## Bayesian Analyses with More Than Two Hypotheses

In some Bayesian analyses, more than two hypotheses must be considered. For example, consider the pedigree in Figure 5-5a, in which a child with clinically typical type I SMA (type I SMA; Werdnig-Hoffman disease; OMIM

#253300) lacks both copies of the SMN1 gene. By dosage analysis, the child's (unaffected) mother has one copy of the SMN1 gene and therefore carries one copy of the SMN1 gene on one chromosome 5, and zero copies of the SMN1 gene on the other chromosome 5, called the "1 + 0" genotype. However, the child's (unaffected) father has two copies of the SMN1 gene and therefore could have one of three possible genotypes: (1) two copies of the SMN1 gene on one chromosome 5 and zero copies of the SMN1 gene on the other chromosome 5 (the "2 + 0" genotype), (2) one copy of the SMN1 gene on one chromosome 5 and a subtle mutation in the SMN1 gene on the other chromosome 5 (the "1 + 1D" genotype, where "1D" stands for a "1-copy-disease" allele), or (3) one copy of the SMN1 gene on each chromosome 5 (the "1 + 1" noncarrier genotype), in which case he passed a de novo deletion of the SMN1 gene to his affected child. Because the relative frequencies of the various SMN1 alleles and genotypes in the general population are known,14 as well as the paternal and maternal de novo deletion rates (pp = 2.11 x 10-4 and pm = 4.15 x 10-5, respectively), the probability that the father is a carrier can be calculated, which obviously has important implications for recurrence risk.

The Bayesian analysis for the father's carrier risk is shown in Figure 5-5b. There are three hypotheses for the father's genotype: 2 + 0,1 + 1, and 1 + 1D. The prior probabilities are the relative population frequencies for these genotypes.14 The conditional probabilities are the probabilities that the father passes a 0-copy allele to his child under each hypothesis. For the 2 + 0 genotype, the conditional probability of passing a 0-copy allele is 0.5, whereas for the 1 + 1 and 1 + 1D genotypes, the conditional probability of passing a 0-copy allele is the de novo deletion rate of pp. As in the generalized Bayesian analysis shown in Table 5-1, the joint probability for each hypothesis is the

2 copies

1 copy

Hypothesis (Father's Genotype) | |||

2 + 0 |
1 + 1 |
1 + 1D | |

Prior Probability (relative probability) |
1.00 x 10-3 |
9.00 x 10-1 |
4.58 x 10-4 |

Conditional Probability (of passing a 0-copy allele) |
0.5 |
IV> |
Vp |

Joint Probability |
5.0 x 10-4 |
1.9 x 10-4 |
9.7 x 10-8 |

Posterior Probability |
0.72 |
0.28 |
0.00014 |

b |

Figure 5-5. (a) Pedigree of a family with an individual affected with type I SMA,with the SMA carrier test results indicated below each individual (see text).(b) Bayesian analysis for carrier risk of the father of the affected child in Figure 5-5a. pp, paternal de novo mutation rate.

Figure 5-5. (a) Pedigree of a family with an individual affected with type I SMA,with the SMA carrier test results indicated below each individual (see text).(b) Bayesian analysis for carrier risk of the father of the affected child in Figure 5-5a. pp, paternal de novo mutation rate.

b product of the prior and conditional probabilities for that hypothesis, and the posterior probability for each hypothesis is the joint probability for that hypothesis divided by the sum of all the joint probabilities. The father's carrier risk is the sum of the posterior probabilities of the first (2 + 0) and third (1 + 1D) columns, or approximately 0.72. The third column contributes little to the carrier risk because the frequency of the 1 + 1D genotype is low and the conditional probability of a de novo deletion is also low. In contrast, although the frequency of the 2 + 0 genotype is much lower than that of the 1 + 1 genotype, this is counterbalanced by the higher conditional probability of passing a 0-copy allele under the former hypothesis.

Suppose that the father's parents, the paternal grandfather and grandmother of the affected child, are tested and found to have three copies and one copy of the SMN1 gene, respectively (Figure 5-6a). What is the father's carrier risk? The Bayesian analysis for this scenario is shown in Figure 5-6b. Again, there are three hypotheses for the father's genotype: 2 + 0,1 + 1, and 1 + 1D. However, in this scenario, the father's prior probabilities derive from the prior and conditional probabilities of his parents. Because the grandfather has three copies of the SMN1 gene, his genotype is either 2 + 1 (columns A and C) or 2 + 1D (columns B and D), and his prior probabilities are the relative population frequencies for these genotypes.14 Because the (unaffected) grandmother has one copy of the SMN1 gene, her genotype is 1 + 0, and her prior probability is the relative population frequency of the 1 + 0 genotype for type I SMA in the general population, which is the carrier frequency of 1/38 (2.50 x 10-2).14 (Note that because the grandmother must be 1 + 0, simply a prior probability of 1 could be used; as noted above, the absolute values of the conditional probabilities are unimportant, as long as the ratio between them is correct.) The four columns (A through D) show the four possible permutations of grandparental genotypes (prior probabilities) with passage of particular alleles to the father (conditional probabilities) so that he would have a 2-copy SMA carrier test result. Under the hypothesis that the father has a 2 + 0 genotype, he could have inherited a 2-copy "allele" (two copies of SMN1 on one chromosome 5) from the grandfather (2 + 1) at a probability of 0.5 and a 0-copy allele from the grandmother (1 + 0) at a probability of 0.5 (column A), or he could have inherited a 2-copy allele from the grandfather (2 + 1D) at a probability of 0.5 and a 0-copy allele from the grandmother (1 + 0) at a probability of 0.5 (column B). Under the hypothesis that the father has a 1 + 1 genotype, he could have inherited a 1-copy allele from the grandfather (2 + 1) at a probability of 0.5 and a 1-copy allele from the grandmother (1 + 0) at a probability of 0.5 (column C). Under the hypothesis that the father has a 1 + 1D genotype, he could have inherited a 1D allele from the grandfather (2 + 1D) at a probability of 0.5 and a 1-copy allele from the grandmother (1 + 0) at a probability of 0.5 (column D).

The father's prior probabilities are the products of the prior and conditional probabilities for the grandparents for each column or permutation. Under the hypothesis that the father has a 2 + 0 genotype, the conditional probability of passing a 0-copy allele to his child is 0.5 (columns A and B), whereas under the hypothesis that the father has a 1 + 1 or 1 + 1D genotype, the conditional probability of passing a 0-copy allele to his child is the de novo deletion rate of ^p (columns C and D). As in the generalized Bayesian analysis shown in Table 5-1, the joint probability for each column is the product of the prior and conditional probabilities for that column, and the posterior probability for each column is the joint probability for that column divided by the sum of all the joint probabilities. The father's carrier risk is the sum of the posterior probabilities of columns A (2 + 0), B (2 + 0), and D (1 + 1D), or approximately 0.999. The father's increased carrier risk in this scenario derives almost entirely from the probability that he has the 2 + 0 genotype; this is unsurprising since the grandfather's 3-copy test result demonstrates the presence of a 2-copy allele in the family. (Note that because the grandmother's prior and conditional probabilities are the same in every column, excluding her data from the analysis will not change the result.)

Suppose instead that the father's parents, the paternal grandfather and grandmother of the affected child, are tested and each is found to have two copies of the SMN1 gene (Figure 5-7a). What is the father's carrier risk? The Bayesian analysis for this scenario is shown in Figure 5-7b. Again, there are three hypotheses for the father's genotype:

0 copies a

0 copies a

Hypothesis (Father's Genotype) | ||||

2 + 0 |
1 + 1 |
1 + 1D | ||

A |
B |
C |
D | |

Grandfather's Genotype |
2 + 1 |
2 + 1D |
2 + 1 |
2 + 1D |

Prior probability (relative probability) |
7.18 x 10-2 |
1.15 x 10-5 |
7.18 x 10-2 |
1.15 x 10-5 |

Conditional probability (of passing a 2-copy allele) |
0.5 |
0.5 | ||

Conditional probability (of passing a 1 or 1D allele) |
0.5 |
0.5 | ||

Grandmother's Genotype |
1 + 0 |
1 + 0 |
1 + 0 |
1 + 0 |

Prior probability |
2.50 x 10-2 |
2.50 x 10-2 |
2.50 x 10-2 |
2.50 x 10-2 |

Conditional probability (of passing a 0-copy allele) |
0.5 |
0.5 | ||

Conditional probability (of passing a 1-copy allele) |
0.5 |
0.5 | ||

Father's Genotype |
2 + 0 |
1 + 1 |
1 + 1D | |

Conditional probability (of passing a 0-copy allele) |
0.5 0.5 |
1% |
ft | |

Joint Probability |
2.24 x 10-4 |
3.59 x 10-8 |
9.47 x 10-8 |
1.52 x 10-11 |

Posterior Probability |
0.999 |
0.00016 |
0.00042 |
0.000000068 |

Figure 5-6. (a) Pedigree of a family with an individual affected with type I SMA with the SMA carrier test results indicated below each individual (see text).(b) Bayesian analysis for carrier risks of the father of the affected child in Figure 5-6a.

0 copies

Hypothesis (Father's Genotype) | ||||||||||||||

2 + 0 |
1 + 1 |
1 + 1" | ||||||||||||

A |
B |
c |
D |
E |
F |
G |
H |
I |
J |
K |
L |
M |
N | |

Grandfather's Genotype |
2+0 |
2+0 |
2+0 |
2+0 |
1 + 1 |
1 + 1D |
1 + 1 |
1 + 1 |
1 + 1D |
1 + 1D |
1 + 1 |
1 + 1D |
1 + 1D |
1 + 1D |

Prior Probability |
1.0 x 10"3 |
1.0 x 10"3 |
1.0 x 10"3 |
1.0 x 10"3 |
0.90 |
4.6 x 10"4 |
0.90 |
0.90 |
4.6 x 10"4 |
4.6 x 10"4 |
0.90 |
4.6 x 10"4 |
4.6 x 10"4 |
4.6 x 10"4 |

Conditional Probability (of passing a 2-copy allele) |
0.5 |
0.5 |
0.5 | |||||||||||

Conditional Probability (of passing a 0-copy allele) |
0.5 |
2.1 x10"4 |
2.1x10"4 | |||||||||||

Conditional Probability (of passing a 1-copy allele) |
1 |
1 |
0.5 |
0.5 |
1 |
0.5 | ||||||||

Conditional Probability (of passing a 1D allele) |
0.5 |
0.5 | ||||||||||||

Grandmother's Genotype |
2+0 |
2+0 |
1 + 1 |
1 + 1D |
2+0 |
2+0 |
1 + 1 |
1 + 1D |
1 + 1 |
1 + 1D |
1 + 1D |
1 + 1 |
1 + 1D |
1 + 1D |

Prior Probability |
1.0 x 10"3 |
1.0 x 10"3 |
0.90 |
4.6 x 10"4 |
1.0 x 10"3 |
1.0 x 10"3 |
0.90 |
4.6 x 10"4 |
0.90 |
4.6 x 10"4 |
4.6 x 10"4 |
0.90 |
4.6 x 10"4 |
4.6 x 10"4 |

Conditional Probability (of passing a 2-copy allele) |
0.5 |
0.5 |
0.5 | |||||||||||

Conditional Probability (of passing a 0-copy allele) |
0.5 |
4.2 x 10"! |
4.2 x 10"! | |||||||||||

Conditional Probability (of passing a 1-copy allele) |
1 |
0.5 |
1 |
0.5 |
1 |
0.5 | ||||||||

Conditional Probability (of passing a 1D allele) |
0.5 |
0.5 | ||||||||||||

Father's Genotype |
2+0 |
1 + 1 |
1 + 1D | |||||||||||

Conditional Probability (of passing a 0-copy allele) |
0.5 |
0.5 |
0.5 |
0.5 |
0.5 |
0.5 |
2.1 x 10"4 |
2.1 x10"4 |
2.1 x 10"4 |
2.1 x 10"4 |
2.1 x 10"4 |
2.1 x 10"4 |
2.1 x 10"4 |
2.1 x 10"4 |

Joint Probability |
1.3 x 10"' |
1.3 x 10"7 |
9.3 x 10"'' |
4.8 x 10"" |
4.8 x 10"' |
2.4 x 10"" |
1.7 x 10"4 |
4.3 x 10"' |
4.3 x 10"' |
1.1 x 10J1 |
4.3 x 10"' |
4.3 x 10"' |
1.1 x 10"" |
1.1 x 10"" |

Posterior Probability |
7.3 x 10"4 |
7.3 x 10"4 |
5.4 x 10" |
2.8 x 10"' |
2.8 x 10"4 |
1.4 x 10"7 |
0.997 |
2.5 x 10"4 |
2.5 x 10"4 |
6.5 x 10"' |
2.5 x 10"4 |
2.5 x 10"4 |
6.5 x 10"' |
6.5 x 10"' |

Grandfather's Genotype (2 copies and asymptomatic) |
1+1 |
2+0 |
1+1D | |||||||||

Prior Probability |
0.9 |
0.001 |
0.00046 | |||||||||

Grandmother's Genotype (2 copies and asymptomatic) |
1+1 |
2+0 |
1+1D |
1+1 |
2+0 |
1+1D |
1+1 |
2+0 |
1+1D | |||

Prior Probability |
0.9 |
0.001 |
0.00046 |
0.9 |
0.001 |
0.00046 |
0.9 |
0.001 |
0.00046 | |||

Father's Genotype |
1+1 |
2+0 |
1+1 |
1+1D |
2+0 |
2+0 |
2+0 |
1+1 |
1+1D |
2+0 |
1+1 |
1+1D |

(2 copies and asymptomatic) |
NC |
c |
NC |
c |
c |
c |
c |
NC |
c |
c |
NC |
c |

Conditional Probability (of receiving alleles) |
1 |
0.5 Hp |
0.5 |
0.5 |
0.5 Hm |
0.5 |
0.5 Hm |
0.5 |
0.5 |
0.5 HP |
0.25 |
0.5 |

Conditional Probability (of passing a 0-copy allele) |
Hp |
0.5 |
HP |
Hp |
0.5 |
0.5 |
0.5 |
HP |
HP |
0.5 |
HP |
Hp |

Joint Probability |
1.71x10^ |
4.75x10"' |
4.35x10"' |
4.35x10"' |
9.34x10"' |
2.5x10"' |
4.75xlO"1J |
4.35x10"' |
4.35x10"' |
2.42x10"" |
1.1 lxl 0"11 |
2.22x10"" |

Posterior Probability |
0.997 |
2.77xl04 |
2.54x10^ |
2.54x104 |
5.45x10"' |
0.00146 |
2.77x10"' |
2.54x10^ |
2.54xl0"4 |
1.41xl0"7 |
6.47x10"s |
1.29xl0"7 |

Column |
A |
B |
c |
D |
E |
F |
G |
H |
I |
J |
K |
L |

Figure 5-7. (a) Pedigree of a family with an individual affected with type I SMA(see text), (b) Bayesian analysis for the father of the affected child in Figure 5-7a. (In the interests of space, only two significant digits are shown.) (c) Alternative Bayesian analysis for the father of the affected child in Figure 5-7a. NC, noncarrier; C, carrier.

2 + 0, 1 + 1, and 1 + 1D. However, in this scenario, the number of possible permutations of grandparental genotypes (prior probabilities) with passage of particular alleles to the father (conditional probabilities) is dramatically increased. This is because each grandparent could have a 2 + 0, 1 + 1, or 1 + 1D genotype, and the father could have received a 2-copy allele, a 0-copy allele, a 1-copy allele, or a 1D allele from either grandparent, in most cases by direct Mendelian inheritance and in some cases from de novo deletions. The organization of the Bayesian analysis in Figure 5-7b is guided by the possible genotypes of the father, which determine the grandparental genotype permutations that need to be considered. Under the hypothesis that the father has the 2 + 0 genotype, he could have received a 2-copy allele from one (2 + 0) grandparent and a 0-copy allele from the other (2 + 0) grandparent, both by direct inheritance (columns A and B), or he could have received a 2-copy allele from one (2 + 0) grandparent by direct inheritance and a de novo deletion allele from the other (1 + 1 or 1 + 1D) grandparent (columns C, D, E, and F). Under the hypothesis that the father has the 1 + 1 genotype, he must have received a 1-copy allele from each (1 + 1 or 1 + 1D) grandparent (columns G, H, I, and J). Under the hypothesis that the father has the 1 + 1D genotype, he must have received a 1-copy allele from one (1 + 1 or 1 + 1D) grandparent and a 1D allele from the other (1 + 1 or 1 + 1D) grandparent (columns K, L, M, and N).

More specifically, under the hypothesis that the father has the 2 + 0 genotype, column A shows the prior probability that the grandfather is 2 + 0 (1.00 x 10-3), the conditional probability that he passes a 2-copy allele to the father (0.5), the prior probability that the grandmother is 2 + 0 (1.00 x 10-3), and the conditional probability that she passes a 0-copy allele to the father (0.5). Under the hypothesis that the father has the 1 + 1 genotype, column G shows the prior probability that the grandfather has a 1 + 1 genotype (0.90), the conditional probability that he passes a 1-copy allele to the father (1), the prior probability that the grandmother has a 1 + 1 genotype (0.90), and the conditional probability that she passes a 1-copy allele to the father (1). Under the hypothesis that the father has the 1 + 1D genotype, column K shows the prior probability that the grandfather has a 1 + 1 genotype (0.90), the conditional probability that he passes a 1-copy allele to the father (1), the prior probability that the grandmother has a 1 + 1D genotype (4.58 x 10-4), and the conditional probability that she passes a 1D allele to the father (0.5).

Again, the father's prior probabilities are the products of the prior and conditional probabilities for the grandparents for each column or permutation. Under the hypothesis that the father has a 2 + 0 genotype, the conditional probability of passing a 0-copy allele to his child is 0.5 (columns A through F), whereas under the hypothesis that the father has a 1 + 1 or 1 + 1D genotype, the conditional probability of passing a 0-copy allele to his child is the paternal de novo deletion rate of pp (columns G through L). As in the generalized Bayesian analysis shown in Table

5-1, the joint probability for each column is the product of the prior and conditional probabilities for that column, and the posterior probability for each column is the joint probability for that column divided by the sum of all the joint probabilities. The father's carrier risk is the sum of the posterior probabilities of columns A through F (2 + 0), and K and L (1 + 1D), or approximately 1/400. Relative to the previous scenario (Figure 5-6), in which the father also had two copies of SMN1 but the grandparents had different copy numbers, the father's dramatically decreased carrier risk in this scenario derives from the much lower probability that a 2-copy allele is present in his family, and illustrates the importance of integrating all available genetic testing information into risk assessment calculations.

An alternative organization of the Bayesian analysis shown in Figure 5-7b is shown in Figure 5-7c and is guided by the three hypotheses for the grandparental genotypes: 1 + 1,2 + 0, and 1 + 1D. For example, under the hypothesis that both of the grandparents have a 1 + 1 genotype, column A shows the prior probabilities that the grandfather has a 1 + 1 genotype (0.9) and that the grandmother has a 1 + 1 genotype (0.9), the conditional probability that the father received 1-copy alleles from both of the grandparents (1), and the conditional probability that the father passed a 0-copy allele to the affected child (by de novo deletion, pp). Under the hypothesis that the grandfather has a 1 + 1 genotype and that the grandmother has a 2 + 0 genotype, column B shows the prior probabilities that the grandfather has a 1 + 1 genotype (0.90) and that the grandmother has a 2 + 0 genotype (0.001), the conditional probability that the father received a 2-copy allele from one of the grandparents (the grandmother in this case) (0.5) and a 0-copy allele from the other grandparent (the grandfather in this case by de novo deletion, pp), and the conditional probability that the father passed a 0-copy allele to the affected child (0.5). Under the hypothesis that the grandfather has the 1 + 1 genotype and that the grandmother has the 1 + 1D genotype, column C shows the prior probabilities that the grandfather has a 1 + 1 genotype (0.90) and that the grandmother has a 1 + 1D genotype (0.00046), the conditional probability that the father received 1-copy alleles from both grandparents (0.5), and the conditional probability that the father passed a 0-copy allele to the affected child (by de novo deletion, pp). Under the hypothesis that the grandfather has the 1 + 1 genotype and that the grandmother has the 1 + 1D genotype, column D shows the prior probabilities that the grandfather has a 1 + 1 genotype (0.90) and that the grandmother has a 1 + 1D genotype (0.00046), the conditional probability that the father received a 1D allele from one of the grandparents (the grandmother in this case) and a 1-copy allele from the other grandparent (the grandfather in this case) (0.5), and the conditional probability that the father passed a 0-copy allele to the affected child (by de novo deletion, pp). The father's carrier risk is the sum of the posterior probabilities of columns B, D through G, I, J, and L, or approximately 1/400.

In both approaches (Figures 5-7b and 5-7c), the use of one comprehensive Bayesian analysis table incorporating all necessary information allows simultaneous calculations of the carrier risks of the father, grandfather, and grandmother. Such a comprehensive approach is necessary because the 2-copy test results for the grandparents influence the carrier risk of the father, and the 2-copy test result for the father influences the carrier risks of the grandparents. Using Figure 5-7b, the posterior carrier risk of the grandfather is the sum of the posterior probabilities of columns A through D, F, I, J, and L through N, or approximately 0.0020 (1/500), and the carrier risk of the grandmother is the sum of the posterior probabilities of columns A, B, D through F, H, J,K,M, and N, or approximately 0.0022 (1/450). The posterior probability that all three of them are carriers is the sum of the posterior probabilities of columns A, B, D, F, M, and N, or approximately 0.0015 (1/600). Using Figure 5-7c, the carrier risk of the grandfather is the sum of the posterior probabilities of columns E through L, or approximately 0.0020 (1/500), and the carrier risk of the grandmother is the sum of the posterior probabilities of columns B through D, F, G, and J through L, or approximately 0.0022 (1/450). The posterior probability that all three of them are carriers is the sum of the posterior probabilities of columns F, G, J, and L, or approximately 0.0015 (1/600).

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