Question

List the different gametes that could be made by a pea plant heterozygous for seed color, seed shape, and pod shape (YyRrIi). How large a punnett square would you need to predict the offspring of a self-pollination of this "trihybrid"?

Answer 1

A pea plant heterozygous for seed color (Yy), seed shape (Rr), and pod shape (Ii) can produce 8 different gametes. To predict the offspring of a self-pollination, a 4x4 punnett square would need to be created.

Step-by-step explanation:

For a pea plant heterozygous for seed color (Yy), seed shape (Rr), and pod shape (Ii), there are several different gametes that could be formed. Each gene has two possible alleles, so for each gene there are 2^1 = 2 possible gametes. Therefore, the number of different gametes that can be made is 2^3 = 8. These gametes would be: YRI, YrI, yRI, yrI, YrI, Yri, yRi, and yri.

To predict the offspring of a self-pollination of this trihybrid, a punnett square would need to be created. Since there are 8 possible gametes, a 4x4 punnett square would be required. The columns and rows would represent the different gametes that could be produced, and the cells of the punnett square would show the potential offspring genotypes.

Answer 2

The pea plant can make 8 varying gametes and they are

1. YRI

2. YRi

3. YrI

4. Yri

5. yRI

6. yRi

7. yrI

8. yri

In order to fit all these gametes in a self-pollination, it is required that a punnett square have 8 columns and 8 rows. This would enable it to have spaces for the entire 64 possible unions of gametes in its offspring. Step-by-step explanation: