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a student crosses two pea plants. in this species, the allele for axial…

Question

a student crosses two pea plants. in this species, the allele for axial flowers is dominant to the allele for terminal flowers. one of the plants that the student uses in the cross is homozygous dominant for axial flowers. the other is heterozygous, which means that it has one allele for axial flowers and one allele for terminal flowers.
if the student examines 200 offspring pea plants from this cross, which of the following is a reasonable result?
○ a. 98 with axial flowers, 102 with terminal flowers
○ b. 47 with axial flowers, 153 with terminal flowers
○ c. 149 with axial flowers, 51 with terminal flowers
○ d. 200 with axial flowers, 0 with terminal flowers

Explanation:

Step1: Determine the genotypes of parents

Let the dominant allele (axial flowers) be \( A \) and the recessive allele (terminal flowers) be \( a \). The homozygous dominant plant has genotype \( AA \), and the heterozygous plant has genotype \( Aa \).

Step2: Perform the Punnett square

When crossing \( AA \) and \( Aa \), the possible gametes from \( AA \) are all \( A \), and from \( Aa \) are \( A \) and \( a \). The Punnett square gives the following offspring genotypes:

  • \( AA \) (from \( A \) of \( AA \) and \( A \) of \( Aa \))
  • \( Aa \) (from \( A \) of \( AA \) and \( a \) of \( Aa \))

So all offspring will have at least one dominant allele \( A \), meaning all offspring should have axial flowers (since the dominant allele determines the phenotype).

Step3: Analyze the options

Since all offspring should have axial flowers, the only reasonable result is 200 with axial flowers and 0 with terminal flowers.

Answer:

D. 200 with axial flowers, 0 with terminal flowers