Feather colour in some breeds of chicken is an example of codominance in birds.
In this example, a chicken with white feathers and a chicken with black feathers can breed to produce offspring that have white feathers and black feathers.
The allele for black feathers is \(F^{B}\) and the allele for white feathers is \(F^{W}\).
Fig. 5.1 is a pedigree diagram of a family of chickens.
(a) State the type of variation that is caused by genes only.
(b) State all the possible genotypes of chicken 2 and chicken 7 in Fig. 5.1.
(c) Complete the genetic diagram to show the results of a cross between chicken 5 and chicken 6 in Fig. 5.1.
offspring genotypes ………… .
expected offspring phenotype percentage ………. .
(d) State one example of codominance in humans.
▶️ Answer/Explanation
(a) Discontinuous variation.
Discontinuous variation is caused solely by genetic differences, resulting in distinct categories (e.g., feather color in chickens).
(b) Chicken 2: \(F^{B}F^{B}\) or \(F^{B}F^{W}\) (black or black-and-white feathers).
Chicken 7: \(F^{W}F^{W}\), \(F^{B}F^{W}\), or \(F^{B}F^{B}\) (white, black-and-white, or black feathers).
Chicken 2 must carry at least one \(F^{B}\) allele (black phenotype). Chicken 7 can inherit any combination since its parents are \(F^{B}F^{W}\) and \(F^{W}F^{W}\).
(c) Offspring genotypes: 50% \(F^{B}F^{B}\) (black) and 50% \(F^{B}F^{W}\) (black-and-white).
Phenotype percentage: 50% black, 50% black-and-white.
Parental cross: \(F^{B}F^{W}\) × \(F^{B}F^{B}\). Gametes: \(F^{B}\), \(F^{W}\) × \(F^{B}\), \(F^{B}\). Punnett square yields equal probabilities for both genotypes.
(d) Example: ABO blood groups (codominance of \(I^{A}\) and \(I^{B}\) alleles).
In humans, codominance is seen when both alleles (e.g., \(I^{A}\) and \(I^{B}\)) are fully expressed in the phenotype (AB blood type).
Fig. 5.1 is a pedigree diagram showing the inheritance of blood group in one family.
(a) State the number of people in Fig. 5.1 with:
XY chromosomes ………… .
only one \(I ^{A}\) allele. …………. .
(b) Explain how Fig. 5.1 shows that blood group is an example of discontinuous variation.
(c) State one example of discontinuous variation in plants.
(d) Explain why the inheritance of blood group is an example of codominance.
(e) A person with the genotype \(I ^{A}I^{o}\) has a child with a person with the genotype \(I ^{B}I^{o}\).
Complete the genetic diagram to determine the probability of the offspring having the blood group AB.
probability of the offspring having the blood group AB ………….. .
▶️ Answer/Explanation
(a) 4 (XY chromosomes); 3 (only one \(I^{A}\) allele).
In the pedigree diagram, 4 individuals are male (XY chromosomes), and 3 individuals have only one \(I^{A}\) allele (heterozygous \(I^{A}I^{o}\)).
(b) Blood group shows discontinuous variation because there are distinct phenotypes (A, B, AB, O) with no intermediates.
Discontinuous variation is characterized by clear-cut categories, unlike continuous variation (e.g., height).
(c) Example: Seed shape (round/wrinkled) or Flower color (e.g., red/white).
Discontinuous variation in plants includes traits with distinct, non-overlapping phenotypes.
(d) Blood group inheritance is codominant because \(I^{A}\) and \(I^{B}\) alleles both express fully in the phenotype (AB blood group).
Codominance occurs when neither allele is recessive, and both contribute equally to the phenotype.
(e) Probability = 25% (0.25).
Gametes: \(I^{A}\), \(I^{o}\) × \(I^{B}\), \(I^{o}\). Punnett square yields 4 combinations: \(I^{A}I^{B}\) (AB), \(I^{A}I^{o}\) (A), \(I^{B}I^{o}\) (B), \(I^{o}I^{o}\) (O). Only 1 out of 4 is AB.