BIOCHEMISTRY (LOOSELEAF)-W/ACCESS
9th Edition
ISBN: 9781319425784
Author: BERG
Publisher: Macmillan Higher Education
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Question
Chapter 10, Problem 4P
Interpretation Introduction
Interpretation:
The
Concept introduction:
Enzymes being the catalysts increase the
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[s] mM
V (M/s)
Uninhibited
0.333
1.65 x 107
1.05 x 107
V (M/s) x 10'
Inhibitor A
V (M/s) x 107
Inhibitor B
0.794 x 107
0.40
1.86 x 107
1.21 x 107
0.893 x 107
0.50
2.13 x 107
1.43 x 107
1.02 x 107
0.666
2.49 x 107
1.74 x 107
1.19 x 107
1.0
2.99 x 107
2.22 x 107
1.43 x 107
2.0
3.72 x 107
3.08 x 107
1.79 x 107
For a Michaelis-Menten reaction, k₁-5 x 10'/M-s, k.-2 x 10%/s, and k₂-4 x 10²/s.
a)
Calculate the Ks and KM for this reaction.
b)
Does substrate binding achieve equilibrium or steady state?
Assume that an enzyme-catalyzed reaction follows the scheme shown:
E+S SES
→E + P
k₁ = 1 x 109/M-s k-1=2.5 x 10%/s k₂ = 3.4 x 107/s
What is the dissociation constant for the enzyme-substrate, K,?
What is the Michaelis constant, Km, for this enzyme?
What is the turnover number, Keat, for this enzyme?
What is the catalytic efficiency for the enzyme?
If the initial Et concentration is 0.25mM, what is Vmax?
Chapter 10 Solutions
BIOCHEMISTRY (LOOSELEAF)-W/ACCESS
Ch. 10 - Prob. 1PCh. 10 - Prob. 2PCh. 10 - Prob. 3PCh. 10 - Prob. 4PCh. 10 - Prob. 5PCh. 10 - Prob. 6PCh. 10 - Prob. 7PCh. 10 - Prob. 8PCh. 10 - Prob. 9PCh. 10 - Prob. 10P
Ch. 10 - Prob. 11PCh. 10 - Prob. 12PCh. 10 - Prob. 13PCh. 10 - Prob. 14PCh. 10 - Prob. 15PCh. 10 - Prob. 16PCh. 10 - Prob. 17PCh. 10 - Prob. 18PCh. 10 - Prob. 19PCh. 10 - Prob. 20PCh. 10 - Prob. 21PCh. 10 - Prob. 22PCh. 10 - Prob. 23PCh. 10 - Prob. 24PCh. 10 - Prob. 25PCh. 10 - Prob. 26PCh. 10 - Prob. 27PCh. 10 - Prob. 28PCh. 10 - Prob. 29PCh. 10 - Prob. 30PCh. 10 - Prob. 31PCh. 10 - Prob. 32PCh. 10 - Prob. 33PCh. 10 - Prob. 34PCh. 10 - Prob. 35PCh. 10 - Prob. 36PCh. 10 - Prob. 37PCh. 10 - Prob. 38PCh. 10 - Prob. 39PCh. 10 - Prob. 40PCh. 10 - Prob. 41PCh. 10 - Prob. 42PCh. 10 - Prob. 43P
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