# Number of homozygous recessive offspring

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?

• Why do you assume the alleles are in a "cis arrangement"? We typically talk about alleles being linked, but you need to specify which alleles are linked (e.g.s 'T is linked to R' or 'T is linked to r'). Is the former what you mean by "cis arrangement"? Also, have you included the full text of the question? It is quite possible that this is (yet another) example of a poorly worded question ... ——— Also note that this is a "homework" question (see help center for details about what that means. You've shown the required effort, but you must also include the homework tag on these questions. Jul 13, 2021 at 19:33
• @tyersome Thanks for the suggestion; I have included the homework tag now. What I mean by "cis arrangement" is that the 'T' and 'R' alleles are present on one chromosome, and 't' and 'r' on the other. Yes, I've included the full text of the question. Jul 13, 2021 at 19:41
• You're welcome. Can you please also edit in answers to the questions I've asked? With the information you've given I'm not sure that any of those answers is correct ... Jul 13, 2021 at 19:49
• @ShishirMaharana, "enough to invalidate it." -- invalidate the answer? No, the trans calculation results in an expected number between 0 and 1, so the "1" answer would be the best answer in that case. Invalidate the question? Well, as shown, the question presentation doesn't have enough information to select an answer, so that's crappy. Jul 14, 2021 at 18:23
• I wonder if the creators wrongly modified a simpler question and botched the modification? If height and color are unlinked, the number of ttrr offspring should be 100/1600. Perhaps they wrongly decided "we'll make the loci 1cM apart, so the new result will just be the original result x 1%, or 1." Jul 15, 2021 at 2:51

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.

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

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