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4 years ago

Please Help ASAP!!

Mathematics

1 answer:

Brilliant_brown [7]4 years ago

4 0

Answer:

The probably genotype of individual #4 if 'Aa' and individual #6 is 'aa'.

Step-by-step explanation:

In a non sex-linked, dominant trait where both parents carry and show the trait and produce children that both have and don't have the trait, they would each have a genotype of 'Aa' which would produce a likelihood of 75% of children that carry the dominant traint and 25% that don't. Since the child of #1 and #2, #5, does not exhibit the trait, nor does the significant other (#6), then they both must have the 'aa' genotype. However, since #4 displays the dominant trait received from the parents, it is more likely they would have the 'Aa' genotype as by the punnet square of 'Aa' x 'Aa', 50% of their children would have the 'Aa' phenotype.

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3 years ago

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Step-by-step explanation:

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3 years ago

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Step-by-step explanation:

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Binomial probability distribution

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$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$