Principles of Biochemistry, Volume 3 - A. Lehninger 1985

Answers

Chapter 24

1. 8.0 kcal; 0.60 mol

2. Pancreatic secretion contains a high concentration of HCO₃; consequently, mixing with gastric juice establishes the optimal pH for enzymatic activity.

3. Because ß-casein has a poorly defined tertiary Structure and its native conformation resembles a random coil, all PARTS OF THE protein chain are readily accessible to Pepsin and pancreatic Enzymes. In contrast, a-Keratins possess a highly organized secondary and tertiary structure with numerous —S—S— bridges; therefore, the Denaturation of these Proteins at the low pH of The Stomach is limited, the accessibility of peptide bonds to digestive enzymes is hindered, and Hydrolysis proceeds slowly.

4. a) In the acidic environment of the stomach, pepsinogen activation occurs spontaneously. The resulting pepsin catalyzes the activation of the remaining pepsinogen molecules. b) Pepsinogen is synthesized and stored in the chief Cells of the gastric mucosa (a mildly alkaline environment) as an inactive precursor. The fragment cleaved from pepsinogen binds tightly to the Active Site of pepsin, acting as an inhibitor.

5. Proteolytic Enzymes secreted by the Pancreas do not destroy the epithelial cells of the Small Intestine because: a) the concentration of ingested dietary protein is higher than the protein concentration On the surface of the epithelial cells; b) pancreatic enzymes undergo autodigestion upon completion of Digestion; c) the secretion of pancreatic enzymes is regulated by Hormones, with secretion occurring only when high-protein food enters the stomach; d) the epithelial cells of the small intestine secrete a layer of mucus that protects The Cell surface.

6. During digestion, proteins are broken down into smaller Polypeptides by pancreatic Endopeptidases. Consequently, in the final stages of digestion, the efficiency of carboxypeptidase increases due to the greater number of available COOH terminals.

7. The substance responsible for the symptoms is lactose. The ability to secrete lactase—the enzyme that hydrolyzes lactose—declines with age. Unabsorbed lactose passes through the small intestine into the Large Intestine, where it is fermented by intestinal Bacteria.

8. If Lipids containing unhydrolyzed triacylglycerols are detected in the feces, a probable cause of steatorrhea is insufficient pancreatic lipase secretion. If hydrolyzed triacylglycerols are present in the fecal lipids, a probable cause of steatorrhea may be inadequate Bile secretion, as bile salts are essential for nutrient absorption.

9. Elevated concentrations of Alanine and glutamine highlight The Importance of the glucose-alanine cycle and The transport of ammonia mediated by glutamine.

10. Nearly two-thirds of the ATP synthesized in the Kidneys is utilized for ion transport.

11. Antimycin A and iodoacetate inhibit Electron Transport and Glycolysis, respectively. The energy source for contraction is the pool of creatine phosphate.

12. A constant ATP level is maintained through phosphate transfer from creatine phosphate. 1-Fluoro-2,4-dinitrobenzene inhibits creatine kinase.

13. Ammonia is highly toxic to Nervous Tissue, especially the Brain. Excess ammonia is removed by converting glutamate into glutamine, which is transported to The Liver and subsequently converted into urea. Additional glutamine is formed via The conversion of glucose to glutamate through a-ketoglutarate, with ammonia being removed both during this process and upon the conversion of Glu to Gln.

14.

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15. The amount of circulating albumin depends on total plasma volume. Albumin loss in patients with renal pathology leads to an osmotic pressure gradient between Blood Plasma and the extracellular fluid, causing Water to shift from the cells into the extracellular space.

16. Pancreatic secretion depends on the composition and amount of food, particularly its protein content. Reductions in fluid, calorie, and electrolyte intake are compensated for by the intravenous administration of physiological saline with glucose. Irritation of pancreatic tissue suppresses Insulin secretion, leading to hyperglycemia.

17. Strenuous physical exertion requires increased ATP production, which is met by an elevated rate of oxygen consumption. During a sprint, Muscles convert a portion of Glycogen into lactate. Following the sprint, lactate is transported to the liver, where it is reconverted into glucose and glycogen. This process requires ATP and, consequently, an additional supply of oxygen compared to the resting state.

18. To maintain urinary neutrality, The excretion of phosphate (PO^3-₄) must be accompanied by the excretion of positive counterions (Na⁺ or H⁺). To minimize Na⁺ loss during phosphate excretion, sodium is exchanged for a proton in the distal tubules. As a result, the proton concentration in the urine increases.

19. a) During Drug poisoning, shallow and irregular breathing leads to the accumulation of CO₂ in the Lungs. An increased CO₂ concentration shifts the reaction below to the right, lowering blood plasma pH:

СО₂ + Н_2О ⇄ Н₂СО₃⇄ Н⁺ + НСО^-₃

b) Mechanical ventilation reduces CO₂ concentration and shifts the equilibrium to the left. Consequently, blood plasma pH in the lungs increases, which enhances Hemoglobin's affinity for oxygen. c) Bicarbonate raises pH and increases the buffering capacity of blood plasma.

20. Glucose serves as the primary fuel for the brain. A key reaction in Glucose Catabolism is the thiamine pyrophosphate-dependent Oxidative Decarboxylation of Pyruvate to yield acetyl-CoA. Therefore, thiamine deficiency impairs glucose utilization by the brain.

21. During reabsorption in the renal tubules, active Na⁺ transport is mediated by Mg²⁺-dependent Na⁺, K⁺-ATPase. In this coupled process, the uptake of three Na⁺ ions occurs simultaneously with the extrusion of two K⁺ ions.

22. If a large amount of seawater is ingested, the overall electrolyte concentration in the extracellular fluid becomes significantly higher than that inside the cells. Such a concentration gradient triggers an efflux of water from the cells into the extracellular space. As cellular dehydration progresses, delicate intracellular Organelles (such as Mitochondria) shrink and ultimately suffer irreversible damage.

Chapter 25

1. 1.87∙10² m² (approximately two football fields).

2. Rapid inactivation serves as a mechanism to quickly reduce hormone concentration. The constancy of insulin concentration is maintained by balancing its rates of Synthesis and degradation. Additionally, hormone levels in the blood can be regulated by altering the release rate of stored hormones, as well as the transport and conversion rates of prohormone into the active hormone.

3. Due to their low lipid solubility, water-soluble hormones cannot cross The cell membrane. Instead, they bind to a receptor on the cell surface. In the case of adrenaline, this receptor is an enzyme that catalyzes the intracellular formation of a second messenger, cyclic AMP (cAMP). Conversely, fat-soluble hormones readily penetrate the hydrophobic interior of the cell membrane. Once inside the cell, they can interact directly with intracellular receptors.

4. a) Since adenylate cyclase is a membrane-associated protein, centrifugation of preparations sediments adenylate cyclase activity in the particulate fraction.

б) Adrenaline stimulates The formation of cAMP, a soluble substance that activates Glycogen phosphorylase.

в) cAMP is a heat-stable compound. It can be obtained by treating ATP with barium hydroxide.

5. Unlike cAMP, dibutyryl-cAMP crosses the cell membrane.

6. Cholera toxin elevates cAMP levels in intestinal epithelial cells. These facts indicate that cAMP regulates membrane permeability to Na⁺ ions. The Treatment of cholera consists of replenishing fluid and electrolyte losses in the body.

7. a) Heart and Skeletal Muscle lack the enzyme glucose-6-phosphatase. Consequently, every molecule of glucose-6-phosphate formed is channeled into The Glycolytic Pathway and, under oxygen-deficient conditions, is converted to lactate. a) The electrical charge characteristic of all phosphorylated intermediates prevents them from leaving the cell, as the membrane is impermeable to charged molecules. In a stress situation, the concentration of glycolytic precursors in Muscle tissue must be high to sustain impending muscular work. The accumulation of precursors in a phosphorylated state prevents their leakage from the muscle tissue. The liver, conversely, supplies glucose, maintaining its required level in the blood. Therefore, in a stress situation, glucose must rapidly pass from liver cells into the blood, a process ensured by the dephosphorylation reaction catalyzed by glucose-6-phosphatase.

8. Excessive insulin secretion by the pancreas promotes enhanced utilization of blood glucose by the liver; this leads to hypoglycemia. Furthermore, high insulin levels result in the slowdown of Amino Acid and fatty acid catabolism. Thus, patients' blood contains insufficient METABOLISM/26.html">Energy Metabolism substrates required for ATP formation. If the hyperinsulinemic state persists for a long time, brain cell damage occurs, since glucose serves as the primary energy source for the brain.

9. The fact that incubation of liver tissue with thyroxine increases The rate of Respiration and heat production without altering ATP concentration is consistent with the assertion that thyroxine is an uncoupler of Oxidative Phosphorylation. Uncoupling agents lower the P/O ratio in tissues, which forces them to increase respiration intensity to meet their ATP demand. The observed heat release could also be caused by an increased rate of ATP utilization by the thyroxine-stimulated tissue. In such tissue, the elevated ATP demand is met by increasing the level of oxidative phosphorylation (respiration), which is accompanied by heat generation. Despite numerous studies, the details of thyroid hormone regulation of aerobic metabolism rates remain elusive.

10. Female Sex Hormones (estrogens) promote The Development of secondary female sex characteristics. One such characteristic is the GROWTH AND DEVELOPMENT of the mammary gland. Since carcinoma affects breast tissue, any intervention that inhibits the growth of this tissue slows down the Development of the carcinoma. Inhibition of mammary tissue growth occurs when estrogen concentration is decreased, which is achieved by ovariectomy. Sometimes, treatment involves an antagonist hormone—that is, a hormone with an opposite effect to estrogens, such as testosterone.

11. The Adrenal Glands, located directly above the kidneys, are essentially an extension of The Nervous system from which they receive signals. Consequently, it is not surprising that endorphins are synthesized by both the brain and The adrenal medulla.

12. Certain polypeptide hormones, namely insulin and Glucagon, are synthesized as inactive precursors whose polypeptide chains are longer than those of the active hormones themselves. The formation of a prohormone offers the advantage that, being inactive, the prohormone can be stored in large quantities within secretory granules and rapidly activated in response to an appropriate signal via proteolytic Cleavage.

13. Adrenaline is used to treat patients suffering from severe asthma because this hormone relieves the spasm of the smooth musculature surrounding the pulmonary bronchioles by stimulating cAMP formation in target cells. However, cAMP is degraded upon hydrolysis by phosphodiesterase. Since purine derivatives—caffeine, theophylline, and aminophylline—inhibit phosphodiesterase, the administration of these medications prolongs the action of adrenaline and enhances its activity by reducing the rate of cAMP cleavage.

14. These observations indicate that phosphodiesterase activity is stimulated by Ca²⁺ ions, an effect mediated by calmodulin. This agrees with the well-established fact that calmodulin is a Ca²⁺-binding protein. The binding of the calmodulin-Ca²⁺ complex to phosphodiesterase stimulates its hydrolytic activity toward cAMP.

Chapter 26

1. For safety assurance.

2. 0.21 mol ATP/g glucose, 0.50 mol ATP/g palmitic acid; 4.2 kcal/g glucose, 9.5 kcal/g palmitic acid; comparison of the two results

shows that The ratio of the amount of ATP produced to the heat released is identical.

3. During the first few hours after the onset of starvation, blood glucose levels begin to drop as glucose is taken up by tissues. To maintain blood glucose concentration at the required level, The Human Body mobilizes the catabolism of glucogenic Amino Acids. This process is coupled with the excretion of nitrogen in the form of urea and is accompanied by the simultaneous elimination of a large amount of water from the body. Over the next few days of fasting, the Organism shifts from Amino Acid Catabolism to fatty acid catabolism to meet its energy demands, thereby substantially reducing daily water loss.

4. Obesity is the result of consuming an excessive number of calories. Given equal amounts, a fat-rich diet is more likely to lead to obesity than a high-carbohydrate diet, as the former has a higher caloric density per gram. However, as a rule, obesity is caused by consuming too many total calories. Since eating sweets brings pleasure, foods with a very high sugar content pose the primary hazard for overweight individuals.

5. a) 3.16 kcal/g; b) 4.21 kcal/g. c) Based on the available data, it can be concluded that wheat flakes are a mixture of proteins and CARBOHYDRATES. Elemental analysis indicates a low nitrogen content in the sample, suggesting that carbohydrates are the primary component of the flakes. d) Oxidation in the organism yields the same amount of energy as oxidation in a calorimeter only if the food is completely digested and assimilated.

6. Her diet should not exceed the following: 3.2 baked potatoes; 19.2 medium-sized fried potato slices; 26.3 medium-sized potato chips; 4.8 slices of white bread; 6 pats of butter; 1.6 cans of beer.

7. If the diet lacks even a single essential amino acid, Protein Synthesis will continue only until the supply of that amino acid is depleted. To replenish this supply, either additional intake of an unbalanced amino acid mixture or The breakdown of body proteins is required. This inevitably leads to an excess of Other Amino Acids that must undergo catabolism, resulting in Nitrogen Excretion and, consequently, a negative nitrogen balance.

8. A rat is placed on a carefully controlled diet containing all amino acids except Phe. To maintain a constant total nitrogen intake throughout the series of experiments, a corresponding amount of urea is added to the food. Because the rat lacks one of the amino acids, a negative nitrogen balance will be observed—i.e., the amount of dietary nitrogen ingested will be less than the amount excreted (see Question 7). Phe is then added to the diet, and the nitrogen balance is measured once more. The minimum daily requirement equals the amount of Phe needed to establish a normal (or slightly positive) balance.

9. Young children suffering from kwashiorkor typically present with a distended abdomen, caused by fluid retention in the extracellular space due to insufficient serum albumin levels in blood plasma. Upon switching to a protein-adequate diet, serum albumin returns to normal, the osmotic pressure gradient reverses, and excess fluid is cleared from the body. Consequently, the initial weight loss during treatment is attributable to water loss.

10. Patients with renal failure require regular dialysis to remove toxic waste products—primarily urea and uric acid—from the blood. To minimize the patient's dependence on this Procedure, nitrogen excretion must be kept to a minimum. This is achieved through a diet balanced in both total amino acid content and their relative proportions; in other words, the food must consist of proteins with a biological value close to 100. Because eggs contain all Essential Amino Acids and have a higher biological value than grain, they are nutritionally better balanced than grain and serve as a superior food source for such patients.

11. Vitamin B₆ (pyridoxine) is utilized in the Synthesis of the coenzyme Pyridoxal phosphate. This coenzyme is essential for transaminases, the enzymes that catalyze the initial step of amino acid catabolism.

12. Gastrointestinal bacteria supply the host organism with a significant portion of its required Vitamins (which the bacteria synthesize themselves). Any diet that reduces the population of intestinal bacteria can therefore lead to vitamin deficiencies.

13. Humans require A balanced diet to meet their biochemical energy demands and to provide the body with building blocks and Cofactors for biosynthetic processes. These requirements can be met using foods from A wide variety of sources; there is no evidence to suggest that any specific food (such as milk) is uniquely necessary for proper Nutrition.

14. a) The diet must contain sufficient fuel reserves to sustain heavy labor and maintain body Temperature, b) Foods that facilitate glycogen storage—namely, easily digestible carbohydrates (honey, dried fruit, chocolate, pancakes, etc.). c) Proteins and fats, d) Potentially, Na⁺ and K⁺ ions.

15. Mammals, including humans, lack the ability to convert two-carbon compounds (acetyl-CoA) into the three-carbon compounds (pyruvate) required for Gluconeogenesis. Therefore, although ethanol is converted into acetyl-CoA, the latter cannot be converted into glucose.

16. 7.2 kg.

17. a) Protein Hydrolysis in the small intestine yields a mixture of amino acids. These amino acids fall into two categories: glucogenic (whose carbon skeletons yield pyruvate during catabolism) and ketogenic (whose carbon skeletons yield acetyl-CoA during catabolism). Because pyruvate is converted to acetyl-CoA, utilizing the carbon skeletons of all Amino acids can lead to the formation of Fatty acids and, consequently, the deposition of triacylglycerols. b) Amino acid catabolism is accompanied by the excretion of nitrogen in the form of urea, as well as the consumption and elimination of large amounts of water.

Chapter 27

1. One DNA molecule contains 32% A, 32% T, 18% G, and 18% C; the other contains 17% A, 17% T, 33% G, and 33% C. This indicates that both DNA molecules are double-stranded. The DNA containing 33% G and 33% C likely belongs to a thermophilic bacterium, as such DNA is more resistant to heat denaturation. G–C Base Pairs feature three Hydrogen Bonds and are therefore more stable than A–T base pairs, which have only two.

2. (5')GAATGCATACGGCAT(3').

3. 0.94 mg.

4. 372 base pairs; the actual length is likely much greater, given that nearly all eukaryotic genes contain introns, which can be longer than exons. Furthermore, most eukaryotic genes also encode a leader or signal sequence for their protein product.

5. 200,000 base pairs; the total length of T2 phage DNA is 68,000 nm, whereas the diameter of the phage HEAD is only 100 nm.

6. The ratio of DNA length to diameter is 6.8 for a nucleosome and 10⁶ for a core.

Within The Nucleus, DNA is packed in a much more compact manner.

7. This is highly unlikely, as such a palindrome lacks the maximum possible number of base pairs. In the chromosome of an intact cell, however, there may be DNA regions where the negative charges of the phosphate backbone are neutralized by Histones, increasing DNA flexibility and allowing it to adopt a cruciform structure.

8. M13 phage DNA is single-stranded, because the amount of A does not match T, nor is the amount of G equivalent to C.

9. a) A single-stranded linear DNA molecule whose size equals the length of the nicked RFII strand, and a single-stranded circular DNA molecule representing the intact RFII strand. b) Three single-stranded linear DNA molecules will be detected: one representing the unbroken RFII strand consisting of 5,386 bases, and two smaller ones derived from the nicked strand.

10. Without labeling the 5' end of the DNA, there would be no reference point required to reconstruct the original base sequence.

11. No; the two strands of a double-stranded DNA molecule differ in nucleotide composition (see Fig. 27-12).

12. The exons of this Gene consist of 3 x 12 = 576 base pairs. The remaining 864 nucleotide pairs are part of the introns and possibly the leader, or signal, sequence.

13. Equilibrium centrifugation of these two DNAs in a cesium chloride density gradient will show that the E. coli DNA, which has a higher G-C content, is denser and will therefore reach equilibrium at a lower point in the gradient than the sea urchin DNA (see Table 27-3).

14. The "eyes" were formed as a result of the unwinding of double-stranded DNA regions rich in A-T pairs. These regions are less thermally stable than G-C pairs. This procedure is useful for determining differences in base composition along a double-stranded DNA molecule.

15. Based on the fact that Homologous proteins from different animal species exhibit Amino Acid Sequence Homology.

b) This is a twofold axis of Symmetry, since a 180° rotation around this point leads to the same nucleotide base sequence. c) Apparently, restriction endonuclease cleavage produced protruding ends that are held together by their inherent "stickiness." Therefore, the DNA remains in a circular form. d) An alkaline environment causes the circular double-stranded DNA to unwind and the sticky ends to separate, resulting in two linear single strands.

17. No; the Central dogma of molecular biology states that Genetic information is transferred from DNA to RNA and subsequently from RNA to protein. In Introduction/7.html">RNA-containing Viruses, genetic information is also transmitted in the direction of RNA → protein, albeit without the participation of DNA.

Chapter 28

1. If Replication occurred via a dispersive mechanism, the density of the DNA after the first doubling would correspond to that observed in the experiment, and such DNA would occupy an intermediate position between the "heavy" and "light" fractions. However, after the second doubling, all DNA molecules would have the same density and form a single band located midway between the band observed after the first doubling and the "light" DNA band.

2. a) Since thymidine residues are present in DNA and absent in RNA, labeled RNA did not interfere with the experiment. b) Guanosine and adenosine are unsuitable because they are present in both DNA and all RNAs. c) Thymidine is phosphorylated by ATP and converted in several steps to thymidine-5'-triphosphate.

3. 400,000 revolutions.

4. a) 44 min. b) One possibility is that the E. coli chromosome is replicated by four replication forks originating from two replication initiation sites.

5. a) 0.42. b) The number of base pairs in the introns and in the leader, or signal, sequence.

6. a) (5') CTAATGCAACGTTGCAAGCT (3') b)

(5') AGCUUGCAACGUUGCAUUAG (3')

7. a) A—21%, U—21%, G—29%, C—29%.

b) It may turn out to be the same as in part (a), but not necessarily.

8. The new DNA synthesized on this template contains 32.7% A, 18.5% G, 24.1% C, and 24.7% T. The new DNA produced on the complementary template contains 24.7% A, 24.1% G, 18.5% C, and 32.7% T. The combined Nucleotide Composition of the two new DNAs is: A—28.7%, G—21.3%, C—21.3%, and T—28.7%. It must be assumed that replication went to completion on both template strands.

9. RNAs are transcribed from only one strand of the double-stranded DNA.

10. a) About 2,000. b) Okazaki fragments are synthesized by DNA polymerase III on the DNA template with the aid of An RNA primer. Because Okazaki fragments in E. coli are about 2,000 bases long, they are tightly bound to the template strand through complementary interactions. Each fragment is rapidly joined to the lagging strand by the sequential action of DNA polymerase I and DNA ligase, thereby ensuring the correct order of the fragments. This is why normal replication does not yield a mixture of various Okazaki fragments dissociated from the template.

11. Leading strand

Precursors: dATP, dGTP, dCTP, dTTP

Enzymes: DNA gyrase, helicase, DNA-binding proteins, DNA polymerase III, inorganic pyrophosphatase. Cofactors: Zn²⁺, Mg²⁺

Lagging strand

Precursors: ATP, GTP, CTP, UTP

dATP, dGTP, dCTP, dTTP

Enzymes: DNA gyrase, helicase, DNA-binding proteins, primase, DNA polymerase III, DNA polymerase I, DNA ligase, pyrophosphatase

Cofactors: Zn²⁺, Mg²⁺, NAD⁺

12. a) First, Watson-Crick base-pairing interactions; second, hydrophobic stabilization of stacked base pairs; third, Enzymatic hydrolysis of the pyrophosphate generated in the DNA polymerase reaction, which drives the reaction virtually to completion at each step; fourth, removal of incorrectly incorporated NUCLEOTIDES via the 3'-exonuclease activity of DNA polymerase III.

b) Since the four factors ensuring replication fidelity operate on both the leading and lagging strands,

it is generally assumed that both strands are synthesized with equal accuracy. However, because a greater number of discrete Chemical Reactions are involved in the assembly of the lagging strand, one might expect that its replication presents more opportunities for errors to occur.

13. a) This property ensures that foreign intact circular DNA cannot replicate (e.g., in E. coli cells) unless its origin of replication is identical to that of E. coli. b) DNA replicase must replicate the E. coli chromosome starting from a region called THE ORIGIN OF replication, which possesses a characteristic base sequence.

14. a) The involvement of NTPs in the RNA polymerase reaction leads to the formation of pyrophosphate, which is subsequently cleaved by pyrophosphatase; this pulls the reaction forward to completion and ensures high fidelity of synthesis. In contrast, the polynucleotide phosphorylase reaction utilizes NDPs, thereby releasing free phosphate. Since the intracellular phosphate concentration is relatively high, it is quite likely that polynucleotide phosphorylase operates in reverse, i.e., degrading RNA. b) RNA polymerase requires all four NTPs and a template, whereas polynucleotide phosphorylase does not require all four NDPs or a template, further supporting the Conclusion that its primary role is RNA degradation.

15. A single-base error during DNA replication, if uncorrected, will result in one of the two daughter cells, as well as all its progeny, carrying the altered chromosome. A single-base error made by RNA polymerase, on the other hand, will lead to the synthesis of a limited number of incorrect copies of a single protein. Moreover, because the cellular mRNA pool turns over rapidly, the majority of the molecules of this protein will be normal, and the progeny of such a cell will also be entirely normal.

Chapter 29

1. a) Gly—Gln—Ser—Leu—Leu—Ile;

b) Leu—Asp—Ala—Pro; c) His—Asp—Ala—Cys—Cys—Tyr; d) Met—Asp—Glu in eukaryotes; fMet—Asp—Glu in prokaryotes.

2. Since Almost all amino acids are specified by multiple codons (e.g., six for leucine), any given polypeptide can be encoded by A large number of different nucleotide sequences.

3. UUAAUGUAU, UUGAUGUAU, CUUAUGUAU, CUCAUGUAU, CUAAU- GUAU, CUGAUGUAU, UUAAUGUAC, UUGAUGUAC, CUUAUGUAC, CUCAU- GUAC, CUAAUGUAC, CUGAUGUAC.

4. a) (5) CGACGGCGCGAAGUCAGGG- GUGUIJAAG (3').

b) Arg—Arg—Arg—Glu—Val—Arg - Gly—Val—Lys.

c) No, because complementary antiparallel strands in double-stranded DNA have different base sequences in the 5'→3' direction. RNA is transcribed from only one specific strand of double-stranded DNA; therefore, RNA polymerase must recognize and bind to the correct strand.

5. There are two tRNAs for Methionine: one is the initiator tRNA^fMet, and the other is tRNA^Met, which delivers internal methionine residues into the polypeptide chain. Catalyzed by methionyl-tRNA synthetase, tRNA^fMet reacts with methionine to form methionyl-tRNA^fMet. The amino group of this methionine is then formylated by N₁₀-formyltetrahydrofolate to yield N-formylmethionyl-tRNA^fMet. Free methionine or methionyl-tRNA^Met cannot be formylated. Only N-formylmethionyl-tRNA^fMet can bind to the initiation codon AUG on mRNA because it recognizes a special initiation signal. This signal consists of a region of six or more A and G residues located upstream of AUG. Methionyl-tRNA^Met is unable to recognize this signal. AUG codons at internal positions of mRNA can bind only methionyl-tRNA.

6. Polynucleotide phosphorylase should be added to a mixture of UDP and CDP in which UDP is present, say, in a fivefold excess over CDP. This will yield an RNA-like polymer rich in UUU triplets (encoding Phe) along with smaller amounts of UUC (Phe), UCU (Ser), and CUU (Leu), as well as (in much lower amounts) UCC (also Ser), CCU (Pro), and CUC (Leu).

7. At least 583 high-energy phosphate groups; the actual number may be higher depending on the number of errors proofread and corrected by Aminoacyl-tRNA synthetases. The correction of each error consumes two high-energy phosphate groups.

8. To synthesize a protein from amino acids, a Eukaryotic Cell must carry out the synthesis of at least 20 activating enzymes, 70 ribosomal proteins, 4 Ribosomal RNAs, no fewer than 20 tRNAs, and at least 10 auxiliary enzymes. In contrast, the synthesis of an α(1→4) glycogen chain from glucose requires only 4–5 enzymes.

9. Glycine codons are recognized by anticodons

a) At the 5' end and in the middle of the anticodon.

b) Wobble base pairs will be formed with their codons by the anticodons (5')GCC, ICC, and UCC. c) In pairs involving the anticodons (5')ACG, GCC, UCC, and CCC. d) The Use of pairs involving the anticodons specified in (c) is least likely because tRNAs containing these anticodons, due to the tight binding of all three anticodon bases, will dissociate from the complex at a lower rate than other Gly tRNAs.

10. It will cause a reading frame shift during Translation, starting from the attachment site of this unusual aminoacyl-tRNA.

11. (a), (c), (e), and (g); substitutions (b), (d), and (f) cannot result from a single-base change; (b) and (f) require a two-base substitution, and (d) requires the substitution of all three bases.

12. In DNA, there are two codons for Glu, (5')TTC and (5')CTC, and four codons for Val, (5')TAC, (5')CAC, (5')AAC, and (5')GAC. The amino acid substitution in sickle cell hemoglobin can be caused by a single-base change: (5')TTC(Glu) → (5')TAC (Val) or (5')CTC(Glu) → (5')CAC (Val). Two-base changes are much less likely: (5')TTC → (5')CAC, (5')AAC, and (5')GAC, or (5')CTC → (5')TAC, (5')AAC, and (5')GAC.

Chapter 30

1. a) The damage must be repaired by DNA polymerase I, which acts only in the 5'→3' direction. b) mRNA is single-stranded, meaning it lacks the template strand necessary for proofreading. c) Double-strand breaks caused by X-rays cannot be repaired because the broken ends are unlikely to rejoin properly.

2. 1) No change, if the mutation results in a synonymous codon for the same amino acid; 2) no change, if the mutation occurs in a functionally insignificant region within an intron; 3) an amino acid substitution that leads to the synthesis of an enhanced, unaltered, less active, or inactive protein; 4) protein truncation resulting from a mutation that converts a sense codon into a termination codon, or protein elongation due to a mutation that converts a termination signal into an amino acid codon.

3. The following products will be obtained (sticky ends generated by Eco RI digestion are indicated):

4. UAA will cause termination of polypeptide synthesis only at positions 334-336 of the prokaryotic mRNA. At positions 330-332 and 338-340, UAA will not lead to termination because it will be out of the correct reading frame. In eukaryotic mRNA, termination will occur at positions 334-336 if there are no intervening sequences preceding UAA. If UAA is located within an intron, it will probably have no effect. If UAA is in an exon (other than the first exon), The Effect of UAA on the protein product will depend on how many nucleotides were removed with the preceding introns.

5. a) This is a single- or double-stranded circle.

b) This is a double-stranded circle, since Eco RI would not cleave single-stranded DNA.

c) A = 1.3×10⁶; B = 0.8×10⁶; C = 0.7×10⁶; D = 0.6×10⁶. d) Fragments C and D are located adjacent to each other. From part (c), we know that fragments B and D are adjacent. e) The restriction site at the boundary between fragments A and B is damaged.

6. Exons likely encode functional Protein domains. A specific exon encoding one domain of a given protein may combine with an exon responsible for the synthesis of a domain from another protein, thereby generating a gene that encodes a novel protein containing the domains of both predecessors.

7. a) 5 mg/L; b) 10⁵ molecules; c) 0.5 g.

8. a) It is necessary to cleave two DNA molecules with a restriction endonuclease, mix and heat them to inactivate the enzyme, and separate the sticky ends at the cleavage sites. Allow the mixture to cool so that non-covalent recombinants form, and then covalently join them using DNA ligase. b) In addition to the desired circular duplex combining the original circular DNAs, side products are formed: the original circular DNAs (small and large) and two other circular DNAs, one resulting from the recombination of two large DNAs and the other from the recombination of two small DNAs. Furthermore, the mixture will contain linear recombinants formed by two large DNAs, two small DNAs, as well as Two Types of recombinants consisting of a small and a large DNA joined in two different combinations.