Question

Two genes of flower, one controlling blue(B) versus white(b) petals and the other controlling round(R) versus oval(r) stamens, are 1000 map units apart . You cross a homozygous blue-oval plant with a homozygous white-round plant. The resulting F1 progeny are crossed with homozygous white oval palnts, and 1000 F progeny are obtained . How many of each of the four phenotypes do you expect?

Answer 1

Answer:

Given in the question that Blue flower is dominant over the white flower and Round stamen is dominant over oval stamen

Now, from the question let Parent A be: Blue-oval = BBrr and let Parent B be: White-round = bbRR

Upon crossing both, i.e, BBrr X bbRR = BbRr (we obtain this result from the crossing of both the parents)

Now, the resulting F1 progeny is crossed with homozygous white oval plants, i.e., When BbRr X bbrr

( I think the question has to be that the two genes are 10mu rather than 1000mu apart since two genes if apart more than 50%, then we couldn't be sure of whether the genes are on same chromosomes,kindly cross check on the question. Taking it to be 10mu apart which means that recombination frequency is 10% )

Then we obtain the following results,

Br/br = blue-oval = which is Parental = 45%

bR/br = white-round = which is Parental = 45%

BR/br = blue-round =which is a  Recombinant = 5%

br/br = white-oval =which is a Recombinant = 5%

Hence, out of the 1000 f2 plants,(given in question), the number of:

blue-oval=450

white-round=450

blue-round=50

white-oval=50

Answer 2

Homozygous parents BbRr can be arranged as probabilities BR, Br, bR, br which then crosses with homozygous BbRr. These results an F2 BbRr (P), BbRr (Recon), bbRr (Recon), and BbRr (P). So we end up with 2Recon=10% *1000, and 2P=90%*1000. So we end up with 450 parentals and 50 recombinants