Pearson Edexcel International A Level Biology: Decomposition, PCR, Electrophoresis and Forensic Biology (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 6.16 to 6.20 of the Pearson Edexcel International Advanced Subsidiary/Advanced Level Biology specification (YBI11), Issue 2, in Topic 6. Topic 6 is in Unit 4, 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.

Questions

1. Explain how a saprotrophic fungus decomposes a fallen log, and how the carbon in the log returns to the atmosphere. [4]

2. Litter bags containing 18.0 g of dry leaves were buried for 90 days in two woodlands. In woodland A the leaves finally weighed 13.5 g. In woodland B they weighed 15.3 g.

(a) Calculate the percentage mass lost and the mean rate of mass loss per day in woodland A. [3] (b) Suggest two reasons why decomposition was slower in woodland B. [2]

3. The polymerase chain reaction (PCR) is carried out in a thermocycler.

(a) Describe what happens at each of the three stages of one cycle, giving a typical temperature for each. [3] (b) Explain why Taq polymerase is used. [1] (c) State the role of primers. [1]

4. A sample contains 12 target DNA molecules. Each PCR cycle takes 2.5 minutes. Assume every cycle doubles the number of target molecules.

(a) Calculate the number of copies after 22 cycles. [2] (b) Find the minimum number of whole cycles needed to exceed 1 000 000 000 copies, and the time this takes. [3]

5. Explain how gel electrophoresis separates DNA fragments of different lengths. [4]

6. A DNA ladder run on an agarose gel gave bands of 3200 bp at 14 mm, 1600 bp at 24 mm, 800 bp at 34 mm and 400 bp at 44 mm, measured from the wells. Each 10 mm of travel halves the length. Two fragments from one sample moved 29 mm (fragment P) and 19 mm (fragment Q).

(a) Estimate the lengths of P and Q. [4] (b) The two fragments came from one linear piece of DNA cut once. Estimate its original length. [1]

7. Hair roots found at a crime scene were profiled. The table shows repeat numbers at three loci.

Locus	Scene	Suspect 1	Suspect 2	Suspect 3
1	9, 12	9, 12	9, 12	9, 11
2	15, 15	15, 15	15, 16	15, 15
3	7, 10	7, 10	7, 10	7, 10

(a) State which suspects can be excluded and give the evidence. [3] (b) Explain why the match with Suspect 1 does not prove that Suspect 1 was at the scene. [2]

8. A seedling S grew from seed on a known mother tree M. Three trees (P1, P2 and P3) could be the pollen parent. Repeat numbers at two loci:

Locus	M	S	P1	P2	P3
1	6, 8	6, 11	9, 11	8, 12	11, 11
2	14, 17	17, 20	14, 20	20, 22	17, 23

(a) Identify the pollen parent, showing your reasoning. [4] (b) Give one other use of DNA profiling in animals or plants. [1]

9. A dead sheep is found at 23:00 with a body temperature of 29.0 °C. It was 37.0 °C at death and cooled at a constant 1.25 °C per hour (data given). The jaw and neck are rigid but the limbs are not.

(a) Calculate the estimated time of death. [3] (b) Explain the biological cause of the stiffness. [3]

10. A body is found in woodland at 09:00. The air temperature has averaged 18 °C. The body is at air temperature, there is no rigidity, the tissues are discoloured and swollen, and the oldest insects are larvae of a blowfly species that fed for 4 days after hatching. At 18 °C this species takes 1.5 days to hatch from eggs, 7 days as larvae and 11 days as pupae.

(a) Calculate the minimum time since death suggested by the larvae. [2] (b) Explain how the three other observations support a long interval. [3] (c) The real air temperature averaged 22 °C. State and explain the effect on your estimate. [2] (d) Explain why the larvae give a minimum interval. [2]

11. DNA from one hair root is amplified by PCR before profiling.

(a) Explain why PCR is needed. [2] (b) Explain why contamination by a worker’s skin cell could give a misleading result. [2]

Answers

1. The fungus secretes enzymes such as cellulase onto the log (extracellular digestion) through its hyphae [1]. The enzymes hydrolyse large molecules to small soluble ones [1]. The fungus absorbs the products [1]. Some are respired, and respiration releases carbon dioxide into the atmosphere [1]. [4] Examiner insight: Answers tend to gain credit when they name the process (respiration) that releases carbon dioxide, not just say that carbon “returns”.

2. (a) Mass lost = 18.0 − 13.5 = 4.5 g [1]. Percentage = 4.5 / 18.0 × 100 = 25.0% [1]. Rate = 4.5 / 90 = 0.050 g per day [1]. (b) Any two from (one mark each): colder, so slower enzyme activity; waterlogged or low oxygen, limiting aerobic respiration of saprotrophs; more acidic soil, slowing enzyme activity; drier conditions [2]. [5] Examiner insight: A reason such as “colder” gains more when linked to an effect, for example slower enzyme activity or fewer saprotrophs.

3. (a) Denaturation at about 95 °C: hydrogen bonds break and the strands separate [1]. Annealing at about 50 to 65 °C: primers bind to complementary sequences on each strand [1]. At about 72 °C (extension) Taq polymerase lengthens each primer with free nucleotides, building new complementary strands [1]. (b) It is heat-stable, so it is not denatured at about 95 °C [1]. (c) Primers bind to the ends of the target sequence and give the polymerase a start point [1]. [5] Examiner insight: Answers on stages usually need both the temperature and the event for each stage, in the correct order.

4. (a) Copies = 12 × 2²² [1] = 50 331 648 [1]. (b) After 26 cycles: 12 × 2²⁶ = 805 306 368, which is too few [1]. After 27 cycles: 12 × 2²⁷ = 1 610 612 736, so 27 cycles [1]. Time = 27 × 2.5 = 67.5 minutes [1]. [5] Examiner insight: Showing the cycle just below the target, as well as the one above it, makes a minimum-number answer easy to follow.

5. The DNA is loaded into wells in an agarose gel covered with buffer, which conducts current [1]. Its phosphate groups make DNA negatively charged, so the fragments run towards the positive electrode once the current is switched on [1]. The gel is a mesh, so shorter fragments move more easily and travel further [1]. Fragments of the same length form a band, and band positions are compared with a ladder of known lengths [1]. [4] Examiner insight: Link each fact to its cause: charge to direction, and fragment length to distance moved.

6. (a) P: 5 mm past the 1600 bp band, so length = 1600 × 2^(−5/10) [1] = 1131, about 1100 bp [1]. Q: 5 mm before the 1600 bp band, so length = 1600 × 2^(5/10) [1] = 2263, about 2300 bp [1]. (b) 1131 + 2263 = 3394, about 3400 bp [1]. [5] Examiner insight: Because distance relates to the logarithm of length, a straight-line reading between two bands is likely to lose credit; state the halving rule you use.

7. (a) Suspect 2 is excluded because at locus 2 the suspect has 15, 16 but the scene sample has 15, 15 [1]. Suspect 3 is excluded because at locus 1 the suspect has 9, 11 but the scene sample has 9, 12 [1]. A single mismatch shows the samples did not come from the same person, so Suspect 1 is the only one not excluded [1]. (b) Only three loci were compared, so an unrelated person could share the profile by chance [1]. Hair could have been transferred innocently or the sample contaminated, or the DNA does not show when the hair was left [1]. [5] Examiner insight: Answers that quote the locus and the repeat numbers that differ are easier to credit than “the bands are different”.

8. (a) Locus 1: S has 6 and 11; M has 6, so 11 came from the pollen parent [1]. Locus 2: S has 17 and 20; M has 17, so 20 came from the pollen parent [1]. P2 lacks 11 at locus 1 and P3 lacks 20 at locus 2, so both are excluded [1]. P1 has 11 and 20, so it can be the pollen parent [1]. (b) Any one: checking parentage in a breeding programme, choosing unrelated mates in captive animals, identifying a crop variety, or tracing illegally traded wildlife or timber [1]. [5] Examiner insight: Work out the child’s allele from each parent first, then test each candidate; a candidate lacking either required allele is excluded.

9. (a) Fall in temperature = 37.0 − 29.0 = 8.0 °C [1]. Time = 8.0 / 1.25 = 6.4 hours = 6 hours 24 minutes [1]. Death at 23:00 − 6 h 24 min = 16:36 [1]. (b) After death no more ATP is made [1]. Myosin heads need ATP to detach from actin, so they stay attached [1]. Cross-bridges remain locked, so the muscles are rigid, starting in small muscles such as the jaw [1]. [6] Examiner insight: Time-of-day answers need a clock time; stopping at 6.4 hours may leave the question unanswered.

10. (a) Time = 1.5 + 4 [1] = 5.5 days [1]. (b) Body at air temperature means cooling is complete, so death was not recent [1]. No rigidity means rigor has passed as muscle proteins broke down [1]. Discolouration and swelling show putrefaction by gut bacteria [1]. (c) The estimate would be too high; the true time since death is shorter [1], because warmer conditions make insects develop faster, so the larvae reached this stage in less than 5.5 days [1]. (d) Insects arrive after death, so some time passed before the eggs were laid [1]. The larvae therefore show the time since egg laying, so death was at least this long ago [1]. [9] Examiner insight: Answers on temperature effects usually need the direction of the change and its cause, not just “it would be different”.

11. (a) A single hair root contains very little DNA [1]. PCR makes many copies, enough for profiling by electrophoresis [1]. (b) PCR amplifies any DNA present, including the worker’s [1]. The extra DNA adds bands, so the profile could be a mixture or match the wrong person [1]. [4] Examiner insight: Answers that say PCR amplifies “any DNA present” explain contamination better than “the sample gets dirty”.

Where marks are usually lost

• Describing decomposers as eating the body without extracellular enzymes and absorption.
• Not naming respiration as the route by which carbon dioxide is released.
• Giving PCR temperatures without saying what happens at each stage.
• Not saying Taq polymerase is heat-stable and so not denatured.
• Reading fragment sizes from a ladder by straight-line interpolation.
• Saying DNA moves to the negative electrode, or longer fragments move further.
• Ruling suspects in or out without quoting the loci that differ.
• Stopping at an elapsed time when asked for a time of death, or treating one indicator as exact.

Next steps

• Revision notes
• Study guide
• Edexcel A Level Biology hub and 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: Topic 6, statements 6.16 to 6.20.