Verified answer

Your cross grid should look like this because both parents have one white allele and one red allele.

-----|--R-----|---W

R---|--RR---|---RW

W--|--RW--|----WW

The Ratio is likely to be 1:1:2

The answer is B, 1/2 will be pink.

a. properly if 2 prevalent father and mother have this, and the ailment purely shows up with homozygous recessive, and they have a daughter laid low with the condition (xx) then they both must be Xx. otherwise they doesn't be able to genetically produce a baby it extremely is xx. So the prospect they could have a common son who's a service of the recessive allele must be 50%. because - (Xx)(Xx) produces proper right here genotype possibilities: a million/4 XX; a million/2 Xx; a million/4 xx b. this can be yet another (Xx)(Xx) flow - because the mother ought to have exceeded the recessive x to the daughter (even notwithstanding she's prevalent, because it calls for homozygosity to ensue phenotypically). So the share of having 2 little ones which have the ailment is a million/4*a million/4 it extremely is a million/16. (First baby threat * second baby threat) the second one one i'm too lazy to envision, sorry.

you mean the cross between two pinks then C is the answer. If you meant a red x white cross then B.