A research group discovers a new version of happyase which t

A research group discovers a new version of happyase, which they call happyase*, that catalyzes the chemical reaction HAPPY. -SAD The researchers begin to characterize the enzyme (a) In the first experiment, with [EJ at 5 nM, they find that the Vmax is 2.61 LM s. Based on this experiment, what is the kcat for happyase*? b In another experiment, with [Etl at 1 nM and [HAPPY] at 40 M, the researchers find that Vo = 389 nM s-1 What is the measured Kn of happyase for its substrate HAPPY? Number Number kcat = 11522 K,\" =| | 13.68 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 M s-1, and the measured Km in (b) is now 15.3 values of and For the inhibitor ANGER, calculate the Number Number Please scroll

Solution

Ans. #a. Kcat = Vmax / [E]_t

= 2.61 uM s^-1 / 7 nm                                                ; [1 uM = 10³ nM]

= (2.37 x 10³ nM s^-1) / 5 nM

= 522 s^-1

Hence, Kcat for Happyase = 338.57 s^-1

#b. The Vmax of a reaction mixture is always proportional to enzyme concentrations.

We have,        Vmax at [E] of 5 nM = 2.61 uM s^-1

Have to calculate Vmax at [E] = 1 nM

Now,

            Vmax at [E] of 1 nM = Vmax at [E] of 5 nM x ([E] of 1 nM)

                                                = (2.61 uM s^-1 / 5 nM E) x 1 nM E

                                                = 0.522 uM s^-1

Hence, at [E] of 1 nM, the Vmax = 0.522 uM s^-1 = 522.0 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= (522nM s^-1 x 40 uM / 389 nM s^-1) – 40 uM

Or, Km = 13.679 uM

Hence, Km of Happyase = 13.68 uM

#c. So far we have-

            Vmax (in presence of Anger, #a) = 2.61 uM s^-1 = Vmax,app

Km (in presence of Anger, #b) = 13.68 uM = Km,app

Vmax (Inhibitor free) = 7.30 uM s^-1

Km (Inhibitor free) = 15.3 uM

# Now, using the formula-

            a’ = Vmax / Vmax,app = Vmax (In presence of inhibitor)/ Vmax (Inhibitor free)

            Or, a’ = 7.30 uM s^-1 / 2.61 uM s^-1

            Hence, a’ = 2.797

# Now, using the formula (for a mixed inhibitor)-

            a = (a’ x Km,app) / Km = (2.797 x 13.68 uM) / 15.3 uM = 2.50