The PAP test data comprise seven data files (trip.tst, header.tst1, header.tst2, header.tst3, header.tst4, phen.tst, ascer.tst) and 54 parameter files (listed in section A.5). File papout.tst contains the output produced by successively executing papdr using each of the 53 parameter files. In this section I hand compute the likelihood for each example; Section IV describes each example.
The test data apply to the four 9- member pedigrees shown below. Pedigree 1 contains a large sibship; Pedigree 2 contains two ancestral couples; Pedigree 3 contains an inbreeding loop; and Pedigree 4 contains a multiple marriage. File trip.tst specifies these pedigree structures for input to PAP; each ID number in trip.tst equals the pedigree number followed by the individual number, so that each individual has a unique number. For descriptions of the contents and format of file trip.dat see section II.1 and section B.1, respectively.
Pedigree 1 Pedigree 2 [1]----------------(2) [1]-----------------(2) [3]-----------(4) | | | ___________________|_______________________________________ | | | | | | | | | | | [3] (4) [5] (6) [7] (8) [9] [5]-------------------------------------- (6) | _ _______|______ | | | (7) [8] (9) Pedigree 3 [1]----------------(2) | ______ |___________ Pedigree 4 | | (3)-----------------[4] [5]-----------------(6) [1]-----------------(2) | | | | | _____________|_________________ | | | | (7)--------------------------------------[8] [3] (4)-----------------[5]----------------- (6) | | | | _ | _______|______ | | | | [9] (7) (8) (9)
_
The test data comprise phenotypes for markers and quantitative and discrete trait traits as shown below. Each ID number equals the pedigree number followed by the individual number corresponding to the ID numbers in trip.tst. File phen.tst contains the listed phenotypes. For descriptions of the contents and format of phen.dat see section II.4 and section B.4, respectively.
ID sex age genotype disease quantit marital marker 11 1 20 3 2 5.5 1 2 2 12 2 40 2 1 - 0.5 1 1 2 13 1 40 2 1 0.0 - 9999 1 2 14 2 40 2 1 0.5 - 9999 1 2 15 1 20 3 2 5.0 - 9999 2 2 16 2 20 3 2 4.5 - 9999 2 2 17 1 40 2 1 - 0.5 - 9999 1 2 18 2 20 3 2 4.5 - 9999 2 2 19 1 40 2 1 0.0 - 9999 1 2 21 1 40 2 1 0.5 1 1 2 22 2 40 2 1 - 0.5 1 1 2 23 1 40 1 1 - 5.5 2 1 1 24 2 40 1 1 - 5.0 2 1 1 25 1 20 3 2 5.0 3 2 2 26 2 40 1 1 - 4.5 3 1 1 27 2 40 2 1 0.0 - 9999 1 2 28 1 40 2 1 0.5 - 9999 1 2 29 2 40 2 1 - 0.5 - 9999 1 2 31 1 20 3 2 5.5 1 2 2 32 2 40 1 1 - 5.5 1 1 1 33 2 40 1 1 - 5.0 2 1 1 34 1 40 2 1 0.0 2 1 2 35 1 40 2 1 0.5 3 1 2 36 2 40 1 1 - 4.5 3 1 1 37 2 40 2 1 - 0.5 4 1 2 38 1 40 2 1 0.0 4 1 2 39 1 20 3 2 4.5 - 9999 2 2 41 1 40 2 1 0.5 1 1 2 42 2 40 2 1 - 0.5 1 1 2 43 1 20 3 2 5.0 - 9999 2 2 44 2 40 1 1 - 5.5 - 9999 1 1 45 1 20 3 2 5.5 2 2 2 46 2 40 1 1 - 5.0 2 1 1 47 2 40 2 1 0.0 - 9999 1 2 48 2 40 2 1 0.5 - 9999 1 2 49 2 40 2 1 - 0.5 - 9999 1 2
We assume each pedigree was ascertained through a single affected proband. File ascer.tst contains the four entries listed below; missing values were assigned for variables other than the discrete trait designated DISEASE. For descriptions of the contents and format of ascer.dat see section II.4 and section B.4, respectively.
ID sex age genotype disease quantit marital marker 11 1 20 - 9999 2 - 9999 - 9999 - 9999 - 9999 25 1 20 - 9999 2 - 9999 - 9999 - 9999 - 9999 31 1 20 - 9999 2 - 9999 - 9999 - 9999 - 9999 43 1 20 - 9999 2 - 9999 - 9999 - 9999 - 9999
Each individual's phenotypes were assigned to reflect the same autosomal genotype in order to simplify the following hand computation of the likelihoods. Each is computed as the log10 likelihood; you can obtain -2 ln likelihoods by multiplying by -2 loge (10) or approximately -4.6.
For the execution option examples, the genetic model comprises 1 locus with 2 alleles; the variable comprises the marker GENOTYPE. Since GENOTYPE is a codominant marker, each individual's phenotype corresponds to his/her genotype at the locus. Since there are no missing values, the genotypes of all pedigree members are known. The likelihood equals the product of the genotype frequency for each founder and the genotype transmission probability for each nonfounder. Since all probands have missing values for GENOTYPE, the ascertainment correction equals 1.
Letting p = frequency of allele 1, q = 1 - p, and assuming Hardy-Weinberg equilibrium, the probability for each individual and pedigree equals:
Pedigree Number ID 1 2 3 4 1 q2 2pq q2 2pq 2 2pq 2pq p2 2pq 3 _ p2 p2 _ 4 _ p2 1 p2 5 _ _ 1 _ 6 _ 1 p2 p2 7 _ 1 _ 1 8 _ 1 _ 1 9 _ 1 _ 1 Total pq3/26 p6q2 p6q2/24 p6q2/4 Total likelihood = p19q9/212
D.1.1. Single likelihood (papfqhw, paptcms, dmlpr0, qmlpr0, papwg, papen, papcr, m_1_1.dat, option 1)
For p = 0.7, q = 0.3, the log10 likelihood of each pedigree and the total equals:
Pedigree Log10 Likelihood 1 - 3.5297 2 - 1.9752 3 - 3.1793 4 - 2.5772 Total - 11.2614
D.1.2. Genotypic probabilities (papfqhw, paptcms, dmlpr0, qmlpr0, papwg, papen, papcr, m_1_2.dat, option 2, response y and y)
For each individual, the probability equals 1 for the genotype specified in phen.dat and 0 for the other 2 genotypes.
D.1.3. Simulation and output to a file (papfqhw, paptcms, dmlpr0, qmlpr0, papwg, papen, papcr, m_1_3_1.dat, option 3, response 2 and 1)
There is no single correct result for simulated phenotypes. That parents and offspring have compatible genotypes is confirmed through the computation of a nonzero likelihood (using file m_1_3_2.dat).
D.1.4. Maximization of a parameter (papfqhw, paptcms, dmlpr0, qmlpr0, papwg, papen, papcr, m_1_4.dat, option 4, response n and n)
Setting the derivative of the ln likelihood to 0 and solving the equation produces the maximum likelihood estimator.
ln likelihood = 19 ln (p) + 9 ln (1 - p) d ln likelihood/dp = 19/p - 9/(1 - p) = 0 = 19/28 = 0.6786
Computing the total log10 likelihood using produces the maximum likelihood.
log10 likelihood = - 11.2483
D.1.5. Standard error of a parameter (papfqhw, paptcms, dmlpr0, qmlpr0, papwg, papen, papcr, m_1_5.dat, option 5, response n)
Since the allele frequency is a proportion, the variance of the estimate equals /n. Therefore the standard error of equals:
Sqrt(/n) = 0.08826
Since papdr approximates the standard error, it may differ more from the theoretical value than occurs in this case.
D.1.6. Simulation and estimation (papfqhw, paptcms, dmlpr0, qmlpr0, papwg, papen, papcr, m_1_6.dat, option 6, responses 2, 10, and n)
Again, there is no single correct answer. The mean estimate across the replicates should approximately equal the simulated value of 0.7, and the standard deviation of the estimate should be approximately 0.145. However, expect a lot of variation for only 10 replicates.
D.1.7. Expected lod score (papfqhw, paptcal, dmlpr0, qmlpr0, papwg, papen, papcr, m_1_7.dat, option 7, responses n, 2, 10, n)
Once again, there is no single correct answer, but the expected lod score should approximately equal the value given in papout.tst, but again expect a lot of variation because of the small number of replicates.
D.1.8. Grid on a parameter (papfqhw, paptcms, dmlpr0, qmlpr0, papwg, papen, papcr, m_1_8.dat, option 8)
Dividing the range from 0.625 to 0.75 into five equal intervals produces six values of p with corresponding log10 likelihoods equal to:
p Log10 Likelihood 0.625 - 11.3244 0.65 - 11.2704 0.675 - 11.2486 0.7 - 11.2614 0.725 - 11.3119 0.75 - 11.4047
For the transmission parameter examples, the genetic model comprises 1 autosomal locus with 2 alleles; the variable comprises the marker GENOTYPE. Designating the two alleles as A and a, we represent the probability that a parent transmits allele A to an offspring as t1 = Pr{AA_A}, t2 = Pr{Aa_A}, t3 = Pr{aa_A} for the three genotypes. As in D.1, the likelihood equals the product of the genotype frequency for each founder and the genotype transmission probability for each nonfounder, where the transmission probabilities now depend on t1, t2, and t3. Since all the probands have missing values for GENOTYPE, the ascertainment correction equals 1. The probability for each individual equals:
Pedigree ID 1 2 3 4 1 q2 2pq q2 2pq 2 2pq 2pq p2 2pq 3 t3(1- t2)+t2(1- t3) p2 p2 (1- t2)2 4 t3(1- t2)+t2(1- t3) p2 t1(1- t3)+t3(1- t1) p2 5 (1- t2)(1- t3) (1- t2)2 t1(1- t3)+t3(1- t1) (1- t2)2 6 (1- t2)(1- t3) t21 p2 p2 7 t3(1- t2)+t2(1- t3) t1(1- t3)+t3(1- t1) t1(1- t2)+t2(1- t1) t1(1- t3)+t3(1- t1) 8 (1- t2)(1- t3) t1(1- t3)+t3(1- t1) t1(1- t2)+t2(1- t1) t1(1- t3)+t3(1- t1) 9 t3(1- t2)+t2(1- t3) t1(1- t3)+t3(1- t1) (1- t2)2+t3(1- t1) t1(1- t3)
And the likelihood of each pedigree equals:
Pedigree Likelihood 1 2pq3 (1- t2)3 (1- t3)3 [t2(1- t3) + t3(1- t2)]4 2 4p6q2 t21 (1- t2)2 [t1(1- t3) + t3(1- t1)]3 3 p6q2 (1- t2)2 [t1(1- t3)+t3(1- t1)]2 [t1(1- t2)+t2(1- t1)]2 4 4p6q2 (1- t2)4 [t1(1- t3) + t3(1- t1)]3
D.2.1. Mendelian segregation (papfqhw, paptctp, dmlpr0, qmlpr0, papwg, papen, papcr, m_2_1.dat, option 1)
For t1 = 1, t2 = _, t3 = 0, p = 0.7, q = 0.3, the log10 likelihood of each pedigree and the total equals:
Pedigree Log10 Likelihood 1 - 3.5297 2 - 1.9752 3 - 3.1793 4 - 2.5772 Total - 11.2614
D.2.2. Environmental nontransmission (papfqhw, paptctp, dmlpr0, qmlpr0, papwg, papen, papcr, m_2_1.dat, option 1)
For t1 = 0.7, t2 = 0.7, t3 = 0.7, p = 0.7, q = 0.3, the log10 likelihood of each pedigree and the total equals:
Pedigree Log10 Likelihood 1 - 6.0668 2 - 3.8589 3 - 4.5279 4 - 4.5949 Total - 19.0485
D.2.3. General transmission model (papfqhw, paptctp, dmlpr0, qmlpr0, papwg, papen, papcr, m_2_1.dat, option 1)
For t1 = 0.9, t2 = 0.4, t3 = 0.1, p = 0.7, q = 0.3, the log10 likelihood of each pedigree and the total equals:
Pedigree Log10 Likelihood 1 - 3.7323 2 - 2.1669 3 - 3.0644 4 - 2.5191 Total - 11.4827 D.3. Recombination Parameters
For the recombination parameter examples, the genetic model comprises 2 loci with 2 alleles each; the variables comprise the markers GENOTYPE and MARKER. Designating the alleles for GENOTYPE as A and a, and the alleles for MARKER as B and b, there are four haplotypes and 10 genotypes, numbered as follows:
AB Ab aB ab AB 1 Ab 2 3 aB 4 5 6 ab 7 8 9 10
For each individual, phenotypes for GENOTYPE and MARKER fix the identical 1-locus genotype and all 2-locus genotypes except 5 and 7, the double heterozygotes. The possible genotypes include 1 set for Pedigree 3, 2 sets for Pedigree 1, and four sets for Pedigrees 2 and 4. Since all the probands have missing values for both GENOTYPE and MARKER, the ascertainment correction equals 1.
Subscripting f to represent the corresponding haplotype frequency and letting _ = recombination probability, the probability by pedigree and individual equals:
Pedigree 1: Two possible sets of genotypes
ID Genotype Probability Genotype Probability 1 ab/ab f10 ab/ab f10 2 Ab/aB f5 AB/ab f7 3 AB/ab _/2 AB/ab (1- _)/2 4 AB/ab _/2 AB/ab (1- _)/2 5 ab/ab _/2 ab/ab (1- _)/2 6 ab/ab _/2 ab/ab (1- _)/2 7 AB/ab _/2 AB/ab (1- _)/2 8 ab/ab _/2 ab/ab (1- _)/2 9 AB/ab _/2 AB/ab (1- _)/2
Likelihood = f10 [f5 _7 + f7 (1- _)7]/27
Pedigree 2: Four possible sets of genotypes
ID Genotype Probability Genotype Probability Genotype Probability Genotype Probability 1 Ab/aB f5 Ab/aB f5 AB/ab f7 AB/ab f7 2 Ab/aB f5 AB/ab f7 Ab/aB f5 AB/ab f7 3 AB/AB f1 AB/AB f1 AB/AB f1 AB/AB f1 4 AB/AB f1 AB/AB f1 AB/AB f1 AB/AB f1 5 ab/ab _2/4 ab/ab _(1- _)/4 ab/ab _(1- _)/4 ab/ab (1- _)2/4 6 AB/AB 1 AB/AB 1 AB/AB 1 AB/AB 1 7 AB/ab 1 AB/ab 1 AB/ab 1 AB/ab 1 8 AB/ab 1 AB/ab 1 AB/ab 1 AB/ab 1 9 AB/ab 1 AB/ab 1 AB/ab 1 AB/ab 1
Likelihood = f21 [f5 _ + f7 (1- _)]2/4
Pedigree 3: One possible set of genotypes ID Genotype Probability 1 AB/AB f10 2 ab/ab f1 3 ab/ab f1 4 ab/AB 1 5 ab/AB 1 6 ab/ab f1 7 ab/AB (1- _)/2 8 ab/AB (1- _)/2 9 AB/AB (1- _)2/4 Likelihood = f31 f10 (1- _)4/16
Pedigree 4: Four possible sets of genotypes ID Genotype Probability Genotype Probability Genotype Probability Genotype Probability 1 Ab/aB f5 Ab/aB f5 AB/ab f7 AB/ab f7 2 Ab/aB f5 AB/ab f7 Ab/aB f5 AB/ab f7 3 ab/ab _2/4 ab/ab _(1- _)/4 ab/ab _(1- _)/4 ab/ab (1- _)2/4 4 AB/AB f1 AB/AB f1 AB/AB f1 AB/AB f1 5 ab/ab _2/4 ab/ab _(1- _)/4 ab/ab _(1- _)/4 ab/ab (1- _)2/4 6 AB/AB f1 AB/AB f1 AB/AB f1 AB/AB f1 7 AB/ab 1 AB/ab 1 AB/ab 1 AB/ab 1 8 AB/ab 1 AB/ab 1 AB/ab 1 AB/ab 1 9 AB/ab 1 AB/ab 1 AB/ab 1 AB/ab 1 Likelihood = f21 [f5 _2 + f7 (1- _)2]2/42
D.3.1. Unlinked (papfqhw, paptcal, dmlpr0, qmlpr0, papen, papcr, m_3_1.dat, option 1)
For _ = _, f1 = 0.3136, f5 = 0.0672, f7 = 0.0672, f10 = 0.0036, the log10 likelihood of each pedigree and the total equals:
Pedigree Log10 Likelihood 1 - 7.5297 2 - 3.9546 3 - 6.3628 4 - 5.1587 Total - 23.0058
D.3.2. Linked (papfqhw, paptcal, dmlpr0, qmlpr0, papen, papcr, m_3_2.dat, option 1)
For _ = 0.0, f1 = 0.3136, f5 = 0.0672, f7 = 0.0672, f10 = 0.0036, the log10 likelihood of each pedigree and the total equals:
Pedigree Log10 Likelihood 1 - 5.7235 2 - 3.9546 3 - 5.1587 4 - 4.5566 Total - 19.3934
For the frequency parameter examples, the genetic model comprises 2 unlinked loci with 2 alleles each; the variables comprise the markers GENOTYPE and MARKER. Setting _ = _, fh = f5 + f7 in the expression in section D.3, the likelihood of each pedigree equals:
Pedigree Likelihood 1 fhf10/47 2 f21fh2/16 3 f31f10/44 4 f21fh2/44D.4.1. Allele frequencies
Let pA represent the frequency of allele A, pB represent the frequency of allele B, D represent disequilibrium, qa = 1 - pA, and qb = 1 - pB. Then the genotype frequencies equal:
f1 = (pApB + D)2 f5 = 2(pAqb - D)(pBqa - D) f7 = 2(pApB + D)(qaqb + D) f10 = (qaqb + D)2
D.4.1.1. Equilibrium (papfqh2, paptcms, dmlpr0, qmlpr0, papwg, papen, papcr, m_4_1_1.dat, option 1)
Linkage equilibrium is defined as D = 0. Setting pA = 0.7, qa = 0.3, pB = 0.8, qb = 0.2, then f1 = 0.3136, f5 = 0.0672, f7 = 0.0672, f10 = 0.0036, and the likelihood of each pedigree and the total equals:
Pedigree Log10 Likelihood 1 - 7.5297 2 - 3.9546 3 - 6.3628 4 - 5.1587 Total - 23.0058
D.4.1.2. Disequilibrium (papfqh2, paptcms, dmlpr0, qmlpr0, papwg, papen, papcr, m_4_1_2.dat, option 1)
Letting D = - 0.02, and pA = 0.7, qa = 0.3, pB = 0.8, qb = 0.2, then f1 = 0.2916, f5 = 0.0832, f7 = 0.0432, f10 = 0.0016, and the likelihood of each pedigree and the total equals:
Pedigree Log10 Likelihood 1 - 7.9086 2 - 4.0711 3 - 6.8098 4 - 5.2752 Total - 24.0645D.4.2. Conditional allele frequencies
Letting pA|B represent the frequency of allele A conditional on allele B, pA|b represent the frequency of allele A conditional on allele b, qA|B = 1 - pA|B, qA|b = 1 - pA|b, pB = frequency of allele B, qb = 1 - pB the genotype frequencies equal:
f1 = (pA|BpB)2 fh = pA|BpB qA|bqb + pA|bqb qA|BpB f10 = (qA|bqb)2
D.4.2.1. Equilibrium (papfqa2, paptcms, dmlpr0, qmlpr0, papwg, papen, papcr, m_4_2_1.dat, option 1)
Letting pA|B = 0.7, qA|B = 0.3, pA|b = 0.7, qA|b = 0.3, pB = 0.8, qb = 0.2, then f1 = 0.3136, f5 = 0.0672, f7 = 0.0672, f10 = 0.0036. See section D.4.1.1 for the likelihood computation.
D.4.2.2. Disequilibrium (papfqa2, paptcms, dmlpr0, qmlpr0, papwg, papen, papcr, m_4_2_2.dat, option 1)
Letting pA|B = 0.675, qA|B = 0.325, pA|b = 0.8, qA|b = 0.2, pB = 0.8, qb = 0.2, then f1 = 0.2916, f5 = 0.0832, f7 = 0.0432, f10 = 0.0016. See section D.4.1.2 for the likelihood computation.
D.4.3. Haplotype frequencies
Letting hAB represent the frequency of haplotype AB, hAb represent the frequency of haplotype Ab, haB represent the frequency of haplotype aB, and hab represent the frequency of haplotype ab, the genotype frequencies equal:
f1 = h2AB fh = hABhab + hAbhaB f10 = h2ab
D.4.3.1. Equilibrium (papfqh, paptcms, dmlpr0, qmlpr0, papwg, papen, papcr, m_4_3_1.dat, option 1)
Letting hAB = 0.56, hAb = 0.14, haB = 0.24, hab = 0.06, then f1 = 0.3136, f5 = 0.0672, f7 = 0.0672, f10 = 0.0036. See section D.4.1.1 for the likelihood computation.
D.4.3.2. Disequilibrium (papfqh, paptcms, dmlpr0, qmlpr0, papwg, papen, papcr, m_4_3_2.dat, option 1)
Letting hAB = 0.54, hAb = 0.16, haB = 0.26, hab = 0.04, then f1 = 0.2916, f5 = 0.0832, f7 = 0.0432, f10 = 0.0016. See section D.4.1.2 for the likelihood computation.
D.4.4. Genotype frequencies
D.4.4.1. Equilibrium (papfqg, paptcms, dmlpr0, qmlpr0, papwg, papen, papcr, m_4_4_1.dat, option 1)
Let f1 = 0.3136, f5 = 0.0672, f7 = 0.0672, f10 = 0.0036. See section D.4.1.1 for the likelihood computation.
D.4.4.2. Disequilibrium (papfqg, paptcms, dmlpr0, qmlpr0, papwg, papen, papcr, m_4_4_2.dat, option 1)
Let f1 = 0.2916, f5 = 0.0832, f7 = 0.0432, f10 = 0.0016. See section D.4.1.2 for the likelihood computation.
For the discrete trait examples, the genetic model comprises 1 autosomal locus with 2 alleles; the variable comprises the discrete trait DISEASE. In this case, the genotype cannot be unequivocally inferred from the phenotype. However, a rare allele frequency and recessive inheritance with affection probabilities of 0 and 1 combine to produce a high probability for the assigned genotype for each individual. The small probability of other genotypes is ignored in the hand likelihood computation, making it differ slightly from the likelihood computed by papdr, but identical to the computation in section D.1. Therefore, the likelihood of each pedigree and the total equals:
Pedigree Likelihood 1 pq3/26 2 p6q2 3 p6q2/24 4 p6q2/4 Total p19q9/212The proband in each pedigree is affected. Therefore, the ascertainment correction for each pedigree equals the probability that a random individual is affected or q2.
D.5.1. Affection probability (papfqhw, paptcms, dmlprpn, qmlpr0, papwgml, papend, papcr, m_5_1.dat, option 1)
Letting p = 0.999, q = 0.001, the log10 likelihood of each pedigree and the total equals:
Pedigree Log10 Likelihood Corrected 1 -10.8066 -4.8066 2 -6.0026 -0.0026 3 -7.2067 -1.2067 4 -6.6047 -0.6047 Total -30.6206 -6.6206
D.5.3. Prevalence (papfqhw, paptcms, dmlprpr, qmlpr0, papwgml, papend, papcr, m_5_3.dat, option 1)
Let the prevalence equal 1 X 10-6, dominance equal 0, and displacement equal 10. See section D.5.1 for the likelihood calculation.
D.5.4. Incidence (papfqhw, paptcms, dmlprin, qmlpr0, papwgml, papend, papcr, m_5_4.dat, option 1)
Let the incidence equal 1 X 10-6, dominance equal 0, and displacement equal 10. See section D.5.1 for the likelihood calculation.
D.5.5. Severity classes (papfqhw, paptcms, dmlprsv, qmlpr0, papwgml, papend, papcr, m_5_5.dat, option 1)
Let the incidence equal 1 X 10-6, dominance equal 0, and displacement equal 10. See section D.5.1 for the likelihood calculation.
For the quantitative trait examples, the genetic model comprises 1 locus with 2 alleles; the variable comprises the quantitative trait QUANTIT. Again the genotype cannot be unequivocally inferred from the phenotype. However, a rare allele frequency and recessive inheritance with a small standard deviation combine to produce a high probability for the assigned genotype for each individual. The likelihood equals the product of the genotype frequency for each founder, the transmission probability for each nonfounder, and the penetrance for each measured individual. Assuming the assigned genotype, the penetrance for each individual equals:
Pedigree ID 1 2 3 4 1 N((5.5-µaa)/saa,saa) N((.5-µAa)/sAa,sAa) N((5.5-µaa)/saa,saa) N((.5-µAa)/sAa,sAa) 2 N((-.5-µAa)/sAa,sAa) N((-.5-µAa)/sAa,sAa) N((-5.5-µAA)/sAA,sAA) N((- .5-µAa)/sAa,sAa) 3 N((0-µAa)/sAa,sAa) N((- 5.5-µAA)/sAA,sAA) N((- 5.-µAA)/sAA,sAA) N((5.-µaa)/saa,saa) 4 N((.5-µAa)/sAa,sAa) N((- 5.-µAA)/sAA,sAA) N((0-µAa)/sAa,sAa) N((- 5.5-µAA)/sAA,sAA) 5 N((5.-µaa)/saa,saa) N((5.-µaa)/saa,saa) N((.5-µAa)/sAa,sAa) N((5.5-µaa)/saa,saa) 6 N((4.5-µaa)/saa,saa) N((- 4.5-µAA)/sAA,sAA) N((- 4.5-µAA)/sAA,sAA) N((- 5.-µAA)/sAA,sAA) 7 N((- .5-µAa)/sAa,sAa) N((0-µAa)/sAa,sAa) N((- .5-µAa)/sAa,sAa) N((0-µAa)/sAa,sAa) 8 N((4.5-µaa)/saa,saa) N((.5-µAa)/sAa,sAa) N((0-µAa)/sAa,sAa) N((.5-µAa)/sAa,sAa) 9 N((0-µAa)/sAa,sAa) N((- .5-µAa)/sAa,sAa) N((4.5-µaa)/saa,saa) N((- .5-µAa)/sAa,sAa)
where N([[partialdiff]],s) represents the height of a normal density for deviation [[partialdiff]] and standard deviation s, the number is the individual's quantitative phenotype, µi and si represents the mean and standard deviation, respectively, of the assigned genotype, i = AA, Aa, or aa. The log10 likelihood of each pedigree and the total equals:
Pedigree Log10 Likelihood 1 3 n(0) + 6 n(.5) - 10.8066 = - 14.7243 2 3 n(0) + 6 n(.5) - 6.0026 = - 9.9203 3 3 n(0) + 6 n(.5) - 7.2067 = - 11.1244 4 3 n(0) + 6 n(.5) - 6.6047 = - 10.5224 Total = - 46.2914
where n(0) = log10 N(0) = - .3991 and n(.5) = log10 N(.5) = log10 N(- .5) = - .4534, and the log10 likelihood of the product of the frequencies and transmission probabilities comes from section D.5.1. The small probability of other genotypes than the assigned genotype is ignored in the hand likelihood computation, making it differ slightly from the likelihood computed by papdr.
D.6.1. Means/Standard Deviations (papfqhw, paptcms, dmlpr0, qmlprmv, papwgml, papenq, papcr, m_6_1.dat, option 1)
Setting p = 0.999, q = 0.001, µAA = - 5., µAa = 0., µaa = 5., and sAA = sAa = saa = 1 produces the above log10 likelihood. Since all the probands have missing values for QUANTIT, the ascertainment correction equals 1.
D.6.2. Dominance/Displacement (papfqhw, paptcms, dmlpr0, qmlprdd, papwgml, papenq, papcr, m_6_2.dat, option 1)
Setting the total mean to -4.99, the total standard deviation to 1.0246710, dominance to 0.5 and displacement to 10 produces the same parameter values and consequently the same log10 likelihood as section D.6.1. Since all the probands have missing values for QUANTIT, the ascertainment correction equals 1.
D.6.3. Means/Standard Deviations/Threshold (papfqhw, paptcms, dmlpr0, qmlprmvt, papwgml, papendq, papcr, m_6_3.dat, option 1)
Setting µ1 = - 5., µ2 = 0., µ3 = 5., s1 = s2 = s3 = 1, and threshold = 5 produces the same log10 likelihood for the pedigrees as in section 6.1. With the threshold equal to µ3, the affection probability equal 0.5 which when multiplied by the genotypic frequency of 0.000001 equals a log10 likelihood for the ascertainment correction of -6.3.
For the within-genotype examples, the genetic model comprises 1 locus with 2 alleles; the variable comprises the quantitative trait QUANTIT. Again the genotype cannot be unequivocally inferred from the phenotype. However, a rare allele frequency and recessive inheritance with a small standard deviation combine to produce a high probability for the assigned genotype for each individual. As in section D.6, the likelihood equals the product of the genotype frequency for each founder, the transmission probability for each nonfounder, and the penetrance for each measured individual. However, the penetrances differ from those in section D.6 because of the correlation between individuals. Let d1 and d2 represent the deviations from the genotypic mean and s1 and s2 represent the within-genotype standard deviations for individuals 1 and 2, and r represent the correlation between them. As before, the penetrance for individual 1 equals N(d1, s1) where N([[partialdiff]], s) represents the normal density with deviation [[partialdiff]] and standard deviation s. The penetrance for individual 2 equals N(d2', s2') where d2' = (d2 - _ d1)/(1 - r2) and s2' = s2 (1 - r2). See section VI.10 for extension beyond two individuals.
D.7.1. Exact Variance Components
Heritability h2 represents the proportion of the within-genotype variance attributed to polygenes and parameter m2 represents the proportion of the within-genotype variance attributed to a spouse effect. Variable MARITAL identifies spouses through assignment of the same number to both except that in pedigree 4, individual 5 shares a marital effect only with individual 6 and not individual 4.
D.7.1.1. Marital Effect (papfqhw, paptcms, dmlpr0, qmlprmv, papwgvc, papenqe, papcrqe, m_7_1_1.dat, option 1)
Setting h2 = 0 and m2 = 0.1, produces a correlation r = 0.1 between spouse pairs (except individuals 4 and 5 in pedigree 4) and r = 0 between all other pairs. The penetrance for each individual equals:
Pedigree ID 1 2 3 4 1 N(.5, 1|-.5) N(.5. 1|-.5) N(.5, 1|-.5) N(.5, 1|-.5) 2 N(-.5, 1) N(-.5, 1) N(-.5, 1) N(- .5, 1) 3 N(0, 1) N(- .5, 1|0) N(0, 1|0) N(0, 1) 4 N(.5, 1) N(0, 1) N(0, 1) N(- .5, 1) 5 N(0, 1) N(0, 1|.5) N(.5, 1|.5) N(.5, 1|0) 6 N(-.5, 1) N(.5, 1) N(.5, 1) N(0, 1) 7 N(- .5, 1) N(0, 1) N(- .5, 1|0) N(0, 1) 8 N(-.5, 1) N(.5, 1) N(0, 1) N(.5) 9 N(0, 1) N(- .5, 1) N(-.5, 1) N(- .5, 1)
where N([[partialdiff]],s) represents the height of a normal density for deviation [[partialdiff]] and standard deviation s, and N([[partialdiff]],s|_) represents the height of a normal density for deviation [[partialdiff]] and standard deviation s after conditioning on deviation _. Note that the husband in each spouse pair is conditioned on his wife's deviation. Consequently, the log10 likelihood of each pedigree and the total equals:
Pedigree log10 likelihood
1 3 n(0) + 5 n(.5) + n(.5|-.5) -10.8066 -14.7348
2 2 n(0) + 4 n(.5) + n(.5|-.5) + n(-.5|0) + n(0|.5) -6.0026 -9.9233
3 2 n(0) + 3 n(.5) + n(-.5|0) + n(.5|.5) + n(0|0) + n(.5|-.5) -7.2067 -11.1200
4 3 n(0) + 4 n(.5) + n(.5|0) + n(.5|-.5) -6.6047 -10.5318
Total -46.3099
where n(d) = log10 N(d,1), n(d1|d2) = log10 N(d1,1|d2), n(0) = -.3991, n(.5) = -.4534, n(.5|-.5) = -.4639, n(-.5|0) = -.4523, n(0|.5) = -.3975, n(.5|.5) = -.4418, n(0|0) = -.3969, n(.5|0) = -.4523, and the log10 likelihood of the product of the frequencies and transmission probabilities comes from section D.5.1. The small probability of other genotypes than the assigned genotype is ignored in the hand likelihood computation, making it differ slightly from the likelihood computed by papdr.
D.7.1.2. Marital Effect and Heritability (papfqhw, paptcms, dmlpr0, qmlprmv, papwgvc, papenqe, papcrqe, m_7_1_2.dat, option 1)
Setting h2 = 0.2 and m2 = 0.1 produces correlations between all pairs in each pedigree except individuals 4 and 5 in pedigree 4. To compute the penetrance requires successively conditioning each individual's deviation on all previous individuals' deviations, a tedious calculation by hand. Consequently, a hand-calculated likelihood is not given.
D.7.2. Approximate Variance Components
As in section D.7.1, heritability h2 represents the proportion of the within-genotype variance attributed to polygenes and parameter m2 represents the proportion of the within-genotype variance attributed to a spouse effect. Variable MARITAL identifies spouses through assignment of the same number to both. Within-genotype correlations between individuals preclude recursive computation of the likelihood. The exact likelihood equals the sum, over all sets of genotypes, of the probability of each set of genotypes for all pedigree members. Approximation [Hasstedt 1993] allows recursive computation and much greater speed.
D.7.2.1. Marital Effect (papfqhw, paptcms, dmlpr0, qmlprmv, papwgvc, papenqa, papcrqa, m_7_2_1.dat, option 1)
Setting h2 = 0 and m2 = 0.1 produces exactly the same log10 likelihood as in section D.7.1.1, despite approximation in the calculation. Only spouses share the within-genotype effect, and recursive calculation considers the genotypes of spouse pairs simultaneously.
D.7.2.2. Marital Effect and Heritability (papfqhw, paptcms, dmlpr0, qmlprmv, papwgvc, papenqa, papcrqa, m_7_2_2.dat, option 1)
Setting h2 = 0.2 and m2 = 0.1 produces an approximate likelihood. Compare to the likelihood obtained in D.7.1.2.
D.7.3. Approximate Familial Correlations
The gender-specific familial correlations comprise husband-wife _hw, mother-daughter _md, mother-son _ms, father-daughter _fd, father-son _fs, sister-sister _ss, sister-brother _sb, and brother-brother _bb. Outside the nuclear family, zero correlation is assumed for all relative pairs.
D.7.3.1. Husband-Wife Correlation (papfqhw, paptcms, dmlpr0, qmlprmv, papwgfc, papenqa, papcrqa, m_7_3_1.dat, option 1)
Setting _hw = 0.1, _md = _ms = _fd = _fs = _ss = _sb = _bb = 0 produces a likelihood identical to the likelihood from section D.7.1.1 for pedigrees 1 through 3. The likelihood for pedigree 4 differs because this parameterization includes a correlation between individual 5 and both spouses; section D.7.1.2 excluded the correlation between individuals 5 and 4.
D.7.3.2. Spouse, Parent-offspring and Sibling Correlations (papfqhw, paptcms, dmlpr0, qmlprmv, papwgfc, papenqa, papcrqa, m_7_3_2.dat, option 1)
Setting _hw = _md = _ms = _fd = _fs = _ss = _sb = _bb = 0.1 produces a likelihood identical to the likelihood from section D.7.2.2 for pedigree 1 since parent-offspring and sibling correlations of 0.1 corresponds to a heritability of 0.2. Pedigrees 2-4 contain more distant relatives for whom this parameterization assumes zero correlation but the variance components parameterization does not.
D.8.1. Measured genotype
The variables comprise the quantitative trait QUANTIT and the marker GENOTYPE. Since all the probands have missing values for QUANTIT and GENOTYPE, the ascertainment correction equals 1.
D.8.1.1. 1-locus (papfqhw, paptcms, dmlpr0, qmlprmv, papwgml, papenq, papcr, m_8_1_1.dat, option 1)
The genetic model comprises 1 locus with 2 alleles. The genotypes assigned by GENOTYPE are equivalent to the genotypes used in previous quantitative trait likelihood computations. Therefore, see section D.6 for the likelihood computation.
D.8.1.2. 2-locus (papfqhw, paptcms, dmlpr0, qmlprmv, papwgml, papenq, papcr, m_8_1_2.dat, option 1)
The genetic model comprises 2 loci each with 2 alleles; QUANTIT is assigned to both loci and GENOTYPE is assigned to locus 1. Means for QUANTIT are assigned to reflect the genotype at locus 1. Therefore, see section D.6 for the likelihood computation.
D.8.2. Bivariate
The variables comprise the quantitative trait QUANTIT and the discrete trait DISEASE. Parameter _ represents the correlation between QUANTIT and DISEASE. Since each proband is affected, the ascertainment correction equals the disease frequency or 10-6.
D.8.2.1. 1-locus (papfqhw, paptcms, dmlprpn, qmlprmv, papwgml, papendq, papcr, m_8_2_1.dat, option 1)
The genetic model comprises 1 autosomal locus with 2 alleles. The genotypes assigned by DISEASE are equivalent to the genotypes used in previous quantitative trait likelihood computations and _ = 0. Therefore, see section D.6 for the likelihood computation.
D.8.2.2. 2-locus (papfqhw, paptcms, dmlprpn, qmlprmv, papwgml, papendq, papcr, m_8_2_2.dat, option 1)
The genetic model comprises 2 autosomal loci with 2 alleles each; QUANTIT is assigned to both loci and DISEASE is assigned to locus 1. The genotypes assigned by DISEASE are equivalent to the genotypes used in previous quantitative trait likelihood computations, the means for QUANTIT correspond to locus 1 genotypes, and _ = 0. Therefore, see section D.6 for the likelihood computation.
For the genotype-assignment code examples, the genetic model comprises 1 or 2 loci with 2 alleles; the variable comprises the discrete trait DISEASE.
D.9.1. Autosomal (papfqhw paptcms, dmlprpn, qmlpr0, papwgml, papend, papcr, m_9_1.dat, option 1)
See section D.5.1 for the likelihood computation.
D.9.2. X- linked (papfqhw paptcms, dmlprpn, qmlpr0, papwgml, papend, papcr, m_9_2.dat, option 1)
For p = allele frequency, q = 1 - p, the probability by individual and pedigree equals:
Pedigree Individual 1 2 3 4 1 q p q p 2 2pq 2pq p2 2pq 3 _ p 2pq _ 4 _ p2 1 p2 5 _ _ 1 _ 6 _ 1 p2 p2 7 _ 1 _ 1 8 _ 1 1 1 9 _ 1 _ 1 Total pq2/26 p4q p5q2/2 p6q/2Since p = 0.999, q = 0.001, the ascertainment correction = -3 for each pedigree, and the log10 likelihood by pedigree equals:
Pedigree Log10 Likelihood Corrected 1 -7.8066 -4.8066 2 -3.0017 -0.0017 3 -6.3032 -3.3032 4 -3.3036 -0.3036 Total -20.4152 -8.4152
D.9.3. Parent-specific autosomal
D.9.3.1. Equal penetrances (papfqhw paptcms, dmlprpn, qmlpr0, papwgml, papend, papcr, m_9_3_1.dat, option 1)
For equal penetrance in heterozygotes regardless of the parental origin of each allele, the model is equivalent to the standard autosomal model. See section D.9.1 for the likelihood computation.
D.9.3.2. Unequal penetrances (papfqhw paptcms, dmlprpn, qmlpr0, papwgml, papend, papcr, m_9_3_2.dat, option 1)
For pen(Aa) = 1, pen(aA) = 0, where the alleles are ordered as maternal/paternal origin, the likelihood by individual and pedigree equals:
Pedigree Individual 1 2 3 4 1 pq p2 pq p2 2 pq pq p2 pq 3 _ p2 p2 _ 4 _ p2 _ p2 5 _ _ 1 _ 6 _ 1 p2 p2 7 _ 1 _ 1 8 _ 1 1 1 9 _ 1 _ 1 Total p2q2/27 p7q/2 p7q/23 p7q/22For p = 0.999, q = 0.001, the ascertainment correction = -3, and the log10 likelihood by pedigree equals:
Pedigree Log10 Likelihood Corrected 1 -8.1081 -5.1081 2 -3.3041 -0.3041 3 -3.9061 -0.9061 4 -3.6051 -0.6051 Total -18.9234 -6.9234
D.9.4. X- linked parental origin
D.9.4.1. Equal penetrances (papfqhw paptcms, dmlprpn, qmlpr0, papwgml, papend, papcr, m_9_4_1.dat, option 1)
For equal penetrance in heterozygotes regardless of the parental origin of each allele, the model is equivalent to the standard X-linked model. See section D.9.2 for the likelihood computation.
D.9.4.2. Unequal penetrances (papfqhw paptcms, dmlprpn, qmlpr0, papwgml, papend, papcr, m_9_4_2.dat, option 1)
For pen(Aa) = 1, pen(aA) = 0, where the alleles are ordered as maternal/paternal origin, the likelihood by individual and pedigree equals:
Pedigree Individual 1 2 3 4 1 q p q p 2 pq pq p2 pq 3 _ p p2 _ 4 _ p2 1 p2 5 _ _ 1 _ 6 _ 1 p2 p2 7 _ 1 1 1 8 _ 1 1 1 9 _ 1 0 1 Total pq2/27 p5q/2 0 p6q/22
For p = 0.999, q = 1 - p, the ascertainment correction = -3 for each pedigree, and the log10 likelihood by pedigree equals:
Pedigree Log10 Likelihood Corrected 1 -8.1076 -5.1076 2 -3.3032 -0.3032 3 -infinity -infinity 4 -3.6047 -0.6047 Total -infinity -infinityThe model is impossible in pedigree 3 because: (1) for 9 to be affected he must receive the allele from his mother 7; (2) for 7 to be unaffected she must receive the allele from her father 4; (3) for 4 to be unaffected, he must not have the allele.
D.9.5. Category-specific autosomal inheritance
D.9.5.1. Equal penetrances (papfqhw paptcms, dmlprpn, qmlpr0, papwgml, papend, papcr, m_9_5_1.dat, option 1)
For equal affection probabilities regardless of sex, the model is equivalent to the standard autosomal model. See section D.9.1 for the likelihood computation.
D.9.5.2. Unequal penetrances (papfqhw paptcms, dmlprpn, qmlpr0, papwgml, papend, papcr, m_9_5_2.dat, option 1)
For pen(Aa) = 1 in males, pen(Aa) = 0 in females, the likelihood by individual and pedigree equals:
Pedigree Individual 1 2 3 4 1 2pq p2 2pq p2 2 2pq 2pq p2 2pq 3 _ p2 2pq _ 4 _ p2 _ p2 5 _ _ _ _ 6 _ 1 p2 p2 7 _ 1 _ 1 8 _ _ 1 1 9 _ 1 _ 1 Total p2q232/46 p7q/2 p6q2/22 p7q/2
For p = 0.999, q = 1 - p, the ascertainment correction equals 1 - p2 = -2.6992, and the log10 likelihood by pedigree equals:
Pedigree Log10 Likelihood Corrected
1 -8.6590 -5.9598
2 -3.3041 -0.6049
3 -6.6047 -3.9055
4 -3.3041 -0.6049
Total -21.8718 -11.0751
D.9.6. Autosomal/X- linked mixed model
In the autosomal/X-linked mixed model, tf represents father-daughter transmission and tm represents father-son transmission.
D.9.6.1. Autosomal (papfqxa paptctp, dmlprpn, qmlpr0, papwgml, papend, papcr, m_9_6_1.dat, option 1)
For tm = tf = 0.5, the model is equivalent to the standard autosomal model. See section D.9.1 for the likelihood computation.
D.9.6.2. X- linked (papfqxa paptctp, dmlprpn, qmlpr0, papwgml, papend, papcr, m_9_6_2.dat, option 1)
For tf = 0, tm = 1, the model is equivalent to the standard X-linked model. See section D.9.2 for the likelihood computation.
D.9.6.3. Mixed (papfqxa paptctp, dmlprpn, qmlpr0, papwgml, papend, papcr, m_9_6_3.dat, option 1)
For tf = 0.5, tm = 1, the likelihood by individual and pedigree equals:
Pedigree
Individual 1 2 3 4
1 q p q p
2 2pq 2pq p2 2pq
3 _ p p2 _
4 _ p2 1 p2
5 _ _ 1 _
6 _ _ _ p2 p2
7 _ _ 1 _ 1
8 _ _ 1 1 1
9 _ _ 1 _ 1
Total pq2/26 p5q/2(1+1/23) p6q/4 p6q/2
For p = 0.999, q = 1 - p, the ascertainment correction = -3, and the log10 likelihood by pedigree and total equals:
Pedigree Log10 Likelihood Corrected 1 -7.8066 -4.8066 2 -3.2521 -0.2521 3 -3.6047 -0.6047 4 -3.3036 -0.3036 Total -17.9670 -5.9670
D.9.7. Autosomal/X- linked admixture
In the autosomal/X-linked admixture model, qx represents the frequency of the X chromosome allele and qa represents the frequency of the autosomal allele.
D.9.7.1. Autosomal allele only (papfqhw paptcms, dmlprpn, qmlpr0, papwgml, papend, papcr, m_9_7_1.dat, option 1)
For qx = 0, the model is equivalent to the standard autosomal model. See section D.9.1 for the likelihood computation.
D.9.7.2. X- linked allele only (papfqhw paptcms, dmlprpn, qmlpr0, papwgml, papend, papcr, m_9_7_2.dat, option 1)
For qa = 0, the model is equivalent to the standard X-linked model. See section D.9.2 for the likelihood computation.
D.9.7.3. Both alleles (papfqhw paptcms, dmlprpn, qmlpr0, papwgml, papend, papcr, m_9_7_3.dat, option 1)
The likelihood of the admixture model can be computed by summing the likelihood of each set of genotypes for affected individuals. In pedigrees 1, 2, and 4, only exclusively autosomal or exclusively X-linked genotypes require fewer than three alleles. Pedigree 3 also requires only two disease alleles if individual 1 has the X-linked form and individual 9 has the autosomal form. The approximate likelihood by pedigree equals:
Pedigree Autosomal X-linked X/Autosomal Likelihood
1 10-10.8066 + 10-7.8066 = 10-7.8062
2 10-6.0026 + 10-3.0017 = 10-3.0013
3 10-7.2048 + 10-6.3032 + 10-7.2098 = 10-6.2519
4 10-6.6047 + 10-3.3036 = 10-3.3034
Total 10-30.6170 + 10-20.4152 = 10-20.4152