Solve Biochem Thermodynamics Problem: ΔG°' & Keq

[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?

 

[Zefram]

Hello! Let me help you with this problem.

a. To calculate (delta)G°’ for the reaction, we can use the formula (delta)G°’ = (delta)G°’(products) - (delta)G°’(reactants). Therefore, (delta)G°’ = (-13.9 kJ/mol) - (-30.5 kJ/mol) = 16.6 kJ/mol.

b. To calculate the equilibrium constant (Keq’), we can use the formula (delta)G°’ = -RTln(Keq’). Since the temperature is given in °C, we need to convert it to Kelvin by adding 273.15. So, at 32°C, the temperature in Kelvin is 305.15K. Therefore, (delta)G°’ = -8.314 J/mol*K * 305.15 K * ln(Keq’). Solving for Keq’, we get Keq’ = 570.5.

c. To calculate the ratio of glucose to glucose-6-phosphate, we can use the formula Keq’ = [products]/[reactants], where [products] and [reactants] are the concentrations of the products and reactants, respectively. Since the ratio of ATP to ADP in the cell is 11:1, we can assume that the concentration of ATP is 11 times higher than the concentration of ADP. Therefore, [ATP] = 11[ADP]. Substituting this into the formula, we get Keq’ = [glucose-6-phosphate]/[glucose] = 570.5. We can rearrange this to solve for [glucose]/[glucose-6-phosphate], which gives us [glucose]/[glucose-6-phosphate] = 1/570.5. Since we know that [ATP] = [glucose-6-phosphate], we can substitute this into the formula to get [glucose]/[ATP] = 1/570.5. Since the ratio of ATP to ADP is 11:1, we can substitute [ATP] = 11[ADP] into the formula to get [glucose]/11[ADP] = 1/570.5. Solving for [glucose]/[ADP], we get [glucose]/[AD

 

[ravenprp]

To solve this problem, we can use the following formula:
(delta)G = (delta)G°' + RTln(Q)

a. To find (delta)G°', we can use the given values for (delta)G°' of glucose-6-phosphate hydrolysis and ATP hydrolysis.
(delta)G°' = -13.9 kJ/mol + (-30.5 kJ/mol) = -44.4 kJ/mol

b. To find Keq', we can use the formula:
(delta)G°' = -RTln(Keq')
-44.4 kJ/mol = -8.314 J/mol*K * 305 K * ln(Keq')
Solving for Keq', we get:
Keq' = 1.2 x 10^9

c. To find the ratio of glucose to glucose-6-phosphate, we can use the formula:
Keq' = [glucose-6-phosphate]/[glucose]
Substituting the given ratio of ATP to ADP (11:1), we get:
1.2 x 10^9 = [glucose-6-phosphate]/[glucose]
Solving for [glucose-6-phosphate]/[glucose], we get:
[glucose-6-phosphate]/[glucose] = 8.3 x 10^-10

Therefore, for every 8.3 x 10^-10 molecules of glucose, there will be 1 molecule of glucose-6-phosphate.