Characteristics and Traits

Free Response

The gene for flower position in pea plants exists as axial or terminal alleles. Given that axial is dominant to terminal, list all of the possible F1 and F2 genotypes and phenotypes from a cross involving parents that are homozygous for each trait. Express genotypes with conventional genetic abbreviations.

Hint:

Because axial is dominant, the gene would be designated as A. F1 would be all heterozygous Aa with axial phenotype. F2 would have possible genotypes of AA, Aa, and aa; these would correspond to axial, axial, and terminal phenotypes, respectively.

Use a Punnett square to predict the offspring in a cross between a dwarf pea plant (homozygous recessive) and a tall pea plant (heterozygous). What is the phenotypic ratio of the offspring?

Hint:

The Punnett square would be 2 × 2 and will have T and T along the top, and T and t along the left side. Clockwise from the top left, the genotypes listed within the boxes will be Tt, Tt, tt, and tt. The phenotypic ratio will be 1 tall:1 dwarf.

Can a human male be a carrier of red-green color blindness?

Hint:

No, males can only express color blindness. They cannot carry it because an individual needs two X chromosomes to be a carrier.

Why is it more efficient to perform a test cross with a homozygous recessive donor than a homozygous dominant donor? How could the same information still be found with a homozygous dominant donor?

Hint:

Using a homozygous recessive donor is more efficient because the genotype of the unknown parent can be determined in a single generation. If a homozygous dominant donor was used, the unknown genotype could still be determined. Instead of knowing the unknown genotype through the F1 phenotype, the F1 offspring would have to be self-crossed (as Mendel allowed his pea plants to self-pollinate) and the F2 generation phenotypes would be used to determine the unknown F0 genotype.