The first reaction in glycolysis is the phosphorylation of glucose: Pi + Glucose  Glucose-6-Phosphate This is a thermodynamically unfavorable process, with ΔG°' = + 14 kJ/mol. a. (5 pts) In a liver cell at 37°C the concentrations of both phosphate and glucose are normally maintained at about 5 mM each. What would the equilibrium concentration of glucose-6-phosphate be, according to the above data?

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Biology

Adam Tran

14 June, 13:19

The first reaction in glycolysis is the phosphorylation of glucose: Pi + Glucose  Glucose-6-Phosphate This is a thermodynamically unfavorable process, with ΔG°' = + 14 kJ/mol. a. (5 pts) In a liver cell at 37°C the concentrations of both phosphate and glucose are normally maintained at about 5 mM each. What would the equilibrium concentration of glucose-6-phosphate be, according to the above data?

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Gabriela · Answer 1

Keq = 0.00442714664

Explanation:

Glycolysis is the process by which glucose molecule at the end of this process splits into two molecules of pyruvate with the release of energy.

In this question we have to calculate equilibrium concentration of glucose-6-phosphate

This is the first step in Glycolysis

Given dа ta:

Change in Gibbs free energy = ΔG°' = + 14 kJ/mol = 14000 J/mol

Temperature = T = 37°C

Temperature must be converted into kelvin scale for this question we get

We use formula 37°C + 273.15 = 310.15K

Value of Gas constant = R = 8.3145 J/mol·K (This value is already known)

Equilibrium concentration of glucose-6-phosphate = Keq = ?

Using Formula

Keq = e-ΔG/RT (e is subscript and - ΔG/RT is superscript of e)

Puting the values we get

Keq = e - 14000/8.3145 X 310.15

we get

Keq = e - 1683.80 X 310.15

Keq = e - 5.42 (As - 5.42 is the power of e) we get

Keq = 0.00442714664