class: center, middle, inverse, title-slide .title[ # Pedigreed Population ] .author[ ### Jinliang Yang ] .date[ ### Feb. 26, 2024 ] --- # Inbreeding coefficient - The probability of any individual being an identical homozygous - Or the probability that a pair of haplotypes within an individual are IBD -- ### Inbreeding coefficent at generation `\(t\)` `\begin{align*} F_{t} =\frac{1}{2N_e} + (1 - \frac{1}{2N_e})F_{t-1} \end{align*}` > From F&M Ch3 without considering mutation This equation estimates the average inbreeding coefficient of __all individuals__ of a generation! -- ### How about the inbreeding coefficient for a given individual `\(F_X\)`? - Pedigree information can be leveraged to get more accurate results --- # Pedigreed population .pull-left[ <div align="center"> <img src="ped.png" width=400> </div> ] -- .pull-right[ - A “common ancestor” is an individual who could pass copies of haplotypes through both sides of the pedigree to some future offspring. - Letters for the haplotypes reflect the parent source. ] --- # Pedigreed population .pull-left[ <div align="center"> <img src="ped2.png" width=400> </div> ] .pull-right[ - A “common ancestor” is an individual who could pass copies of haplotypes through both sides of the pedigree to some future offspring. - Letters for the haplotypes reflect the parent source. - Haplotypes in genotypes ( `\(a_1a_1\)` ) and ( `\(a_2a_2\)` ) are identical copies of haplotypes from A. - Haplotypes in genotypes ( `\(a_1a_2\)` ) may be identical by descent if parents of A are related. - The probability that `\(a_1 \equiv a_2 = F_A\)`. - here, `\(\equiv\)` means IBD ] --- # Pedigreed population Question: what is the inbreeding of individual dog X? .pull-left[ <div align="center"> <img src="ped2.png" width=400> </div> ] -- .pull-right[ `\begin{align*} F_X = & Pr_X(a_1a_1) + Pr_X(a_2a_2) \\ & + Pr_X(a_1a_2)Pr(a_1 \equiv a_2) \\ \end{align*}` ] --- # Pedigreed population Question: what is the inbreeding of individual dog X? .pull-left[ <div align="center"> <img src="ped3.png" width=400> </div> ] .pull-right[ `\begin{align*} F_x = & Pr_X(a_1a_1) + Pr_X(a_2a_2) \\ & + Pr_X(a_1a_2)Pr(a_1 \equiv a_2) \\ = & (\frac{1}{2})^4 + (\frac{1}{2})^4 + 2(\frac{1}{2})^4F_A \\ = & (\frac{1}{2})^3 + (\frac{1}{2})^3F_A \\ = & (\frac{1}{2})^3(1 + F_A) \end{align*}` ] --- # Pedigreed population Question: what is the inbreeding of individual dog X? .pull-left[ <div align="center"> <img src="ped3.png" width=400> </div> ] .pull-right[ `\begin{align*} F_x = & Pr_X(a_1a_1) + Pr_X(a_2a_2) \\ & + Pr_X(a_1a_2)Pr(a_1 \equiv a_2) \\ = & (\frac{1}{2})^4 + (\frac{1}{2})^4 + 2(\frac{1}{2})^4F_A \\ = & (\frac{1}{2})^3 + (\frac{1}{2})^3F_A \\ = & (\frac{1}{2})^3(1 + F_A) \end{align*}` - Look for “paths” connecting the parents and the common ancestor - Or for example, __C A D = 3__ individuals (let this be n) ] --- # Individual Inbreeding Now we can write an expression for `\(F_X\)`: `\begin{align*} F_X = (\frac{1}{2})^n(1 + F_A) \end{align*}` -- - In the previous example, `\(F_X = (\frac{1}{2})^3(1 + F_A) = 1/8(1 + F_A)\)` -- - If there are more generations between the common ancestor and parents – change `\(n\)`. -- -------------- .pull-left[ <div align="center"> <img src="ped4.png" width=200> </div> ] -- .pull-right[ - Path: D B __A__ C E - `\(F_X = (\frac{1}{2})^5(1 + F_A)\)` ] --- # It is never be that simple ... .pull-left[ #### Inbreeding of X if `\(F_A=F_B=F_E=0\)`? <div align="center"> <img src="ped5.png" width=400> </div> ] -- .pull-right[ ### Paths for `\(F_X\)` - M K __G__ N - M K G D __B__ E H L N - M K C __A__ D G N - M J C __A__ D G N - M K G __E__ H L N ] --- # It is never be that simple ... .pull-left[ #### Inbreeding of X if `\(F_A=F_B=F_E=0\)`? <div align="center"> <img src="ped5.png" width=400> </div> ] .pull-right[ ### Paths for `\(F_X\)` - M K __G__ N (n=4) - M K G D __B__ E H L N (n=9) - M K C __A__ D G N (n=7) - M J C __A__ D G N (n=7) - M K G __E__ H L N (n=7) #### Summing all these: - Note: `\(F_G=1/8\)` `\begin{align*} F_X = (\frac{1}{2})^4(1 + F_G) + (\frac{1}{2})^9 + (\frac{1}{2})^7 \times 3 \end{align*}` - `\(F_{X}= 0.096\)` or `\(9.6\%\)` ] -- X is expected to have __9.6% fewer heterozygous loci__ than a non-inbreeding individual in the same population. --- # What does it mean for `\(F_X=9.6\%\)`? ### Assuming 4 loci segregating in the population #### The frequencies: - `\(p_{A1}=0.05\)`, `\(p_{B1}=0.1\)`, `\(p_{C1}=0.15\)`, `\(p_{D1}=0.45\)` -- #### `\(H_{exp}\)` in the population = `\(\frac{1}{m}\sum2pq\)` ```r h <- function(p){return(2*p*(1-p))} (h(p=0.05)+h(p=0.1)+h(p=0.15)+h(p=0.45))/4 ``` ``` ## [1] 0.25625 ``` - Avg across 4 loci = 0.26 -- - A non-inbreeding individual is expected to be heterozygous at 26% of its loci - Individual X ( `\(F_X = 0.096\)` ) is expected to be heterozygous at `\((1-0.096) \times 0.26 = 23.5\%\)` of the loci