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Secondary 4 Pure Biology Genetics Inheritance Quiz
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Questions
Secondary 4 Pure Biology Quiz - Genetics Inheritance
Name: ___________________________
Class: ___________________________
Date: ___________________________
Score: _____ / 40
Duration: 45 minutes
Total Marks: 40
Instructions:
- Answer all questions in the spaces provided.
- The number of marks is given in brackets [ ] at the end of each question or part question.
- Write your answers clearly and legibly.
- For calculation questions, show all working clearly.
- Diagrams are not drawn to scale unless stated otherwise.
Section A: Multiple Choice Questions (10 marks)
Answer all questions. For each question, choose the correct answer and write the letter (A, B, C, or D) in the box provided.
1. In a monohybrid cross between two heterozygous tall pea plants (Tt × Tt), what is the expected phenotypic ratio of the offspring? [1]
☐ A. 1 tall : 3 dwarf
☐ B. 3 tall : 1 dwarf
☐ C. 1 tall : 2 heterozygous : 1 dwarf
☐ D. All tall
2. Which of the following statements about alleles is correct? [1]
☐ A. An allele is a different gene that occupies the same locus
☐ B. Alleles are alternative forms of the same gene that occupy the same locus on homologous chromosomes
☐ C. A dominant allele is always more common in a population than a recessive allele
☐ D. Alleles are found only on sex chromosomes
3. A man with blood group A and a woman with blood group B have a child with blood group O. What are the genotypes of the parents? [1]
☐ A. IᴬIᴬ and IᴮIᴮ
☐ B. IᴬIᴬ and Iᴮi
☐ C. Iᴬi and Iᴮi
☐ D. IᴬIᴮ and ii
4. In humans, the allele for brown eyes (B) is dominant to the allele for blue eyes (b). A brown-eyed man whose mother had blue eyes marries a blue-eyed woman. What is the probability that their first child will have blue eyes? [1]
☐ A. 0%
☐ B. 25%
☐ C. 50%
☐ D. 75%
5. Which of the following describes codominance? [1]
☐ A. The heterozygous phenotype is intermediate between the two homozygous phenotypes
☐ B. Both alleles are expressed equally in the heterozygous phenotype
☐ C. One allele completely masks the expression of the other allele
☐ D. The gene is located on the X chromosome
6. A genetic cross between two plants with red flowers produces offspring in the ratio 3 red : 1 white. If a red-flowered offspring is crossed with a white-flowered plant, what phenotypic ratio is expected in the offspring? [1]
☐ A. All red
☐ B. 3 red : 1 white
☐ C. 1 red : 1 white
☐ D. All white
7. In Drosophila (fruit flies), the gene for wing shape is located on the X chromosome. Normal wings (N) are dominant to vestigial wings (n). A homozygous normal-winged female is crossed with a vestigial-winged male. What percentage of the male offspring will have vestigial wings? [1]
☐ A. 0%
☐ B. 25%
☐ C. 50%
☐ D. 100%
8. Which of the following is an example of a discontinuous variation in humans? [1]
☐ A. Height
☐ B. Skin colour
☐ C. Blood group
☐ D. Body mass
9. In a dihybrid cross involving two independently assorting genes (AaBb × AaBb), what fraction of the offspring will be homozygous recessive for both traits? [1]
☐ A. 1/4
☐ B. 1/8
☐ C. 1/16
☐ D. 1/32
10. A pedigree diagram shows a genetic condition that affects only males and is passed from affected fathers to all their sons. What is the most likely mode of inheritance? [1]
☐ A. Autosomal dominant
☐ B. Autosomal recessive
☐ C. X-linked dominant
☐ D. Y-linked
Section B: Structured Questions (20 marks)
Answer all questions in the spaces provided.
11. In garden peas, the allele for round seeds (R) is dominant to the allele for wrinkled seeds (r). A plant breeder crosses a pure-breeding round-seeded plant with a pure-breeding wrinkled-seeded plant.
(a) State the genotype of the F₁ generation. [1]
(b) The F₁ plants are self-pollinated to produce the F₂ generation. Using a genetic diagram, show the expected genotypic and phenotypic ratios of the F₂ generation. [3]
(c) The plant breeder obtains 480 F₂ seeds. Calculate the expected number of wrinkled seeds. Show your working. [2]
12. In humans, the ability to taste phenylthiocarbamide (PTC) is controlled by a single gene with two alleles. The ability to taste (T) is dominant to non-tasting (t).
A man who can taste PTC marries a woman who cannot taste PTC. They have four children: two can taste PTC and two cannot.
(a) State the genotype of the woman. [1]
(b) Deduce the genotype of the man. Explain your reasoning. [2]
(c) What is the probability that their next child will be a non-taster? [1]
13. Snapdragons (Antirrhinum) show incomplete dominance for flower colour. The allele for red flowers (Cᴿ) and the allele for white flowers (Cᵂ) are codominant, producing pink flowers in the heterozygous condition (CᴿCᵂ).
(a) A red-flowered snapdragon is crossed with a white-flowered snapdragon. State the phenotype and genotype of the F₁ offspring. [2]
(b) Two pink-flowered snapdragons are crossed. Using a genetic diagram, determine the phenotypic ratio of the offspring. [3]
14. The diagram below shows a pedigree for a rare genetic condition in a family.
<image_placeholder> id: Q14-fig1 type: diagram linked_question: Q14 description: Pedigree diagram showing a family tree with 3 generations. Generation I: two individuals (male and female), both unaffected. Generation II: four offspring - two males (one affected, one unaffected) and two females (both unaffected). Generation III: offspring of the affected male (married to unaffected female) - two sons (both affected) and one daughter (unaffected). Offspring of unaffected male (married to unaffected female) - one son and one daughter, both unaffected. Offspring of the two unaffected females (married to unaffected males) - all unaffected children. labels: Squares = males, circles = females; shaded = affected, unshaded = unaffected; Roman numerals for generations; Arabic numerals for individuals within generation values: Generation I: I-1 (male, unaffected), I-2 (female, unaffected). Generation II: II-1 (male, affected), II-2 (male, unaffected), II-3 (female, unaffected), II-4 (female, unaffected). Generation III: III-1 (male, affected), III-2 (male, affected), III-3 (female, unaffected) - children of II-1; III-4 (male, unaffected), III-5 (female, unaffected) - children of II-2; III-6, III-7 (unaffected) - children of II-3; III-8, III-9 (unaffected) - children of II-4. must_show: Clear pedigree symbols with shading for affected individuals, generation labels, individual numbers, marriage lines, and descent lines </image_placeholder>
(a) Using evidence from the pedigree, state whether the condition is likely to be dominant or recessive. Explain your reasoning. [2]
(b) State whether the condition is likely to be autosomal or sex-linked. Explain your reasoning. [2]
(c) Individual II-3 marries an unaffected man. What is the probability that their first child will be affected? [1]
15. In cats, coat colour is controlled by a gene on the X chromosome. The allele for orange fur (Xᴼ) is codominant with the allele for black fur (Xᴮ). Heterozygous females (XᴼXᴮ) have tortoiseshell coats.
A tortoiseshell female is crossed with an orange male.
(a) State the genotypes of the parents. [1]
(b) Using a genetic diagram, determine the phenotypes and genotypes of all possible offspring, distinguishing between male and female offspring. [4]
Section C: Data Analysis and Extended Response (10 marks)
16. A student investigated the inheritance of seed shape and seed colour in peas. The alleles for round (R) and wrinkled (r) seeds, and yellow (Y) and green (y) seeds, assort independently.
The student crossed a pure-breeding round, yellow-seeded plant with a pure-breeding wrinkled, green-seeded plant. The F₁ plants were all round and yellow. The F₁ plants were then self-pollinated, and 640 F₂ seeds were collected and classified.
The observed results were:
- Round, yellow: 362
- Round, green: 118
- Wrinkled, yellow: 108
- Wrinkled, green: 52
(a) State the expected phenotypic ratio for a dihybrid cross involving independent assortment. [1]
(b) Calculate the expected number of seeds in each phenotypic category for 640 total seeds. Show your working. [2]
(c) The student concludes that the observed results do not match the expected ratio exactly. Suggest two reasons why the observed results might differ from the expected ratio. [2]
17. The table below shows data on the inheritance of a genetic condition in a population over three generations.
| Generation | Number of Affected Individuals | Total Population Size |
|---|---|---|
| 1 | 2 | 100 |
| 2 | 8 | 200 |
| 3 | 18 | 300 |
(a) Calculate the frequency of affected individuals in each generation. Express your answers as percentages to one decimal place. [2]
(b) The condition is caused by a recessive allele. Assuming Hardy-Weinberg equilibrium, calculate the frequency of the recessive allele (q) in Generation 3. Show your working. [2]
(c) State two assumptions of the Hardy-Weinberg equilibrium that may not be met in this population. [2]
18. Haemophilia is an X-linked recessive disorder. A normal male marries a female carrier. They have a son who has haemophilia.
(a) Explain why males are more frequently affected by X-linked recessive disorders than females. [2]
(b) Using a genetic diagram, show the possible genotypes and phenotypes of the children from this marriage. [3]
(c) If the couple has another son, what is the probability that he will have haemophilia? [1]
19. In a certain species of plant, flower colour is controlled by two genes that interact (epistasis). Gene A controls pigment production: the dominant allele (A) allows pigment production, while the recessive allele (a) prevents pigment production (resulting in white flowers). Gene B controls pigment colour: the dominant allele (B) produces purple pigment, while the recessive allele (b) produces red pigment. Plants with genotype aa__ have white flowers regardless of Gene B alleles.
A plant with genotype AaBb is self-pollinated.
(a) Determine the phenotypic ratio of the offspring. [3]
(b) State the genotype(s) that would produce red flowers. [1]
20. Cystic fibrosis is an autosomal recessive genetic disorder. The allele for normal CFTR protein (F) is dominant to the allele for defective CFTR protein (f).
A couple, both of whom are carriers, seek genetic counselling. They already have one child with cystic fibrosis.
(a) Explain what is meant by the term "carrier" in the context of an autosomal recessive disorder. [1]
(b) What is the probability that their next child will have cystic fibrosis? Explain why the fact that they already have an affected child does not change this probability. [2]
(c) In a population where the frequency of the cystic fibrosis allele (f) is 0.02, calculate the expected frequency of carriers (heterozygotes) assuming Hardy-Weinberg equilibrium. Show your working. [2]
End of Quiz
Answers
Secondary 4 Pure Biology Quiz - Genetics Inheritance (Answer Key)
Total Marks: 40
Section A: Multiple Choice Questions (10 marks)
1. Answer: B [1]
Explanation: In a monohybrid cross Tt × Tt, the genotypic ratio is 1 TT : 2 Tt : 1 tt. Since T (tall) is dominant to t (dwarf), the phenotypic ratio is 3 tall : 1 dwarf.
Marking note: Option C shows the genotypic ratio, not phenotypic.
2. Answer: B [1]
Explanation: Alleles are alternative forms of the same gene that occupy the same locus on homologous chromosomes. They arise by mutation and differ in DNA sequence.
Common trap: Option A incorrectly states "different gene" – alleles are variants of the same gene. Option C is false – dominance does not determine population frequency. Option D is false – alleles exist on all chromosomes.
3. Answer: C [1]
Explanation: Blood group O has genotype ii. For a child to be ii, they must inherit i from each parent. Parent with blood group A must be Iᴬi (heterozygous) to pass on i. Parent with blood group B must be Iᴮi (heterozygous) to pass on i.
Marking note: Iᴬ and Iᴮ are codominant; i is recessive.
4. Answer: C [1]
Explanation: Brown-eyed man with a blue-eyed mother must be heterozygous (Bb) – he inherited b from his mother. Blue-eyed woman is bb. Cross: Bb × bb → 50% Bb (brown), 50% bb (blue). Probability = 50%.
5. Answer: B [1]
Explanation: Codominance = both alleles expressed equally in heterozygote (e.g., AB blood group, roan cattle). Option A describes incomplete dominance. Option C describes complete dominance. Option D describes sex-linkage.
6. Answer: C [1]
Explanation: 3 red : 1 white ratio indicates both parents are heterozygous (Rr × Rr). Red-flowered offspring could be RR or Rr (2/3 chance Rr, 1/3 chance RR). But crossing a red-flowered offspring with white (rr): if offspring is RR → all red; if Rr → 1 red : 1 white. Since we don't know which red offspring, but the question implies a typical red from the cross (which is 2/3 likely heterozygous), the expected test cross ratio for a heterozygous dominant is 1:1. Standard exam interpretation: the red-flowered plant from a 3:1 cross is assumed heterozygous for test cross → 1 red : 1 white.
7. Answer: A [1]
Explanation: Cross: XᴺXᴺ (female) × XⁿY (male). Female gametes: all Xᴺ. Male gametes: Xⁿ or Y. Male offspring inherit Y from father and Xᴺ from mother → all XᴺY (normal wings). 0% vestigial.
8. Answer: C [1]
Explanation: Discontinuous variation = distinct categories with no intermediates (e.g., blood groups A, B, AB, O). Height, skin colour, and body mass show continuous variation (polygenic + environmental influence).
9. Answer: C [1]
Explanation: Independent assortment: AaBb × AaBb. Probability of aa = 1/4. Probability of bb = 1/4. Probability of aabb = 1/4 × 1/4 = 1/16.
10. Answer: D [1]
Explanation: Condition affects only males, passed from affected fathers to all sons. Y-linked inheritance: father passes Y chromosome to all sons. Autosomal would affect both sexes. X-linked dominant would pass from father to all daughters, not sons. X-linked recessive would not pass from father to son (father gives Y to son).
Section B: Structured Questions (20 marks)
11. (a) Genotype of F₁: Rr [1]
Explanation: Pure-breeding round = RR. Pure-breeding wrinkled = rr. All gametes from RR carry R; all from rr carry r. All F₁ are heterozygous Rr.
(b) Genetic diagram for F₁ self-pollination (Rr × Rr): [3]
Parental genotypes: Rr × Rr
Gametes: R, r × R, r
Punnett square:
| R | r | |
|---|---|---|
| R | RR | Rr |
| r | Rr | rr |
Genotypic ratio: 1 RR : 2 Rr : 1 rr [1]
Phenotypic ratio: 3 round : 1 wrinkled [1]
Marking notes: 1 mark for correct gametes, 1 mark for correct genotypes in Punnett square, 1 mark for correct ratios stated clearly. Must show both genotypic and phenotypic ratios.
(c) Expected wrinkled seeds: [2]
Working: Phenotypic ratio wrinkled = 1/4 of total.
Expected number = 480 × 1/4 = 120 wrinkled seeds
Marking: 1 mark for correct fraction (1/4), 1 mark for correct calculation and answer with units.
12. (a) Woman's genotype: tt [1]
Explanation: Non-taster phenotype is recessive, so must be homozygous recessive.
(b) Man's genotype: Tt [2]
Reasoning: Man can taste PTC, so has at least one T allele (genotype TT or Tt). They have non-taster children (tt), who must inherit t from each parent. Therefore the man must carry a t allele → genotype Tt.
Marking: 1 mark for correct genotype, 1 mark for explanation referencing non-taster children inheriting t from father.
(c) Probability next child is non-taster: 50% (or 1/2, 0.5) [1]
Explanation: Cross Tt × tt → 1/2 Tt (taster) : 1/2 tt (non-taster). Each pregnancy is an independent event.
13. (a) F₁ phenotype: Pink flowers. Genotype: CᴿCᵂ [2]
Explanation: Red parent = CᴿCᴿ. White parent = CᵂCᵂ. All gametes from red carry Cᴿ; all from white carry Cᵂ. All F₁ are heterozygous CᴿCᵂ → pink (incomplete dominance/codominance).
Marking: 1 mark for phenotype, 1 mark for genotype.
(b) Cross: CᴿCᵂ × CᴿCᵂ [3]
Gametes: Cᴿ, Cᵂ × Cᴿ, Cᵂ
Punnett square:
| Cᴿ | Cᵂ | |
|---|---|---|
| Cᴿ | CᴿCᴿ (red) | CᴿCᵂ (pink) |
| Cᵂ | CᴿCᵂ (pink) | CᵂCᵂ (white) |
Phenotypic ratio: 1 red : 2 pink : 1 white [1]
Marking: 1 mark for correct gametes, 1 mark for correct Punnett square/genotypes, 1 mark for correct phenotypic ratio.
14. (a) Condition is dominant. [2]
Evidence: Affected individual (II-1) has unaffected parents (I-1 and I-2). If recessive, two unaffected parents could produce affected offspring only if both are carriers. But the condition appears in every generation (I → II → III), and affected father (II-1) passes condition to all his sons (III-1, III-2). This pattern (vertical transmission, affected parent → affected offspring) is characteristic of dominant inheritance.
Marking: 1 mark for "dominant", 1 mark for correct evidence (affected child of unaffected parents + vertical transmission).
(b) Condition is Y-linked. [2]
Evidence: Only males affected. Affected father (II-1) passes condition to all sons (III-1, III-2) but no daughters (III-3 unaffected). Affected males in Generation III (III-1, III-2) would pass to all their sons. Unaffected males (II-2, II-4's husbands, etc.) have no affected children. This is the hallmark of Y-linked inheritance (father-to-son transmission only).
Marking: 1 mark for "Y-linked" (or "sex-linked, Y chromosome"), 1 mark for evidence (only males affected, father-to-son transmission to all sons, no father-to-daughter transmission).
(c) Probability = 0% [1]
Explanation: Individual II-3 is an unaffected female. Y-linked conditions are only carried on the Y chromosome, which females do not have. She cannot pass a Y-linked allele to any child. Her husband is unaffected (no Y-linked allele). No child can be affected.
15. (a) Parents' genotypes:
Female (tortoiseshell): XᴼXᴮ [½]
Male (orange): XᴼY [½]
Marking: ½ mark each.
(b) Genetic diagram: [4]
Parental genotypes: XᴼXᴮ × XᴼY
Female gametes: Xᴼ, Xᴮ
Male gametes: Xᴼ, Y
Punnett square:
| Xᴼ (sperm) | Y (sperm) | |
|---|---|---|
| Xᴼ (egg) | XᴼXᴼ (orange female) | XᴼY (orange male) |
| Xᴮ (egg) | XᴼXᴮ (tortoiseshell female) | XᴮY (black male) |
Offspring phenotypes and genotypes:
- Female offspring: 50% XᴼXᴼ (orange), 50% XᴼXᴮ (tortoiseshell)
- Male offspring: 50% XᴼY (orange), 50% XᴮY (black)
Marking: 1 mark for correct gametes, 1 mark for correct Punnett square, 1 mark for correct female offspring genotypes/phenotypes, 1 mark for correct male offspring genotypes/phenotypes. Must distinguish sexes.
Section C: Data Analysis and Extended Response (10 marks)
16. (a) Expected phenotypic ratio: 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green [1]
Explanation: Standard dihybrid cross ratio for independently assorting genes with complete dominance.
(b) Expected numbers for 640 seeds: [2]
Working: Total ratio parts = 9 + 3 + 3 + 1 = 16
- Round yellow: 640 × 9/16 = 360
- Round green: 640 × 3/16 = 120
- Wrinkled yellow: 640 × 3/16 = 120
- Wrinkled green: 640 × 1/16 = 40
Marking: 1 mark for correct method (divide by 16, multiply by ratio parts), 1 mark for all four correct values.
(c) Two reasons for discrepancy: [2]
- Random sampling variation / chance – small sample sizes deviate from theoretical ratios due to random segregation of alleles during meiosis and random fertilisation.
- Non-random mating / selection bias – e.g., some seeds may not have germinated or been counted, or there could be differential survival of genotypes.
- Genes may not assort independently – linkage (though question states they do).
- Human error in classification – misidentifying phenotypes.
Marking: 1 mark each for any two valid reasons. Must be biological/statistical, not just "experimental error" without specification.
17. (a) Frequency of affected individuals: [2]
Generation 1: (2/100) × 100% = 2.0%
Generation 2: (8/200) × 100% = 4.0%
Generation 3: (18/300) × 100% = 6.0%
Marking: 1 mark for correct calculations, 1 mark for all three correct percentages to 1 d.p.
(b) Frequency of recessive allele (q) in Generation 3: [2]
Working: Affected frequency = q² = 18/300 = 0.06
q = √0.06 = 0.245 (or 0.24 to 2 d.p.)
Marking: 1 mark for q² = 0.06, 1 mark for correct square root and answer.
(c) Two Hardy-Weinberg assumptions not met: [2]
- No natural selection – affected frequency increases across generations (2% → 4% → 6%), suggesting selection or drift.
- No mutation / no migration / large population / random mating – any two valid assumptions.
Common valid answers: No selection, no mutation, no migration (gene flow), infinitely large population, random mating.
Marking: 1 mark each for two distinct, correctly stated assumptions.
18. (a) Why males more frequently affected by X-linked recessive disorders: [2]
Explanation: Males have only one X chromosome (XY). If they inherit a recessive allele on their single X, they express the condition because there is no second X chromosome with a dominant allele to mask it. Females have two X chromosomes (XX); they need two recessive alleles (homozygous) to express the condition, which is much less probable.
Marking: 1 mark for "males have only one X chromosome", 1 mark for "no second allele to mask recessive allele / hemizygous expression".
(b) Genetic diagram: [3]
Parental genotypes: Male (normal) = XᴴY; Female (carrier) = XᴴXʰ
Gametes: Male: Xᴴ, Y; Female: Xᴴ, Xʰ
Punnett square:
| Xᴴ (sperm) | Y (sperm) | |
|---|---|---|
| Xᴴ (egg) | XᴴXᴴ (normal female) | XᴴY (normal male) |
| Xʰ (egg) | XᴴXʰ (carrier female) | XʰY (haemophiliac male) |
Phenotypes:
- 25% normal female (XᴴXᴴ)
- 25% carrier female (XᴴXʰ)
- 25% normal male (XᴴY)
- 25% haemophiliac male (XʰY)
Marking: 1 mark for correct parental genotypes and gametes, 1 mark for correct Punnett square/offspring genotypes, 1 mark for correct phenotypes with sexes distinguished.
(c) Probability next son has haemophilia: 50% (1/2) [1]
Explanation: Each son inherits Y from father and one X from mother. Mother is carrier (XᴴXʰ), so 50% chance of passing Xʰ. Each pregnancy is independent.
19. (a) Phenotypic ratio from AaBb × AaBb (epistasis): [3]
Gene A (pigment production): A = pigment, a = no pigment (white)
Gene B (pigment colour): B = purple, b = red
Epistasis: aa masks Gene B → white regardless of B/b
Dihybrid cross AaBb × AaBb → 16 combinations:
- A_B_ (9/16): pigment produced + purple → purple
- A_bb (3/16): pigment produced + red → red
- aaB_ (3/16): no pigment → white
- aabb (1/16): no pigment → white
Phenotypic ratio: 9 purple : 3 red : 4 white (or 9:3:4)
Marking: 1 mark for identifying epistatic interaction (aa = white), 1 mark for correct genotype groupings, 1 mark for correct final ratio 9:3:4.
(b) Genotype(s) for red flowers: A_bb (AAbb or Aabb) [1]
Explanation: Must have at least one A allele (for pigment production) and two b alleles (for red pigment).
Marking: Accept "A_bb" or "AAbb and Aabb".
20. (a) Definition of carrier: [1]
Explanation: A carrier is an individual who is heterozygous (Ff) for an autosomal recessive disorder. They have one normal allele and one disease allele, do not show symptoms (phenotypically normal), but can pass the disease allele to offspring.
Marking: Must mention heterozygous, phenotypically normal, can transmit allele.
(b) Probability next child has cystic fibrosis: 25% (1/4) [2]
Explanation: Both parents are carriers (Ff × Ff). Each pregnancy: 1/4 FF (normal), 1/2 Ff (carrier), 1/4 ff (affected). The genotype of each child is determined by independent assortment and random fertilisation – previous children's genotypes do not influence future probabilities (independent events).
Marking: 1 mark for 1/4 or 25%, 1 mark for explanation of independence / Mendelian segregation each meiosis.
(c) Frequency of carriers (heterozygotes) if q = 0.02: [2]
Working: Hardy-Weinberg: p + q = 1 → p = 1 - 0.02 = 0.98
Carrier frequency = 2pq = 2 × 0.98 × 0.02 = 0.0392 (or 3.92%)
Marking: 1 mark for p = 0.98, 1 mark for 2pq calculation and correct answer.
End of Answer Key