Answer:
1/4 or 25%
Step-by-step explanation:
This question illustrates a cross involving two genes coding for two desirable traits in a tomato plant. Let's assume that the desirable traits are coded for by the dominant alleles B and C.
According to the question, the female parent is heterozygous for both genes B and C i.e. will have a genotype, BbCc while the father is heterozygous for Gene B and homozygous for the desirable trait in gene C I.e. will have a genotype, BbCC
N.B: Dominant alleles B and C are responsible for the desirable traits.
In a cross between parents: BbCc and BbCC, each parent will produce gametes according to Mendel's law of segregation. The possible gamete combinations are:
BbCc - BC, Bc, bC and bc
BbCC - BC, BC, bC and bC
Using these gametes in a punnet square (see attached image), 16 possible seed offsprings will be produced. However, out of the 16 possible offsprings, 4 of them will likely be heterozygous for both genes i.e. possess a genotype, BbCC.
Hence, 4/16 = 1/4 is the probability of a given seed to be heterozygous for both genes.