Question

The following data were obtained for the variation of initial velocity and substrate for a reaction catalyzed by chymotrypsin (5nM) at pH 8, 37 °C.

1. Using a linear plot, determine the KM and Vmax of the chymotrypsin given the data below: (12)

[S] (μM) V0 (μmol/min)

0    0
0.5 200

1.0    400
1.5    580

2.0    750
2.5    840

3.0    860
4.0    875

6.0    890

The kcat/KM parameter is a measure of catalytic efficiency.

2. Under what experimental conditions can this parameter be determined? Show the mathematical basis to your answer (5)
3. Select the appropriate data from the table above, and plot a new graph to determine the kcat/KM of chymotrypsin. (15)
How would the rate of product formation change in the following scenarios:
4.1 the substrate concentration were doubled? (2)
4.2. the enzyme concentration were doubled? (2)
4.3. temperature of the reaction was halved? (2)

Answer 1

Ans:

From the plot V₀ vs [S] below, K_m and V_max are calulated

V_max of the chymotrypsin = 890μmol/min

K_M of the chymotrypsin = 1.2 μM

2. Under what experimental conditions can this parameter be determined? Show the mathematical basis to your answer (5)

Ans:

K_cat = V_max/E_t

V_{max =} 890μmol/min

E_{t =} 5nM or 5 x 10^-3 μmol

Substituting values in equation below

K_cat = V_max/E_t

K_cat = 890μmol/min / 5 x 10^-3 μmol

K_cat = 178 x 10³ min^-1

How would the rate of product formation change in the following scenarios:
4.1 the substrate concentration were doubled? (2)

Ans: Inreases rate of product formation

4.2. the enzyme concentration were doubled? (2)

Ans: Inreasing enzyme concentration does not alter the rate of product formation, since the reaction has already achieved max velocity.

4.3. temperature of the reaction was halved? (2)

Ans: Deacresing temperature slows down the rate of product formation as maximum product formation is achieved at 37⁰C.