Number of homozygous recessive offspring

Question

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?

Answer 1

Taking your assumption of a cis arrangement of alleles to be true, here is how I would solve the problem.

A diploid plant heterozygous for height and flower color is selfed.

$TR / tr \times TR / tr$

A map distance of 1 cM means that 1% of meioses will result in recombinant gametes, so we can calculate the expected distribution of gamete genotypes assuming equal probability of the strictly dominant or strictly recessive gametes in the absence of recombination.

$p(TR) = 0.5(1 - 0.01) = 0.495$
$p(tr) = 0.5(1 - 0.01) = 0.495$
$p(Tr) = 0.5(0.01) = 0.005$
$p(tR) = 0.5(0.01) = 0.005$

The expected proportion of homozygous recessive $tr/tr$ offspring is given by the product of the gamete probabilities.

$p(tr) \times p(tr) = 0.495^2 \approx 0.245$
$0.245 \times 1600 = 392\space\texttt{offspring}$

So it seems that your answer is correct and the given answer is wrong.

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Since the loci of interest are very close together, a quick way to think about this is to treat the height and flower color loci as a single locus with dominant and recessive alleles $A$ and $a$ (i.e. independent assortment never occurs). This allows us to model the problem as a monohybrid cross of heterozygous plants.

$Aa \times Aa$

The phenotypic ratio of dominant to recessive progeny from a monohybrid cross is 3:1, so you can expect 25% of offspring to be recessive.

$0.25 \times 1600 = 400\space\texttt{offspring}$