## Bayesian Analysis Using Pedigree Information

In the pedigree shown in Figure 5-1a, the two brothers of the consultand (indicated by the arrow) have Kennedy disease (X-linked spinal and bulbar muscular atrophy; Online Mendelian Inheritance in Man [OMIM; database online] #300377), which is caused by a CAG trinucleotide expansion in the androgen receptor (AR) gene (OMIM #310200). Because both of the consultand's brothers are affected, we can assume that the consultand's mother is an obligate carrier. Before taking into account her three unaffected sons, the consultand's carrier risk is 1/2, since there is a 1/2 chance that she inherited the mutant X chromosome from her mother. If we take into account that the con sultand has three unaffected sons,how does her carrier risk change?

Bayesian analysis starts with mutually exclusive hypotheses. In this example, there are two: that the consultand is a carrier, and that the consultand is a noncarrier. Setting up a table with separate columns for each hypothesis facilitates Bayesian analyses, as shown in Figure 5-1b for this case. The first row of the table comprises the prior probability for each hypothesis. In this example, the prior probabilities are the probability that the consultand is a carrier (1/2), and the probability that she is a noncarrier (also 1/2), prior to taking into account the subsequent information that she has three unaffected sons.

The second row of the table comprises the conditional probability for each hypothesis. The conditional probability for each hypothesis is the probability that the subsequent information would occur if we assume that each hypothesis is true. In this example, the subsequent information is that the consultand has three unaffected sons. Thus, the conditional probabilities are the probability that the consultand would have three unaffected sons under the assumption (or condition) that she is a carrier, and the probability that she would have three unaffected sons under the assumption (or condition) that she is a noncar-rier. If we assume that she is a carrier, the probability that she would have three unaffected sons is 1/2 x 1/2 x 1/2 = 1/8. This is because she would have to have passed the normal X chromosome three times in succession, each time with a probability of 1/2. If we assume that she is a non-carrier, the probability that she would have three unaffected sons approximates 1, since only in the event of a rare de novo mutation would a noncarrier have an affected son. Thus, the conditional probabilities in this example are 1/8 and 1 (Figure 5-1b).

The third row of the table comprises the joint probability for each hypothesis, which is the product of the prior and conditional probabilities for each hypothesis. For the first hypothesis in this example, that the consultand is a carrier, the joint probability is the prior probability that she is a carrier, multiplied by the conditional probability that a

 Hypothesis Carrier Non-carrier Prior Probability 1/2 1/2 Conditional Probability (of three normal sons) 1/8 ~1 Joint Probability 1/16 1/2 Posterior Probability (1/16) / (1/16 +1/2) = 1/9 (1/2) / (1/16 + 1/2) = 8/9
 Carrier (1/2) à à à mu à ù à mu ùùù à ù ù ùàù ùùà ù ù ù

Figure 5-1. (a) Pedigree of a family with individuals affected with Kennedy disease (see text). Consultand is indicated by an arrow. (b) Bayesian analysis for the consultand in Figure 5-1a. (c) Schematic representation of the Bayesian analysis of Figure 5-1b. Pedigrees shown in the rectangles represent all possible disease status outcomes for the third generation of the pedigree in Figure 5-1a, given the carrier or noncarrier status of the consultand. Each small rectangle to the left represents one sixteenth of the total area. (See text for full description.)

Figure 5-1. (a) Pedigree of a family with individuals affected with Kennedy disease (see text). Consultand is indicated by an arrow. (b) Bayesian analysis for the consultand in Figure 5-1a. (c) Schematic representation of the Bayesian analysis of Figure 5-1b. Pedigrees shown in the rectangles represent all possible disease status outcomes for the third generation of the pedigree in Figure 5-1a, given the carrier or noncarrier status of the consultand. Each small rectangle to the left represents one sixteenth of the total area. (See text for full description.)

carrier would have three normal sons, which in this case is 1/2 x 1/8 = 1/16 (Figure 5-1b). For the second hypothesis in this example, that the consultand is a noncarrier, the joint probability is the prior probability that she is a non-carrier, multiplied by the conditional probability that a noncarrier would have three normal sons, which in this case is 1/2 x 1 = 1/2 (Figure 5-1b).

The fourth row of the table comprises the posterior probability for each hypothesis. The posterior probability for each hypothesis is the probability that each hypothesis is true after (or posterior to) taking into account both prior and subsequent information. The posterior probability for each hypothesis is calculated by dividing the joint proba bility for that hypothesis by the sum of all the joint probabilities. In this example, the posterior probability that the consultand is a carrier is the joint probability for the first hypothesis (1/16), divided by the sum of the joint probabilities for both hypotheses (1/16 + 1/2 = 9/16), or 1/16 9/16 = 1/9. The posterior probability that the consultand is a noncarrier is the joint probability for the second hypothesis (1/2 = 8/16), divided by the sum of the joint probabilities for both hypotheses (1/16 + 1/2 = 9/16), or 8/16 + 9/16 = 8/9. Thus, taking into account the prior family history, and the subsequent information that the consultand has three unaffected sons, the probability that the consultand is a carrier is 1/9 (Figure 5-1b).

The preceding example is illustrated graphically in Figure 5-1c. The total area represents the total prior probabilities. The left half represents the prior probability that the consultand is a carrier (1/2), and the right half represents the prior probability that she is a noncarrier (also 1/2). Under the hypothesis that the consultand is a carrier, there are eight possibilities, comprising all the permutations of zero, one, two, or three affected sons. The area of the small rectangle that contains three unshaded squares (for three unaffected sons) comprises one eighth of the left half and represents the conditional probability of three normal sons under the hypothesis that the consultand is a carrier. The area of this small rectangle is one sixteenth of the total area and therefore also represents the joint probability that the consultand is a carrier (1/2), and that as a carrier she would have three normal sons (1/8), or 1/2 x 1/8 = 1/16.

Under the hypothesis that the consultand is a noncarrier, there is essentially only one possibility, which is that all three sons are unaffected. The area of the larger rectangle that contains the pedigree with three unshaded squares (for three unaffected sons) comprises all of the noncarrier half and represents the conditional probability of three normal sons under the hypothesis that the consultand is a noncarrier. The area of this larger rectangle is one half of the total area and therefore also represents the joint probability that the consultand is a noncarrier (1/2), and that as a noncarrier she would have three normal sons (~1), or 1/2 x 1 = 1/2. The reverse-L-shaped box, which is demarcated by a bold line, represents the sum of the joint probabilities, or nine sixteenths of the total area.

Because the consultand has three unaffected sons, the area of the reverse-L-shaped box represents the only component of the prior probabilities needed to determine the posterior probability that the consultand is a carrier. Taking into account that all three of the consultand's sons are unaffected, Bayesian analysis allows us to exclude 7/16 of the prior probabilities, those that include one or more affected sons, from consideration. (Note that this explains why the joint probabilities sum to less than 1.) The posterior probability that the consultand is a carrier is therefore the area of the small rectangle with three unshaded squares (for three unaffected sons) divided by the area of the entire reverse-L-shaped box, which represents the only probabil a c ities relevant to the consultand's risk, or 1/16 + 9/16 = 1/9. Likewise, the posterior probability that the consultand is a noncarrier is the area of the larger rectangle with three unshaded squares (for three unaffected sons) divided by the area of the entire reversed-L-shaped box, or 8/16 + 9/16 = 8/9.