Punnett square? how do i figure this out, cystic Fibrosis? ? - cystic fibrosis genotype
A person who inherits a copy of the allele will be affected, however, if these two people have a child, the child may inherit two copies of allele .....
CC
CC
Legend: normal allele of C / C allele of the disease
1. A healthy man and a healthy woman has a child with cystic fibrosis. Identify the genotypes of this function for the parents and their children.
2. This man and this woman wants a child. Calculate the probability that the second child has cystic fibrosis.
3. Calculate the probability that a third child with cystic fibrosis.
4. The majority of serious genetic diseases are caused by recessive and dominant alleles. Explain why this is the case.
I know how to make a Punnett Square I do not know how to calculate the valuation, instant messaging, do not know what a genotype
Sunday, January 10, 2010
Cystic Fibrosis Genotype Punnett Square? How Do I Figure This Out, Cystic Fibrosis? ?
2 comments:
1. So that the child has cystic fibrosis, the parents must allele as a carrier of the disease. This means that the child has a genotype C / C, and the parents are heterozygous with C / C genotypes.
2. It is a 1 / 16 the possibility that parents have a second child with cystic fibrosis. Because you have a child with cystic fibrosis, and they are healthy, they now know they are carriers of the disease. A square simply insert two C / C have shown that genotype is 1 / 4 Kids c / c, giving them the disease. Because the questions on the likelihood that he had two children with CF, then take 1 / 4 X 1 / 4 First 1 / 16.
3. Same principle. Take the opportunity to get a child with cystic fibrosis and three times more for. It is 1 / 4 times get 1 / 4 times on 1 / 64.
4. If severe disease is dominant as anyone living in the population in the age of reproduction. This is the reason that diseases are recessive lethal. However, you can have serious diseases that are dominant, but the population would be especiallyconsisting of persons recessive trait.
Hope this helps.
1. So that the child has cystic fibrosis, the parents must allele as a carrier of the disease. This means that the child has a genotype C / C, and the parents are heterozygous with C / C genotypes.
2. It is a 1 / 16 the possibility that parents have a second child with cystic fibrosis. Because you have a child with cystic fibrosis, and they are healthy, they now know they are carriers of the disease. A square simply insert two C / C have shown that genotype is 1 / 4 Kids c / c, giving them the disease. Because the questions on the likelihood that he had two children with CF, then take 1 / 4 X 1 / 4 First 1 / 16.
3. Same principle. Take the opportunity to get a child with cystic fibrosis and three times more for. It is 1 / 4 times get 1 / 4 times on 1 / 64.
4. If severe disease is dominant as anyone living in the population in the age of reproduction. This is the reason that diseases are recessive lethal. However, you can have serious diseases that are dominant, but the population would be especiallyconsisting of persons recessive trait.
Hope this helps.
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