The oxidation of malate by NAD+ to form oxaloacetate is a highly endergonic reaction under standard conditions (DG’o = +7.0 kcal/mol). The reaction proceeds readily under physiological conditions:a) Why?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?

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.

i think it is dna, i'm not totally sure but i think it might be.

the correct answer is sex-linked recessive.

i just took the test