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Chapter 14: Q 28P (page 476)

Question: If intracellular [ATP] = 5mM, [ADP] = 0.5mM, and Pi=1.0mM , calculate the concentration of AMP at pH 7 and 25掳Cunder the condition that the adenylate kinase reaction is at equilibrium.

Short Answer

Expert verified

Answer

The concentration of AMP is 0.23 mM.

Step by step solution

01

Step-by-step-solutionStep 1: Calculating equilibrium constant

To calculate equilibrium constant(K) for the reaction of 2ADP鈫�ATP + AMP at its equilibrium using the following equation, where c marks concentrations:

In this case, K is determined using the following formula:K =c(ATP)c(AMP)c(ADP)2

02

Combining the given equations

To compute the concentration of AMP, we'll combine the previous and subsequent reactions of the step two into one:

In this equation stands for standard Gibbs free energy, represents gas constant and T is temperature.

03

Calculating the concentration

After combining those two equations we calculate the concentration of AMP at pH 7 and 25掳C (298K) under the condition that the adenylate kinase reaction is at equilibrium,

c(ATP)c(AMP)a(ADP)2=e-鈭�蟽BT

c(AMP) = 0.23mM

Hence, the concentration of AMP is 23mM.

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