Question

. A geneticist crosses a pure-breeding strain of peas producing yellow, wrinkled seeds with one that...

. A geneticist crosses a pure-breeding strain of peas producing yellow, wrinkled seeds with one that is pure-breeding for green, round seeds. (Remember that yellow and round are the dominant alleles).

a. Use a Punnett square to predict the F2 progeny that would be expected if the F1 are allowed to self-fertilize.

b. What proportion of the F2 progeny are expected to have yellow seeds? Wrinkled seeds? Green seeds? Round seeds?

c. What is the expected phenotype distribution (ratio) among the F2 progeny?

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Answer #1

Ans. As the round and yellow coloured characteristics are the dominant ones it would be expressed in the F1 generation. On self fertilization the F2 progeny would be havind dominant round and yellow allele as well as recessive wrinkled and green allele. Considering the all four possible diallelic combinations the probability would lead to 1/4.

As there are 4 possible gamete types in each parent which lead to the combination of 4 x 4 = 16. Hence 16 genotypes would be formed.

Considering the punnett square ratio the combination would be

9 round yellow : 3 wrinkled yellow : 3 round green : 1 wrinkled green

The phenotypic distribution ratio would be 9:3:3:1

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