I would really appreciate some help with this problem. I will rate complete answ
ID: 1036891 • Letter: I
Question
I would really appreciate some help with this problem. I will rate complete answers.
1o 3. An enzyme (MW-20,250) catalyzes the hydrolysis of a newly discovered substrate. The k cat was found to be 2000 sec and Km0.05 M. Using this information answer the following questions. (12 points) answer the Golowring questis. 2poima a) What is the Vmax of this enzyme when enzyme concentration equal to 0.3 mg ml-1? b) When the above reaction was carried out in the presence of 2.5 ?? inhibitor, the Vmax is 0.03 M sec What kind of inhibitor is this? Justify your answer using calculated values and proper explanation.Explanation / Answer
Ans. #a. Step 1 : Calculate [E]
Given, [E] = 0.3 mg/ mL
= 0.0003 g / mL ; [1 mg = 0.001 g]
= (0.0003 g / 20250 g mol-1) / mL ; [Moles = Mass / MW]
= 1.4815 x 10-8 mol / mL
= 1.4815 x 10-8 mol / 0.001 L ; [1 mL = 0.001 L]
= 1.4815 x 10-5 mol / L
= 1.4815 x 10-5 M
# Step 2: Kcat = Vmax / [E]
Or, 2000 sec-1 = Vmax / (1.4815 x 10-5 M)
Or, Vmax = 2000 sec-1 x 1.4815 x 10-5 M
Hence, Vmax = 0.02963 M sec-1 = 0.03 M sec-1
#b. Given, Vmax in presence of inhibitor = 0.03 M sec-1
Vmax in absence of inhibitor = 0.03 M sec-1 (calculated in #a above)
# Comparing the two values, Vmax remains unaffected of the presence of inhibitor. It is the characteristic of a competitive inhibitor.
Therefore, the inhibitor is a competitive inhibitor.