Question

In Mendel's peas the alleles for flower color are P and p. The alleles for seed color are Y and y. A plant has the genotype Pp Yy. Which of the following genotypes are possible in a gamete from this plant?

1) PY
2) Pp
3) Pp Yy
4) yy

Answer 1

The correct gametes from a Pp Yy plant are single allele combinations like PY or yy; Pp or Pp Yy are incorrect. In a PpYY x ppyy dihybrid cross, a 4x4 Punnett square is required to determine the genotypes and phenotypes of the offspring.

Step-by-step explanation:

The student's question about Mendelian genetics pertains to the possible genotypes of gametes. Considering a pea plant with a genotype of Pp Yy, the possible gametes it can produce will contain one allele for each gene. Gametes are haploid, meaning they carry only one allele per gene. Hence, the correct answers would be options that reflect single alleles for each trait, such as PY or yy. Options involving double letters, like Pp or Pp Yy, would be incorrect as they represent a diploid genotype, not haploid.

In a dihybrid cross between PpYY and ppyY pea plants, we must examine both flower color and pea color genes to determine the possible genotypes and phenotypes of the offspring. Creating a Punnett square, we can visualize all the potential allele combinations. With a dominant allele P for purple flowers and a dominant allele Y for yellow peas, predicted phenotypes can involve purple flowers with yellow peas, among other combinations, based on the specific parent genotypes.

To conduct a Punnett square analysis for this cross, a 4x4 Punnett square is necessary because each parent plant has two heterozygous traits, resulting in four different types of gametes from each parent. This will create 16 squares in the Punnett square for all possible combinations of alleles from both parents.