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1. The third step in glycolysis is the conversion of fructose-6-phosphate to fru

ID: 576301 • Letter: 1

Question

1. The third step in glycolysis is the conversion of fructose-6-phosphate to fructose-1,6- bisphosphate (F16BP) catalyzed by the enzyme PFK-1. This reaction is the "committed step" in glycolysis. Considering this, answer the following questions (please show all your work): A) If the reaction occurs as follows what will the Ksq be (1 point)? Assume R 8.315 J/mol and temp is 298K. fructose-6-phosphate + Pi F16BP Go'z +16.3 kJ/mol B) For the reaction above if we assume the [Pl to be 1.65mM what will the molar ratio of [F16BPl:[fructose-6-phosphate] be at equilibrium (I point)? C) Let's assume that the hexokinase reaction proceeds in the following manner as opposed to what we see in A) and B). Given this what is the new Keq (0.5 point)? fructose-6-phosphate + ATP Fi6BP+ ADP Gor=-14.2 kJ/mol

Explanation / Answer

Biochemistry

1. for the given reaction,

A) Using,

dGo = -RTlnKeq

with,

dGo = +16.3 kJ/mol = 16.3 x 10^3 J/mol

R = gas constant

T = 298 K

we get,

16.3 x 10^3 = -8.314 x 298 lnKeq

Keq = 1.4 x 10^-3

B) when [Pi] = 1.65 mM

Keq = [F16BP]/[fructose-6-phosphate][Pi]

1.4 x 10^-3 = [F16BP]/[fructose-6-phosphate]1.65

ratio [F16BP]/[fructose-6-phosphate] = 2.31 x 10^-3

C) when the given transformation occurs in this way,

dGo = -RTlnKeq

with,

dGo = -14.2 kJ/mol = -14.2 x 10^3 J/mol

we get,

-14.2 x 10^3 = -8.314 x 298 lnKeq

So,

Keq = 308.41

D) when [ATP] is 10x times greater than [ADP]

Keq = [F16BP][ADP]/[ATP][fructose-6-phosphate]

308.41 = [F16BP]/10[fructose-6-phosphate]

ratio [F16BP]/[fructose-6-phosphate] = 3084.1

E) If the cellular concentration of fructose-6-phosphate is 0.028 mM and F16BP is 0.014 mM,

it means that the PFK-1 equilibrium will shift towards F16BP until the equilibrium Keq reaches the above value in D.

We expect a large flux control for the reaction of PFK-1 due to large Keq value.