Question

I'm stuck on this question, not sure where to start. can you show detail and explain so that it helps me to understand it. help please, thanks.

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

Answer 1

This 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.

Answer 2

Now, 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.

Answer 3

I had problems posting, which required that I do it in two parts, which required some time.