Video Discription |
In this video, Dr. Korf talks about calculating genetic risks based on Mendelian Inheritance and Bayes’ Theorem.
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Genetic Risk Assessment, by Dr. Bruce Korf. In collaboration with the University of Alabama at Birmingham.
My name is Bruce Korf. I'm a medical geneticist at University of Alabama at Birmingham. In this lecture, we'll consider principles of genetic risk assessment. We will calculate genetic risks based on Mendelian inheritance, and utilize Bayes' theorem in genetic risk calculations.
We'll begin with autosomal recessive inheritance. Remember that both parents are carriers, and they have a one in four chance of having an affected homozygous child. So if a couple are both carriers, which you know because they've had an affected child, their risk of having another affected child is simply one in four. Consider a couple where one partner has a sibling with an autosomal recessive trait, and her partner has no known history of the condition.
We can estimate that her risk of being a carrier is 2/3, and we know this because there were four possibilities for her-- that she would be homozygous affected, that she would be a carrier having inherited the mutation from her father, a carrier having inherited the mutation from her mother, or homozygous unaffected. We know that she is not affected, and so there are only three possibilities left, of which two would make her a carrier, hence the 2/3.
To calculate the risk for her partner, we need to use data on the population frequency of carrier risk. If this were a pedigree with cystic fibrosis, and he were of northern European descent, the risk of his being a carrier would be about 1 in 25. So this couple's risk, then, of having an affected child would be 2/3, her risk of being a carrier times 1 in 25, his times 1 in 4. That is that they're both carriers, and then have an affected child. And so the total risk is one in 150.
Now let's consider autosomal dominant inheritance. Remember here that individuals who are heterozygous would be affected, and they have a 50% chance of transmitting the trait to any offspring. And so this individual, who is known to be affected, would face a 50% chance of having another affected child. For an x-linked recessive, remember that carrier females transmit the trait to half their males and carrier status to half their daughters.
So if this individual is known to be a carrier, her risk of having another affected child would be 50% if the child is male and essentially zero if the child is female, although she could end up being a carrier. Well, now let's consider the instance of incomplete penetrance. For example, here where this individual has transmitted the trait telling you she must have inherited it, and yet is phenotypically unaffected.
Well, let's assume, first of all, that there is 80% penetrance and 80% probability that a person who inherits a mutation will show the phenotype. And consider this couple and this couple. Now in this case, she is known to be affected, and hence, their risk of having an affected child is one half that they transmit the mutation times 0.8, which is the probability that a child who inherits the mutation will show the phenotype, which comes to 0.4.
Now in this case, this individual is not phenotypically affected. That means either he didn't inherit the mutation, or he did, and he's non-penetrant. The chance that he inherited the mutation is one half. The chance that he's non-penetrant is one minus 0.8, which is 0.2. Then the chance, then, that he transmits the mutation is another one half, and the chance that the child would be phenotypically affected is 0.8, and this comes to 0.04. ZqUxZomTRUU |