Free practice final
Biochemistry practice final (free, with answers)
A typical one-semester biochemistry final covers amino acids and protein structure, enzyme kinetics and inhibition, and the core energy pathways: glycolysis, the citric acid cycle, oxidative phosphorylation, and fatty acid oxidation. These 15 questions mix concepts with the calculations that usually show up, such as Michaelis–Menten rates, buffer pH, and ATP yields. Try each one on your own before opening the answer.
Topics
- Amino Acids and Buffers
- Protein Structure
- Enzyme Kinetics
- Enzyme Inhibition
- Bioenergetics
- Glycolysis and Gluconeogenesis
- Citric Acid Cycle
- Oxidative Phosphorylation
- Lipid Metabolism
15 questions
Try each one before you open its answer.
1. Aspartic acid has pKa values of 2.1 (α-carboxyl), 3.9 (side-chain carboxyl), and 9.8 (α-amino). What is its isoelectric point (pI)?
- A 3.0
- B 5.3
- C 6.0
- D 6.9
Show answer
Answer: A. 3.0
Going from low to high pH, the net charge runs +1 → 0 → −1 → −2, and the neutral form exists between the first two pKa values. The pI is their average: (2.1 + 3.9) / 2 = 3.0.
2. An enzyme follows Michaelis–Menten kinetics with Km = 2.0 mM and Vmax = 100 μmol/min. What is the initial rate when [S] = 6.0 mM?
Show answer
Answer: 75 μmol/min
v = Vmax[S] / (Km + [S]) = (100 × 6.0) / (2.0 + 6.0) = 75 μmol/min. At [S] = 3 Km the enzyme runs at three quarters of Vmax.
3. Which interaction is mainly responsible for stabilizing an α-helix?
- A Disulfide bonds between cysteine side chains one turn apart
- B Hydrogen bonds between backbone groups on separate, side-by-side strands
- C Hydrophobic interactions between side chains that point into the center of the helix
- D Hydrogen bonds between the backbone C=O of residue n and the backbone N–H of residue n + 4
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Answer: D. Hydrogen bonds between the backbone C=O of residue n and the backbone N–H of residue n + 4
In an α-helix every backbone carbonyl hydrogen-bonds to the backbone N–H four residues further along the same chain. Side chains point outward from the helix, and H-bonds between separate strands describe a β-sheet.
4. Which molecule is a potent allosteric activator of phosphofructokinase-1 (PFK-1)?
- A Citrate
- B ATP at high concentration
- C Fructose 2,6-bisphosphate
- D H⁺ (low pH)
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Answer: C. Fructose 2,6-bisphosphate
Fructose 2,6-bisphosphate, made by PFK-2 in the fed state, raises PFK-1's affinity for fructose 6-phosphate and relieves ATP inhibition. Citrate, high ATP, and low pH all inhibit PFK-1.
5. On a Lineweaver–Burk plot (1/v versus 1/[S]), adding an inhibitor makes the line steeper but leaves the y-intercept unchanged. What does this show?
- A The inhibitor is uncompetitive: apparent Km and Vmax both decrease
- B The inhibitor is competitive: apparent Km increases and Vmax is unchanged
- C The inhibitor is competitive: Vmax decreases and Km is unchanged
- D The inhibitor irreversibly destroys a fraction of the enzyme's active sites
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Answer: B. The inhibitor is competitive: apparent Km increases and Vmax is unchanged
The y-intercept is 1/Vmax, so Vmax is unchanged, and the slope is Km/Vmax, so a steeper line means a higher apparent Km. That pattern is competitive inhibition, which enough substrate can overcome.
6. What is the pH of a buffer containing 0.10 M acetic acid and 0.20 M sodium acetate? Use pKa = 4.76.
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Answer: pH ≈ 5.06
Henderson–Hasselbalch: pH = pKa + log([A⁻]/[HA]) = 4.76 + log(0.20/0.10) = 4.76 + 0.30 = 5.06. More conjugate base than acid puts the pH above the pKa.
7. Respiring mitochondria are treated with oligomycin, an inhibitor of ATP synthase, and their O₂ consumption drops. The uncoupler 2,4-dinitrophenol (DNP) is then added. What happens next?
- A O₂ consumption stays low, because ATP synthase is still blocked
- B O₂ consumption rises again, but ATP synthesis stays blocked
- C O₂ consumption rises again, and ATP synthesis resumes
- D Electron transport runs in reverse and NADH is regenerated
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Answer: B. O₂ consumption rises again, but ATP synthesis stays blocked
With ATP synthase blocked, the proton gradient builds up and slows electron transport. DNP carries protons back into the matrix, collapsing the gradient, so electron transport and O₂ use resume while ATP synthase stays inhibited.
8. Glucose + Pi → glucose 6-phosphate + H₂O has ΔG°' = +13.8 kJ/mol, and ATP + H₂O → ADP + Pi has ΔG°' = −30.5 kJ/mol. What is ΔG°' for glucose + ATP → glucose 6-phosphate + ADP?
- A −16.7 kJ/mol
- B +16.7 kJ/mol
- C −44.3 kJ/mol
- D +44.3 kJ/mol
Show answer
Answer: A. −16.7 kJ/mol
The coupled reaction is the sum of the two reactions, so its ΔG°' is the sum: +13.8 + (−30.5) = −16.7 kJ/mol. Coupling to ATP hydrolysis makes an unfavorable phosphorylation favorable.
9. Which citric acid cycle enzyme is embedded in the inner mitochondrial membrane and passes electrons directly into the electron transport chain?
- A Isocitrate dehydrogenase
- B α-Ketoglutarate dehydrogenase
- C Succinate dehydrogenase
- D Malate dehydrogenase
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Answer: C. Succinate dehydrogenase
Succinate dehydrogenase is Complex II: it oxidizes succinate to fumarate, reducing its bound FAD, and hands the electrons to ubiquinone. The other three are matrix enzymes that reduce NAD⁺.
10. During intense exercise, muscle reduces pyruvate to lactate. What does this reaction do that lets glycolysis keep running?
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Answer: It regenerates NAD⁺ from NADH, which the glyceraldehyde-3-phosphate dehydrogenase step needs.
Glyceraldehyde-3-phosphate dehydrogenase consumes NAD⁺. When glycolysis runs faster than mitochondria can reoxidize NADH, lactate dehydrogenase uses NADH to reduce pyruvate and frees up NAD⁺.
11. An enzyme follows Michaelis–Menten kinetics with Km = 5 μM. At what substrate concentration does the initial rate equal 90% of Vmax?
- A 4.5 μM
- B 9.0 μM
- C 45 μM
- D 450 μM
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Answer: C. 45 μM
Set [S] / (Km + [S]) = 0.9, which gives [S] = 9 Km = 9 × 5 μM = 45 μM. Getting close to Vmax takes a substrate concentration many times Km.
12. Malonyl-CoA, made in the committed step of fatty acid synthesis, also keeps newly made fatty acids from being oxidized. Which enzyme does it inhibit?
- A Acetyl-CoA carboxylase
- B Carnitine palmitoyltransferase I
- C HMG-CoA reductase
- D Acyl-CoA dehydrogenase
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Answer: B. Carnitine palmitoyltransferase I
Carnitine palmitoyltransferase I, on the outer mitochondrial membrane, attaches fatty acyl groups to carnitine so they can enter the matrix for β-oxidation. Inhibiting it with malonyl-CoA keeps synthesis and breakdown from running at the same time.
13. Assume 2.5 ATP per NADH, 1.5 ATP per FADH₂, 10 ATP per acetyl-CoA oxidized in the citric acid cycle, and a cost of 2 ATP equivalents to activate the fatty acid. What is the net ATP yield from complete oxidation of one palmitate (16:0)?
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Answer: 106 ATP
Palmitate goes through 7 rounds of β-oxidation, giving 7 FADH₂ (10.5 ATP), 7 NADH (17.5 ATP), and 8 acetyl-CoA (80 ATP), for 108. Subtracting the 2 ATP equivalents for activation gives 106.
14. In gluconeogenesis, which pair of enzymes bypasses the irreversible pyruvate kinase step of glycolysis?
- A Fructose 1,6-bisphosphatase and glucose 6-phosphatase
- B Pyruvate dehydrogenase and citrate synthase
- C Hexokinase and phosphofructokinase-1
- D Pyruvate carboxylase and phosphoenolpyruvate carboxykinase
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Answer: D. Pyruvate carboxylase and phosphoenolpyruvate carboxykinase
Pyruvate carboxylase turns pyruvate into oxaloacetate in the mitochondria, and PEP carboxykinase turns oxaloacetate into phosphoenolpyruvate. The two phosphatases bypass the other irreversible steps, at PFK-1 and hexokinase.
15. An enzyme has Km = 1.0 mM. A competitive inhibitor with Ki = 10 μM is added at 20 μM. What is the apparent Km?
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Answer: 3.0 mM
For competitive inhibition, apparent Km = Km(1 + [I]/Ki) = 1.0 mM × (1 + 20/10) = 3.0 mM. Vmax is unchanged.