypurinol, inhibit xanthine oxidase of purine degradation. The amidotransferase is the regulated enzyme of purine synthesis. Its activity is decreased by purine nucleotides and increased by PRPP.
Answer = B. Methotrexate inhibits dihydrofolate reductase, decreasing the availability of N5,N10-methylene tetrahydrofolate needed for synthesis of deoxythymidine monophosphate (dTMP) from deoxyuridine monophosphate (dUMP) by thymidylate synthase (see figure at right). Carbamoyl phosphate synthetase (CPS) II is the regulated enzymic activity of pyrimidine biosynthesis in humans. CPS I is an enzyme of the urea cycle. Orotic aciduria is a rare pathology of pyrimidine synthesis caused by a deficiency in one or both enzymic activities of bifunctional uridine monophosphate synthase. Pyrimidine nucleotide synthesis, like purine synthesis and salvage, requires PRPP.
Increased activity of PRPP synthetase results in increased synthesis of PRPP. This results in an increase in purine nucleotide synthesis beyond need. The excess purine nucleotides get degraded to uric acid, thereby causing hyperuricemia.
Case 8: No Bowel Movement with Cystic Fibrosis
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Answer = A. The clinical manifestations of cystic fibrosis (CF) are the consequence of chloride retention with increased water absorption that causes mucus on an epithelial surface to be excessively thick and sticky. The result is pulmonary and gastrointestinal problems such as respiratory infection and impaired exocrine and endocrine pancreatic functions (pancreatic insufficiency). Impaired endocrine pancreatic function can result in diabetes with associated hyperglycemia. The genetic testing technique described, and one used in the diagnosis of CF, is the use of allele-specific oligonucleotides (ASO). Some mutations do result in increased degradation of the CF transmembrane conductance regulator (CFTR) protein, but degradation is initiated by tagging the protein with ubiquitin. Frameshift mutations alter the reading frame through the addition or deletion of nucleotides by a number not divisible by three. Because the ∆F509 mutation is caused by the loss of three nucleotides that code for phenylalanine (F) at position 509 in the CFTR protein, it is not a frameshift mutation. Answer = A. Targeting of proteins destined to function as components of the plasma membrane is an example of cotranslational targeting. It involves the initiation of translation on cytosolic ribosomes; recognition of the amino (N)terminal signal sequence in the protein by the signal recognition particle; movement of the protein-synthesizing complex to the outer face of the membrane of the endoplasmic reticulum (ER); and continuation of protein synthesis, such that the protein is threaded into the lumen of the ER and packaged into vesicles that travel to and through the Golgi and eventually fuse with the plasma membrane. The N-terminal signal sequence is removed by a peptidase in the lumen of the ER. Mannose 6-phosphate is the signal that cotranslationally targets proteins to the matrix of the lysosome where they function as acid hydrolases.
The pancreatic insufficiency seen in some patients with CF results in a decreased ability to digest food, and digestion is required for absorption. Dietary fats move through the intestine and are excreted in the stool (see figure at right), which is foul-smelling and bulky and may float. Patients are at risk for malnutrition and deficiencies in fat-soluble vitamins. Oral supplementation of pancreatic enzymes is the treatment.
Case 9: Hyperammonemia with a Urea Cycle Defect
Answer = B. Argininosuccinate lyase (ASL) cleaves argininosuccinate to arginine (Arg) and fumarate. The increase in argininosuccinate and citrulline and the decrease in Arg seen in RL indicate a deficiency in ASL (see figure below). With arginase deficiency, Arg would be increased, not decreased. Additionally, with arginase deficiency, the hyperammonemia would be less severe because two nitrogens are excreted. Deficiency of argininosuccinate synthetase (ASS) would also cause an increase in citrulline, but argininosuccinate would be low to absent. Deficiency of carbamoyl phosphate synthetase (CPS) I is characterized by low levels of Arg and citrulline. Deficiency of ornithine transcarbamoylase (OTC), the only X-linked enzyme of the urea cycle, would result in low levels of Arg and citrulline and elevated levels of urinary orotic acid. [Note: The orotic acid is elevated because the carbamoyl phosphate (CP) substrate of OTC is being used in the cytosol as a substrate for pyrimidine synthesis.]
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Arg supplementation is helpful because the Arg will be hydrolyzed to urea + ornithine by arginase. The ornithine will be combined with CP to form citrulline (see figure above). With ASL (and ASS) deficiency, citrulline accumulates and is excreted, thereby carrying waste nitrogen out of the body. Answer = D. In individuals with milder (partial) deficiencies in the enzymes of the urea cycle, hyperammonemia may be triggered by physiologic stress (for example, an illness or prolonged fasting) that decreases the insulin/counterregulatory hormone ratio. [Note: The degree of the hyperammonemia is usually less severe than that seen in the neonatal onset forms.] The shift in the ratio results, in part, in skeletal muscle proteolysis, and the amino acids that are released get degraded. Degradation involves transamination by pyridoxal phosphate–requiring aminotransferases that generate the α-keto acid derivative of the amino acid + glutamate. The glutamate undergoes oxidative deamination to α-ketoglutarate and ammonia (NH3) by glutamate dehydrogenase (GDH; see figure at right). [Note: GDH is unusual in that it uses both nicotinamide adenine dinucleotide (NAD) and nicotinamide adenine dinucleotide phosphate (NADP) as coenzymes.]
The NH3, which is toxic, can be transported to the liver as glutamine (Gln) and alanine (Ala). The Gln is generated by the amination of glutamate by ATP-requiring glutamine synthetase. In the liver, the enzyme glutaminase removes the NH3, which can be converted to urea by the urea cycle or excreted as ammonium (NH4+) (see figure at right). Gln, then, is a nontoxic vehicle of NH3 transport in the blood. Ala is generated in skeletal muscle from the catabolism of the branched-chain amino acids (BCAA). In the liver, Ala is transaminated by alanine transaminase (ALT) to pyruvate (used in gluconeogenesis) and glutamate. Thus, Ala carries nitrogen to the liver for conversion to urea (see figure below). Therefore, defects in the urea cycle would result in an elevation in NH3, Gln, and Ala. The elevated NH3 drives respiration, and the hyperventilation causes a rise in pH (respiratory alkalosis). [Note: Hyperammonemia is toxic to the nervous system. Although the exact mechanisms are not completely understood, it is known that the metabolism of large amounts of NH3 to Gln (in the astrocytes of the brain) results in osmotic effects that cause the brain to swell. Additionally, the rise in Gln decreases the availability of glutamate, an excitatory neurotransmitter.]
Case 10: Swollen, Painful Calf with Deep Venous Thrombosis
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Lippincott's Biochemistry
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Answer = D. Thrombin, a serine protease, is activated by the prothrombinase complex of factor (F)Xa + FVa. Once formed, activated thrombin (FIIa) proteolytically activates components of the extrinsic (FVII) and intrinsic (FXI, FVIII) pathways, generating FXa. Thrombin can also activate FV, FI, and FXIII of the common pathway (see figure below). Hemophilia A is caused by a deficiency in FVIII. FIX deficiency results in hemophilia B. FIII, also known as tissue factor (TF), is a transmembrane glycoprotein of the vascular endothelium. It functions as an accessory protein and not a protease. Formation of the platelet plug is primary hemostasis, and formation of the fibrin meshwork is secondary hemostasis. Vitamin K is required for the activation (γ-carboxylation) of FII, FVII, FIX, and FX (proteases that require calcium [Ca2+] and phospholipids [PL]) but not for FI (fibrinogen).
Answer = C. FV Leiden is a mutant form of FV that is resistant to proteolysis by the activated protein C complex. Decreased ability to degrade FV allows continued production of activated thrombin and leads to an increased risk of clot formation or thrombophilia. Antithrombin III (ATIII) and protein S are proteins of anticoagulation. Increased, not decreased, production of prothrombin would result in thrombophilia. Deficiency of von Willebrand factor causes a coagulopathy or a deficiency in clotting through effects on FVIII and platelets. Heparin and warfarin are anticoagulants. Heparin, a glycosaminoglycan, increases the affinity of ATIII for thrombin. Binding of ATIII removes thrombin from the blood and prevents it from converting fibrinogen to fibrin. Warfarin, a synthetic analog of vitamin K, inhibits vitamin K epoxide reductase and prevents the regeneration of the functional hydroquinone form of the vitamin that is required for the γ-carboxylation of glutamate residues to γ-carboxyglutamate (Gla) residues in FII, FVII, FIX, and FX (see figures below).
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The surface of our planet is populated by living things—curious, intricately organized chemical factories that take in matter from their surroundings and use these raw materials to generate copies of themselves. These living organisms appear extraordinarily diverse. What could be more different than a tiger and a piece of seaweed, or a bacterium and a tree? Yet our ancestors, knowing nothing of cells or DNA, saw that all these things had something in common. They called that something “life,” marveled at it, struggled to define it, and despaired of explaining what it was or how it worked in terms that relate to nonliving matter.
The discoveries of the past century have not diminished the marvel—quite the contrary. But they have removed the central mystery regarding the nature of life. We can now see that all living things are made of cells: small, membrane-enclosed units filled with a concentrated aqueous solution of chemicals and endowed with the extraordinary ability to create copies of themselves by growing and then dividing in two.
Because cells are the fundamental units of life, it is to cell biology—the study of the structure, function, and behavior of cells—that we must look for answers to the questions of what life is and how it works. With a deeper understanding of cells and their evolution, we can begin to tackle the grand historical problems of life on Earth: its mysterious origins, its stunning diversity, and its invasion of every conceivable habitat. Indeed, as emphasized long ago by the pioneering cell biologist E. B. Wilson, “the key to every biological problem must finally be sought in the cell; for every living organism is, or at some time has been, a cell.”
Despite their apparent diversity, living things are fundamentally similar inside. The whole of biology is thus a counterpoint between two themes: astonishing variety in individual particulars; astonishing constancy in fundamental mechanisms. In this first chapter, we begin by outlining the universal features common to all life on our planet. We then survey, briefly, the diversity of cells. And we see how, thanks to the common molecular code in which the specifications for all living organisms are written, it is possible to read, measure, and decipher these specifications to help us achieve a coherent understanding of all the forms of life, from the smallest to the greatest.
The unIveRsAl FeATuRes oF cells on eARTh
The dIveRsITy oF Genomes And The TRee oF lIFe
The unIveRsAl FeATuRes oF cells on eARTh
It is estimated that there are more than 10 million—perhaps 100 million—living species on Earth today. Each species is different, and each reproduces itself faithfully, yielding progeny that belong to the same species: the parent organism hands down information specifying, in extraordinary detail, the characteristics that the offspring shall have. This phenomenon of heredity is central to the definition of life: it distinguishes life from other processes, such as the growth of a crystal, or the burning of a candle, or the formation of waves on water, in which orderly structures are generated but without the same type of link between the peculiarities of parents and the peculiarities of offspring. Like the candle flame, the living organism must consume free energy to create and maintain its organization. But life employs the free energy to drive a hugely complex system of chemical processes that are specified by hereditary information.
Most living organisms are single cells. Others, such as ourselves, are vast multicellular cities in which groups of cells perform specialized functions linked by intricate systems of communication. But even for the aggregate of more than 1013 cells that form a human body, the whole organism has been generated by cell divisions from a single cell. The single cell, therefore, is the vehicle for all of the hereditary information that defines each species (Figure 1–1). This cell includes the machinery to gather raw materials from the environment and to construct from them a new cell in its own image, complete with a new copy of its hereditary information. Each and every cell is truly amazing.
All cells store Their hereditary Information in the same linear chemical code: dnA
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Computers have made us familiar with the concept of information as a measurable quantity—a million bytes (to record a few hundred pages of text or an image from a digital camera), 600 million bytes for the music on a CD, and so on. Computers have also made us well aware that the same information can be recorded in many different physical forms: the discs and tapes that we used 20 years ago for our electronic archives have become unreadable on present-day machines. Living
Figure 1–1 The hereditary information in the fertilized egg cell determines the nature of the whole multicellular organism. Although their starting cells look superficially similar, as indicated: a sea urchin egg gives rise to a sea urchin (A and B). A mouse egg gives rise to a mouse (c and d). An egg of the seaweed Fucus gives rise to a Fucus seaweed (e and F). (A, courtesy of david mcclay; B, courtesy of m. Gibbs, oxford scientific Films; c, courtesy of Patricia calarco, from G. martin, Science 209:768–776, 1980. With permission from AAAs; d, courtesy of o. newman, oxford scientific Films; e and F, courtesy of colin Brownlee.) (A) building block of DNA (D) double-stranded DNA (C) templated polymerization of new strand cells, like computers, store information, and it is estimated that they have been evolving and diversifying for over 3.5 billion years. It is scarcely to be expected that they would all store their information in the same form, or that the archives of one type of cell should be readable by the information-handling machinery of another. And yet it is so. All living cells on Earth store their hereditary information in the form of double-stranded molecules of DNA—long, unbranched, paired polymer chains, formed always of the same four types of monomers. These monomers, chemical compounds known as nucleotides, have nicknames drawn from a four-letter alphabet—A, T, C, G—and they are strung together in a long linear sequence that encodes the genetic information, just as the sequence of 1s and 0s encodes the information in a computer file. We can take a piece of DNA from a human cell and insert it into a bacterium, or a piece of bacterial DNA and insert it into a human cell, and the information will be successfully read, interpreted, and copied. Using chemical methods, scientists have learned how to read out the complete sequence of monomers in any DNA molecule—extending for many millions of nucleotides—and thereby decipher all of the hereditary information that each organism contains.
The mechanisms that make life possible depend on the structure of the double-stranded DNA molecule. Each monomer in a single DNA strand—that is, each nucleotide—consists of two parts: a sugar (deoxyribose) with a phosphate group attached to it, and a base, which may be either adenine (A), guanine (G), cytosine (C), or thymine (T) (Figure 1–2). Each sugar is linked to the next via the phosphate group, creating a polymer chain composed of a repetitive sugar-phosphate backbone with a series of bases protruding from it. The DNA polymer is extended by adding monomers at one end. For a single isolated strand, these monomers can, in principle, be added in any order, because each one links to the next in the same way, through the part of the molecule that is the same for all of them. In the living cell, however, DNA is not synthesized as a free strand in isolation, but on a template formed by a preexisting DNA strand. The bases protruding from the
Figure 1–2 DNA and its building blocks.
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(A) dnA is made from simple subunits, called nucleotides, each consisting of a sugar-phosphate molecule with a nitrogen-containing side group, or base, attached to it. The bases are of four types (adenine, guanine, cytosine, and thymine), corresponding to four distinct nucleotides, labeled A, G, c, and T. (B) A single strand of dnA consists of nucleotides joined together by sugar-phosphate linkages. note that the individual sugar-phosphate units are asymmetric, giving the backbone of the strand a definite directionality, or polarity. This directionality guides the molecular processes by which the information in dnA is interpreted and copied in cells: the information is always “read” in a consistent order, just as written english text is read from left to right. (c) Through templated polymerization, the sequence of nucleotides in an existing dnA strand controls the sequence in which nucleotides are joined together in a new dnA strand; T in one strand pairs with A in the other, and G in one strand with c in the other. The new strand has a nucleotide sequence complementary to that of the old strand, and a backbone with opposite directionality: corresponding to the GTAA... of the original strand, it has ...TTAc. (d) A normal dnA molecule consists of two such complementary strands. The nucleotides within each strand are linked by strong (covalent) chemical bonds; the complementary nucleotides on opposite strands are held together more weakly, by hydrogen bonds. (e) The two strands twist around each other to form a double helix—a robust structure that can accommodate any sequence of nucleotides without altering its basic structure (see movie 4.1).
existing strand bind to bases of the strand being synthesized, according to a strict rule defined by the complementary structures of the bases: A binds to T, and C binds to G. This base-pairing holds fresh monomers in place and thereby controls the selection of which one of the four monomers shall be added to the growing strand next. In this way, a double-stranded structure is created, consisting of two exactly complementary sequences of As, Cs, Ts, and Gs. The two strands twist around each other, forming a DNA double helix (Figure 1–2E).
The bonds between the base pairs are weak compared with the sugar-phosphate links, and this allows the two DNA strands to be pulled apart without breakage of their backbones. Each strand then can serve as a template, in the way just described, for the synthesis of a fresh DNA strand complementary to itself—a fresh copy, that is, of the hereditary information (Figure 1–3). In different types of cells, this process of DNA replication occurs at different rates, with different controls to start it or stop it, and different auxiliary molecules to help it along. But the basics are universal: DNA is the information store for heredity, and templated polymerization is the way in which this information is copied throughout the living world.
All cells Transcribe Portions of Their hereditary Information into the same Intermediary Form: RnA
To carry out its information-bearing function, DNA must do more than copy itself. It must also express its information, by letting the information guide the synthesis of other molecules in the cell. This expression occurs by a mechanism that is the same in all living organisms, leading first and foremost to the production of two other key classes of polymers: RNAs and proteins. The process (discussed in detail in Chapters 6 and 7) begins with a templated polymerization called transcription, in which segments of the DNA sequence are used as templates for the synthesis of shorter molecules of the closely related polymer ribonucleic acid, or RNA. Later, in the more complex process of translation, many of these RNA molecules direct the synthesis of polymers of a radically different chemical class—the proteins (Figure 1–4).
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In RNA, the backbone is formed of a slightly different sugar from that of DNA— ribose instead of deoxyribose—and one of the four bases is slightly different—uracil (U) in place of thymine (T). But the other three bases—A, C, and G—are the same, and all four bases pair with their complementary counterparts in DNA—the A, U, C, and G of RNA with the T, A, G, and C of DNA. During transcription, the RNA monomers are lined up and selected for polymerization on a template strand of DNA, just as DNA monomers are selected during replication. The outcome is a polymer molecule whose sequence of nucleotides faithfully represents a portion of the cell’s genetic information, even though it is written in a slightly different alphabet—consisting of RNA monomers instead of DNA monomers.
The same segment of DNA can be used repeatedly to guide the synthesis of many identical RNA molecules. Thus, whereas the cell’s archive of genetic information in the form of DNA is fixed and sacrosanct, these RNA transcripts are
Figure 1–3 The copying of genetic information by DNA replication. In this process, the two strands of a dnA double helix are pulled apart, and each serves as a template for synthesis of a new complementary strand.
Figure 1–4 From DNA to protein.
Genetic information is read out and put to use through a two-step process. First, in transcription, segments of the dnA sequence are used to guide the synthesis of molecules of RnA. Then, in translation, the RnA molecules are used to guide the synthesis of molecules of protein.
strand used as a template to direct RNA synthesis many identical RNA transcripts mass-produced and disposable (Figure 1–5). As we shall see, these transcripts function as intermediates in the transfer of genetic information. Most notably, they serve as messenger RNA (mRNA) molecules that guide the synthesis of proteins according to the genetic instructions stored in the DNA.
RNA molecules have distinctive structures that can also give them other specialized chemical capabilities. Being single-stranded, their backbone is flexible, so that the polymer chain can bend back on itself to allow one part of the molecule to form weak bonds with another part of the same molecule. This occurs when segments of the sequence are locally complementary: a ...GGGG... segment, for example, will tend to associate with a ...CCCC... segment. These types of internal associations can cause an RNA chain to fold up into a specific shape that is dictated by its sequence (Figure 1–6). The shape of the RNA molecule, in turn, may enable it to recognize other molecules by binding to them selectively—and even, in certain cases, to catalyze chemical changes in the molecules that are bound. In fact, some chemical reactions catalyzed by RNA molecules are crucial for several of the most ancient and fundamental processes in living cells, and it has been suggested that an extensive catalysis by RNA played a central part in the early evolution of life (discussed in Chapter 6).
All cells use Proteins as catalysts
Protein molecules, like DNA and RNA molecules, are long unbranched polymer chains, formed by stringing together monomeric building blocks drawn from a standard repertoire that is the same for all living cells. Like DNA and RNA, proteins carry information in the form of a linear sequence of symbols, in the same way as a human message written in an alphabetic script. There are many different protein molecules in each cell, and—leaving out the water—they form most of the cell’s mass.
Figure 1–5 How genetic information is broadcast for use inside the cell.
each cell contains a fixed set of dnA molecules—its archive of genetic information. A given segment of this dnA guides the synthesis of many identical RnA transcripts, which serve as working copies of the information stored in the archive. many different sets of RnA molecules can be made by transcribing different parts of a cell’s dnA sequences, allowing different types of cells to use the same information store differently.
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Figure 1–6 The conformation of an RNA molecule. (A) nucleotide pairing between different regions of the same RnA polymer chain causes the molecule to adopt a distinctive shape. (B) The three-dimensional structure of an actual RnA molecule produced by hepatitis delta virus; this RnA can catalyze RnA strand cleavage. The blue ribbon represents the sugar-phosphate backbone and the bars represent base pairs (see movie 6.1). (B, based on A.R. Ferré-d’Amaré, k. Zhou, and J.A. doudna, Nature 395:567–574, 1998. With permission from macmillan Publishers ltd.)
Figure 1–7 How a protein molecule acts as a catalyst for a chemical reaction. (A) In a protein molecule, the polymer chain folds up into a specific shape defined by its amino acid sequence. A groove in the surface of this particular folded molecule, the enzyme lysozyme, forms a catalytic site.
(B) A polysaccharide molecule (red)—a polymer chain of sugar monomers—binds to the catalytic site of lysozyme and is broken apart, as a result of a covalent bond-breaking reaction catalyzed by the amino acids lining the groove (see movie 3.9). (PdB code: 1lyd.)
The monomers of protein, the amino acids, are quite different from those of DNA and RNA, and there are 20 types instead of 4. Each amino acid is built around the same core structure through which it can be linked in a standard way to any other amino acid in the set; attached to this core is a side group that gives each amino acid a distinctive chemical character. Each of the protein molecules is a polypeptide, created by joining its amino acids in a particular sequence. Through billions of years of evolution, this sequence has been selected to give the protein a useful function. Thus, by folding into a precise three-dimensional form with reactive sites on its surface (Figure 1–7A), these amino-acid polymers can bind with high specificity to other molecules and can act as enzymes to catalyze reactions that make or break covalent bonds. In this way they direct the vast majority of chemical processes in the cell (Figure 1–7B).
Proteins have many other functions as well—maintaining structures, generating movements, sensing signals, and so on—each protein molecule performing a specific function according to its own genetically specified sequence of amino acids. Proteins, above all, are the main molecules that put the cell’s genetic information into action.
Thus, polynucleotides specify the amino acid sequences of proteins. Proteins, in turn, catalyze many chemical reactions, including those by which new DNA molecules are synthesized. From the most fundamental point of view, a living cell is a self-replicating collection of catalysts that takes in food, processes this food to derive both the building blocks and energy needed to make more catalysts, and discards the materials left over as waste (Figure 1–8A). A feedback loop that connects proteins and polynucleotides forms the basis for this autocatalytic, self-reproducing behavior of living organisms (Figure 1–8B).
All cells Translate RnA into Protein in the same Way
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How the information in DNA specifies the production of proteins was a complete mystery in the 1950s when the double-stranded structure of DNA was first revealed as the basis of heredity. But in the intervening years, scientists have discovered the elegant mechanisms involved. The translation of genetic information from the 4-letter alphabet of polynucleotides into the 20-letter alphabet of proteins is a complex process. The rules of this translation seem in some respects neat and rational but in other respects strangely arbitrary, given that they are (with minor exceptions) identical in all living things. These arbitrary features, it is thought, reflect frozen accidents in the early history of life. They stem from the chance properties of the earliest organisms that were passed on by heredity and have become so deeply embedded in the constitution of all living cells that they cannot be changed without disastrous effects.
It turns out that the information in the sequence of a messenger RNA molecule is read out in groups of three nucleotides at a time: each triplet of nucleotides, or codon, specifies (codes for) a single amino acid in a corresponding protein. Since the number of distinct triplets that can be formed from four nucleotides is 43, there are 64 possible codons, all of which occur in nature. However, there are only 20 naturally occurring amino acids. That means there are necessarily many cases in which several codons correspond to the same amino acid. This genetic code is read out by a special class of small RNA molecules, the transfer RNAs (tRNAs). Each type of tRNA becomes attached at one end to a specific amino acid, and displays at its other end a specific sequence of three nucleotides—an anticodon— that enables it to recognize, through base-pairing, a particular codon or subset of codons in mRNA. The intricate chemistry that enables these tRNAs to translate a specific sequence of A, C, G, and U nucleotides in an mRNA molecule into a specific sequence of amino acids in a protein molecule occurs on the ribosome, a large multimolecular machine composed of both protein and ribosomal RNA. All of these processes are described in detail in Chapter 6.
DNA molecules as a rule are very large, containing the specifications for thousands of proteins. Special sequences in the DNA serve as punctuation, defining where the information for each protein begins and ends. And individual segments of the long DNA sequence are transcribed into separate mRNA molecules, coding for different proteins. Each such DNA segment represents one gene. A complication is that RNA molecules transcribed from the same DNA segment can often be processed in more than one way, so as to give rise to a set of alternative versions of a protein, especially in more complex cells such as those of plants and animals. In addition, some DNA segments—a smaller number—are transcribed into RNA molecules that are not translated but have catalytic, regulatory, or structural functions; such DNA segments also count as genes. A gene therefore is defined as the segment of DNA sequence corresponding to a single protein or set of alternative protein variants or to a single catalytic, regulatory, or structural RNA molecule.
In all cells, the expression of individual genes is regulated: instead of manufacturing its full repertoire of possible proteins at full tilt all the time, the cell adjusts the rate of transcription and translation of different genes independently, according to need. Stretches of regulatory DNA are interspersed among the segments that code for protein, and these noncoding regions bind to special protein molecules that control the local rate of transcription. The quantity and organization of the regulatory DNA vary widely from one class of organisms to another, but the basic strategy is universal. In this way, the genome of the cell—that is, the totality of its genetic information as embodied in its complete DNA sequence— dictates not only the nature of the cell’s proteins, but also when and where they are to be made.
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Figure 1–8 Life as an autocatalytic process. (A) The cell as a self-replicating collection of catalysts. (B) Polynucleotides (the nucleic acids dnA and RnA, which are nucleotide polymers) provide the sequence information, while proteins (amino acid polymers) provide most of the catalytic functions that serve—through a complex set of chemical reactions—to bring about the synthesis of more polynucleotides and proteins of the same types.
A living cell is a dynamic chemical system, operating far from chemical equilibrium. For a cell to grow or to make a new cell in its own image, it must take in free energy from the environment, as well as raw materials, to drive the necessary synthetic reactions. This consumption of free energy is fundamental to life. When it stops, a cell decays toward chemical equilibrium and soon dies.
Genetic information is also fundamental to life, and free energy is required for the propagation of this information. For example, to specify one bit of information—that is, one yes/no choice between two equally probable alternatives— costs a defined amount of free energy that can be calculated. The quantitative relationship involves some deep reasoning and depends on a precise definition of the term “free energy,” as explained in Chapter 2. The basic idea, however, is not difficult to understand intuitively.
Picture the molecules in a cell as a swarm of objects endowed with thermal energy, moving around violently at random, buffeted by collisions with one another. To specify genetic information—in the form of a DNA sequence, for example—molecules from this wild crowd must be captured, arranged in a specific order defined by some preexisting template, and linked together in a fixed relationship. The bonds that hold the molecules in their proper places on the template and join them together must be strong enough to resist the disordering effect of thermal motion. The process is driven forward by consumption of free energy, which is needed to ensure that the correct bonds are made, and made robustly. In the simplest case, the molecules can be compared with spring-loaded traps, ready to snap into a more stable, lower-energy attached state when they meet their proper partners; as they snap together into the bonded arrangement, their available stored energy—their free energy—like the energy of the spring in the trap, is released and dissipated as heat. In a cell, the chemical processes underlying information transfer are more complex, but the same basic principle applies: free energy has to be spent on the creation of order.
To replicate its genetic information faithfully, and indeed to make all its complex molecules according to the correct specifications, the cell therefore requires free energy, which has to be imported somehow from the surroundings. As we shall see in Chapter 2, the free energy required by animal cells is derived from chemical bonds in food molecules that the animals eat, while plants get their free energy from sunlight.
All cells Function as Biochemical Factories dealing with the same Basic molecular Building Blocks
Because all cells make DNA, RNA, and protein, all cells have to contain and manipulate a similar collection of small molecules, including simple sugars, nucleotides, and amino acids, as well as other substances that are universally required. All cells, for example, require the phosphorylated nucleotide ATP (adenosine triphosphate), not only as a building block for the synthesis of DNA and RNA, but also as a carrier of the free energy that is needed to drive a huge number of chemical reactions in the cell.
Although all cells function as biochemical factories of a broadly similar type, many of the details of their small-molecule transactions differ. Some organisms, such as plants, require only the simplest of nutrients and harness the energy of sunlight to make all their own small organic molecules. Other organisms, such as animals, feed on living things and must obtain many of their organic molecules ready-made. We return to this point later.
Another universal feature is that each cell is enclosed by a membrane—the plasma membrane. This container acts as a selective barrier that enables the cell to concentrate nutrients gathered from its environment and retain the products it
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Figure 1–9 Formation of a membrane by amphiphilic phospholipid molecules. Phospholipids have a hydrophilic (water-loving, phosphate) head group and a hydrophobic (water-avoiding, hydrocarbon) tail. At an interface between oil and water, they arrange themselves as a single sheet with their head groups facing the water and their tail groups facing the oil. But when immersed in water, they aggregate to form bilayers enclosing aqueous compartments, as indicated.
synthesizes for its own use, while excreting its waste products. Without a plasma membrane, the cell could not maintain its integrity as a coordinated chemical system.
The molecules that form a membrane have the simple physicochemical property of being amphiphilic—that is, consisting of one part that is hydrophobic (water-insoluble) and another part that is hydrophilic (water-soluble). Such molecules placed in water aggregate spontaneously, arranging their hydrophobic portions to be as much in contact with one another as possible to hide them from the water, while keeping their hydrophilic portions exposed. Amphiphilic molecules of appropriate shape, such as the phospholipid molecules that comprise most of the plasma membrane, spontaneously aggregate in water to create a bilayer that forms small closed vesicles (Figure 1–9). The phenomenon can be demonstrated in a test tube by simply mixing phospholipids and water together; under appropriate conditions, small vesicles form whose aqueous contents are isolated from the external medium.
Although the chemical details vary, the hydrophobic tails of the predominant membrane molecules in all cells are hydrocarbon polymers (–CH2–CH2–CH2–), and their spontaneous assembly into a bilayered vesicle is but one of many examples of an important general principle: cells produce molecules whose chemical properties cause them to self-assemble into the structures that a cell needs.
The cell boundary cannot be totally impermeable. If a cell is to grow and reproduce, it must be able to import raw materials and export waste across its plasma membrane. All cells therefore have specialized proteins embedded in their membrane that transport specific molecules from one side to the other. Some of these membrane transport proteins, like some of the proteins that catalyze the fundamental small-molecule reactions inside the cell, have been so well preserved over the course of evolution that we can recognize the family resemblances between them in comparisons of even the most distantly related groups of living organisms.
The transport proteins in the membrane largely determine which molecules enter the cell, and the catalytic proteins inside the cell determine the reactions that those molecules undergo. Thus, by specifying the proteins that the cell is to manufacture, the genetic information recorded in the DNA sequence dictates the entire chemistry of the cell; and not only its chemistry, but also its form and its behavior, for these too are chiefly constructed and controlled by the cell’s proteins.
A living cell can exist with Fewer Than 500 Genes
The basic principles of biological information transfer are simple enough, but how complex are real living cells? In particular, what are the minimum requirements? We can get a rough indication by considering a species that has one of the smallest known genomes—the bacterium Mycoplasma genitalium (Figure 1–10). This organism lives as a parasite in mammals, and its environment provides it with many of its small molecules ready-made. Nevertheless, it still has to make all the large molecules—DNA, RNAs, and proteins—required for the basic processes of heredity. It has about 530 genes, about 400 of which are essential. Its genome of 580,070 nucleotide pairs represents 145,018 bytes of information—about as much as it takes to record the text of one chapter of this book. Cell biology may be complicated, but it is not impossibly so.
The minimum number of genes for a viable cell in today’s environments is probably not less than 300, although there are only about 60 genes in the core set that is shared by all living species.
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The individual cell is the minimal self-reproducing unit of living matter, and it consists of a self-replicating collection of catalysts. Central to this reproduction is the transmission of genetic information to progeny cells. Every cell on our planet stores its genetic information in the same chemical form—as double-stranded DNA. The cell replicates its information by separating the paired DNA strands and using each as a template for polymerization to make a new DNA strand with a complementary sequence of nucleotides. The same strategy of templated polymerization is used to transcribe portions of the information from DNA into molecules of the closely related polymer, RNA. These RNA molecules in turn guide the synthesis of protein molecules by the more complex machinery of translation, involving a large multi-molecular machine, the ribosome. Proteins are the principal catalysts for almost all the chemical reactions in the cell; their other functions include the selective import and export of small molecules across the plasma membrane that forms the cell’s boundary. The specific function of each protein depends on its amino acid sequence, which is specified by the nucleotide sequence of a corresponding segment of the DNA—the gene that codes for that protein. In this way, the genome of the cell determines its chemistry; and the chemistry of every living cell is fundamentally similar, because it must provide for the synthesis of DNA, RNA, and protein. The simplest known cells can survive with about 400 genes.
The dIveRsITy oF Genomes And The TRee oF lIFe
The success of living organisms based on DNA, RNA, and protein has been spectacular. Life has populated the oceans, covered the land, infiltrated the Earth’s crust, and molded the surface of our planet. Our oxygen-rich atmosphere, the deposits of coal and oil, the layers of iron ores, the cliffs of chalk and limestone and marble—all these are products, directly or indirectly, of past biological activity on Earth.
Living things are not confined to the familiar temperate realm of land, water, and sunlight inhabited by plants and plant-eating animals. They can be found in the darkest depths of the ocean, in hot volcanic mud, in pools beneath the frozen surface of the Antarctic, and buried kilometers deep in the Earth’s crust. The creatures that live in these extreme environments are generally unfamiliar, not only because they are inaccessible, but also because they are mostly microscopic. In more homely habitats, too, most organisms are too small for us to see without special equipment: they tend to go unnoticed, unless they cause a disease or rot the timbers of our houses. Yet microorganisms make up most of the total mass of living matter on our planet. Only recently, through new methods of molecular analysis and specifically through the analysis of DNA sequences, have we begun to get a picture of life on Earth that is not grossly distorted by our biased perspective as large animals living on dry land.
In this section, we consider the diversity of organisms and the relationships among them. Because the genetic information for every organism is written in the universal language of DNA sequences, and the DNA sequence of any given organism can be readily obtained by standard biochemical techniques, it is now possible to characterize, catalog, and compare any set of living organisms with reference to these sequences. From such comparisons we can estimate the place of each organism in the family tree of living species—the “tree of life.” But before describing what this approach reveals, we need first to consider the routes by which cells in different environments obtain the matter and energy they require to survive and proliferate, and the ways in which some classes of organisms depend on others for their basic chemical needs.
cells can Be Powered by a variety of Free-energy sources
Living organisms obtain their free energy in different ways. Some, such as animals, fungi, and the many different bacteria that live in the human gut, get it by feeding on other living things or the organic chemicals they produce; such organisms 0.2 µm
Figure 1–10 Mycoplasma genitalium.
scanning electron micrograph showing the irregular shape of this small bacterium, reflecting the lack of any rigid cell wall.
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cross section (transmission electron micrograph) of a Mycoplasma cell. of the 530 genes of Mycoplasma genitalium, 43 code for transfer, ribosomal, and other non-messenger RnAs. Functions are known, or can be guessed, for 339 of the genes coding for protein: of these, 154 are involved in replication, transcription, translation, and related processes involving dnA, RnA, and protein; 98 in the membrane and surface structures of the cell; 46 in the transport of nutrients and other molecules across the membrane; 71 in energy conversion and the synthesis and degradation of small molecules; and 12 in the regulation of cell division and other processes. note that these categories are partly overlapping, so that some genes feature twice. (A, from s.
Razin et al., Infect. Immun. 30:538–546, 1980. With permission from the American society for microbiology; B, courtesy of Roger cole, in medical microbiology, 4th ed. [s. Baron ed.]. Galveston: university of Texas medical Branch, 1996.) are called organotrophic (from the Greek word trophe, meaning “food”). Others derive their energy directly from the nonliving world. These primary energy converters fall into two classes: those that harvest the energy of sunlight, and those that capture their energy from energy-rich systems of inorganic chemicals in the environment (chemical systems that are far from chemical equilibrium). Organisms of the former class are called phototrophic (feeding on sunlight); those of the latter are called lithotrophic (feeding on rock). Organotrophic organisms could not exist without these primary energy converters, which are the most plentiful form of life.
Phototrophic organisms include many types of bacteria, as well as algae and plants, on which we—and virtually all the living things that we ordinarily see around us—depend. Phototrophic organisms have changed the whole chemistry of our environment: the oxygen in the Earth’s atmosphere is a by-product of their biosynthetic activities.
Lithotrophic organisms are not such an obvious feature of our world, because they are microscopic and mostly live in habitats that humans do not frequent— deep in the ocean, buried in the Earth’s crust, or in various other inhospitable environments. But they are a major part of the living world, and they are especially important in any consideration of the history of life on Earth.
Some lithotrophs get energy from aerobic reactions, which use molecular oxygen from the environment; since atmospheric O2 is ultimately the product of living organisms, these aerobic lithotrophs are, in a sense, feeding on the products of past life. There are, however, other lithotrophs that live anaerobically, in places where little or no molecular oxygen is present. These are circumstances similar to those that existed in the early days of life on Earth, before oxygen had accumulated.
The most dramatic of these sites are the hot hydrothermal vents on the floor of the Pacific and Atlantic Oceans. They are located where the ocean floor is spreading as new portions of the Earth’s crust form by a gradual upwelling of material from the Earth’s interior (Figure 1–11). Downward-percolating seawater is heated and driven back upward as a submarine geyser, carrying with it a current of chemicals from the hot rocks below. A typical cocktail might include H2S, H2, CO,
Mn2+, Fe2+, Ni2+, CH2, NH4 , and phosphorus-containing compounds. A dense
SEA dark cloud of hot, mineral-rich water 350°C percolation contour of seawater
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Figure 1–11 The geology of a hot hydrothermal vent in the ocean floor. As indicated, water percolates down toward the hot molten rock upwelling from the earth’s interior and is heated and driven back upward, carrying minerals leached from the hot rock. A temperature gradient is set up, from more than 350°c near the core of the vent, down to 2–3°c in the surrounding ocean. minerals precipitate from the water as it cools, forming a chimney. different classes of organisms, thriving at different temperatures, live in different neighborhoods of the chimney. A typical chimney might be a few meters tall, spewing out hot, mineral-rich water at a flow rate of 1–2 m/sec.
multicellular animals, e.g., tube worms population of microbes lives in the neighborhood of the vent, thriving on this austere diet and harvesting free energy from reactions between the available chemicals. Other organisms—clams, mussels, and giant marine worms—in turn live off the microbes at the vent, forming an entire ecosystem analogous to the world of plants and animals that we belong to, but powered by geochemical energy instead of light (Figure 1–12).
To make a living cell requires matter, as well as free energy. DNA, RNA, and protein are composed of just six elements: hydrogen, carbon, nitrogen, oxygen, sulfur, and phosphorus. These are all plentiful in the nonliving environment, in the Earth’s rocks, water, and atmosphere. But they are not present in chemical forms that allow easy incorporation into biological molecules. Atmospheric N2 and CO2, in particular, are extremely unreactive. A large amount of free energy is required to drive the reactions that use these inorganic molecules to make the organic compounds needed for further biosynthesis—that is, to fix nitrogen and carbon dioxide, so as to make N and C available to living organisms. Many types of living cells lack the biochemical machinery to achieve this fixation; they instead rely on other classes of cells to do the job for them. We animals depend on plants for our supplies of organic carbon and nitrogen compounds. Plants in turn, although they can fix carbon dioxide from the atmosphere, lack the ability to fix atmospheric nitrogen; they depend in part on nitrogen-fixing bacteria to supply their need for nitrogen compounds. Plants of the pea family, for example, harbor symbiotic nitrogen-fixing bacteria in nodules in their roots.
Living cells therefore differ widely in some of the most basic aspects of their biochemistry. Not surprisingly, cells with complementary needs and capabilities have developed close associations. Some of these associations, as we see below, have evolved to the point where the partners have lost their separate identities altogether: they have joined forces to form a single composite cell.
The Greatest Biochemical diversity exists Among Prokaryotic cells
From simple microscopy, it has long been clear that living organisms can be classified on the basis of cell structure into two groups: the eukaryotes and the
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Figure 1–12 Organisms living at a depth of 2500 meters near a vent in the ocean floor. close to the vent, at temperatures up to about 120°c, various lithotrophic species of bacteria and archaea (archaebacteria) live, directly fueled by geochemical energy. A little further away, where the temperature is lower, various invertebrate animals live by feeding on these microorganisms. most remarkable are these giant (2 meter) tube worms, Riftia pachyptila, which, rather than feed on the lithotrophic cells, live in symbiosis with them: specialized organs in the worms harbor huge numbers of symbiotic sulfur-oxidizing bacteria. These bacteria harness geochemical energy and supply nourishment to their hosts, which have no mouth, gut, or anus. The tube worms are thought to have evolved from more conventional animals, and to have become secondarily adapted to life at hydrothermal vents. (courtesy of monika Bright, university of vienna, Austria.) rod-shaped cells the smallest cells e.g., Escherichia coli, e.g., Mycoplasma, Vibrio cholerae Spiroplasma prokaryotes. Eukaryotes keep their DNA in a distinct membrane-enclosed intracellular compartment called the nucleus. (The name is from the Greek, meaning “truly nucleated,” from the words eu, “well” or “truly,” and karyon, “kernel” or “nucleus.”) Prokaryotes have no distinct nuclear compartment to house their DNA. Plants, fungi, and animals are eukaryotes; bacteria are prokaryotes, as are archaea—a separate class of prokaryotic cells, discussed below.
Most prokaryotic cells are small and simple in outward appearance (Figure 1–13), and they live mostly as independent individuals or in loosely organized communities, rather than as multicellular organisms. They are typically spherical or rod-shaped and measure a few micrometers in linear dimension. They often have a tough protective coat, called a cell wall, beneath which a plasma membrane encloses a single cytoplasmic compartment containing DNA, RNA, proteins, and the many small molecules needed for life. In the electron microscope, this cell interior appears as a matrix of varying texture without any discernible organized internal structure (Figure 1–14).
Prokaryotic cells live in an enormous variety of ecological niches, and they are astonishingly varied in their biochemical capabilities—far more so than eukaryotic cells. Organotrophic species can utilize virtually any type of organic molecule as food, from sugars and amino acids to hydrocarbons and methane gas. Photo-trophic species (Figure 1–15) harvest light energy in a variety of ways, some of them generating oxygen as a by-product, others not. Lithotrophic species can feed on a plain diet of inorganic nutrients, getting their carbon from CO2, and relying on H2S to fuel their energy needs (Figure 1–16)—or on H2, or Fe2+, or elemental sulfur, or any of a host of other chemicals that occur in the environment.
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Figure 1–14 The structure of a bacterium. (A) The bacterium Vibrio cholerae, showing its simple internal organization. like many other species, Vibrio has a helical appendage at one end—a flagellum—that rotates as a propeller to drive the cell forward. It can infect the human small intestine to cause cholera; the severe diarrhea that accompanies this disease kills more than 100,000 people a year. (B) An electron micrograph of a longitudinal section through the widely studied bacterium Escherichia coli (E. coli). The cell’s dnA is concentrated in the lightly stained region. Part of our normal intestinal flora, E. coli is related to Vibrio, and it has many flagella distributed over its surface that are not visible in this section. (B, courtesy of e. kellenberger.) spherical cells e.g., Streptococcus spiral cells e.g., Treponema pallidum
Figure 1–13 Shapes and sizes of some bacteria. Although most are small, as shown, measuring a few micrometers in linear dimension, there are also some giant species. An extreme example (not shown) is the cigar-shaped bacterium Epulopiscium fishelsoni, which lives in the gut of a surgeonfish and can be up to 600 μm long.
Much of this world of microscopic organisms is virtually unexplored. Traditional methods of bacteriology have given us an acquaintance with those species that can be isolated and cultured in the laboratory. But DNA sequence analysis of the populations of bacteria and archaea in samples from natural habitats—such as soil or ocean water, or even the human mouth—has opened our eyes to the fact that most species cannot be cultured by standard laboratory techniques. According to one estimate, at least 99% of prokaryotic species remain to be characterized. Detected only by their DNA, it has not yet been possible to grow the vast majority of them in laboratories.
The Tree of life has Three Primary Branches: Bacteria, Archaea, and eukaryotes
The classification of living things has traditionally depended on comparisons of their outward appearances: we can see that a fish has eyes, jaws, backbone, brain, and so on, just as we do, and that a worm does not; that a rosebush is cousin to an apple tree, but is less similar to a grass. As Darwin showed, we can readily interpret such close family resemblances in terms of evolution from common ancestors, and we can find the remains of many of these ancestors preserved in the fossil record. In this way, it has been possible to begin to draw a family tree of living organisms, showing the various lines of descent, as well as branch points in the history, where the ancestors of one group of species became different from those of another.
When the disparities between organisms become very great, however, these methods begin to fail. How do we decide whether a fungus is closer kin to a plant or to an animal? When it comes to prokaryotes, the task becomes harder still: one microscopic rod or sphere looks much like another. Microbiologists have therefore sought to classify prokaryotes in terms of their biochemistry and nutritional requirements. But this approach also has its pitfalls. Amid the bewildering variety of biochemical behaviors, it is difficult to know which differences truly reflect differences of evolutionary history.
Genome analysis has now given us a simpler, more direct, and much more powerful way to determine evolutionary relationships. The complete DNA sequence of an organism defines its nature with almost perfect precision and in exhaustive detail. Moreover, this specification is in a digital form—a string of let-ters—that can be entered straightforwardly into a computer and compared with the corresponding information for any other living thing. Because DNA is subject to random changes that accumulate over long periods of time (as we shall see shortly), the number of differences between the DNA sequences of two organisms can provide a direct, objective, quantitative indication of the evolutionary distance between them.
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This approach has shown that the organisms that were traditionally classed together as “bacteria” can be as widely divergent in their evolutionary origins as is any prokaryote from any eukaryote. It is now clear that the prokaryotes comprise two distinct groups that diverged early in the history of life on Earth, before the eukaryotes diverged as a separate group. The two groups of prokaryotes are called the bacteria (or eubacteria) and the archaea (or archaebacteria). Detailed genome analyses have recently revealed that the first eukayotic cell formed after a
Figure 1–15 The phototrophic bacterium Anabaena cylindrica viewed in the light microscope. The cells of this species form long, multicellular filaments. most of the cells (labeled v) perform photosynthesis, while others become specialized for nitrogen fixation (labeled h) or develop into resistant spores (labeled s). (courtesy of dave G. Adams.)
Figure 1–16 A lithotrophic bacterium.
Beggiatoa, which lives in sulfurous environments, gets its energy by oxidizing h2s and can fix carbon even in the dark. note the yellow deposits of sulfur inside the cells. (courtesy of Ralph W. Wolfe.)
ARCHAEA BACTERIA EUKARYOTES common ancestor cellAquifex Thermotoga cyanobacteria Bacillus E. coli Aeropyrum Sulfolobus Haloferax Methano-thermobacter Methanococcus Paramecium Dictyostelium Euglena Trypanosoma maize yeast human 1 change/10 nucleotides frst eukaryote Giardia Trichomonas particular type of ancient archaeal cell engulfed an ancient bacterium (see Figure 12–3). Thus, the living world today is considered to consist of three major divisions or domains: bacteria, archaea, and eukaryotes (Figure 1–17).
Archaea are often found inhabiting environments that we humans avoid, such as bogs, sewage treatment plants, ocean depths, salt brines, and hot acid springs, although they are also widespread in less extreme and more homely environments, from soils and lakes to the stomachs of cattle. In outward appearance they are not easily distinguished from bacteria. At a molecular level, archaea seem to resemble eukaryotes more closely in their machinery for handling genetic information (replication, transcription, and translation), but bacteria more closely in their apparatus for metabolism and energy conversion. We discuss below how this might be explained.
Both in the storage and in the copying of genetic information, random accidents and errors occur, altering the nucleotide sequence—that is, creating mutations. Therefore, when a cell divides, its two daughters are often not quite identical to one another or to their parent. On rare occasions, the error may represent a change for the better; more probably, it will cause no significant difference in the cell’s prospects. But in many cases, the error will cause serious damage—for example, by disrupting the coding sequence for a key protein. Changes due to mistakes of the first type will tend to be perpetuated, because the altered cell has an increased likelihood of reproducing itself. Changes due to mistakes of the second type—selectively neutral changes—may be perpetuated or not: in the competition for limited resources, it is a matter of chance whether the altered cell or its cousins will succeed. But changes that cause serious damage lead nowhere: the cell that suffers them dies, leaving no progeny. Through endless repetition of this cycle of error and trial—of mutation and natural selection—organisms evolve: their genetic specifications change, giving them new ways to exploit the environment more effectively, to survive in competition with others, and to reproduce successfully.
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Some parts of the genome will change more easily than others in the course of evolution. A segment of DNA that does not code for protein and has no significant regulatory role is free to change at a rate limited only by the frequency of random errors. In contrast, a gene that codes for a highly optimized essential protein or RNA molecule cannot alter so easily: when mistakes occur, the faulty cells are almost always eliminated. Genes of this latter sort are therefore highly conserved. Through 3.5 billion years or more of evolutionary history, many features of the genome have changed beyond all recognition, but the most highly conserved genes remain perfectly recognizable in all living species.
Figure 1–17 The three major divisions (domains) of the living world. note that the word bacteria was originally used to refer to prokaryotes in general, but more recently has been redefined to refer to eubacteria specifically. The tree shown here is based on comparisons of the nucleotide sequence of a ribosomal RnA (rRnA) subunit in the different species, and the distances in the diagram represent estimates of the numbers of evolutionary changes that have occurred in this molecule in each lineage (see Figure 1–18). The parts of the tree shrouded in gray cloud represent uncertainties about details of the true pattern of species divergence in the course of evolution: comparisons of nucleotide or amino acid sequences of molecules other than rRnA, as well as other arguments, can lead to somewhat different trees. As indicated, the nucleus of the eukaryotic cell is now thought to have emerged from a sub-branch within the archaea, so that in the beginning the tree of life had only two branches—bacteria and archaea.
GTTCCGGGGGGAGTATGGTTGCAAAGCTGAAACTTAAAGGAATTGACGGAAGGGCACCACCAGGAGTGGAGCCTGCGGCTTAATTTGACTCAACACGGGAAACCTCACCC human GCCGCCTGGGGAGTACGGTCGCAAGACTGAAACTTAAAGGAATTGGCGGGGGAGCACTACAACGGGTGGAGCCTGCGGTTTAATTGGATTCAACGCCGGGCATCTTACCA Methanococcus ACCGCCTGGGGAGTACGGCCGCAAGGTTAAAACTCAAATGAATTGACGGGGGCCCGC.ACAAGCGGTGGAGCATGTGGTTTAATTCGATGCAACGCGAAGAACCTTACCT E. coli GTTCCGGGGGGAGTATGGTTGCAAAGCTGAAACTTAAAGGAATTGACGGAAGGGCACCACCAGGAGTGGAGCCTGCGGCTTAATTTGACTCAACACGGGAAACCTCACCC human
These latter genes are the ones we must examine if we wish to trace family relationships between the most distantly related organisms in the tree of life. The initial studies that led to the classification of the living world into the three domains of bacteria, archaea, and eukaryotes were based chiefly on analysis of one of the rRNA components of the ribosome. Because the translation of RNA into protein is fundamental to all living cells, this component of the ribosome has been very well conserved since early in the history of life on Earth (Figure 1–18).
Natural selection has generally favored those prokaryotic cells that can reproduce the fastest by taking up raw materials from their environment and replicating themselves most efficiently, at the maximal rate permitted by the available food supplies. Small size implies a large ratio of surface area to volume, thereby helping to maximize the uptake of nutrients across the plasma membrane and boosting a cell’s reproductive rate.
Presumably for these reasons, most prokaryotic cells carry very little superfluous baggage; their genomes are small, with genes packed closely together and minimal quantities of regulatory DNA between them. The small genome size has made it easy to use modern DNA sequencing techniques to determine complete genome sequences. We now have this information for thousands of species of bacteria and archaea, as well as for hundreds of species of eukaryotes. Most bacterial and archaeal genomes contain between 106 and 107 nucleotide pairs, encoding 1000–6000 genes.
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A complete DNA sequence reveals both the genes an organism possesses and the genes it lacks. When we compare the three domains of the living world, we can begin to see which genes are common to all of them and must therefore have been present in the cell that was ancestral to all present-day living things, and which genes are peculiar to a single branch in the tree of life. To explain the findings, however, we need to consider a little more closely how new genes arise and genomes evolve.
The raw material of evolution is the DNA sequence that already exists: there is no natural mechanism for making long stretches of new random sequence. In this sense, no gene is ever entirely new. Innovation can, however, occur in several ways (Figure 1–19): 1.
Intragenic mutation: an existing gene can be randomly modified by changes in its DNA sequence, through various types of error that occur mainly in the process of DNA replication.
2.
Gene duplication: an existing gene can be accidentally duplicated so as to create a pair of initially identical genes within a single cell; these two genes may then diverge in the course of evolution.
3.
DNA segment shuffling: two or more existing genes can break and rejoin to make a hybrid gene consisting of DNA segments that originally belonged to separate genes.
4.
Horizontal (intercellular) transfer: a piece of DNA can be transferred from the genome of one cell to that of another—even to that of another species. This process is in contrast with the usual vertical transfer of genetic information from parent to progeny.
Each of these types of change leaves a characteristic trace in the DNA sequence of the organism, and there is clear evidence that all four processes have frequently
Figure 1–18 Genetic information conserved since the days of the last common ancestor of all living things.
A part of the gene for the smaller of the two main rRnA components of the ribosome is shown. (The complete molecule is about 1500–1900 nucleotides long, depending on species.) corresponding segments of nucleotide sequence from an archaean (Methanococcus jannaschii), a bacterium (Escherichia coli), and a eukaryote (Homo sapiens) are aligned. sites where the nucleotides are identical between species are indicated by a vertical line; the human sequence is repeated at the bottom of the alignment so that all three two-way comparisons can be seen. A dot halfway along the E. coli sequence denotes a site where a nucleotide has been either deleted from the bacterial lineage in the course of evolution or inserted in the other two lineages. note that the sequences from these three organisms, representative of the three domains of the living world, still retain unmistakable similarities.
occurred. In later chapters, we discuss the underlying mechanisms, but for the present we focus on the consequences.
Gene duplications Give Rise to Families of Related Genes Within a single cell
A cell duplicates its entire genome each time it divides into two daughter cells. However, accidents occasionally result in the inappropriate duplication of just part of the genome, with retention of original and duplicate segments in a single cell. Once a gene has been duplicated in this way, one of the two gene copies is free to mutate and become specialized to perform a different function within the same cell. Repeated rounds of this process of duplication and divergence, over many millions of years, have enabled one gene to give rise to a family of genes that may all be found within a single genome. Analysis of the DNA sequence of prokaryotic genomes reveals many examples of such gene families: in the bacterium Bacillus subtilis, for example, 47% of the genes have one or more obvious relatives (Figure 1–20).
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When genes duplicate and diverge in this way, the individuals of one species become endowed with multiple variants of a primordial gene. This evolutionary process has to be distinguished from the genetic divergence that occurs when one species of organism splits into two separate lines of descent at a branch point in the family tree—when the human line of descent became separate from that of chimpanzees, for example. There, the genes gradually become different in the course of evolution, but they are likely to continue to have corresponding functions in the two sister species. Genes that are related by descent in this way—that is, genes in two separate species that derive from the same ancestral gene in the last common ancestor of those two species—are called orthologs. Related genes that have resulted from a gene duplication event within a single genome—and
Figure 1–19 Four modes of genetic innovation and their effects on the DNA sequence of an organism. A special form of horizontal transfer occurs when two different types of cells enter into a permanent symbiotic association. Genes from one of the cells then may be transferred to the genome of the other, as we shall see below when we discuss mitochondria and chloroplasts.
283 genes in families with 38–77 gene members 764 genes in families with 4–19 gene members 2126 genes with no family relationship273 genes in families with 3 gene members 568 genes in families with 2 gene members are likely to have diverged in their function—are called paralogs. Genes that are related by descent in either way are called homologs, a general term used to cover both types of relationship (Figure 1–21).
Genes can Be Transferred Between organisms, Both in the laboratory and in nature
Prokaryotes provide good examples of the horizontal transfer of genes from one species of cell to another. The most obvious tell-tale signs are sequences recognizable as being derived from viruses, those infecting bacteria being called bacteriophages (Figure 1–22). Viruses are small packets of genetic material that have evolved as parasites on the reproductive and biosynthetic machinery of host cells. Although not themselves living cells, they often serve as vectors for gene transfer. A virus will replicate in one cell, emerge from it with a protective wrapping, and then enter and infect another cell, which may be of the same or a different species. Often, the infected cell will be killed by the massive proliferation of virus particles inside it; but sometimes, the viral DNA, instead of directly generating these particles, may persist in its host for many cell generations as a relatively innocuous passenger, either as a separate intracellular fragment of DNA, known as a plasmid, or as a sequence inserted into the cell’s regular genome. In their travels, viruses can accidentally pick up fragments of DNA from the genome of one host cell and ferry them into another cell. Such transfers of genetic material are very common in prokaryotes.
Horizontal transfers of genes between eukaryotic cells of different species are very rare, and they do not seem to have played a significant part in eukaryote evolution (although massive transfers from bacterial to eukaryotic genomes have occurred in the evolution of mitochondria and chloroplasts, as we discuss below).
genes GA and GB are orthologs genes G1 and G2 are paralogs
Figure 1–20 Families of evolutionarily related genes in the genome of Bacillus subtilis. The largest gene family in this bacterium consists of 77 genes coding for varieties of ABc transporters—a class of membrane transport proteins found in all three domains of the living world. (Adapted from F. kunst et al., Nature 390:249–256, 1997. With permission from macmillan Publishers ltd.)
Figure 1–21 Paralogous genes and orthologous genes: two types of gene homology based on different evolutionary pathways. (A) orthologs.
(B) Paralogs.
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In contrast, horizontal gene transfers occur much more frequently between different species of prokaryotes. Many prokaryotes have a remarkable capacity to take up even nonviral DNA molecules from their surroundings and thereby capture the genetic information these molecules carry. By this route, or by virus-mediated transfer, bacteria and archaea in the wild can acquire genes from neighboring cells relatively easily. Genes that confer resistance to an antibiotic or an ability to produce a toxin, for example, can be transferred from species to species and provide the recipient bacterium with a selective advantage. In this way, new and sometimes dangerous strains of bacteria have been observed to evolve in the bacterial ecosystems that inhabit hospitals or the various niches in the human body. For example, horizontal gene transfer is responsible for the spread, over the past 40 years, of penicillin-resistant strains of Neisseria gonorrhoeae, the bacterium that causes gonorrhea. On a longer time scale, the results can be even more profound; it has been estimated that at least 18% of all of the genes in the present-day genome of E. coli have been acquired by horizontal transfer from another species within the past 100 million years.
sex Results in horizontal exchanges of Genetic Information Within a species
Horizontal gene transfer among prokaryotes has a parallel in a phenomenon familiar to us all: sex. In addition to the usual vertical transfer of genetic material from parent to offspring, sexual reproduction causes a large-scale horizontal transfer of genetic information between two initially separate cell lineages—those of the father and the mother. A key feature of sex, of course, is that the genetic exchange normally occurs only between individuals of the same species. But no matter whether they occur within a species or between species, horizontal gene
Figure 1–22 The viral transfer of DNA into a cell. (A) An electron micrograph of particles of a bacterial virus, the T4 bacteriophage. The head of this virus contains the viral dnA; the tail contains the apparatus for injecting the dnA into a host bacterium. (B) A cross section of an
E. coli bacterium with a T4 bacteriophage latched onto its surface. The large dark objects inside the bacterium are the heads of new T4 particles in the course of assembly. When they are mature, the bacterium will burst open to release them. (c–e) The process of dnA injection into the bacterium, as visualized in unstained, frozen samples by cryoelectron microscopy. (c) Attachment begins.
Attached state during dnA injection.
virus head has emptied all of its dnA into the bacterium. (A, courtesy of James Paulson; B, courtesy of Jonathan king and erika hartwig from G. karp, cell and molecular Biology, 2nd ed. new york: John Wiley & sons, 1999. With permission from John Wiley & sons; c–e, courtesy of Ian molineux, university of Texas at Austin and Jun liu, university of Texas health science center, houston.) transfers leave a characteristic imprint: they result in individuals who are related more closely to one set of relatives with respect to some genes, and more closely to another set of relatives with respect to others. By comparing the DNA sequences of individual human genomes, an intelligent visitor from outer space could deduce that humans reproduce sexually, even if it knew nothing about human behavior.
Sexual reproduction is widespread (although not universal), especially among eukaryotes. Even bacteria indulge from time to time in controlled sexual exchanges of DNA with other members of their own species. Natural selection has clearly favored organisms that can reproduce sexually, although evolutionary theorists dispute precisely what that selective advantage is.
The Function of a Gene can often Be deduced from Its sequence
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Family relationships among genes are important not just for their historical interest, but because they simplify the task of deciphering gene functions. Once the sequence of a newly discovered gene has been determined, a scientist can tap a few keys on a computer to search the entire database of known gene sequences for genes related to it. In many cases, the function of one or more of these homo-logs will have been already determined experimentally. Since gene sequence determines gene function, one can frequently make a good guess at the function of the new gene: it is likely to be similar to that of the already known homologs.
In this way, it is possible to decipher a great deal of the biology of an organism simply by analyzing the DNA sequence of its genome and using the information we already have about the functions of genes in other organisms that have been more intensively studied.
more Than 200 Gene Families Are common to All Three Primary Branches of the Tree of life
Given the complete genome sequences of representative organisms from all three domains—archaea, bacteria, and eukaryotes—we can search systematically for homologies that span this enormous evolutionary divide. In this way we can begin to take stock of the common inheritance of all living things. There are considerable difficulties in this enterprise. For example, individual species have often lost some of the ancestral genes; other genes have almost certainly been acquired by horizontal transfer from another species and therefore are not truly ancestral, even though shared. In fact, genome comparisons strongly suggest that both lineage-specific gene loss and horizontal gene transfer, in some cases between evolutionarily distant species, have been major factors of evolution, at least among prokaryotes. Finally, in the course of 2 or 3 billion years, some genes that were initially shared will have changed beyond recognition through mutation.
Because of all these vagaries of the evolutionary process, it seems that only a small proportion of ancestral gene families has been universally retained in a recognizable form. Thus, out of 4873 protein-coding gene families defined by comparing the genomes of 50 species of bacteria, 13 archaea, and 3 unicellular eukaryotes, only 63 are truly ubiquitous (that is, represented in all the genomes analyzed). The great majority of these universal families include components of the translation and transcription systems. This is not likely to be a realistic approximation of an ancestral gene set. A better—though still crude—idea of the latter can be obtained by tallying the gene families that have representatives in multiple, but not necessarily all, species from all three major domains. Such an analysis reveals 264 ancient conserved families. Each family can be assigned a function (at least in terms of general biochemical activity, but usually with more precision). As shown in Table 1–1, the largest number of shared gene families are involved in translation and in amino acid metabolism and transport. However, this set of highly conserved gene families represents only a very rough sketch of the common inheritance of all modern life. A more precise reconstruction of the gene complement of the last universal common ancestor will hopefully become feasible with further genome sequencing and more sophisticated forms of comparative analysis.
mutations Reveal the Functions of Genes
Without additional information, no amount of gazing at genome sequences will reveal the functions of genes. We may recognize that gene B is like gene A, but how do we discover the function of gene A in the first place? And even if we know the function of gene A, how do we test whether the function of gene B is truly the same as the sequence similarity suggests? How do we connect the world of abstract genetic information with the world of real living organisms?
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The analysis of gene functions depends on two complementary approaches: genetics and biochemistry. Genetics starts with the study of mutants: we either find or make an organism in which a gene is altered, and then examine the effects on the organism’s structure and performance (Figure 1–23). Biochemistry more directly examines the functions of molecules: here we extract molecules from an organism and then study their chemical activities. By combining genetics and biochemistry, it is possible to find those molecules whose production depends on a given gene. At the same time, careful studies of the performance of the mutant organism show us what role those molecules have in the operation of the organism as a whole. Thus, genetics and biochemistry used in combination with cell biology provide the best way to relate genes and molecules to the structure and function of an organism.
In recent years, DNA sequence information and the powerful tools of molecular biology have accelerated progress. From sequence comparisons, we can often identify particular subregions within a gene that have been preserved nearly unchanged over the course of evolution. These conserved subregions are likely to be the most important parts of the gene in terms of function. We can test their individual contributions to the activity of the gene product by creating in the laboratory mutations of specific sites within the gene, or by constructing artificial hybrid genes that combine part of one gene with part of another. Organisms can be engineered to make either the RNA or the protein specified by the gene in large quantities to facilitate biochemical analysis. Specialists in molecular structure can determine the three-dimensional conformation of the gene product, revealing the exact position of every atom in it. Biochemists can determine how each of the
Figure 1–23 A mutant phenotype reflecting the function of a gene. A normal yeast (of the species Schizosaccharomyces pombe) is compared with a mutant in which a change in a single gene has converted the cell from a cigar shape (left) to a T shape (right). The mutant gene therefore has a function in the control of cell shape. But how, in molecular terms, does the gene product perform that function? That is a harder question, and it needs biochemical analysis to answer it. (courtesy of kenneth sawin and Paul nurse.) parts of the genetically specified molecule contributes to its chemical behavior. Cell biologists can analyze the behavior of cells that are engineered to express a mutant version of the gene.
There is, however, no one simple recipe for discovering a gene’s function, and no simple standard universal format for describing it. We may discover, for example, that the product of a given gene catalyzes a certain chemical reaction, and yet have no idea how or why that reaction is important to the organism. The functional characterization of each new family of gene products, unlike the description of the gene sequences, presents a fresh challenge to the biologist’s ingenuity. Moreover, we will never fully understand the function of a gene until we learn its role in the life of the organism as a whole. To make ultimate sense of gene functions, therefore, we have to study whole organisms, not just molecules or cells.
molecular Biology Began with a spotlight on E. coli
Because living organisms are so complex, the more we learn about any particular species, the more attractive it becomes as an object for further study. Each discovery raises new questions and provides new tools with which to tackle general questions in the context of the chosen organism. For this reason, large communities of biologists have become dedicated to studying different aspects of the same model organism.
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In the early days of molecular biology, the spotlight focused intensely on just one species: the Escherichia coli, or E. coli, bacterium (see Figures 1–13 and 1–14). This small, rod-shaped bacterial cell normally lives in the gut of humans and other vertebrates, but it can be grown easily in a simple nutrient broth in a culture bottle. It adapts to variable chemical conditions and reproduces rapidly, and it can evolve by mutation and selection at a remarkable speed. As with other bacteria, different strains of E. coli, though classified as members of a single species, differ genetically to a much greater degree than do different varieties of a sexually reproducing organism such as a plant or animal. One E. coli strain may possess many hundreds of genes that are absent from another, and the two strains could have as little as 50% of their genes in common. The standard laboratory strain
E. coli K-12 has a genome of approximately 4.6 million nucleotide pairs, contained in a single circular molecule of DNA that codes for about 4300 different kinds of proteins (Figure 1–24).
In molecular terms, we know more about E. coli than about any other living organism. Most of our understanding of the fundamental mechanisms of life— for example, how cells replicate their DNA, or how they decode the instructions represented in the DNA to direct the synthesis of specific proteins—initially came from studies of E. coli. The basic genetic mechanisms have turned out to be highly conserved throughout evolution: these mechanisms are essentially the same in our own cells as in E. coli.
Prokaryotes (cells without a distinct nucleus) are biochemically the most diverse organisms and include species that can obtain all their energy and nutrients from inorganic chemical sources, such as the reactive mixtures of minerals released at hydrothermal vents on the ocean floor—the sort of diet that may have nourished the first living cells 3.5 billion years ago. DNA sequence comparisons reveal the family relationships of living organisms and show that the prokaryotes fall into two groups that diverged early in the course of evolution: the bacteria (or eubacteria) and the archaea. Together with the eukaryotes (cells with a membrane-enclosed nucleus), these constitute the three primary branches of the tree of life.
Most bacteria and archaea are small unicellular organisms with compact genomes comprising 1000–6000 genes. Many of the genes within a single organism show strong family resemblances in their DNA sequences, implying that they originated from the same ancestral gene through gene duplication and divergence. Family resemblances (homologies) are also clear when gene sequences are compared between different species, and more than 200 gene families have been so highly (A) 4,639,221 nucleotide pairs Escherichia coli K-12 origin of replication terminus of replication (B)
Figure 1–24 The genome of E. coli. (A) A cluster of E. coli cells. (B) A diagram of the genome of
E. coli strain k-12. The diagram is circular because the dnA of E. coli, like that of other prokaryotes, forms a single, closed loop. Protein-coding genes are shown as yellow or orange bars, depending on the dnA strand from which they are transcribed; genes encoding only RnA molecules are indicated by green arrows. some genes are transcribed from one strand of the dnA double helix (in a clockwise direction in this diagram), others from the other strand (counterclockwise). (A, courtesy of dr. Tony Brain and david Parker/Photo Researchers; B, adapted from F.R. Blattner et al., Science 277:1453–1462, 1997.) conserved that they can be recognized as common to most species from all three domains of the living world. Thus, given the DNA sequence of a newly discovered gene, it is often possible to deduce the gene’s function from the known function of a homologous gene in an intensively studied model organism, such as the bacterium
E. coli.
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Eukaryotic cells, in general, are bigger and more elaborate than prokaryotic cells, and their genomes are bigger and more elaborate, too. The greater size is accompanied by radical differences in cell structure and function. Moreover, many classes of eukaryotic cells form multicellular organisms that attain levels of complexity unmatched by any prokaryote.
Because they are so complex, eukaryotes confront molecular biologists with a special set of challenges that will concern us in the rest of this book. Increasingly, biologists attempt to meet these challenges through the analysis and manipulation of the genetic information within cells and organisms. It is therefore important at the outset to know something of the special features of the eukaryotic genome. We begin by briefly discussing how eukaryotic cells are organized, how this reflects their way of life, and how their genomes differ from those of prokaryotes. This leads us to an outline of the strategy by which cell biologists, by exploiting genetic and biochemical information, are attempting to discover how eukaryotic organisms work.
eukaryotic cells may have originated as Predators
By definition, eukaryotic cells keep their DNA in an internal compartment called the nucleus. The nuclear envelope, a double layer of membrane, surrounds the nucleus and separates the DNA from the cytoplasm. Eukaryotes also have other features that set them apart from prokaryotes (Figure 1–25). Their cells are, typically, 10 times bigger in linear dimension and 1000 times larger in volume. They have an elaborate cytoskeleton—a system of protein filaments crisscrossing the cytoplasm and forming, together with the many proteins that attach to them, a system of girders, ropes, and motors that gives the cell mechanical strength, controls its shape, and drives and guides its movements (Movie 1.1). And the nuclear envelope is only one part of a set of internal membranes, each structurally similar to the plasma membrane and enclosing different types of spaces inside the cell, many of them involved in digestion and secretion. Lacking the tough cell wall of most bacteria, animal cells and the free-living eukaryotic cells called protozoa can change their shape rapidly and engulf other cells and small objects by phagocytosis (Figure 1–26).
How all of the unique properties of eukaryotic cells evolved, and in what sequence, is still a mystery. One plausible view, however, is that they are all reflections of the way of life of a primordial cell that was a predator, living by capturing other cells and eating them (Figure 1–27). Such a way of life requires a large cell with a flexible plasma membrane, as well as an elaborate cytoskeleton to support
Figure 1–25 The major features of eukaryotic cells. The drawing depicts a typical animal cell, but almost all the same components are found in plants and fungi as well as in single-celled eukaryotes such as yeasts and protozoa. Plant cells contain chloroplasts in addition to the components shown here, and their plasma membrane is surrounded by a tough external wall formed of cellulose.
and move this membrane. It may also require that the cell’s long, fragile DNA molecules be sequestered in a separate nuclear compartment, to protect the genome from damage by the movements of the cytoskeleton.
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A predatory way of life helps to explain another feature of eukaryotic cells. All such cells contain (or at one time did contain) mitochondria (Figure 1–28). These small bodies in the cytoplasm, enclosed by a double layer of membrane, take up oxygen and harness energy from the oxidation of food molecules—such as sugars—to produce most of the ATP that powers the cell’s activities. Mitochondria are similar in size to small bacteria, and, like bacteria, they have their own genome in the form of a circular DNA molecule, their own ribosomes that differ from those elsewhere in the eukaryotic cell, and their own transfer RNAs. It is now generally accepted that mitochondria originated from free-living oxygen-metabolizing (aerobic) bacteria that were engulfed by an ancestral cell that could otherwise make no such use of oxygen (that is, was anaerobic). Escaping digestion, these bacteria evolved in symbiosis with the engulfing cell and its progeny, receiving
Figure 1–27 A single-celled eukaryote that eats other cells. (A) Didinium is a carnivorous protozoan, belonging to the group known as ciliates. It has a globular body, about 150 μm in diameter, encircled by two fringes of cilia—sinuous, whiplike appendages that beat continually; its front end is flattened except for a single protrusion, rather like a snout. (B) A Didinium engulfing its prey. Didinium normally swims around in the water at high speed by means of the synchronous beating of its cilia. When it encounters a suitable prey (yellow), usually another type of protozoan, it releases numerous small paralyzing darts from its snout region. Then, the Didinium attaches to and devours the other cell by phagocytosis, inverting like a hollow ball to engulf its victim, which can be almost as large as itself. (courtesy of d. Barlow.)
Figure 1–26 Phagocytosis. This series of stills from a movie shows a human white blood cell (a neutrophil) engulfing a red blood cell (artificially colored red) that has been treated with an antibody that marks it for destruction (see movie 13.5). (courtesy of stephen e. malawista and Anne de Boisfleury chevance.) shelter and nourishment in return for the power generation they performed for their hosts. This partnership between a primitive anaerobic predator cell and an aerobic bacterial cell is thought to have been established about 1.5 billion years ago, when the Earth’s atmosphere first became rich in oxygen.
As indicated in Figure 1–29, recent genomic analyses suggest that the first eukaryotic cells formed after an archaeal cell engulfed an aerobic bacterium. This would explain why all eukaryotic cells today, including those that live as strict anaerobes show clear evidence that they once contained mitochondria.
Many eukaryotic cells—specifically, those of plants and algae—also contain another class of small membrane-enclosed organelles somewhat similar to mitochondria—the chloroplasts (Figure 1–30). Chloroplasts perform photosynthesis, using the energy of sunlight to synthesize carbohydrates from atmospheric carbon dioxide and water, and deliver the products to the host cell as food. Like mitochondria, chloroplasts have their own genome. They almost certainly originated as symbiotic photosynthetic bacteria, acquired by eukaryotic cells that already possessed mitochondria (Figure 1–31).
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A eukaryotic cell equipped with chloroplasts has no need to chase after other cells as prey; it is nourished by the captive chloroplasts it has inherited from its ancestors. Correspondingly, plant cells, although they possess the cytoskeletal equipment for movement, have lost the ability to change shape rapidly and to engulf other cells by phagocytosis. Instead, they create around themselves a tough, protective cell wall. If the first eukaryotic cells were predators on other organisms, we can view plant cells as cells that have made the transition from hunting to farming.
Fungi represent yet another eukaryotic way of life. Fungal cells, like animal cells, possess mitochondria but not chloroplasts; but in contrast with animal cells and protozoa, they have a tough outer wall that limits their ability to move rapidly
Figure 1–28 A mitochondrion. (A) A cross section, as seen in the electron microscope.
(B) A drawing of a mitochondrion with part of it cut away to show the three-dimensional structure (Movie 1.2). (c) A schematic eukaryotic cell, with the interior space of a mitochondrion, containing the mitochondrial dnA and ribosomes, colored. note the smooth outer membrane and the convoluted inner membrane, which houses the proteins that generate ATP from the oxidation of food molecules. (A, courtesy of daniel s. Friend.)
Figure 1–29 The origin of mitochondria.
archaeon) is thought to have engulfed the bacterial ancestor of mitochondria, initiating a symbiotic relationship. clear evidence of a dual bacterial and archaeal inheritance can be discerned today in the genomes of all eukaryotes.
or to swallow up other cells. Fungi, it seems, have turned from hunters into scavengers: other cells secrete nutrient molecules or release them upon death, and fungi feed on these leavings—performing whatever digestion is necessary extracellularly, by secreting digestive enzymes to the exterior.
The genetic information of eukaryotic cells has a hybrid origin—from the ancestral anaerobic archaeal cell, and from the bacteria that it adopted as symbionts. Most of this information is stored in the nucleus, but a small amount remains inside the mitochondria and, for plant and algal cells, in the chloroplasts. When mitochondrial DNA and the chloroplast DNA are separated from the nuclear DNA and individually analyzed and sequenced, the mitochondrial and chloroplast genomes are found to be degenerate, cut-down versions of the corresponding bacterial genomes. In a human cell, for example, the mitochondrial genome consists of only 16,569 nucleotide pairs, and codes for only 13 proteins, 2 ribosomal RNA components, and 22 transfer RNAs.
Figure 1–30 Chloroplasts. These organelles capture the energy of sunlight in plant cells and some single-celled eukaryotes. (A) A single cell isolated from a leaf of a flowering plant, seen in the light microscope, showing the green chloroplasts (Movie 1.3 and see movie 14.9). (B) A drawing of one of the chloroplasts, showing the highly folded system of internal membranes containing the chlorophyll molecules by which light is absorbed. (A, courtesy of Preeti dahiya.) early eukaryotic cell eukaryotic cell capable of photosynthesis
Many of the genes that are missing from the mitochondria and chloroplasts have not been lost; instead, they have moved from the symbiont genome into the DNA of the host cell nucleus. The nuclear DNA of humans contains many genes coding for proteins that serve essential functions inside the mitochondria; in plants, the nuclear DNA also contains many genes specifying proteins required in chloroplasts. In both cases, the DNA sequences of these nuclear genes show clear evidence of their origin from the bacterial ancestor of the respective organelle.
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Natural selection has evidently favored mitochondria with small genomes. By contrast, the nuclear genomes of most eukaryotes seem to have been free to enlarge. Perhaps the eukaryotic way of life has made large size an advantage: predators typically need to be bigger than their prey, and cell size generally increases in proportion to genome size. Whatever the reason, aided by a massive accumulation of DNA segments derived from parasitic transposable elements (discussed in Chapter 5), the genomes of most eukaryotes have become orders of magnitude larger than those of bacteria and archaea (Figure 1–32).
The freedom to be extravagant with DNA has had profound implications. Eukaryotes not only have more genes than prokaryotes; they also have vastly more DNA that does not code for protein. The human genome contains 1000 times as many nucleotide pairs as the genome of a typical bacterium, perhaps 10 times as
Figure 1–31 The origin of chloroplasts.
An early eukaryotic cell, already possessing mitochondria, engulfed a photosynthetic bacterium (a cyanobacterium) and retained it in symbiosis. Present-day chloroplasts are thought to trace their ancestry back to a single species of cyanobacterium that was adopted as an internal symbiont (an endosymbiont) over a billion years ago.
Figure 1–32 Genome sizes compared.
Genome size is measured in nucleotide pairs of dnA per haploid genome, that is, per single copy of the genome. (The cells of sexually reproducing organisms such as ourselves are generally diploid: they contain two copies of the genome, one inherited from the mother, the other from the father.) closely related organisms can vary widely in the quantity of dnA in their genomes, even though they contain similar numbers of functionally distinct genes. (data from
W.h. li, molecular evolution, pp. 380–383. sunderland, mA: sinauer, 1997.) many genes, and a great deal more noncoding DNA (~98.5% of the genome for a human does not code for proteins, as opposed to 11% of the genome for the bacterium E. coli). The estimated genome sizes and gene numbers for some eukaryotes are compiled for easy comparison with E. coli in Table 1–2; we shall discuss how each of these eukaryotes serves as a model organism shortly.
Much of our noncoding DNA is almost certainly dispensable junk, retained like a mass of old papers because, when there is little pressure to keep an archive small, it is easier to retain everything than to sort out the valuable information and discard the rest. Certain exceptional eukaryotic species, such as the puffer fish, bear witness to the profligacy of their relatives; they have somehow managed to rid themselves of large quantities of noncoding DNA. Yet they appear similar in structure, behavior, and fitness to related species that have vastly more such DNA (see Figure 4–71).
Even in compact eukaryotic genomes such as that of puffer fish, there is more noncoding DNA than coding DNA, and at least some of the noncoding DNA certainly has important functions. In particular, it regulates the expression of adjacent genes. With this regulatory DNA, eukaryotes have evolved distinctive ways of controlling when and where a gene is brought into play. This sophisticated gene regulation is crucial for the formation of complex multicellular organisms.
The Genome defines the Program of multicellular development
The cells in an individual animal or plant are extraordinarily varied. Fat cells, skin cells, bone cells, nerve cells—they seem as dissimilar as any cells could be (Figure 1–33). Yet all these cell types are the descendants of a single fertilized egg cell, and all (with minor exceptions) contain identical copies of the genome of the species.
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The differences result from the way in which the cells make selective use of their genetic instructions according to the cues they get from their surroundings in the developing embryo. The DNA is not just a shopping list specifying the molecules that every cell must have, and the cell is not an assembly of all the items on the list. Rather, the cell behaves as a multipurpose machine, with sensors to receive environmental signals and with highly developed abilities to call different sets of genes into action according to the sequences of signals to which the cell has been exposed. The genome in each cell is big enough to accommodate the
Figure 1–33 Cell types can vary enormously in size and shape. An animal nerve cell is compared here with a neutrophil, a type of white blood cell. Both are drawn to scale.
information that specifies an entire multicellular organism, but in any individual cell only part of that information is used.
A large number of genes in the eukaryotic genome code for proteins that regulate the activities of other genes. Most of these transcription regulators act by binding, directly or indirectly, to the regulatory DNA adjacent to the genes that are to be controlled, or by interfering with the abilities of other proteins to do so. The expanded genome of eukaryotes therefore not only specifies the hardware of the cell, but also stores the software that controls how that hardware is used (Figure 1–34).
Cells do not just passively receive signals; rather, they actively exchange signals with their neighbors. Thus, in a developing multicellular organism, the same control system governs each cell, but with different consequences depending on the messages exchanged. The outcome, astonishingly, is a precisely patterned array of cells in different states, each displaying a character appropriate to its position in the multicellular structure.
many eukaryotes live as solitary cells
Many species of eukaryotic cells lead a solitary life—some as hunters (the protozoa), some as photosynthesizers (the unicellular algae), some as scavengers (the unicellular fungi, or yeasts). Figure 1–35 conveys something of the astonishing variety of the single-celled eukaryotes. The anatomy of protozoa, especially, is often elaborate and includes such structures as sensory bristles, photoreceptors, sinuously beating cilia, leglike appendages, mouth parts, stinging darts, and musclelike contractile bundles. Although they are single cells, protozoa can be as intricate, as versatile, and as complex in their behavior as many multicellular organisms (see Figure 1–27, Movie 1.4, and Movie 1.5).
In terms of their ancestry and DNA sequences, the unicellular eukaryotes are far more diverse than the multicellular animals, plants, and fungi, which arose as three comparatively late branches of the eukaryotic pedigree (see Figure 1–17). As with prokaryotes, humans have tended to neglect them because they are microscopic. Only now, with the help of genome analysis, are we beginning to understand their positions in the tree of life, and to put into context the glimpses these strange creatures can offer us of our distant evolutionary past.
A yeast serves as a minimal model eukaryote
The molecular and genetic complexity of eukaryotes is daunting. Even more than for prokaryotes, biologists need to concentrate their limited resources on a few selected model organisms to unravel this complexity.
Figure 1–34 Genetic control of the program of multicellular development.
The role of a regulatory gene is demonstrated in the snapdragon Antirrhinum. In this example, a mutation in a single gene coding for a regulatory protein causes leafy shoots to develop in place of flowers: because a regulatory protein has been changed, the cells adopt characters that would be appropriate to a different location in the normal plant. The mutant is on the left, the normal plant on the right. (courtesy of enrico coen and Rosemary carpenter.)
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To analyze the internal workings of the eukaryotic cell without the additional problems of multicellular development, it makes sense to use a species that is unicellular and as simple as possible. The popular choice for this role of minimal model eukaryote has been the yeast Saccharomyces cerevisiae (Figure 1–36)—the same species that is used by brewers of beer and bakers of bread.
S. cerevisiae is a small, single-celled member of the kingdom of fungi and thus, according to modern views, is at least as closely related to animals as it is to plants. It is robust and easy to grow in a simple nutrient medium. Like other fungi, it has a tough cell wall, is relatively immobile, and possesses mitochondria but not chloroplasts. When nutrients are plentiful, it grows and divides almost as rapidly as a bacterium. It can reproduce either vegetatively (that is, by simple cell division), or sexually: two yeast cells that are haploid (possessing a single copy of the genome) can fuse to create a cell that is diploid (containing a double genome); and the diploid cell can undergo meiosis (a reduction division) to produce cells that are once again haploid (Figure 1–37). In contrast with higher plants and animals, the yeast can divide indefinitely in either the haploid or the diploid state, and the process leading from one state to the other can be induced at will by changing the growth conditions.
In addition to these features, the yeast has a further property that makes it a convenient organism for genetic studies: its genome, by eukaryotic standards, is exceptionally small. Nevertheless, it suffices for all the basic tasks that every eukaryotic cell must perform. Mutants are available for essentially every gene,
Figure 1–35 An assortment of protozoa: a small sample of an extremely diverse class of organisms. The drawings are done to different scales, but in each case the scale bar represents 10 μm. The organisms in (A), (c), and (G) are ciliates; is a heliozoan; (d) is an amoeba; is a dinoflagellate; and (F) is a euglenoid. (From m.A. sleigh, Biology of Protozoa. cambridge, uk: cambridge university Press, 1973.)
Figure 1–36 The yeast Saccharomyces cerevisiae. (A) A scanning electron micrograph of a cluster of the cells. This species is also known as budding yeast; it proliferates by forming a protrusion or bud that enlarges and then separates from the rest of the original cell. many cells with buds are visible in this micrograph.
(B) A transmission electron micrograph of a cross section of a yeast cell, showing its nucleus, mitochondrion, and thick cell wall. (A, courtesy of Ira herskowitz and eric schabatach.)
Figure 1–37 The reproductive cycles of the yeast S. cerevisiae.
depending on environmental conditions and on details of the genotype, cells of this species can exist in either a diploid (2n) state, with a double chromosome set, or a haploid (n) state, with a single chromosome set. The diploid form can either proliferate by ordinary cell-division cycles or undergo meiosis to produce haploid cells. The haploid form can either proliferate by ordinary cell-division cycles or undergo sexual fusion with another haploid cell to become diploid. meiosis is triggered by starvation and gives rise to spores—haploid cells in a dormant state, resistant to harsh environmental conditions.
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and studies on yeasts (using both S. cerevisiae and other species) have provided a key to many crucial processes, including the eukaryotic cell-division cycle—the critical chain of events by which the nucleus and all the other components of a cell are duplicated and parceled out to create two daughter cells from one. The control system that governs this process has been so well conserved over the course of evolution that many of its components can function interchangeably in yeast and human cells: if a mutant yeast lacking an essential yeast cell-division-cycle gene is supplied with a copy of the homologous cell-division-cycle gene from a human, the yeast is cured of its defect and becomes able to divide normally.
The expression levels of All the Genes of An organism can Be monitored simultaneously
The complete genome sequence of S. cerevisiae, determined in 1997, consists of approximately 13,117,000 nucleotide pairs, including the small contribution (78,520 nucleotide pairs) of the mitochondrial DNA. This total is only about 2.5 times as much DNA as there is in E. coli, and it codes for only 1.5 times as many distinct proteins (about 6600 in all). The way of life of S. cerevisiae is similar in many ways to that of a bacterium, and it seems that this yeast has likewise been subject to selection pressures that have kept its genome compact.
Knowledge of the complete genome sequence of any organism—be it a yeast or a human—opens up new perspectives on the workings of the cell: things that once seemed impossibly complex now seem within our grasp. Using techniques described in Chapter 8, it is now possible, for example, to monitor, simultaneously, the amount of mRNA transcript that is produced from every gene in the yeast genome under any chosen conditions, and to see how this whole pattern of gene activity changes when conditions change. The analysis can be repeated with mRNA prepared from mutant cells lacking a chosen gene—any gene that we care to test. In principle, this approach provides a way to reveal the entire system of control relationships that govern gene expression—not only in yeast cells, but in any organism whose genome sequence is known.
Arabidopsis has Been chosen out of 300,000 species As a model Plant
The large multicellular organisms that we see around us—the flowers and trees and animals—seem fantastically varied, but they are much closer to one another in their evolutionary origins, and more similar in their basic cell biology, than the great host of microscopic single-celled organisms. Thus, while bacteria and archaea are separated by perhaps 3.5 billion years of evolution, vertebrates and insects are separated by about 700 million years, fish and mammals by about 450 million years, and the different species of flowering plants by only about 150 million years.
Because of the close evolutionary relationship between all flowering plants, we can, once again, get insight into the cell and molecular biology of this whole class of organisms by focusing on just one or a few species for detailed analysis. Out of the several hundred thousand species of flowering plants on Earth today, molecular biologists have chosen to concentrate their efforts on a small weed, the common Thale cress Arabidopsis thaliana (Figure 1–38), which can be grown indoors in large numbers and produces thousands of offspring per plant after 8–10 weeks. Arabidopsis has a total genome size of approximately 220 million nucleotide pairs, about 17 times the size of yeast’s (see Table 1–2).
The World of Animal cells Is Represented By a Worm, a Fly, a Fish, a mouse, and a human
Multicellular animals account for the majority of all named species of living organisms, and for the largest part of the biological research effort. Five species have emerged as the foremost model organisms for molecular genetic studies. In order of increasing size, they are the nematode worm Caenorhabditis elegans, the fly Drosophila melanogaster, the zebrafish Danio rerio, the mouse Mus musculus, and the human, Homo sapiens. Each has had its genome sequenced.
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Caenorhabditis elegans (Figure 1–39) is a small, harmless relative of the eel-worm that attacks crops. With a life cycle of only a few days, an ability to survive in a freezer indefinitely in a state of suspended animation, a simple body plan, and an unusual life cycle that is well suited for genetic studies (described in Chapter 21), it is an ideal model organism. C. elegans develops with clockwork precision from a fertilized egg cell into an adult worm with exactly 959 body cells (plus a variable number of egg and sperm cells)—an unusual degree of regularity for an animal. We now have a minutely detailed description of the sequence of events by which this occurs, as the cells divide, move, and change their character according to strict and predictable rules. The genome of 130 million nucleotide pairs codes for about 21,000 proteins, and many mutants and other tools are available for the testing of gene functions. Although the worm has a body plan very different from our own, the conservation of biological mechanisms has been sufficient for the worm to be a model for many of the developmental and cell-biological processes that occur in the human body. Thus, for example, studies of the worm have been critical for helping us to understand the programs of cell division and cell death that determine the number of cells in the body—a topic of great importance for both developmental biology and cancer research.
studies in Drosophila Provide a key to vertebrate development
The fruit fly Drosophila melanogaster (Figure 1–40) has been used as a model genetic organism for longer than any other; in fact, the foundations of classical genetics were built to a large extent on studies of this insect. Over 80 years ago, it provided, for example, definitive proof that genes—the abstract units of hereditary information—are carried on chromosomes, concrete physical objects whose behavior had been closely followed in the eukaryotic cell with the light microscope, but whose function was at first unknown. The proof depended on one of the many features that make Drosophila peculiarly convenient for genetics—the giant chromosomes, with characteristic banded appearance, that are visible in
Figure 1–38 Arabidopsis thaliana, the plant chosen as the primary model for studying plant molecular genetics.
(courtesy of Toni hayden and the John Innes Foundation.) 0.2 mm
Figure 1–39 Caenorhabditis elegans, the first multicellular organism to have its complete genome sequence determined. This small nematode, about 1 mm long, lives in the soil. most individuals are hermaphrodites, producing both eggs and sperm. (courtesy of maria Gallegos, university of Wisconsin, madison.) some of its cells (Figure 1–41). Specific changes in the hereditary information, manifest in families of mutant flies, were found to correlate exactly with the loss or alteration of specific giant-chromosome bands.
In more recent times, Drosophila, more than any other organism, has shown us how to trace the chain of cause and effect from the genetic instructions encoded in the chromosomal DNA to the structure of the adult multicellular body. Drosophila mutants with body parts strangely misplaced or mispatterned provided the key to the identification and characterization of the genes required to make a properly structured body, with gut, limbs, eyes, and all the other parts in their correct places. Once these Drosophila genes were sequenced, the genomes of vertebrates could be scanned for homologs. These were found, and their functions in vertebrates were then tested by analyzing mice in which the genes had been mutated. The results, as we see later in the book, reveal an astonishing degree of similarity in the molecular mechanisms that govern insect and vertebrate development (discussed in Chapter 21).
The majority of all named species of living organisms are insects. Even if Drosophila had nothing in common with vertebrates, but only with insects, it would still be an important model organism. But if understanding the molecular genetics of vertebrates is the goal, why not simply tackle the problem head-on? Why sidle up to it obliquely, through studies in Drosophila?
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Drosophila requires only 9 days to progress from a fertilized egg to an adult; it is vastly easier and cheaper to breed than any vertebrate, and its genome is much smaller—about 200 million nucleotide pairs, compared with 3200 million for a human. This genome codes for about 15,000 proteins, and mutants can now be obtained for essentially any gene. But there is also another, deeper reason why genetic mechanisms that are hard to discover in a vertebrate are often readily revealed in the fly. This relates, as we now explain, to the frequency of gene duplication, which is substantially greater in vertebrate genomes than in the fly genome and has probably been crucial in making vertebrates the complex and subtle creatures that they are.
The vertebrate Genome Is a Product of Repeated duplications
Almost every gene in the vertebrate genome has paralogs—other genes in the same genome that are unmistakably related and must have arisen by gene duplication. In many cases, a whole cluster of genes is closely related to similar clusters present elsewhere in the genome, suggesting that genes have been duplicated in linked groups rather than as isolated individuals. According to one hypothesis, at an early stage in the evolution of the vertebrates, the entire genome underwent duplication twice in succession, giving rise to four copies of every gene.
The precise course of vertebrate genome evolution remains uncertain, because many further evolutionary changes have occurred since these ancient events.
Figure 1–40 Drosophila melanogaster.
molecular genetic studies on this fly have provided the main key to understanding how all animals develop from a fertilized egg into an adult. (From e.B. lewis, Science 221:cover, 1983. With permission from AAAs.)
Figure 1–41 Giant chromosomes from salivary gland cells of Drosophila.
Because many rounds of dnA replication have occurred without an intervening cell division, each of the chromosomes in these unusual cells contains over 1000 identical dnA molecules, all aligned in register. This makes them easy to see in the light microscope, where they display a characteristic and reproducible banding pattern. specific bands can be identified as the locations of specific genes: a mutant fly with a region of the banding pattern missing shows a phenotype reflecting loss of the genes in that region. Genes that are being transcribed at a high rate correspond to bands with a “puffed” appearance. The bands stained dark brown in the micrograph are sites where a particular regulatory protein is bound to the dnA. (courtesy of B. Zink and R. Paro, from
R. Paro, Trends Genet. 6:416–421, 1990. With permission from elsevier.)
Genes that were once identical have diverged; many of the gene copies have been lost through disruptive mutations; some have undergone further rounds of local duplication; and the genome, in each branch of the vertebrate family tree, has suffered repeated rearrangements, breaking up most of the original gene orderings. Comparison of the gene order in two related organisms, such as the human and the mouse, reveals that—on the time scale of vertebrate evolution—chromosomes frequently fuse and fragment to move large blocks of DNA sequence around. Indeed, it is possible, as discussed in Chapter 4, that the present state of affairs is the result of many separate duplications of fragments of the genome, rather than duplications of the genome as a whole.
There is, however, no doubt that such whole-genome duplications do occur from time to time in evolution, for we can see recent instances in which duplicated chromosome sets are still clearly identifiable as such. The frog genus Xenopus, for example, comprises a set of closely similar species related to one another by repeated duplications or triplications of the whole genome. Among these frogs are X. tropicalis, with an ordinary diploid genome; the common laboratory species X. laevis, with a duplicated genome and twice as much DNA per cell; and
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X. ruwenzoriensis, with a sixfold reduplication of the original genome and six times as much DNA per cell (108 chromosomes, compared with 36 in X. laevis, for example). These species are estimated to have diverged from one another within the past 120 million years (Figure 1–42).
The Frog and the Zebrafish Provide Accessible models for vertebrate development
Frogs have long been used to study the early steps of embryonic development in vertebrates, because their eggs are big, easy to manipulate, and fertilized outside of the animal, so that the subsequent development of the early embryo is easily followed (Figure 1–43). Xenopus laevis, in particular, continues to be an important model organism, even though it is poorly suited for genetic analysis (Movie 1.6 and see Movie 21.1).
The zebrafish Danio rerio has similar advantages, but without this drawback. Its genome is compact—only half as big as that of a mouse or a human—and it has a generation time of only about three months. Many mutants are known, and genetic engineering is relatively easy. The zebrafish has the added virtue that it is transparent for the first two weeks of its life, so that one can watch the behavior of individual cells in the living organism (see Movie 21.2). All this has made it an increasingly important model vertebrate (Figure 1–44).
The mouse Is the Predominant mammalian model organism
Mammals have typically two times as many genes as Drosophila, a genome that is 16 times larger, and millions or billions of times as many cells in their adult bodies. In terms of genome size and function, cell biology, and molecular mechanisms, mammals are nevertheless a highly uniform group of organisms. Even anatomically, the differences among mammals are chiefly a matter of size and proportions; it is hard to think of a human body part that does not have a counterpart in elephants and mice, and vice versa. Evolution plays freely with quantitative features, but it does not readily change the logic of the structure.
Figure 1–42 Two species of the frog genus Xenopus. X. tropicalis, above, has an ordinary diploid genome; X. laevis, below, has twice as much dnA per cell. From the banding patterns of their chromosomes and the arrangement of genes along them, as well as from comparisons of gene sequences, it is clear that the large-genome species have evolved through duplications of the whole genome. These duplications are thought to have occurred in the aftermath of matings between frogs of slightly divergent Xenopus species. (courtesy of e. Amaya, m. offield, and R. Grainger, Trends Genet. 14:253– 255, 1998. With permission from elsevier.)
Figure 1–43 Stages in the normal development of a frog. These drawings show the development of a Rana pipiens tadpole from a fertilized egg. The entire process takes place outside of the mother, making the mechanisms tail bud involved readily accessible for experimental studies. (From W. shumway, Anat. Rec. 78:139–147, 1940.)
For a more exact measure of how closely mammalian species resemble one another genetically, we can compare the nucleotide sequences of corresponding (orthologous) genes, or the amino acid sequences of the proteins that these genes encode. The results for individual genes and proteins vary widely. But typically, if we line up the amino acid sequence of a human protein with that of the orthologous protein from, say, an elephant, about 85% of the amino acids are identical. A similar comparison between human and bird shows an amino acid identity of about 70%—twice as many differences, because the bird and the mammalian lineages have had twice as long to diverge as those of the elephant and the human (Figure 1–45).
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The mouse, being small, hardy, and a rapid breeder, has become the foremost model organism for experimental studies of vertebrate molecular genetics. Many naturally occurring mutations are known, often mimicking the effects of corresponding mutations in humans (Figure 1–46). Methods have been developed, moreover, to test the function of any chosen mouse gene, or of any noncoding portion of the mouse genome, by artificially creating mutations in it, as we explain later in the book.
Just one made-to-order mutant mouse can provide a wealth of information for the cell biologist. It reveals the effects of the chosen mutation in a host of different contexts, simultaneously testing the action of the gene in all the different kinds of cells in the body that could in principle be affected.
As humans, we have a special interest in the human genome. We want to know the full set of parts from which we are made, and to discover how they work. But even
Figure 1–44 Zebrafish as a model for studies of vertebrate development. These small, hardy tropical fish are convenient for genetic studies. Additionally, they have transparent embryos that develop outside of the mother, so that one can clearly observe cells moving and changing their character in the living organism throughout its development. (A) Adult fish. (B) An embryo 24 hours after fertilization. (A, with permission from steve Baskauf; B, from m. Rhinn et al., Neural Dev. 4:12, 2009.)
Figure 1–45 Times of divergence of different vertebrates. The scale on the left shows the estimated date and geological era of the last common ancestor of each specified pair of animals. each time estimate is based on comparisons of the amino acid sequences of orthologous proteins; the longer the animals of a pair have had to evolve independently, the smaller the percentage of amino acids that remain identical. The time scale has been calibrated to match the fossil evidence showing that the last common ancestor of mammals and birds lived 310 million years ago.
The figures on the right give data on protein—the α chain of hemoglobin. note that although there is a clear general trend of increasing divergence with increasing time for this protein, there are irregularities that are thought to reflect the action of changes of hemoglobin sequence when the organisms experienced special physiological demands. some proteins, subject to stricter functional constraints, evolve much more slowly than hemoglobin, time in millions of years others as much as five times faster. All this gives rise to substantial uncertainties in estimates of divergence times, and some experts believe that the major groups of mammals diverged from one another as if you were a mouse, preoccupied with the molecular biology of mice, humans would be attractive as model genetic organisms, because of one special property: through medical examinations and self-reporting, we catalog our own genetic (and other) disorders. The human population is enormous, consisting today of some 7 billion individuals, and this self-documenting property means that a huge database of information exists on human mutations. The human genome sequence of more than 3 billion nucleotide pairs has been determined for thousands of different people, making it easier than ever before to identify at a molecular level the precise genetic change responsible for any given human mutant phenotype.
By drawing together the insights from humans, mice, fish, flies, worms, yeasts, plants, and bacteria—using gene sequence similarities to map out the correspondences between one model organism and another—we are enriching our understanding of them all.
much as 60 million years more recently than shown here. (Adapted from s. kumar and s.B. hedges, Nature 392:917–920, 1998. With permission from macmillan Publishers ltd.)
Figure 1–46 Human and mouse: similar genes and similar development. The human baby and the mouse shown here have similar white patches on their foreheads because both have mutations in the same gene (called Kit), required for the development and maintenance of pigment cells. (courtesy of R.A. Fleischman.)
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What precisely do we mean when we speak of the human genome? Whose genome? On average, any two people taken at random differ in about one or two in every 1000 nucleotide pairs in their DNA sequence. The genome of the human species is, properly speaking, a very complex thing, embracing the entire pool of variant genes found in the human population. Knowledge of this variation is helping us to understand, for example, why some people are prone to one disease, others to another; why some respond well to a drug, others badly. It is also providing clues to our history—the population movements and minglings of our ancestors, the infections they suffered, the diets they ate. All these things have left traces in the variant forms of genes that survive today in the human communities that populate the globe.
To understand cells and organisms Will Require mathematics, computers, and Quantitative Information
Empowered by knowledge of complete genome sequences, we can list the genes, proteins, and RNA molecules in a cell, and we have methods that allow us to begin to depict the complex web of interactions between them. But how are we to turn all this information into an understanding of how cells work? Even for a single cell type belonging to a single species of organism, the current deluge of data seems overwhelming. The sort of informal reasoning on which biologists usually rely seems totally inadequate in the face of such complexity.
In fact, the difficulty is more than just a matter of information overload. Biological systems are, for example, full of feedback loops, and the behavior of even the simplest of systems with feedback is remarkably difficult to predict by intuition alone (Figure 1–47); small changes in parameters can cause radical changes in outcome. To go from a circuit diagram to a prediction of the behavior of the system, we need detailed quantitative information, and to draw deductions from that information we need mathematics and computers.
Such tools for quantitative reasoning are essential, but they are not all-powerful. You might think that, knowing how each protein influences each other protein, and how the expression of each gene is regulated by the products of others, we should soon be able to calculate how the cell as a whole will behave, just as an astronomer can calculate the orbits of the planets, or a chemical engineer can calculate the flows through a chemical plant. But any attempt to perform this feat for anything close to an entire living cell rapidly reveals the limits of our present knowledge. The information we have, plentiful as it is, is full of gaps and uncertainties. Moreover, it is largely qualitative rather than quantitative. Most often, cell biologists studying the cell’s control systems sum up their knowledge in simple schematic diagrams—this book is full of them—rather than in numbers, graphs, and differential equations.
To progress from qualitative descriptions and intuitive reasoning to quantitative descriptions and mathematical deduction is one of the biggest challenges for contemporary cell biology. So far, the challenge has been met only for a few very simple fragments of the machinery of living cells—subsystems involving a handful of different proteins, or two or three cross-regulatory genes, where theory and experiment go closely hand in hand. We discuss some of these examples later in the book and devote the entire final section of Chapter 8 to the role of quantitation in cell biology.
Knowledge and understanding bring the power to intervene—with humans, to avoid or prevent disease; with plants, to create better crops; with bacteria, to turn them to our own uses. All these biological enterprises are linked, because the genetic information of all living organisms is written in the same language. The new-found ability of molecular biologists to read and decipher this language has already begun to transform our relationship to the living world. The account of cell biology in the subsequent chapters will, we hope, equip the reader to understand, and possibly to contribute to, the great scientific adventure of the twenty-first century.
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