Question

A cross is done between two mutant T4 phage (a- b+ c+) and (a+ b-c)(no order implied by these genotypes). The genotypes of the progeny from this cross and their relative frequencies are given below: a-b+ c+ (30%) a+D c. (30%) a+ b+ c-(2%) a+ b+ c+ (8%) a- b-e+(2%) a-b-c-(8%) e+b-c+ (10%) a-b+ c-(10%) What is the distance between "a" and "b? Please give your answer in % recombination, but L number please! EAVE OUT the units (NO % sign), such as 20, or 3,

Answer 1

Since, the relative frequency is incorrectly mentioned in the question. The correct relative frequency table is attached below.

Answer:

20.

Explanation:

The genotype with the least recombinant frequency will represent the double cross overs.

The double cross overs progeny are a+ b+ c-(2%) and a- b-c+(2%).

The percentage of the recombinant frequency determined the distance between the genes. In the double crossovers, the b gene gas been exchanged and present in the middle.

The single recombinant crossovers as compared with the parents are a+ b+ c+ (8%) and  a-b-c-(8%).

Distance between the gene a and b = Single cross overs + double cross overs

Distance between the gene a and b = 8 + 8 +2 +2 = 20.

Therefore, the distance distance between the gene a and b is 20 centi morgan.