Biochemistry, 4th Edition P111 ppsx

Abbreviated Answers to Problems A-3 6. a. K eq (AC) ϭ (0.02 ϫ 1000) ϭ 20. b. ⌬G°(AB) ϭ 10.1 kJ/mol. ⌬G°(BC) ϭϪ17.8 kJ/mol. ⌬G°(AC) ϭϪ7.7 kJ/mol. K eq ϭ 20. 7. See The Student Solutions Manual, Study Guide and Problems Book for resonance structures. 8. K eq ϭ [Cr][P i ]/[CrP][H 2 O]. K eq ϭ 3.89 ϫ 10 7 . 9. CrP in the amount of 135.3 moles would be required per day to provide 5860 kJ energy. This corresponds to 17,730 g of CrP per day. With a body content of 20 g CrP, each molecule would recycle 886 times per day. Similarly, 637 moles of glycerol-3-P, or 108,300 g of glycerol-3-P, would be required. Each molecule would recycle 5410 times/day. 10. ⌬G ϭ Ϫ46.1 kJ/mol. 11. The hexokinase reaction is a sum of the reactions for hydrolysis of ATP and phosphorylation of glucose: ATP ϩ H 2 O 34 ADP ϩ P i The free energy change for the hexokinase reaction can thus be obtained by summing the free energy changes for the first two reactions listed here. ⌬G°Ј for hexokinase ϭϪ30.5 kJ/mol ϩ 13.9 kJ/mol ϭ Ϫ16.6 kJ/mol 12. Comparing the acetyl group of acetoacetyl-CoA and the methyl group of acetyl-CoA, it is reasonable to suggest that the acetyl group is more electron-withdrawing in nature. For this reason, it tends to destabilize the thiol ester of acetoacetyl-CoA, and the free energy of hydrolysis of acetoacetyl-CoA should be somewhat larger than that of acetyl-CoA. In fact, ⌬G°ЈϭϪ43.9 kJ/mol, compared with Ϫ31.5 kJ/mol for acetyl-CoA. 13. Carbamoyl phosphate should have a somewhat larger free energy of hydrolysis than acetyl phosphate, at least in part because of greater opportunities for resonance stabilization in the products. In fact, the free energy of hydrolysis of carbamoyl phosphate is Ϫ51.5 kJ/mol, compared with Ϫ43.3 kJ/mol for acetyl phosphate. 14. The denaturation of chymotrypsinogen is spontaneous at 58°C, because the ⌬G° at this temperature is negative (at approximately Ϫ2.8 kJ/mol). The native and denatured forms are in equilibrium at approximately 56.6°C. 15. The positive values for ⌬C p for the protein denaturations described in Table 3.1 reflect an increase in motion of the peptide chain in the denatured state. This increased motion provides new ways to store heat energy. 16. K eq ϭ 2.04 ϫ 10 Ϫ2 at 298K; K eq ϭ 9 at 320K. ⌬H°Јϭ219 kJ/mol ⌬G°Јϭ9.64 kJ/mol ⌬S°Јϭ702 J/mol и K 17. The value of ⌬G° is determined at standard state, which includes 1M [H ϩ ]. Using Equation 3.23, a value for ⌬G°of Ϫ3.36 kJ/mol can be calculated. This value applies at pH 2, 7, and 12 because, of course, ⌬G° is pH-independent. However, Equation 3.23 can also be used to determine that ⌬G°Ј is Ϫ14.77 kJ/mol at pH 2 and Ϫ71.82 kJ/mol at pH 12. Finally, ⌬G°Ј for enolase is not pH-dependent, because the enolase re- Glucose ϩ P i 34 G-6-P ϩ H 2 O Glucose ϩ ATP 34 G-6-P ϩ ADP action (see Figure 3.11) neither consumes nor produces protons. ⌬G° ϭ ⌬G°Јϭ1.8 kJ/mol. 18. The magnitude of ⌬G°Ј for ATP hydrolysis is sufficiently great that it provides ample energy to drive the conversion of A to B, even though the reaction is unfavorable. The result is that the equilibrium ratio of B to A is more than 10 8 greater when the reaction is coupled to ATP hydrolysis. If A→B were coupled instead to 1,3-bisphosphoglycerate hydrolysis, the ratio of B to A would be even greater, since the ⌬G°Ј for hydrolysis of 1,3-bisphosphoglycerate is substantially greater than that of ATP. Using the concentrations for 1,3-BPG and 3-PG in Table 18.2, and repeating the calculations on page 67 yields a ratio of [B eq ]/[A eq ] ϭ 4.14 ϫ 10 9 , an even greater ratio than that calculated on page 67. 19. This exercise is left to the student. Use Figure 3.8 as a guide. 20. Without pyrophosphate cleavage, the acyl-CoA synthetase reaction is only slightly favorable, with a ⌬G°Ј of 0.8 kJ/mol. With pyrophosphate cleavage included, the net ⌬G°Ј for the reaction is Ϫ33.6 kJ/mol, a far more favorable value. Chapter 4 1. Structures for glycine, aspartate, leucine, isoleucine, methio- nine, and threonine are presented in Figure 4.3. 2. Asparagine ϭ Asn ϭ N. Arginine ϭ Arg ϭ R. Cysteine ϭ Cys ϭ C. Lysine ϭ Lys ϭ K. Proline ϭ Pro ϭ P. Tyrosine ϭ Tyr ϭ Y. Tryptophan ϭ Trp ϭ W. 3. Phenylalanine dissociation: Alanine dissociation: OOC A H 2 N CH 2 A OHOC A COO Ϫ H 3 N ϩ A OHOC A COO Ϫ H 2 N CH 3 A OHOC A COOH H 3 N ϩ CH 3 A OHOC A COO Ϫ H 3 N ϩ CH 3 A CH 2 A A COOH CH 2 A A COOH CH 2 A OHOC A COO Ϫ H 3 N ϩ A CH 2 A OHOC A COO Ϫ H 2 N A CH 2 A CH 2 COO Ϫ COO Ϫ Glutamate dissociation: A OHOCH 3 N ϩ A COOH CH 2 A A OHOCH 3 N ϩ A COOH A COO Ϫ A OHOC CH 2 COO Ϫ A A (CH 2 ) 4 H OOCH 2 N COO Ϫ A A H Histidine dissociation: A O N OCH 2 N COO Ϫ A A CH 2 CH HC NH CH A OHOCH 3 N ϩ A COOH H ϩ N A CH 2 CH HC NH CH A OHOCH 3 N ϩ COO Ϫ A H ϩ N A CH 2 CH HC NH CH H A O N OCH 3 N ϩ COO Ϫ A A CH 2 CH HC NH CH H H 3 N ϩ (CH 2 ) 4 NH 3 ϩ A NH 3 ϩ (CH 2 ) 4 A NH 3 ϩ (CH 2 ) 4 Lysine dissociation: OHOCH 3 N ϩ A COOH A OHOC COO Ϫ A A OOCH 2 N COO Ϫ A A HH 3 N ϩ NH 2 CH 2 CH 2 A-4 Abbreviated Answers to Problems 4. The proximity of the ␣-carboxyl group lowers the pK a of the ␣-amino group. 5. pH pK a ϭ 2.1 10 9 1 8 7 6 5 4 3 2 0 0 123 13 12 11 ␣ -COOH ϩ pK a ϭ 9.8 ␣ -NH 3 R group pK a ϭ 3.9 Equivalence point E quivalents 6. Denoting the four histidine species as His 2ϩ , His ϩ , His 0 , and His Ϫ , the concentrations are: pH 2: [His 2ϩ ] ϭ 0.097 M, [His ϩ ] ϭ 0.153 M, [His 0 ] ϭ 1.53 ϫ 10 Ϫ5 M, [His Ϫ ] ϭ 9.6 ϫ 10 Ϫ13 M. pH 6.4: [His 2ϩ ] ϭ 1.78 ϫ 10 Ϫ4 M, [His ϩ ] ϭ 0.071 M, [His 0 ] ϭ 0.179 M, [His Ϫ ] ϭ 2.8 ϫ 10 Ϫ4 M. pH 9.3: [His 2ϩ ] ϭ 1.75 ϫ 10 Ϫ12 M, [His ϩ ] Ϫ 5.5 ϫ 10 Ϫ5 M, [His 0 ] ϭ 0.111 M, [His Ϫ ] ϭ 0.139 M. 7. pH ϭ pK a ϩ log (2/1) ϭ 4.3 ϩ 0.3 ϭ 4.6. The ␥-carboxyl group of glutamic acid is 2/3 dissociated at pH ϭ 4.6. 8. pH ϭ pK a ϩ log (1/4) ϭ 10.5 ϩ (Ϫ0.6) ϭ 9.9. 9. a. The pH of a 0.3 M leucine hydrochloride solution is approximately 1.46. b. The pH of a 0.3 M sodium leucinate solution is approximately 11.5. c. The pH of a 0.3 M solution of isoelectric leucine is approximately 6.05. 10. The sequence of reactions shown would demonstrate that L(Ϫ)-serine is related stereochemically to L(Ϫ)-glyceraldehyde: A O H O S A COO Ϫ CH 2 H 3 N ϩ O D -Cystine OSO NH 3 A CO ϩ H A COO Ϫ O A A COO Ϫ CH 2 OCO CH 2 H 3 N ϩ O SOSO A A COO Ϫ CH 2 O COHHNH 3 ϩ C Mirror plane meso-Cystine O A CH A COO Ϫ O CH 2 H 3 N ϩ OSOSO A A COO Ϫ CH 2 CHNH 3 ϩ Mirror plane L -Cystine O O Straight arrows indicate reactions that occur with retention of configuration. Looped arrows indicate inversion of configuration. (From Kopple, K. D., 1966. Peptides and Amino Acids. New York: Benjamin Co.) CH 3 COOH H L-(Ϫ)-Glyceraldehyde L-Glyceric acid CH 2 OH HO CHO H HO COOH H 2-Bromopropanoic acid CH 3 H COOH Br L-(Ϫ)-Serine L-(Ϫ)-Alanine CH 2 OH H 2 N COOH H CH 3 H 2 N COOH H CH 2 OH HO 2-Hydroxypropanoic acid 11. Cystine (disulfide-linked cysteine) has two chiral carbons, the two ␣-carbons of the cysteine moieties. Each chiral center can exist in two forms, so there are four stereoisomers of cystine. However, it is impossible to distinguish the difference between L-cysteine/D-cysteine and D-cysteine/L-cysteine conjugates. So three distinct isomers are formed: 12. A OO A C CH 2 H 3 N ϩ B P C A COO Ϫ A O O H A I Ϫ CH 2 H 3 N ϩ C A COO Ϫ A A A NH 2 O CH 2 I S CH 2 C O NH 2 O OO S Ϫ H 13. There are eight Tyr residues in the protein. 14. A water is removed from each amino acid when it is incorpo- rated into a protein, so the “molecular weight” of a residue is lower by 18 units. Also, most proteins have relatively more small side chains (Gly, Ala) and fewer Trp side chains than a statistical average would predict. 15. Aspartame is composed of aspartic acid and phenylalanine, with a carboxymethyl cap. These amino acids are linked by a peptide Abbreviated Answers to Problems A-5 (amide) bond in aspartame. Heating can cleave amide linkages. For this reason, drinks such as coffee and hot chocolate must be consumed relatively quickly after preparation. Aspartame kept hot for several hours is quite bitter tasting (based on the experience of one of the authors). 16. Phenylketonuria is an autosomal recessive genetic disease caused by a deficiency or absence of the enzyme phenylalanine hydroxylase (PAH), an essential enzyme that converts phenylalanine to tyrosine. Without sufficient PAH activity, phenylalanine accumulates and is converted to phenylpyruvate (which can be detected in the urine). Without treatment, phenylketonurics eventually experience progressive mental retardation and seizures. Phenylketonuria can be controlled by eliminating phenylalanine from the diet, and phenylketonurics should be advised not to use aspartame. 17. The process for distinguishing R- and S-configurations of chiral molecules is described in the text on page 84. Enzymes discrimi- nate between isomers of chiral molecules thanks to the asym- metric arrangement of amino acid residues in the enzyme active site. 18. Appropriate ranges for buffering: Alanine—1.4–3.4, 8.7–10.7 Histidine—0.8–2.8, 5.0–7.0, 8.2–10.2 Aspartic acid—1.1–4.9, 8.8–10.8 Lysine—1.2–3.2, 8–11.5 19. With pK a of 8.3, cysteine would make a useful buffer except that cystine, the disulfide of cysteine, can form readily in this pH range. For every cystine formed, two cysteine-SH groups are eliminated, making the buffering capacity of cysteine of limited usefulness. Also, the cysteine sulfhydryl group is the most potent nucleophile among the side chains of the 20 common amino acids. 20. L-threonine is (2S, 3R)-threonine. D-threonine is (2R, 3S)-threonine. L-allothreonine is (2S, 3S)-threonine. D-allothreonine is (2R, 3R)-threonine. Chapter 5 1. Nitrate reductase is a dimer (2 Mo/240,000 M r ). 2. Phe-Asp-Tyr-Met-Leu-Met-Lys. 3. Tyr-Asn-Trp-Met-(Glu-Leu)-Lys. Parentheses indicate that the relative positions of Glu and Leu cannot be assigned from the information provided. 4. Ser-Glu-Tyr-Arg-Lys-Lys-Phe-Met-Asn-Pro. 5. Ala-Arg-Met-Tyr-Asn-Ala-Val-Tyr or Asn-Ala-Val-Tyr-Ala-Arg-Met- Tyr sequences both fit the results. (That is, in one-letter code, either ARMYNAVY or NAVYARMY.) 6. Gly-Arg-Lys-Trp-Met-Tyr-Arg-Phe. 7. Actually, there are four possible sequences: NIGIRVIA, GINIRVIA, VIRNIGIA, and of course, VIRGINIA. 8. Gly-Trp-Arg-Met-Tyr-Lys-Gly-Pro. 9. 10. Alanine, attached to a solid-phase matrix via its ␣-carboxyl group, is reacted with diisopropylcarbodiimide-activated lysine. Both the ␣-amino and ⑀-amino groups of the lysine must be blocked with 9-fluorenyl-methoxycarbonyl (Fmoc) groups. To Leu-Met-Cys-Val-Tyr-Arg-Cys-Gly-Pro. SS add leucine to Lys-Ala to form a linear tripeptide, precautions must be taken to prevent the incoming Leu ␣-carboxyl group from reacting inappropriately with the Lys ⑀-amino group instead of the Lys ␣-amino group. 11. The mass of the myoglobin chain is calculated to be 16,947 Ϯ 1 daltons. 12. Unlike any amino acid side chain, the phosphate group (or more appropriately, the phosphoryl group) bears two equiva- lents of negative charge at physiological pH. Furthermore, replacing an H atom on an S, T, or Y side chain with a phosphoryl group introduces a very bulky substituent into the protein structure where none existed before. 13. A graph of ␯ versus [L] reveals that at ␯ ϭ 0.5, [L] ϭ K D ϭ 2.4 mM. 14. IRS-1 has 1242 amino acids. Its average molecular mass is 131,590.97. The amino acid sequence of the tryptic peptide of IRS-1 of mass of 1741.9629 is LNSEAAAVVLQLMNIR. The sequence of the tryptic fragment containing the SHPTP-2 site is LCGAAGGLENGLNYIDLDLVK. 15. Nucleophilic attack by the hydroxyl O of the active-site serine on the carbonyl carbon of a peptide bond. 16. a. Amino acid changes in mutant hemoglobins that appear on the surface of the folded globin chains may affect quaternary structure. b. Amino acid substitutions on the surface on the quaternary hemoglobin structure that create hydrophobic patches might lead to polymerization. Such amino acids would include all of the hydrophobic amino acids. Chapter 6 1. The central rod domain of keratin is composed of distorted ␣-helices, with 3.6 residues per turn, but a pitch of 0.51 nm, compared with 0.54 nm for a true ␣-helix. (0.51 nm/turn)(312 residues)/(3.6 residues/turn) ϭ 44.2 nm ϭ 442 Å. For an ␣-helix, the length would be: (0.54 nm/turn)(312 residues)/(3.6 residues/turn) ϭ 46.8 nm ϭ 468 Å. The distance between residues is 0.347 nm for antiparallel ␤-sheets and 0.325 nm for parallel ␤-sheets. So 312 residues of antiparallel ␤-sheet amount to 1083 Å and 312 residues of parallel ␤-sheet amount to 1014 Å. 2. The collagen helix has 3.3 residues per turn and 0.29 nm per residue, or 0.96 nm/turn. Then: (4 in/year)(2.54 cm/in)(10 7 nm/cm)/(0.96 nm/turn) ϭ 1.06 ϫ 10 8 turns/year. (1.06 ϫ 10 8 turns/year)(1 year/365 days)(1 day/24 hours) (1 hour/60 minutes) ϭ 201 turns/minute. 3. Asp: The ionizable carboxyl can participate in ionic and hydrogen bonds. Hydrophobic and van der Waals interactions are negligible. Leu: The leucine side chain does not participate in hydrogen bonds or ionic bonds, but it will participate in hydrophobic and van der Waals interactions. Tyr: The phenolic hydroxyl of tyrosine, with a relatively high pK a , will participate in ionic bonds only at high pH but can both donate and accept hydrogen bonds. Uncharged tyrosine is capable of hydrophobic interactions. The relatively large size of A-6 Abbreviated Answers to Problems the tyrosine side chain will permit substantial van der Waals interactions. His: The imidazole side chain of histidine can act as both an ac- ceptor and donor of hydrogen bonds and, when protonated, can participate in ionic bonds. Van der Waals interactions are expected, but hydrophobic interactions are less likely in most cases. 4. As an imino acid, proline has a secondary nitrogen with only one hydrogen. In a peptide bond, this nitrogen possesses no hydrogens and thus cannot function as a hydrogen-bond donor in ␣-helices. On the other hand, proline stabilizes the cis-configuration of a peptide bond and is thus well suited to ␤-turns, which require the cis-configuration. 5. For a right-handed crossover, moving in the N-terminal to C-terminal direction, the crossover moves in a clockwise direction when viewed from the C-terminal side toward the N-terminal side. The reverse is true for a left-handed crossover; that is, movement from N-terminus to C-terminus is accompa- nied by counterclockwise rotation. 6. The Ramachandran plot reveals allowable values of ␾ and ␺ for ␣-helix and ␤-sheet formation. The plots consider steric hin- drance and will be somewhat specific for individual amino acids. For example, peptide bonds containing glycine can adopt a much wider range of ␾ and ␺ angles than can peptide bonds containing tryptophan. 7. The protein appears to be a tetramer of four 60-kD subunits. Each of the 60-kD subunits in turn is a heterodimer of two peptides, one of 34 kD and one of 26 kD, joined by at least one disulfide bond. 8. Hydrophobic interactions frequently play a major role in subunit–subunit interactions. The surfaces that participate in subunit–subunit interactions in the B 4 tetramer are likely to possess larger numbers of hydrophobic residues than the corre- sponding surfaces of protein A. 9. The length is given by (53 residues)ϫ(0.15 nm run/residue) ϭ 7.95 nm. The number of turns in the helix is given by (53 residues)/(3.6 residues/turn) ϭ 14.7 turns. There are 49 hydrogen bonds in this helix. 10. Glycines are essential components of tight turns (␤-turns) and thus are often essential for maintenance of protein structure. 11. Asp, Glu, Ser, Thr, His, and perhaps also Asn, Gln, Cys, Arg, Lys. 12. The ability of poly-Glu to form ␣-helices requires that the glutamate carboxyls be protonated. Deprotonation produces a polyanionic peptide that is not amenable to helix formation. 13. A coiled-coil formed from ␣-helices with 3.5 residues per turn would form a symmetrical seven-residue-repeating structure that would place the first and fourth residues of the seven-residue repeat at the same positions about the helix axis in every seven- residue repeat. This would allow the two helices of a coiled-coil structure to lie side by side with no twist about the coiled-coil axis. Such a structure would probably not be as stable as the twisted structure of coiled coils formed from ␣-helices with 3.6 residues per turn. 14. a. The third sequence would place hydrophobic residues on both sides of a ␤-strand and could thus be found in a parallel ␤-sheet. b. The second sequence would place hydrophobic residues on just one side of a ␤-strand and could thus be found in an anti-parallel ␤-sheet. c. The sixth sequence consists of GPX repeats (where X is any amino acid) and could thus be part of a tropocollagen molecule. d. The first sequence consists of seven-residue repeats, with first and fourth residues hydrophobic, and could thus be part of a coiled coil structure. 15. The solution to this exercise is to be completed by the student. 16. ⌬G°ЈϭϪ34.23 kJ/mol, a number that corresponds to one to two H bonds. Chapter 7 1. See structures in The Student Solutions Manual, Study Guide and Problems Book. 2. See structures and titration curve in The Student Solutions Manual, Study Guide and Problems Book. 3. The systematic name for stachyose is ␤- D-fructofuranosyl-O-␣-D- galactopyranosyl-(1 ⎯→ 6)-O-␣- D-galactopyranosyl-(1 ⎯→ 6)-␣-D- glucopyranoside. 4. Glycated hemoglobin can be separated from ordinary hemoglobin on the basis of charge difference (by ion-exchange chromatography, high-performance liquid chromatography [HPLC] electrophoresis, and isoelectric focusing) or on the basis of structural difference (by affinity chromatography). 5. The systematic name for trehalose is ␣- D-glucopyranosyl- (1 ⎯→ 1)-␣- D-glucopyranoside. Trehalose is not a reducing sugar. Both anomeric carbons are occupied in the disaccharide linkage. 6. See structures in The Student Solutions Manual, Study Guide and Problems Book. 7. A sample that is 0.69 g ␣- D-glucose/mL and 0.31 g ␤- D-glucose/mL will produce a specific rotation of 83°. 8. A 0.2 g sample of amylopectin corresponds to 0.2 g/162 g/mole or 1.23 ϫ 10 Ϫ3 mole glucose residues; 50 ␮mole is 0.04 of the total sample or 4% of the residues. Methylation of such a sample should yield 1,2,3,6-tetramethylglucose for the glucose residues on the reducing ends of the sample. The amylopectin sample contains 1.2 ϫ 10 18 reducing ends. 9. There are several target sites for trypsin and chymotrypsin in the extracellular sequence of glycophorin, and it would be reasonable to expect that access to these sites by trypsin and chymotrypsin would be restricted by the presence of oligosaccharides in the extracellular domain of glycophorin. 10. Energy yield upon combustion (whether by metabolic pathways or other reactions) depends on the oxidation level. Carbohy- drate and protein are at approximately the same oxidation level, and both of these are significantly less than that of fat. 11. This mechanism could involve either an S N 1 or S N 2 mechanism. In the former, protonation of the bridging oxygen would result in dissociation to produce a carbo-cation intermediate, which could be attacked by the phosphate nucleophile. The observa- tion of retention of configuration at the anomeric carbon favors this mechanism. An S N 2 mechanism would presumably involve water attack at the anomeric carbon, with dissociation of the oxygen of the carbohydrate chain. This would be followed by S N 2 attack by phosphate. Two S N 2 attacks would result in retention of configuration at the anomeric carbon atom. 12. Laetrile contains a cyanide group. Breakdown of laetrile and release of this cyanide function in the body would be highly toxic. Abbreviated Answers to Problems A-7 13. Chondroitin and glucosamine are amino sugar components of cartilage and connective tissue. Dietary supplement with these substances could help replenish the cartilage matrix proteo- glycan, relieving pain and restoring the proper function of connective tissue structures. 14. Two of the sugar units in stachyose are glucose and fructose, and the bond joining them is easily cleaved by stomach enzymes. However, the other two sugar units in stachyose are galactoses in ␤(1 ⎯→ 6)-linkages, which are not broken down by human enzymes. The result is that stachyose loses a fructose in the stomach but the resulting trisaccharide passes into the intestines, where bacterial enzymes degrade it, producing intestinal gas in the process. Beano contains an enzyme that hydrolyzes ␤(1 ⎯→6)-galactose linkages. Taking several Beano tablets before a meal of beans or legumes facilitates complete breakdown of stachyose and other related oligosaccharides in the stomach, avoiding the production of intestinal gas. The active enzyme in Beano is referred to as a ␤(1 ⎯→6)-galactosidase. 15. ␤- D-Fructofuranosyl-O-␣-D-galactopyranosyl-(1 ⎯→6)-O-␣-D- galactopyranosyl-(1 ⎯→ 6)-␣- D-glucopyranoside 16. Starch phosphorylase cleaves glucose units, one at a time, from starch chains until a (1 ⎯→ 6)-branch is encountered. Thus, limit dextrins are amylopectin fragments with a glucose in (1 ⎯→ 6)-linkage at each nonreducing end. The mechanism of the reaction that cleaves these glucose units is essentially the same as that for starch phosphorylase, except that it occurs at a glucose unit that is (1 ⎯→ 6)-linked. 17. In the beer-making process, the mash fermented by yeast contains starch, which is partially broken down by the amylase from the malt to produce limit dextrins (see problem 16). These limit dextrins add significant caloric content to regular beers. Joseph Owades used the enzyme amyloglucosidase, which hydrolyzes both (1 ⎯→ 4) and (1 ⎯→ 6) linkages in starch, thus breaking down the limit dextrins to simple glucose, which is fermented normally. (This raises the alcohol content of the beer above what would normally be produced, and water is typically added to adjust the alcohol content.) 18. As described in problem 14, Beano is a ␤(1 ⎯→ 6)-galactosidase. This enzyme can also hydrolyze (1 ⎯→ 6)-glucose linkages, although somewhat more slowly than for its natural substrate. Thus, Beano can slowly convert the limit dextrins produced in the beer-brewing process into simple glucose units, much like the enzyme used by Joseph Owades (see problem 17). Amateur brewers have learned to use Beano to make their own light beer, often referred to as Beano beer. 19. Assuming each glucose unit contributes 0.55 nm to the growing cellulose polymer (and the plant length), a growth rate of 1 foot per day corresponds to a rate for cellulose synthase of 6,400 glucose units per second. 20. Basic amino acid side chains (Arg, His, Lys) in antithrombin III present positively charged side chains for ionic interactions with sulfate functions on heparin; H-bond donating amino acid side chains could form H bonds with O atoms in OOH groups on the heparin carbohydrate residues; H-bond accepting amino acid side chains could form H bonds with H atoms in OOH groups of heparin. 21. Because these glycosaminoglycans are rich in hydroxyl groups, amine groups, and anionic functions (carboxylates, sulfates), they interact strongly with water. The heavily hydrated proteogly- cans formed from these glycosaminoglycans are reversibly dehy- drated in response to the pressure imposed on a joint during normal body movements. This dehydration has a cushioning effect on the joint; when the pressure is relieved, the glycosaminoglycans are spontaneously rehydrated due to their affinity for water. Chapter 8 1. Because the question specifically asks for triacylglycerols that contain stearic acid and arachidonic acid, we can discount the triacylglycerols that contain only stearic or only arachidonic acid. In this case, there are six possibilities: 2. See The Student Solutions Manual, Study Guide and Problems Book for a discussion. 3. a. Phosphatidylethanolamine and phosphatidylserine have a net positive charge at low pH. b. Phosphatidic acid, phosphatidylglycerol, phosphatidylinosi- tol, phosphatidylserine, and diphosphatidylglycerol normally carry a net negative charge. c. Phosphatidylethanolamine and phosphatidylcholine carry a net zero charge at neutral pH. 4. Diets high in cholesterol contribute to heart disease and stroke. On the other hand, plant sterols bind to cholesterol receptors in the intestines but are not taken up by the cells containing these receptors. There is substantial evidence that a diet that includes plant sterols can reduce serum cholesterol levels significantly. 5. Former Interior Secretary James Watt was well known during the Reagan administration for his comments on several occasions that trees caused and produced air pollution. As noted in this chapter, it is of course true that trees emit isoprenes that are the cause of the blue–gray haze that is common in still air in mid- to late-summer in the eastern United States. However, these isoprene compounds are not significantly toxic to living things, and it would be misleading to call them pollutants. 6. Louis L’Amour clearly knew his biochemistry. His protagonist knew that fat carries a higher energy content than protein or carbohydrate. If meals are going to be scarce (as for a person living in the wild and on the run), it is wise to consume fat rather than protein or carbohydrate. The same reasoning applies for migratory birds in the weeks preceding their long flights. 7. Phospholipase A 2 from snake bites operates without regulation or control, progressively breaking down cell membranes. Phos- pholipase A 2 action to produce cell signals is under precise control and is carefully regulated to produce just the right amounts of required cell signals. 8. The lethal dose for 50% of animals is referred to as the LD 50 . The LD 50 for dogs is approximately 3 mg/kg of body weight. Thus, for a 40-lb dog (18.2 kg), consumption of approximately 55 mg of warfarin would be lethal for 50% of animals. 9. See the Student Solution Guide and www.cengage.com/login for the solutions to this problem. 10. Humans require approximately 2000 kcal per day. Seal blubber is predominantly triglycerides, which yield approximately 9 kcal per gram. This means that a typical human would need to consume 222 grams of seal blubber per day (about half a SSA SAS ASS SAA ASA AAS A-8 Abbreviated Answers to Problems pound) in order to obtain all of his or her calories from this energy source. 11. Results for this problem will depend on the particular cookies chosen by the student. 12. The only structural differences between cholesterol and stigmasterol are the double bond of stigmasterol at C 22 –C 23 , and the ethyl group at C 24 . 13. Androgens mediate the development of sexual characteristics and sexual function in animals. Glucocorticoids participate in the control of carbohydrate, protein, and lipid metabolism. Mineralocorticoids regulate salt (Na ϩ , K ϩ , and Cl Ϫ ) balances in tissues. 14. The answers to this question will depend on the household products chosen and the isoprene substances identified. 15. Hydroxide ions (in lye) catalyze the breakdown of triglycerides to produce fatty acid salts such as sodium stearate and sodium palmitate and leaving glycerol as a byproduct. Micelle formation by these “soaps” leads to emulsification of fats and other non- polar substances. 16. Amphipathic phosphatidylcholine from egg yolk exerts a deter- gentlike action on mixtures of vegetable oil and water. Micelles made primarily of egg phosphatidylcholine emulsify the vegetable oil, forming a stable suspension that persists indefinitely. (Thus, mayonnaise does not need to be mixed or shaken before use, whereas oil and vinegar mixtures do.) The micellar parti- cles of mayonnaise are large enough to scatter visible light, so mayonnaise appears milky white, whereas the separated layers of oil and water in containers of oil and vinegar appear clear. 17. Stanol esters function in cholesterol reduction by binding to cholesterol receptors in the intestines. However, each serving of stanol esters consumed only blocks a fraction of all intestinal receptors. Regular consumption of stanol esters eventually blocks all or most available receptors. Binding of stanol esters is tight and long lasting, but not indefinite, so stanol ester consumption must be continued to maintain the beneficial results. The graph on page 236 shows that reductions of serum cholesterol levels can approach 15%; given that the typical diet accounts for about 15% of total serum cholesterol, it may be concluded that stanol esters are highly effective in preventing uptake of dietary cholesterol. 18. Serum cholesterol is partly derived from diet and partly from synthesis in the liver. Stanol esters prevent uptake of dietary cholesterol but do not affect synthesis in the liver. Cholesterol- lowering drugs, on the other hand, block cholesterol synthesis but have no effect on uptake from diet. Thus, the effects of these two classes of cholesterol-lowering agents are additive. 19. Research has shown that consumption of substantial quantities of plant fats can lead to significant reduction of serum cholesterol. The content of so-called phytosterols varies depending on the source. However, a sterol- or stanol-fortified spread like Benecol probably provides the highest concentration of dietary agents for cholesterol lowering. 20. Tetrahydrogestrinone (THG) is synthesized by the catalytic hy- drogenation of gestrinone (which has an acetylene group in place of the ethyl group on the D-ring of tetrahydrogestrinone). Patrick Arnold, known as the “father of prohormones,” is cred- ited with the synthesis of THG from gestrinone and the promo- tion of THG as an anabolic steroid in preparations such as The Clear and The Cream. These substances were marketed aggres- sively to top athletes in several sports by Arnold and Victor Conte of the Bay Area Laboratory Co-Operative. THG (at that time) was undetectable by existing laboratory analyses, and its use was not specifically prohibited in athletic competitions by the World Anti-Doping Code. Use of THG has since been prohibited by nearly all sports regulatory agencies, and effective methods for its detection have been developed. Numerous athletes have by now admitted to use of THG or have been con- victed for perjury in denial of its use, and many have been stripped of medals and awards for athletic accomplishments aided by THG. 21. Most obviously, a diet of triglycerides (from the blubber of seals, the polar bears’ favorite food) provides high energy and material that can be reprocessed into other triglycerides and used as insulation under the skin. Less obvious is the need of the polar bear to stay warm and conserve water. (Polar bears cannot afford to eat snow or ice [they are too cold], and they cannot drink seawater [it is too salty].) The polar bear is adapted to conserve body water and stay warm, and thus it does not urinate for months at a time. (Urination would give up both water and heat.) To achieve this, it must consume little or no nitrogen, because a diet rich in nitrogen would require urination and defecation. Triglycerides contain no nitrogen and are thus ideal food for the adult polar bear. Juvenile polar bears, which have not yet reached their full adult body size, must consume protein in order to make their own proteins. Once the bear reaches its full size, it changes its diet and no longer consumes protein! 22. Snake venom phospholipase A 2 cleaves fatty acids from phospholipids at the C-2 position. The fatty acids behave as deter- gents and form micelles (see Chapter 9) that can remove lipids and proteins from the membrane and disrupt membrane structure, causing pores and eventually rupturing the cell itself. Chapter 9 1. Glycerophospholipids with an unsaturated chain at the C-1 position and a saturated chain at the C-2 position are rare to nonex- istent. Glycerophospholipids with two unsaturated chains, or with a saturated chain at C-1 and an unsaturated chain at C-2, are commonly found in biomembranes. 2. The phospholipid/protein molar ratio in purple patches of H. halobium is 10.8. 3. See The Student Solutions Manual, Study Guide and Problems Book for plots of sucrose solution density versus percent by weight and by volume. The plot in terms of percent by weight exhibits a greater curve, because less water is required to form a solution that is, for example, 10% by weight (10 g sucrose/100 g total) than to form a solution that is 10% by volume (10 g sucrose/ 100 mL solution). 4. r ϭ (4Dt) 1/2 . According to this equation, a phospholipid with D ϭ 1 ϫ 10 Ϫ8 cm 2 will move approximately 200 nm in 10 milliseconds. 5. Fibronectin: For t ϭ 10 msec, r ϭ 1.67 ϫ 10 Ϫ7 cm ϭ 1.67 nm. Rhodopsin: r ϭ 110 nm. All else being equal, the value of D is roughly proportional to (M r ) Ϫ1/3 . Molecular weights of rhodopsin and fibronectin are 40,000 and 460,000, respectively. The ratio of diffusion coeffi- cients is thus expected to be (40,000) Ϫ1/3 /(460,000) Ϫ1/3 ϭ 2.3. On the other hand, the values given for rhodopsin and fibronectin give an actual ratio of 4286.The explanation is that fibro- Abbreviated Answers to Problems A-9 nectin is anchored in the membrane via interactions with cytoskeletal proteins, and its diffusion is severely restricted compared with that of rhodopsin. 6. a. Divalent cations increase T m . b. Cholesterol broadens the phase transition without significantly changing T m . c. Distearoylphosphatidylserine should increase T m , due to increased chain length and also to the favorable interactions between the more negative PS head group and the more positive PC head groups. d. Dioleoylphosphatidylcholine, with unsaturated fatty acid chains, will decrease T m . e. Integral proteins will broaden the transition and could raise or lower T m , depending on the nature of the protein. 7. ⌬G ϭ RT ln([C 2 ]/[C 1 ]). ⌬G ϭ ϩ4.0 kJ/mol. 8. ⌬G ϭ RT ln([C out ]/[C in ]) ϩ ZᏲ⌬␺. ⌬G ϭ ϩ4.08 kJ/mol Ϫ2.89 kJ/mol. ⌬G ϭ ϩ1.19 kJ/mol. The unfavorable concentration gradient thus overcomes the favorable electrical potential and the outward movement of Na ϩ is not thermodynamically favored. 9. One could solve this problem by going to the trouble of plotting the data in v vs. [S], 1/v vs. 1/[S], or [S]/v vs. [S] plots, but it is simpler to examine the value of [S]/v at each value of [S]. The Hanes–Woolf plot makes clear that [S]/v should be constant for all [S] for the case of passive diffusion. In the present case, [S]/v is a constant value of 0.0588 (L/min) Ϫ1 . It is thus easy to recognize that this problem describes a system that permits passive diffusion of histidine. 10. This is a two-part problem. First calculate the energy available from ATP hydrolysis under the stated conditions; then use the answer to calculate the maximal internal fructose concentration against which fructose can be transported by coupling to ATP hydrolysis. Using a value of Ϫ30.5 kJ/mol for the ⌬G°Ј of ATP and the indicated concentrations of ATP, ADP, and P i , one finds that the ⌬G for ATP hydrolysis under these conditions (and at 298 K) is Ϫ52.0 kJ/mol. Putting the value of ϩ52.0 kJ/mol into Equation 9.1 and solving for C 2 yields a value for the maximum possible internal fructose concentration of 1300 M! Thus, ATP hydrolysis could (theoretically) drive fructose transport against internal fructose concentrations up to this value. (In fact, this value is vastly in excess of the limit of fructose solubility.) 11. Each of the transport systems described can be inhibited (with varying degrees of specificity). Inhibition of the rhamnose transport system by one or more of these agents would be consistent with involvement of one of these transport systems with rhamnose transport. Thus, nonhydrolyzable ATP analogs should inhibit ATP-dependent transport systems, ouabain should specifically block Na ϩ (and K ϩ ) transport, uncouplers should inhibit proton gradient–dependent systems, and fluoride should inhibit the PTS system (via inhibition of enolase). 12. N-myristoyl lipid anchors are found linked only to N-terminal Gly residues. Only the peptide in (e) of this problem contains an N-terminal Gly residue. 13. Only the peptide in (a) possesses a CAAX sequence (where C ϭ Cys, A ϭ aliphatic (Ala, Val, Leu, Ile), and X ϭ any amino acid). 14. The hydropathy plot of a soluble protein should show no substantial stretches of hydrophobic residues, except for the signal sequence. 15. Prolines are destabilizing to ␣-helices, and a short helix with a proline would not be likely to be stable. Where a proline occurs in an ␣-helix, the helix is bent or kinked. Helices with a proline kink tend to be longer than average, presumably because longer helices are more stable and more able to tolerate the loss of H bonds in the kink region. 16. The structural consequences of a proline-induced kink in an ␣-helix are described in detail in the references provided in this problem. 17. Porin proteins typically consist of 18-stranded ␤-barrels. With 9 to 11 residues per strand, about 180–200 residues would be required to form the barrel, with a roughly equivalent number required to form the loops between strands. (The maltoporin chains from E. coli consists of about 420 residues.) The transmembrane domain of Wza consists of eight ␣-helical segments with about 25 residues per helix, about 200 residues in all. Thus, the number of residues needed to create the transmembrane pore in these two proteins is about the same, even though the number of residues per membrane spanning segment is less for a ␤-strand (9–11) than for an ␣-helix (21–25). 18. From Figure 9.29, the area within a typical “fenced” area is about 0.3 ␮m ϫ 0.3 ␮m, or 9 ϫ 10 6 Å 2 . Dividing by 60 Å 2 , it appears that there are about 150,000 phospholipids in a monolayer-fenced area in the membrane (assuming a membrane of pure phospholipid, with no cholesterol, etc.). 19. The lysine side chain N is about 5.5 Å from the ␣-carbon (about 1.1 Å per bond). If a Lys side chain was reoriented toward the membrane center, the maximum change in position relative to the membrane center would be 10–11 Å (assuming the position of the ␣-carbon did not change). Figure 9.15 indicates that the energy of a Lys side chain would change by 4kT over the 15-Å distance from the membrane surface to the membrane center. A movement of 10 Å would thus correspond to 2/3 of 4kT, or 8kT/3—almost twice the average translation kinetic energy of a molecule in the gas phase (3kT/2). 20. In a hydrophobic environment, the aspartate carboxyl group will be more stable in the protonated (uncharged) form than in the deprotonated (charged) form. 21. In a hydrophobic environment, the side chains of lysine and arginine would be more stable in their deprotonated (uncharged) forms than in the protonated (charged) forms. As a result, the pK a values of these residues would be lowered significantly in a hydrophobic environment. 22. Based on the discussion in problems 20 and 21 in this chapter, it would be reasonable to imagine that light-induced conformation changes could alter the pK a values of the proton-transferring moieties in the protein. For example, a conformation change that made the environment around an Asp carboxyl more polar would reduce the pK a of that group, promoting dissociation (i.e., proton release). An appropriate sequence of such changes could accomplish transmembrane proton transport. 23. Point (b)—that proteins can be anchored to the membrane via covalent links to lipid molecules—was not part of the Singer– Nicolson fluid mosaic model. A-10 Abbreviated Answers to Problems Chapter 10 1. 2. See Figure 10.15. 3. f A ϭ 0.29; f G ϭ 0.22; f C ϭ 0.25; f T ϭ 0.28. 4. The number of A and T residues ϭ 3.14 ϫ 10 9 each; the number of G and C residues ϭ 2.68 ϫ 10 9 each. 5. 5Ј-TAGTGACAGTTGCGAT-3Ј. 6. 5Ј-ATCGCAACTGTCACTA-3Ј. 7. 5Ј-TACGGTCTAAGCTGA-3Ј. 8. There are two possibilities, a and b. (E ϭ EcoRI site; B ϭ BamHI site.) Note that b is the reverse of a. 9. a. GGATCCCGGGTCGACTGCAG; b. GTCGACCCGGGATCCTGCAG. SmaI products: a. GGATCCC and GGGTCGACTGCAG. b. GTCGACCC and GGGATCCTGCAG. 10. Synthesis of a polynucleotide 100 residues long requires formation of 99 phosphodiester bonds. ⌬G°Ј for phosphodiester synthesis (assuming it is the same magnitude but opposite sign as that for phosphoric anhydride cleavage of an NTP to give NMP ϩ PP i ) is ϩ32.3 kJ/mol. ⌬G°Ј overall ϭ (99)(ϩ32.3) ϭ 3198 kJ/mol. 11. a. Hydrogen bonding and van der Waals interactions between amino acid side chains and DNA, and ionic interactions of amino acid side chains with the nucleic acid backbone phosphate groups. Double-helical DNA does not present hydrophobic regions for interaction with proteins because of base-pair stacking. b. Proteins can recognize specific base sequences if they can fit within the major or minor groove of DNA and “read” the H-bonding pattern presented by the edges of the bases in the groove. The dimensions of an ␣-helix are such that it fits snugly within the major groove of B-DNA; then, depending on the amino acid sequence of the protein, the side chains displayed on the circumference of the ␣-helix have the potential to form H bonds with H-bonding functions provided by the bases. 12. a. The restriction endonuclease must be able to recognize a specific nucleotide sequence in the DNA that has twofold rotational symmetry. b. In order to read a base sequence within DNA, either the restriction endonuclease must interact with the bases by direct a. BEB 1 3 0.5 5.5 b. BE B 5.5 0.5 3 1 OCH 2 HO P Ϫ O ϩ O N N H O O HH OH OH H H 5Ј N N NH 2 H access, for example, by binding in the major groove, or the restriction endonuclease must be able to “read” the base sequence by indirect means, for example, if the base sequence imparts some local variation in the cylindrical surface of the DNA that the enzyme might recognize. c. The restriction endonuclease must be able to cleave both DNA strands, often in a staggered fashion. d. An obvious solution to the requirements listed in (a) and (c) would be a homodimeric subunit organization for restriction endonucleases. 13. a. The ribose group of nucleosides greatly increases the water solubility of the base. b. Ribose has fewer hydroxyl groups and thus less likelihood to undergo unwanted side reactions. c. The absence of the 2-OH in 2-deoxyribose leads to a polynucleotide sugar–phosphate backbone that is more stable because it is not susceptible to alkaline hydrolysis. 14. a. Phosphate groups bear a negative electrical charge at neutral pH. b. Cleavage of phosphoric anhydride bonds is strongly exer- gonic and can provide the thermodynamic driving force for diverse metabolic reactions. c. Cleavage of phosphoric anhydride bonds is strongly exer- gonic and can provide the thermodynamic driving force for phosphodiester bond formation in polynucleotide synthesis. 15. Once in every 4.4 ϫ 10 12 nucleotides. 16. A DNA sequence 16 nucleotides long. 17. The strategy for protein sequencing by Edman degradation, as described on pages 80 and 102, could be adapted, replacing Edman’s reagent with snake venom phosphodiesterase acting on an immobilized nucleotide sequence. 18. 3.59 ϫ 10 12 D. 19. The use of bases as “information symbols” in metabolism allows the cell to allocate portions of its phosphorylation potential (as total NTP) to dedicated tasks, as in GTP for protein synthesis, CTP for phospholipid synthesis, UTP for carbohydrate synthesis, with ATP serving the central role. Enzymes in these pathways se- lectively bind the proper nucleotide through recognition of the particular base. 20. The most prominent structural feature of the DNA double helix in terms of structural complementarity is found in the canonical base-pairing of A with T and G with C. Base pairing is essential to DNA replication (preservation of the genetic information) and transcription (expression of genetic information). Chapter 11 1. T C T A C G G G A A C T Abbreviated Answers to Problems A-11 2. Original nucleotide: 5Ј-GATAGCGCAAAGATCAACCTT. 3. a. 10.5 base pairs per turn; b. ⌬␾ ϭ 34.3°; c (true repeat) ϭ 6.72 nm. 4. 27.3 nm; 122 base pairs. 5. 4.35 ␮m. 6. L 0 ϭ 160. If W ϭ Ϫ12, L ϭ T ϩ W ϭ 160 ϩ (Ϫ12) ϭ 148. ␴ ϭ ⌬L/L 0 ϭϪ12/160 ϭϪ0.075. 7. For 1 turn of B-DNA (10 base pairs): L B ϭ 1.0 ϩ W B . For Z-DNA, 10 base pairs can only form 10/12 turn (0.833 turn), and L Z ϭ 0.833 ϩ W Z . For the transition B-DNA to Z-DNA, strands are not broken, so L B ϭ L Z ; that is, 1.0 ϩ W B ϭ 0.833 ϭ W Z , or W Z Ϫ W B ϭϩ0.167. (In going from B-DNA to Z-DNA, the change in W, the number of supercoils, is positive. This result means that, if B-DNA contains negative supercoils, their number will be reduced in Z-DNA. Thus, all else being equal, negative supercoils favor the B ⎯→ Z transition.) 8. 6 ϫ 10 9 bp/200 bp ϭ 3 ϫ 10 7 nucleosomes. The length of B-DNA 6 ϫ 10 9 bp long ϭ (0.34 nm)(6 ϫ 10 9 ) ϭ 2.04 ϫ 10 9 nm (more than 2 meters!). The height of 3 ϫ 10 7 nucleosomes ϭ (6 nm)(3 ϫ 10 7 ) ϭ 18 ϫ 10 7 nm (0.18 meter). 9. From Figure 11.33: Similarly shaded regions indicate complementary sequences joined via intrastrand hydrogen bonds: 70020 nucleotide no. 40 50 6010 30 CCA endanticodon 10. Increasing order of T m : yeast Ͻ human Ͻ salmon Ͻ wheat Ͻ E. coli. 11. In 0.2 M Na ϩ , T m (°C) ϭ 69.3 ϩ 0.41(%G ϩ C): Rats (%G ϩ C) ϭ 40%, T m ϭ 69.3 ϩ 0.41(40) ϭ 85.7°C Mice (%G ϩ C) ϭ 44%, T m ϭ 69.3 ϩ 0.41(44) ϭ 87.3°C Because mouse DNA differs in GC content from rat DNA, they could be separated by isopycnic centrifugation in a CsCl gradient. 12. GC content ϭ 0.714 (from Table 10.1 and equations used in problem 3, Chapter 10). ␳ϭ1.660 ϩ 0.098(GC) ϭ 1.730 g/mL. 13. See Figure 11.35 and compare your structure with the base pair formed between G18 and ␺55 in yeast phenylalanine tRNA, which has a single H bond between the 2-amino group of G (2-NH 2 Z OP) and the O atom at position 4 in ␺. 14. Assuming the plasmid is in the B-DNA conformation, where each pair contributes 0.34 nm to the length of the molecule, the circumference of a perfect circle formed from pBR322 would be (0.34 nm/bp)(4363 bp) ϭ 1483 nm ϭ 1.48 ␮m. The E. coli K12 chromosome laid out as a perfect circle would have a circumference of (0.34 nm/bp)(4,639,000 bp) ϭ 1,577,260 nm ϭ 1.58 mm. The diameter of a B-DNA molecule is about 2.4 nm (see Table 11.1). For pBR322, the length/diameter ratio would be 1480/2.4 ϭ 616.7. For the E. coli K12 chromosome, the length/diameter ratio would be 1,577,260/2.4 ϭ 657,192. 15. (a) Z-DNA; (b) cruciform; (c) triplex DNA; (d) tRNA; (e) type-II restriction endonuclease site. 16. Rely on the features in Figure 11.33 to draw a tRNA cloverleaf with the anticodon positioned at nucleotides 34–36: CUG. 17. Erythromycin is within the palm of the mitten. 18. (a) ␤-globin; (b) hexosaminidase A (␣-polypeptide); (c) insulin receptor; (d) neutral amino acid transporter. 19. The DNA in such a thermophilic organism would be subjected to very high temperatures that might denature the DNA. Because GϺC base pairs are more heat stable than AϺT base pairs, one might expect DNA from a thermophilic organism to have a high G ϩ C content. 20. DNA is the material of heredity; that is, genetic information. This information is encoded in the sequence of bases in DNA, so this is its most important structural feature. The double- stranded nature of DNA and the complementary base sequence of the two strands are also crucial structural features in the transmission of genetic information through DNA replication. Beyond these points, one might cite the structural features of DNA that impart stability to the double helix and structural aspects that render DNA less susceptible to degradation. Chapter 12 1. Linear and circular DNA molecules consisting of one or more copies of just the genomic DNA fragment; linear and circular DNA molecules consisting of one or more copies of just the vector DNA; linear and circular DNA molecules containing one or more copies of both the genomic DNA fragment and plasmid DNA. 2. 3. 4. N ϭ 3480. 5. N ϭ 10.4 million. 6. 5Ј-ATGCCGTAGTCGATCAT and 5Ј-ATGCTATCTGTCCTATG. 7. -Thr-Met-Ile-Thr-Asn-Ser-Pro-Asp-Pro-Phe-Ile-His-Arg-Arg-Ala-Gly-Ile- Pro-Lys-Arg-Arg-Pro… a. AAGCTTGAGCTCGAGATCTAGATCGAT HindIII XhoI XbaI SacI Bg III ClaI b. Vector: HindIII: 5Ј-A gap . . CGAT-3Ј: ClaI 3Ј-TTCGA. . . gap . TA-5Ј Fragment: HindIII: 5Ј-AGCTT(NNNN-etc-NNNN)AT-3Ј 3Ј-A(NNNN-etc-NNNN)TAGC-5Ј -GAATTCCCGGGGATCCTCTAGAGTCGACCTGCAGGCATGC - GAATTC GGATCC GTCGAC GCATGC EcoRI BamHI SalI SphI CCCGGG TCTAGA CTGCAG SmaI XbaI PstI A-12 Abbreviated Answers to Problems The junction between ␤-galactosidase and the insert amino acid sequence is between Pro and Asp, so the first amino acid encoded by the insert is Asp. (The polylinker itself codes for Asp just at the BamHI site, but in constructing the fusion, this Asp and all of this downstream section of polylinker DNA is dis- placed to a position after the end of the insert.) 8. 5Ј(G)AATTCNGGNATGCAYCCNGGNAAR C T T N Y GCNAGY TGG- TTYGTNGGGAATTCN- (Note: The underlined triplet AGY represents the middle Ser residue. Ser codons are either AGY or TCN [where Y ϭ pyrimidine and N ϭ any base]; AGY was selected here so that the mutagenesis of this codon to a Cys codon [TGY] would involve only an A ⎯→ T change in the nucleotide sequence.) Because the middle Ser residue lies nearer to the 3Ј-end of this EcoRI fragment, the mutant primer for PCR amplification should encompass this end. That is, it should be the primer for the 3 ⎯→ 5Ј strand of the EcoRI fragment: 5Ј-NNNGAATTCCCN c ACR c AACCAR c CAN c GC-3Ј where the mutated Ser ⎯→ Cys triplet is underlined, NNN ϭ several extra bases at the 5Ј-end of the primer to place the EcoRI site internal, N c ϭ the nucleotide complementary to the nucleotide at this position in the 5Ј→3Ј strand, and R c ϭ the pyrimidine complementary to the purine at this position in the 5Ј→3Ј strand. 9. The number of sequence possibilities for a polymer is given by x y , where x is the number of different monomer types and y is the number of monomers in the oligomers. Thus, for RNA oligomers 15 nucleotides long: x y ϭ 4 15 ϭ 1,073,741,824. For pentapeptides, x y ϭ 20 5 ϭ 3,200,000. 10. See Figure 12.17. You would need a GAL4-deficient yeast strain expressing a fusion protein composed of the GAL4 DB domain fused with a cytoskeletal protein (such as actin or tubulin) to serve as the bait. These GAL4 Ϫ cells would be transformed with a cDNA library of human epithelial proteins constructed so that the human proteins were expressed as GAL4 TA fusion proteins; these fused proteins are the target. Interaction of a target protein with the cytoskeletal protein fused to the GAL4 DB domain will lead to GAL4-driven expression of the lacZ gene, whose presence can be revealed by testing for ␤-galactosidase activity. 11. See Figure 12.9 for isolation of mRNA and Figure 12.10 for preparation of a cDNA library from mRNA. The mRNA would be isolated, and the cDNA libraries would be prepared from two batches of yeast cells, one grown aerobically and one grown anaerobically. 12. See Figure 12.11. Differently labeled single-stranded cDNA from separate libraries prepared from aerobically versus anaerobically grown yeast cells would be hybridized with a DNA microarray (gene chip) of yeast genes. 13. An antibody against hexokinase A could be used to screen the yeast cDNA library to identify a yeast colony expressing this protein (see discussion on page 369). Once a sample of the putative yeast hexokinase A protein was isolated, its identity could be confirmed by peptide mass fingerprinting using mass spec- trometry (see page 108). 14. The experiment in problem 12 identified the cDNA clone for the protein, but this clone will not have the regulatory elements (promoter) necessary for the experiment at hand. Using the cDNA clone as a probe, the genomic clone for this nox gene could be identified in a yeast genomic library. Cloning and sequencing a genomic clone would identify putative promoter regions that could be fused to the coding region of green fluo- rescent protein (GFP) to create a reporter gene construct (see pages 371–372) that would “light up” more and more as oxygen levels declined. 15. Go to the NCBI site map at http://www.ncbi.nlm.nih.gov/Sitemap/ index.html. In the alphabetical index, click on Genomes and Maps to go to the Entrez Genome site at http://www.ncbi.nlm.nih .gov/Sitemap/index.html#Genomes. Under Organism Collections, Entrez Genomes section, click on the links for bacteria, archea, and eukaryotes to see: Number of bacterial genomes completed at http://www.ncbi.nlm.nih.gov/PMGifs/Genomes/eub_g.html Number of archeal genomes completed at http://www.ncbi.nlm.nih.gov/PMGifs/Genomes/a_g.html Completed eukaryotic genomes accessible at http://www.ncbi.nlm.nih.gov/PMGifs/Genomes/euk_g.html 16. Insertion of DNA at any of the following restriction sites would render cells harboring the recombinant plasmid ampicillin- sensitive: SspI, ScaI, PvuI, PstI, and PpaI. 17. The respective codons are Trp ϭ UGG; Cys ϭ UGC and UGU. Thus, changing only the third base G in the Trp codon to a pyrimidine (either C or U) would yield a Cys codon. In this case, there are now 6 differences in codon possibilities, so 2 6 ϭ 64 different oligonucleotides must be synthesized. Chapter 13 1. v/V max ϭ 0.8. 2. v ϭ 91 ␮mol/mL и sec. 3. K s ϭ 1.43 ϫ 10 Ϫ5 M; K m ϭ 3 ϫ 10 Ϫ4 M. Because k 2 is 20 times greater than k Ϫ1 , the system behaves like a steady-state system. 4. If the data are graphed as double-reciprocal Lineweaver–Burk plots: a. V max ϭ 51 ␮mol/mL и sec and K m ϭ 3.2 mM. b. Inhibitor (2) shows competitive inhibition with a K I ϭ 2.13 mM. Inhibitor (3) shows noncompetitive inhibition with a K I ϭ 4 mM. 5. (b) v [S] E 1 /2 E (a) v [S] 2E E

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