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
#SPJ1
Answer:
28,436,297,825
30,943,267,101
45,000,000,000
Step-by-step explanation:
ur welcome
Given:
The side of square = 12 in.
Scale factor of enlargement = 3 in : 2 m
To find:
The proportion that is use to solve the side length, x, of the enlarged square.
Solution:
Let, the side of length of enlarged square = x m
In case of enlargement the corresponding sides are proportional.
Divide both sides by 3.
Therefore, the required proportion is and the side length of the square after enlargement is 8 m.
Hello,
==================================
