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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
1:1
Assuming that each parent is a carrier for cystic fibrosis (has the genotype Ff), the probability that their second child will develop cystic fibrosis is one fourth. The probability doesn't change with the number of children they have. For each pregnancy, the chance that the child will have cystic fibrosis (have the genotype ff) is exactly the same.
Calculate the probability of a child having either sickle-cell anemia or cystic fibrosis if parents are each heterozygous for both.
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.
The Punnett square is a diagram that is used to predict the outcome of a particular cross or breeding experiment. It is used by biologists to determine the probability of an offspring having a particular genotype.
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".
The Punnett square is a diagram that is used to predict an outcome of a particular cross or breeding experiment. It is named after Reginald C. Punnett, who devised the approach, and is used by biologists to determine the probability of an offspring having a particular genotype. The Punnett square is a summary of every possible combination of one maternal allele with one paternal allele for each gene being studied in the cross.
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.