3. (8 malate synthase catalyzes conversion of the metabolite late to the pts) Th

Question

3. (8 malate synthase catalyzes conversion of the metabolite late to the pts) The enzyme metabolite malate, using the co-factor acetyl-CoA reaction, below). The biochemical stand free energy change for this reaction is -43.1 kJ/mol. glyoxylate acetyl CoA H20 malate coA H a) Is this reaction spontaneous, under standard-state conditions? Please explain your answer b) Given the standard state free energy, at equilibrium, you can calculate the Keq for this reaction a a temperature of 37 oC. Show all work for full credit c) When the reaction has reached equilibrium at 37 oC, [glyoxylate] 2 mM, [acetyl CoA] mM,

Explanation / Answer

a)
given, G = -43.1 KJ/mol

since G is negative, reaction will be spontaneous.

b)
G = -43.1 KJ/mol = -43100 J/mol
T = 37 oC = (37 + 273)K = 310 K

now use:
G = -R*T*ln Kc
-43100 = -8.314*310*ln Kc
ln Kc = 16.723
Kc = 1.83*10^7

c)
the concentration of H+ should be given
if we assume H+ to be constant concentration.

Kc = [malate][coA] / [glyoxylate][acetyl CoA]
1.83*10^7 = [malate]*(800*10^-6)/{(2*10^-3)(1*10^-3)}
[malate ] = 4.6*10^4 M
Answer: 4.6*10^4 M

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.