Biochemistry
9th Edition
ISBN: 9781319114671
Author: Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Publisher: W. H. Freeman
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Chapter 10, Problem 11P
Interpretation Introduction
Interpretation:
The reason corresponding to the fact that the given analog PALA is used in the X-ray crystallographic study of ATCase instead of actual substrates is to be stated.
Concept introduction:
A molecule that disturbs or inhibits the activity of an enzyme is known as enzyme inhibitor. The enzyme inhibition stops the generation of enzyme-substrate complex whichin turn, inhibits the formation of the product.
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Synthetic stoichiometries. What is the stoichiometry of the synthesis of...
(a) ribose 5-phosphate from glucose 6- phosphate without the concomitant generation of NADPH?
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An enzyme-catalyzes the isomerization of substrate S to product P. The enzyme has a molecular weight of 120,000 g/mol. In assays using 1 μg of enzyme per assay the Km was 3 x 10^-3M and the Vmax was 2.75 μmole per minute. What would be the Kcat (turnover number or molecular activity) of the enzyme under these conditions?
2.75 min^-1?
3,300,000 min^-1?
330,000 s^-1?
19,800,000 min^-1?
5,500 s^-1?
not true about the Michaelis-Menten equation?
The equation that gives the rate, v, of an
the substrate concentration [S] is the Michaelis-Menten equation
= Vmax[S]/(Km + [S]), where V,
enzyme-catalyzed reaction for all values of
max and Km are constants. Which of the following is
a)
for [S] << Km, V = Vmax
applies to most enzymes, but allosteric enzymes have different kinetics
when [S] = Km, then v =
Vmax/2
gives the rate when the enzyme concentration, temperature, pH, and ionic
strength are constant
for very high values of [S], v approaches Vmax
e)
Which is correct about the constant Km in the Michaelis-Menten equation?
also called the catalytic constant or turnover number
equal to the number of product molecules produced per unit time when the
enzyme is saturated with substrate
it is the constant in the first order rate equation v = k[A]
it is the constant in the second order rate equation v =
equal to the substrate concentration at which the velocity or rate of a reaction is
½ the…
Chapter 10 Solutions
Biochemistry
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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