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:
$18.5685 or $18.56
Step-by-step explanation:
The answer is $18.56 because 123.79 multiplied by 15% is $18.56
Answer:
Height of tree= 13.8m
Step-by-step explanation:
When finding the value of a in your working, I suggest to leave the answer to 5 significant figures so that your final answer is more accurate. The value of a should be 26.048m to 5 s.f.
Then multiplying 26.048 with sin 32° would give you 13.8m too when rounded off to 3 significant figures :)
I have attached my working too.
