00:01
Hi, in the given question we have to determine the proportion of offspring with the given genotype when two parents are crossed having the genotype, heterozygous for the first pair, homozygous for the second, then homozygous recessive, then heterozygous crossed with heterozygous for the first pair, heterozygous for second, heterozygous for third, and heterozygous for the fourth.
00:28
So the genotype can be predicted by determining the probability of the individual gene pairs.
00:37
So first of all, for the first pair, that is when a small a is crossed with a small a, then in this case the genotype will be homozygous dominant, heterozygous, again heterozygous and homozygous recessive.
00:59
For the second pair, which is heterozygous crossed with heterozygous, capital b, small b, the genotypes for this will be again homozygous dominant, heterozygous, heterozygous, and homozygous recessive.
01:29
For the third pair, that is, homozygous recessive crossed with heterozygous, small c, small c, capital c, again, so the third pair, that is homozygous recessive, crossed with heterozygous, the genotypes here will be heterozygous, then again heterozygous, homozygous recessive, again homozygous recessive.
01:58
Then for the last pair, heterozygous is crossed with heterozygous, capital d, small d, again capital d, small d.
02:18
First will be homozygous dominant, then hetrozygous, again heterozygous and homozygous and homozygous and homozygous and homozygous recessive.
02:25
Now their individual probabilities, that is, the probability of heterozygous for the first pair will be 1 by 2...