While doing some biology I came across this question:

The following tables present results of plant crosses involving two linked genes: S is a seed-color gene, and L is a plant-height gene. Each gene has two alleles with one allele exhibiting complete dominance over the other allele. Dominant phenotypes are yellow seeds, and tall plants; recessive phenotypes are green, and short, respectively. Assume that crossing-over between two genes occurs once.

table with cross results

Calculate the map unit between gene L and gene S. (One map unit = distance of 1% recombination)

This question confuses me because what I understand from the table is that from crossing over we get short flowers with green seeds(0.01) and tall ones with yellow seeds(that 0.01 from 0.51). So as a result from the fertilization we get a 0.02 frequency, that as I understand also equals the recombination rate. And so the map unit would need to be 0.02*100= 2%. But the response I have to get is 20%, which I do no understand how to get to.


Here's a hint: you're right about 0.01 being the recombinant (crossed-over) double homozygous recessive offspring. I think you're forgetting that each offspring contains 2 gamete haplotypes, and so predicting the percent of offspring with a given genotype involves multiplying the frequencies of the individual gametes that produce that offspring. To calculate the gamete frequencies (and thus the recombination frequency) you need to reverse that multiplication process.

Full answer:

I find it easiest to deal with uniform genotypes first. Of the 4 offspring phenotype categories, only the double homozygous recessive category (short/green) has uniform genotype - llss. If l and s were unlinked, 1/16 of the total offspring would be short/green=llss.

However, many fewer than 1/16 are short/green=llss, so linkage is confirmed and (as you note), ls must represent one of the crossed-over haplotypes. The other crossed-over haplotype is then LS. Thus, the parental (non-crossed-over) haplotypes must be lS and Ls.

Let's call r the recombination distance (in map units) between l and s. Each recombination between l and s produces 2 recombinant gametes: 1 ls and 1 LS. Thus the fraction of ls gametes - f(ls) - will be (r/2)/100 and the fraction of LS gametes - f(LS) - will also be (r/2)/100. Since each F1 offspring consists of 2 gametes, the fraction of offspring with an ls/ls genotype will be f(ls) x f(ls) = (r/2)/100 x (r/2)/100 = r^2/40000 = 0.01 (from F1 data). Solving for r, r^2 = 0.01 x 40000 = 400, so sqrt(r^2) = r = sqrt(400) = 20. r = 20 map units

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    – tyersome
    May 22 at 17:54
  • $\begingroup$ Thanks for your suggestions! $\endgroup$
    – Armand
    May 22 at 19:22

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