biochem Please type answers instead of hand writing for me to be able to read. t
ID: 201843 • Letter: B
Question
biochem
Please type answers instead of hand writing for me to be able to read. thanks!
Electron transport translocates protons across the mitochondrial membrane, establishing a pH gradient. The tendency of protons to diffuse back into the matrix is the driving force for ATP synthesis by ATP synthase. During oxidative phosphorylation by a suspension of liver mitochondria in a medium of pH 7.4, the pH of the matrix was measured as 7.7 and the membrane potential as 0.17 V a) Calculate the [H'] in the external medium and in the matrix under these conditions b) What is the outside : inside ratio of [H']? Under the conditions listed, what is the free energy for the movement of a proton from the intermembrane space to the matrix at 37°C? c) Is this pH gradient alone sufficient enough to drive ATP synthesis based on Table 14-3? d) If not and this gradient represents 1 mol of protons, how many moles of protons are needed to synthesize 1 ATP?Explanation / Answer
“pH” = -log10 [H+]; here pH of the matrix is given = 7.7; and pH of the medium is = 7.4
For the matrix:
7.7 = -log10 [H+]
a^x = n; loga n = x
10^7.7= [H+]
[H+] (Matrix) = 5.01 * 10^7
For the medium:
7.4 = -log10 [H+]
a^x = n; loga n = x
10^7.4= [H+]
[H+] (Medium) = 2.51 * 10^7
b)
[H+] (Medium) / [H+] (Matrix) = (2.51 * 10^7) / (5.01 * 10^7) ~ 0.5
So,
[H+] (Outside) / [H+] (Inside) = 0.5
c)
At equilibrium,
G = Go + RT log K
At equilibrium, G = 0
Putting the given values in the equation,
Go = ?; T = 37 + 273 = 310K; R = 0.001987kcal/mol-K
Go = 2.303 * 0.001987 * 310 log (0.5)
From part b, Keq = 0.5; log (0.5) = -0.301
Go = -0.43kcal/mol
c) Note that table 14.3 is required for this answer.
d) Note that 4[H+] are required for synthesis of one ATP molecule. You need to confirm this answer with figure 14.3.
**Note: Post table 14.3 separately along with part c and d (for more clarification).
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.