Free MCAT-style practice, 1D
MCAT Fatty Acid Oxidation, Ketone Bodies and Lipoprotein Practice Questions
Eight free MCAT-style practice questions on beta oxidation, fatty acid synthesis, ketone bodies, cholesterol synthesis and lipoprotein transport, with calculations and full explanations.
Quick concept primer
Fat metabolism is built from two-carbon units. Beta oxidation in the mitochondrial matrix removes acetyl-CoA from a fatty acyl-CoA two carbons at a time, generating 1 FADH2 and 1 NADH per cycle. Long-chain fatty acids enter the matrix through the carnitine shuttle, whose rate-limiting enzyme CPT-I is inhibited by malonyl-CoA. Fatty acid synthesis is the cytosolic mirror image: acetyl-CoA carboxylase makes malonyl-CoA (needing biotin and ATP), and fatty acid synthase then adds two carbons at a time using NADPH.
When oxaloacetate is scarce and acetyl-CoA is abundant, as in fasting, the liver converts acetyl-CoA to ketone bodies (acetoacetate and beta-hydroxybutyrate) that the brain, heart and muscle can burn. Cholesterol synthesis is controlled at HMG-CoA reductase, the target of statins. Lipoproteins move insoluble lipids in blood: the higher the protein fraction, the denser the particle.
Traps: even-chain fatty acids cannot be converted into net glucose in humans, and the liver makes ketone bodies but cannot use them.
8 practice questions
Question 1
How many rounds of beta oxidation, and how many acetyl-CoA, are required to completely cleave stearic acid (an 18-carbon saturated fatty acid)?
- 9 rounds, 18 acetyl-CoA
- 8 rounds, 8 acetyl-CoA
- 9 rounds, 9 acetyl-CoA
- 8 rounds, 9 acetyl-CoA
Show answer and explanation
Answer: D. 8 rounds, 9 acetyl-CoA
An 18-carbon chain contains nine two-carbon units. The last round cleaves a 4-carbon acyl-CoA into two acetyl-CoA, so the number of rounds is one fewer than the number of acetyl-CoA: 8 rounds, 9 acetyl-CoA. Each round also yields 1 FADH2 and 1 NADH, so stearate gives 8 FADH2 and 8 NADH.
Question 2
A well-fed person is making fatty acids in the liver. Which mechanism prevents the newly made fatty acids from being immediately oxidized?
- NADPH inhibits beta oxidation directly
- Malonyl-CoA inhibits carnitine palmitoyltransferase I, blocking entry of fatty acyl groups into the mitochondria
- Glucagon activates acetyl-CoA carboxylase
- Citrate inhibits acetyl-CoA carboxylase
Show answer and explanation
Answer: B. Malonyl-CoA inhibits carnitine palmitoyltransferase I, blocking entry of fatty acyl groups into the mitochondria
Malonyl-CoA, the first committed intermediate of fatty acid synthesis, inhibits CPT-I, the rate-limiting step of mitochondrial fatty acid import. This keeps synthesis and oxidation from running simultaneously in a futile cycle. Citrate activates (not inhibits) acetyl-CoA carboxylase, and glucagon turns the enzyme off by phosphorylation.
Question 3
Oxidation of a 15-carbon (odd-chain) fatty acid ends with a three-carbon fragment. Conversion of this fragment to succinyl-CoA requires which vitamin as a cofactor for the final step, methylmalonyl-CoA mutase?
- Vitamin K
- Thiamine
- Vitamin B12
- Vitamin C
Show answer and explanation
Answer: C. Vitamin B12
Beta oxidation of odd-chain fatty acids ends with propionyl-CoA. Propionyl-CoA is carboxylated (biotin-dependent) to methylmalonyl-CoA, which methylmalonyl-CoA mutase rearranges to succinyl-CoA using adenosyl-cobalamin (B12). Succinyl-CoA enters the citric acid cycle, so unlike acetyl-CoA from even-chain fatty acids, this carbon skeleton can contribute to net glucose synthesis.
Question 4
Fatty acid synthesis in the cytosol requires all of the following EXCEPT
- FAD as the electron carrier
- malonyl-CoA
- NADPH
- biotin-dependent acetyl-CoA carboxylase
Show answer and explanation
Answer: A. FAD as the electron carrier
Fatty acid synthase reduces its intermediates with NADPH, usually supplied by the pentose phosphate pathway. FAD-based electron carriers are used in beta oxidation, the reverse process in the mitochondria. Acetyl-CoA carboxylase makes malonyl-CoA from acetyl-CoA, bicarbonate and ATP, using biotin.
Question 5
During several days of fasting, which tissue makes ketone bodies but is unable to use them as fuel?
- Heart muscle
- Skeletal muscle
- Brain
- Liver
Show answer and explanation
Answer: D. Liver
Hepatocytes synthesize acetoacetate and beta-hydroxybutyrate from acetyl-CoA in the mitochondrial matrix, but they lack the transferase (succinyl-CoA:acetoacetate CoA transferase) that activates acetoacetate for oxidation. This lets the liver export ketone bodies to other tissues. Brain, heart and skeletal muscle all express it and oxidize ketone bodies, and red blood cells cannot use them because they have no mitochondria.
Question 6
A patient begins taking a statin. Which sequence of events best explains the drop in plasma LDL cholesterol?
- Acyl-CoA cholesterol acyltransferase is activated and cholesterol esters accumulate
- HMG-CoA synthase is activated and more ketone bodies are made
- HMG-CoA reductase is inhibited, hepatocyte cholesterol falls, SREBP signaling rises, and more LDL receptors are placed on the cell surface, clearing more LDL
- LDL receptors are degraded, so plasma LDL falls
Show answer and explanation
Answer: C. HMG-CoA reductase is inhibited, hepatocyte cholesterol falls, SREBP signaling rises, and more LDL receptors are placed on the cell surface, clearing more LDL
Statins competitively inhibit HMG-CoA reductase, the rate-limiting enzyme of cholesterol synthesis. The resulting fall in intracellular cholesterol activates SREBP transcription factors, which raise expression of both HMG-CoA reductase and the LDL receptor. More receptors pull more LDL out of the blood. If receptors were degraded (choice D), plasma LDL would rise.
Question 7
Four lipoprotein fractions were separated by density, and their composition was measured.
| Particle | Protein (% by mass) | Triacylglycerol (% by mass) |
|---|---|---|
| P1 | 2 | 88 |
| P2 | 10 | 55 |
| P3 | 22 | 10 |
| P4 | 50 | 4 |
Which particle is most likely HDL, and which is most likely a chylomicron?
- HDL is P2 and chylomicron is P3
- HDL is P1 and chylomicron is P4
- HDL is P4 and chylomicron is P1
- HDL is P3 and chylomicron is P2
Show answer and explanation
Answer: C. HDL is P4 and chylomicron is P1
Protein is denser than lipid, so a higher protein fraction means a denser particle. HDL is the densest class, with the highest protein content and low triacylglycerol, matching P4. Chylomicrons carry dietary triacylglycerol from the intestine and are the least dense, with the lowest protein and the highest triacylglycerol, matching P1. P2 is VLDL-like and P3 is LDL-like.
Question 8
Why can humans NOT convert the acetyl-CoA from even-chain fatty acid oxidation into a net amount of glucose?
- Acetyl-CoA cannot cross the mitochondrial membrane
- Beta oxidation produces no ATP
- The pyruvate dehydrogenase reaction, which converts pyruvate to acetyl-CoA, is irreversible, and the citric acid cycle releases two carbons as CO2 for every two-carbon acetyl group that enters, so no net oxaloacetate is gained
- Fatty acids are too hydrophobic to be metabolized
Show answer and explanation
Answer: C. The pyruvate dehydrogenase reaction, which converts pyruvate to acetyl-CoA, is irreversible, and the citric acid cycle releases two carbons as CO2 for every two-carbon acetyl group that enters, so no net oxaloacetate is gained
Entry of acetyl-CoA into the citric acid cycle adds two carbons, but two carbons leave as CO2 in the cycle, so there is no net gain of oxaloacetate to feed gluconeogenesis. Because PDH is irreversible, acetyl-CoA cannot go back to pyruvate. Plants and bacteria have the glyoxylate cycle, which bypasses the decarboxylation steps, but animals do not.