Answers to Chapter 6 (in-text & asterisked problems)

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1 Introduction to Bioorganic Chemistry and Chemical Biology 1 Answers to Chapter 6 (intext & asterisked problems) Answer 6.1 olve the equation. Convert temperature in C to K by adding 273. Plug the numbers into ΔG = Δ TΔ. K association = e ΔG/T. This is the equilibrium constant for association, so take the inverse: K d = 1/K association. Protein Protein Δ (kcal mol 1 ) mutant TC β chain 8.2. aureus enterotoxin C3 Δ (cal K 1 mol 1 ) K d at 25 C M p67 phox iso1 cytochrome c ac GTP complex iso1cc peroxidase at 18 C M at 25 C M Answer 6.2 Calculated using K d = k off /k on : Protein mall ligands k on (M 1 s 1 ) k off (s 1 ) K d chymotrypsin proflavin creatine kinase ADP , G3P dehydrog. AD lactate dehydrog. AD , alcohol dehydrog. AD lysozyme (AcGlu) , ribonuclease 3ʹUMP , Protein Large ligands k on (M 1 s 1 ) k off (s 1 ) K d ta er synthetase ta er trypsin protein inhibitor insulin insulin , βlactoglobulin βlactoglobulin αchymotrypsin αchymotrypsin

2 2 Introduction to Bioorganic Chemistry and Chemical Biology: Answers to Chapter 6 Answer 6.3 When the concentration of AD is M, the ratio of bound to unbound alcohol dehydrogenase is 1:1. From there, the ratio of bound to unbound enzyme can easily be estimated at other concentrations of AD. bound enzyme AD K D = M unbound enzyme AD [AD] [enzyme AD] : [enzyme] M 10 : M 1 : M 1 : M 1 : 100 A When [AD] = 3 μm, the ratio of bound to unbound enzyme is 10/11 91%. B When [AD] = 3 nm, the ratio of bound to unbound enzyme is 1/101 1%. Answer 6.4 cancer cells [Curcumin] Dead / Live Percentage (µm) viable : : : : curcumin : Introduction to Bioorganic Chemistry and Chemical Biology 6117 Curcumin has poor bioavailability. At an oral dose of 8 g of curcumin per day, the peak serum concentrations of curcumin reach only 1.8 μm. ypothetically, eating large quantities of curcumin might be effective for colorectal cancer in the GI tract, but not for systemic cancers like leukemias. Answer 6.5 A Galactose binds most tightly because it has the lowest K m ; however, the affinities of all three substrates are within a factor of two. B Galactose is isomerized more than 100 times faster than the other two substrates on the basis of the k cat /K m values: galactose (3700 mm 1 s 1 ), glucose (13 mm 1 s 1 ), xylose (20 mm 1 s 1 ). C If the system is at equilibrium, when the concentration of glucose is 10 K m (340 mm), the ratio of glucose enzyme complex to free enzyme will be 10:1. In a typical mammalian cell, the intracellular glucose concentration is less than 1 mm. f course the amount of free enzyme is likely to be small because galactose and other sugars can occupy the enzyme active site. Answer 6.6 A The substrate with the lowest K m binds most tightly: LALG. B The substrate with the highest k cat is phosphorylated fastest (once it binds): L ALG. C The relative rates of phosphorylation will be proportional to k cat /K m. LALG is better than LAALG by a factor of ubstrate K m (μm) k cat (s 1 ) k cat / K m (M 1 s 1 ) LAALG LALG LALG

3 Introduction to Bioorganic Chemistry and Chemical Biology: Answers to Chapter 6 3 Answer 6.7 Because imine formation is fast and reversible, the following mechanism is reasonable. The mechanism for imine/iminium ion formation was covered in Chapter 2. 2 C A C : 2 C 2 The ratedetermining step for this reaction has not yet been determined, making it difficult to determine the ordering of the various steps. It has been proposed that hemithioacetal formation precedes imine formation. Unfortunately, this proposed mechanism involves the formation of a benzylic cation that is destabilized by the ortho imine substituent. many steps 2 Cys unstable cation Cys Cys Answer 6.8 Carboxamide hydrolysis and βelimination prevent quantitative isolation of Asn, Gln, er, Cys, and Thr. Asn Gln 2 a a 2 2 : A : 2 er Cys Thr a a a Introduction to Bioorganic Chemistry and Chemical Biology A6119

4 Introduction to Bioorganic Chemistry and Chemical Biology: Answers to Chapter 6 Answer 6.9 The enzyme uses two Zn2 ions and an arginine to stabilize the serine alkoxide, the alkoxide leaving group and the anionic phosphorane intermediate, but these are omitted to simplify the problem. er : er P er P P er P : er P er P Answer 6.10 Prostromelysin cannot cleave itself, because Cys75 holds the inhibitory domain in place by coordinating to the Zn2 ion at the active site (see the rendering of prostromelysin in Figure 6.48). Arylmercurials have a high affinity for sulfur. They coordinate to Cys75, opening up the Zn2 active site, which can then proteolytically cleave the inhibitory domain. Arg Arg Arg Zn Zn Zn Introduction to Bioorganic Chemistry and Chemical Biology A6120 Van Vranken & Weiss design by Answer 6.11 trapoxin peptide Ph Ph Zn2 binding element Introduction to Bioorganic Chemistry and Chemical Biology A6121 Van Vranken & Weiss Introduction to Bioorganic Chemistry and Chemical Biology A design by Van Vranken & Weiss Answer 6.12 design by If the two ligands bound with perfect cooperativity, the dissociation constant would be the product of the two Kd values, namely = M. Answer 6.13 : : B 4 C2 C2 B C2

5 Introduction to Bioorganic Chemistry and Chemical Biology: Answers to Chapter 6 5 B C 2 C 2 C 2 C 2 C 2 Introduction to Bioorganic Chemistry and Chemical Biology 6123 Answer 6.14 : : 2 : 3 P 3 P 3 P 2 : : C 2 3 P 3 P 3 P Introduction to Bioorganic Chemistry and Chemical Biology A6124 Answer 6.15 A ({G/C}{A/C}T) 6 B {G/A}{T/C}G{G/C}{G/C}G{G/T}{T/C}G{G/T}{T/C}G{G/A}{A/C}G C {G/C}{C/A}C{G/C}CGG{C/T}G{G/T}CGG{C/A}G *Answer 6.16 A Chorismate binds more tightly because it has the lower K m. B Chorismate is also rearranged more quickly (after it binds) because it has the much larger k cat. C verall, chorismate (k cat /K m = 207 mm 1 s 1 ) is a better substrate than the methyl analog (k cat /K m = 0.29 mm 1 s 1 ) by almost three orders of magnitude. *Answer 6.17 A strained reactive intermediate

6 6 Introduction to Bioorganic Chemistry and Chemical Biology: Answers to Chapter 6 B C : Cl Cl Cl D nonfluorescent 2 fluorescent 2 Lone pair donation of the amino group into the coumarin ring system favors a crossconjugated, nonaromatic form. When aminocoumarin is conjugated to a peptide as an amide, the amino lone pair donates more into the amide carbonyl than into the coumarin ring system. *Answer 6.20 A AcAspGluVal 2 C AcAspGluVal 2 C B AcAspGluVal 2 C B 2 C 2 C 2 C B C 2 C 2 C 2 C B Introduction to Bioorganic Chemistry and Chemical Biology A6128 *Answer 6.22 er : : er er Introduction to Bioorganic Chemistry and Chemical Biology A6130

7 Introduction to Bioorganic Chemistry and Chemical Biology: Answers to Chapter 6 7 *Answer 6.25 The wildtype enzyme processes aspartate times faster. The 292D variant exhibits reversed selectivity, favoring arginine. owever, the 292 variant processes arginine more than about 10 4 times slower than the wildtype enzyme processes aspartate. 2 D292 K P 2 D Introduction to Bioorganic Chemistry and Chemical Biology A6133 *Answer 6.27 Cinnabaramide A is a strained βlactone, structurally similar to salinosporamide. The nucleophilic threonine of the proteasome reversibly attacks the βlactone. C 6 13 Ac C 2 C 6 13 Ac C 2 C 6 13 Ac C 2 C 6 13 : 3 C *Answer 6.28 A G C G Introduction to Bioorganic Chemistry and Chemical Biology A6135 A B Although some of the codons below are susceptible to C to A transversion, those mutations would be silent. Lys (AAA, AAG) t (AUG) Glu (GAA, GAG) Gly (GGU, GGC, GGA, GGG) Trp (UGG) Ile (AUU, AUC, AUA) Val (GUU, GUC, GUA, GUG) TP (UAG, UGA, UAA)

8 8 Introduction to Bioorganic Chemistry and Chemical Biology: Answers to Chapter 6 *Answer 6.30 ne way to approach this problem would be to examine each of the five histidine residues in neuropsin to see which one is close to an Asp and a er residue. The catalytic triad involves residues Asp57, is102, and er