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flower position, stem length, and seed shape are three characters that …

Question

flower position, stem length, and seed shape are three characters that mendel studied. each is controlled by an independently assorting gene and has dominant and recessive expression as follows:
character dominant trait recessive trait
flower position axial (a) terminal (a)
stem length tall (t) dwarf (t)
seed shape round (r) wrinkled (r)
if a plant that is heterozygous for all three characters is allowed to self - fertilize, what proportion of the offspring would you expect to be each of the following? (note: use the rules of probability instead of a huge punnett square.)
match the terms in the left column to the appropriate blanks in the sentences on the right. terms can be used once, more than once, or not at all.
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  1. homozygous for the three dominant traits:
  2. homozygous for the three recessive traits:
  3. heterozygous for all three characters:
  4. homozygous for axial and tall, heterozygous for seed shape:

Explanation:

Step1: Determine probability for each trait

For a heterozygous plant \(AaTtRr\) self - fertilizing:

  • For homozygous dominant (\(AA\)): probability \(P(AA)=\frac{1}{4}\) (from \(Aa\times Aa\): \(AA:Aa:aa = 1:2:1\)). Similarly for \(TT\) and \(RR\), each has probability \(\frac{1}{4}\).
  • For homozygous recessive (\(aa\)): probability \(P(aa)=\frac{1}{4}\). Similarly for \(tt\) and \(rr\), each has probability \(\frac{1}{4}\).
  • For heterozygous (\(Aa\)): probability \(P(Aa)=\frac{1}{2}\). Similarly for \(Tt\) and \(Rr\), each has probability \(\frac{1}{2}\).

Step2: Calculate for each case

  1. Homozygous for three dominant traits (\(AATT RR\)):

Using the multiplication rule (since genes assort independently), \(P=\frac{1}{4}\times\frac{1}{4}\times\frac{1}{4}=\frac{1}{64}\)

  1. Homozygous for three recessive traits (\(aatt rr\)):

\(P=\frac{1}{4}\times\frac{1}{4}\times\frac{1}{4}=\frac{1}{64}\)

  1. Heterozygous for all three characters (\(AaTtRr\)):

\(P=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{8}\)

  1. Homozygous for axial (\(AA\)) and tall (\(TT\)), heterozygous for seed shape (\(Rr\)):

\(P=\frac{1}{4}\times\frac{1}{4}\times\frac{1}{2}=\frac{1}{32}\)

Answer:

  1. \(\frac{1}{64}\)
  2. \(\frac{1}{64}\)
  3. \(\frac{1}{8}\)
  4. \(\frac{1}{32}\)