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38 problems with worked solutions. Try each one first, then open the solution.
pKa values follow Lehninger, Principles of Biochemistry. Your course may use slightly different values; the method is the same.
Acid–base and pI 7 problems
1 Associate Calculate the pI of glycine (pKa 2.34 and 9.60).
Solution. pI = (2.34 + 9.60) ÷ 2 = 5.97.
2 Bachelor Calculate the pI of glutamic acid (pKa 2.19, 4.25, 9.67).
Solution. Acidic side chain: average the two lowest, (2.19 + 4.25) ÷ 2 = 3.22.
3 Bachelor Calculate the pI of arginine (pKa 2.17, 9.04, 12.48).
Solution. Basic side chain: average the two highest, (9.04 + 12.48) ÷ 2 = 10.76.
4 Associate What percentage of aspartate side chains (pKa 3.65) are deprotonated at pH 4.65?
Solution. pH − pKa = 1, so [A⁻]/[HA] = 10. Deprotonated = 10/11 ≈ 91%.
5 Associate At what pH are exactly half of glutamate side chains protonated?
Solution. At pH = pKa = 4.25.
6 Bachelor Blood pH is 7.4 and the pKa of the bicarbonate buffer is 6.1. What is the ratio of bicarbonate to carbonic acid?
Solution. log([HCO₃⁻]/[H₂CO₃]) = 7.4 − 6.1 = 1.3, so the ratio is 10^1.3 ≈ 20 : 1.
7 Bachelor Which amino acid is the best buffer near pH 7, and why?
Solution. Histidine: its side-chain pKa (about 6.0) is closest to 7, and buffers work best within about one unit of their pKa.
Peptides and charge 5 problems
1 Bachelor Estimate the net charge of the peptide Lys-Asp-Gly-Glu at pH 7.
Solution. N-terminus +1, Lys +1, Asp −1, Glu −1, C-terminus −1. Net ≈ −1. Check with the calculator.
2 Associate What is the molecular weight of the tripeptide Gly-His-Lys (GHK)?
Solution. Free amino acids: 75.07 + 155.16 + 146.19 = 376.42. Subtract two waters (2 × 18.02): ≈ 340.4 Da.
3 Bachelor A protein runs toward the positive electrode at pH 8. Is its pI above or below 8?
Solution. It is negatively charged, so the pH is above its pI. The pI is below 8.
4 Associate How many different tripeptides can be made from the 20 standard amino acids?
Solution. 20 × 20 × 20 = 8,000. Order matters: Gly-Ala-Ser differs from Ser-Ala-Gly.
5 Associate You dissolve 5 mg of a 1,000 Da peptide in 1 mL of water. What is the concentration?
Solution. 5 mg ÷ 1,000 g/mol = 5 µmol in 1 mL = 5 mM.
Protein structure 3 problems
1 Bachelor An α-helix has 36 residues. How many turns does it make, and how long is it? (3.6 residues per turn, 1.5 Å rise per residue)
Solution. 36 ÷ 3.6 = 10 turns; 36 × 1.5 Å = 54 Å.
2 Bachelor Why can’t a proline in the middle of an α-helix take part in the normal hydrogen bonding?
Solution. Its backbone nitrogen is part of a ring and has no hydrogen to donate, and the ring restricts the φ angle.
3 Bachelor Hemoglobin has four chains. Which level of structure describes how they fit together?
Solution. Quaternary structure.
Protein folding and energy 3 problems
1 Master A protein’s ΔG of unfolding is +20 kJ/mol at 25 °C. What fraction is unfolded at equilibrium? (R = 8.314 J/mol·K)
Solution. K = e^(−ΔG/RT) = e^(−20,000 ÷ 2,478) = e^(−8.07) ≈ 3.1 × 10⁻⁴. Fraction unfolded = K ÷ (1 + K) ≈ 0.03%. Folded proteins are only modestly stable, yet almost all molecules are folded.
2 Master A reaction has an equilibrium constant of 10 at 25 °C. What is ΔG°?
Solution. ΔG° = −RT ln K = −(8.314)(298)(ln 10) ≈ −5.7 kJ/mol.
3 Bachelor Does heating a protein until it unfolds break its peptide bonds?
Solution. No. Denaturation disrupts 3D structure; the primary sequence stays intact.
Enzyme kinetics 5 problems
1 Bachelor Km = 4 mM and Vmax = 200 µmol/min. What is v at [S] = 12 mM?
Solution. v = 200 × 12 ÷ (4 + 12) = 150 µmol/min.
2 Bachelor A competitive inhibitor raises the apparent Km from 2 to 6 mM (Vmax 100). What is v at [S] = 6 mM?
Solution. v = 100 × 6 ÷ (6 + 6) = 50. Without the inhibitor it would be 75.
3 Bachelor Vmax = 120 µmol/min with 0.02 µmol of enzyme. What is kcat?
Solution. kcat = 120 ÷ 0.02 = 6,000 min⁻¹ = 100 s⁻¹.
4 Bachelor A Lineweaver–Burk plot has a y-intercept of 0.01 min/µmol and an x-intercept of −0.5 mM⁻¹. Find Vmax and Km.
Solution. Vmax = 1 ÷ 0.01 = 100 µmol/min; Km = 1 ÷ 0.5 = 2 mM.
5 Master Enzyme A has kcat/Km = 10⁸ M⁻¹s⁻¹; enzyme B has 10⁴ M⁻¹s⁻¹. Which is more efficient, and what does 10⁸ suggest?
Solution. Enzyme A, by 10,000-fold. Values near 10⁸–10⁹ M⁻¹s⁻¹ approach the diffusion limit: the enzyme works almost every time it meets a substrate.
Lab methods and measurement 5 problems
1 Bachelor A peptide with 1 Trp and 2 Tyr has A₂₈₀ = 0.42 (1 cm). What is its concentration?
Solution. ε = 5,500 + 2 × 1,490 = 8,480 M⁻¹cm⁻¹. c = 0.42 ÷ 8,480 ≈ 4.95 × 10⁻⁵ M ≈ 50 µM.
2 Bachelor Why can’t you measure the concentration of the peptide GGSGG at 280 nm?
Solution. It contains no Trp or Tyr, so it barely absorbs at 280 nm. Use another assay or 205–214 nm.
3 Master A 1,000 Da peptide is detected with a 2+ charge by electrospray mass spectrometry. What m/z do you expect?
Solution. m/z = (1,000 + 2 × 1.007) ÷ 2 = 501.0.
4 Master An HPLC trace shows peaks with areas 950, 30 and 20. What is the area-percent purity of the main peak, and what does it not tell you?
Solution. 950 ÷ 1,000 = 95.0%. It does not confirm identity, sterility, endotoxin levels or exact concentration. See Understanding HPLC.
5 Associate How would you make 50 mL of a 2 mM solution from a 100 mM stock?
Solution. V₁ = (2 × 50) ÷ 100 = 1 mL of stock, made up to 50 mL.
Pharmacology and peptide drugs 3 problems
1 Master A peptide has a 2-minute half-life in blood. What percentage remains after 10 minutes?
Solution. 10 ÷ 2 = 5 half-lives; (½)⁵ = 3.1%. This is why natural peptides make poor drugs. See Peptide hormones and signaling.
2 Master A once-weekly drug has a half-life of about 7 days. Roughly how long until it reaches steady state with weekly dosing?
Solution. Steady state takes about 4–5 half-lives: about 4–5 weeks.
3 Bachelor Why does continuous exposure to a GnRH “superagonist” suppress reproductive hormones?
Solution. The pituitary’s GnRH receptors desensitize and are downregulated under constant stimulation; pulsatile signaling is needed for normal function.
Evidence and statistics 4 problems
1 Master Risk falls from 10% in the control group to 8% with treatment. Give the absolute risk reduction, relative risk reduction and NNT.
Solution. ARR = 2 percentage points; RRR = 2 ÷ 10 = 20%; NNT = 1 ÷ 0.02 = 50.
2 Master A hazard ratio is 0.95 with a 95% confidence interval of 0.85 to 1.10. Is the result statistically significant?
Solution. No. The interval includes 1.0 (no difference).
3 Associate A study reports p = 0.03. Does that mean there is a 97% chance the treatment works?
Solution. No. It means results this extreme would occur about 3% of the time if there were no real effect. It says nothing about the size or importance of the effect.
4 Associate A supplement extended lifespan in mice by 10%. Can you conclude it will do the same in people?
Solution. No. Animal results often fail to carry over to humans; it needs controlled human trials. See The evidence pyramid.
Aging and longevity 3 problems
1 Associate In 2024 U.S. life expectancy was 81.4 years for females and 76.5 for males. What is the gap?
Solution. 81.4 − 76.5 = 4.9 years.
2 Master Adult mortality risk roughly doubles every 8 years. If a 60-year-old’s annual risk of death is 1%, estimate it at 76.
Solution. 76 − 60 = 16 years = 2 doublings: 1% × 2 × 2 = about 4%. This pattern is called the Gompertz law.
3 Bachelor Name the three groups of the hallmarks of aging.
Solution. Primary, antagonistic and integrative. See the 12 hallmarks.
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VALREXA Biosciences. (2026, November 7). Biochemistry practice problems with worked solutions. VALREXA Biosciences. https://valrexa.com/study/practice.html
APA style · Last reviewed November 7, 2026 · Our editorial standardsPut your understanding to work
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