The researchers begin to characterize the enzyme. (a) In the first experiment, w
ID: 544323 • Letter: T
Question
The researchers begin to characterize the enzyme. (a) In the first experiment, wth IEJ at 8 nM, they find (b) in anather experiment wth IEJ at3 nM and that the Vmmx is 3.32 .M s-1. Based on this experiment, what is the Keat for happyase"? [HAPPY] at 25 the researchers find that Vo=694 nM s-1, what is the measured Knof happyase* for its substrate HAPPY? Number Number c) Further research shows that the purified happyase used in the first two experiments was actually contaminated with a reversible inhibitor called ANGER. When ANGER is carefully removed from the happyase preparation, and the two experiments repeated, the measured Vmax in (a) is increased to 7.30 AM s-1, and the measured Km in (b) is now 23.0 For the inhibitor ANGER, calculate the values of d and . Number Number ' = = Please scroll down (d) Based on the information given above, what type of inhibitor is ANGER? O competitive inhibitor 0 uncompetitive inhibitor O mixed inhibitor VHintExplanation / Answer
Ans. #a. Kcat = Vmax / [E]t
= 3.32 uM s-1 / 8 nm ; [1 uM = 103 nM]
= (3.32 x 103 nM s-1) / 8 nM
= 415.0 s-1
Hence, Kcat for Happyase = 356.25 s-1
#b. The Vmax of a reaction mixture is always proportional to enzyme concentrations.
We have, Vmax at [E] of 8 nM = 3.32 uM s-1
Have to calculate Vmax at [E] = 3 nM
Now,
Vmax at [E] of 3 nM = Vmax at [E] of 8 nM x [E] of 3 nM
= (3.32 uM s-1 per 8 nM enzyme) x 3 nM
= (3.32 uM s-1 / 8 nM enzyme) x 3 nM
= 1.245 uM s-1
Hence, at [E] of 3 nM, the Vmax = 1.245 uM s-1 = 1245.0 nM s-1
# Calculate Km using MM equation- Vo = Vmax [S] / (Km + [S])
Re-arranging the above equation-
Km = (Vmax [S] / Vo) – [S]
Putting the values in above equation-
Km= (1245 nM s-1 x 25 uM / 694 nM s-1) – 25 uM
Or, Km = 44.85 uM – 25 uM = 19.85 uM
Hence, Km of Happyase = 19.85 uM
#c. So far we have-
Vmax (in presence of Anger, #a) = 3.32 uM s-1
Km (in presence of Anger, #b) = 19.85 uM
Vmax (Inhibitor free) = 7.30 uM s-1
Km (Inhibitor free) = 23.0 uM
# Now, using the formula-
Vmax (In presence of inhibitor) = (1 / a’) x Vmax (Inhibitor free)
Or, 3.32 uM s-1 = (1/ a’) x 7.30 uM s-1
Or, (1/ a’) = 3.32 uM s-1 / 7.30 uM s-1 = 0.454
Or, a’ = 1/ 0.454 = 2.20
Hence, a’ = 2.20
# Now, using the formula-
Km (in presence of inhibitor) = (a / a’) x Km (inhibitor free)
Or, 19.85 uM = (a/ 2.20) x 23.0 uM
Or, (a / 2.20) = 19.85 uM / 23.0 uM
Or, a = 0.863 x 2.20 = 1.899
Hence, a’ = 1.90
#d. Type of inhibitor: Note the following points-
I. Vmax is decreased in presence of inhibitor.
II. Km is also decreased in presence of inhibitor.
Decrease in both Vmax and Km is a characteristic of uncompetitive inhibition.
Therefore, ANGRY is an uncompetitive inhibitor.