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amid [387]
4 years ago
13

(please help me)

Biology
1 answer:
Schach [20]4 years ago
8 0

Answer:

The answer is the third option; all the pea plants with green seeds contain same alleles for that trait.

Explanation:

This question shows a monohybrid cross involving a single gene coding for seed colour in pea plant. Two traits controlled by two alleles are involved viz: Yellow allele (Y) coding for yellow trait and green allele (y) coding for green trait. Two purebreeding parents are crossed in the P generation i.e. YY (yellow pea) × yy (green pea).

Based on the image, all the F1 offsprings exhibited yellow seeds meaning that the allele for yellow seed color (Y) is dominant over the allele for green seed color (y). Two heterozygous F1 offsprings (Yy) are self-crossed to give an F2 offsprings in a ratio 3 Yellow : 1 green.

Using a punnet square, a Yy × Yy cross in the F1 generation will give rise to four offsprings with genotypes: YY, Yy, Yy and yy.

YY, Yy and Yy are yellow offsprings while yy is a green offspring.

Based on this, all the pea plants with green seeds in the F2 generation contained the same alleles (yy) for the trait.

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