UMD BCHM 461 - 203386054-Lehninger-Solutions-Ch06 (15 pages)

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203386054-Lehninger-Solutions-Ch06



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203386054-Lehninger-Solutions-Ch06

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Pages:
15
School:
University of Maryland, College Park
Course:
Bchm 461 - Biochemistry I
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2608T ch06sm S63 S77 2 1 08 7 34AM Page S 63 ntt 102 WHQY028 Solutions Manual Ch 06 chapter 6 Enzymes 1 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 the sweetness 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 maintains its sweetness What is the biochemical basis for this procedure Answer After an ear of corn has been removed from the plant the enzyme catalyzed conversion of sugar to starch continues Inactivation of these enzymes slows down the conversion to an imperceptible rate One of the simplest techniques for inactivating enzymes is heat denaturation 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 a bacterial cell assume that the cell contains equal concentrations of 1 000 different enzymes in solution in the cytosol and that each protein has a molecular weight of 100 000 Assume also that the bacterial cell is a cylinder diameter 1 0 mm height 2 0 mm that the cytosol specific gravity 1 20 is 20 soluble protein by weight and that the soluble protein consists entirely of enzymes Calculate the average molar 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 1 2 g mL 0 20 0 24 10 5 mol mL 2 4 10 3 M 100 000 g mol Thus for 1 enzyme in 1 000 the enzyme concentration is 2 4 10 3 M 2 4 10 6 M 1000 moles of each enzyme in cell ii The average molar concentration volume of cell in liters pr2h 3 14 0 50 mm 2 2 0 mm 1 6 mm3 1 6 10 12 cm3 1 6 10 12 mL 1 6 10 15 L Amount in moles of each enzyme in cell is Volume of bacterial cytosol 0 20 1 2 g cm3 1 6 mm3 10 12 cm3 mm3 3 8 10 21 mol 100 000 g mol 1000 3 8 10 21 mol Average molar concentration 1 6 10 15 L 2 4 10 6 mol L 2 4 10 6 M S 63 2608T ch06sm S63 S77 S 64 2 1 08 7 34AM Page S 64 ntt 102 WHQY028 Solutions Manual Ch 06 Chapter 6 Enzymes iii Volume of bacterial cytosol pr2h 3 14 0 50 mm 2 2 0 mm 1 6 mm3 1 6 10 12 mL Weight of cytosol specific gravity volume 1 2 g mL 1 6 10 12 mL 1 9 10 12 g Average weight of each protein 1 in 1 000 20 wt wt protein 1 9 10 12 g 0 20 1 000 3 8 10 16 g Average molar concentration of each protein average weight Mr volume 3 8 10 16 g 105 g mol 1 6 10 12 mL 1 L 1000 mL 2 4 10 6 mol L 2 4 10 6 M 3 Rate Enhancement by Urease The enzyme urease enhances the rate of urea hydrolysis at pH 8 0 and 20 C by a factor of 1014 If a given quantity of urease can completely hydrolyze a given quantity of urea in 5 0 min at 20 C and pH 8 0 how long would it take for this amount of urea to be hydrolyzed under the same conditions in the absence of urease Assume that both reactions take place in sterile systems so that bacteria cannot attack the urea Answer Time to hydrolyze urea 5 0 min 1014 60 min hr 24 hr day 365 days yr 9 5 108 yr 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 solution of the enzyme hexokinase incubated at 45 C lost 50 of its activity in 12 min but when incubated at 45 C in the presence of a very large concentration of one of its substrates it lost only 3 of its activity in 12 min Suggest why thermal denaturation of hexokinase was retarded in the presence of one of its substrates Answer One possibility is that the ES complex is more stable than the free enzyme This implies that the ground state for the ES complex is at a lower energy level than that for the free enzyme thus increasing the height of the energy barrier to be crossed in passing from the native to the denatured or unfolded state An alternative view is that an enzyme denatures in two stages reversible conversion of active native enzyme N to an inactive unfolded state U followed by irreversible conversion to inactivated enzyme I 88z U 88n I N y88 If substrate S binds only to N saturation with S to form NS would leave less free N available 88z U equilibrium is perturbed toward N If N but not NS is for conversion to U or I as the N y88 converted to U or I then substrate binding will cause stabilization 5 Requirements of Active Sites in Enzymes Carboxypeptidase which sequentially removes carboxyl terminal amino acid residues from its peptide substrates is a single polypeptide of 307 amino acids The two essential catalytic groups in the active site are furnished by Arg145 and Glu270 a If the carboxypeptidase chain were a perfect a helix how far apart in would Arg145 and Glu270 be Hint see Fig 4 4a b Explain how the two amino acid residues can catalyze a reaction occurring in the space of a few angstroms 2608T ch06sm S63 S77 2 1 08 7 34AM Page S 65 ntt 102 WHQY028 Solutions Manual Ch 06 Chapter 6 Enzymes S 65 Answer a Arg145 is separated from Glu270 by 270 145 125 amino acid AA residues From Figure 4 4a we see that the a helix has 3 6 AA turn and increases in length along the major axis by 5 4 turn Thus the distance between the two residues is 125 AA 5 4 turn 190 3 6 AA turn b Three dimensional folding of the enzyme brings the two amino acid residues into close proximity 6 Quantitative Assay for Lactate Dehydrogenase The muscle enzyme lactate dehydrogenase catalyzes the reaction O CH3 C OH COO NADH H CH3 C COO NAD H Pyruvate Lactate NADH and NAD are the reduced and oxidized forms respectively of the coenzyme NAD Solutions of NADH but not NAD absorb light at 340 nm This property is used to determine the concentration of NADH in solution by measuring spectrophotometrically the amount of light absorbed at 340 nm by the solution Explain how these …


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