Bayesian Analysis Using Genetic Test Results

In the second example, information from a test result modifies the prior risk. In the pedigree shown in Figure 5-2a, the consultand is pregnant with her first child and has a family history of CF (OMIM #219700). CF is caused by mutations in the cystic fibrosis transmembrane conductance regulator gene (CFTR; OMIM #602421). The consultand is an unaffected European Caucasian and her brother died years earlier of complications of CF. She undergoes carrier testing for the 23 mutations recommended by the American College of Medical Genetics (ACMG) CF screening guidelines,10-12 which detects approximately 90% of disease alleles in European Caucasians. The consultand tests negative for all 23 mutations. What is her carrier risk after testing?

As in the first example, the two hypotheses are that the consultand is a carrier and that she is a noncarrier. The prior probability that she is a carrier is 2/3. Because the consultand is unaffected, she could not have inherited disease alleles from both parents. Thus, she either inherited a disease allele from her mother or father, or she inherited only normal alleles; in two of these three scenarios she would be a carrier (shown in Figure 5-2b). The prior probability that the consultand is a noncarrier is 1/3 (Figure 5-2c).

As in the first example, 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 tests negative for all 23 mutations. Thus, the conditional probabilities are the probability that the consultand would test negative under the assumption (or condition) that she is a carrier, and the probability that she would test negative under the assumption (or condition) that she is a noncarrier. If we assume that she is a carrier, the probability that she would test negative is 1/10, since the test detects 90% of European Caucasian disease alleles or carriers. If we assume that she is a noncarrier, the probability that she would test negative approximates 1. Thus, the conditional probabilities in this example are 1/10 and 1 for the carrier and noncarrier hypotheses, respectively (Figure 5-2c).

As in the first case, the joint probability for each hypothesis is the product of the prior and conditional probabilities for that 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 (2/3)

Maternal allele

mt

N

Paternal allele

mt

(mt/mt)

mt/N

N

N/mt

N/N

Hypothesis

Carrier

Non-carrier

Prior Probability

2/3

1/3

Conditional Probability (of a negative test result)

1/10

1

Joint Probability

1/15

1/3

Posterior Probability

(1/15) / (1/15 + 1/3) = 1/6

(1/3) / (1/15 + 1/3) = 5/6

1/10

9/10

1/10

Positive test

2/3 x 9/10=3/5

Negative test

1/3

Negative test

2/3 x 1/10=1/15

Figure 5-2. (a) Pedigree of a family with an individual affected with CF (see text). Consultand is indicated by an arrow.(b) Possible genotypes of the sibling (consultand in this case) of the affected child prior to genetic testing. The mt/mt genotype (in parentheses) is excluded based on the fact that the consultand is unaffected. mt, mutant; N, normal. (c) Bayesian analysis for the consultand in Figure 5-2a. (d) Schematic representation of the Bayesian analysis of Figure 5-2c (see text).

multiplied by the conditional probability that a carrier of European Caucasian ancestry would test negative (1/10), or 2/3 x 1/10 = 1/15 (Figure 5-2c). For the second hypothesis in this example, that the consultand is a noncarrier, the a b d joint probability is the prior probability that she is a non-carrier (1/3) multiplied by the conditional probability that a noncarrier would test negative (1), or 1/3 x 1 = 1/3 (Figure 5-2c).

Finally, the posterior probability is calculated for each hypothesis by dividing the joint probability for that hypothesis by the sum of all the joint probabilities. In this example, the posterior probability that the consultand is a carrier and tests negative for 23 CF mutations is the joint probability for the first hypothesis (1/15) divided by the sum of the joint probabilities for both hypotheses (1/15 + 1/3 = 2/5), or 1/15 + 2/5 = 1/6 (Figure 5-2c). The posterior probability that the consultand is a noncarrier and tests negative for 23 CF mutations is the joint probability for the second hypothesis (1/3) divided by the sum of the joint probabilities for both hypotheses (2/5), or 1/3 + 2/5 = 5/6 (Figure 5-2c).

The preceding example is illustrated graphically in Figure 5-2d. The total area represents the total prior probabilities. The left two thirds represents the prior probability that the consultand is a carrier, and the right third represents the prior probability that the consultand is a noncarrier. Under the hypothesis that the consultand is a carrier, there are two possibilities for the test result: positive or negative. The area of the small rectangle on the lower left comprises one tenth of the 2/3 carrier region of the figure and represents the conditional probability of a normal test result under the hypothesis that the consultand is a carrier. The area of this small rectangle is 1/10 x 2/3 = 1/15 of the total probabilities area and therefore also represents the joint probability that the consultand is a carrier (2/3) and that as a European Caucasian carrier she would test negative for all 23 mutations (1/10), or 2/3 x 1/10 = 1/15 (Figure 5-2d).

Under the hypothesis that the consultand is a noncarrier, there is essentially only one possibility for the test result, which is negative. The area of the rectangle that comprises all of the 1/3 noncarrier region represents the conditional probability of a negative test result under the hypothesis that the consultand is a noncarrier. The area of this rectangle is one third of the total area and therefore also represents the joint probability that the consultand is a noncarrier (1/3), and that as a noncarrier she would test negative (~1), or 1/3 x 1 = 1/3. The reverse-L-shaped box, which is demarcated by a bold line, represents the sum of the joint probabilities, or 2/5 (= 1/3 + 1/15) of the total area.

Because the consultand tested negative, 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 she tested negative, Bayesian analysis allows us to exclude 3/5 of the prior probability, that portion comprising a positive test result, from consideration. (Note, again, 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 at the lower left divided by the area of the reverse-L-shaped box, which represents the only probabilities relevant to the con-sultand's risk, or 1/15 + 2/5 = 1/6. Likewise, the posterior probability that the consultand is a noncarrier is the area of the larger rectangle on the right divided by the area of the reverse-L-shaped box, or 1/3 + 2/5 = 5/6.

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