Question

The weight of the nut of a commercially valuable species was determined to be under the control of two genes. Each locus can be occupied by either an additive or non-additive allele, and the effect of the additive alleles is approximately equal. In a population study, two true-breeding strains were isolated where the weights were 10.0 g and 13.2 g, respectively. Assuming that these strains represent the upper and lower levels of nut weight, predict *

a) the weight of the nuts in F1 plants when the two strains are crossed, and

b) the range and distribution of phenotypes in the F2 plants.

Answer 1

GIVEN THAT:-

The weight of the nut of a commercially valuable species was determined to be under the control of two genes. Each locus can be occupied by either an additive or non-additive allele, and the effect of the additive alleles is approximately equal. In a population study, two true-breeding strains were isolated where the weights were 10.0 g and 13.2 g, respectively. Assuming that these strains represent the upper and lower levels of nut weight, so, we can say that

As the parents are taken as true-breeding One with Low weight (10.0g), crossed with high weight (13.2g), in the F1 generation all the plants with hight weight.

(a):

the weight of the nuts in F1 plants when the two strains are crossed, and

If F1 progeny are self polinated the resulting progeny will be 75% high weight nuts(13.2g) and 25% with low weight nuts are obtained.

(b):

The result of phynotype from additive alleles depends on number of alleles that are present. If more number additive alleles, the nuts obtained with more weight. If no additive alleles obtain with less weight.

If the genes controlling the,

high weight (WW)

low weight (ww)

Parents WW X ww

Gamets W X w ,. Then

F1 generation Ww X Ww if self polinated

F2 generation WW Ww Ww ww The phenotypic ration is 3:1