AQA A-level Biology: Genetic Diversity, Natural Selection and Required Practical 6 (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 cover AQA A-level Biology (7402) section 3.4.4, Genetic diversity and adaptation, with MS 2.5 and required practical 6, from the AQA AS and A-level Biology specification for first teaching 2015. Section 3.4.4 is topic 4, which is AS content as well as A-level content. Teaching is in the study guide and the revision notes. Course pages: AQA A-level Biology hub and the printable checklist.

Questions

1. (a) State what is meant by genetic diversity. [1] (b) Explain why a population with high genetic diversity is more likely to survive the arrival of a new disease. [2]

2. Six animals were genotyped for one gene. Their genotypes were A1A1, A1A2, A2A3, A3A4, A1A4 and A5A5. (a) State the number of different alleles of the gene in this sample. [1] (b) Calculate the frequency of allele A1 in the sample. [2]

3. A mosquito population was sprayed with an insecticide each year. After ten years most mosquitoes survived spraying. Explain how this change occurred. [4]

4. A broth culture contains 4.0 × 10⁸ bacterial cells. One cell in 5.0 × 10⁶ carries an allele for antibiotic resistance. (a) Calculate the number of resistant cells. [1] (b) An antibiotic is added. Only 1 in 10⁴ of the susceptible cells survive. Calculate the number of susceptible survivors, treating the number of susceptible cells as 4.0 × 10⁸. [2] (c) Calculate the percentage of the surviving cells that are resistant, giving your answer to 2 significant figures. [2]

5. The number of bacteria per cm³ in a culture was counted every hour.

Time (h)	0	1	2	3
Count per cm³	2.5 × 10³	2.0 × 10⁴	1.6 × 10⁵	1.28 × 10⁶

(a) Calculate log₁₀ of the count at 2 hours, to 3 decimal places. [1] (b) Give one reason for plotting log₁₀ of the count rather than the count itself. [1] (c) The log₁₀ values at 0 h and 3 h are 3.398 and 6.107. Use them to calculate the generation time in minutes. Doubling adds 0.301 to log₁₀ count. [3]

6. The percentage of weeds resistant to a herbicide in a field was recorded for seven years of spraying.

Year	1	2	3	4	5	6	7
Resistant (%)	2	5	11	24	41	58	72

(a) Describe the trend and name the type of selection shown. [2] (b) Calculate the largest rise in the percentage between consecutive years. [1] (c) Calculate the percentage of susceptible weeds in year 7. [1] (d) A farmer says the data prove the herbicide caused the change. Give one reason for caution. [1]

7. Birth mass was recorded for babies in a hospital. Mean birth mass was 3.2 kg with a standard deviation of 0.55 kg. Among the babies alive after four weeks, the mean was 3.2 kg and the standard deviation was 0.40 kg. (a) Name the type of selection and give two pieces of evidence from the data. [2] (b) Calculate the range of mean ± 2 standard deviations for each group. [2] (c) Calculate the percentage decrease in the width of this range. [1] (d) Suggest one reason why very low birth mass reduces the chance of survival. [1]

8. (a) Classify each adaptation as anatomical, physiological or behavioural. [3] (i) A desert hare has very long ears with many blood vessels near the surface. (ii) A desert rodent produces very concentrated urine. (iii) Male birds perform a display to attract a mate. (b) Give the term for an environmental factor that makes some alleles more advantageous than others. [1]

9. Describe the aseptic precautions you would take when spreading a bacterial culture on an agar plate to test antibiotics. [6]

10. A bacterial lawn was grown on three plates. A paper disc soaked in antibiotic A was placed on each plate. The clear-zone diameters were 16, 18 and 17 mm. The diameters for antibiotic B were 9, 10 and 8 mm (mean 9 mm). (a) Calculate the mean diameter for A. [1] (b) Calculate the area of the mean clear zone for A, in mm² to 3 significant figures. [2] (c) The area for B is 63.6 mm². Calculate how many times larger A’s area is than B’s, to 3 significant figures. [2] (d) State the purpose of a disc soaked in sterile water. [1]

11. The frequency of a resistance allele in an insect pest was measured in two fields for five years. One was sprayed each year and one was not.

Year	0	1	2	3	4	5
Sprayed field	0.01	0.04	0.13	0.34	0.62	0.81
Unsprayed field	0.01	0.01	0.02	0.01	0.02	0.01

(a) Describe the change in the sprayed field. [2] (b) Calculate the largest increase in frequency between consecutive years in the sprayed field. [2] (c) Explain the change in the sprayed field. [3] (d) Suggest why the frequency stayed low in the unsprayed field. [1]

12. (a) A patient stops taking antibiotics after two days because they feel better. Explain why doctors advise finishing the course. [3] (b) Explain how adaptation and selection contribute to the diversity of living organisms. [3]

Answers

1. (a) The number of different alleles of genes in a population [1]. (b) A wider range of alleles means some individuals may carry an allele giving resistance or survival [1]. These survive and reproduce, passing the allele on, so the population does not die out [1]. [3] Examiner insight: Defining diversity as “variety” without alleles tends to lose credit. Link alleles to survival and reproduction.

2. (a) Alleles A1, A2, A3, A4 and A5, so 5 [1]. (b) Total copies = 6 × 2 = 12 [1]. A1 appears in A1A1 (2), A1A2 (1) and A1A4 (1) = 4 copies. Frequency = 4/12 = 0.33 [1]. [3] Examiner insight: Count alleles, not animals. Show the total copies and the copies of the allele.

3. Random mutation produced an insecticide-resistance allele in some mosquitoes [1]. Insecticide is the selection pressure; susceptible mosquitoes died and resistant ones were more likely to survive [1]. Survivors reproduced more and passed the allele to their offspring [1]. Over generations the frequency of the allele increased [1]. [4] Examiner insight: Each step is a separate point. Do not write that mosquitoes “became” resistant or that the insecticide caused the mutation.

4. (a) 4.0 × 10⁸ / 5.0 × 10⁶ = 80 [1]. (b) Susceptible cells survive: 4.0 × 10⁸ × 10⁻⁴ [1] = 4.0 × 10⁴ (40 000) [1]. (c) Resistant percentage = 80 / (40 000 + 80) × 100 [1] = 0.20% [1]. [5] Examiner insight: Add the resistant cells to the survivors when finding the total. Give the percentage to the significant figures asked.

5. (a) log₁₀ (1.6 × 10⁵) = 5.204 [1]. (b) The counts cover several orders of magnitude, so a log scale fits them on one axis [1]. (c) Slope = (6.107 − 3.398) / 3 = 0.903 per hour [1]. Doublings per hour = 0.903 / 0.301 = 3 [1]. Generation time = 60 / 3 = 20 minutes [1]. [5] Examiner insight: Convert hours to minutes at the end and give the unit. “Easier” is weaker than saying the data span orders of magnitude.

6. (a) The percentage of resistant weeds increased every year from 2% to 72% [1]. This is directional selection [1]. (b) Differences are 3, 6, 13, 17, 17 and 14, so the largest is 17 percentage points [1]. (c) 100 − 72 = 28% [1]. (d) Correlation does not prove cause; there may be other factors, and there is no control field without spraying [1]. [5] Examiner insight: Quote start and end figures with units when describing a trend. For a change in a percentage, “percentage points” is the clearer wording.

7. (a) Stabilising selection [1]. The mean is unchanged (3.2 kg) and the standard deviation decreased from 0.55 to 0.40 kg [1]. (b) Before: 3.2 ± 1.10 = 2.10 to 4.30 kg [1]. After: 3.2 ± 0.80 = 2.40 to 4.00 kg [1]. (c) Widths are 2.2 and 1.6 kg; decrease = 0.6 / 2.2 × 100 = 27% [1]. (d) Small babies lose heat quickly (large surface area compared with volume) or have underdeveloped organs [1]. [6] Examiner insight: Referring to both the mean and the spread separates stabilising from directional selection. Show your working for the range.

8. (a) (i) Anatomical [1]. (ii) Physiological [1]. (iii) Behavioural [1]. (b) Selection pressure [1]. [4] Examiner insight: Decide whether the feature is a structure, a process or an action. If two categories fit, the reason you give matters.

9. Any six points, one mark each: wash hands and disinfect the bench [1]; work near a Bunsen flame so rising air carries microbes away [1]; flame the loop or spreader until red hot and cool it before use [1]; use sterile equipment such as pipettes and forceps [1]; open the lid as little as possible [1]; tape the lid so it does not seal fully [1]. Also creditworthy: invert the plate; incubate below body temperature to reduce the growth of pathogens. [6] Examiner insight: Give a reason with each precaution. Repeating one idea in different words adds nothing.

10. (a) (16 + 18 + 17) / 3 = 17 mm [1]. (b) Area = π × 8.5² [1] = 227 mm² [1]. (c) 227 / 63.6 [1] = 3.57 times [1]. (d) To show that the disc and water alone do not inhibit growth [1]. [6] Examiner insight: Use the radius, not the diameter, in the area formula. Comparing areas, not diameters, gives the true size of the effect.

11. (a) The frequency of the allele increased each year [1] from 0.01 to 0.81 [1]. (b) Increases are 0.03, 0.09, 0.21, 0.28 and 0.19 [1], so the largest is 0.28 (year 3 to 4) [1]. (c) Random mutation produced the resistance allele [1]. Spraying was the selection pressure, so resistant insects survived and reproduced more [1]. They passed on the allele, so its frequency rose over generations [1]. (d) Without spraying the allele gives no advantage (and may carry a cost), so resistant insects do not reproduce more than others [1]. [8] Examiner insight: Use the unsprayed field as the control and refer to both fields.

12. (a) Stopping early may leave the most tolerant bacteria alive [1]. These reproduce and pass on the resistance allele [1]. The population becomes more resistant, so the infection is harder to treat [1]. (b) Different environments exert different selection pressures [1]. Different alleles are advantageous, so allele frequencies change differently in each population [1]. Over many generations populations become adapted in different ways, giving a range of different organisms [1]. [6] Examiner insight: Link each part to the selection chain, not to survival alone.

Where marks are usually lost

• Describing genetic diversity as variety between organisms, without alleles.
• Writing that organisms change because they need to.
• Leaving out inheritance of the allele or the rise in allele frequency.
• Forgetting that the mean does not change in stabilising selection.
• Mixing up percentage and percentage points.
• Using diameter instead of radius in area calculations.
• Leaving out units, or giving the wrong number of significant figures.

Next steps

• Revision notes: revision notes
• Study guide: study guide
• AQA A-level Biology hub and printable checklist
• All free 10-minute diagnostics: diagnostics
• Book a free trial class: trial

Official syllabus

This page follows the AQA AS and A-level Biology specification (7401 and 7402), for first teaching 2015, published by AQA, section 3.4.4 and required practical 6.