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, 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 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 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 1/2 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 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 comes 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 course, 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 they'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 distinguish 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 0.984, 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 likelihood 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.
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