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Rufina [12.5K]
3 years ago
6

One parent is a cystic fibrosis carrier, and the other has

Mathematics
1 answer:
Marysya12 [62]3 years ago
5 0

Answer:

A - 0%

B- 50%

C- 50%

D- 100%

Step-by-step explanation:

Cystic fibrosis is inherited in an autosomal recessive form, meaning that a person has to inherit two abnormal genes for the disease to manifest. In the case of this question, one parent is a gene carrier, so his genotype is Aa, while the other does not have the cystic fibrosis gene, so AA.

Performing the cross of Aa x AA, we can see that:

a.) The probability of a child would have cystic fibrosis is 0%, since the disease is recessive and to be affected it should receive a recessive gene from each parent.

b.) The probability of a child would be a carrier is 50%, as 50% of the crossing phenotypes are Aa.

c.) The probability of a child would not have cystic fibrosis and is not a carrier is 50%, as 50% of the child's genotype is AA.

d.) The probability of a child would be healthy is 100%, as of all possible phenotypes none is affected.

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Number of dimes are 18 and number of quarters are 31

<em><u>Solution:</u></em>

Let "d" be the number of dimes

Let "q" be the number of quarters

value of 1 dime = $ 0.10

value of 1 quarter = $ 0.25

A collection of dimes and quarters is worth $9.55

value of 1 dime x number of dimes + value of 1 dime x number of quarters = 9.55

0.10d + 0.25q = 9.55 ---------- eqn 1

If the quarters were dimes and the dimes were quarters, the total value would be 7.60

quarters were dimes means , q = d

dimes were quarters means d = q

0.25d + 0.10q = 7.60 ----- eqn 2

Let us solve eqn 1 and eqn 2 to find "d" and "q"

Multiply eqn 1 by 2.5

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Subtract eqn 2 from eqn 3

0.25d + 0.625q = 23.875

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0.525q = 16.275

q = 31

Substitute q = 31 in eqn 1

0.10d + 0.25q = 9.55

0.10d + 0.25(31) = 9.55

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Answer:

<em>The true statements are:</em>

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upper quartile: 28

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