Suppose that in peas, the number of seeds in a pod were governed by the additive

Question

Suppose that in peas, the number of seeds in a pod were governed by the additive polygenic model presented in class (like Nilsson-Ehle’s wheat kernel color). Each adding allele of each gene adds one pea to the pod. The non-adding alleles do not add peas. If a true-breeding strain with 2 peas per pod is mated to a true-breeding strain with 10 peas per pod...

a. what would be the phenotype of the F1 hybrid?

b. how many genes are responsible for the difference between these strains?

c. If the F1 were self-crossed, what portion of the F2 would have 10 peas per pod?

Explain work and include Punett Squares where applicable.

Explanation / Answer

The genotype of the two plants are

AAbbccddee X AABBCCDDEE

AABbCcDdEe

6 peas per pod

AABbCcDdEe

6 peas per pod

AABbCcDdEe

6 peas per pod

AABbCcDdEe

6 peas per pod

a) Phenotype of F1 will be 6 peas per pod.

b) 5 genes are responsible for the difference between these strains

c) F1 self cross- AABbCcDdEe X AABbCcDdEe

so to have 10 peas per pod the phenotype should be AABBCCDDEE.which is possible by the combination of following gamate-

ABCDE X ABCDE

No of gamates produced from each genotype is measured by the formula 2n where n is the number of heterozygous position.

Here heterozygous position is 4 so no gamates will be 24 = 16

So total no of different type progeny is 16X16=256

So the portion of plants having 10 peas per pods(AABBCCDDEE) will be= 1/256X100= 0.39%

AABbCcDdEe

6 peas per pod

AABbCcDdEe

6 peas per pod

AABbCcDdEe

6 peas per pod

AABbCcDdEe

6 peas per pod

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.