## 2013 Final Exam #2

Michael Lonsway 3O
Posts: 43
Joined: Wed Sep 21, 2016 2:57 pm

### 2013 Final Exam #2

For this problem I'm confused on why we divide by 30.5kj for part a and divide by 30.5kj for part b as well. I was hoping someone could explain this problem in simple terms too.
Attachments  Amy Ko 3C
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Joined: Sat Jul 09, 2016 3:00 am

### Re: 2013 Final Exam #2

Part A
From the problem, we are given that the generation of ATP from ADP is nonspontaneous and must be coupled to a spontaneous reaction, which is 3 NADH + 3/2 O2 + 3 H+ --> 3 H2O +3 NAD+ for 3 moles of NADH, ΔGr = -660.3kJ
We know that the ΔGr of generation of ATP4- is +30.5kJ. For every 3 moles of NADH, we will have 660.3kJ. To generate 1 mole of ATP, we need 30.5kJ. If we have 660.3kJ from NADH, we will have 660.3kJ/30.5kJ = 21.6 moles of ATP generated.

Part B
From the problem, we are given that the phosphorylation of acetic acid was driven by coupling to the hydrolysis of ATP (which is -30.5 kJ from part A). Phosphorylation of acetic acid is ΔGr= +41 kJ. For every mole of acetyl phosphate, we need 41 kJ. If hydrolysis of 1 mole of ATP is 30.5kJ, we need 41 kJ / 30.5kJ = 1.3 mole of ATP to drive the formation of acetyle phosphate.

Anthony_Imperial_1G
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Joined: Wed Sep 21, 2016 2:59 pm

### Re: 2013 Final Exam #2

Why did you switch the signs for hydrolysis of ATP form positive to negative and the formation of acetyl phosphate from negative to positive?

Yamilex Velgara 2I
Posts: 21
Joined: Wed Sep 21, 2016 2:59 pm

### Re: 2013 Final Exam #2

Anthony_Imperial_1G wrote:Why did you switch the signs for hydrolysis of ATP form positive to negative and the formation of acetyl phosphate from negative to positive?

In part 2B, the question states that the phosphorylation of acetic acid is the REVERSE of the hydrolysis of acetyl phosphate (this info is in parentheses). So for the phosphorylation of acetic acid, we must change the signs of the delta Gibbs free energy of the hydrolysis of acetyl phosphate.

As for ATP, the same applies. In part 2A it states that +30.5 kJ is the delta Gibbs free energy for the phosphorylation of ATP, but we need to know the delta Gibbs free energy for the hydrolysis of ATP. So we switch the signs because these are reverse processes.

Just a hint: hydrolysis is an exergonic reaction (releases energy), so the delta G should be negative. Phosphorylation (formation) is an endergonic reaction, which requires energy (+).