Question

approximately how much more free energy is supplied to the electron transport chain by nadh than by fadh₂? 8 kcal/mol 45 kcal/mol 53 kcal/mol 98 kcal/mol they both feed electrons into q, so there is no difference in free energy.

Explanation:

NADH donates electrons at Complex I, while FADH₂ donates at Complex II. The difference in free - energy change ($\Delta G$) can be calculated using the formula $\Delta G=-nF\Delta E_{0}^{\prime}$, where $n$ is the number of electrons, $F = 23\ kcal/mol\cdot V$ (Faraday's constant), and $\Delta E_{0}^{\prime}$ is the difference in reduction potential.

The reduction potential of $NAD^{+}/NADH$ is $E_{0}^{\prime}=- 0.32\ V$, and for $Q/QH_{2}$ (where both NADH and FADH₂ ultimately transfer electrons to $Q$), $E_{0}^{\prime}=+0.045\ V$. For NADH: $\Delta E_{0}^{\prime}=0.045-( - 0.32)=0.365\ V$. Using $\Delta G=-nF\Delta E_{0}^{\prime}$ with $n = 2$ (since NADH donates 2 electrons), $\Delta G=-2\times23\times0.365=-16.79\ kcal/mol$.

For FADH₂, its reduction potential is $E_{0}^{\prime}=-0.18\ V$. $\Delta E_{0}^{\prime}=0.045-( - 0.18)=0.225\ V$. Using $\Delta G=-nF\Delta E_{0}^{\prime}$ with $n = 2$, $\Delta G=-2\times23\times0.225=-10.35\ kcal/mol$.

The difference in free - energy supplied is $(-10.35)-(-16.79)\approx6.44\ kcal/mol$, which is approximately $8\ kcal/mol$ (due to approximations in potential values and rounding in the calculation process).

Answer:

8 kcal/mol