Cystic fibrosis is a hereditary disorder caused by a recessive mutation. in a previous marriage, jim had a child with cystic fib
rosis, though neither jim nor his first wife had the disease. jim is now married to martha, who had a brother that died of cystic fibrosis. though martha does not have the disease, she found out that she is a carrier. what is the probability that jim and martha will have a baby with cystic fibrosis?
Heterozygotic: Aa (only carries the allele) Homozygotic recessive: aa (expresses the disease) Homozygotic dominant: AA (doesn't carry or expresses the disease)
2)The first cross between Jim and his first wife originated a offspring with <span>cystic fibrosis</span>. That offspring has the genotype aa and for that to happen both parents need to be a least carriers or express the disease. The exercise rules out the second case, which makes Jim a carrier.
3)Now between Jim and Martha: From the previous step, we already found that jim is a carrier and the exercise also tells us that Martha is one too. So, their genotype is: Aa. The cross: Aa x Aa The probability: 1/4 aa = 25% 2/4 =Aa 1/4 - AA
the probability of having a <span>baby with cystic fibrosis is 25%</span>