S-63Enzymeschapter61. Keeping the Sweet Taste of Corn The sweet taste of freshly picked corn (maize) is due to the high level of sugar in the kernels. Store-bought corn (several days after picking) is not as sweet,because about 50% of the free sugar is converted to starch within one day of picking. To preserve thesweetness of fresh corn, the husked ears can be immersed in boiling water for a few minutes(“blanched”) then cooled in cold water. Corn processed in this way and stored in a freezer maintainsits sweetness. What is the biochemical basis for this procedure?Answer After an ear of corn has been removed from the plant, the enzyme-catalyzed conver-sion of sugar to starch continues. Inactivation of these enzymes slows down the conversion toan imperceptible rate. One of the simplest techniques for inactivating enzymes is heat denatura-tion. Freezing the corn lowers any remaining enzyme activity to an insignificant level.2. Intracellular Concentration of Enzymes To approximate the actual concentration of enzymes in abacterial cell, assume that the cell contains equal concentrations of 1,000 different enzymes in solutionin the cytosol and that each protein has a molecular weight of 100,000. Assume also that the bacterialcell is a cylinder (diameter 1.0 mm, height 2.0 mm), that the cytosol (specific gravity 1.20) is 20% solu-ble protein by weight, and that the soluble protein consists entirely of enzymes. Calculate the averagemolar concentration of each enzyme in this hypothetical cell.Answer There are three different ways to approach this problem.(i) The concentration of total protein in the cytosol is 0.24 105mol/mL  2.4 103MThus, for 1 enzyme in 1,000, the enzyme concentration is 2.4 106M(ii) The average molar concentration Volume of bacterial cytosol  pr2h  (3.14)(0.50 mm)2(2.0 mm)  1.6 mm3 1.6 1012cm3 1.6 1012mL 1.6 1015LAmount (in moles) of each enzyme in cell is  3.8 1021molAverage molar concentration  2.4 106mol/L  2.4 106M3.8 1021mol1.6 1015L(0.20)(1.2 g/cm3)(1.6 mm3)(1012cm3/mm3)(100,000 g/mol)(1000)moles of each enzyme in cellvolume of cell in liters2.4 103M1000(1.2 g/mL)(0.20)100,000 g/mol2608T_ch06sm_S63-S77 2/1/08 7:34AM Page S-63 ntt 102:WHQY028:Solutions Manual:Ch-06:S-64 Chapter 6 Enzymes(iii) Volume of bacterial cytosol  pr2h (3.14)(0.50 mm)2(2.0 mm)  1.6 mm3 1.6 1012mLWeight of cytosol  (specific gravity)(volume) (1.2 g/mL)(1.6 1012mL)  1.9 1012gAverage weight of each protein (1 in 1,000, 20% wt/wt protein) (1.9 1012g)(0.20)/(1,000)  3.8 1016gAverage molar concentration of each protein (average weight)/(Mr)(volume) (3.8 1016g)/(105g/mol)(1.6 1012mL)(1 L/1000 mL) 2.4 106mol/L  2.4 106M3. Rate Enhancement by Urease The enzyme urease enhances the rate of urea hydrolysis at pH 8.0and 20 C by a factor of 1014. If a given quantity of urease can completely hydrolyze a given quantity ofurea in 5.0 min at 20 C and pH 8.0, how long would it take for this amount of urea to be hydrolyzedunder the same conditions in the absence of urease? Assume that both reactions take place in sterilesystems so that bacteria cannot attack the urea.AnswerTime to hydrolyze urea 9.5 108yr 950 million years!4. Protection of an Enzyme against Denaturation by Heat When enzyme solutions are heated,there is a progressive loss of catalytic activity over time due to denaturation of the enzyme. A solutionof the enzyme hexokinase incubated at 45 C lost 50% of its activity in 12 min, but when incubated at45 C in the presence of a very large concentration of one of its substrates, it lost only 3% of its activ-ity in 12 min. Suggest why thermal denaturation of hexokinase was retarded in the presence of one ofits substrates.Answer One possibility is that the ES complex is more stable than the free enzyme. This im-plies that the ground state for the ES complex is at a lower energy level than that for the freeenzyme, thus increasing the height of the energy barrier to be crossed in passing from thenative to the denatured or unfolded state.An alternative view is that an enzyme denatures in two stages: reversible conversion ofactive native enzyme (N) to an inactive unfolded state (U), followed by irreversible conversionto inactivated enzyme (I):N 88zy88 U 88nIIf substrate, S, binds only to N, saturation with S to form NS would leave less free N availablefor conversion to U or I, as the N 88zy88 U equilibrium is perturbed toward N. If N but not NS isconverted to U or I, then substrate binding will cause stabilization.5. Requirements of Active Sites in Enzymes Carboxypeptidase, which sequentially removescarboxyl-terminal amino acid residues from its peptide substrates, is a single polypeptide of 307 aminoacids. The two essential catalytic groups in the active site are furnished by Arg145and Glu270.(a) If the carboxypeptidase chain were a perfect a helix, how far apart (in Å) would Arg145andGlu270be? (Hint: see Fig. 4–4a.)(b) Explain how the two amino acid residues can catalyze a reaction occurring in the space of a fewangstroms.(5.0 min)(1014)(60 min/hr)(24 hr/day)(365 days/yr)2608T_ch06sm_S63-S77 2/1/08 7:34AM Page S-64 ntt 102:WHQY028:Solutions Manual:Ch-06:Answer(a) Arg145is separated from Glu270by 270  145  125 amino acid (AA) residues. FromFigure 4–4a we see that the a helix has 3.6 AA/turn and increases in length along themajor axis by 5.4 Å/turn. Thus, the distance between the two residues is 190 Å(b) Three-dimensional folding of the enzyme brings the two amino acid residues into closeproximity.6. Quantitative Assay for Lactate Dehydrogenase The muscle enzyme lactate dehydrogenase cat-alyzes the reaction(125 AA)(5.4 Å/turn)3.6 AA/turnChapter 6 Enzymes S-65Pyruvate LactateHNADHCH3C COO O NADCH3COO HCOHNADH and NADare the reduced and oxidized forms, respectively, of the coenzyme NAD. Solutions ofNADH, but not NAD, absorb light at 340 nm. This property is used to determine the concentration ofNADH in solution by measuring spectrophotometrically the amount of light absorbed at 340 nm by thesolution. Explain how these properties of NADH can be used to design a quantitative assay for lactatedehydrogenase.Answer The reaction rate can be measured by following the decrease in absorption