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:
A= 14 B=17 C=20 D=26
Step-by-step explanation:
Not 100% sure this is what you needed but this is my try.
The y-intercept of the trend line is (0,28). The slope of this line is
28-0
m = --------- = 2
14-0
Thus, the equation of the trend line, in variables K and J, is
K = 2J + 28
Answer:
????
Step-by-step explanation:
number 1 could be 2 different answers
Step-by-step explanation:
- In the line 2, only variables were operated with a reduction of similar terms in each side of the equation, giving as result -4x at one side, and 2x at the other side.
- In the line 3, terms with variables were joined together in one side of the equation, and then, they were operate it, given as result -6X.
