Q. 18.98 - An Introduction to General, Organic, and Biological Chemistry

Question

Which of the following molecules will produce the most ATP per mole? $(18.5 , 18.6)$

a. glucose or stearic acid $\left(C_{18}\right)$

b. glucose or two pyruvate

c. two acetyl  $\mathrm{CoA}$ or one palmitic acid $\left(C_{16}\right)$

d. lauric acid $\left(C_{12}\right)$or palmitic acid $\left(C_{16}\right)$

e. $\mathrm{\alpha }‐\mathrm{ketoglutarate}$ or fumarate in one turn of the citric acid cycle.$$" width="9" style="max-width: none;" >

Step-by-Step Solution

Answer

a. Stearic acid $\left({\mathrm{C}}_{18}\right)$ will produce more $\mathrm{ATP}$ as compared to glucose.

b. Glucose will produce more $\mathrm{ATP}$than two pyruvates.

c. Palmitic acid $\left({\mathrm{C}}_{16}\right)$will produce more $\mathrm{ATP}$ than two acetyl- $\mathrm{CoA}$.

d. Palmitic acid will more $\mathrm{ATP}$ compare lauric acid.

e. $\alpha$-ketoglutarate will produce more $\mathrm{ATP}$.

Part (a) step 1: Given Information

We need to find that they will produce the most ATP per mole between glucose or stearic acid $\left(C_{18}\right)$.

Part (a) step 2: Simplify

Here, stearic is a saturated fatty acid and carboxylic acid consisting of $18$carbon atoms. It undergoes the eighth $\mathrm{\beta }- \mathrm{oxidation}$ cycle and generates nine acetyl$-\mathrm{CoA}$. So, we write the calculation for the $\mathrm{ATP}$ production as:

Now, 

The stearic acid will produce more $\mathrm{ATP}$ compared to glucose.

Part (b) step 1: Given Information

We need to find that they will produce the most ATP per mole between glucose or two pyruvate.

Part (b) step 2: Simplify

The complete oxidation of glucose generates $36 \mathrm{ATP}$ . So, the oxidation of $2$pyruvates generates $6\mathrm{ATP}$. Therefore, glucose will produce more than $2$pyruvates.

Part (c) step 1: Given Information

We need to find that they will produce the most ATP per mole between  two acetyl $\mathrm{CoA}$or one palmitic acid  $\left(C_{16}\right)$.

Part (c) step 2: Simplify

Here, each of the two acetyl-$\mathrm{CoA}$ yields $12 \mathrm{ATP}$, so the total for the two acetyl-$\mathrm{CoA}$ is $24 \mathrm{ATP}$.

Now, palmitic acid undergoes seven $\mathrm{\beta }$-oxidation cycles, and generates eight acetyl-$\mathrm{CoA}$ . In each $\mathrm{\beta }$-oxidation cycle, one $\mathrm{NADH}$, one ${\mathrm{FADH}}_2$ and one acetyl-$\mathrm{CoA}$are produced. That $\mathrm{NADH}$ and ${\mathrm{FADH}}_2$ generate three and two $\mathrm{ATP}$, respectively, the activation of the fatty acid requires two $\mathrm{ATP}$.

Therefore, the calculation for the $\mathrm{ATP}$ production as:

So, palmitic acid will produce more $\mathrm{ATP}$ than two acetyl- $\mathrm{CoA}.$

Part (d) step 1: Given Information

We need to find that they will produce the most ATP per mole between  lauric acid $\left(C_{12}\right)$or palmitic acid $\left(C_{16}\right)$.

Part (d) step 2: Simplify

Lauric acid undergoes five $\mathrm{\beta }$-oxidation cycles and generates six acetyl-$\mathrm{CoA}$. In each $\mathrm{\beta }$-oxidation cycle, one$\mathrm{NADH}$, one ${\mathrm{FADH}}_2$, and one acetyl-$\mathrm{CoA}$ are produced. That $\mathrm{NADH}$ and  ${\mathrm{FADH}}_2$generate three and two $\mathrm{ATP}$ , respectively, the activation of the fatty acid requires two $\mathrm{ATP}$.

So, the calculation for the $\mathrm{ATP}$ production is as:

Now, palmitic acid undergoes seven $\mathrm{\beta }$-oxidation cycles, and generates eight acetyl-$\mathrm{CoA}$. In each$\mathrm{\beta }$-oxidation cycle, one$\mathrm{NADH}$, one ${\mathrm{FADH}}_2$ , and one acetyl-$\mathrm{CoA}$ are produced. That $\mathrm{NADH}$ and ${\mathrm{FADH}}_2$  generate three and two$\mathrm{ATP}$, respectively, the activation of the fatty acid requires two.

Therefore, the calculation for the  production is as:

So, palmitic acid will more ATP compared lauric acid.

Part (e) step 1: Given Information

We need to find that they will produce the most ATP per mole between $\mathrm{\alpha }‐\mathrm{ketoglutarate}$ or fumarate in one turn of the citric acid cycle.

Part (e) step 2: Simplify

In one turn of the citric acid cycle, the oxidation of $\mathrm{\alpha }$-ketoglutarate yield one$\mathrm{NADH}$. Each $\mathrm{NADH}$generates $3 \mathrm{ATP}$. So, in the reaction $7$of the citric acid cycle, fumarate is hydrated to form malate. Because no amount of $\mathrm{ATP}$ is generated from fumarate itself. As such, $\mathrm{\alpha }$-ketoglutarate will produce more $\mathrm{ATP}$.