A man is heterozygous for achondroplasia for which the autosomal dominant A allele confers extremely short...

A man is heterozygous for achondroplasia for which the autosomal dominant A allele confers extremely short stature. His wife's genotype is aa and she is normal height. They have five children together. What is the probability that any two of their five children will have the short stature phenotype, in any birth order?

Solutions

Expert Solution

solution:

The probability is 5/16 that any two of their five children will have the short stature phenotype.

Explanation:

As man is heterozygous for achondroplasia (Aa) and and his wife has normal genotype (aa). So there is 50% chance of child having normal genotype and 50% chances of child with achondroplasia.

	a	a
A	Aa	Aa
a	aa	aa

50% or 1/2 = Aa;

50%or 1/2 = aa

To calculate the probability we use binomial theorem:

Let's assign probabilities to each having the achondroplasia, n= 5 no. Of children

a = child having achondroplasia = 1/2 (Aa)

b = normal child =1/2 (aa)

The expanded binomial for the value of n=5

(a+b)5 = a5 + 5a4b+10a3b2+10a2b3+5ab4+b5

As each term represents a possible outcome. The term describing the outcome of two achondroplasia (short stature) children and three normal children, the expression of probability is

p = 10a2b3

= 10(1/2)2(1/2)3

= 10(1/2)5

= 10(1/32) = 10/32 = 5/16

p = 5/16

The probability is 5/16 that two children out of five will have the short stature phenotype.

gladiator answered 1 year ago

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