Q14.1-1CC. 聽Pea plants heterozygous for fl... [FREE SOLUTION]

Chapter 14: Q14.1-1CC. (page 276)

Pea plants heterozygous for flower position and stem length (AaTt) are allowed to self-pollinate, and 400 of the resulting seeds are planted. Draw a Punnett square for this cross. How many offspring would be predicted to have terminal flowers and be a dwarf? (See Table 14.1.)

Short Answer

Expert verified

In the given case, 25 offspring would be predicted to have terminal flowers and dwarf because both alleles are recessive for both genes.

Step by step solution

Dihybrid Cross

A dihybrid cross shows how traits are inherited to the next generation when two pure plants having a difference in only two traits are crossed. So a dihybrid is heterozygous for only two traits. In F2 generation 9:3:3:1 ratio is observed that shows an independent assortment of genes.

Punnet Square

A Punnet square is the representation of expected genotypes obtained by different combinations of gametes produced by two parents. It is in the form of a chart that contains cells having possible combinations of alleles.

Self-crossing in heterozygous pea plants

When a heterozygous pea plant for two traits is self-fertilized, and seeds are grown, the ratio of offspring is found 9:3:3:1.

Out of the total, 9 are dominant phenotypes for both traits, and 6(3+3) offspring have one dominant and one recessive allele for both traits. But one offspring in 16 (9+3+3+1) offspring has both recessive phenotypes for both traits.

According to table 14.1, terminal flower and dwarf plants both are recessive characters. So one in 16 offspring are plants with terminal flower and dwarf plants, and in 400 plants, 25 plants would be with these characters.