Flower position, stem length, and seed shape are three characters that Mendel studied. Each is controlled by an independently assorting gene and has a dominant and recessive expression, as indicated in Table 14.1. If a plant that is heterozygous for all three characters is allowed to self fertilize, what proportion of the offspring would you expect to be each of the following? (Note: Use the rules of probability instead of a huge Punnett square)

(a) homozygous for the three dominant traits

(b) homozygous for the three recessive traits

(c) heterozygous for all three characters

(d) homozygous for axial and tall, heterozygous for seed shape

The two normal genes possessing the same variant are known as homozygous. The two normal genes having the different variants are known as heterozygous. The probability of the homozygous offspring is ¼, and the probability of heterozygous offspring is ½.

The given condition is homozygous for the three dominant traits. The probability for the given condition is as follows:

$\text{Probability}=\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}=\frac{1}{64}$

The given condition is homozygous for the three recessive traits. The probability for the given condition is as follows:

$\text{Probability}=\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}=\frac{1}{64}$

In general, the probability of the heterozygous offspring is ½. The probability of the three heterozygotes is as follows:

$\text{Probability}=\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{8}$

In general, the probability of the heterozygous offspring is ½. The probability of the homozygous for axial and tall, heterozygous for seed shape is as follows:

$\text{Probability}=\frac{1}{4}\times \frac{1}{4}\times \frac{1}{2}=\frac{1}{32}$