Question

Two different enzymes catalyze the same reaction and both exhibit the same Vmax. When Enzyme A was run with a 40 uM substrate, the initial rate (Vo) was 10 uM/min, and when g it was run with a 4 mM substrate, the Vo was 20 uM/min. Estimate the approximate Vmax and Km of Enzyme A. When Enzyme B was run with 80 uM substrate, the initial rate (Vo) was 10 uM per minute. Estimate the Km of Enzyme B.

Answer 1

Answer:

a) k_m = 4.08 uM

  V_{max} = 20.07 uM/min

b) k_m = 8.16 uM

Explanation:

Given that:

For Enzyme A:

the substrate concentration [S] = 40 uM

the initial velocity rate v = 10 uM/min

when it was 4mM, v = 20 uM/min

i.e.

at 4mM = 4000 uM;

Using Michealis -menten equation;

when v = 10

$V = \dfrac{V_{max}[S]}{k_m+[S]}$

∴

$10 = \dfrac{V_{max}\times 40}{k_m + 40}$

$10 (k_m + 40) = V_{max}40$

$40V_{max} -10k_m = 400 --- (1)$

when v= 20

$20= \dfrac{V_{max}\times 4000}{k_m + 4000}$

$20 (k_m + 4000) = V_{max}4000$

$4000V_{max} -20k_m = 8000 --- (2)$

equating equation (1) and (2):

$40V_{max} -10k_m = 400 --- (1)$

$4000V_{max} -20k_m = 8000 --- (2)$

let multiply equation (1) by 100 and equation (2) by 1

4000V_{max} - 1000K_m = 4000

4000V_{max} - 20 k_m = 8000    

  0        -980k_m = 4000

k_m = 4000/-980

k_m = 4.08 uM

replacing the value of k_m into equation (1)

40{V_max } - 10(4.08) = 400

40{V_max } - 40.8 = 400

40{V_max } = 400 +  40.8

40{V_max } = 440.8

V_{max} = 440.8/40

V_{max} = 11.02 uM/min

b)

Since V_{max} of A ie equivalent to that of B; then:

V_{max} of B = 11.02 uM/min

Here;

[S] = 80  uM

V = 10 uM/min

∴

$10 = \dfrac{11.02 \times 80}{k_m + 80 }$

10(k_m +80) = 881.6

10k_m = 881.6  - 800

10k_m = 81.6

k_m = 81.6/10

k_m = 8.16 uM