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Cystic Fibrosis Activity:  Understanding the inheritance and expression of an autosomal recessive trait

GenScope File: CysticFibrosis.gs
This file consists of 2 humans, Jean and Phil, each heterozygous for cystic fibrosis CF/cf. Various children marry.  Steve, a husband in the F1 generation is homozygous normal, CF/CF. Nicole, a wife in the F2 generation, has an unknown genotype for the disease.  One of the problems is to figure out what her genotype is.  FYI, she is heterozygous: CF/cf.  It is possible to view everyone's chromosomes except Jake and Nicole's.

1.      Jean and Phil have a daughter, Kelly, who has cystic fibrosis.

Is it possible that her brother, Jake, has the disease, too?

What are the chances that he does have the disease? (25%, 50%, 75%)

Explain how you know.

The student can inspect the pedigree and see not only that Kelly has the disease, but also that Jenny doesn't.  By doing a Punnett square, s/he can see that the chances of a child having the disease are 1:4.  Her brother could have the disease.

Here is another example of phenotype/genotype thinking, but this time pedigree is involved.  The graphics of the pedigree make it easier to see the relationships.  By drawing a Punnett square, the probabilities of the relationships may be seen.  Between the two approaches -- one a picture of what actually happened in a family and the other a mathematical representation of the statistical probabilities, we hope that the student will have a better understanding of what is involved in figuring out this problem.

First, because Kelly has the disease, the student should be able to recognize that both parents are heterozygous without ever looking at their chromosomes.  Cystic fibrosis is a recessive trait, thus requiring two recessive alleles (cf/cf) for it to appear in an organism.  Since neither of the two parents have the disease, they must be heterozygous, or carriers (CF/cf).

Second, when you do a Punnett square for the two heterozygous parents, the genotypic ratio is 1 heterozygous normal (CF/CF), 2 heterozygous normals (CF/cf) and 1 homozygous recessive, who has the disease (cf/cf).  This means that the chances of having a child with the disease is 1:4 for heterozygous parents and the chances of having a normal child are 3:4. Many students think the Punnett square is a predictor; if a couple has 4 children, then 1 of them will surely have cystic fibrosis.  You may have more trouble convincing them that this is not true than you will convincing them that two normal parents can have a diseased child!  You should stress, whenever explaining Punnett squares, that these are statistical probabilities only -- that with each child born to this couple the odds are the same.
 

2.       If Kelly marries a person who is normal and a non-carrier, will all of her children have the disease?    How do you know?

The student should answer that all of the children will be normal.  The answer to "how do you know?" may be expressed in words, as a Punnett square, etc.  The normal condition is dominant, a non-carrier is homozygous (CF/CF), and therefore, all of the children will be heterozygous (CF/cf) and carriers, but disease free.

In the first problem, the student worked backward from the phenotype and genotype of the child to those of the parents. In this one s/he must work forward from the parent to the children.  S/he must also keep in mind that having the disease means that the person is homozygous for the trait (cf/cf).  If one of the parents is homozygous dominant (a normal non-carrier, CF/CF) and the other homozygous recessive, then all of the children will be normal carriers (CF/cf). Again, the student may do a Punnett square or figure it out by inspection.
 

3.       Scott, from the F2 generation, marries Nicole and they have two children, one with cystic fibrosis and the other normal. What are Nicole's genes for the cystic fibrosis trait?
  The student should answer that Nicole must be heterozygous for cystic fibrosis (CF/cf).  Since neither Scott nor, presumably, Nicole has the disease, but one of their kids does, they must be "carriers".  If Nicole has the disease, then she will have two alleles for the disease (cf/cf).

Again, we are doing the phenotype/genotype tango here, but this time we know what the father and both children's phenotypes are and we have to figure out what the mother's genotype is.  The reasoning, therefore, goes back and between phenotype and genotype and between generations.  Because the child has the disease and her father doesn't, we immediately know that Scott is heterozygous (CF/cf).  For a child to have the disease, she must be homozygous recessive (cf/cf).  Since she gets one of her chromosomes, and therefore one of the alleles, from her father, the other must come from her mother, Nicole.  We are not given any information about Nicole's phenotype or genotype.  If she does not have the disease, then Nicole is heterozygous and is able to contribute a CF as well as a cf to her children.  If she has the disease, then both of her alleles are cf and she is able to contribute only the recessive allele.  Since Scott has both alleles, Stacey, who is normal, could have received a CF from her father and a cf from her mother.  In other words, the only statement we can make about Nicole's genotype is that she has the cf allele.  The other allele is unknown.
 
 
 

4.       If Eric were able to take the medical history of his girl friends, what should he look for in the genetic makeup of a wife to ensure that none of their children would have cystic fibrosis? Can he be sure that none of his grandchildren will have cystic fibrosis?

The student will probably answer that the girl friend should be homozygous dominant, that is CF/CF.  That way she would not be a carrier.

The safest answer without even thinking about it would be that she should be homozygous dominant and it is correct.  This can be answered by an educated guess.  The thought processes needed to reason through to the answer are complicated.  First, you have to work backward to understand what possible alleles Eric has inherited.  You will remember that both of his grandparents were carriers. Eric's sister has the disease.  Therefore, he has a 50% chance of being a carrier himself.  To make sure that none of his children have the disease he must find someone who is homozygous dominant (CF/CF), that is, neither a carrier nor has the trait.  Of course, you can always look at Eric's chromosomes to check out his genetic makeup.  If you do, you will see that he is heterozygous (CF/cF).  Therefore, he must marry someone who is homozygous dominant to ensure that his children will be disease free.  There is no way that he can ensure that his grandchildren won't have the disease since 1/2 of his children will be carriers.
 
 
 
 

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