AQA A-level Biology: Nucleic Acids, DNA Replication and ATP (7402) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow the AQA AS and A-level Biology specification (7401 and 7402, first teaching 2015), sections 3.1.5.1, 3.1.5.2 and 3.1.6. The content is part of both the AS and the A-level. Data values in the questions are supplied for the question and are not figures to memorise.

Read the study guide first if the content is new, or use the revision notes. Course page: AQA A-level Biology. Checklist: AQA A-level Biology checklist. Free 10-minute diagnostics show which topics need more work.

Questions

1. (a) State the three components of an RNA nucleotide. [2] (b) State two differences between a DNA nucleotide and an RNA nucleotide. [2]

2. A single polynucleotide chain is made of 84 nucleotides. (a) Name the type of reaction that joins two nucleotides. [1] (b) Calculate the number of phosphodiester bonds in the chain and the number of water molecules released as it was made. [2]

3. A double-stranded DNA molecule has 2400 base pairs. 35% of its bases are adenine. Calculate the number of adenine, thymine, guanine and cytosine bases in the molecule. Show your working. [3]

4. One strand of a DNA molecule has this base composition: 17% adenine, 26% thymine, 31% guanine and 26% cytosine. (a) State the percentage of each base in the complementary strand. [2] (b) Calculate the percentage of adenine and of guanine in the whole double-stranded molecule. [2]

5. Describe how a DNA molecule is replicated semi-conservatively. [5]

6. A culture of bacteria was grown for many generations in a medium containing light nitrogen (¹⁴N). The bacteria were then moved to a medium containing heavy nitrogen (¹⁵N) and allowed to divide, doubling every 40 minutes. DNA was extracted at intervals and separated by density in a caesium chloride gradient. After 0 minutes there was one light band. After 40 minutes there was one band at an intermediate position. (a) Predict the percentage of DNA in the intermediate band and in the heavy band after 80 minutes and after 120 minutes if replication is semi-conservative. [2] (b) Explain why the result at 40 minutes is not consistent with conservative replication. [2] (c) Dispersive replication also gives one intermediate band at 40 minutes. Describe what the band pattern at 80 minutes would show if replication were dispersive rather than semi-conservative. [2]

7. A bacterial chromosome has 3.6 × 10⁶ base pairs. Replication starts at one point and two replication forks move away from it in opposite directions. Each fork copies 600 base pairs per second. Calculate the time taken to replicate the whole chromosome, in minutes. [3]

8. (a) State the components of an ATP molecule. [2] (b) Name the enzyme that catalyses the hydrolysis of ATP and name the two products. [2]

9. A reaction in a cell needs an input of 14.0 kJ per mole. The hydrolysis of ATP releases 30.5 kJ per mole (data supplied). (a) Calculate the net energy change per mole when the two are coupled. [2] (b) Explain how the phosphate released from ATP can make a compound more reactive. [2]

10. The relative molecular mass of ATP is 507. Calculate the energy released when 2.54 g of ATP is hydrolysed, using 30.5 kJ per mole. Give your answer in kJ to 3 significant figures. [3]

11. (a) Explain why the relative simplicity of DNA led many scientists to doubt that it carried the genetic code. [2] (b) Evaluate the work of Meselson and Stahl in validating the Watson-Crick model of DNA replication. [6]

12. (a) Explain why ATP is described as a nucleotide derivative. [2] (b) Compare the reactions catalysed by ATP hydrolase and ATP synthase. [4]

Answers

1. (a) Ribose (the pentose) [1]; a phosphate group and a nitrogen-containing organic base (adenine, cytosine, guanine or uracil) [1]. [2] (b) DNA has deoxyribose, RNA has ribose [1]; DNA has thymine, RNA has uracil [1]. [2] Examiner insight: Each difference needs both sides stated. Writing only “RNA has uracil” tends to gain less credit than “uracil in RNA, thymine in DNA”.

2. (a) Condensation [1]. [1] (b) A chain of n nucleotides has n − 1 bonds, so 84 − 1 = 83 phosphodiester bonds [1]; one water is released per bond, so 83 water molecules [1]. [2] Examiner insight: Giving 84 as the answer is an easy slip. Showing “n − 1” in your working shows the reasoning even if the arithmetic goes wrong.

3. Total bases = 2 × 2400 = 4800 [1]. A = T = 35% of 4800 = 1680 each [1]. G + C = 100 − 70 = 30%, so G = C = 15% of 4800 = 720 each [1]. [3] Examiner insight: A base pair is two bases. Writing the total of 4800 first makes the method clear, and checking that 1680 + 1680 + 720 + 720 = 4800 catches slips.

4. (a) Complementary strand: adenine 26% (pairs with T of strand 1) [1]; thymine 17%, guanine 26%, cytosine 31% [1]. [2] (b) Adenine = (17 + 26) ÷ 2 = 21.5% [1]. Guanine = (31 + 26) ÷ 2 = 28.5% [1]. [2] Examiner insight: Show the pairing logic. A common error is to give the same percentages for both strands, or to forget that the whole-molecule value is an average.

5. DNA helicase breaks hydrogen bonds between the complementary bases and unwinds the double helix [1]. Each separated strand acts as a template [1]. Free DNA nucleotides are attracted to the exposed bases and base-pair with them (A with T, C with G) [1]. DNA polymerase catalyses the condensation reaction that joins adjacent nucleotides, forming phosphodiester bonds [1]. Each new molecule contains one original strand and one new strand, so replication is semi-conservative [1]. [5] Examiner insight: Name the bond with each enzyme: hydrogen bonds for helicase, condensation for polymerase. Vague wording such as “the enzyme splits the DNA” tends to earn less than precise terms.

6. (a) At 80 minutes (2 generations): 4 molecules, 2 hybrid and 2 heavy, so 50% intermediate and 50% heavy [1]. At 120 minutes (3 generations): 8 molecules, 2 hybrid and 6 heavy, so 25% intermediate and 75% heavy [1]. [2] (b) Conservative replication would give one band of old light DNA and one band of new heavy DNA at 40 minutes [1]. No intermediate band would appear, so a single intermediate band shows every molecule contained both old and new strands [1]. [2] (c) Dispersive replication would give a single band at 80 minutes [1], at a position heavier than intermediate (75% of the nitrogen would now be ¹⁵N), whereas semi-conservative gives two bands, intermediate and heavy, in equal amounts [1]. [2] Examiner insight: This version reverses the usual labelling, so take care which band is old and which is new. Credit tends to follow predictions that name the bands and their amounts, not just “it fits”.

7. Combined rate = 2 × 600 = 1200 base pairs per second [1]. Time = 3.6 × 10⁶ ÷ 1200 = 3000 s [1]. 3000 ÷ 60 = 50 minutes [1]. [3] Examiner insight: Two forks double the rate, and the question asks for minutes. Leaving the answer in seconds, or using one fork, are the two usual losses.

8. (a) Ribose [1]; adenine and three phosphate groups [1]. [2] (b) ATP hydrolase [1]; ADP and inorganic phosphate (Pi) [1]. [2] Examiner insight: ATP hydrolase is the enzyme named in the specification. Writing “ATPase” or “ATP synthase” for hydrolysis is a mismatch that is easy to avoid by learning the pair together.

9. (a) Net = 30.5 − 14.0 [1] = 16.5 kJ per mole released [1]. [2] (b) The inorganic phosphate released is transferred to the compound, which is phosphorylated [1]. A phosphorylated compound is often more reactive, so the reaction is more likely to take place [1]. [2] Examiner insight: State the direction of the net change. Name “phosphorylate” and “more reactive” together: the idea is incomplete if only one appears.

10. Moles of ATP = 2.54 ÷ 507 [1] = 0.00501 mol. Energy = 0.00501 × 30.5 [1] = 0.153 kJ [1]. [3] Examiner insight: Carry the unrounded mole value into the second step, then round once to 3 significant figures. State the unit with the final value.

11. (a) DNA has only four different bases [1]. Proteins contain many more different amino acids, so proteins looked more able to carry a large amount of information and DNA looked too simple [1]. [2] (b) Bacteria were grown in ¹⁵N then moved to ¹⁴N, and DNA was separated by density in a gradient [1]. After one generation there was one intermediate band, which rules out conservative replication [1]. After two generations there were intermediate and light bands in equal amounts, which rules out dispersive replication and fits semi-conservative replication [1]. Strength: the predictions of the three models were different, so the experiment could have failed, and several generations were sampled [1]. Limitation: it was done in bacteria, so other organisms would need further evidence [1]. Limitation: it shows how strands are distributed but not the enzymes involved or the accuracy of the base sequence copied [1]. [6] Examiner insight: An evaluation tends to gain credit when it includes both strengths and limitations, each tied to the evidence. Statements such as “it proved the model” with no band description are weaker.

12. (a) ATP contains ribose and adenine, as an RNA nucleotide does [1], but has three phosphate groups instead of one [1]. [2] (b) ATP hydrolase catalyses hydrolysis of ATP to ADP and Pi [1], using a water molecule [1]. ATP synthase catalyses the condensation of ADP and Pi to form ATP [1], during photosynthesis or respiration [1]. [4] Examiner insight: A comparison needs both enzymes covered in the same terms. Giving the direction of each reaction and the type (hydrolysis or condensation) is clearer than a description of energy alone.

Where marks are usually lost

• Stating “bonds” without saying hydrogen bonds for helicase or phosphodiester bonds for polymerase.
• Counting bases as base pairs in a double-stranded molecule.
• Forgetting that the percentages of all four bases must total 100.
• Treating the percentages of one strand as the percentages of the whole molecule.
• Writing that Meselson-Stahl “proved” the model without describing the bands at each generation.
• Giving only strengths in an evaluation, or only limitations.
• Confusing ATP hydrolase and ATP synthase, or writing “ATP contains energy” instead of linking energy to hydrolysis.
• Leaving a calculated answer without a unit or in the wrong unit.

Next steps

• Review the revision notes.
• Re-read the study guide.
• Course page: AQA A-level Biology.
• Checklist: AQA A-level Biology checklist.
• Take the free 10-minute diagnostics.
• Book a free trial class: trial.

Official syllabus

AQA, AS and A-level Biology (7401 and 7402), specification for first teaching 2015, sections 3.1.5.1, 3.1.5.2 and 3.1.6.