Risk Calculation

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CHAPTER 22 Risk Calculation

One of the most important aspects of genetic counseling is the provision of a risk figure. This is often referred to as a recurrence risk. Estimation of the recurrence risk usually requires careful consideration and takes into account:

Sometimes the provision of a risk figure can be quite easy, but in a surprisingly large number of complicating factors arise that make the calculation very difficult. For example, the mother of a boy who is an isolated case of a sex-linked recessive disorder could very reasonably wish to know the recurrence risk for her next child. This is a very simple question, but the solution may be far from straightforward, as will become clear later in this chapter.

Before proceeding any further, it is necessary to clarify what we mean by probability and review the different ways in which it can be expressed. The probability of an outcome can be defined as the number or, more correctly, the proportion of times it occurs in a large series of events. Conventionally, probability is indicated as a proportion of 1, so that a probability of 0 implies that an outcome will never be observed, whereas a probability of 1 implies that it will always be observed. Therefore, a probability of 0.25 indicates that, on average, a particular outcome or event will be observed on 1 in 4 occasions, or 25%. The probability that the outcome will not occur is 0.75, which can also be expressed as 3 chances out of 4, or 75%. Alternatively, this probability could be expressed as odds of 3 to 1 against, or 1 to 3 in favor of the particular outcome being observed. In this chapter, fractions are used where possible as these tend to be more easily understood than proportions of 1 expressed as decimals.

Probability Theory

To calculate genetic risks it is necessary to have a basic understanding of probability theory. This will be discussed in so far as it is relevant to the skills required for genetic counseling.

Bayes’ Theorem

Bayes’ theorem, which was first devised by the Reverend Thomas Bayes (1702–1761) and published after his death in 1763, is widely used in genetic counseling. Essentially it provides a very valuable method for determining the overall probability of an event or outcome, such as carrier status, by considering all initial possibilities (e.g., carrier or non-carrier) and then modifying or ‘conditioning’ these by incorporating information, such as test results or pedigree information, that indicates which is the more likely. Thus, the theorem combines the probability that an event will occur with the probability that it will not occur. The theorem lay fairly dormant for a long time, but has been enthusiastically employed by geneticists. In recent years its beauty, simplicity and usefulness have been recognized in many other fields—for example, legal work, computing, and statistical analysis—such that it has truly come of age.

The initial probability of each event is known as its prior probability, and is based on ancestral or anterior information. The observations that modify these prior probabilities allow conditional probabilities to be determined. In genetic counseling these are usually based on numbers of offspring and/or the results of tests. This is posterior information. The resulting probability for each event or outcome is known as its joint probability. The final probability for each event is known as its posterior or relative probability and is obtained by dividing the joint probability for that event by the sum of all the joint probabilities.

This is not an easy concept to grasp! To try to make it a little more comprehensible, consider a pedigree with two males, I3 and II1, who have a sex-linked recessive disorder (Figure 22.1). The sister, II2, of one of these men wishes to know the probability that she is a carrier. Her mother, I2, must be a carrier because she has both an affected brother and an affected son (i.e., she is an obligate carrier). Therefore, the prior probability that II2 is a carrier equals 1/2. Similarly, the prior probability that II2 is not a carrier equals 1/2.

The fact that II2 already has three healthy sons must be taken into consideration, as intuitively this makes it rather unlikely that she is a carrier. Bayes’ theorem provides a way to quantify this intuition. These three healthy sons provide posterior information. The conditional probability that II2 will have three healthy sons if she is a carrier is 1/2 × 1/2 × 1/2, which equals 1/8. These values are multiplied as they are independent events, in that the health of one son is not influenced by the health of his brother(s). The conditional probability that II2 will have three healthy sons if she is not a carrier equals 1.

This information is now incorporated into a bayesian calculation (Table 22.1). From this table, the posterior probability that II2 is a carrier equals 1/16/(1/16 + 1/2), which reduces to 1/9. Similarly the posterior probability that II2 is not a carrier equals 1/2/(1/16 + 1/2), which reduces to 8/9. Another way to obtain these results is to consider that the odds for II2 being a carrier versus not being a carrier are 1/16 to 1/2 (i.e., 1 to 8, which equals 1 in 9). Thus, by taking into account the fact that II2 has three healthy sons, we have been able to reduce her risk of being a carrier from 1 in 2 to 1 in 9.

Table 22.1 Bayesian Calculation for II2 in Figure 22.1

Probability II2 is a Carrier II2 is not a Carrier
Prior 1/2 1/2
Conditional
Three healthy sons (1/2)3 = 1/8 (1)3 = 1
Joint 1/6 1/2 (= 8/16)
Expressed as odds 1 to 8
Posterior 1/9 8/9

Perhaps by now the use of Bayes’ theorem will be a little clearer. Try to remember that the basic approach is to draw up a table showing all of the possibilities (e.g., carrier, not a carrier), then establish the background (prior) risk for each possibility, next determine the chance (conditional possibility) that certain observed events (e.g., healthy children) would have happened if each possibility were true, then work out the combined (joint) likelihood for each possibility, and finally weigh up each of the joint probabilities to calculate the exact (posterior) probability for each of the original possibilities. If this is still confusing, some of the following examples may bring more clarity.

Autosomal Dominant Inheritance

For someone with an autosomal dominant disorder, the risk that each of his or her children will inherit the mutant gene equals 1 in 2. This will apply whether the affected individual inherited the disorder from a parent or developed the condition as the result of a new mutation. Therefore the provision of risks for disorders showing autosomal dominant inheritance is usually straightforward as long as there is a clear family history, the condition is characterized by being fully penetrant, and there is a reliable means of diagnosing heterozygotes. However, if penetrance is incomplete or there is a delay in the age of onset so that heterozygotes cannot always be diagnosed, the risk calculation becomes more complicated. Two examples will be discussed to illustrate the sorts of problem that can arise.

Reduced Penetrance

A disorder is said to show reduced penetrance when it has clearly been demonstrated that individuals who must possess the abnormal gene, who by pedigree analysis must be obligate heterozygotes, show absolutely no manifestations of the condition. For example, if someone who was completely unaffected had both a parent and a child with the same autosomal dominant disorder, this would be an example of non-penetrance. Penetrance is usually quoted as a percentage (e.g., 80%) or as a proportion of 1 (e.g., 0.8). This would imply that 80% of all heterozygotes express the condition in some way.

For a condition showing reduced penetrance, the risk that the child of an affected individual will be affected equals 1/2—i.e., the probability that the child will inherit the mutant allele, × P, the proportion of heterozygotes who are affected. Therefore, for a disorder such as hereditary retinoblastoma, an embryonic eye tumor (p. 215), which shows dominant inheritance in some families with a penetrance of P = 0.8, the risk that the child of an affected parent will develop a tumor equals 1/2 × 0.8, which equals 0.4.

A more difficult calculation arises when a risk is sought for the future child of someone who is healthy but whose parent has, or had, an autosomal dominant disorder showing reduced penetrance (Figure 22.2).

Let us assume that the penetrance, P, equals 0.8. Calculation of the risk that III1 will be affected can be approached in two ways. The first simply involves a little logic. The second uses Bayes’ theorem.

1 Imagine that I2 has 10 children. On average, five children will inherit the gene, but because P = 0.8, only four will be affected (Figure 22.3). Therefore, six of the 10 children will be unaffected, one of whom has the mutant allele, with the remaining five having the normal allele. II1 is unaffected, so there is a probability of 1 in 6 that she is, in fact, a heterozygote. Consequently, the probability that III1 will both inherit the mutant gene and be affected equals 1/6 × 1/2 × P, which equals 1/15 if P is 0.8.
2 Now consider II1 in Figure 22.2. The prior probability that she is a heterozygote equals 1/2. Similarly, the prior probability that she is not a heterozygote equals 1/2. Now a bayesian table can be constructed to determine how these prior probabilities are modified by the fact that II1 is not affected (Table 22.2).

Table 22.2 Bayesian Calculation for II1 in Figure 22.2

Probability II1 Is Heterozygous II1 Is Not Heterozygous
Prior 1/2 1/2
Conditional
Not affected 1 − P 1
Joint 1/2 (1 − P) 1/2

The posterior probability that II1 is a heterozygote equals 1/2(1 − P)/[1/2(1 − P) + 1/2], which reduces to {1 − P/2 − P}. Therefore, the risk that III1 will both inherit the mutant allele and be affected equals ({1 − P/2 − P}) × 1/2 × P, which reduces to {(P − P2)/(4 − 2P)}. If P equals 0.8, this expression equals 1/15 or 0.067.

By substituting different values of P in the above expression, it can be shown that the maximum risk for III1