00:01
All right, so 125 progeny were produced from the falling cross.
00:03
Little a, little a, big b, little b, big c, little c, big d, little d.
00:07
I'm crossed with big a, little a, big b, big b, little c, big d, big d.
00:13
Assume all the genes are unlinked and that the upperclutter represent dominant alleles.
00:18
How many offspring are expected to express the dominant phenotype for each gene? how many offspring are expected to have the genotype little a, little a, big b, big b, big b, big b, little c, big d, little d, how many offspring are expected to have the genotype, a little a big b little b big c little c big d little d so to start off you can find the likelihood of a dominant phenotype for each of the four independent components of a monohybrid cross will be determined therefore we'll have a equal one -half times 125 divided by two and this will be equal to 62 .5 and we'll have b equal to 125 times three divide by four this will give us 93 .7.
01:13
C will have one -a -half times 125 divided by two again is 62 .5 and then d will be one times 125.
01:27
So we'll take 125 divided by 2 .75 divided by 2 .75, divide by 2, and we'll get 23 offsprings are expected to express the dominant phenotype for each gene.
01:54
So we'll write dominant.
02:02
So then that's the answer for a, for b.
02:07
How many offspring are expected to have this genotype? so the quadric hybrid cross given here is simplified by breaking it.
02:15
Down into four separate mono -hybrid cross components...