Q.52 In maize, the genes for colored seed and round seed are dominant over the genes for colorless seed and shrunken seed. Pure breeding strains of the double dominant variety were crossed with the double recessive variety and a test cross of the F1 generation produced the following: Phenotypes Number of seeds Colored, round seed 380 Colorless, shrunken seed 396 Colored, shrunken seed 14 Colorless, round seed 10 For the above, the distance between the genes for seed color and seed shape on the chromosomes would be _____ centimorgan units.

Q.52

In maize, the genes for colored seed and round seed are dominant over the genes for colorless seed and shrunken seed. Pure breeding strains of the double dominant variety were crossed with the double recessive variety and a test cross of the F1 generation produced the following:

Phenotypes Number of seeds
Colored, round seed 380
Colorless, shrunken seed 396
Colored, shrunken seed 14
Colorless, round seed 10

For the above, the distance between the genes for seed color and seed shape on the chromosomes would be _____ centimorgan units.

The distance between the genes for seed color and seed shape in maize is 3.0 centimorgans (cM).

Concept of recombination frequency

In maize, colored seed (C) and round seed (R) are dominant over colorless (c) and shrunken (r). In the cross described, a pure-breeding double dominant parent (CCRR) is crossed with a pure-breeding double recessive parent (ccrr), producing F₁ plants with genotype CcRr. These F₁ plants are test-crossed with ccrr, so every gamete from the F₁ is directly visible in the offspring phenotypes. The four phenotypic classes and their counts are: 380 colored round, 396 colorless shrunken, 14 colored shrunken, and 10 colorless round. The two largest classes represent parental combinations (CR and cr), while the two smallest classes represent recombinant combinations (Cr and cR).

Stepwise recombination calculation

Total offspring in the test cross equal 380+396+14+10=800. Recombinant progeny are those with new combinations of traits, namely colored shrunken (Cr, 14) and colorless round (cR, 10), giving a total of 14+10=24 recombinants. Recombination frequency is therefore 24800×100=3%. Because 1% recombination corresponds to 1 centimorgan, the map distance between the loci controlling seed color and seed shape is 3.0 cM.

SEO‑oriented problem discussion

This is a typical gene mapping question where test-cross progeny counts are used to estimate recombination frequency and thus genetic distance in centimorgans. The larger parental classes (colored round and colorless shrunken) show that the original coupling of dominant and recessive alleles is mainly preserved, while the much smaller recombinant classes reveal rare crossover events between the color and shape loci. In exam contexts, common mistakes include dividing by parental total instead of overall progeny or misidentifying which phenotypes are recombinant.

 

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