The probability of a specific combination of offspring = { n! / (w! x!)} multipl
ID: 61325 • Letter: T
Question
The probability of a specific combination of offspring = { n! / (w! x!)} multiplied by pw multiplied by q x. Where n, w and x refer to numbers of observations (e.g. progeny) and p and q refer to the probability of the respective observation.
A man and a woman are heterozygous for cystic fibrosis, an autosomal recessive disorder: ie., both are carriers (genotype Aa).
_______: (2 pts) what is the probability that, if they have 5 children by separate conceptions and births, 1 will have CF and 4 will be normal? Show your work.
Explanation / Answer
Given, both man and woman are heterozygous for cystic fibrosis. It means both are carriers and have the genotype Aa. Then the different possible genotypes in children are AA, Aa, and aa. From normal cross, we know that 1 child will be affected and other 3 will be normal phenotypically. This shows that there is 1/4 or 25% chance of getting affected child in every birth. Thus, if the parents have 5 children (each time 4 chances), then out of them the probability of having 1 affected child is 5/20 = 1/4.
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