OxfordAQA International AS and A-level Biology: Recombinant DNA technology (9610) – 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 cover section 3.4.10 (Recombinant DNA technology) of the OxfordAQA International AS and A-level Biology (9610) specification, Version 5.1, Unit 4. Read the study guide and revision notes first. Course pages: hub and checklist. Unsure where to start? Try the free diagnostics.

Questions

1. Explain why a human gene can be transcribed and translated in a bacterium. [3]

2. A researcher makes the gene for a hormone by starting with mRNA from the gland cells that make that hormone in large amounts. Explain why these cells are chosen, and describe how the gene is made from the mRNA. [4]

3. A circular plasmid of 6,800 base pairs (bp) has BamHI recognition sites at 1,150 bp and 3,900 bp.

(a) Calculate the lengths of the fragments after digestion with BamHI. [2] (b) State the number and length of fragments if the plasmid had only one BamHI site. [1] (c) State which fragment from part (a) lies closer to the wells on a gel, and why. [1]

4. Explain why a restriction endonuclease cuts DNA only at one recognition sequence, and why a gene and a vector are cut with the same enzyme. [4]

5. A synthetic gene is built for a peptide of 62 amino acids.

(a) Calculate the number of bases in the coding sequence, including one stop codon. [2] (b) Give one advantage of artificial gene synthesis over cutting the gene from the genome. [1] (c) Explain why the synthetic sequence may differ from the human sequence yet give the same peptide. [1]

6. A PCR tube starts with 5 template molecules. One cycle takes 70 s. The reaction runs for 20 cycles.

(a) Calculate the number of copies if every cycle copies perfectly. [2] (b) Calculate the time taken, in minutes to 1 decimal place. [2]

7. Describe what happens in each of the three stages of one PCR cycle, and explain why a heat-stable DNA polymerase is used. [5]

8. DNA from three patients (A, B, C) was tested with primers for a gene of a bacterium. Products were run on a gel. Lane 1: size markers. Lane 2 (A): a band at 420 bp. Lane 3 (B): no band. Lane 4 (C): a band at 420 bp. Lane 5: tube with no template, no band. Lane 6: DNA known to contain the gene, a band at 420 bp.

(a) State which patients’ samples contain the gene, and what the band shows. [2] (b) Explain the purpose of lane 5 and of lane 6. [2]

9. The gene made in question 2 is to be put into a plasmid and cultured in bacteria.

(a) Explain the roles of a promoter and of a terminator. [2] (b) Explain why the promoter must be one that the bacterium recognises. [1] (c) Describe how the gene is inserted into the plasmid. [3]

10. 0.025 µg of plasmid was added to bacteria, which recovered in 500 µl of broth. The plasmid carries an ampicillin-resistance gene and a gene for green fluorescent protein (GFP). A new gene was inserted into the GFP gene. A 50 µl sample on ampicillin agar gave 66 colonies, of which 45 fluoresced under UV light.

(a) Calculate the transformation efficiency in colonies per µg of plasmid. [3] (b) Explain why the colonies that do not fluoresce are the recombinant ones. [2] (c) Calculate the percentage of colonies that are recombinant. [1] (d) Cells with no plasmid were spread on ampicillin agar. State what no growth shows. [1]

11. A protein drug is extracted from animal tissue at 48 currency units per dose, with 120,000 doses a year. A company proposes making it in genetically modified microorganisms at 19 units per dose, with 2,000,000 doses a year. Development would cost 90 million units, and the company would hold the patent. Evaluate the proposal, considering financial, ethical and social issues. [8]

Answers

1. The genetic code is universal [1], so a codon specifies the same amino acid in human and bacterium [1]. The mechanisms of transcription and translation are also universal, so the bacterium’s RNA polymerase and ribosomes can read the gene (given a promoter it recognises) [1]. [3] Examiner insight: “The code is the same” is weaker than saying it is universal and linking codons to amino acids. Answers gain credit for stating both the code and the mechanisms.

2. These cells contain a lot of mRNA for the hormone, so there is plenty of template [1]. The mRNA has already had its introns removed [1]. Reverse transcriptase makes a single strand of DNA complementary to the mRNA [1]. DNA polymerase makes the second strand, giving double-stranded cDNA [1]. [4] Examiner insight: Name the enzymes and the template. “The mRNA is turned into DNA” does not say how.

3. (a) 3,900 − 1,150 = 2,750 bp [1]. 6,800 − 3,900 + 1,150 = 4,050 bp [1]. (b) One linear fragment of 6,800 bp [1]. (c) The 4,050 bp fragment, because larger fragments move more slowly through the gel [1]. [4] Examiner insight: Check that your fragments add up to the total. A circular molecule cut twice gives two fragments, not three.

4. The active site is complementary in shape to the recognition sequence only [1], so the enzyme does not fit other sequences [1]. The same enzyme leaves complementary sticky ends on the gene and the vector [1], which pair by hydrogen bonds between bases so that ligase can join them [1]. [4] Examiner insight: Explain specificity through the shape of the active site. Saying the enzyme “recognises” the sequence adds little.

5. (a) 62 × 3 = 186 [1]. 186 + 3 = 189 bases [1]. (b) No source organism or mRNA is needed, or the codons can be chosen to suit the host [1]. (c) The code is degenerate, so an amino acid can have more than one codon [1]. [4] Examiner insight: A calculation about a gene needs the stop codon only if the question says to include it. Read the wording.

6. (a) 2²⁰ = 1,048,576 [1]. 5 × 1,048,576 = 5,242,880 copies [1]. (b) 20 × 70 = 1,400 s [1]. 1,400 ÷ 60 = 23.3 minutes [1]. [4] Examiner insight: Show the power of 2 and the multiplication by the starting number separately, so a slip in one step does not hide the other.

7. Denaturation at about 95 °C: heat breaks the hydrogen bonds between the strands [1]. Annealing at about 50 to 65 °C: primers pair with their complementary sequences on the strands [1]. Extension at about 72 °C: polymerase extends each primer from its 3′ end, copying the template [1]. A heat-stable polymerase keeps its shape at about 95 °C [1], so it does not need replacing each cycle [1]. [5] Examiner insight: Give the event with each stage, not only the temperature. Mention of the primers at annealing is needed to show you know what is pairing.

8. (a) Patients A and C [1]. The band at 420 bp matches the product from the known gene, so the primers’ target sequence is present [1]. (b) Lane 5 shows that the reagents are not contaminated with DNA, since no band formed without template [1]. Lane 6 shows that the PCR conditions and primers work, so the absence of a band for B is a true negative [1]. [4] Examiner insight: For a gel, quote the band size and compare it with the control. A control lane needs both what it contains and what its result shows.

9. (a) RNA polymerase attaches at the promoter so that transcription can start [1]. The terminator stops transcription [1]. (b) If the bacterial RNA polymerase cannot bind, the gene is not transcribed [1]. (c) Cut the plasmid and the gene with the same restriction endonuclease [1], so that sticky ends are complementary and pair [1]. DNA ligase forms phosphodiester bonds between the sugar-phosphate backbones [1]. [6] Examiner insight: Say which end each region sits at, before and after the gene, and say that ligase joins backbones, not bases.

10. (a) Colonies in all the broth: 66 × (500 ÷ 50) = 660 [1]. 660 ÷ 0.025 [1] = 26,400 colonies per µg [1]. (b) The new gene was inserted in the GFP gene, which disrupts it [1], so cells with the insert cannot make GFP and do not glow, but they still resist ampicillin [1]. (c) (66 − 45) ÷ 66 × 100 = 21 ÷ 66 × 100 = 31.8% [1]. (d) No growth shows that the cells without plasmid are not resistant, so growth on ampicillin means a plasmid was taken up [1]. [7] Examiner insight: Scale the sample up to the whole volume before dividing by the mass of DNA, and give units. Explain a marker in terms of what happens to the gene, not just the colour.

11. Data: cost per dose falls by (48 − 19) ÷ 48 × 100 = 60.4% [1]. Supply rises 2,000,000 ÷ 120,000 = 16.7 times [1]. Financial: cheaper doses and much larger supply, but 90 million units of development cost must be recovered [1]. A patent may keep prices high [1]. Ethical: no animals are needed for tissue [1], but the transfer of a gene between species or the use of GM organisms concerns some people [1]. Social: more patients could be treated, though access depends on price and on who holds the patent [1]. Judgement: the proposal is worthwhile if the lower cost is passed on so that more people can afford the drug [1]. [8] Examiner insight: An evaluation answer gains credit when it uses the numbers and weighs one issue against another. A list with no judgement stays low.

Where marks are usually lost

• Calling the genetic code “the same” rather than universal.
• Leaving out introns when explaining why cDNA is used.
• Describing restriction endonucleases as scissors that cut anywhere.
• Forgetting that a circular molecule gives as many fragments as cut sites.
• Writing that ligase joins bases or hydrogen bonds.
• Missing the primers, or the temperatures, in the three stages of PCR.
• Dividing colonies by the mass of DNA without scaling up to the full volume.
• Giving only benefits or only risks in an evaluation.

Next steps

• Review with the revision notes and the study guide.
• Find your weak areas with the free diagnostics, and tick off topics on the checklist.
• Back to the course hub.
• Book a free trial class: trial.

Official syllabus

The OxfordAQA International AS and A-level Biology (9610) specification, Version 5.1, for International A-level exams May/June 2018 onwards, published by OxfordAQA.