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Cystic fibrosis is a hereditary disorder caused by a recessive mutation. in a previous marriage, jim had a child with cystic fibrosis, 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?

Respuesta :

underV
1) Data:
dominant allele- A
recessive allele- a

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
cystic fibrosis. 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 baby with cystic fibrosis is 25%

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