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Stella [2.4K]
3 years ago
8

g In the opium poppy, leaves can either be normal-shaped or lacerate (with rough edges). Leaf shape is determined by two loci, A

and B. A single dominant allele at either gene (i.e. either A/- OR B/-) is sufficient to get normal-shaped leaves. A pure breeding plant with normal leaves was crossed with a pure breeding plant with lacerate leaves to generate an F1. If this F1 is selfed, what will be the ratios of normal:lacerate leaved plants in the F2 generation

Biology
1 answer:
Katen [24]3 years ago
4 0

Answer:

15:1

Explanation:

Normal shape leaf is controlled by A/_ or B/_.

Pure breeding normal leaf shape plant would have the genotype AABB.

A pure breeding lacerate leaf plant would have the genotype aabb.

              AABB    x    aabb

                         AaBb

At F1, all the offspring would have normal shape leaves. If F1 is selfed:

        AaBb    x    AaBb

A_B_  Normal leaved plants = 15/16

aabb Lacerate leaved plants =  1/16 <em>(See the attached image for the Punnet's square)</em>

<em>Hence, the ratio of normal:lacerate leaved plants at F2 would be 15:1</em>

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