Question

Answer the following questions for the data set below.

A species of plant 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

SHOW ALL WORK!!!

Answer 1

Ans:

1. a) Yes, the gene appears to be linked.

b) If the calculated recombination frequency is less than 50 %, then the genes occur in the same chromosome i. e. linked. The calculated recombination frequency in this experiment is 6.41%. So, the genes are linked.

2. a) The phenotypes whose frequency is maximum is called parental phenotypes and whose frequency is minimum is called recombinant frequency. So, short hair wildtype, short tail and long hair, long tail wild type are the parental phenotypes and short hair wildtype, long tail wild type and long hair, short tail are the recombinant phenotypes. The calculated recombinant frequency is 6.41%.

b) Since the calculated recombinant frequency is 6.41%, therefore the genes are in same chromosome. So, there is a linkage between them. The genes are in repulsion phase because a wild type and a mutant genes are in same chromosome.

3. The genotypes of F1 individuals are h+t/ht+ and ht/ht.

4. The genotypes of F1 female is either h+t/ht+ or ht/ht because it can not determine that which plant is heterozygous dominant and which plant is recessive.

All the calculation is given below-