Question

Question: If intracellular $\text{[ATP] = 5mM, [ADP] = 0.5mM}$, and $\left[{\text{P}}_{\text{i}}\right]\text{=}$$\text{1.0mM}$ , calculate the concentration of AMP at pH 7 and ${25}^{°}C$under the condition that the adenylate kinase reaction is at equilibrium.

Short answer

Answer

The concentration of AMP is 0.23 mM.

Step-by-step solution

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

To calculate equilibrium constant$\text{(K)}$ for the reaction of $\text{2ADP}\to \text{ATP + AMP}$ at its equilibrium using the following equation, where c marks concentrations:

In this case, K is determined using the following formula$\text{:K =}\frac{\text{c(ATP)c(AMP)}}{{\text{c(ADP)}}^{\text{2}}}$

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.

Calculating the concentration

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

$\frac{\text{c(ATP)c(AMP)}}{a{\left(ADP\right)}^2}=e^{\raisebox{1ex}{$-∆\sigma $}\!\left/ \!\raisebox{-1ex}{$BT$}\right.}$

$\text{c(AMP) = 0.23mM}$

Hence, the concentration of AMP is 23mM.