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Imagine that you are a genetic counselor, and a couple planning to start a family comes to you for information. Charles was married once before, and he and his first wife had a child with cystic fibrosis. The brother of his current wife, Elaine, died of cystic fibrosis. What is the probability that Charles and Elaine will have a baby with cystic fibrosis

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Answer:

1/6

Explanation:

Assuming the disorder is not sex linked, Charles is a carrier so he will be Ff.

His wife Elaine can be a carrier or not because her parents must have been both Ff

Possible F1 generation for her was FF, Ff, Ff and ff. She does not have the disease.

Therefore the total chance of her having FF was 1/3 whereas Ff will be  2/3

Probability of the kid having the disease will be calculated by;

(2/3×1/4) +(1/3×0)= 1/6

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The probability will be equal to 1/6.

We can achieve this result because:

  • Cystic fibrosis is a genetic disease linked to a dominant allele, in this case, we can say that it presents in individuals with the AA and Aa allele.
  • This disease is not sex-linked, for this reason, we can consider that Charles is the carrier of the gene, having received a dominant allele from his parents, therefore being Ff.
  • We can also consider that the gene representing the disease is in Elaine's family and therefore we can consider that she may also have the Ff alleles.

The cross between Elaine and Charles would be Ff x Ff and this would result in offspring FF, Ff, Ff, ff, that is, 1/3 FF, 2/3 Ff, and 1/3 ff.

Based on this, we can calculate the probability as follows:

[tex](FF*Ff)+(ff*0)=\\(2/3*1/4) +(1/3*0)= 1/6[/tex]

Therefore, we can state that the probability of one of Elaine and Charles' children possessing the FF or Ff alleles and having cystic fibrosis is 1/6.

More information:

https://brainly.com/question/14132766?referrer=searchResults

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