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Secondary 3 Biology Genetics Inheritance Quiz
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Questions
Secondary 3 Biology Quiz – Genetics Inheritance
Name: ________________________________
Class: ________ Date: _____________
Score: _______ / 40
Duration: 45 minutes
Total Marks: 40
Instructions:
- Answer all questions.
- Write your answers in the spaces provided.
- Marks are shown in brackets at the end of each question or part-question.
- For multiple choice (Section A), circle the letter of the one best answer.
Section A: Multiple Choice (5 × 1 mark = 5 marks)
1. In pea plants, tall (T) is dominant over dwarf (t). A homozygous tall plant is crossed with a dwarf plant.
What proportion of the offspring will be tall?
A 0%
B 25%
C 50%
D 100%
2. The genetic makeup of an organism is known as its
A phenotype
B genotype
C allele
D chromosome
3. In humans, which combination of sex chromosomes results in a female?
A XX
B XY
C YY
D XO
4. Which of the following is an example of discontinuous variation?
A height
B weight
C skin colour
D ability to roll tongue
5. A test cross is carried out to determine
A the phenotype of a dominant organism
B the genotype of a recessive organism
C the genotype of a dominant organism
D the number of chromosomes
Section B: Short Answer (10 × 2 marks = 20 marks)
6. Define the term “allele”.
7. Distinguish between homozygous and heterozygous.
8. In fruit flies, long wings (L) are dominant over vestigial wings (l).
State the genotype of a fruit fly that is heterozygous for wing length.
9. A red-flowered snapdragon (RR) is crossed with a white-flowered snapdragon (WW) and all offspring have pink flowers.
Explain why no red or white flowers appear in the F₁ generation.
10. List two factors that contribute to continuous variation in a population.
11. Human blood groups A, B, AB and O are controlled by three alleles: Iᴬ, Iᴮ and i.
State the genotype of a person with blood group O.
12. What is the function of the SRY gene located on the Y chromosome?
13. A man with normal vision (Xᴺ Y) and a woman who is a carrier for colour blindness (Xᴺ Xⁿ) have a child.
State the probability that a son will be colour blind.
14. Explain why a mutation in a somatic cell is not passed on to offspring, but a mutation in a gamete can be.
15. In guinea pigs, black coat (B) is dominant to white coat (b).
If a heterozygous black guinea pig is crossed with a white guinea pig, what is the expected phenotypic ratio of the offspring?
Section C: Structured / Data Interpretation (5 × 3 marks = 15 marks)
16. The alleles for flower colour in a certain plant are R (red) and r (white). A homozygous red plant (RR) was crossed with a homozygous white plant (rr).
(a) Draw a Punnett square to show the cross.
(b) State the genotype(s) and phenotype(s) of the F₁ offspring.
Genotype(s): _________________
Phenotype(s): _________________
(c) If two F₁ plants are crossed, what will be the phenotypic ratio of the F₂ generation? Show your working.
Phenotypic ratio: _________________
Working:
17. The pedigree chart below shows the inheritance of a recessive genetic disorder in a family. Filled symbols indicate affected individuals.
1 □──○ 2
│
┌┴┐
3 ○ 4 ■ 5 □
│
┌─┴─┐
7 □ 8 □
(married to 6 □)
Individual 5 □ married to 9 ○
│
10 ●
(a) Using the allele D for normal and d for the disorder, state the genotype of individual 1.
(b) Explain why individual 4 must be homozygous recessive.
(c) Calculate the chance that individual 9 is a carrier.
18. In a certain breed of cattle, coat colour is controlled by codominant alleles: Cᴿ (red) and Cᵂ (white). Heterozygotes have roan (red and white hairs intermixed).
(a) State the genotype of a roan cow.
(b) If a roan cow is mated with a roan bull, what proportion of the calves are expected to be white? Show your working.
Proportion white: _________________
Working:
19. A geneticist isolated a new mutation in mice that causes a bent tail. When a bent‑tailed male was mated with a normal‑tailed female, all offspring had normal tails. When two of these F₁ offspring were mated, approximately 25 % of the F₂ had bent tails.
(a) Is the bent‑tail allele dominant or recessive? Explain your answer.
(b) Using B for normal and b for bent, state the genotype of the original bent‑tailed male.
20. Variation in human blood pressure is influenced by both genetic and environmental factors.
Explain, using blood pressure as an example, what is meant by “continuous variation” and give one environmental factor that can affect it.
END OF QUIZ
Answers
Secondary 3 Biology Quiz – Genetics Inheritance
Answer Key and Marking Scheme
Section A: Multiple Choice (1 mark each)
| Question | Answer | Explanation/Notes |
|---|---|---|
| 1 | D | Homozygous tall (TT) × dwarf (tt) → all offspring Tt (tall). |
| 2 | B | Genotype is the genetic constitution. |
| 3 | A | Females have XX; males have XY. |
| 4 | D | Tongue rolling is a discrete trait (you can or can’t roll). Others show a range of values. |
| 5 | C | A test cross (crossing with homozygous recessive) reveals the genotype of a dominant‑looking individual. |
Section B: Short Answer (2 marks each)
6.
An alternative form of a gene that occupies the same position (locus) on homologous chromosomes. [2]
7.
- Homozygous: an individual has two identical alleles for a given gene (e.g., TT or tt). [1]
- Heterozygous: an individual has two different alleles for a given gene (e.g., Tt). [1]
8.
Ll [2]
9.
- The red allele (R) and white allele (W) show incomplete dominance (or partial dominance). [1]
- In the heterozygous offspring (RW), neither allele is fully expressed; the combined effect produces a pink phenotype. [1]
(Accept equivalent explanation of blending / lack of complete dominance.)
10.
Any two from:
- polygenic inheritance (many genes involved)
- environmental influences (e.g., nutrition, temperature)
- additive effects of alleles
[1 mark each, max 2]
11.
ii (or I⁰I⁰) [2] – must indicate homozygosity for the i allele.
12.
- The SRY gene triggers male sex determination (development of testes). [2]
(Accept: “causes the embryo to develop into a male”.)
13.
50% (or ½)
Explanation: The woman’s gametes are Xᴺ or Xⁿ; the man’s gametes are Xᴺ or Y. Sons receive Y from father and either Xᴺ or Xⁿ from mother → 1 out of 2 sons receives Xⁿ and is colour blind. [2]
14.
- Somatic cells are body cells; their mutations affect only that organism and are not transmitted to offspring because they are not involved in gamete formation. [1]
- Gametes (sperm/egg) carry genetic information that will form the next generation; a mutation in a gamete can be passed to the zygote and every cell of the offspring. [1]
15.
Heterozygous black (Bb) × white (bb) → offspring: ½ Bb (black) : ½ bb (white).
Phenotypic ratio = 1 black : 1 white (or 50% black, 50% white). [2]
Section C: Structured / Data Interpretation (3 marks each)
16.
(a) Punnett square:
| R | R | |
|---|---|---|
| r | Rr | Rr |
| r | Rr | Rr |
[1] – correct gametes and filling.
(b) Genotype: Rr (all heterozygous).
Phenotype: red (dominant trait). [1]
(c) F₁ cross (Rr × Rr):
Punnett square gives RR, Rr, rR, rr → 3 red (RR, Rr) : 1 white (rr).
Phenotypic ratio = 3 red : 1 white. [1]
17.
(a) Dd [1] – Individual 1 is unaffected but must be a carrier because he and his wife produced an affected child (individual 4).
(b) The disorder is recessive; to be affected (filled symbol), an individual must possess two copies of the recessive allele (dd). Hence individual 4 is homozygous recessive. [1]
(c) Individual 9 is the spouse of individual 5. Individual 5 is unaffected (son 5 son of carriers 1 and 2), so his genotype is either DD or Dd with ⅔ probability of being a carrier. They have an affected daughter (10), therefore individual 5 must be Dd. For the couple to produce an affected child, individual 9 must also be at least a carrier (Dd). Since the daughter is affected (dd), individual 9 must be a carrier (Dd).
Therefore the chance that individual 9 is a carrier = 100% (certain). [1]
(If reasoning shown that 9 could be DD only if daughter were impossible, but since daughter is dd, 9 must be Dd, so probability = 1. Accept correct logic.)
18.
(a) Cᴿ Cᵂ [1]
(b) Cross: Cᴿ Cᵂ × Cᴿ Cᵂ.
Punnett square:
| Cᴿ | Cᵂ | |
|---|---|---|
| Cᴿ | CᴿCᴿ | CᴿCᵂ |
| Cᵂ | CᴿCᵂ | CᵂCᵂ |
Phenotypes: 1 red (CᴿCᴿ) : 2 roan (CᴿCᵂ) : 1 white (CᵂCᵂ).
Proportion of white calves = ¼ (or 25%). [2]
Working must be shown.
19.
(a) Recessive.
All F₁ had normal tails, so the normal allele is dominant over bent. The bent trait skipped the F₁ generation and reappeared in 25 % of the F₂, characteristic of a recessive trait. [2]
(b) The original bent‑tailed male must be homozygous recessive: bb. [1]
20.
- Continuous variation describes a trait that shows a smoothly graded range of values rather than distinct categories. Blood pressure can vary continuously from low to high, with many intermediate values, influenced by multiple genes and environmental factors. [2]
- One environmental factor: diet (e.g., high salt intake), or stress, or exercise level. [1]
(Accept any plausible environmental factor with brief explanation.)
Total: 40 marks