A cross 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 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:

320 a⁺b⁺c⁺
308 a b c
102 a⁺ b c⁺
112 a b⁺ c
66 a⁺ b⁺ c
59 a b c⁺
18 a⁺ b c
15 a b⁺ c⁺

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⁺.

1st. Determine the gene order

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.

2nd Determine the single crossover gametes

The F1 mother that generated all 8 types of gametes had the genotype b⁺a⁺c⁺/bac (correct order of genes).

The single crossover (SCO) gametes resulting from recombination between genes b and a are b⁺ac and ba⁺c⁺.
The single crossover (SCO) gametes resulting from recombination between genes a and c are b⁺a⁺c and bac⁺.

3) Calculate the recombination frequencies between genes

Recombination frequency (RF) = #Recombinants/Total progeny

RF [b-a]= (102+112+18+15)/1000= 0.247
RF [f-br]= (66+59+18+15)/1000= 0.158

4) Calculate the distance in map units

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

The gene map therefore looks like:

b------------24.7 mu--------------------------a---------15.8 mu-----------c