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In peas, axial (A) flower position is dominant to terminal (a), tall (L) is dominant to short (l), and yellow (Y) is dominant to green (y). If a plant that is heterozygous for all three traits is allowed to self-fertilize, how many of the offspring would be dominant for all three traits? In peas, axial (A) flower position is dominant to terminal (a), tall (L) is dominant to short (l), and yellow (Y) is dominant to green (y). If a plant that is heterozygous for all three traits is allowed to self-fertilize, how many of the offspring would be dominant for all three traits? 3/64 32/64 27/64 64/64 9/64

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namordu

Answer:

27/64

Explanation:

  • Axial flowers (A) are dominant to terminal (a)
  • Tall (L) is dominant to short (l)
  • Yellow (Y) is dominant to green (y)

The genotype of the heterozygous plant can be written as AaLlYy and the cross would be AaLlYy x AaLlYy.

If the three genes are independent, we can analyze each gene individually, obtain the proportion of offpsring with the dominant allele for each gene, and then multiply those probabilities to obtain the probability of having A_L_Y_ offspring, because the rules of probability state that the probability of two independent events occurring at the same time can be calculated as the multiplication of the probabilities of occurrence of each event.

Aa x Aa

  • 3/4 A_
  • 1/4 aa

Ll x Ll

  • 3/4 L_
  • 1/4 ll

Yy x Yy

  • 3/4 Y_
  • 1/4 yy

So the total proportion of offspring with dominant phenotype for all traits will be 3/4 × 3/4 × 3/4 = 27/64

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