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April 18, 2014

April 18, 2014

Posted by **frank** on Saturday, February 9, 2013 at 6:12pm.

The value of delta G for the conversion of 3-phosphoglycerate to 2-phosphoglycerate (2PG) is +4.4kj/mol. If the concentration of 3-phosphoglycerate at equilibrium is 1.75mM, what is the concentration of 2-phosphoglycerate. Assume temperature of 25C. Constant R = 0.0083145 kJ/mol

Also another question I have (I think I have it right but not sure)

consider: A + B double arrow C + D

which of the following would decrease delta G, which would increase delta G, which would have no effect on delta G?

Decreasing A and B

Adding a Catalyst

Coupling with ATP hydrolysis

Decreasing C and D

I think Adding a catalyst would have no effect

Decreasing A and B would decrease delta G

Coupling with ATP hydrolysis would also decrease delta G

and decreasing C and D would increase delta G

is that correct?

thanks for the help

- biochem -
**Devron**, Sunday, February 10, 2013 at 2:14amThis is a bioenergetics problem, so I am going to be doing some assuming. I believe that you know by now that there are a lot of different ΔG values, and they have all these different variations and rules associated with them.

The reaction is as followed:

3-phosphoglycerate (3PG) ---> 2-phosphoglycerate (2PG)

ΔGo’=4.4kj/mol

The following equation must be used to solve the problem.

ΔG=ΔGo’+RTlnQ

Since the reaction is at equilibrium, ΔG=0.

0=ΔGo’+RTlnQ

-ΔGo'=RTlnQ

-ΔGo'/RT=lnQ

e^(-ΔGo'/RT)=Q

Q=(products/reactants)=2PG/1.75mM

R=0.0083145 kJ/mol

T=273.15+25=298.15K

ΔGo'=+4.4kj/mol

3PG=1.75mM

2PG=?

Plug in your values and solve for 2PG.

e^(-4.4kj*mol-1/[(0.0083145 kJ/mol)(298.15K)])*1.75mM=2PG

2PG=0.297mM or 0.30mM depending on how many significant figures that your professor/teacher wants you to report.

- biochem -
**Devron**, Sunday, February 10, 2013 at 2:29amNow, for the second part, you only need to know the equation to use and how lnQ varies with an increase or a decrease. I am going to let Q=2/1 and 1/2=0.5 to illustrate.

ln(2)=0.7

ln(0.5)= -0.7

∆G=∆Go'+lnQ

lets look at the reaction in question.

1. Decreasing A and B; so from what we see with our fictional numbers, when reactants increase lnQ increases. So our equation shows us that ∆G will increase.

2. Adding a catalyst; yes this will decrease ∆G, but will it decrease the ∆G in question? No!!! However, it will cause a decrease in ∆G‡, which is equal to the difference in free energy between the transition state and the substrate (i.e., activation energy). So, ∆G will not change.

3. Coupling the reaction with ATP hydrolysis; the hydrolysis of ATP has a ∆Go'= -31.kj/mol. So, if ∆Go' decreased then ∆G decreases since you will have to add both ∆Go's together since they are coupled.

4. Decreasing C and D; I did number 1, so I will let you tackle this one.

- biochem -
**Devron**, Sunday, February 10, 2013 at 2:30amI had problems posting, which required that I do it in two parts, which required some time.

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