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
[tex]V = \dfrac{V_{max}[S]}{k_m+[S]}[/tex]
∴
[tex]10 = \dfrac{V_{max}\times 40}{k_m + 40}[/tex]
[tex]10 (k_m + 40) = V_{max}40[/tex]
[tex]40V_{max} -10k_m = 400 --- (1)[/tex]
when v= 20
[tex]20= \dfrac{V_{max}\times 4000}{k_m + 4000}[/tex]
[tex]20 (k_m + 4000) = V_{max}4000[/tex]
[tex]4000V_{max} -20k_m = 8000 --- (2)[/tex]
equating equation (1) and (2):
[tex]40V_{max} -10k_m = 400 --- (1)[/tex]
[tex]4000V_{max} -20k_m = 8000 --- (2)[/tex]
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
∴
[tex]10 = \dfrac{11.02 \times 80}{k_m + 80 }[/tex]
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
