Coancestry

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# Coancestry
### Jinliang Yang
### Sept. 28, 2022

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# Individual Inbreeding: `\(F_X\)`

`\begin{align*}
F_{X} = (\frac{1}{2})^n(1 + F_{A})
\end{align*}`

- Where `\(n\)` is the number of individuals in the path from the individual’s sire (dad), through the common ancestor, to the dam (mom).

- If multiple common ancestors, must sum the individual estimates.

--
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The inbreeding coefficient of an individual X depends on the amount of common ancestry in its two parents.

# Coancestry (Kinship): `\(f_{XY}\)`

- Coancestry of two parents = inbreeding of their offspring

- Probability of IBD of two haplotypes, one drawn from each parent X and Y, symbolized by `\(f_{XY}\)`

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# Coancestry

Coancestry ( `\(f_{XY}\)` ) is the probability of two gametes, one from each parent (X and Y), will contain haplotypes that are IBD

### Example: `\(X \times Y\)` mating

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<img src="dog1.png" width=280>
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### Coancestry `\(f_{XY}\)`
`\begin{align*}
f_{XY} = & \frac{1}{4}Pr(x_1 	\equiv y_1) \\
& + \frac{1}{4}Pr(x_1 	\equiv y_2) \\
& + \frac{1}{4}Pr(x_2 	\equiv y_1) \\
& + \frac{1}{4}Pr(x_2 	\equiv y_2) \\
\end{align*}`
]

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# Case study 1: selfing

### Coefficient of coancestry with one’s self.

- If __ `\(x1x2\)`__ are sampled, the probability they are IBD depends on whether or not the individual could have obtained these gametes from a common ancestor

- This is reflected by the inbreeding coefficient of the individual __ `\(F_X\)`__

- If inbreeding = 0, this combination of gametes does NOT contribute to the possibility of IBD.

---

# Case study 1: selfing

### Coefficient of coancestry with one’s self.

`\begin{align*}
f_{XX} = & \frac{1}{4}Pr(x_1 	\equiv x_1) + \frac{1}{4}Pr(x_1 	\equiv x_2)  + \frac{1}{4}Pr(x_2 	\equiv x_1)  + \frac{1}{4}Pr(x_2 	\equiv x_2) \\
= & \frac{1}{2}(1 + F_X) \\
\end{align*}`

This result suggests that `\(f_{XX} = \frac{1}{2}\)` if individual X is a non-inbred with `\(F_X=0\)`.

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# Case study 2: Parent-Offspring

Draw gametes at random (one from each individual X and its progeny A), what is probability they are IBD (or `\(f_{XA}\)`)?

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# Case study 2: Parent-Offspring

Draw gametes at random (one from each individual X and its progeny A), what is probability they are IBD (or `\(f_{XA}\)`)?

#### Eight possible combinations

- 1) `\(x_1x_1\)` 2) `\(x_1x_2\)` 3) `\(x_2x_1\)` 4) `\(x_2x_2\)`

- 5) `\(x_1y_1\)` 6) `\(x_1y_2\)` 7) `\(x_2y_1\)` 8) `\(x_2y_2\)`

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# Case study 2: Parent-Offspring

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<img src="dog2.png" width=200>
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.pull-right[
#### Eight possible combinations
- 1) `\(x_1x_1\)` 2) `\(x_1x_2\)` 3) `\(x_2x_1\)` 4) `\(x_2x_2\)`

- 5) `\(x_1y_1\)` 6) `\(x_1y_2\)` 7) `\(x_2y_1\)` 8) `\(x_2y_2\)`

]

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- If gamete from X is `\(x_1\)`, and gamete from A is `\(y_1\)`, the probability of IBD depends on if there is a relationship between X and Y (the parents of A)

- This is quantified by the coefficient of coancestry between those parents `\(f_{XY}\)`

- In total, of the 8 possible combinations, 4 include a gamete from parent Y where this relationship holds

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# Case study 2: Parent-Offspring

.pull-left[
<div align="center">
<img src="dog2.png" width=200>
</div>
]

.pull-right[
#### Eight possible combinations
- 1) `\(x_1x_1\)` 2) `\(x_1x_2\)` 3) `\(x_2x_1\)` 4) `\(x_2x_2\)`

- 5) `\(x_1y_1\)` 6) `\(x_1y_2\)` 7) `\(x_2y_1\)` 8) `\(x_2y_2\)`
]

------
`\begin{align*}
f_{XA} = & \frac{1}{8}[Pr(x_1 	\equiv x_1) + Pr(x_1 	\equiv x_2)  + Pr(x_2 	\equiv x_1)  + Pr(x_2 	\equiv x_2) )\\
& + Pr(x_1 	\equiv y_1) + Pr(x_1 	\equiv y_2)  + Pr(x_2 	\equiv y_1)  + Pr(x_2 	\equiv y_2)] \\
= & \frac{1}{8}(1 + F_X+ F_X+1 + 4f_{XY})\\
= & \frac{1}{4}(1 + F_X + 2f_{XY})\\
\end{align*}`

Assuming no inbreeding, i.e., `\(F_X = 0\)` and `\(f_{XY}=0\)`

Coefficient of coancestry of parent-offspring, `\(f_{XA}=\frac{1}{4}\)`

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# Relationship vs Coancestry

### Coefficient of coancestry:

Pick an allele from X. The probability that you will __pick__ that allele in Y is the coefficient of coancestry.

### Coefficient of relationship:

Pick an allele from X. The probability that Y __has__ that allele is the coefficient of relationship

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Therefore: __coancestry = 1⁄2 relationship__

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# Relationship vs Coancestry

|  | Coancestry   | Relationship    | 
| :-------: | : ------ : | :-------: | 
| Parent-offspring  | 0.25    | 0.5    | 
| Full-siblings     | 0.25    | 0.5    |
| Half-siblings     | 0.125   | 0.25    |

- Full-sibs have two parents in common

- Half-sibs have one parent in common

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# The basic rule relating coancestries

#### What is coancestry of X and Y (or `\(f_{XY}=Pr(x \equiv y)\)`)?

Coancestry between X and Y depends on relationship among parents

`\begin{align*}
f_{XY} & = Pr(x \equiv y) \\
& = Pr(a 	\equiv c) + Pr(a 	\equiv d)  + Pr(b 	\equiv c)  + Pr(b 	\equiv d )
\end{align*}`

Consider one part of this equation: `\(Pr(a \equiv c)\)`
- This is the probability that a random allele from X is derived from individual A 
- And a randomly chosen allele from Y is from individual C 
- And these are IBD

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# The basic rule relating coancestries

`\begin{align*}
f_{XY} & = Pr(x \equiv y) \\
& = Pr(a 	\equiv c) + Pr(a 	\equiv d)  + Pr(b 	\equiv c)  + Pr(b 	\equiv d )
\end{align*}`

### Consider one part of this equation: `\(Pr(a \equiv c)\)`

`\begin{align*}
Pr(a 	\equiv c) & =  \frac{1}{4}\frac{1}{4}Pr(a_1 	\equiv c_1) + \frac{1}{4}\frac{1}{4}Pr(a_1 	\equiv c_2)  + \frac{1}{4}\frac{1}{4}Pr(a_2 	\equiv c_1)  + \frac{1}{4}\frac{1}{4}Pr(a_2 	\equiv c_2) \\
& = \frac{1}{16}(Pr(a_1 	\equiv c_1) + Pr(a_1 	\equiv c_2) + Pr(a_2 	\equiv c_1) + Pr(a_2 	\equiv c_2)) \\
& = \frac{1}{16}(4f_{AC}) = \frac{1}{4}(f_{AC})
\end{align*}`

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# The basic rule relating coancestries

### Repeat for the 3 other terms

`\begin{align*}
f_{XY} & = Pr(x \equiv y) \\
& = Pr(a 	\equiv c) + Pr(a 	\equiv d)  + Pr(b 	\equiv c)  + Pr(b 	\equiv d ) \\
& = \frac{1}{4}(f_{AC} + f_{AD} + f_{BC}+ f_{BD})
\end{align*}`

In summary, `\(f_{XY}\)` is equal to the average coefficient of coancestry between the parents of X and Y
- Avg coancestry between parents: AC, AD, BC, and BD

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# Relationship vs Coancestry

- Full-sibs have two parents in common

`\begin{align*}
f_{XY} & = Pr(x \equiv y) \\
& = \frac{1}{4}(f_{AA} + f_{AB} + f_{BA}+ f_{BB})
\end{align*}`

With no previous inbreeding or relationship, and `\(f_{AA}=f_{BB}\)`=1/2, `\(f_{XY}=1/4\)`