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OlgaM077 [116]
3 years ago
10

Hemophilia a is an x-linked recessive disorder. if a normal man marries a woman who is a carrier, what fraction of their daughte

rs will have hemophilia
Biology
1 answer:
RSB [31]3 years ago
3 0
Daughters get one X gene from each parent.
If the father is a normal male, he carries only a normal X-gene.
Therefore the daughter will always get a normal gene from the father, and a 50% probability getting an affected gene from the mother, therefore 50% chance of becoming a carrier.  The other 50% she will inherit a normal X-gene from each parent, thus a healthy female.

In conclusion, no daughter will have haemophilia from a carrier mother and a normal male.
(however, sons will have a 50% chance of inheriting affected X-gene and hence will have haemophilia).
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Humerus bones from the same species of animal have approximately the same length-to-width ratios. It is known that Species A has
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Answer:

The unearthed Humerous bones don't belong to the species A.

Explanation:

Hello!

You are studying the Humerus bones species A, who is known to have a mean ratio of 8,5. This value corresponds to the population mean of the length-to-with ratio of bones of species A, symbolized as μ.

The hypothesis you want to study is "The Humorous bones unearthed belong to the species A" if you assume this to be true, then the mean of the length-to-with ratio should be equal to the known population mean of the length-to-with ratio.

Symbolized:

H₀: μ = 8,5

H₁: μ ≠ 8,5

Significance level: α: 0,05

You are asked to use a Z-test, since you don't know the value of the population variance, but have the sample values, the sample size is big enough (more than n=30). Assuming that the sample values are independent, the statistic test of choice is the approximation:

Z= \frac{x_bar-μ}{S\sqrt[]{n} }≅N(0;1)

The critical region, in this case, it's a two-tailed test (remember the type is determined by the null hypothesis) so you'll have two critical values.

Left value [/tex]Z_{\alpha/2} = Z_{0,025} = -1,96

Right value Z_{1-\alpha/2} = Z_{0,975} = 1,96So you'll reject the null hyphotesis if the calculated [tex]Z_{obs} value is ≤-1,96 or ≥1,96 and you'll support it if -1,96<Z_{obs}<1,96

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x_bar = 9,26

S = 1,20

n = 41

Z_{obs}= \frac{x_bar-μ}{S\sqrt[]{n} }

Z_{obs}= \frac{9,26-8,5}{1,20\sqrt[]{41} }

Z_{obs}= 4,0553

Since the calculated value falls in the rejection region, this means, you have statistically significant results. In other words you can reject the null hipothesis (H₀: μ = 8,5) and asume that the unearthed Humerous bones don't belong to the species A.

I hope you have a SUPER day!

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