Background:

Consider the marker locus having two alleles M₁ and M₂, with population frequencies g and g′ = 1 − g, respectively. Note that M₁ and M₂ may represent two groups of alleles. It is convenient to use 0, 1, and 2 to denote the genotypes M₂M₂, M₁M₂, and M₁M₁, respectively, where the number indicates the number of marker allele M₁ a genotype has. Denote the two alleles at a DSL as D and d, with population frequencies p and q = 1 − p, respectively. To specify the source of alleles D and d, let d/d, D/d, d/D, and D/D denote the four ordered genotypes of a child at a DSL. The allele on the left side of the slash (/) is paternal and the one on the right side is maternal. The corresponding risks of being affected are denoted by ϕ_d/d, ϕ_D/d, ϕ_d/D, and ϕ_D/D, respectively. The risk for an individual with one copy of D is assumed to be less than that with two copies of D and be greater than that with no copies of D. The population disease prevalence is then given as ϕ = p²ϕ_D/D + pqϕ_D/d + pqϕ_d/D + q²ϕ_d/d. Strauch et al. (2000) introduced the degree of imprinting I = (ϕ_D/d − ϕ_d/D)/2, ranging from (ϕ_d/d − ϕ_D/D)/2 to (ϕ_D/D − ϕ_d/d)/2, to measure parent-of-origin effects. Generally, I < 0 signifies paternal imprinting or maternal expression, and I > 0 signifies maternal imprinting or paternal expression. I = 0 implies that the two heterozygote risks are identical, i.e., no imprinting. There are two extreme cases, one is completely paternal imprinting or complete maternal expression, i.e., ϕ_D/d = ϕ_d/d and ϕ_d/D = ϕ_D/D, and the other is completely maternal imprinting or complete paternal expression, i.e., ϕ_D/d = ϕ_D/D and ϕ_d/D = ϕ_d/d.

For a simple representation of some of our results later, let R denote the difference between the sum of two homozygote risks and the sum of two heterozygote risks, i.e., R = ϕ_D/D − ϕ_D/d − ϕ_d/D + ϕ_d/d, and Δ denote the difference between two ratios P(D | affected child)/P(D | random man) and P(d | affected child)/P(d | random man), where P(D | affected child) represents the probability that a chromosome of an affected child has a disease allele D at a DSL, and the other probabilities P(D | random man), P(d | affected child), and P(d | random man) are similarly defined. It is derived that Δ = (p(2ϕ_D/D − ϕ_D/d − ϕ_d/D) + q(ϕ_D/d + ϕ_d/D − 2ϕ_d/d))/(2ϕ) and this ratio difference Δ is a positive quantity according to the relative magnitude of the four risk parameters.

Denote the three genotypic relative risks (Risch and Merikangas 1996) as γ_1p = ϕ_D/d/ϕ_d/d, γ_1m = ϕ_d/D/ϕ_d/d, and γ₂ = ϕ_D/D/ϕ_d/d. Denote γ₁ = (γ_1p + γ_1m)/2, the average of the two heterozygous genotypic relative risks. It follows immediately that 1 ≤ γ_1p, γ_1m ≤ γ₂, and 1 ≤ γ₁ ≤ γ₂. On the basis of those three genotypic relative risks, we have Δ = [pγ₂ + (q − p)γ₁ − q]/(p²γ₂ + 2pqγ₁ + q²), I/ϕ = (γ_1p − γ_1m)/[2(p²γ₂ + 2pqγ₁ + q²)], and R/ϕ = (γ₂ − 2γ₁ + 1)/(p²γ₂ + 2pqγ₁ + q²). It is noted in the following text that the power approximation using a normal distribution depends on the values of three genotypic relative risks γ_1p, γ_1m, and γ₂, not directly on the original four risks ϕ_d/d, ϕ_D/d, ϕ_d/D, and ϕ_D/D. This facilitates the choice of the risks in the simulation studies. For simplicity, the homozygote risk ϕ_D/D is set to 0.8 in the simulation study.

For a given γ₂, γ₁ can take any value ranging from 1 to γ₂. Denote an arbitrary γ₁ in [1, γ₂] as 1 + β(γ₂ − 1) with β ∈ [0, 1]. β = 0 (γ₁ = 1) means that the mode of inheritance is recessive, (γ₁ = (1 + γ₂)/2) and I = 0 mean that it is additive (Knapp 1999), and β = 1 (γ₁ = γ₂) means that it is dominant. Either the recessive or the dominant mode implies that the degree of imprinting is zero. For illustration, we choose β to be , , and or equivalently γ₁ to be (3 + γ₂)/4, (1 + γ₂)/2, and (1 + 3γ₂)/4, equally spaced in the range of 1 and γ₂. When γ₁ (β) is given, we can derive the range of the degree of imprinting I from the diamond of inheritance (Strauch et al. 2000) as follows: [ϕ_d/dmin(β, 1 − β)(1 − γ₂), ϕ_d/dmin(β, 1 − β)(γ₂ − 1)]. Particularly, when γ₁ = (3 + γ₂)/4 () or γ₁ = (1 + 3γ₂)/4 (), ϕ_d/d(1 − γ₂)/4 ≤ I ≤ ϕ_d/d(γ₂ − 1)/4; when γ₁ = (1 + γ₂)/2, ϕ_d/d(1 − γ₂)/2 ≤ I ≤ ϕ_d/d(γ₂ − 1)/2.

Let θ_f and θ_m be, respectively, the female and male recombination rates in meioses between the marker locus and a DSL and θ = (θ_f + θ_m)/2 denote the sex-averaged recombination rate. It is obvious that the null hypothesis of no linkage is equivalent to ; i.e., . It is reported that θ_f > θ_m for most chromosomal regions, and the recombination rate for human females is greater than that for human males by an average of 60% (Fann and Ott 1995; Broman et al. 1998). The coefficient of LD between the marker locus and a DSL is , where is the frequency of haplotype DM₁. The four population haplotype frequencies are then, respectively, , , , and . The marker allele M₁ in this article is taken to be positively associated with the disease allele D; i.e., δ > 0. Note that the replacement of M₁ by M₂ will change the sign of δ but not its magnitude.

Hardy–Weinberg equilibrium in the parental generation is assumed throughout this article, and the frequencies of three genotypes 0, 1, and 2 at the marker locus are then g′², 2gg′, and g², respectively. For n independent families, each one is characterized by the marker genotype trio FMC, where F, M, and C represent the marker genotype of the father, mother, and affected child, respectively. On the basis of the marker genotype trio, all the families are classified to 15 categories. Let N_FMC denote the number of families that fall into category FMC (see Table 1 for more details). We are interested in the conditional probability of the family in the jth category given that the child is a case, which is denoted by s_j (1 ≤ j ≤ 15). Bayes' theorem is employed to get these conditional probabilities and the detailed expressions of s_j (1 ≤ j ≤ 15) are reported in Zhou et al. (2007). It is noted that {N_FMC} follows a multinomial distribution with parameters (n, s₁,…, s₁₅).

For 1 ≤ j ≤ 15, let us consider an arbitrary family in the jth category. It is feasible to count the times marker allele M₁ was transmitted and not transmitted from the heterozygous parents to their affected child, which are denoted, respectively, by u_j and v_j. Moreover, let denote the number of times that marker allele M₁ was transmitted/not transmitted from the heterozygous father (mother) to the affected offspring, and let be 1 if the father carries more copies of M₁ than the mother does and 0 otherwise and be 1 if the mother carries more copies of M₁ than the father does and 0 otherwise. The detailed values of , , , , , and (1 ≤ j ≤ 15) are shown in Table 1.

TDT with imprinting:

The total numbers of times of allele M₁ transmitted and not transmitted from heterozygous parents to their offspring among n independent case–parents trios can be expressed as u^TN and v^TN, respectively, where , v = , and N = .

So the TDT statistic can be expressed as(1)Note that the conventional TDT (Spielman et al. 1993) is actually the square of the right side of Equation 1. It is concluded in the appendix that the asymptotic distribution of the TDT (1) is a normal distribution with mean and variance σ²; see Corollary 1 in the appendix for the detailed expressions of the mean μ and variance σ². Furthermore, Corollary 2 shows that this asymptotic distribution is a standard normal under the null hypothesis of no linkage. So, in the presence of linkage disequilibrium, the TDT (1) can still be used to test for linkage when the genes are imprinted and when the recombination fractions are sex specific. The validity of the TDT will be verified in the simulation study by checking the values of the empirical type I error rates.

For a given significance level α, the rejection region of the two-sided test of the null hypothesis of no linkage is |TDT| > z_α/2, where z_α/2 is the upper α/2 point of a standard normal distribution (e.g., when α = 0.05, z_α/2 = 1.96). Thus we have the power approximation formula,(2)where Z is a standard normal random variable. The accuracy of the power approximation Equation 2 will be validated in the simulation study. Note that the principle of the power approximation Equation 2 is the same as that in the first approximation method developed in Knapp (1999). We can also employ the second approximation method in Knapp (1999) and the χ²-approximation method in Deng and Chen (2001) to estimate the power of the TDT. Our simulation results show that these three methods have equally good performance.

Imprinting effects on the power of the TDT:

It is interesting to find from the expressions of the mean and variance of the TDT (1) that the degree of imprinting I is always accompanied by the difference between the sex-specific recombination fractions θ_f − θ_m, which implies that this difference is vital in the evaluation of imprinting effects on the power of the TDT. One direct consequence is that the power of the TDT depends only on the sum of two heterozygote risks ϕ_D/d + ϕ_d/D, but not on their difference, or equivalently the degree of imprinting I, when the female and male recombination rates are identical (θ_f = θ_m).

To investigate extensively the imprinting effects on the power of the TDT, various sets of parameter values have been selected for power calculations. According to the numerical results, the variances σ² are very close to one and the value of |μ| is so large that either the first or the second probability on the right-hand side of Equation 2 is almost zero. So the required sample size n to achieve a given power, e.g. 80%, assuming no imprinting can be solved through 2δ²Δ²(1 − 2θ)²n = (z_0.025 + z_0.2)² (2gg′ + δΔ(1 − 2g)). It follows immediately that the analytical power of the TDT with this sample size n when the degree of imprinting I varies is(3)where Φ(·) is the cumulative probability function of a standard normal random variable. It is clear from Equation 3 that the power of the TDT is monotonic in I.

Note thatand the range of the degree of imprinting I is [ϕ_d/dmin (β, 1 − β)(1 − γ₂), ϕ_d/dmin(β, 1 − β)(γ₂ − 1)]. So the range of I/(ϕΔ) in Equation 3 is independent of γ₂, δ, and g, which implies that the pattern of the analytical power of the TDT associated with different values of γ₂, δ, and g vs. the degree of imprinting I remains almost the same except for the different scaling of I. This property is also observed while plotting the graphs of the analytical power (calculated from Equation 2), having different parameter values, against the degree of imprinting. Thus, it could be sufficient to show only the graphs corresponding to, for example, γ₂ = 4, δ = 0.9δ_max = 0.9 min(p(1 − g), g(1 − p)) (Deng and Chen 2001), and g = 0.4. We compare the powers calculated, respectively, by Equations 2 and 3 with various sets of parameter values and find that they are almost identical.

Moreover, when γ₁ = (1 + γ₂)/2 (), we have ϕΔ = ϕ_d/d(γ₂ − 1)/2, which is independent of p. So the power of the TDT is independent of p if γ₁ = (1 + γ₂)/2 (the additive mode of inheritance). Generally, for any given γ₂ and γ₁ = 1 + β(γ₂ − 1), the power of the TDT, as a function of the degree of imprinting I ranging from ϕ_d/dmin(β, 1 − β)(1 − γ₂) to ϕ_d/dmin(β, 1 − β)(γ₂ − 1), attains its minimum and maximum, Φ(z_0.2 − min(β, 1 − β)(θ_f − θ_m)(z_0.025 + z_0.2)/((1 − 2θ)(βq + (1 − β)p))) and Φ(z_0.2 + min(β, 1 − β)(θ_f − θ_m)(z_0.025 + z_0.2)/((1 − 2θ)(βq + (1 − β)p))) at the two boundary points of I, respectively, under the situation of θ_f > θ_m. It can be proved that min(β, 1 − β)/(βq + (1 − β)p) is a monotonically increasing function of β for and is a monotonically decreasing function of β when . It follows at once that the minimum and maximum powers over β and I are attained at and I = ±ϕ_d/d(1 − γ₂)/2 and are Φ(z_0.2 − (θ_f − θ_m)(z_0.025 + z_0.2)/(1 − θ_f − θ_m)) and Φ(z_0.2 + (θ_f − θ_m)(z_0.025 + z_0.2)/(1 − θ_f − θ_m)), respectively, which depend only on the two sex-specific recombination fractions. Note that and I = ϕ_d/d(1 − γ₂)/2 if and only if and , i.e., complete paternal imprinting, and and I = ϕ_d/d(γ₂ − 1)/2 if and only if and , i.e., complete maternal imprinting. Thus the analytical power attains its minimum and maximum in the cases of complete paternal imprinting and complete maternal imprinting, respectively.

The interval [Φ(z_0.2 − (θ_f − θ_m)(z_0.025 + z_0.2)/(1 − θ_f − θ_m)), Φ(z_0.2 + (θ_f − θ_m)(z_0.025 + z_0.2)/(1 − θ_f − θ_m))] is termed the maximum range of imprinting effects on the power of the TDT. For example, when θ_f = 0.146 and θ_m = 0.084, which are the sex-specific recombination fractions between ABO and the locus for the nail-patella syndrome (NPS1) (Ott 1999), the maximum range of imprinting effects on the power of the TDT is [0.7311, 0.8571]; when θ_f = 0.05 and θ_m = 0.02, the maximum range of imprinting effects is [0.7737, 0.8243]; when θ_f = 0.01 and θ_m = 0.006, the maximum range is [0.7968, 0.8032]. The parent-of-origin effect on the power of the TDT is negligible in the last case.

In short, to assess thoroughly the parent-of-origin effects on the power of the TDT, it is first advised to draw the graph of the power (Equation 3) vs. the degree of imprinting when parameters γ₂, β, and p are taken as 4, , and , respectively, which exhibits the maximum range of imprinting effects on the power of the TDT. If it is observed from the graph that the imprinting effect on the power of the TDT is substantial, then more graphs for specific parameter values may be needed. This greatly facilitates the assessment of the effect of genomic imprinting on the power of the TDT since lots of parameters are involved. The parent-of-origin effects on the power depend largely on the female and male recombination fractions and are negligible when their difference is small, e.g., both rates <1 cM.

Improve the power of the test:

The TDT (1) essentially tests for the equality of the expected number of transmissions of allele M₁ from heterozygous parents to the affected children and that of nontransmissions (Spielman et al. 1993). Taking parent-of-origin effects into account, it is natural to stratify the transmission/nontransmission numbers according to whether the father or the mother is the source.

In fact, the difference between the probability of the marker allele M₁ being transmitted from the heterozygous father to the affected child and the probability of the marker allele M₁ not being transmitted (equivalent to M₂ being transmitted) from the heterozygous father to the affected child can be given as(4)and similarly we have(5)The details of the derivation of Equations 4 and 5 are sketched in the appendix. It follows immediately from Equations 4 and 5 that the difference between the probabilities of M₁ being transmitted and not being transmitted is(6)

When the two heterozygous genotypic risks ϕ_D/d and ϕ_d/D vary between ϕ_d/d and ϕ_D/D and the other parameter values remain unchanged, it is obvious that when the degree of imprinting I changes from the leftmost point to the rightmost point, i.e., from the case of complete paternal imprinting to the case of complete maternal imprinting, the quantity p(ϕ_D/D − ϕ_d/D) + q(ϕ_D/d − ϕ_d/d) changes from 0 to the maximum ϕ_D/D − ϕ_d/d, while the quantity p(ϕ_D/D − ϕ_D/d) + q(ϕ_d/D − ϕ_d/d) changes from the maximum ϕ_D/D − ϕ_d/d to 0. Equivalently speaking, in the case of complete paternal imprinting, the difference between the numbers of marker allele M₁ being transmitted and nontransmitted from the heterozygous fathers to the children would provide strong evidence for linkage while the difference between the numbers of marker allele M₁ being transmitted and nontransmitted from the heterozygous mothers to the children would provide weak evidence for linkage. In the case of completely maternally imprinting, the situation is the reverse. So it is expected that the TDT_p, the paternal version of the TDT, would provide more evidence for linkage when the positive value of the degree of imprinting is large. Similarly, the TDT_m, the maternal version of the TDT, would provide more evidence for linkage when the negative value of the degree of imprinting is large. Detailed expressions of the TDT_p and the TDT_m are given in the following section. A suitable combination of the TDT, the TDT_p, and the TDT_m is expected to be more powerful than the TDT.

When there is little imprinting, the TDT is expected to be powerful in conducting linkage analysis. It is also understood from Equation 6 that when θ_f = θ_m, the difference between probabilities of the marker allele M₁ being transmitted and not being transmitted from heterozygous parents to offspring is independent of the difference of two heterozygous genotype risks ϕ_D/d and ϕ_d/D, but depends on the sum of these two risks. So it again implies that the power of the TDT is independent of the degree of imprinting I when θ_f = θ_m.

Proposed test statistic:

Note that the numbers of paternal transmission and paternal nontransmission of the marker allele M₁ are, respectively, T_p = N₁₀₁ + N₁₂₂ + N₁₁₂ and NT_p = N₁₀₀ + N₁₂₁ + N₁₁₀, and the numbers of maternal transmission and maternal nontransmission of the marker allele M₁ are, respectively, T_m = N₂₁₂ + N₀₁₁ + N₁₁₂ and NT_m = N₂₁₁ + N₀₁₀ + N₁₁₀; see Table 1 for details. The cell N₁₁₁ is ignored in counting T_p, NT_p, T_m, and NT_m because the origin of M₁ in the affected child is ambiguous. The numbers of total transmission and nontransmission are then, respectively, T = T_p + T_m + N₁₁₁ and NT = NT_p + NT_m + N₁₁₁. Using the constant vectors , , , and with components given in Table 1, the TDT_p and the TDT_m can be expressed, respectively, as(7)(8)The asymptotic distributions of the TDT_p and the TDT_m are given in the appendix.

Zhou et al. (2007) proposed the parent-of-origin effects test statistic (POET) to test for parent-of-origin effects, which essentially tests for the equality of the expected numbers of two groups of families, the first one comprising the families in which the father carries more copies of M₁ than the mother does, and the second group comprising the families in which the mother carries more copies of M₁ than the father does. The POET (Zhou et al. 2007) can be expressed as(9)where and with components given in Table 1. The asymptotic distribution of the POET is given in Equation A4 and is a standard normal distribution under the null hypothesis of no imprinting. On the basis of Equation A4, a large positive value of the POET may infer maternal imprinting, or equivalently paternal expression, and a large negative value of the POET may infer paternal imprinting, or equivalently maternal expression.

In conducting linkage analysis, it is discussed in the previous section that the TDT_p could be more powerful than the TDT for a large positive value of I, and the TDT_m could be more powerful than the TDT for a large negative value of I. When the degree of imprinting is small or does not exist, we keep closely to the proposed statistic for the TDT, which is commonly used and is powerful for testing for linkage. Thus, we propose an extension of the transmission disequilibrium test incorporating imprinting (TDTI)(10)where I_{{comparison statement}} = 1 when the comparison statement holds and is 0 otherwise, and α′ is the significance level in testing for imprinting by the POET. It is proved in the appendix that all the joint distributions of the POET and the TDT_m/TDT/TDT_p are asymptotically normal. So the asymptotic distribution of the TDTI is a mixture of three two-dimensional normal distributions. Furthermore, under the null hypothesis of no linkage, the POET and the TDT_m/TDT/TDT_p are asymptotically independent and TDTI .

For a significance level α, the rejection region about the null hypothesis of no linkage in the presence of linkage disequilibrium, i.e., , can be expressed as |TDTI| > z_α/2 and the power of the TDTI is P(|TDTI| > z_α/2). This significance level α may be different from α′ on the right side of Equation 10. For simplicity, α and α′ are taken to be 0.05 in our simulation study. The power approximation formula for the TDTI is then(11)which is asymptotically the sum of three two-dimensional normal distributions.

Sizes of the TDT and the TDTI:

To investigate the validity of the TDT and the TDTI, it is necessary to check the sizes of the TDT and the TDTI, i.e., the proportion of rejecting the null hypothesis of no linkage when θ_f = θ_m = 0.5. For illustration purposes, we choose the homozygote risk ϕ_d/d = 0.1; the disease allele frequency p = 0.01, 0.1, 0.5, and 0.8; and the sample size n = 100, to evaluate the sizes of the TDT and the TDTI. For the completeness of this investigation, we choose the following 13 representative pairs of and , which are scattered uniformly in the diamond (Strauch et al. 2000) composed of {(, ) | 1 ≤ , ≤ γ₂}: (, ) = (γ₂, γ₂), ((1 + γ₂)/2, γ₂), ((1 + 3γ₂)/4, (1 + 3γ₂)/4), (γ₂, (1 + γ₂)/2), (1, γ₂), ((3 + γ₂)/4, (1 + 3γ₂)/4), ((1 + γ₂)/2, (1 + γ₂)/2), ((1 + 3γ₂)/4, (3 + γ₂)/4), (γ₂, 1), (1, (1 + γ₂)/2), ((3 + γ₂)/4, (3 + γ₂)/4), ((1 + γ₂)/2, 1), (1, 1). It is noted that (, ) = (γ₂, γ₂) corresponds to the common dominant mode of inheritance, (, ) = ((1 + γ₂)/2, (1 + γ₂)/2) corresponds to the additive mode of inheritance, (, ) = (1, 1) corresponds to the common recessive mode of inheritance, (, ) = (γ₂, 1) indicates complete maternal imprinting, and (, ) = (1, γ₂) indicates complete paternal imprinting.

For those 13 pairs of values of (, ), we estimated the actual sizes of the TDT and the TDTI. The results are listed in Table 2, where and are expressed equivalently as γ₁ and I. All the entries in Table 2 show that the sizes of the TDT and the TDTI are consistent with the nominal 0.05, which signifies the validity of the TDT and the TDTI. Due to the uniform distribution of those 13 pairs of (, ), the simulated sizes of the TDT and the TDTI listed in Table 2 show that it is valid to use the TDT and the TDTI to test for linkage when the disease gene is imprinted and the recombination fractions are sex specific.

In fact, we have selected a number of other parameter values involved in evaluating the TDT and TDTI. The patterns of sizes of the TDT and the TDTI are similar to those reported above, and these results are omitted for brevity.

Power approximation formulas of the TDT and the TDTI:

The precision of the power approximation formulas (2) and (11) is evaluated by simulation and their comparisons are made on the basis of the simulation results. To investigate the accuracy of Equations 2 and 11 in calculating the powers of the TDT and the TDTI, various choices of the disease allele frequency p and the homozygote risk ϕ_d/d are taken, and the following three scenarios on imprinting are considered: (a) complete paternal imprinting, (b) general paternal imprinting, and (c) no imprinting, while the other parameter values remain unchanged. For a given set of parameter values, the sample sizes necessary to achieve 80% power in the TDT and the TDTI are calculated by Equations 2 and 11, respectively. The actual powers of the TDT and the TDTI corresponding to these particular sample sizes are then obtained by simulation with 10,000 replicates.

Table 3 lists the results when the female and male recombination fractions are 0.02 and 0.01, respectively. All the entries of Table 3 show a strong agreement between the analytical powers and the actual powers. Table 3 demonstrates that both Equations 2 and 11 perform well. It is also observed from Table 3 that the sample sizes needed by the TDTI are always smaller than those needed by the TDT in the cases of complete/general paternal imprinting, and conversely the sample sizes for the TDT are smaller in the case of no imprinting. This may imply that the power of the TDTI is greater than that of the TDT when the genes are paternally imprinted, and the power of the TDTI is slightly smaller than that of the TDT when the genes are not imprinted.

It is observed from Table 3 that the sample sizes necessary to gain 80% power in the TDT decrease in the order of complete paternal imprinting, general paternal imprinting, and no imprinting. Actually, we also investigate the required sample sizes in the case of complete/general maternal imprinting (results not shown for brevity) and find that these sample sizes decrease in the order of no imprinting, general maternal imprinting, and complete maternal imprinting. However, for the TDTI, it is observed that the sample size necessary to gain 80% power in the cases of complete/general imprinting is smaller than that in the case of no imprinting. This is equivalent to saying that the power of the TDTI in the case of complete/general imprinting is greater than that in the case of no imprinting when the sample size is fixed.

We also obtained similar findings when the parameters take other values, and they are not shown here for brevity. The power comparison is conducted further in the following two sections.

Powers of the TDT and the TDTI:

To compare the powers of the TDT and the TDTI, a simulation with a large number of different parameter values was conducted. For simplicity, we show just the simulation results with 10,000 replicates in Table 4 when the sample size n = 100/200 with three scenarios concerning θ_f and θ_m—(a) θ_f = θ_m = 0, (b) θ_f = θ_m = 0.001, and (c) θ_f = 0.001 and θ_m = 0.0006—and three scenarios concerning imprinting—(a) complete paternal imprinting, (b) general paternal imprinting, and (c) no imprinting. Columns 4 and 5, 6 and 7, and 8 and 9 in Table 4 exhibit the powers of the TDT and the TDTI with complete paternal imprinting genes, general paternal imprinting genes, and no imprinting genes, respectively. Table 4 shows that for complete/general paternal imprinting genes, the TDTI has a greater power than the TDT for testing for linkage, while the TDT has a slightly higher power than the TDTI for no imprinting genes. For example, when p = 0.1, ϕ_d/d = 0.1, n = 100, θ_f = 0.001, and θ_m = 0.0006, the powers of the TDT and the TDTI are 60.50% vs. 70.77% for complete paternal imprinting genes and 60.25% vs. 64.02% for general paternal imprinting genes; in both cases the TDTI has higher power than the TDT, while for no imprinting genes the TDT (power = 60.98%) has <1% higher power than the TDTI (power = 60.22%). Note that the TDT is <1% more powerful than the TDTI in all except one case of no imprinting genes in Table 4. The similar findings in the case of complete/general maternal imprinting are omitted for brevity.

Power comparisons of the TDT and the TDTI:

The previous section exhibits the powers of the TDT and the TDTI for a variety of parameter values and shows that the TDTI outperforms the TDT in testing for linkage in the cases of complete/general imprinting, where the degree of imprinting or sample size is fixed at some values. In this section, we continue to perform power comparisons of the TDT and the TDTI and focus on the pattern of the powers vs. various degrees of imprinting or sample sizes when all other parameter values remain unchanged.

We fix the parameter values as follows: disease allele frequency p = 0.1, the homozygote risk ϕ_d/d = 0.12, and two recombination fractions θ_f = 0.001 and θ_m = 0.0006. Figure 1a exhibits the simulated powers of the TDT and the TDTI with 100,000 replicates when the degree of imprinting I varies in the interval [(ϕ_d/d − ϕ_D/D)/2, (ϕ_D/D − ϕ_d/d)/2] and the sample size n = 200. Figure 1a illustrates that the power of the TDTI is greater than that of the TDT for a moderate to large degree of imprinting, and the improvement may be substantial in the case of complete paternal/maternal imprinting. For example, in the case of complete maternal imprinting, the powers of the TDTI and the TDT are, respectively, 89.40 and 79.54%. Also noted is that for a small degree of imprinting, the TDT is slightly more powerful to detect linkage than the TDTI. For example, in the case of no imprinting, the powers of the TDT and the TDTI are, respectively, 79.30 and 78.54%.

• Download figure
• Open in new tab
• Download powerpoint

Figure 1.—

The actual powers of the TDT and TDTI are plotted against (a) the degree of imprinting I ∈ [(ϕ_d/d − ϕ_D/D)/2, (ϕ_D/D − ϕ_d/d)/2] when the sample size n = 200 and (b) the sample size n ∈ [100, 250] in increments of 15 in the case of general paternal imprinting, having p = 0.1, g = 0.4, δ = 0.054, ϕ_D/D = 0.8, ϕ_d/d = 0.12, I = −0.3, θ_f = 0.001, and θ_m = 0.0006. Powers are based on 100,000 replicates and assessed at the 5% level.

Overall, the proposed TDTI for testing for linkage outperforms the conventional TDT when the degree of imprinting is moderate to large. When the degree of imprinting is negligible, there is a slight loss of power from using the TDTI.

Figure 1b plots the powers of the TDT and the TDTI vs. the sample size n ∈ [100, 250] in increments of 15 and the degree of imprinting I = 4(ϕ_d/d − ϕ_D/D)/9 (general paternal imprinting). Figure 1b illustrates that the power of the TDTI is greater than that of the TDT. For example, the powers of the TDT/TDTI are, respectively, 69.52/77.19% for n = 160, 73.72/81.42% for n = 175, and 77.34/84.86% for n = 190. It is noted that for small/large sample sizes, the powers of both the TDT and the TDTI are small/large, and so the improvement of the TDTI over the TDT would be less significant. The reason is that both powers are close to 0/1 under those situations. The pattern of the powers of the TDT/TDTI in the case of general maternal imprinting is similar and the details are not reported here for brevity.

Derivation of Equations 4 and 5:

For illustration, we give the details of the derivation of Equation 4 (the counterpart for Equation 5 is skipped):where the haplotype on the left side of the slash (/) is paternal and the one on the right side is maternal, the asterisk denotes an arbitrary allele, and, for example, C_p = DM₁ indicates that the paternal haplotype of the child is DM₁. The last equality is based on the following four equations: P(F = M₁M₂, C_p = DM₁) = pgg′ + g′δ − δθ_m, P(F = M₁M₂, C_p = DM₂) = pgg′ − gδ + δθ_m, P(F = M₁M₂, C_p = dM₁) = qgg′ − g′δ + δθ_m, and P(F = M₁M₂, C_p = dM₂) = qgg′ + gδ − δθ_m. The principle of their derivation is referred to in Zhou et al. (2007). Thus we have Equation 4.

Joint distribution of two TDT-like statistics:

We first give a theorem to describe the joint distribution of two TDT-like statistics and two corollaries to describe the asymptotic distributions of the TDT, the TDT_p, and the TDT_m. Then we provide a theorem that shows the asymptotic distributions of the TDTI.

Theorem 1. Let u₁ = (u₁₁,…, u_1m)^T, v₁ = (v₁₁,…, v_1m)^T, u₂ = (u₂₁,…, u_2m)^T, v₂ = (v₂₁,…, v_2m)^T be four constant vectors of length m and (N₁,…, N_m, N_m+1)^T be a multinomially distributed random variable of size and with parameters (r₁,…, r_m, r_m+1), where . Denote N = (N₁,…, N_m)^T; then we have in lawwhere , b₁ = , , , σ₁₁ = β^T(u₁, v₁)^TΣ(u₁, v₁)β, σ₁₂ = σ₂₁ = β^T(u₁, v₁)^TΣ(u₂, v₂)γ, σ₂₂ = γ^T(u₂, v₂)^TΣ(u₂, v₂)γ, Σ = diag(r) − rr^T, r = (r₁,…, r_m)^T, β = ((x₁ + 3y₁)/(2(x₁ + y₁)^3/2), −(3x₁ + y₁)/(2(x₁ + y₁)^3/2))^T, and γ = ((x₂ + 3y₂)/(2(x₂ + y₂)^3/2), −(3x₂ + y₂)/(2(x₂ + y₂)^3/2))^T are evaluated at x₁ = a₁, y₁ = b₁, x₂ = a₂, and y₂ = b₂, respectively. Particularly, we have

Proof. Note that N/n is the maximum-likelihood estimator of the parameter vector r and from the asymptotic normality of the maximum-likelihood estimator, we have in law(Rao 1973). It follows immediately thatLetthen we havewhereNote that the derivatives are evaluated at x₁ = a₁, y₁ = b₁, x₂ = a₂, and y₂ = b₂. After some transformations, we complete the proof.

Remark. By matrix multiplication, we find that(A1)where , , and . Moreover, when and for all j, we have(A2)Note that Equation A2 in this situation is just the theorem in Zhou et al. (2007). We also find thatIn particular, when a₁ = b₁ and a₂ = b₂, we have(A3)

Let u = , , , , , , and with components listed in Table 1; by using Theorem 1 and Equations A1 and A2, we have the following two corollaries after tedious algebra:

Corollary 1. For the test statistic (1), we have in lawwhere

For the test statistic (7), we have in law where

For the test statistic (8), we have in law where

For the test statistic (9), we have in law (A4) (Zhou et al. 2007), where

Corollary 2. When , we have , ,and ; when I = 0, we have .

Proof. When , it is obvious that a = b, d₁ = a + b, a_p = b_p, and a_m = b_m. When I = 0, we have a_I = b_I. Furthermore, . Thus we have proven Corollary 2.

Theorem 2 (distribution of the TDTI). Under the null hypothesis of no linkage, i.e., , the POET and the TDT/TDT_p/TDT_m are asymptotically independent, and .

Proof. When , we have that a = b, a_p = b_p, and a_m = b_m. So by Equation A3 and the eight constant vectors u_I, v_I, u, v, u_p, v_p, u_m, and v_m with components listed in Table 1, the covariances of the POET and the TDT_p, the TDT_m, and the TDT are, respectively,After tedious algebra, we haveIt follows immediately that Cov(POET, TDT) = Cov(POET, TDT_p) = Cov(POET, TDT_m) = 0 when . So the POET and the TDT/TDT_p/TDT_m are asymptotically independent. Moreover, by Corollary 2, it is easy to show that under the null hypothesis of no linkage. Hence the theorem is proved.