Across is made between homozygous wild-type female drosophila (a^+ a^+ b^+ b^+ c^+ c^+) and triple-mutant males (aa bb cc) (the order here is arbitrary). the f1 females are test crossed back to the triple-mutant males and the f2 phenotypic ratios are as follows: a^+ b c = 18 a b^+ c = 112 a b c = 308 a^+ b^+ c = 66 a b c^+ = 59 a^+ b^+ c^+ = 320 a^+ b c^+ = 102 a b^+ c^+ = 15 total = 1000 map these gene to a chromosome in a correct order and determine the map distance between them. show all your work.

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

b24.7 mua15.8 muc

i think the answer is

mass