Free energy! Assume that you have a solution of 0.1 M glucose 6-phosphate. To th

Question

Free energy! Assume that you have a solution of 0.1 M glucose 6-phosphate. To this solution, you add the enzyme phosphoglucomutase, which catalyzes the following reaction: T

he ?G°? for the reaction is +7.5 kJ mol?1 (+1.8 kcal mol?1).

(a) Does the reaction proceed as written? If so, what are the final concentrations of glucose 6-phosphate and glucose 1-phosphate? (b) Under what cellular conditions could you produce glucose 1-phosphate at a high rate?

Explanation / Answer

G° =+7.5 kJ mol1 (+1.8 kcal mol1)

Answer) Glucose-6-phosphate + phosphoglucomutase=Glucose-1-phosphate

G=-RTln keq

    +7.5KJ/mol   =-8.314J/K mol (298K)ln keq

Or,+7500J/mol=2477.57 J/mol ln keq

Or,ln keq=7500/2477.57=3.03

Keq=e^3.03

Keq=21

Keq=[G6p]/[G1p]=21

For every molecule of G1p there are 21 molecule of G6P.As G1p=0.1M.

[G1p]=1/22 (0.1M)=0.0045M

And [G6P]=21/22 (0.1M)=0.096M

The reaction does not proceed to a significant extent as very low product conc is formed.

b) If the value of the ratio of [G6p]/[G1p] is kept large, then G becomes more negative and the product yield increases.This can be done by supplying G6P at high rate or removing G1p at high rate by other simultaneous reactions.

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