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 the F2 would have 10 peas per pod?

The answer for b. is 4 genes and c. is 1/256 but im not sure how to get to that answer.

Explanation / Answer

If multiple genes are involved, and single allele adds a single pea. Suppose that you have a plant with recessive alleles on all genes that do not add any pea, that plant will have 2 pea in its seed as shown in the question for true breading strain.

On the other hand the one that has all the dominant alleles have 10 peas so the increase in number of peas are 10-2=8. And if each allele increase the number by 1, then the number of genes would be 8/2=4, 2 being the number of alleles possible for a single gene.

Now that the number of genes are 4, lets label them A,B,C and D.

the f1 would have all AaBbCcDd genotype and crossing them would give 28 = 256 possible genotypes (8 being the number of possible alleles) and only one of them could be all dominant (AABBCDD) so answer is 1/256.

Feel free to leave a comment down below for any furhter query. Good rating would be appretiated if you find my answer helpful. Thank you.

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