Chapter 19, Problem 16P

Interpretation Introduction

Interpretation:

The number of molecules of oxaloacetate in the mitochondrion and definition of pOAA should be determined.

Concept Introduction:

The expression used to determine the standard free energy is:

  $\Delta G=\Delta G^{0\text{'}}+RT\ln \frac{\left[\text{Oxaloacetate}\right]\left[\text{NADH}\right]}{\left[\text{Malate}\right]\left[{\text{NAD}}^{+}\right]}$

Here $\Delta G$ is the observed free energy change in J, ${\text{ΔG}}^{0\text{'}}$ is the standard free energy change in J, R is the gas constant, T is the temperature in K and [Malate] is the concentration of Malate in M.

Explanation of Solution

The enzyme which converts malate into oxaloacetate is malate dehydrogenase. The reaction is as shown:

  ${\text{Malate + NAD}}^{+}\to {\text{ Oxaloacetate + NADH + H}}^{+}$

${\text{ΔG}}^0$ = 29.7 kJ/mol

Below reaction is used to determine the standard free energy:

  $\Delta G=\Delta G^{0\text{'}}+RT\ln \frac{\left[\text{Oxaloacetate}\right]\left[\text{NADH}\right]}{\left[\text{Malate}\right]\left[{\text{NAD}}^{+}\right]}$

Here $\Delta G$ is the observed free energy change in J, ${\text{ΔG}}^{0\text{'}}$ is the standard free energy change in J, R is the gas constant, T is the temperature in K and [Malate] is the concentration of Malate in M.

At equilibrium $\Delta G$ =0 at equilibrium. Thus, the equation will be:

  $\Delta G=RT\ln \frac{\left[\text{Oxaloacetate}\right]\left[\text{NADH}\right]}{\left[\text{Malate}\right]\left[{\text{NAD}}^{+}\right]}$

After rearranging the equation, we will get:

  $\begin{array}{l}\frac{\left[\text{Oxaloacetate}\right]\left[\text{NADH}\right]}{\left[\text{Malate}\right]\left[{\text{NAD}}^{+}\right]}=e^{\frac{\Delta G^{0\text{'}}}{RT}}\\ \left[\text{Oxaloacetate}\right]=\left[\text{Malate}\right]\frac{\left[\text{NADH}\right]}{\left[{\text{NAD}}^{+}\right]}{e}^{\frac{\Delta G^{0\text{'}}}{RT}}\end{array}$

Plugging in the values to determine the oxaloacetate concentration:

  $\begin{array}{l}\left[\text{Oxaloacetate}\right]=\left(0.22\times {10}^{-3}\right)\left(20\right)e^{\frac{-29.700}{8.314\times \left(273+25\right)}}\\ \text{ = 2}\text{.14}\times {\text{10}}^{-8}\text{ M}\end{array}$

Concentration of oxaloacetate is 0.214 µM.

Below formula will be used to determine the volume:

  $V=\pi {\left(\frac{d}{2}\right)}^{2}l$

Here, V = Volume, d =diameter, l = length

Plugging the values in the above equation to determine the volume:

  $\begin{array}{l}V=3.14\left(\frac{1\times {10}^{-6}m}{2}\right)\left(2\times {10}^{-6}m\right)\\ =1.57\times {10}^{-18}{m}^3\times \frac{1000000c{m}^3}{m^3}\times \frac{1\text{ L}}{1000{\text{ cm}}^3}\\ =1.57\times {10}^{-15}\text{ L}\end{array}$

Now we have to calculate concentration and volume now determine the oxaloacetate molecules:

  $\begin{array}{l}N=\left(2.14\times {10}^{-8}\frac{mol}{L}\right)\left(1.57\times {10}^{-15}L\right)\left(6.02\times {10}^{23}\frac{molecules}{mol}\right)\\ =20.2\text{ molecules}\\ \text{ = 20 molecules}\end{array}$

There are 20 molecules of oxaloacetate in the mitochondrion.

The pOAA can be determined from the negative log of oxaloacetate concentration:

  $\begin{array}{l}\text{pOAA =}-\text{log}\left[\text{oxaloacetate}\right]\\ \text{ =}-\log \left(2.14\times {10}^{-8}\text{M}\right)\\ =-\left(-7.67\right)\\ =7.67\end{array}$