How Does Temperature Affect the Equilibrium Constant in Biochemical Reactions?

[rp6655321]

can someone help me with this problem

8. The first step in using glucose as a source of energy is a priming reaction that
consumes ATP:
alpha-D-glucose + ATP4 --> alpha-D-glucose-6-phosphate2- + ADP3- + H+

a.
What is the (delta)G°’ for this reaction if (delta)G°’ for glucose-6-phosphate
hydrolysis is –13.9 kJ/mol and (delta)G°’ for ATP hydrolysis (in the presence
of excess Mg2+) is –30.5 kJ/mol?


b. From this value, calculate the equilibrium constant (Keq’)for the reaction at
32°C.


c. If the ratio of ATP to ADP in the cell were 11:1, what would the ratio of
glucose to glucose-6-phosphate be?



for part A I got -16.6kJ/mol. Then using 37°C as the standard temp. I calculated the Keq to be 626.9. With that I got 0.00015 for part C. I can't figure out how to get part B though. Do I just use the -16.6kJ/mol for part B and substitute the new temperature into the equation (delta)G=-RTln(Keq)? Wouldn't the (delta)G be different at 32°C than at 37°C?

 

[anathema]

Hello!

Thank you for reaching out for help with this problem. Let me walk you through the steps to solve this problem and hopefully it will help you understand how to approach similar problems in the future.

a. To calculate the (delta)G°' for this reaction, we can use the equation:
(delta)G°' = (delta)G°' of products - (delta)G°' of reactants
Substituting the values given in the problem, we get:
(delta)G°' = -13.9 kJ/mol - (-30.5 kJ/mol) = 16.6 kJ/mol

b. To calculate the equilibrium constant (Keq') for the reaction at 32°C, we can use the equation:
(delta)G°' = -RT ln(Keq')
Substituting the value of (delta)G°' calculated in part a and the given temperature of 32°C (305 K), we get:
16.6 kJ/mol = -(8.314 J/K/mol)(305 K) ln(Keq')
Solving for Keq', we get:
Keq' = 3.12 x 10^7

c. To calculate the ratio of glucose to glucose-6-phosphate, we can use the equation:
Keq' = [products]/[reactants]
Substituting the values given in the problem, we get:
3.12 x 10^7 = ([glucose-6-phosphate]/[glucose]) x ([ADP]/[ATP])
Since the ratio of ADP to ATP is given as 1:11, we can substitute [ADP]/[ATP] with 1/11.
3.12 x 10^7 = ([glucose-6-phosphate]/[glucose]) x (1/11)
Solving for [glucose-6-phosphate]/[glucose], we get:
[glucose-6-phosphate]/[glucose] = 3.72 x 10^8

I hope this helps! Let me know if you have any further questions. Good luck with your studies!

 

[quicknote]

a. The (delta)G°’ for the reaction can be calculated using the equation (delta)G°’ = (delta)G°’f(products) - (delta)G°’f(reactants), where (delta)G°’f is the standard free energy of formation. Using the values given, we can calculate:
(delta)G°’ = [(-13.9 kJ/mol) + (-30.5 kJ/mol)] - [(-16.6 kJ/mol) + (0 kJ/mol) + (0 kJ/mol)] = -16.6 kJ/mol

b. To calculate the equilibrium constant (Keq’) at 32°C, we need to use the Van't Hoff equation:
ln(Keq’) = -(delta)H°/RT + (delta)S°/R
where (delta)H° is the standard enthalpy change and (delta)S° is the standard entropy change. We can assume that (delta)H° and (delta)S° do not change significantly with temperature, so we can use the values given at 37°C. Substituting the values into the equation, we get:
ln(Keq’) = -(-30.5 kJ/mol)/(8.314 J/mol*K * 305 K) + (0 kJ/mol)/(8.314 J/mol*K) = 0.000117
Taking the exponent of both sides, we get:
Keq’ = e^(0.000117) = 1.00012

c. The ratio of glucose to glucose-6-phosphate can be calculated using the equilibrium constant (Keq’) and the concentrations of the reactants and products. From the reaction equation, we can see that for every 1 molecule of ATP consumed, 1 molecule of ADP and 1 molecule of H+ are produced. Therefore, the ratio of ATP to ADP is 11:1, and the concentration of ATP is 11 times higher than the concentration of ADP. Using the equilibrium constant equation:
Keq’ = [glucose-6-phosphate]/[glucose]*[ADP]/[ATP]*[H+]
Substituting in the values given, we get:
1.00012 = [glucose-6-phosphate]/[glucose]*(1/11)*(1/11)
Solving for [glucose-6