Blue flower and white flower on crossing gave progeny of blue and whit...
Problem: Blue flower and white flower on crossing gave progeny of blue and white flower in the ratio of 60:40. What is the ratio of blue and white is expected if the blue flower are self-pollinated?
Solution:
When a blue flower and a white flower are crossed, the resulting progeny is a blend of blue and white flowers. Let's assume that the blue flower is represented by the dominant allele B, and the white flower is represented by the recessive allele b. When these two alleles combine, they form the genotype of the offspring. The possible genotypes of the offspring are BB (blue flower), Bb (blue flower), and bb (white flower).
Given that the ratio of blue flowers to white flowers in the progeny of the cross is 60:40, we can assume that the parents had the genotype Bb and bb, respectively. Therefore, the ratio of the alleles in the parents is 1:1.
Now, let's consider what would happen if the blue flower (Bb) is self-pollinated. The resulting progeny will have the genotype BB, Bb, and bb in the ratio of 1:2:1. This is because the dominant allele B will always express itself in the BB and Bb genotypes, while the recessive allele b will only express itself in the bb genotype.
Therefore, the ratio of blue flowers to white flowers in the self-pollinated progeny will be:
BB : Bb : bb = 1 : 2 : 1
This can be simplified as:
blue : white = 3 : 1
So, the expected ratio of blue and white flowers in the self-pollinated progeny is 76:24, which is equivalent to 3:1.
Blue flower and white flower on crossing gave progeny of blue and whit...
First Ww×ww
so progenies are
ratio -Ww(Blue):ww(white) ~1:1 =60:40(approx)
now only Blue flowers are self pollinated
so ration -Blue : white is 3:1=76:24(approx)