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KS3 · Biology

K4D Reproduction and Inheritance: Exam Drill

AQA AQA KS3 Biology - Reproduction and Inheritance · Calculator allowed · about 75 minutes

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

State one reason why living organisms reproduce.

(1)

Name the male gamete and the female gamete in humans.

(a)Male gamete(1)

(b)Female gamete(1)

Match each part of the female reproductive system to its main role. 1) Ovary 2) Fallopian tube 3) Uterus

(1)Ovary - role(1)

(2)Fallopian tube - role(1)

(3)Uterus - role(1)

Describe what happens during fertilisation in humans. Your answer should include where fertilisation usually happens and what the immediate result is.

(4)

A pea plant with purple flowers (P) is crossed with a pea plant with white flowers (p). Purple is dominant. Use a simple genetic cross to answer the parts below. Show the parents and the possible genotypes and phenotypes of the offspring.

(a)If the purple parent is heterozygous (Pp) and the white parent is homozygous recessive (pp), write the genotypes of the two parents.(1)

(b)Complete a Punnett square and give the genotypes of the offspring with their probabilities.(2)

(c)Give the expected flower colours of the offspring and their probabilities.(1)

Extended response: A student proposes an investigation to show how a particular environmental factor affects plant variation. The student plans to grow two groups of genetically identical radish plants: Group A in full sunlight and Group B in partial shade. They will measure the length of the main root after 4 weeks. Discuss how the student should design a fair investigation, what results might show about variation caused by environment, and one limitation of the investigation. You will be assessed on using scientific ideas, planning a fair test and evaluating the method.

(6)

State whether the following are examples of genetic variation or environmental variation. 1) Eye colour in humans 2) Scars from a cut 3) Height differences between two siblings raised in different diets

(1)Eye colour(1)

(2)Scars from a cut(1)

(3)Height differences due to different diets(1)

Study the family tree information below. A family has a genetic condition caused by a recessive allele r. Parents X and Y have no symptoms but they have a child Z who displays the condition. Use this information to answer the parts.

Part a: What must the genotype of child Z be? Part b: What are the possible genotypes of parents X and Y? Part c: If X and Y have another child, what is the probability this child will be a carrier but not display the condition? Assume both parents are carriers.

(a)Genotype of child Z(1)

(b)Possible genotypes of parents X and Y(2)

(c)Probability the next child is a carrier but not affected (from two carriers Rr x Rr)(1)

A student measured the height of 6 seedlings each day for 5 days. The mean heights (in cm) are shown in the table: Day 1: 2.0, Day 2: 3.1, Day 3: 4.0, Day 4: 4.8, Day 5: 5.3.

a) Calculate the average growth rate in cm per day between Day 1 and Day 5. Give your answer to two decimal places. b) What type of graph would be best to show this change over time?

(a)Average growth rate between Day 1 and Day 5 in cm per day(3)

(b)Best type of graph to show change over time(1)

A student wants to test whether watering frequency affects the number of flowers on a bean plant. Identify one variable the student should keep the same to make the test fair and explain why keeping it the same is important.

(3)

Explain how both mutations and environmental factors can influence variation in a population. Give one example for a mutation and one example for an environmental factor.

(5)

In peas, round seeds (R) are dominant to wrinkled seeds (r). A pea plant that shows round seeds is crossed with a plant showing wrinkled seeds. The offspring are 50% round and 50% wrinkled. What was the genotype of the round parent? Explain your answer briefly.

(3)

Mark scheme · K4D Reproduction and Inheritance: Exam Drill

Question 1

B1 to produce more organisms / to make offspring cao
Answer: To produce more organisms (to make offspring).

Question 2

(a) B1 sperm (sperm cell) cao
(a) Answer: Sperm
(b) B1 egg (ovum) cao
(b) Answer: Egg (ovum)

Question 3

(1) B1 produces eggs / releases ova cao
(1) Answer: Produces eggs / releases ova
(2) B1 carries egg to the uterus / site where fertilisation often occurs (accept either)
(2) Answer: Carries egg to the uterus; site where fertilisation often occurs
(3) B1 nurtures and supports a developing embryo / site of foetus development cao
(3) Answer: Nurtures and supports a developing embryo; site of foetus development

Question 4

M1 sperm meets egg / sperm fuses with egg in the Fallopian tube
A1 fertilisation usually in the Fallopian tube / oviduct cao
A1 nuclei combine or sperm nucleus enters egg (zygote formed) cao
A1 result is a fertilised egg / zygote that begins to divide (embryo) cao
Answer: Sperm meets and fuses with the egg in the Fallopian tube; the nuclei combine to form a fertilised egg (zygote) that begins to divide to form an embryo.

Question 5

(a) B1 Pp and pp cao
(a) Answer: Pp and pp
(b) M1 Punnett square showing Pp and pp offspring (P from one parent on one axis, p and p on the other)
(b) A1 offspring genotypes: 50% Pp, 50% pp cao
(b) Answer: Offspring genotypes: 50% Pp, 50% pp
(c) B1 50% purple (Pp), 50% white (pp) cao
(c) Answer: 50% purple, 50% white

Question 6

Level 3 (5-6): Detailed answer that describes a fair experimental design controlling variables, predicts clear results linking environment to variation, and evaluates a limitation with a suggestion to improve. Uses correct scientific ideas about environmental variation and reproducibility.
Level 2 (3-4): Answer gives a reasonable plan with some controls, a simple prediction about differences caused by light, and mentions at least one limitation. Scientific ideas are partly developed.
Level 1 (0-2): Basic or incomplete ideas: may state a simple prediction or an uncontrolled method with little explanation.
Indicative content:
- Fair test: use genetically identical seeds, same soil type, same watering schedule, same pot size, same temperature, random assignment to groups, sufficient sample size (e.g. 10+ plants each) and repeat the experiment.
- Measurements: measure main root length after 4 weeks with a ruler to nearest mm, same person measuring or blind measurement to reduce bias, record individual values and calculate mean and range.
- Prediction: plants in partial shade may have longer roots or shorter roots depending on whether they invest in root growth to search for water; more likely shade causes longer stems and possibly shallower roots but environmental response must be justified.
- Data interpretation: compare means and ranges, consider whether differences are larger than natural variation; use standard presentation (table and bar chart) and consider anomalous results.
- Limitation and improvement: small sample size, uncontrolled pests or microclimate; improve by increasing replicates, rotating pot positions, or measuring additional traits like shoot length.
- Scientific ideas: environmental variation causes differences in phenotype without changes in genotype; results cannot show inheritance changes.

Question 7

(1) B1 genetic variation cao
(1) Answer: Genetic variation
(2) B1 environmental variation / acquired characteristic cao
(2) Answer: Environmental variation
(3) B1 environmental variation cao
(3) Answer: Environmental variation

Question 8

(a) B1 rr cao
(a) Answer: rr
(b) M1 both parents must carry at least one r allele
(b) A1 parents genotypes are most likely Rr and Rr (both carriers) cao
(b) Answer: Rr and Rr (both carriers)
(c) B1 50% (1/2) cao
(c) Answer: 50%

Question 9

(a) M1 use growth = 5.3 - 2.0 = 3.3 cm and divide by time 4 days
(a) A1 0.825 rounded to 0.83 cm/day cao
(a) A1 units cm/day stated
(a) Answer: 0.83 cm/day
(b) B1 line graph / line chart cao
(b) Answer: Line graph

Question 10

M1 identifies a correct control variable e.g. light level, pot size, plant variety, soil type, or temperature
A1 explains that keeping it constant prevents it affecting the number of flowers so the effect of watering can be isolated
B1 gives an example reason such as more light could increase flowering so it must be the same between groups

Question 11

M1 states that mutation is a change in DNA that can create new alleles
A1 gives an example mutation effect, e.g. a mutation causing albinism or altered coat colour in animals
M1 states that environmental factors change the way genes are expressed or affect phenotype without changing DNA
A1 gives an example environmental effect, e.g. nutrition affecting height or sunlight affecting leaf size
A1 links both to population variation by saying these create differences between individuals and can affect survival/reproduction

Question 12

M1 states the round parent genotype is Rr (heterozygous)
A1 explains that crossing Rr with rr gives 1/2 Rr (round) and 1/2 rr (wrinkled)
B1 conclusion that this matches the observed 50% round and 50% wrinkled
Answer: Rr; because Rr x rr gives 50% Rr (round) and 50% rr (wrinkled).

Reproduction and Inheritance: Exam Drill - Worksheets, Questions and Revision