A question from the KVPY-SX examination, organised by the Indian Institute of Science (IISc) in Bengaluru, India, held on 3rd November, 2019.
A plant heterozygous for height and flower colour (TtRr) are selfed and 1600 of the resulting seeds are planted. If the distance between the loci controlling height and flower colour is 1 centimorgan, then how many offspring are expected to be short with white flower (ttrr)?
(A) 1 (B) 10 (C) 100 (D) 400
The answer given by the organising committee, of the KVPY, is (A).
My problem with this answer is that if the alleles are in a cis arrangement*, and the distance between them is 1 cM, then the recombination frequency will be 1% (a good enough approximation in this case, in my opinion). So, on selfing of TtRr, we'll get; TR - 49.5% Tr - 0.5% tR - 0.5% tr - 49.5%
So, the probability of getting a homozygous recessive genotype, which we obtain from tr × tr, is (0.495)^2 = 0.245, and hence, we get (0.245)(1600) = 392 offspring approximately, the answer closest to which, is (D).
*By "cis arrangement" of alleles, I'm assuming that the 'T' & 'R' alleles are present on one chromosome, and 't' and 'r' alleles are present on another. (An NCBI webpage explaining this terminology)
Where am I going wrong?
What's the right solution to this problem?