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If both parents are carriers on cystic fibrosis, just because two children may have the disease it doesn't mean that say,a further 3 children in future will be "normal". Inheriting genes is like a lottery. I can say though that because cystic fibrosis is recessive, every offspring born under carriers will always have a 75% chance of being phenotypically normal.

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Two parents who were phenotypically normal had to children The oldest was also phenotypically normal but the younger child had cystic fibrosis what is happening in the cross?

Cystic fibrosis is an autosomal recessive genetic disorder, meaning that a child must inherit two copies of the mutated gene (one from each parent) to express the condition. In this scenario, both parents are likely carriers of the cystic fibrosis allele (genotype Cc), where "C" is the normal allele and "c" is the mutated allele. The older child is phenotypically normal, which could mean they are either homozygous dominant (CC) or a carrier (Cc), while the younger child inherited two copies of the mutated allele (cc) from both parents, resulting in cystic fibrosis. This cross illustrates Mendelian inheritance patterns, where two carrier parents can have a child with a recessive condition.


What is the probability of two healthy people where one is a carrier being parents to a child born with cystic fibrosis?

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.


Do carriers have to have cystic fibrosis?

Carriers of cystic fibrosis do not have the disease themselves, as they possess one normal copy and one mutated copy of the CFTR gene. However, they can pass the mutated gene to their children. For a child to develop cystic fibrosis, they must inherit two copies of the mutated gene, one from each parent. Therefore, while carriers do not have cystic fibrosis, they play a crucial role in its inheritance.


Is it possible for a mother to have two kids with different father be born with cystic fibrosis?

Depends. If your mother has cystic fibrosis and your father is a carrier, there is a 50% chance that any of their children will have cystic fibrosis. If the father is not a carrier, no children will have cystic fibrosis, but they will all be carriers.


What is the probability of a child having cystic fibrosis if one of her parents was homozygous dominant and the other was a carrier for cystic fibrosis?

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.


What percentage of Caucasian Americans are carriers of the Cystic Fibrosis?

Approximately 1 in 31 Caucasian Americans are carriers of the cystic fibrosis gene, which equates to around 3.2% of the population.


What is a punnett square for 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.


Why does a person with Cystic Fibrosis have the genotype cc?

Cystic Fibrosis is inherited as a recessive trait. This means that both parents have to carry the allele for the defect. Both parents are Cc which means that 25% of their children will be CC (homozyous dominant and will not be able to pass the recessive allele in subsequent generations, 50% will be Cc(carriers) and 25% will be cc (homozygous reccessive and have Cystic Fibrosis).


To be a cystic fibrosis carrier do your parents have to have a cystics fibrosis gene?

Yes, since the disease is a recessive inherted trait BOTH parents must carry the gene but will not have the disease itself. Approximately 30,000 people in the United States have cystic fibrosis. An additional ten million more-or about one in every 31 Americans-are carriers of the defective CF gene, but do not have the disease. The disease is most common in Caucasians, but it can affect all races.


What is the genotype of an individual that is not affected with cystic fibrosis?

Let's call the gene "C." Capital C means no cystic fibrosis; lower c means cystic fibrosis, since it is a recessive gene. CC is a person who does not have cystic fibrosis and also is not a carrier. Cc indicated a carrier. cc shows a person with cystic fibrosis. In order for a child to have cystic fibrosis, its parents must be: 1. cc and cc (both have cystic fibrosis, so every child will as well.) 2. Cc and Cc (both carriers; 25% chance of having a child with cystic fibrosis) 3. Cc and cc (one parent is a carrier and one has cystic fibrosis; there is a 50% chance that the children will have cystic fibrosis.)


How disorder is inherited cystic fibrosis?

Cystic Fibrosis, is a recessive disease (meaning that both parents must be carriers of the cystic fibrosis gene, for the offspring to have a chance of being born with it). A child has a 25% chance (1/4) of being born with Cystic Fibrosis.


What would be the best way to predict the probability of a baby having Christic fibrosis?

The best way to predict the probability of a baby having cystic fibrosis (CF) is through genetic testing of the parents, specifically looking for mutations in the CFTR gene. If both parents are carriers of the CF gene mutation, there is a 25% chance with each pregnancy that their child will inherit cystic fibrosis. Additionally, prenatal testing methods like chorionic villus sampling (CVS) or amniocentesis can be used to test the fetus directly for CF. Genetic counseling can also provide valuable information and support for prospective parents.