Can someone explain this genetics question to me?
angeleyez2936 asked:
A couple who are both carriers of the gene for cystic fibrosis have two children who have cystic fibrosis. What is the probability that their next child will be phenotypically normal?
A couple who are both carriers of the gene for cystic fibrosis have two children who have cystic fibrosis. What is the probability that their next child will be phenotypically normal?
The correct answer is 75% but I don’t see how?
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The point is that the probability doesn’t depend on previous children. So, if both parents are normal and carry a copy of cystic fibrosis gene, then every time they have a kid they have 50% x 50% (25%) chance of it getting both copies of the cystic fibrosis gene. Therefore, 100-25 = 75% chance they will have a normal child every time..
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Both children have 100% enforcement of the cystic fibrosis gene.
If they each had children by someone who did not have the gene then their child would have three genetic enforcements of the gene plus one that isn’t. The father, mother, child and childs’ mate. Therefore there would be a 75% probability of cystic fibrosis. Four people with one free of the gene.
Cystic Fibrosis
Simple: each child has an equal chance of getting cystic fibrosis, regardless of how many children a couple has.
See, the mother and the father each have one allele for CF. That means that they each have a 50% chance of giving it to their children. When multiplied together, there is a 25% chance that their child will have CF.
So, they have their first baby, and they have CF. They have a second baby. They still both have the same alleles, so they each have a 50% chance of passing on that allele (and multiplied together, the chance of the child recieving both alleles is again, 25%). So, their second child also has a 25% chance of getting CF, just like the first, and a 75% chance of being phenotypically normal. It will be the same for their third, fourth, and fifth child.
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Probability? third child’s probabilty is independent, that means info abt “two children who have cystic fibrosis” is irrevalent
if you toss a coin, n result was 2 times tail, n wats is the probabilty of 5th time tail? probabilty is still 50 %. i mean the previous results will not influence the future.
so here probabilty of any child witout desease is always 75 %
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The male and female are both carriers meaning they carry a gene that causes cystic fibrosis. This means that cystic fibrosis is a homozygous recessive trait since their kids have it.
Lets say F is for not having cystic fibrosis (normal) –> dominant
and f is for having cystic fibrosis. (not normal) –> recessive
Since the parents carry these gene that means they are heterozygous….
So their genotype would be Ff (with one normal and one affected)
When you cross them ( do a punnett square)
Ff x Ff you get —-> FF Ff Ff ff
So from the 4 results you see the F on 3/4 meaning theres 3/4 of a chance that their next kid will be phenotypically normal.
3/4 = 75%
Get it?