Secondary 4 Pure Biology Genetics Inheritance Quiz

Secondary 4 Pure Biology Quiz - Genetics Inheritance

Answer Key

Section A: Multiple Choice Questions

1. B — A different form of a gene [1]
Note: An allele is a variant form of a gene at a specific locus. Option A confuses alleles with chromosomes; Option C describes a gene, not an allele; Option D describes phenotype.

2. B — Brown eyes [1]
Note: Since B (brown) is dominant over b (blue), the heterozygous genotype Bb expresses the dominant phenotype (brown eyes).

3. A — All Rr [1]
Note: A cross between RR and rr produces 100% heterozygous offspring (Rr) in the F₁ generation. Option B describes the F₂ ratio from a monohybrid cross between two heterozygotes.

4. B — Meiosis I [1]
Note: Independent assortment occurs during metaphase I of meiosis, when homologous chromosome pairs align randomly at the cell equator, leading to different combinations of maternal and paternal chromosomes in gametes.

5. C — Males are more likely to express the trait than females [1]
Note: Males have only one X chromosome (XY), so a single recessive allele on the X chromosome will be expressed. Females need two copies of the recessive allele (XᵇXᵇ) to express the trait.

6. B — Crossing an individual of unknown genotype with a homozygous recessive individual [1]
Note: A test cross determines whether an individual showing a dominant phenotype is homozygous dominant or homozygous recessive by crossing with a homozygous recessive partner.

7. B — 9 : 3 : 3 : 1 [1]
Note: This is the classic dihybrid phenotypic ratio for a cross between two heterozygous parents (RrYy × RrYy), representing 9 dominant for both : 3 dominant first/recessive second : 3 recessive first/dominant second : 1 recessive for both.

8. B — A blood group A parent and a blood group B parent can have a child with blood group AB [1]
Note: In codominance, both alleles are fully expressed in the heterozygote. In blood group AB, both A and B antigens are expressed on the red blood cells. Options A and D describe incomplete dominance.

9. C — 50% [1]
Note: The cross is XᴮXᵇ × XᴮY. Sons inherit the X chromosome from the mother and Y from the father. There is a 50% chance the son inherits Xᵇ (colour-blind) and a 50% chance he inherits Xᴮ (normal).

10. C — It is influenced by both genetic and environmental factors [1]
Note: Continuous variation (e.g., height, skin colour) is polygenic and influenced by environmental factors. Options A, B, and D describe discontinuous variation.

Section B: Structured Questions

11.
(a) Heterozygous tall parent: Tt; Short parent: tt [2]
Award 1 mark for each correct genotype.

(b) Punnett square:

[2]
Award 2 marks for all four boxes correct. Award 1 mark for two or three correct boxes.

(c) 1 tall : 1 short (or 50% tall, 50% short) [1]

(d) Test cross [1]
Note: A test cross involves crossing an individual of unknown/dominant phenotype with a homozygous recessive individual to determine its genotype.

12.
(a) The disorder is autosomal recessive. [1]
Reasoning: Individual II-3 is unaffected but has an affected son (III-1), meaning II-3 must be a carrier. If the disorder were X-linked recessive, an unaffected female carrier (XᴬXᵃ) married to an unaffected male (XᴬY) could produce an affected son, but the fact that II-1 is an affected female rules out X-linked recessive (as the father I-1 would also need to be affected). Since both parents (I-1 and I-2) are unaffected but produced an affected child (II-1), the disorder must be recessive. The affected female II-1 confirms it is autosomal (not X-linked). [1]

(b) Genotype of II-3: Aa (heterozygous carrier) [1]
Explanation: II-3 is unaffected but has an affected son (III-1, genotype aa). The affected son must have inherited one recessive allele from each parent. Since II-3 is unaffected, she must be heterozygous (Aa). [1]

(c) Cross: Aa (II-3) × Aa (II-4, deduced as carrier since they have an affected son)
Probability of affected child (aa) = 1/4
Probability of daughter = 1/2
Probability of affected daughter = 1/4 × 1/2 = 1/8 [2]
Award 1 mark for correct parental genotypes, 1 mark for correct final answer with working.

13.
(a) A carrier is an individual who carries one copy of a recessive allele for a genetic disorder but does not show symptoms of the disease because the dominant allele masks its effect. [1]

(b) Cross: Ss × Ss

Genotypic ratio: 1 SS : 2 Ss : 1 ss
Probability of child with sickle cell anaemia (ss) = 1/4 (25%) [3]
Award 1 mark for correct parental cross setup, 1 mark for correct Punnett square, 1 mark for correct probability.

(c) In malaria-endemic regions, heterozygous carriers (Ss) have a survival advantage because the presence of some sickle-shaped red blood cells provides resistance to malaria infection. This means carriers are more likely to survive and reproduce, passing on the s allele to the next generation. This phenomenon, called heterozygote advantage, maintains the sickle cell allele in the population despite the harmful effects in homozygous (ss) individuals. [2]
Award 1 mark for identifying heterozygote advantage/survival advantage, 1 mark for linking it to malaria resistance.

14.
(a) Codominance occurs when both alleles in a heterozygous individual are fully and equally expressed in the phenotype, resulting in a phenotype that shows both traits simultaneously rather than an intermediate blend. [2]
Award 1 mark for the concept of both alleles being expressed, 1 mark for stating that both traits appear in the phenotype.

(b) Parental phenotypes: roan female × white male
Parental genotypes: CᴿCᵀ × CᵀCᵀ

Gametes from female: Cᴿ, Cᵀ
Gametes from male: Cᵀ, Cᵀ

Offspring genotypes: 2 CᴿCᵀ : 2 CᵀCᵀ (or 1 CᴿCᵀ : 1 CᵀCᵀ)
Offspring phenotypes: 2 roan : 2 white
Phenotypic ratio: 1 roan : 1 white [3]
Award 1 mark for correct gametes, 1 mark for correct offspring genotypes, 1 mark for correct phenotypic ratio.

15.
(a) Tongue rolling is a dominant trait controlled by a single gene, and the ability to roll the tongue is more common than the inability in the population. [1]
Accept any reasonable hypothesis related to the inheritance pattern of tongue rolling.

(b) A sample size of 40 is relatively small and may not be representative of the general population. [1] Additionally, the class may not be a random sample — students may share similar genetic backgrounds (e.g., same ethnic group), which could bias the results. A larger, more diverse sample would be needed to draw reliable conclusions. [1]

(c) Nutrition (for traits like height) / Sunlight exposure (for skin colour) / Exercise (for muscle mass) [1]
Accept any valid environmental factor affecting a continuously variable trait.

Section C: Free Response Questions

16.
Meiosis is significant because it produces genetically unique gametes, which leads to genetic variation in offspring. This variation is essential for evolution by natural selection. [1]

Two specific events during meiosis that contribute to genetic diversity:

1. Independent assortment (Metaphase I): During metaphase I, homologous chromosome pairs line up randomly at the cell equator. The orientation of each pair is independent of other pairs, resulting in 2ⁿ possible combinations of chromosomes in gametes (where n = haploid number). In humans (n = 23), this produces over 8 million possible combinations. [1.5]

2. Crossing over (Prophase I): During prophase I, homologous chromosomes pair up and exchange segments of DNA at points called chiasmata. This recombination creates new combinations of alleles on each chromosome, increasing genetic variation beyond what independent assortment alone can produce. [1.5]

Award marks as indicated. Accept other valid points such as random fertilisation (bonus point, not required for full marks). Maximum 4 marks.

17.
(a) Males have only one X chromosome (XY). If the X chromosome carries the recessive haemophilia allele (Xʰ), the male will express the disorder because there is no corresponding allele on the Y chromosome to mask it. [1.5] Females have two chromosomes (XX), so they need two copies of the recessive allele (XʰXʰ) to express haemophilia. A female with only one copy (XᴴXʰ) is a carrier and does not show symptoms. Therefore, haemophilia is much more common in males. [1.5]

(b) Cross: XᴴXʰ (carrier female) × XʰY (haemophiliac male)

Gametes from female: Xᴴ, Xʰ
Gametes from male: Xʰ, Y

Daughters: XᴴXʰ (carrier, unaffected) and XʰXʰ (haemophiliac)
Probability that a daughter has haemophilia = 1/2 (50%) [3]
Award 1 mark for correct gametes, 1 mark for correct offspring genotypes, 1 mark for correct probability. Note: The question asks specifically about daughters, so only the XᴴXʰ and XʰXʰ outcomes are relevant.

18.

Biological significance:

• Mitosis produces genetically identical cells, which is important for growth, tissue repair, and asexual reproduction, ensuring that all body cells carry the same genetic information. [1]
• Meiosis produces genetically diverse gametes with half the chromosome number, which is essential for sexual reproduction. When two gametes fuse during fertilisation, the diploid number is restored. The genetic variation produced by meiosis provides the raw material for natural selection and evolution. [1]

Award 1 mark for each distinct difference (up to 3 marks) and 1 mark for each significance (up to 2 marks). Maximum 5 marks.

19.
(a) The red allele (R) is dominant. [1] When two red-flowered plants are crossed and produce white-flowered offspring, this indicates that both parents carry the recessive allele (r) and are heterozygous (Rr). The appearance of the recessive phenotype (white) in the offspring confirms that red is dominant over white. [1]

(b) Expected ratio from Rr × Rr cross: 3 red : 1 white [1]
Observed: 72 red : 24 white = 3 : 1
The observed ratio matches the expected Mendelian ratio exactly, so there is no significant difference. The results are consistent with a monohybrid cross between two heterozygotes, and any minor variation in other experiments would likely be due to chance (random fertilisation). [1]

(c) Both parent plants are Rr (heterozygous red) [1]

20.
(a) Both parents are heterozygous carriers (Cc). [1] Cystic fibrosis is autosomal recessive, so affected children must have the genotype cc. Each affected child inherits one recessive allele (c) from each parent. Since the parents are unaffected but have produced affected children, they must both be carriers (Cc). [1]

(b) Cross: Cc × Cc

Genotypic ratio: 1 CC : 2 Cc : 1 cc
Probability of being a carrier (Cc) = 2/4 = 1/2 (50%) [2]
Award 1 mark for correct cross/Punnett square, 1 mark for correct probability.

(c) Advantage: Genetic counselling can inform the couple of the probability of having another affected child, helping them make informed family planning decisions. It can also provide information about available treatments, support services, and prenatal testing options. [1]

Limitation: Genetic counselling cannot change the genetic risk — each pregnancy still has a 25% chance of producing an affected child. It may also cause emotional distress or raise ethical dilemmas about reproductive choices. Additionally, it cannot predict the severity of the condition in an affected child. [1]

Award 1 mark for a valid advantage and 1 mark for a valid limitation.

END OF ANSWER KEY