Pearson Edexcel International A Level Biology: Recombinant DNA, Genetic Engineering and Bioinformatics (YBI11) – 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 statements 8.17 to 8.22 of the Pearson Edexcel International Advanced Subsidiary/Advanced Level Biology specification (YBI11), Issue 2, in Topic 8. Topic 8 is in the International A2 (IA2) part of the course. Revise from the study guide and the revision notes, and see the Edexcel A Level Biology hub and checklist. The free diagnostics are another way to find weak spots.

Questions

1. Describe the roles of a restriction endonuclease and of DNA ligase in making recombinant DNA. [4]

2. A linear DNA molecule of 7,400 base pairs (bp) is cut by one restriction endonuclease at two sites, 1,500 bp and 5,200 bp from one end.

(a) Calculate the sizes of the fragments. [2] (b) State which fragment would travel furthest in gel electrophoresis, and why. [2]

3. A plasmid of 3,900 bp is cut once and a gene of 1,800 bp is ligated into it.

(a) Calculate the size of the recombinant plasmid. [1] (b) Describe how cutting the recombinant plasmid with the same enzyme could show that the gene is present. [2]

4. Explain why the human gene for a protein hormone is often obtained as cDNA before it is put into bacteria, and describe how the cDNA is made. [4]

5. Describe how recombinant DNA is inserted into:

(a) bacterial cells [3] (b) plant cells [2]

6. Bacteria were mixed with 0.005 µg of plasmid DNA and allowed to recover in 500 µl of broth. A 50 µl sample was spread on agar containing an antibiotic, and 60 colonies grew. Calculate the transformation efficiency in colonies per µg of DNA. [3]

7. A plasmid carries genes for resistance to ampicillin and to kanamycin. A gene was inserted into the kanamycin-resistance gene. After transformation, cells were spread on agar with ampicillin and 212 colonies grew. These were copied by replica plating onto agar containing both antibiotics, and 47 colonies grew.

(a) Calculate the number of colonies with the recombinant plasmid. [2] (b) Calculate this as a percentage of the colonies on the ampicillin plate. [1] (c) Explain the reasoning behind your answer to (a). [2]

8. A herd of genetically modified goats secretes a human protein in its milk at 1.5 g per dm³. The herd produces 400 dm³ of milk per day.

(a) Explain why the gene was joined to a promoter that is active only in mammary gland cells. [2] (b) Calculate the mass of protein available from 30 days of milk. [2] (c) Give one advantage and one disadvantage of using goats rather than bacteria. [2]

9. A microarray compares gene activity in tumour tissue and healthy tissue. Tumour cDNA is labelled red and healthy cDNA is labelled green.

Gene	Red intensity	Green intensity
1	5,600	700
2	900	880
3	150	1,200
4	2,400	1,200

(a) Describe how the labelled cDNA is prepared and how it identifies active genes on the array. [3] (b) Calculate the red to green ratio for each gene and state what each shows. [4]

10. (a) State what is meant by bioinformatics. [1] (b) Two aligned 20-base sequences of the same gene from different species are:

species P: ATGCCGTTAGGCATCCGATA species Q: ATGCCCTTAGGAATCCGTTA

Calculate the percentage of bases that are identical. [2] (c) Suggest what a higher percentage identity suggests about two species, and give one other use of bioinformatics. [2]

11. A trial of a GM crop gave the following results.

	Non-GM	GM
Yield in tonnes per hectare	5.2	6.5
Seed cost per hectare	30	45
Herbicide applications per season	4	1

Evaluate the use of this crop, using the data and your own knowledge of risks and benefits. [8]

12. Compare antibiotic-resistance marker genes with a gene for green fluorescent protein as ways of identifying transformed bacteria. [3]

Answers

1. Restriction endonuclease: cuts DNA at a specific recognition sequence [1], hydrolysing the sugar-phosphate backbone, often leaving sticky ends [1]. DNA ligase: forms phosphodiester bonds between the backbones [1], permanently joining the gene to the vector once sticky ends have paired [1]. [4] Examiner insight: Name the bond that ligase makes; “joins the pieces” is a description of the result, not of the enzyme’s action.

2. (a) First fragment 1,500 bp [1]. Middle fragment 5,200 − 1,500 = 3,700 bp, and end fragment 7,400 − 5,200 = 2,200 bp [1]. [2] (b) The 1,500 bp fragment [1] because it is the smallest and moves most easily through the gel pores [1]. [2] Examiner insight: A linear molecule cut twice gives three fragments; stating only two sizes is a common slip, so check that your fragments add to the total.

3. (a) 3,900 + 1,800 = 5,700 bp [1]. [1] (b) The enzyme should cut at the two ends of the insert, releasing it [1]; on a gel there should be a band at 1,800 bp and one at 3,900 bp [1]. [2] Examiner insight: Give the expected band sizes in a “show it is present” answer; “there will be two bands” alone gives no way to check.

4. The gene in genomic DNA contains introns [1]; bacteria cannot splice out introns, so the protein would not be made correctly [1]. mRNA for the protein is extracted from cells that make it [1], and reverse transcriptase copies it into DNA, then DNA polymerase makes the second strand [1]. [4] Examiner insight: Both halves of a “why and how” question need marks; a full method with no reason for using cDNA leaves two marks uncollected.

5. (a) Mix the bacteria with the plasmids in calcium chloride solution, chilled [1]. Apply a brief heat shock, about 42 °C [1], making the membrane permeable so the plasmids enter [1]. Electroporation is an alternative. [3] (b) Put the gene into the Ti plasmid of Agrobacterium tumefaciens [1], which infects plant cells and transfers the DNA into the plant chromosomes [1]. A gene gun is an alternative. [2] Examiner insight: “Heat shock” alone does not explain anything; link it to membrane permeability.

6. Total colonies in 500 µl = 60 × (500 ÷ 50) = 600 [1]. Efficiency = 600 ÷ 0.005 [1] = 120,000 colonies per µg [1]. [3] Examiner insight: Scaling up from the sample plated to the whole volume is the step most often missed; state it clearly and include units.

7. (a) Colonies with a plasmid = 212 [1]. Recombinant = 212 − 47 = 165 [1]. [2] (b) 165 ÷ 212 × 100 = 77.8% [1]. [1] (c) Growth on ampicillin shows the cell has the plasmid [1]. Failure to grow on kanamycin shows the kanamycin-resistance gene was disrupted by the insert, so these cells carry the gene [1]. [2] Examiner insight: Explain insertional inactivation in terms of the gene being broken; stating that colonies are “recombinant” repeats the question.

8. (a) The promoter switches on transcription only in mammary gland cells [1], so the protein is made in the milk, where it is easily collected, and not in other tissues [1]. [2] (b) Per day: 1.5 × 400 = 600 g [1]. Over 30 days: 600 × 30 = 18,000 g = 18.0 kg [1]. [2] (c) Advantage: large volumes of protein with complex folding made by an animal cell, or a continuous supply [1]. Disadvantage: slow to produce a new herd, expensive, or animal welfare concerns [1]. [2] Examiner insight: Give a comparative point against bacteria; comments on goats alone do not answer “rather than”.

9. (a) Extract mRNA from each tissue and use reverse transcriptase to make cDNA, labelled with a fluorescent dye [1]. Apply to the array so that cDNA binds to complementary DNA probes [1]. Wash off unbound cDNA and scan; brighter fluorescence shows more mRNA, so a more active gene [1]. [3] (b) Gene 1: 5,600 ÷ 700 = 8.0, much more active in tumour [1]. Gene 2: 900 ÷ 880 = 1.02, about equal [1]. Gene 3: 150 ÷ 1,200 = 0.125, more active in healthy tissue [1]. Gene 4: 2,400 ÷ 1,200 = 2.0, twice as active in tumour [1]. [4] Examiner insight: Interpret each ratio in words as well as stating the number, and say which tissue is more active.

10. (a) The use of computers and software to store, analyse and compare biological data, such as DNA and protein sequences [1]. [1] (b) Count identical positions: 17 of 20 [1]; 17 ÷ 20 × 100 = 85% [1]. [2] (c) Species are more closely related, with a more recent common ancestor [1]. Other use: finding genes in a genome, or linking variants with disease [1]. [2] Examiner insight: Align and count positions carefully; miscounting by one changes the percentage and the answer to the following part.

11. Benefits, any three with data where possible: yield rises (6.5 − 5.2) ÷ 5.2 × 100 = 25.0% [1]; herbicide applications fall by 75%, saving labour and cost [1]; less herbicide may mean less pollution [1]. Also relevant: weeds controlled without damaging the crop. Risks, any three: seed costs 50% more [1]; gene flow to wild relatives could create herbicide-resistant weeds [1]; reliance on one herbicide may select for resistant weeds [1]. Also relevant: effects on biodiversity, or concerns about patents. Judgement that weighs the data, such as “worthwhile if the extra yield pays for the dearer seed” [1], with a condition or limit such as monitoring for resistance or the trial being limited [1]. [8] Examiner insight: An evaluation needs a conclusion that weighs the data; lists of benefits and risks with no weighing do not evaluate.

12. Antibiotic resistance: only transformed cells grow on agar with the antibiotic, so they are selected directly [1]. Fluorescent protein: transformed cells glow under UV light, so they can be identified without being killed or treated [1]. Comparison: the antibiotic method adds resistance genes, which may spread, while fluorescent markers need UV equipment [1]. [3] Examiner insight: “Compare” needs both methods in the same answer with a link between them; describing each method on its own, with no link between them, does not compare.

Where marks are usually lost

• Naming the enzymes but not the bond or the specific sequence involved.
• Forgetting that a linear molecule cut twice gives three pieces.
• Stating a gene sequence is “copied” without naming reverse transcriptase.
• Calling the heat shock a way of killing or weakening bacteria.
• Not scaling colony counts from the volume plated to the whole volume.
• Leaving out units such as bp and µg.
• Quoting a microarray ratio without saying which sample is more active.
• Giving risks or benefits without a final judgement in an evaluation.

Next steps

• Revision notes
• Study guide
• Edexcel A Level Biology hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Biology (YBI11), Specification, Issue 2, February 2021, Pearson Education Limited.