Question

In cells, ATP can be regenerated by the transfer of a phosphate group from creatinephosphate (Cre-P) to ADP. The reaction is the following: Cre-P + ADP <----> Cre + ATP Additional information: Temperature of the reaction 37°C R = 8.31 J.mol-1 .K-1

1) Calculate the change of free energy under standard conditions (DG°’) of this reaction assuming the following: Hydrolysis (dephosphorylation) of Cre-P: DG°’1 = -42 kJ.mol-1 Hydrolysis of ATP: DG°’2 = -30 kJ.mol-1 Show your calculations.

2) In a resting muscle cell, the concentrations of the reactants are: Creatine phosphate: 25.00 mmol.L-1 Creatine: 12.00 mmol.L-1 ADP: 0.02 mmol.L-1 ATP: 4 mmol.L-1 Calculate the change in free energy (DG) of the reaction in resting muscle cells. Show your calculations.

Answer 1

summary:

for solving the first part , we write reaction for hydrolysis of Cre-P and hydrolysis of ATP. After writing the reaction, we arranged these reactions in such way that we got final given reaction.

For solving second part of problem, we simply used following equation

G = G° + RT lnQ

Where

Q = reaction quotient which is equal to ratio of concentrations of product to concentrations of reactant

R = 8.31J/mol .k

T = temperature of reaction in Kelvin

Remember:

In this equation, we put G° value in j/mol because value of R has unit J/mol.k.

Since concentration of product and reactant is given. So we calculate Q. Since G° is already calculated in first part. So we used above equation to calculate value of G in rest muscle with given concentration.

Remember:

The value of G° for a reaction is same whether reaction is done in one step or multiple steps. This is called Hess's law. So first part of the problem is based on this law.