Question

Assume that genes X and Y are linked and are 20 cM apart. Also assume that an individual homozygous for X and y is cross to an individual homozygous for x and Y to produce an F₁ individual.

True/False: 40% of the gametes produced by this F₁ individual should be x Y gametes.

Assume that this F1 individual is crossed with an xx yy homozygote.

True/False: 10% of the progeny should be heterozygous for both genes.

Answer 1

As both the individuals are homozygous for their gene combination which means that both will give only one type of gamete. which means F1 generation will have genotype Xx/Yy.

Gamete formation by F1 generation will have 80% non-recombinant or parental type that gamete will be Xy(40%) and xY(40%) and 20% will be the Recombinant type (As gene are 20cM apart) that gamete will be XY(10%) and xy(10%).

so the given statement is true

for the 2nd problem: progeny will be heterozygous for both the gene when the F1 generation will provide gamete XY which has 10% probability as the 2nd individual will always provide xy with 100% gamete formation.

So, chances of Heterozygous progeny for both the gene will be=

10% X 100% = 10%

so the given statement is true