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## Main Question or Discussion Point

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?

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?