In: Biology
2. In butterflies, round eyes (re) and white body (wb) are encoded by two mutant alleles that are recessive to those that produce wild-type traits (re+ and wb+); these two genes are on the same chromosome. A butterfly homozygous for round eyes and white body is crossed with a butterfly homozygous for the wild-type traits. The F1 have normal eyes and normal body. The F1 are crossed with butterflies that have round eyes and white body in a testcross. The progeny of this testcross are:
wild-type eyes, wild-type body 418
round eyes, wild-type body 19
wild-type eyes, white body 16
round eyes, white body 426
a. What phenotypic proportions would be expected if the genes for round eyes and white body were located on different chromosomes?
b. What is the percent recombination between the genes for round eyes and white body?
2. Parental cross rere wbwb x re+re+ wb+wb+
Parental gametes re wb (From rere wbwb) & re+ wb+ (From re+re+ wb+wb+)
F1 re+re wb+wb
Test cross re+re wb+wb x rere wbwb
Test cross gametes re+ wb+, re+ wb, re wb+ & re wb (From re+re wb+wb); re wb (From rere wbwb)
a) When genes are located on the different chromosomes, they will assort independently. In this case, F1 heterozygous individuals are testcrossed & genes are assorted independently. As a result, we will expect to observe 1:1:1:1 phenotypic ratio for the given four phenotypes- (wild type eyes, wild type body), (round eyes, wild type body), (wild type eyes, white body) & (round eyes, white body).
b) Parental progeny are always highest in numbers. So, (wild type eyes, wild type body) & (round eyes, white body) are parental (non-recombinant) traits; (round eyes, wild type body) & (wild type eyes, white body) are recombinant traits. It can be also seen from the test cross gametes.
Now, recombination frequency = (Number of recombinant progeny / Total number of progeny) x 100%
So, recombination frequency between the round eyes & white body genes = [(19+16) / (418+19+16+426)] x 100% = 3.98% (Up to 2 decimals)