In: Biology
Suppose colourblindness is an X-linked recessive trait, while Huntington's disease is an autosomal dominant trait. Assume both traits are rare. A colourblind man decides to have children with a women who is heterozygous for both genes. What is the probability that they will have a child that is colourblind and does not have Huntington's?
The gene for colourblindness is located on X-chromosome while that of Huntington's disease is located on an autosome. The two events or disorders are thus independent i.e. having colourblindness will not affect the susceptibility of a person towards Huntington's disease or vice versa.
To calculate the probability of independent events, product rule is applied. So, probability of a child that is colorblind and does not have Huntington's disease will be the product of probability having colourblindness and probability of not having Huntington's.
In simple terms, for two independent events P and Q,
probablity of P and Q = (probability of P) x (probability of Q).
Step 1 : To determine probability of colourblindness trait among progeny.
Parents - ♂ x ◙
genotypes - XCY x XCX
gametes - XC , Y XC, X
f1 generation - XCXC , XCX , XCY , XY
XCXC - affected,
XCX - normal
XCY - affected
XY - normal
Since colourblindness is a recessive trait, probability of having a colourblind child = 2/4 or 1/2.
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Step 2 : To determine probability of Huntington's disease among progeny.
Let's denote the allele for Huntington's disease with "A" and normal allele with "a".
Parents - ♂ x ◙
genotypes - aa x Aa
gametes - a , a A , a
f1 generation - Aa , aa , Aa , aa
Aa - affected, and
aa - normal.
Since Huntington's disease is an autosomal dominant trait, probability of not having Huntington's (i.e. genotype "aa") = 2/4 or 1/2.
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Probability of a child that is colorblind and does not have Huntington's disease will be 1/2 x 1/2 = 1/4