Question

In pea plants, yellow seeds (Y) are dominant to green seeds (y) and round seeds (R) are dominant to wrinkled seeds (r). The genes for seed color and seed shape are on different chromosomes. Two true-breeding parents, one with yellow round peas and the other with green wrinkled peas, are crossed. What are the genotypes of the parents, and what kind of gametes will they produce?

a. The genotype of the plant with yellow round seeds will be YYRR and produce Y gametes and R gametes, while the genotype of the plant with green wrinkled seeds will be yyrr and produce y gametes and r gametes.
b. The genotype of the plant with yellow round seeds will be yyrr and produce yr gametes, while the genotype of the plant with green wrinkled seeds will be YYRR and produce YR gametes.
c. The genotype of the plant with yellow round seeds will be YYRR and produce YR gametes, while the genotype of the plant with green wrinkled seeds will be yyrr and produce yr gametes.
d. The genotype of the plant with yellow round seeds can be YYRR or YyRr and will produce YR, Yr, yR and yr gametes, while the genotype of the plant with green wrinkled seeds will be yyrr and produce yr gametes.
e. The genotype of the plant with yellow round seeds will be YR and produce Y and R gametes, while the genotype of the plant with green wrinkled seeds will be yr and produce y and r gametes.

Answer 1

The genotype for the parent pea plant with yellow round peas is YYRR, producing YR gametes, and for the parent with green wrinkled peas is yyrr, producing yr gametes. The correct answer is option c. A Punnett square analysis of this cross requires 4 squares.

Step-by-step explanation:

The genotype of the plant with yellow round peas in a pea plant is YYRR because yellow (Y) is dominant over green (y) and round (R) is dominant over wrinkled (r). As the plant is true-breeding, it would only produce one type of gamete for each gene, thus the yellow round pea plant will produce YR gametes.

The plant with green wrinkled peas has the genotype yyrr, also being true-breeding, hence it will produce yr gametes. Therefore, the correct answer is option c: The genotype of the plant with yellow round seeds will be YYRR and produce YR gametes, while the genotype of the plant with green wrinkled seeds will be yyrr and produce yr gametes.

When considering a Punnett square analysis for such a cross, we would need 4 squares because each parent can produce two different types of gametes and they are combining two traits (seed color and seed shape).

Answer 2

c. The genotype of the plant with yellow round seeds will be YYRR and produce YR gametes, while the genotype of the plant with green wrinkled seeds will be yyrr and produce yr gametes.

Step-by-step explanation:

A true breeding plant always has identical alleles of a gene and is homozygous in nature.

The genotype of the true-breeding parents with yellow round peas: YYRR. The genotype of the pure breeding parent plant with green wrinkled peas: yyrr.

Gametes are always pure which means that a gamete carries only one allele of a gene. Since both the parent plants are true breeding, there will not be any segregation of alleles during gamete formation. Therefore, the parent plant with yellow round peas "YYRR" would form all the gametes with "YR" alleles while the parent plant with green wrinkled peas "yyrr" would form all the gametes with "yr" alleles.