Level 2 — RecallExtensions of Mendelian Genetics

Extensions of Mendelian Genetics

⏱ 30 minutes40 marksprintable — key stays hidden on paper

Level 2 — Recall & Standard Problems

Time Limit: 30 minutes
Total Marks: 40

Q1. Define the following terms clearly. (4 marks) (a) Incomplete dominance (b) Codominance

Q2. In snapdragons, red flower colour ( $C^R C^R$ ) is incompletely dominant to white ( $C^W C^W$ ), producing pink heterozygotes ( $C^R C^W$ ). A pink-flowered plant is crossed with another pink-flowered plant. State the expected genotypic and phenotypic ratios of the offspring. (4 marks)

Q3. The ABO blood group system is controlled by three alleles: $I^A$ , $I^B$ , and $i$ . (6 marks) (a) Name the type of inheritance shown between $I^A$ and $I^B$ , and between $I^A$ and $i$ . (2) (b) A woman with blood group AB marries a man with blood group O. Determine the possible blood groups of their children (show a Punnett square). (4)

Q4. Define multiple alleles and give one example. (3 marks)

Q5. Distinguish between pleiotropy and polygenic inheritance, giving one example of each. (4 marks)

Q6. Haemophilia is an X-linked recessive disorder. A carrier woman ( $X^H X^h$ ) marries a normal man ( $X^H Y$ ). (6 marks) (a) Draw a Punnett square for this cross. (3) (b) State the probability that a son will be haemophilic and the probability that a daughter will be a carrier. (3)

Q7. (a) State what is meant by epistasis. (2) (b) In a dihybrid cross of two heterozygotes, name the modified $F_2$ phenotypic ratio associated with recessive epistasis. (1) (3 marks)

Q8. Two genes are located on the same chromosome. In a testcross, 800 offspring were produced, of which 160 were recombinant types. (4 marks) (a) Calculate the recombination frequency. (2) (b) State the map distance between the two genes in map units (centimorgans). (2)

Q9. In a three-point linkage study, the recombination frequencies are: A–B = 8%, B–C = 12%, A–C = 20%. Construct a simple linkage map showing the order and distances of the three genes. (3 marks)

Q10. Give three examples of environmental effects on phenotype. (3 marks)

End of Paper

Answer keyMark scheme & solutions

Q1. (4 marks) (a) Incomplete dominance: a condition in which the heterozygote shows an intermediate phenotype (blend) between the two homozygous parental phenotypes; neither allele is fully dominant. (2) (b) Codominance: a condition in which both alleles in a heterozygote are fully and simultaneously expressed, so both phenotypes appear together (e.g., AB blood group). (2)

Q2. (4 marks) Cross: $C^R C^W \times C^R C^W$ . Punnett square gives: $1\ C^R C^R : 2\ C^R C^W : 1\ C^W C^W$ . (2) Because of incomplete dominance, genotype = phenotype: Phenotypic ratio = 1 red : 2 pink : 1 white. (2)

Q3. (6 marks) (a) $I^A$ and $I^B$ show codominance. $I^A$ (or $I^B$ ) and $i$ show complete dominance ( $I^A$ / $I^B$ dominant over $i$ ). (2) (b) Mother AB = $I^A I^B$ ; Father O = $ii$ . (1 for genotypes)

	$I^A$	$I^B$
$i$	$I^A i$	$I^B i$
$i$	$I^A i$	$I^B i$

Offspring: ½ $I^A i$ (group A) : ½ $I^B i$ (group B). (2) Possible blood groups: A and B only (no AB, no O). (1)

Q4. (3 marks) Multiple alleles: the existence of three or more alternative forms (alleles) of a single gene within a population, although any one diploid individual carries only two of them. (2) Example: ABO blood group system ( $I^A$ , $I^B$ , $i$ ) / coat colour in rabbits. (1)

Q5. (4 marks) Pleiotropy: one single gene affects multiple, seemingly unrelated phenotypic traits. Example: sickle-cell allele affects blood, spleen, pain, growth. (2) Polygenic inheritance: a single trait is controlled by many genes, giving continuous variation. Example: human height / skin colour. (2)

Q6. (6 marks) Cross: $X^H X^h \times X^H Y$ .

	$X^H$	$X^h$
$X^H$	$X^H X^H$	$X^H X^h$
$Y$	$X^H Y$	$X^h Y$

(3) Offspring: ¼ $X^H X^H$ (normal daughter), ¼ $X^H X^h$ (carrier daughter), ¼ $X^H Y$ (normal son), ¼ $X^h Y$ (haemophilic son). (b) Probability a son is haemophilic = ½ of sons = 1/2 of sons (1/4 of all offspring). (1.5) Probability a daughter is a carrier = 1/2 of daughters (1/4 of all offspring). (1.5)

Q7. (3 marks) (a) Epistasis: an interaction in which one gene (epistatic gene) masks or suppresses the phenotypic expression of another gene (hypostatic gene) at a different locus. (2) (b) Recessive epistasis modified ratio = 9 : 3 : 4. (1)

Q8. (4 marks) (a) Recombination frequency = (recombinants / total) × 100 = $\frac{160}{800}\times100 = 20\%$ . (2) (b) Map distance = 20% = 20 map units (20 cM). (2)

Q9. (3 marks) A–C (20%) is the largest, so A and C are the outer genes, B is in the middle. Check: A–B (8) + B–C (12) = 20 = A–C ✓ (1) Map: A —8 cM— B —12 cM— C, total 20 cM. (2)

Q10. (3 marks) (any three, 1 each)

Temperature affecting fur colour (Himalayan rabbits, Siamese cats).
Sunlight/UV affecting human skin pigmentation.
Diet/nutrition affecting body size or height.
Altitude affecting red blood cell count.
Soil pH affecting hydrangea flower colour.

[
  {"claim":"Q2 phenotypic ratio red:pink:white sums correctly and equals 1:2:1", "code":"red=1; pink=2; white=1; total=red+pink+white; result = (total==4 and pink==2*red and white==red)"},
  {"claim":"Q3 AB x OO gives only A and B in equal ratio, no AB or O", "code":"offspring=['Ai','Bi','Ai','Bi']; groups=set(offspring); result = (groups=={'Ai','Bi'} and offspring.count('Ai')==2 and offspring.count('Bi')==2)"},
  {"claim":"Q8 recombination frequency is 20 percent = 20 cM", "code":"rf=Rational(160,800)*100; result = (rf==20)"},
  {"claim":"Q9 gene order gives A-B + B-C equal to A-C", "code":"AB=8; BC=12; AC=20; result = (AB+BC==AC)"}
]