In fruit flies, the yellow gene controls pigmentation. The yellow allele (y) giv

ID: 181667 • Letter: I

Question

In fruit flies, the yellow gene controls pigmentation. The yellow allele (y) gives flies a yellow colored body and is recessive to the wildtype allele (y^+) which yields a brown body color. Another gene, crossveinless, determines the wing vein pattern. The crossveinless allele (cv) is recessive to the wildtype allele (cv^+). A heterozygous female, with mutant alleles in the repulsion (trans) configuration, is test crossed to a male homozygous recessive for both genes. The two genes are approximately 14 m.u. apart. Fill in the table below with the predicted phenotypic frequency for the progeny from this cross. Color blindness in humans is an X-linked recessive trait. Approximately 8% of the men in a particular population are color blind. If mating is random for the color-blind locus, what is the frequency of the WILDTYPE (non-colorblind) allele in this population? What proportion of women in this population would be expected to be carriers for colorblindness?

Explanation / Answer

Wikd type brown color y+ , recessive yellow color y

Wild type wings cv+, recessive crossveinless cv

In repulsion configuration the hetroygous genotype of female is y+y cv+cv

y+y cv+cv x yy cvcv

Gametes

y cv

y+ cv+

y+y cv+cv brown, wild type wings =39%

y+cv

y+y cvcv brown, crossveinless wings =11%

ycv+

yy cv+cv yellow, wild type wings =11%

ycv

yy cvcv yellow, crossveinless wings = 39%

Because two genes are 14m.u apart so there is 14% linkage is present. The 14% is always count for 50% linkage. That mean 14% progenies will have linked genotype than the other 36%.

Mendel’s test cross was given the result into 1:1:1:1 for double cross, [in this case there is 50% linkage]. So, for every type of genotype will get 25% progenies [¼ x 100=25%] if there is no linkage [under Mendel’s law]

But here we have 14 % linkage, so the two genotypes which have been crossed were get 14% more progenies than other. That are brown, wild type wings ( y+ cv+) and yellow, crossveinless wings (y cv). The recombinants who get 14% less progenies are brown, crossveinless wings and yellow, wild type wings.

So linked genotype will get= general percentage is 25% and exceeding percentage is 14% so total is 39%

The other two genotypes will get [100-78 (39+39)]/2 = 22/2 = 11% each