N-acetyltyrosine ethyl ester may likely have a higher Vmax.
<h3>How are the Km and Vmax of an enzyme and substrate related?</h3>
Km is inversely proportional enzyme affinity for substrate. The lower the Km, the higher the enzyme affinity for substrate and vice versa.
Vmax is the maximum velocity of the reaction when the enzyme is fully saturated with substrate.
Based on the Km values, chymotrypsin has higher affinity for N-acetyltyrosine ethyl ester, thus, N-acetyltyrosine ethyl ester may likely have a higher Vmax.
In conclusion, the Vmax is the maximum velocity of an enzyme-catalyzed reaction.
<em>Note that the complete question is given below:</em>
<em>The Km for the reaction of chymotrypsin with N-acetylvaline ethyl ester is 8.8 x 10-2 M, and the Km for the reaction of chymotrypsin with N-acetyltyrosine ethyl ester is 6.6 x 10-4 M.</em>
<em>Which of the following substrates is likely to give a higher value for Vmax?</em>
Learn more about Km and Vmax at: brainly.com/question/16108691
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Answer:
5 apples
Step-by-step explanation:
20 - 15 = 5
The regression equation is in the form y = mx+b where m = 938 is the slope
The slope of 938 means that each year the tuition is increasing by roughly $938.
The answer is choice C.
You have to do whats in the parentheses from left to right then subtract the answer you get by 20
(9x10) - (30+30)
90 - 60 = 30
8 x (12+5) -7^2
17
17-49
18 x -32
= -576.
You have to use pemdas for your check off
P=parentheses
E=exponet
M=multiplication
D=division
A=add
S=subtrabt
HOPE IM CORRECT
Answer:
B
Step-by-step explanation:
