Suppose a and b are recessive alleles of genes located 20 ma

Suppose a and b are recessive alleles of genes located 20 map units apart on the same chromosome in pea plants. True-breeding a + plants are mated to true-breeding + b plants, and the heterozygous a +/+ b plants that are produced are then test-crossed.

What proportion of plants with the double recessive a b phenotype are expected in the progeny of this testcross?

Solution

AS THE GENES ARE PRESENT IN 20 MAP UNITS DISTANCE, THEY WILL PRODUCE TOTAL RECOMBINANTS AT 20%.

AS THE PARENTS ARE a+/+b,
GENOTYPES a+/ab and +b/ab ARE THE PARENTAL COMBINANTIONS. WHICH WILL OCCUPY 80% OF THE TOTAL PROGENY.

GENOTYPES ab/ab and ++/ab ARE THE RECOMBINANTAL CATERGORIES, WHICH WILL OCCUPY 20 % OF THE TOAL POPULATION.

THE PROPORTION OF AB/AB IS 20/2 = 10% OR 0.1.

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