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Say there is a situtaiton when Parent 1 is a heteroyzgote for gene A. Parent 2...

Say there is a situtaiton when Parent 1 is a heteroyzgote for gene A. Parent 2 is homozygous recessive for gene A. What is the probability that these parents will have 4 children who are all heterozygotes? Assume no multiple births

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Answer #1

Answer: 0.0625 (6.25%)

Explanation:

  • Parent 1: Heterozygote: Aa
  • Parent 2: homozygous recessive: aa

- Following would be the cross: (Aa x aa)

A a
a Aa aa
a Aa aa

- Genotypic frequency

Genotype Chances Percentage
Aa 2/4 = 1/2 50%
aa 2/4 = 1/2 50%

- So, there are 50% (1/2) chances of child to be heterozygotes.

- Probability of all 4 child to be heterozygote will be,

= 0.0625 (6.25%)

- Brief Note: As these are independent events, we will use the product rule.

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