Question

In beans the pod texture and growth form genes are located on different chromosomes. There are two varieties of beans in a farmer’s field. One has tough pods (T) and a bush-type growth form (B). The second has tender pods (t) and also has a climbing-type (b) growth form. A farmer starts a breeding program with these two homozygous (= true breeding) varieties. The F1 plants are then self-crossed to produce an F2 generation.
a. What is the genotype and phenotype of the F1?

b. What proportion of the F2 are heterozygous for bush-type growth form?

c. What proportion of the F2 have tender pods?

d. What proportion of the F2 are heterozygous for growth form and have tender pods on the same plants?

Answer 1

Answer:

a. Genotype in F1 - TtBb

   Phenotype in F1 - tough pods, bush-type growth

b. Proportion of F2 that are heterozygous for bush-type growth form = 5/8

c. Proportion of the F2 with tender pods =  1/4

d. Proportion of F2 that are heterozygous for growth form and have tender     pods = 1/8

Explanation:

Since the two traits are located on different chromosomes, they will independently assort during gametes formation. T is dominant over t and B is dominant over b.

Two true-breeding varieties were crossed at the beginning of the program:

TTBB   x   ttbb

All F1 plants will have the genotype TtBb with tough pods and bush-type growth form.

Advancing the cross to F2:

TtBb   x   TtBb

Progeny

1 TTBB   -   tough pods, bush-type

2 TTBb     -  tough pods, bush-type

2 TtBB    -   tough pods, bush-type

4 TtBb     -   tough pods, bush-type

1 TTbb     -    tough pods, climbing-type

2 Ttbb     -    tough pods, climbing-type

1 ttBB       -    tender pods, bush-type

2 ttBb     -      tender pods, bush-type

1 ttbb      -      tender pods, climbing-type

Proportion of F2 that are heterozygous for bush-type growth form = 10/16 = 5/8

Proportion of the F2 with tender pods = 4/16 = 1/4

Proportion of F2 that are heterozygous for growth form and have tender pods = 2/16 = 1/8