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Andreyy89
3 years ago
11

Genetic Inheritance: Calculate probability of two children having the same genotype for achrondoplasia Achondroplasia is a commo

n form of dwarfism caused by autosomal dominant mutation in the fibroblast growth factor receptor 3 (Fgfr3) gene. If a person with achondroplasia (Aa) married and had children with a person of normal height, what is the probability that both their first child and second child would have achondroplasia?
Mathematics
1 answer:
dimaraw [331]3 years ago
6 0

Answer:

The correct answer is - 1/2 or 50% for first and second child to be affected.

Step-by-step explanation:

Achondroplasia is an autosomal dominant disorder. Autosomal dominant disorder refers to the presence of a single copy of the defective gene that is enough to lead to dwarfness.

A cross of achondroplasia (Aa) parent to a person of normal height (aa) result in half of their children will be affected with dwarfism and the other half will be normal.

a  cross between affected or dwarf  and normal parent

     Aa X aa

Punnett square:

         a a

A  Aa Aa

a aa aa

Aa- dwarfness

aa- normal height

The probability that both their first child and second child would have achondroplasia is

2/4 =1/2 or 50%.

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Answer: The tailor made 8 complete hats

Step-by-step explanation:

Amount of fabrics available to make hat = 6 2/3 yards

1 hat requires = 5/6 yards

Number of hat to obtain from the available fabric = Amount of fabrics available to make hat / requirement of 1 hat

=(6 2/3) / (5/6)

(20/3) / (5/6)

20/3 x 6/5 = cancelling out, we are left with

4x2= 8 hats

or

(20/3) / (5/6)

6.6666 /0.83333

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6 0
3 years ago
Cali had $5 to spend on food. If she wants a hamburger that cost $3.50 which of the following can she afford?
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She has $5.
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Now she has $5 - $3.50 = $1.50
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3 years ago
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navik [9.2K]

Answer:

\large\boxed{-y-3(-3y+5)=8y-15}

Step-by-step explanation:

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Answer:

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Step-by-step explanation: HOPE THIS HELPS

99 = 3 x 3 x 11, which can also be written 99 = 3² x 11.

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