# Biochemistry

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1. Given the following equation:
Glycerol-3-phosphate glycerol + Pi Go’ = -9.7 kJ/mol
At equilibrium the concentrations of both glycerol and inorganic phosphate are 1 mM. Under these conditions, calculate the final concentration of glycerol-3-phosphate. Remember to convert kJ to J.

2. Using the data for the reaction in question 1, calculate the G value (Go’ = -13.8 kJ/mol).

• Biochemistry -

I'm not sure if you typed in the reaction correctly. If the reaction is as followed:

Glycerol-3-phosphate----> Glycerol + Pi
∆G0'= -9.7kj

Then this what you I believe you should do:

The removal of the phosphate is the only way that I can see the reaction being energetically favorable, and at equilibrium, ∆G =0.

R=0.0083145 kJ/mol
T=273.15+25=298.15K
∆G=0
∆Go'= -9.7kj/mol
Pi=1mM
Glycerol= 1mM
Q=reactants/products
Glycerol-3-phosphate=G3P=?
Q=reactants/products=[1mM][1mM]/G3P

Plug in your values and solve for Glycerol 3-phosphate

∆G=∆G0'+RTlnQ

0=∆G0'+RTlnQ

-∆G0'=RTlnQ

-∆G0'/RT=lnQ

10^(-∆G0'/RT)=Reactants/Products

Reactants/10^(-∆G0'/RT)=Products

[1mM][1mM]/10^(-∆G0'/RT)=G3P

[1mM][1mM]/(9.7kj/mol)/[(0.0083145 kJ/mol)(298.15K)]=G3P

For the second part, use the value that you obtained for G3P, and the values that are given to you for Glycerol, ∆G0' (-13.8 kJ/mol), and Pi and plug into
the equation below and solve.

∆G=∆G0'+RTlnQ

• Biochemistry -

Last part of the equation for the first step should be,

[1mM][1mM]/10^[(9.7kj/mol)/[(0.0083145 kJ/mol)(298.15K)]=G3P

• Biochemistry -

I have a bad typo in my original post that may lead to confusion: replace the words reactants with products and vis versa. Everything else is okay except for that.

• Biochemistry -

Meaning, Q=products/reactants, not the other way around. for 10^(-∆G0'/RT)=Reactants/Products part of the equation manipulation.