Biochemistry Practice Exam — Free

Identify the inhibition type from which kinetic constant changes. A competitive inhibitor binds only the free enzyme, at the active site, so a high enough substrate concentration can always outcompete it: $V_{max}$ is unchanged, but more substrate is needed to reach half of it, so the apparent $K_m$ rises. That is exactly the pattern observed. A non-competitive inhibitor lowers $V_{max}$ and leaves $K_m$ unchanged; an uncompetitive inhibitor lowers both. So the inhibition is competitive.

For competitive inhibition the apparent Michaelis constant is

Substitute the measured values and solve for $K_i$:

Back-substitute to confirm: $2.0\times(1 + 1.0/0.50) = 2.0\times3 = 6.0\,\text{mM}$, matching the apparent $K_m$.

Competitive inhibition with $K_i = 0.50\,\text{mM}$ (option A).

Option B is the common slip of writing $K_m^{app}/K_m = [I]/K_i$ and forgetting the 1 in the factor $(1 + [I]/K_i)$, which gives 0.33 mM. Option C has the right $K_i$ but the wrong type: non-competitive inhibition would have lowered $V_{max}$. Option D is ruled out because uncompetitive inhibition decreases, rather than increases, the apparent $K_m$.

List every ionizable group, because only those carry charge; the peptide bonds and the glycine residue contribute none. The tetrapeptide has five: the N-terminal amino group, the C-terminal carboxyl group, two lysine $\varepsilon$-amino side chains and one aspartate carboxyl side chain. Decide the charge of each by comparing pH 7.0 with its $\text{p}K_a$: a group whose $\text{p}K_a$ is two or more units above the pH is protonated, and one whose $\text{p}K_a$ is two or more units below the pH is deprotonated.

• N-terminal $\alpha$-$\mathrm{NH_3^+}$ ($\text{p}K_a$ 9.0): pH is 2 units below, so protonated, charge $+1$.
• C-terminal $\alpha$-$\mathrm{COO^-}$ ($\text{p}K_a$ 2.0): pH is 5 units above, so deprotonated, charge $-1$.
• Two Lys side chains ($\text{p}K_a$ 10.5): protonated, $+1$ each, total $+2$.
• Asp side chain ($\text{p}K_a$ 3.9): deprotonated, charge $-1$.

Sum the charges:

Because the peptide carries a net positive charge, in an electric field it migrates toward the negatively charged electrode, the cathode. Two common mistakes are to count the glycine or the peptide backbone as charged, and to overlook that the C-terminal lysine contributes two separate groups: its side-chain amine ($+1$) and the backbone carboxylate that forms the C-terminus ($-1$).

Net charge at pH 7.0 is +1; the peptide moves toward the cathode.

Step 1, count the rounds. A $\mathrm{C_{16}}$ chain is cut into 8 two-carbon acetyl-CoA units, which needs only 7 cleavages because the final round releases two acetyl-CoA at once. So there are 7 rounds of $\beta$-oxidation and 8 acetyl-CoA.

Step 2, cofactors from $\beta$-oxidation. Each round has one FAD-dependent dehydrogenation (acyl-CoA dehydrogenase) and one $\mathrm{NAD^+}$-dependent dehydrogenation (3-hydroxyacyl-CoA dehydrogenase): $7\ \mathrm{FADH_2}$ and $7\ \text{NADH}$.

Step 3, cofactors from the citric acid cycle. Eight acetyl-CoA give $8\times3 = 24\ \text{NADH}$, $8\ \mathrm{FADH_2}$ and $8\ \text{GTP}$.

Step 4, totals: $31\ \text{NADH}$, $15\ \mathrm{FADH_2}$ and $8\ \text{GTP}$.

Step 5, convert to ATP with the stated P/O ratios:

Step 6, subtract the activation cost. Acyl-CoA synthetase converts ATP to AMP + $\mathrm{PP_i}$, and the pyrophosphate is then hydrolyzed, so two high-energy phosphate bonds are spent:

Because both $\beta$-oxidation and the citric acid cycle run inside the mitochondrial matrix, no shuttle is needed for the NADH, which is why the yield is not reduced the way cytosolic NADH from glycolysis can be. The 106 ATP from a 16-carbon fatty acid, compared with about 32 from six-carbon glucose, is the quantitative reason fat is the denser fuel.

Net yield: 106 ATP per palmitate (7 rounds of β-oxidation give 7 NADH + 7 FADH₂; eight turns of the cycle give 24 NADH + 8 FADH₂ + 8 GTP).