Question

A species of mouse F1 testcross gave rise to the F2 phenotypes:

Short hair wildtype, short tail 625 Short hair wildtype, long tail wildtype 48 Long hair, long tail wildtype 645 Long hair, short tail 39

Q1. a) Do the genes appear to be linked? Answer yes, no, or maybe. 1 mark b) Explain your answer. 2 marks

Q2. a) What do the phenotype numbers tell you about recombination frequencies of these alleles during meiosis? 1 mark

b) How does this relate to linkage? 2 marks

Q3. What is the genotype of the testcross individual? 2 marks

Q4. What is the genotype of the F1 female? 2 marks

Answer 1

1.

a. Yes. The genes are linked

b. Recombination frequency (RF) is used to determine whether the two genes are linked (present on the same chromosome). It can be calculated by:

Recombination frequency = (Number of Recombinant / Total number of offspring) × 100

To determine the recombination frequency between two genes, the first step is to identify the parent type and recombinant type offspring. Since recombinant offspring are under-expressed, therefore out of the given offspring black, Short hair, long tail, and Long hair, short tail are recombinant types. Rest Short hair, short tail and Long hair, long tail are parental type offspring. Thereby recombinant frequency can be calculated as below:

Linked genes have a recombination frequency that is less than 50%. Therefore, the genes are linked.

2.

a. This experiment revealed linkage between the H and T alleles and the h and t alleles. The frequency of H occurring together with L and with h occurring together with l is greater than that of the recombinant Ht and hT.

b. When two genes are located on the same chromosome, the possibility of a crossover that produces recombination between the genes is depending on the distance between the two genes. Thus, recombination frequencies are used to develop linkage maps.

3.

A test cross involves mating an unknown genotypic individual with a known homozygous recessive. Therefore, the genotype of testcross individual is hhtt.

4.

A true-breeding involves the mating of two homozygous individuals (HHTT and hhtt). This will produce all males and females with similar genotype that is HhTt. Therefore, the genotype of the F1 female is HhTt.