Question

Kernel color in corn is determined by two genes (A and B), with two alleles of each (A1, A2, B1, B2). Each 1 allele contributes a single dose of color, and each 2 allele contributes no color. The greater the number of 1 alleles, the darker the color. If double heterozygotes are selfed (A1A2 x B1B2), what proportion of the progeny will have exactly three doses of color?

Answer 1

For doing this you need to solve as a Punnett square.

You have all the alleles so you can cross them.

A1A2 x B1B2 = A1B1 A1B2 A2B1 A2B2

A1B1 will be two color

A1B2 will be one color, A color

A2B1 will be one collor, B color

A2B2 will be no color