If Two Parents, Are Heterozygous Carriers of the Autosomal Recessive Gene Causing Cystic Fibrosis,?
Thursday, September 2nd, 2010 at
3:31 am
Microbiology2011 asked:
If two parents, are heterozygous carriers of the autosomal recessive gene causing cystic fibrosis,
have 5 children, what is the probability that: (8pts)
a. three will be normal?
b. four will have cystic fibrosis?
c. all are normal?
Tagged with: Autosomal Recessive Gene • Cystic Fibrosis • Probability
Filed under: Cystic Fibrosis
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You’ve got a 50% chance of a carrier, 25% chance of one effected, and 25% of a normal offspring.
Chance of three being normal= .25x.25x.25… so 1.5625%
Chance of four having CF= .25x.25x.25x.25= .039%
chance of five being normal=.25x.25x.25x.25x.25= .097%