Answer:
a is the middle gene.
Distance [b-a]= 24.7 mu
Distance [a-c]= 15.8 mu
Distance [b-a} = 40.5 mu
Explanation:
A homozygous wild-type female drosophila (a⁺b⁺c⁺/a⁺b⁺c⁺) is crossed with a homozygous recessive male (abc/abc). The order of the genes here is arbitrary.
The F1 is heterozygous for the three genes (a⁺b⁺c⁺/abc). The F1 females were test crossed (crossed with abc/abc males).
The F2 shows the following phenotypic ratios:
Total = 1000
The male parent is homozygous recessive for the 3 genes, so the observed phenotypes of the offspring correspond to the gametes received from the mother.
Recombination during meiosis is a rare event, so the most abundant gametes are always the parentals: a⁺b⁺c⁺ and abc.
The least abundant gametes, following the same logic, are the double crossovers (DCO): a⁺bc and ab⁺c⁺.
Compare the parental and the DCO gametes. The allele that is switched corresponds to the middle gene. In this case, gene a is in the middle of the other two.
The F1 mother that generated all 8 types of gametes had the genotype b⁺a⁺c⁺/bac (correct order of genes).
Recombination frequency (RF) = #Recombinants/Total progeny
Distance (mu) = RF x 100
Distance [b-a]= 0.247 × 100 = 24.7 mu
Distance [a-c]= 0.158 × 100 = 15.8 mu
Distance [b-a} = 24.7 mu + 15.8 mu = 40.5 mu
b------------24.7 mu--------------------------a---------15.8 mu-----------c
