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Mnenie [13.5K]
4 years ago
10

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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Step-by-step explanation:

As per given , we have

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Margin of error (maximum error): E=7\%=0.07

Formula to find the sample size :

n=p(1-p)(\dfrac{z_{\alpha/2}}{E})^2\\\\ n=0.20(1-0.20)(\dfrac{1.96}{0.07})^2\\\\ n=125.44\approx126

Therefore , the required minimum sample size = 126

5 0
3 years ago
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