Laws of Addition and Multiplication

When considering the probability of two different events or outcomes, it is essential to clarify whether they are mutually exclusive or independent. If the events are mutually exclusive, then the probability that either one or the other will occur equals the sum of their individual probabilities. This is known as the law of addition.

If, however, two or more events or outcomes are independent, then the probability that both the first and the second will occur equals the product of their individual probabilities. This is known as the law of multiplication.

As a simple illustration of these laws, consider parents who have embarked upon their first pregnancy. The probability that the baby will be either a boy or a girl equals 1—i.e., 1/2 + 1/2. If the mother is found on ultrasonography to be carrying twins who are non-identical, then the probability that both the first and the second twin will be boys equals 1/4—i.e., 1/2 × 1/2.

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.

FIGURE 22.1 Pedigree showing sex-linked recessive inheritance. When calculating the probability that II₂ is a carrier, it is necessary to take into account her three unaffected sons.

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

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.

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).

FIGURE 22.2 I₂ has an autosomal dominant disorder that shows reduced penetrance. The probability that III₁ will be affected has to take into account the possibility that his mother (II₁) is a non-penetrant heterozygote.

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.

FIGURE 22.3 Expected genotypes and phenotypes in 10 children born to an individual with an autosomal dominant disorder with penetrance equal to 0.8.

Table 22.2 Bayesian Calculation for II₁ in Figure 22.2

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 being affected equals 0.086, approximately 1/12, which is obtained when P equals 0.6. This maximal risk figure can be used when counseling people at risk for late-onset autosomal disorders with reduced penetrance and who have an affected grandparent and unaffected parents.

Delayed Age of Onset

Many autosomal dominant disorders do not present until well into adult life. Healthy members of families in which these disorders are segregating often wish to know whether they themselves will develop the condition or pass it on to their children. Risks for these individuals can be calculated in the following way.

Consider someone who has died with a confirmed diagnosis of Huntington disease (Figure 22.4). This is a late-onset autosomal dominant disorder. The son of I₂ is entirely healthy at age 50 years and wishes to know the probability that his 10-year-old daughter, III₁, will develop Huntington disease in later life. In this condition, the first signs usually appear between the ages of 30 and 60 years, and approximately 50% of all heterozygotes have shown signs by the age of 50 years (Figure 22.5).

FIGURE 22.4 I₂ had an autosomal dominant disorder showing delayed age of onset. When calculating the probability that III₁ will develop the disorder, it is necessary to determine the probability that II₁ is a heterozygote who is not yet clinically affected.

FIGURE 22.5 Graph showing age of onset in years of clinical expression in Huntington disease heterozygotes. Approximately 50% show clinical signs or symptoms by age 50 years.

(Data from Newcombe RG 1981 A life table for onset of Huntington’s chorea. Ann Hum Genet 45:375–385.)

To answer the question about the risk to III₁, it is first necessary to calculate the risk for II₁ (if III₁ was asking about her own risk, her father might be referred to as the dummy consultand). The probability that II₁ has inherited the gene, given that he shows no signs of the condition, can be determined by a simple bayesian calculation (Table 22.3).

Table 22.3 Bayesian Calculation for II₁ in Figure 22.4

The posterior probability that II₁ is heterozygous equals 1/4/(1/4 + 1/2), which equals 1/3. Therefore the prior probability that his daughter III₁ will have inherited the disorder equals 1/3 × 1/2, or 1/6.

There is a temptation when doing calculations such as these to conclude that the overall risk for II₁ being a heterozygote simply equals 1/2 × 1/2—i.e., the prior probability that he will have inherited the mutant gene times the probability that a heterozygote will be unaffected at age 50 years, giving a risk of 1/4. This is correct in as much as it gives the joint probability for this possible outcome, but it does not take into account the possibility that II₁ is not a heterozygote. Consider the possibility that I₂ has four children. On average, two will inherit the mutant allele, one of whom will be affected by the age of 50 years. The remaining two children will not inherit the mutant allele. By the time these children have grown up and reached the age of 50 years, on average one will be affected and three will not. Therefore, on average, one-third of the healthy 50-year-old offspring of I₂ will be heterozygotes. Hence the correct risk for II₁ is 1/3 and not 1/4.

Carrier Risks for the Extended Family

When both parents are heterozygotes, the risk that each of their children will be affected is 1 in 4. On average three of their four children will be unaffected, of whom, on average, two will be carriers (Figure 22.6). Therefore the probability that the healthy sibling of someone with an autosomal recessive disorder will be a carrier equals 2/3. Carrier risks can be derived for other family members, starting with the assumption that both parents of an affected child are carriers (Figure 22.7).

FIGURE 22.6 Possible genotypes and phenotypes in the offspring of parents who are both carriers of an autosomal recessive disorder. On average, two of three healthy offspring are carriers.

FIGURE 22.7 Autosomal recessive inheritance. The probabilities that various family members are carriers are indicated as fractions.

When calculating risks in autosomal recessive inheritance the underlying principle is to establish the probability that each prospective parent is a carrier, and then multiply the product of these probabilities by 1/4, this being the risk that any child born to two carriers will be affected. Therefore, in Figure 22.7, if the sister, III₃, of the affected boy was to marry her first cousin, III₄, the probability that their first baby would be affected would equal 2/3 × 1/4 × 1/4—i.e., the probability that III₃ is a carrier times the probability that III₄ is a carrier times the probability that a child of two carriers will be affected. This gives a total risk of 1/24.

If this same sister, III₃, was to marry a healthy unrelated individual, the probability that their first child would be affected would equal 2/3 × 2pq × 1/4—i.e., the probability that III₃ is a carrier times the carrier frequency in the general population (p. 132) times the probability that a child of two carriers will be affected. For a condition such as cystic fibrosis, with a disease incidence of approximately 1 in 5000, q² = 1/2500 and therefore q = 1/50 and thus 2pq = 1/25. Therefore the final risk would be 2/3 × 1/25 × 1/4, or 1 in 150.

Modifying a Carrier Risk by Mutation Analysis

Population screening for cystic fibrosis has been introduced in the United Kingdom after pilot studies (p. 321). More than 1500 different mutations have been identified in the cystic fibrosis gene, so that carrier detection by DNA mutation analysis is not straightforward. However, a relatively simple test has been developed for the most common mutations, which enables about 90% of all carriers of western European origin to be detected. What is the probability that a healthy individual who has no family history of cystic fibrosis, and who tests negative on the common mutation screen, is a carrier?

The answer is obtained, once again, by drawing up a simple bayesian table (Table 22.4). The prior probability that this healthy member of the general population is a carrier equals 1/25; therefore the prior probability that he or she is not a carrier equals 24/25. If this individual is a carrier, then the probability that the common mutation test will be normal is 0.10 as only 10% of carriers do not have a common mutation. The probability that someone who is not a carrier will have a normal common mutation test result is 1.

Table 22.4 Bayesian Table for Cystic Fibrosis Carrier Risk if Common Mutation Screen Is Negative

This gives a joint probability for being a carrier of 1/250 and for not being a carrier of 24/25. Therefore the posterior probability that this individual is a carrier equals 1/250/(1/250 + 24/25), which equals 1/241. Thus, the normal result on common mutation testing has reduced the carrier risk from 1/25 to 1/241.

The Isolated Case

If a woman has only one affected son, then in the absence of a positive family history there are three possible ways in which this can have occurred.

In practice it is often very difficult to distinguish among these three possibilities unless reliable tests are available for carrier detection. If a woman is found to be a carrier, then risk calculation is straightforward. If the tests indicate that she is not a carrier, the recurrence risk is probably low, but not negligible because of the possibility of gonadal mosaicism.

For example, in Duchenne muscular dystrophy (DMD), it has been estimated that among the mothers of isolated cases approximately two-thirds are carriers, 5% to 10% are gonadal mosaics, and in the remaining 25% to 30% the disorder has arisen as a new mutation in meiosis.

Leaving aside the complicating factor of gonadal mosaicism, risk calculation in the context of an isolated case (Figure 22.9) is possible, but may require calculation of the risk for a dummy consultand within the pedigree as well as taking account of the mutation rate, or µ. For a fuller understanding of µ, the student is referred to one of the more detailed texts listed at the end of the chapter.

FIGURE 22.9 In this pedigree III₁ is affected by Duchenne muscular dystrophy and is an isolated case (i.e., there is no history of the condition in the wider family). The consultand, II₅ (arrow), wishes to know whether she is at risk of having affected sons. To calculate her risk, the risk that her mother, I₂, is a carrier is first calculated; this requires consideration of the mutation rate, µ. I₂ is the dummy consultand in this scenario.

Incorporating Carrier Test Results

Several biochemical tests are available for detecting carriers of sex-linked recessive disorders. Unfortunately, there is often overlap in the values obtained for controls and women known to be carriers (i.e., obligate carriers). Although an abnormal result in a potential carrier would suggest that she is likely to be a carrier, a normal test result does not exclude a woman from being a carrier. Although for many sex-linked recessive disorders this problem can be overcome by using linked DNA markers, the difficulties presented by overlapping biochemical test results arise sufficiently often to justify further consideration.

For example, in DMD, the serum creatine kinase level is raised in approximately two out of three obligate carriers (see Figure 20.1; p. 314). Therefore, if a possible carrier such as II₂ in Figure 22.1 is found to have a normal level of creatine kinase, this would provide further support for her not being a carrier. The test result therefore provides a conditional probability, which is included in a new bayesian calculation (Table 22.5).

Table 22.5 Bayesian Calculation for II₂ in Figure 22.1

The posterior probability that II₂ is a carrier equals 1/48/(1/48 + 1/2), or 1/25. Consequently, by first taking into account this woman’s three healthy sons, and second her normal creatine kinase test result, it has been possible to reduce her carrier risk from 1 in 2 to 1 in 9 and then to 1 in 25.

Multifactorial Disorders

One of the basic principles of multifactorial inheritance is that the risk of recurrence in first-degree relatives, siblings and offspring, equals the square root of the incidence of the disease in the general population (p. 146)—i.e., , where P equals the general population incidence. For example, if the general population incidence equals 1/1000, then the theoretical risk to a first-degree relative equals the square root of 1/1000, which approximates to 1 in 32 or 3%. The theoretical risks for second- and third-degree relatives can be shown to approximate to and , respectively. Therefore, if there is strong support for multifactorial inheritance, it is reasonable to use these theoretical risks when counseling close family relatives.

However, when using this approach it is important to remember that the confirmation of multifactorial inheritance will often have been based on the study of observed recurrence risks. Consequently, it is generally more appropriate to refer back to the original family studies and counsel on the basis of the risks derived in these (Table 22.7).

Ideally, reference should be made to local studies as recurrence risks can differ quite substantially in different communities, ethnic groups, and geographical locations. For example, the recurrence risk for neural tube defects in siblings used to be quoted as 4% (before the promotion of periconceptional maternal folate intake). This, essentially, was an average risk. The actual risk varied from 2% to 3% in southeast England up to 8% in Northern Ireland, and also showed an inverse relationship with the family’s socioeconomic status, being greatest for mothers in poorest circumstances.

Unfortunately, empiric risks are rarely available for families in which there are several affected family members, or for disorders with variable severity or different sex incidences. For example, in a family where several members have been affected by cleft lip/palate, the empiric risks based on population data may not apply—the condition may appear to be segregating as an autosomal dominant trait with a high penetrance. In the absence of a syndrome diagnosis being made and genetic testing being possible, the clinical geneticist has to make the best judgement about recurrence risk.

Conditions Showing Causal Heterogeneity

Many referrals to genetic clinics relate to a clinical phenotype rather than to a precise underlying diagnosis (Table 22.8). In these situations, great care must be taken to ensure that all appropriate diagnostic investigations have been undertaken before resorting to the use of empiric risk data (p. 346).

It is worth emphasizing that the use of empiric risks for conditions such as sensorineural hearing loss in childhood is at best a compromise, as the figure quoted to an individual family will rarely be the correct one for their particular diagnosis. Severe sensorineural hearing loss in a young child is usually caused either by single-gene inheritance, most commonly autosomal recessive, but occasionally autosomal dominant or sex-linked recessive, or by an environmental condition such as rubella embryopathy. Therefore, for most families the correct risk of recurrence will be either 25% or 0%. In practice, it is often not possible to establish the precise cause, so that the only option available is to offer the family an empiric or ‘average’ risk.