Suppose you have an EMP. You install the EMP to a temporary data distiller. The data you get back reads simply static.
I believe it relates because as we know the cell membrane is permeable meaning that allows oxygen, carbohydrates, food to enter the cell but it doesn't allow the entrance of large molecules that harm the cells.
After all the movement around it, matters because for example without food the mitochondria won't be able to create the energy and then with that energy, the nucleus to use it to complete the main the activities of the cell.
Hopefully this what you meant when you asked the question. Have a nice day/evening
Touch,pressure,temperature, and pain.
Answer: through somatic cells or inherited through the germaine.
Explanation: My best guess
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)
∴



when v= 20



equating equation (1) and (2):


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(k_m +80) = 881.6
10k_m = 881.6 - 800
10k_m = 81.6
k_m = 81.6/10
k_m = 8.16 uM