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17 Nov 2019
An enzymatic reaction was analyzed for rate versus substrate concentration. Roughly estimate the Vmax and Km from these measurements. (Hint: Slope = rise/run and remember what Km represents!)
vo(nM/s) [S] (nM)
30 100,000
29.5 10,000
29 1,000
28 200
25 150
20 100
10 50
5 20
a. Vmax = 10 nM/s, Km = 20 nM
b. Vmax = 20 nM/s , Km = 50 nM
c. Vmax = 40 nM/s , Km = 150 nM
d. Vmax = 70 nM/s, Km = 250 nM
ANSWER is (c), how?
An enzymatic reaction was analyzed for rate versus substrate concentration. Roughly estimate the Vmax and Km from these measurements. (Hint: Slope = rise/run and remember what Km represents!)
vo(nM/s) [S] (nM)
30 100,000
29.5 10,000
29 1,000
28 200
25 150
20 100
10 50
5 20
a. Vmax = 10 nM/s, Km = 20 nM
b. Vmax = 20 nM/s , Km = 50 nM
c. Vmax = 40 nM/s , Km = 150 nM
d. Vmax = 70 nM/s, Km = 250 nM
ANSWER is (c), how?
Bunny GreenfelderLv2
22 Sep 2019