Secondary 3 Biology Genetics Inheritance Quiz

Secondary 3 Biology Quiz - Genetics Inheritance

Answer Key

Section A: Multiple Choice

1. C. bb
[1 mark]
Reasoning: Blue eyes is the recessive phenotype, so the individual must be homozygous recessive (bb).
Common mistake: Selecting Bb — students forget that a heterozygous individual shows the dominant phenotype.

2. C. 1 tall : 1 short
[1 mark]
Working: Tt × tt → offspring: Tt, Tt, tt, tt → 2 tall : 2 short = 1 : 1 ratio.
Common mistake: Selecting B (3:1) — this ratio applies to a cross between two heterozygotes (Tt × Tt).

3. B. An allele that is expressed in the phenotype even when only one copy is present.
[1 mark]
Common mistake: Selecting A — dominance does not mean the allele is always expressed in the population, only that it is expressed when present.

4. B. 25%
[1 mark]
Working: Aa × Aa → offspring: AA, Aa, aA, aa → 1 out of 4 = 25% homozygous recessive.

5. C. 50%
[1 mark]
Working: X^CX^c × X^CY → sons inherit X from mother: X^CY (normal) or X^cY (colour-blind) → 1 out of 2 = 50%.
Common mistake: Selecting B (25%) — this would be the probability for any child (including daughters), but sons only receive one X from the mother.

Section B: Structured Response

6.
(a) Allele — An alternative form of a gene that occupies the same locus on homologous chromosomes. [1 mark]
(b) Genotype — The genetic makeup of an organism; the combination of alleles an individual possesses for a particular gene. [1 mark]
(c) Phenotype — The observable physical or biochemical characteristics of an organism, determined by its genotype and environmental influences. [1 mark]

7.
(a) Parent 1: PP; Parent 2: pp [1 mark]
(b) Punnett square:

[2 marks] — 1 mark for correct gametes, 1 mark for correct offspring genotypes.
(c) Genotypic ratio: All Pp (or 100% heterozygous) [1 mark]
(d) Phenotypic ratio: All purple (or 4 purple : 0 white) [1 mark]

8.
(a) The condition is caused by a recessive allele. [1 mark] Reasoning: Individual I-1 and I-2 are unaffected but have an affected child (II-2), which means both parents must be carriers of a recessive allele. A dominant condition would require at least one affected parent to pass on the allele. [1 mark]
(b) Individual II-3 is unaffected but has an affected sibling, so he could be either AA or Aa. [1 mark]
Marking note: Accept "AA or Aa" or "homozygous dominant or heterozygous". Award 0 if only one genotype is given.

9.
(a) The genotype of the short-furred parent is Ss (heterozygous). [1 mark] If the short-furred parent were homozygous dominant (SS), all offspring would have short fur. The 1:1 ratio indicates a test cross between a heterozygote and a homozygous recessive individual. [1 mark]
(b) This is a test cross. [1 mark]

10.
(a) The man's genotype is I^AI^B. [1 mark]
(b) Punnett square:

Possible blood groups of children: A and B (in equal proportion). [2 marks] — 1 mark for correct Punnett square, 1 mark for correct blood groups.
(c) No, this couple cannot have a child with blood group O. [1 mark] The child would need to inherit the i allele from both parents, but the father (I^AI^B) does not carry the i allele.

Section C: Data Interpretation & Extended Response

11. A recessive allele can be "hidden" in carriers (heterozygous individuals) [1 mark] because the dominant allele masks its expression in the phenotype. Carriers do not show the disorder but can pass the recessive allele to their offspring. If two carriers have children, there is a 25% chance the child will be homozygous recessive and express the disorder. [1 mark]
Marking note: Award 1 mark for mentioning carriers/heterozygotes, 1 mark for explaining how the allele persists without being expressed.

12.
(a) Round is dominant. [1 mark] In Cross 1, two round-seeded parents produce some wrinkled-seeded offspring (approximately 1:3 ratio), which means both parents are heterozygous and round is the dominant trait. If wrinkled were dominant, two wrinkled parents could not produce round offspring. [1 mark]
(b) Parent 1: Rr; Parent 2: Rr [1 mark]
(c) Punnett square for Cross 3 (Rr × rr):

[2 marks] — 1 mark for correct Punnett square setup, 1 mark for explaining that the 1:1 ratio (90 round : 90 wrinkled) matches the expected result of a test cross.
Explanation: The round parent is heterozygous (Rr) and the wrinkled parent is homozygous recessive (rr). The offspring are 50% Rr (round) and 50% rr (wrinkled), giving a 1:1 phenotypic ratio.

13.
(a) A carrier (Ss) has one copy of the recessive allele but does not show the full disease phenotype. [1 mark] They can pass the recessive allele to their offspring while appearing phenotypically normal (or mostly normal).
(b) Punnett square (Ss × Ss):

[2 marks] — 1 mark for correct Punnett square, 1 mark for stating the probability: 25% (1 in 4) chance of the child having sickle cell anaemia (ss).
(c) Carriers (Ss) have some resistance to malaria because the malaria parasite has difficulty infecting red blood cells that contain some abnormal haemoglobin. [1 mark]
Marking note: Accept any valid explanation linking the sickle cell carrier state to malaria resistance.

14.
(a) Both parents must be Ff (heterozygous carriers). [1 mark]
(b) Both parents are carriers (Ff) and are phenotypically normal because the dominant allele (F) masks the recessive allele (f). [1 mark] When both parents pass on the recessive allele (f) to their child, the child has genotype ff and expresses cystic fibrosis. There is a 25% chance of this occurring with each pregnancy. [1 mark]
Marking note: Award marks for explaining that each parent contributes one recessive allele, resulting in a homozygous recessive child.

15.
The student's claim is not necessarily true. [1 mark] The frequency of an allele in a population depends on factors such as natural selection, genetic drift, mutation, and gene flow — not simply whether the allele is dominant or recessive. [1 mark] For example, the recessive allele for sickle cell anaemia (s) is more common than the normal allele (S) in some populations because being a carrier (Ss) provides resistance to malaria, giving carriers a selective advantage. Therefore, a recessive allele can be more common than a dominant allele in certain environments. [1 mark]
Marking note: Award 1 mark for disagreeing with the claim, 1 mark for explaining that allele frequency depends on other factors, 1 mark for providing a valid example or further explanation.

Section D: Application & Analysis

16.
(a) Punnett square (Bb × Bb):

[2 marks] — 1 mark for correct Punnett square, 1 mark for identifying both parents as Bb (heterozygous). The 6:2 (3:1) ratio of black to white offspring confirms both parents are heterozygous.
(b) The probability that the next offspring will have white fur is 25% (1 in 4). [1 mark]
(c) The black-furred offspring could be either BB (1/3 chance) or Bb (2/3 chance). [1 mark] If crossed with a white-furred guinea pig (bb):

• If the black offspring is BB: all offspring would be Bb (black) → 0% white
• If the black offspring is Bb: 50% Bb (black) and 50% bb (white) → 50% white
  Overall probability = (2/3) × (1/2) = 1/3 or approximately 33%. [1 mark]
  Marking note: Award 1 mark for identifying the possible genotypes of the black offspring, 1 mark for calculating the overall probability.

17.
(a) Haemophilia is more common in males because males have only one X chromosome (XY). [1 mark] If a male inherits the recessive allele (X^h) on his single X chromosome, he will express the disorder. Females have two X chromosomes, so they need two copies of the recessive allele (X^hX^h) to express haemophilia; with one copy (X^HX^h) they are carriers but unaffected. [1 mark]
(b) Punnett square (X^HX^h × X^HY):

Daughters: X^HX^H (normal) or X^HX^h (carrier). The probability that their daughter will be a carrier is 50%. [1 mark]

18.
(a) The possible phenotypes of the offspring are red, pink, and white. [1 mark]
(b) Punnett square (Rr × Rr):

[2 marks] — 1 mark for correct Punnett square, 1 mark for stating the phenotypic ratio: 1 red : 2 pink : 1 white.
Explanation: RR = red flowers, Rr = pink flowers, rr = white flowers. The phenotypic ratio is 1:2:1.

19.
(a) Yellow is dominant. [1 mark] When two yellow-seeded plants produce some green-seeded offspring, this indicates that both parents are heterozygous (Yy) and yellow is the dominant trait. If green were dominant, two green parents could not produce yellow offspring.
(b) If the yellow-seeded parent plants (Yy) are crossed with a green-seeded plant (yy), the expected phenotypic ratio is 1 yellow : 1 green. [1 mark]
Working: Yy × yy → offspring: Yy, Yy, yy, yy → 2 yellow : 2 green = 1:1 ratio.

20.
(a) The condition is autosomal (not sex-linked). [1 mark] If it were sex-linked recessive, an affected female (II-1) would need to have an affected father, but her father (I-1) is unaffected. The pattern is consistent with an autosomal recessive condition where both unaffected parents (I-1 and I-2) are carriers and have affected children. [1 mark]
(b) Individual II-1 is affected, so her genotype must be dd (homozygous recessive). [1 mark]
(c) Individual III-2 is unaffected. Since her mother (II-1) is dd and her father (II-2) is unaffected (Dd), III-2 must be a carrier (Dd). [1 mark] If III-2 (Dd) marries an unaffected male (dd), the cross is Dd × dd:

• Offspring: Dd (unaffected carrier) or dd (affected) in a 1:1 ratio
  The probability that their first child will be affected is 50% (1 in 2). [1 mark]
  Marking note: Award 1 mark for determining III-2's genotype, 1 mark for the probability calculation.

Total: 40 marks