Question

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

Answer 1

15:1

Step-by-step 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 (See the attached image for the Punnet's square)

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