Biochemistry Practice Questions

These 15 questions cover the core of a high school or first-year biochemistry course: biomolecules, enzymes, energy, photosynthesis and nucleic acids. They get harder as you go, and include short calculations. Try each one on paper before checking the answer key at the end — the working matters as much as the final answer. If you get stuck, the biochemistry study guide has links to every topic.

Useful data: relative atomic masses H = 1.01, C = 12.01, N = 14.01, O = 16.00, P = 30.97. Avogadro constant = 6.022 × 10²³ mol⁻¹.

Questions

Q1. Name the type of bond that joins: (a) two amino acids, (b) two monosaccharides, (c) glycerol and a fatty acid, (d) the two strands of DNA.

Q2. Glucose has the formula C₆H₁₂O₆. (a) Calculate its molar mass. (b) Two glucose molecules join by a condensation reaction to form maltose. Give the molecular formula and molar mass of maltose.

Q3. State three ways the structure of starch makes it suitable as an energy store in plants.

Q4. Explain why olive oil is liquid at room temperature while butter is solid. Refer to structure and intermolecular forces.

Q5. A student adds Benedict’s reagent to solutions A, B and C and heats them. A turns brick-red, B stays blue, C turns green. Solution B is then boiled with dilute hydrochloric acid, neutralised and retested; it turns orange. (a) Rank A and C by reducing sugar concentration. (b) Suggest what B contains and explain the second result.

Q6. A protein is heated to 70 °C and loses its function, although its primary structure is unchanged. (a) Name this process. (b) Which bonds are broken, and which are not?

Q7. An enzyme’s activity is measured at different temperatures. Activity rises from 10 °C to 40 °C, then falls sharply to near zero at 60 °C. Explain both parts of the curve.

Q8. A competitive inhibitor and a non-competitive inhibitor are each added to separate enzyme reactions. (a) For each, state whether adding a large excess of substrate can restore the maximum rate. (b) Explain your answer for the non-competitive inhibitor.

Q9. Write a balanced equation for the complete aerobic respiration of glucose, with state symbols (assume glucose is dissolved in water).

Q10. In aerobic respiration, state where each of the following occurs: (a) glycolysis, (b) the Krebs cycle, (c) the electron transport chain. (d) What is the final electron acceptor?

Q11. Human muscle cells and yeast cells both respire anaerobically. Compare the products and explain why regenerating NAD⁺ is essential in both.

Q12. A DNA sample contains 22 % adenine. Calculate the percentages of thymine, guanine and cytosine. State the rule you used.

Q13. An mRNA sequence reads 5′-AUG GCU UUU UAA-3′. Using AUG = Met, GCU = Ala, UUU = Phe, UAA = stop: (a) Give the peptide produced. (b) How many peptide bonds does it contain? (c) A single base is deleted from the second codon (GCU → GU…). Explain why this is likely to have a large effect.

Q14. Hydrolysis of ATP to ADP releases about 30.5 kJ mol⁻¹ under standard conditions. Aerobic respiration yields about 30 ATP per glucose, and complete oxidation of glucose releases about 2,870 kJ mol⁻¹. (a) Calculate the energy captured in ATP per mole of glucose. (b) Calculate the efficiency of respiration as a percentage. (c) What happens to the rest of the energy?

Q15. A leaf absorbs light at 680 nm. Calculate the energy of one mole of photons at this wavelength. (h = 6.626 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹.) Comment on how this compares with the energy stored in one mole of ATP.

Answer key

A1. (a) Peptide bond. (b) Glycosidic bond. (c) Ester bond. (d) Hydrogen bonds between complementary bases. (See peptide bonds and DNA structure.)

A2. (a) M = 6(12.01) + 12(1.01) + 6(16.00) = 72.06 + 12.12 + 96.00 = 180.18 g mol⁻¹ (180.16 with more precise atomic masses). (b) Condensation releases one H₂O: C₁₂H₂₂O₁₁. M = 2(180.18) − 18.02 = 342.34 g mol⁻¹. Check it with the molar mass calculator.

A3. Any three: insoluble, so it doesn’t affect water potential (osmosis) in cells; compact, because amylose coils into a helix and amylopectin is branched; large, so it can’t diffuse out of cells; easily hydrolysed to glucose when needed, with branches giving many ends for enzymes to act on.

A4. Olive oil is rich in unsaturated fatty acids with cis C=C double bonds, which put kinks in the chains. Kinked chains can’t pack closely, so the London (dispersion) forces between them are weaker and less energy is needed to separate them — a lower melting point. Butter contains more saturated fatty acids, whose straight chains pack closely with stronger dispersion forces, so it’s solid (see fatty acids).

A5. (a) A has more reducing sugar than C (brick-red indicates more Cu₂O precipitate than green). (b) B contains a non-reducing sugar, most likely sucrose. Boiling with acid hydrolyses the glycosidic bond, releasing glucose and fructose, which are reducing sugars. Neutralising is needed because Benedict’s test only works in alkaline conditions (see reducing sugars test).

A6. (a) Denaturation. (b) Hydrogen bonds, ionic bonds and hydrophobic interactions holding the secondary and tertiary structure are disrupted; the peptide bonds (primary structure) are not broken. Disulfide bridges usually survive moderate heating too.

A7. From 10 °C to 40 °C, molecules have more kinetic energy, so enzyme and substrate collide more often and more collisions have enough energy to react — the rate rises. Above the optimum, vibrations break hydrogen and ionic bonds holding the enzyme’s tertiary structure, the active site changes shape and substrate no longer fits. The enzyme is denatured, and the rate falls (see factors affecting enzymes).

A8. (a) Competitive: yes — excess substrate outcompetes the inhibitor, so Vmax is restored (Km appears higher). Non-competitive: no. (b) The non-competitive inhibitor binds elsewhere (an allosteric site) and changes the shape of the enzyme, so the inhibited enzymes can’t work however much substrate is present. Vmax is lowered (see enzyme inhibition).

A9. C₆H₁₂O₆(aq) + 6O₂(g) → 6CO₂(g) + 6H₂O(l)

A10. (a) Cytoplasm. (b) Mitochondrial matrix. (c) Inner mitochondrial membrane (cristae). (d) Oxygen, which is reduced to water.

A11. Muscle: pyruvate → lactate (no CO₂). Yeast: pyruvate → ethanol + CO₂. In both, glycolysis reduces NAD⁺ to NADH. Without oxygen, the electron transport chain can’t reoxidise NADH, so NAD⁺ would run out and glycolysis — the only source of ATP — would stop. Converting pyruvate to lactate or ethanol reoxidises NADH to NAD⁺ (see aerobic vs anaerobic respiration).

A12. A pairs with T, so T = 22 %. A + T = 44 %, so G + C = 56 %, and since G = C, each is 28 %. Rule: Chargaff’s rule / complementary base pairing (A = T, G = C in double-stranded DNA).

A13. (a) Met–Ala–Phe. (b) Three amino acids → 2 peptide bonds. (c) Deleting one base causes a frameshift: every codon after the deletion is read in a different frame, so the amino acid sequence changes completely from that point and a stop codon may be reached early or missed (see the genetic code).

A14. (a) 30 × 30.5 = 915 kJ per mole of glucose. (b) Efficiency = 915 / 2,870 × 100 = 31.9 % (about 32 %). (c) The rest is released as heat, which helps maintain body temperature in mammals and birds. (In cells, ATP hydrolysis actually releases more than 30.5 kJ mol⁻¹ because concentrations aren’t standard, so real efficiency is higher, perhaps around 40 % or more.)

A15. Energy of one photon: E = hc/λ = (6.626 × 10⁻³⁴ × 3.00 × 10⁸) / (680 × 10⁻⁹) = 2.92 × 10⁻¹⁹ J. Per mole: 2.92 × 10⁻¹⁹ × 6.022 × 10²³ = 1.76 × 10⁵ J = 176 kJ mol⁻¹. This is almost six times the ~30.5 kJ mol⁻¹ released by ATP hydrolysis, so a mole of red photons carries enough energy, in principle, to make several moles of ATP. In reality, photosynthesis captures only part of it, because energy is lost at each step of the light-dependent reactions (see light-dependent reactions and photons and energy levels).

How did you do?

• 13–15 correct: excellent — move on to harder problems in metabolism practice.
• 9–12: solid; revisit the topics you missed.
• Below 9: work back through the study guide and try again in a few days.

Key takeaways

• Always name the specific bond (peptide, glycosidic, ester, hydrogen).
• Link structure to function and intermolecular forces in “explain” questions.
• For enzymes, talk about active site shape, collisions and denaturation.
• Practise calculations: molar masses, base percentages, efficiencies and photon energies.