Domestic rabbits vary in the length and colour of their fur. Fig. 6.1 shows a domestic rabbit with short fur and a fur colour pattern called Himalayan. [Figure 6.1] The two genes that determine the length and colour of the fur of this rabbit occur at the A / a locus and the B / bh / b locus. These two gene loci are on separate chromosomes. • The allele A results in short fur. • The allele a results in long fur. • A is dominant to a. • The allele B results in black fur all over the body. • The allele bh results in black fur on the nose, ears, paws and tail of the rabbit, and white fur on the rest of the body (Himalayan pattern). • The allele b results in white fur all over the body (albino). • B is dominant to bh and bh is dominant to b.
- (a)(i)List the four possible genotypes of the rabbit shown in Fig. 6.1.[2]
- (a)(ii)One phenotype of rabbit always breeds true. This means that when it is mated to a rabbit that looks the same as itself, all the offspring look the same as the parents. Describe the phenotype of the rabbit that breeds true.[1]
- (b)A rabbit with long, black fur all over the body that was homozygous at both loci was crossed with a rabbit with short, white fur that was homozygous at both loci. The F1 offspring had short, black fur. These F1 rabbits were mated together to become the parents of the F2 generation. Draw a genetic diagram to predict the F2 offspring genotypes and the ratio of F2 offspring phenotypes. F1 genotypes: gametes: F2 offspring genotypes: ratio of F2 offspring phenotypes:[5]
- (c)(i)A rabbit breeder performed multiple crosses of the type described in (b). This gave enough data to test the prediction that the genes for fur length and fur colour show independent assortment. Explain why the two genes assort independently.[3]
- (c)(ii)The rabbit breeder placed the results in a table and started to calculate the χ² value. Table 6.1 shows the results and some of the calculations made. Calculate the expected numbers and write them in the shaded column in Table 6.1.[1]
- (c)(iii)Use the formula provided and the figures in Table 6.1 to calculate the χ² value. χ² = Σ (O – E)² / E χ² =[1]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.
- AAbhbh
- Aabhbh
- AAbhb
- Aabhb;;
- long, white / albino ;
- 1 F1 parent genotypes: AaBb X AaBb ;
- 2 gametes: (drawing of gametes, e.g. AB, Ab, aB, ab for both parents);
- 3 F2 offspring genotypes: (Punnett square or list of genotypes showing all 16 combinations correctly filled in);
- 4 evidence of deduction of phenotypes from (min 9 different) genotypes ;
- 5 F2 offspring phenotype ratio: 9 : 3 : 3 : 1 black short : white short : black long : white long ;
- 1 because they are, on separate chromosomes / not linked ;
- 2 each, bivalent / pair of homologous chromosomes, orients itself separately ;
- 3 (at equator) in metaphase I of meiosis ;
- 4 gives, four / different / new, combinations of alleles / gametes ;
- 5 parental allele combinations are not, preserved / fixed ;
- 6 AVP;
- short, black fur, expected number (E): 27
- long, black fur, expected number (E): 9
- short, white fur, expected number (E): 9
- long, white long, expected number (E): 3;
- 1.036 ;
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