"Genetic Risk Assessment" by Bruce Korf for OPENPediatrics

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 1 in 4

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 1 in 4. 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, t

hat 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 1 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 0 if the child is female, although she could end up being a carrier. Well, now let's consider the instance of incomplete penetrance. For ex

ample, 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 1/2 that they transmit the mutation times 0.8, which is the probability th

at 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 1/2. The chance that he's non-penetrant is 1 minus 0.8, which is 0.2. Then the chance, then, that he transmits the mutation is another 1/2, and the chance that the child would be phenotypically affected is 0.8, and this co

mes to 0.04. Well, now let's consider the instance of age-dependent penetrance, remember, where the probability of being phenotypically affected increases with age. To do this, we need to use Bayes' Theorem. Let's consider the scenario where this individual who has an affected father, and for that matter an affected brother, wants to know if she has, in fact, inherited the mutation. Let's assume that at her age, 30% of individuals who have inherited the gene would show the phenotype, and hence,

the probability of getting to her age and not showing the phenotype is 0.7. Well, we have to consider two competing hypotheses, one that she is a carrier, or two, that she's not. Based on Mendel's laws, the prior probability of each of these as 0.5 that she inherits the mutation from her father. Now let's consider, what's the probability that she would reach adulthood so far, and not be phenotypically affected? Well, if I said that 30% of people her age are affected, the probability then of her

not being affected is 1 minus that, which is 0.7. If she's not a carrier, the probability of being unaffected, of course, would be 1. So what's the probability that she is a carrier and is unaffected? It's 0.5 times 0.7, which is 0.35, or that she's not a carrier and is unaffected, 0.5 times 1, which is 0.5. Now we look at what is the relative likelihood of these two scenarios. We call this the posterior probability that she is a carrier is 0.35 over 0.35 plus 0.5, which comes to 0.41. And of co

urse, the probability of her not being a carrier, 0.5 over 0.35 plus 0.5, which is 0.59. We can also use Bayes' Theorem in this autosomal recessive example that I showed you earlier, except now, this couple have had three unaffected children. Now you might imagine that this would be evidence that maybe one or both of them is not a carrier, but it's possible, of course, to have three unaffected children and still have both parents be carriers and still at risk. Well, they're either at risk or the

y're not at risk. We're going to try to compute the relative likelihood of each of these two scenarios. The prior probability that they're at risk is 2/3 times 1 in 25, which is 0.026, and then 1 minus that is the probability that they're not at risk, which is 0.974. The conditional probability is that they would have three unaffected children given that they are at risk or that they're not. Now if they are at risk, the probability of an unaffected child is 3/4, 1 minus 1/4. Now this does not di

stinguish whether the child is a carrier or homozygous unaffected. All we know is they're unaffected. So 3/4 times 3/4 times 3/4, which is 27/64 versus 1, since if they're not at risk, none of their children are expected to be affected. So the probability that they're at risk and have three unaffected children is 0.026 times 0.42, which is about 0.01. And the probability that they're not at risk and have three unaffected children is 0.974 times 1. The posterior probability then is 0.01 over 0.98

4, and that they're not at risk, 0.974 over 0.984. Well, finally, let's consider the use of Bayes' Theorem in an x-linked recessive pedigree. So here, this individual is an obligate carrier, because she's had two affected sons. Question is, is her daughter a carrier? And you see she's had two healthy sons, which might lead you to believe that maybe she's not. But of course, it's quite possible that she still is. Her prior probability, this individual's prior probability, of being a carrier is 0.

5, based on Mendelian genetics, or not being a carrier is 0.5. The conditional probability here is having two healthy sons given that she is a carrier or not. Here it would be 1/2 times 1/2, which is 0.25, versus 1 if she's not a carrier. You expect both sons to be unaffected. The joint probability, then, of being a carrier and having two healthy sons is 0.5 times 0.25, which is 0.125, versus 0.5 times 1, which is 0.5 that she's not a carrier and has two healthy sons. And so the relative likelih

ood or posterior probability for her to be a carrier is 0.125 over 0.125 plus 0.5, which is 0.625, and that comes to 0.2. Posterior probability that she's not a carrier is 0.5 over 0.625, which is 0.8. In conclusion, simple Mendelian risks can be calculated based on recessive, dominant, or x-linked inheritance, and Bayes' Theorem can be used to take account of incomplete penetrance or to refine risks based on additional family data. Please help us improve the content by providing us with some fe

edback.