All methodology is from the first section of the appendix of the linked paper. The paper cited pages 81-87 of Genetics and Analysis of Quantitative Traits. I read from chapter 4 up until those pages to understand the method conceptually. Every niceness assumption is made except for “no shared environment.” For example, “no assortative mating.”
Changing some notation: rMZ=S+1, i.e. we normalize so that the total “variance due to genes” is 1. We assume that the variance due to shared environment is the same for twins and non-twins, SMZ=SDZ=S. This is a standard assumption in ACE, and it seems reasonable.R2 will represent the “nonlinear part of the effect due to genes,” i.e. that due to epistasis and dominance.VA=1−R2 is the effect due to alleles, what you called s.
Facts:
2(rMZ−rDZ−12)<R2≤4(rMZ−rDZ−12) always. (1)
When S=0:
2(12−rDZrMZ)<R2≤4(12−rDZrMZ). (2)
Note that R2≤1, so the upper bound becomes trivial when either score is 0.25.
Explanation:
We can decompose rMZ=S+VA+R2. We can decompose this further into R2=∑i,j≥0,(i,j)≠(1,0)VAiDj. That is, we’re taking the nonlinear part and decomposing it into interactions involving alleles across i loci and dominance effects in j loci. To understand dominance effects, note that a locus can have 0, 1, or 2 instances of an allele. The respective phenotypes resulting from these might not be produced by any linear function on alleles, because not every three points are colinear. The dominance term is the error resulting from a linear regression. If we were haploid, we wouldn’t have to deal with this. So for example, VA5D3 refers to phenotype effects that only appear when there is a specific combination of two alleles at three separate loci, and are multilinear in the alleles occurring at five other loci.
Interpreting this in context, rDZ=SDZ+12VA+∑i,j≥0,(i,j)≠(1,0)2−i−2jVAiDj. Now we can justify our conclusions. Note that the third term is at most 14R2 but has no lower bound. When we have to write it out, we’ll call it X. Remember that R2 is exactly the proportion of variance due to genes that cannot be captured by a polygenic score, the “phantom heritability.” The paper is concerned with how substantial R2 means VA<1, so that if the polygenic score is close to VA people will assume there is missing heritability when it reality the polygenic score is perfect and the heritability is simply nonlinear.
The ACE Estimate:
2(rMZ−rDZ)=VA+∑i,j≥0,(i,j)≠(1,0)(2−2−i−2j+1)VAiDj. This disagrees with the figure in the appendix of the paper. I believe they made an arithmetic error, but it is possible I made a conceptual error. Recalling VA=1−R2, rMZ−rDZ−12=∑i,j≥0,(i,j)≠(1,0)(12−2−i−2j)VAiDj. Those constant terms in the sum go as low as 14 and arbitrarily close to 12, so by taking the bounds and dividing we recover (1).
The Rule of Thumb:
rDZrMZ=S+12VA+X2S+VA+R2=SS+1+12(S+1)(1−R2)+1S+1X. Remember the bounds on X, we can write X=αR2, where α∈(0,14]. Combining this with the middle −R2 term we have 12+S2S+1−αR2 where α∈[14(S+1),12(S+1)). Doing the arithmetic
I don’t yet rigorously understand how R2 is decomposed into epistasis and dominance. The book gives only an intuition and not a proof. It is very ad hoc. Edit: As of yesterday, I now understand.
All methodology is from the first section of the appendix of the linked paper. The paper cited pages 81-87 of Genetics and Analysis of Quantitative Traits. I read from chapter 4 up until those pages to understand the method conceptually. Every niceness assumption is made except for “no shared environment.” For example, “no assortative mating.”
Changing some notation: rMZ=S+1, i.e. we normalize so that the total “variance due to genes” is 1. We assume that the variance due to shared environment is the same for twins and non-twins, SMZ=SDZ=S. This is a standard assumption in ACE, and it seems reasonable.R2 will represent the “nonlinear part of the effect due to genes,” i.e. that due to epistasis and dominance.VA=1−R2 is the effect due to alleles, what you called s.
Facts:
2(rMZ−rDZ−12)<R2≤4(rMZ−rDZ−12) always. (1)
When S=0:
2(12−rDZrMZ)<R2≤4(12−rDZrMZ). (2)
Note that R2≤1, so the upper bound becomes trivial when either score is 0.25.
Explanation:
We can decompose rMZ=S+VA+R2. We can decompose this further into R2=∑i,j≥0,(i,j)≠(1,0)VAiDj. That is, we’re taking the nonlinear part and decomposing it into interactions involving alleles across i loci and dominance effects in j loci.
To understand dominance effects, note that a locus can have 0, 1, or 2 instances of an allele. The respective phenotypes resulting from these might not be produced by any linear function on alleles, because not every three points are colinear. The dominance term is the error resulting from a linear regression. If we were haploid, we wouldn’t have to deal with this.
So for example, VA5D3 refers to phenotype effects that only appear when there is a specific combination of two alleles at three separate loci, and are multilinear in the alleles occurring at five other loci.
Interpreting this in context, rDZ=SDZ+12VA+∑i,j≥0,(i,j)≠(1,0)2−i−2jVAiDj.
Now we can justify our conclusions. Note that the third term is at most 14R2 but has no lower bound. When we have to write it out, we’ll call it X.
Remember that R2 is exactly the proportion of variance due to genes that cannot be captured by a polygenic score, the “phantom heritability.” The paper is concerned with how substantial R2 means VA<1, so that if the polygenic score is close to VA people will assume there is missing heritability when it reality the polygenic score is perfect and the heritability is simply nonlinear.
The ACE Estimate:
2(rMZ−rDZ)=VA+∑i,j≥0,(i,j)≠(1,0)(2−2−i−2j+1)VAiDj. This disagrees with the figure in the appendix of the paper. I believe they made an arithmetic error, but it is possible I made a conceptual error.
Recalling VA=1−R2, rMZ−rDZ−12=∑i,j≥0,(i,j)≠(1,0)(12−2−i−2j)VAiDj. Those constant terms in the sum go as low as 14 and arbitrarily close to 12, so by taking the bounds and dividing we recover (1).
The Rule of Thumb:
rDZrMZ=S+12VA+X2S+VA+R2=SS+1+12(S+1)(1−R2)+1S+1X. Remember the bounds on X, we can write X=αR2, where α∈(0,14]. Combining this with the middle −R2 term we have 12+S2S+1−αR2 where α∈[14(S+1),12(S+1)). Doing the arithmetic
R2∈(2(S+1)(12+S2S+1−rDZrMZ),4(S+1)(12+S2S+1−rDZrMZ)]. Picking S=0 yields (2).
Comments:
I don’t yet rigorously understand how R2 is decomposed into epistasis and dominance. The book gives only an intuition and not a proof. It is very ad hoc.
Edit: As of yesterday, I now understand.