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.
Assuming that the man who has normal vision is homozygous for normal vision, the couple's daughter will either be homozygous for normal vision or heterozygous (normal vision but carrier for color blindness) for normal vision. In light of this, the couple's daughter will not be color blind.
You might need to indicate what the disorder is.
50%
A calculator for the Bivariate Normal At the bottom of the page at the link, under "A calculator for cumulative probabilities from the bivariate normal distribution" there's a little binary applet, which can be downloaded, and which calculates the percentage chances of being in various parts of the volume of the distribution.
Let me guess, from an on line test? Chances are the test was a fake. If however this test was administered and scored by a trained psychologist, than it depends on which test you took. The normal average is 100 for any age.
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.
The chances are one half. You will most likley have 2 children without the gene, and 2 with one gene. Since the normal parent has the dominate allele, the Cystic Fiborisis will not be present.
Sure! When both parents are heterozygous carriers (Cc) for cystic fibrosis, the Punnett Square would result in a 25% chance of having a child with cystic fibrosis (cc), a 50% chance of having a child who is a carrier (Cc), and a 25% chance of having a child who is not a carrier and does not have the disease (CC).
If both parents are carriers of the cystic fibrosis gene, each child has a 25% chance of inheriting two copies of the gene and having cystic fibrosis, a 50% chance of being a carrier like the parents, and a 25% chance of not inheriting any copies of the gene. Therefore, the percentage of a child not having cystic fibrosis with carrier parents is 25%.
The genotype would have to be homozygous recessive if the child was completely effected by the disease. The "normal" paretns would have to have heterozygous recessive genotypes. This makes sense since the allele that causes sickle cell shows incomplete dominance when present with a normal allele in a pair. The "normal" parents actually would have a mixture of sickle cell shaped red blood cells combined with normal shaped ones. The carrier parents does not display symptoms of the disease since the regular red blood cells alone can fill the body's need for oxygen under normal circumstances. The only time the cArrier would notice would be under times of extreme oxygen demand, such as a sprint.
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.
Individuals with cystic fibrosis have a mutation in the CFTR gene. Being heterozygous for the mutant CFTR gene means having one normal and one mutant copy of the gene, resulting in a carrier state. Carriers are typically unaffected by cystic fibrosis but can pass the mutant gene on to their offspring.
Cystic Fibrosis is an example of a disease inherited as an autosomal recessive genetic disorder. Both parents of a baby who has Cystic Fibrosis have one abnormal recessive gene and one normal dominant gene and are therefore, Cystic Fibrosis carriers. Carriers are said to be heterozygous because they contain one normal copy of the gene and one abnormal copy. A baby who inherits cystic fibrosis has inherited the abnormal gene from both parents and is therefore said to be homozygous recessive.
Cystic fibrosis is a genetic disorder that gene therapy may correct by introducing a healthy copy of the CFTR gene into cells to restore normal function in affected individuals.
In order to have a child with cystic fibrosis, both parents must be carriers (assuming neither of the parents actually have CF; then obviously they will have symptoms). In CF, a carrier has no symptoms, unlike sickle cell anemia, where a carrier will have mild symptoms of the disease. A carrier of CF has only one defected 7th chromosome, and the other unaffected one makes the CFTR gene work correctly. When both are defected, the result is the CF disease, and normal CFTR genes are impossible to make, causing the person to have the disease.
Cystic fibrosis is a genetic disorder that affects the lungs and digestive system. It is caused by a mutation in the CFTR gene, which leads to the production of thick and sticky mucus that can clog the airways and prevent proper digestion. This can result in respiratory infections and difficulty breathing, as well as problems with nutrient absorption and digestion.
By "another pregnancy", I assume that the parents have had one child with cystic fibrosis. As cystic fibrosis is recessive and neither parent suffers from it, they must both be carriers. That means the chances of the next baby having CF is 25%, or a one in four chance. Assuming both parents have one CF and one non-CF gene, the combinations work out: non-CF x non-CF (normal) non-CF x CF (carrier) CF x non-CF (carrier) CF x CF (cystic fibrosis sufferer) Therefore their chances of having a: normal child = 25% child who is a CF carrier = 50% child who suffers from CF = 25%