Well-adjusted barometer can help you predict changing weather day to day because it detects minute changes in the air pressure.
The single type of bacteria is not adequate to make any conclusions.
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]}](https://tex.z-dn.net/?f=V%20%3D%20%5Cdfrac%7BV_%7Bmax%7D%5BS%5D%7D%7Bk_m%2B%5BS%5D%7D)
∴
![10 = \dfrac{V_{max}\times 40}{k_m + 40}](https://tex.z-dn.net/?f=10%20%3D%20%5Cdfrac%7BV_%7Bmax%7D%5Ctimes%2040%7D%7Bk_m%20%2B%2040%7D)
![10 (k_m + 40) = V_{max}40](https://tex.z-dn.net/?f=10%20%28k_m%20%2B%2040%29%20%3D%20V_%7Bmax%7D40)
![40V_{max} -10k_m = 400 --- (1)](https://tex.z-dn.net/?f=40V_%7Bmax%7D%20-10k_m%20%3D%20400%20---%20%281%29)
when v= 20
![20= \dfrac{V_{max}\times 4000}{k_m + 4000}](https://tex.z-dn.net/?f=20%3D%20%5Cdfrac%7BV_%7Bmax%7D%5Ctimes%204000%7D%7Bk_m%20%2B%204000%7D)
![20 (k_m + 4000) = V_{max}4000](https://tex.z-dn.net/?f=20%20%28k_m%20%2B%204000%29%20%3D%20V_%7Bmax%7D4000)
![4000V_{max} -20k_m = 8000 --- (2)](https://tex.z-dn.net/?f=4000V_%7Bmax%7D%20-20k_m%20%3D%208000%20---%20%282%29)
equating equation (1) and (2):
![40V_{max} -10k_m = 400 --- (1)](https://tex.z-dn.net/?f=40V_%7Bmax%7D%20-10k_m%20%3D%20400%20---%20%281%29)
![4000V_{max} -20k_m = 8000 --- (2)](https://tex.z-dn.net/?f=4000V_%7Bmax%7D%20-20k_m%20%3D%208000%20---%20%282%29)
let multiply equation (1) by 100 and equation (2) by 1
4000V_{max} - 1000K_m = 4000
<u>4000V_{max} - 20 k_m = 8000 </u>
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 }](https://tex.z-dn.net/?f=10%20%3D%20%5Cdfrac%7B11.02%20%5Ctimes%2080%7D%7Bk_m%20%2B%2080%20%7D)
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