Q: 2. What would a Punnett square that shows how the son in generation IV (marked with an A) inherited cystic fibrosis look like?

A: 1. In order for the child to have cystic fibrosis, the parents must both be carriers of the disease allele. This means that the child has a genotype of c/c and the parents are both heterozygous with C/c genotypes.
2. [ There is a 1/16 chance that the parents will have a second child with cystic fibrosis. Because they have a child that has cystic fibrosis, and they are healthy, they now know

that they are carriers for the disease. A simple punnet square of two C/c genotypes will show that 1/4 of the children will be c/c which gives the disease. Therefore, because the questions is asking what is the chance that they will have two children with CF, then you take 1/4 times 1/4 getting 1/16.
3. Same principle. You take the chance of getting a child with CF and multiply it three times. So it's 1/4 times 1/4 times to get 1/64.
4. If the serious diseases were dominant than no one in the population would live to reproduction age. This is why that lethal diseases are recessive. However, you can have serious diseases that are dominant, but then the population would be mostly composed of recessive individuals. ]

Marvel2|Points 484|

Rating

There are no new answers.

26,528,028 questions answered

There are no comments.