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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

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3 answers
  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.

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  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.

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

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