Answer:
A. Why?
The amount of oxaloacetate is minimal and is not in equilibrium being pinched into citrate when the citric acid cycle is being induced in the classic “forward” direction.
b. Assuming an [NAD+]/[NADH] ratio of 8 and a pH of 7, what is the lowest [malate]/[oxaloacetate] ratio at which oxaloacetate can be formed from malate
Given,
ΔG°′ = + 7 kcal mol-1
T=310K
R = 1.9872 cal/(mole K)
The ΔG is calculated using
ΔG = ΔG°′+ RTln(([oxaloacetate][NADH])/([malate][NAD+]))
0 = ΔG°′+ RTln(([oxaloacetate][NADH])/([malate][NAD+])) - being positioned at the tipping stage
ΔG°′= - RTln(([oxaloacetate][NADH])/([malate][NAD+]))
7 kJ mol-1 = - 0.00831451(310)ln(([1][oxaloacetate])/([malate][8]))
11.36 = - ln(([oxaloacetate])/(8*[malate]))
e11.36 = (8*[malate]/[oxaloacetate])
86081 = (8*[malate]/[oxaloacetate])
[malate]/[oxaloacetate] > 1.075x10^4
The actual values will differ slightly based on how it is rounded up - the vital point is to be in the proximity of 1x10^4.