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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. 47 with axial flowers, 153 with terminal flowers B. 149 with axial flowers, 51 with terminal flowers C. 98 with axial flowers, 102 with terminal flowers D. 200 with axial flowers, 0 with terminal flowers

User Cyva
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2 Answers

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I would have to say it would be c
hope its correct
let me know if it's not
User Jdobry
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Answer:

The correct answer would be option D.

It can be explained with the help of Punnett square.

Let A and a be the alleles of the gene responsible for the position of the flowers.

A is the dominant allele which codes for axial flower and a be the recessive allele that codes for terminal flowers.

The genotype of a plant homozygous dominant axial flower would be AA.

The genotype of a plant heterozygous dominant axial flower would be Aa.

All the offspring produced by this cross would have axial flowers as the genotype of the offspring would either be AA or Aa, that is, the dominant phenotype in all the offsprings.

A student crosses two pea plants. In this species, the allele for axial flowers is-example-1
User Wirling
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