The KM for the reaction of chymotrypsin with Substrate A is 8.8 x 10-4 M, while theKM for the reaction of chymotrypsin with Substrate B is 8.7 x 10-3 M. Which of thefollowing statements are likely true?Chymotrypsin has a higher apparent affinity for Substrate A.The Vmax would be higher in the presence of Substrate A.The kcat would be higher with Substrate A.The Vmaxwould be higher in the presence of Substrate B.Two of the above are true.

The Michaelis Menten constant (KM) can be used as a measure of the enzyme's affinity for its…

Answered: The KM for the reaction of chymotrypsin…

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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An enzyme that follows simple Michaelis–Menten kinetics has an initial reaction velocity of 10 µmol⋅min−110 µmol⋅min−1 when the substrate concentration is five times greater than the ?M.KM. What is the ?maxVmax of this enzyme?
Which substrate is the best substrate for chymotrypsin? O N-acetylphenylalanine ethyl ester with Km of 8.2 x 102 M and keat of 2.1 x 102 sec 1. O N-acetylmethionine ethyl ester with Km of 12.3 x 102 M and keat of 2.1 x 1o sec1. O N-acetylleucine ethyl ester with Km of 8.2 x 10o2 M and kcat of 4.1 x 102 sec1. O N-acetylalanine ethyl ester with Km of 4.1 x 101 M and kcat of 4.1 x 101 sec.
TABLE 3-LACTATE PRODUCTION IN FORTIFIED HEMOLYSATES OF HUMAN ERYTHROCYTES* Substrate Glucose Glucose Lactate production† No. of experiments pH 6 7.1 2.03 ± 0.91 6 7.8 4.76 ± 1.09 7-1 10-73 +1-88 5 7.8 12.34 ±2.92 5 7.0 7-15±0.73 5 7-7 (b)( ) In mature erythrocytes (red blood cells) the end product of glycolysis is lactate because of the absence of mitochondria. On the right is a table comparing the rate of lac- tate production in hemolysates (lysed cells) of human RBCs as a function of pH with dif- ferent substrates introduced into the glyco- lytic pathway. The hemolysate was fortified with 30 μmoles substrate, 7.5 μmoles MgCl2, 10 μmoles disodium phosphate, 1.5 μmoles NAD and 5 μmoles ATP in a volume of 5 mL. The rate of lactate production is given as μmoles of lactate/g Hb/hr at 37° C, buffered to either pH 7.1 or 7.8, as indicated. According to the results in the table which glycolytic enzyme is rate-limiting? Explain. Glucose-6-phosphate Glucose-6-phosphate Fructose-1,6-diphosphate…
. The following data, presented by G. Bowes and W. L. Ogre in J. Biol. Chem. (1972) 247:2171–2176, describe the relative rates of incorpo- ration of CO, by rubisco under N, and under pure O,. Decide whether O, is a competitive or uncompetitive inhibitor. [CO,] (mM) Under N2 Under O, 0.20 16.7 10 0.10 12.5 5.6 0.067 8.3 4.2 0.050 7.1 3.2
Aerobic degradation of an organic compound by mixed cultureof organism in wastewater can be represented by following reaction. C3H6O3 + a O2 + b NH3 → c C5H7NO2 + d H2o + e CO2 A. Determine a, b, c, d and e, if YX/S = 0.4 d X/g S. B. Determine the yield coefficients YX/O2 and YX/NH3. C. Determine the degree of reductions for the substrate, bacteria and RQ for the organisms
Acetazolamide is a drug which inhibits carbonic anhydrase. Carbonic anhydrase participates in regulation of the pH and bicarbonate content of a number of body fluids. Figure 2 shows the experimental curve of initial reaction velocity (as percentage of Vma) versus [S] (concentration) for the carbonic anhydrase reaction. The graph also shows the curve in the presence of acetazolamide. 100 No inhibitor 50 Acetazolamide 0.2 0.4 0.6 0.8 [S] (mM) Figure 2 (i) Compare the maximal velocities and Michaelis Menten constants of the enzyme in the absence and the presence of the inhibitor acetazolamide. Determine the nature of inhibition by acetazolamide. Explain your answer. (ii) Name TWO (2) other types of inhibitions besides the inhibition shown by acetazolamide. Sketch a graph of V versus [S] showing curves in the absence of an inhibitor and in the presence of the types of inhibitors not shown by acetazolamide. ("AJO %) A
A particular reaction has a ΔG‡ of 30.0 kJ mol-1 at 25.0 °C. In the presence of an enzyme, the same reaction has a ΔG‡ of 1.50 kJ mol-1 at the same temperature. Calculate the rate enhancement of this enzyme. (R = 8.3145 J mol-1 K-1)
An inhibitor "I" is added to the enzymatic reaction at a concentration of 1.0 g/L. The data obtained are shown in the table below. No Inhibitor With Inhibitor [S](g/L) v ([E0] = 0,015 g/L) (g/L.min) v ([E0] = 0,015 g/L)(g/L.min) 40 2,28 1,818 20 1,74 1,316 13,4 1,4 0,986 10 1,18 0,76 8 1 0,62 (a)What kind of inhibition occurred?(b)Determine Ki (in g/L) for the inhibition of the previous exercise.
8L.3.4
Conversion of F1,6BP to GA3P and DHAP by aldolase is striking in that it is strongly unfavorable when all species are at 1 M (∆G°’ = +22.8 kJ/mol) but favorable when the species are at their physiological conditions (∆G = -5.9 kJ/mol). Based on ∆G°’ and ∆G of aldolase and TIM, and using a reasonable physiological value of 15 mM for the concentration of F1,6BP, calculate the concentrations of GA3P and DHAP under physiological conditions (use T = 37°C). Why is the reaction so much more favorable under these conditions than when all species are at 1 M?
The Lineweaver - Burk plot (Figure 1) shows an enzyme-catalyzed reaction in the absence and presence of 0.1µM inhibitor (ketoconazole). O Estimate Vmax and Km in the absence and presence of the inhibito.. (ii) Determine the type of inhibition shown by the inhibitor. Explain. 0.1- 0.08 0.06 - With inhibitor 0.04 0.02 Without inhibitor 0.4 -0.2 -0.02 0.2 0.4 0.6 0.8 1.0 1/{S\(&M-1) 1/vo (pmol-11 min)
Identify the enzymes in Table, whose catalytic effi ciencies are near the diff usion-controlled limit.

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