Question

2. The first step in glycolysis is catalyzed by the enzyme herokinase: glucose + ATP glucose-6-P + ADP This reaction can be thought of as two coupled reactions (P is phosphate): (1) glucose + P + glucose-6-P (2) ATP + ADP + P For the overall reaction, AG" = -16.7 kJ/mol. For the hydrolysis of ATP (equation 2), AG = -310 kJ/mol. All reactions are performed at 37°C. (a) What is AGº for reaction 1? Iglucose - 6-11 if AG 0? Iglucose)(phosphate) (c) Starting with 1M glucose and 1 M phosphate, what are the equilibrium concentrations of glucose, phosphate, and glucose-6-phosphate if only reaction l occurs? (b) What is glucose - 6-P)

Answer 1

Answer a)

∆G°_overall = ∆G₁° + ∆G₂°

-16.7 kJ/mol = ∆G₁° + (-31.0 kJ/mol)

∆G₁° = (-16.7 + 31.0)kJ/mol

= +13.3 kJ/mol

Answer b)

If ∆G = 0,

Then ∆G° = -RTln(Keq)

Here, Keq = [glucose-6-phosphate]/[glucose][Pi]

T = 37°C = 37 + 273 = 310 K

R = 8.314 J/K mol

∆G° = +13.3 kJ/mol = +13300 J/mol

Substituting values,

+13300 J/mol = -(8.314 J/K mol)(310 K)ln(Keq)

ln(Keq) = +13300/(-2577.34) = -5.160

Keq = e^-5.160 = 5.74 x 10^-3

Answer c)

If only reaction 1 occurs, assuming reaction of 'x' mol each of glucose and phosphate, ICE table will be:

	[Glucose]	[phosphate]	[glucose-6-phosphate]
Initial (M)	1	1	---
Change (M)	-x	-x	+x
Equilibrium (M)	1 - x	1 - x	x

Keq = (x)/(1-x)(1-x) = 5.74 x 10^-3

x/(1-x)² = 0.00574

x = 0.00574 - 0.01148x - 0.00574x²

0.00574x² + 1.01148x + 0.00574 = 0

On solving, x = 0.00567 M

So, the concentrations are:

[Glucose]= [phosphate] = 1-x

= 1- 0.00567 M = 0.994 M

[Glucose-6-phosphate] = x = 0.00567 M