Question

1. You are interested in how weight of brazil nuts was shown 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 each additive allele on nut weight is approximately equal. In a population study, two true-breeding strains were isolated. The nuts from one of the strains weighed 10.0 g and the nuts from the other strain weighed 15.0 g. Assuming, that these strains represent the upper and lower limits of nut weight, answer the following questions.

a. If the two strains are crossed, what would be the weight of the nuts produced by the F1 plants? Explain your answer

b. If two F1 plants are crossed, what range and distribution of nut weight would you expect to see among 64, F2 progeny? Be sure to list the classes and the # of additive alleles for each class, as well as the predicted # of individuals in each class

Answer 1

a. Ans:-

Solution) This is the case of two genes with additive alleles and the effect of each additive allele on nut weight is approximately equal. Let’s say the gene 1 has a dominant allele as A and the recessive allele as a. and the gene 2 has a dominant allele as B and the recessive allele as b.

We are having two true-breeding parents, one has the upper weight limit as 15g and the other has a lower weight limit of 10g. so the two parents must be AABB and aabb parents.

We represented that each additive allele is giving equal contribution,

So AABB = 15g, as each A, and each B allele contributes 1/4th (=3.75g to total nut weight)

And aabb = 10g, as each a and each b allele contributes 1/4th (=2.5g to total nut weight).

Now the F1 progenies will be AaBb (as a cross between AABB and aabb; parent 1 gives AB and parent 2 gives ab gametes to all F1 progenies)

So, the weight of F1 progenies (AaBb) = 3.75 + 2.5 + 3.75 + 2.5 =12.5g

Now, solve the outcomes and their classes by the punnet square method;

b. Ans :-

If we consider total 100 individuals in F2 progenies, then the number of individuals in each class;

Class 1 = 1/16 x 64 = 4

Class 2 = 4/16 x 64 = 16

Class 3 = 6/16 x 64 = 6

Class 4 = 4/16 x 64 = 16

Class 5 = 1/16 x 64 = 4