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If only one person is a carrier of cystic fibrosis than there is no chance of having a child with it. Both parents have to be carriers and even then there is only a 25% chance. If only one carries than there is a 50% chance that their children will carry but will not have cystic fibrosis.
1-.015 = .985
Zero. Cystic fibrosis is an autosomal recessive disease. This means that both parents must carry a mutated gene and have a 1 in 4 (25%) chance of having a child with CF.
probability is used in genetics to determine the possibilities of offspring having a certain trait
Genotype is used to determine the probability of having specific offspring from two known parents.
The probability of a child having cystic fibrosis in this scenario is 50%. If one parent is homozygous dominant (no cystic fibrosis allele) and the other is a carrier (heterozygous), there is a 50% chance of passing on the cystic fibrosis allele to their child, resulting in the child having cystic fibrosis.
The addition rule is used when calculating the probability of two mutually exclusive events occurring together. For example, when calculating the probability of rolling a 2 or a 6 on a six-sided die, you would use the addition rule.
A Punnett square is a diagram used to predict the probability of offspring having specific genotypes based on the genotypes of the parents. It helps visualize the potential genetic outcomes of a mating.
If only one person is a carrier of cystic fibrosis than there is no chance of having a child with it. Both parents have to be carriers and even then there is only a 25% chance. If only one carries than there is a 50% chance that their children will carry but will not have cystic fibrosis.
A Punnett square for cystic fibrosis would involve crossing two parents who are carriers of the recessive allele for the disease (denoted as "cf"). The square would show the probability of having a child with cystic fibrosis (25%), a carrier (50%), or unaffected by the disease (25%). Each parent would have a genotype of "Cf" (carrier) for the Punnett square.
same probable as the first child having it. By doing the punnet square, they are both recessive for the disease. There is a 25% chance that the child will get the disease.
because 95% of men with cystic fibrosis are sterile
phenotypes are decided by the alleles for that particular characteristic, by a dominant or two recessive alleles. For example, cystic fibrosis has a recessive allele so the phenotype of cystic fibrosis would only appear if there were two of the recessive allele, one from each parent, were present. A heterozygous carrier of the cystic fibrosis allele would show the phenotype of not having cystic fibrosis. So to determine the phenotype simply find out which allele is dominant and find what alleles each parent has the the probability of each phenotype can be calculated
1-.015 = .985
50%
Equiprobable, but I would stick with simplicity of communication and go with "having the same probability".
If both parents are carriers then the child has a 25% chance of having cystic fibrosis. If one parent has CF and the other the other was just a carrier then the child has a 50% chance of having CF. If one parent has CF and the other has two normal genes then there is no chance of the child having CF. If one parent is a carrier and the other has two normal genes then there is no chance of the child having CF. If both parents have CF then there is a 100% chance that the child will also have CF.