Hardy-Weinberg Equilibrium

class: center, middle, inverse, title-slide

.title[
# Hardy-Weinberg Equilibrium
]
.author[
### Jinliang Yang
]
.date[
### Jan. 24, 2024
]

---

# Allele and genotype frequencies

.pull-left[
<div align="center">
<img src="dogs.jpg" height=150>
</div>
]

.pull-right[
- dog 1: AA AA __TT__ CC GG 
- dog 2: AA AA __CC__ CC GG
- dog 3: AA AA __CT__ CC GG 
- dog 4: AA AA __CT__ CC GG
- dog 5: AA AA __CC__ CC GG
]

Consider a diploid locus segregating for two alleles ( `$A_1$` and `$A_2$` ).
We usually define the less frequent allele (or minor allele) as the `$A_1$` allele.

### Allele frequency ( `$p$` and `$q$` ) 
Frequency/proportion of alleles of a particular identity at one locus

### Genotype frequency ( `$f_{11}$`, `$f_{12}$`, `$f_{22}$` )
Proportion of individuals with a specific genotype (combination of alleles)

---

# Allele and genotype frequencies

.pull-left[
<div align="center">
<img src="dogs.jpg" height=150>
</div>
]

.pull-right[
- dog 1: AA AA __TT__ CC GG 
- dog 2: AA AA __CC__ CC GG
- dog 3: AA AA __CT__ CC GG 
- dog 4: AA AA __CT__ CC GG
- dog 5: AA AA __CC__ CC GG
]

Let `$n$` be the total number of individuals in the population. Genotype frequency of __TT__ is:

```r
n = 5
n11 = 1
f11 = n11/n
f11
```

```
## [1] 0.2
```

The frequency of minor allele `$A_1$` in the population is then given by

`\begin{align*}
p = \frac{2n_{11}+n_{12}}{2n} = f_{11} + \frac{1}{2}f_{12}
\end{align*}`

---

# Hardy-Weinberg Equilibrium

#### A population is in HWE if it has constant allele and genotype frequencies __from generation to generation__

- Allele frequencies in parents predict allele and genotype frequencies in offspring

### Caveats

- A large, randomly-mating population for diploid species

- No selection, no mutation, no migration, etc.
  
---

# In Hardy-Weinberg Equilibrium

If allele frequencies in parent population = `$p$` & `$q$`, 
then, genotype frequencies in progeny after random mating are:

`\begin{align*}
(p + q)^2 = p^2 + 2pq + q^2
\end{align*}`

- The array of genotypes in progeny equals the square of the parental gametic array.

- (sum of allele frequencies) `$^2$` = (sum of genotype frequencies)

---
# For HWE to hold true

### NO

- Migration
- Mutation 
- Selection

### YES

- Random Mating
- Large Population Size

--------

- If any condition is not met, allele and genotype frequencies in the population may (will) change.

- With respect to one autosomal locus, __HWE is attained after one generation of random mating__.

---

# Proof of HWE

### 1. From allele frequency in parents to allele frequency in gametes

In the parent generation:

| Allele/Genotype | `$A_1$`   | `$A_2$`   | `$A_1A_1$`   | `$A_1A_2$`   | `$A_2A_2$` |
| :-------: |: ------- :| :-------: | :-------: | :-------: | :-------: |
| Freq  | `$p$`   | `$q$`    | `$p^2$`    | `$2pq$`    | `$q^2$`    |

---

# Proof of HWE

### 2. From allele frequency in gametes to genotype frequencies in zygotes

Random union among their gametes:

.pull-left[
<div align="center">
<img src="hwe.png" height=250>
</div>
]

### For heterozygotes

The union of A1 eegs with A2 sperms need to be distinguished from that of A2 eggs with A1 sperms.

---

# Proof of HWE

### 3. From zygotes to adults

Genotype frequencies would be observable unless the zygotes survive equally well.

- Equal viability
- Same fitness and no selection

### 4. From genotype freq to allele freq in progeny

Given above genotype freq, allele freq for A1 allele in progeny can be calculated as:

`\begin{align*}
f_1 = & \frac{2n_{11}+n_{12}}{2n} = f_{11} + \frac{1}{2}f_{12} \\
= & p^2 + pq = p(p+q) =p \\
\end{align*}`

---

# Relationship between allele and genotype frequencies

```r
p = seq(0, 1, by=0.01)
q = 1 - p
f11 <- p^2
f12 <- 2*p*q
f22 <- q^2

plot(f11, type="l", col="black", xlab="p", ylab="Genotype Frequency")
lines(f12, col="red")
lines(f22, col="blue")
```

![](week1_c2_files/figure-html/unnamed-chunk-2-1.png)

---

# Relationship between allele and genotype frequencies

![](week1_c2_files/figure-html/unnamed-chunk-3-1.png)

- Maximum frequency of heterozygotes occurs when p = q = 0.50

- Frequency of heterozygotes cannot be greater than 0.50

- When allele frequency is low, that allele occurs predominantly in heterozygotes

---

# How do we determine if a locus is in HWE?

### First, why do we care?

- One Locus?

- Across multiple Loci?

### Is your answer different if it is just one locus vs multiple loci?

- One Locus --- Estimate freq of recessive alleles 
  - How many people carry a disease

- Across multiple Loci
  - Which locus is affected by selection (or something else)?

---

# Chi-squared test of HWE

`\begin{align*}
\chi^2 = \sum{\frac{(Obs - Exp)^2}{Exp}}
\end{align*}`

### Example: M-N blood groups

| Genotype | `$MM$`   | `$MN$`   | `$NN$`   | Total  |
| :-------: |: ------- :| :-------: | :-------: | :-------: | 
| Observed  | `$84$`   | `$131$`    | `$216$`    | `$431$`    |

#### Using HWE to calculate expected values

`\begin{align*}
f_M = & \frac{2n_{MM}+n_{MN}}{2n} = \frac{2 \times 84 + 131}{2 \times 431} = 0.35 = p \\
f_N = & 0.65 = q \\
\end{align*}`

| Genotype | `$MM$`   | `$MN$`   | `$NN$`   | Total  |
| :-------: |: ------- :| :-------: | :-------: | :-------: | 
| Observed  | `$84$`   | `$131$`    | `$216$`    | `$431$`    | 
| Expected  | `$p^2 \times 431=53$`   | `$2pq \times 431=196$`    | `$q^2 \times 431=182$`    | `$431$`    |

---

# Chi-squared test of HWE

### Null hypothesis (H0): population is in HWE

`\begin{align*}
\chi^2 = & \sum{\frac{(Obs - Exp)^2}{Exp}} \\
= & \frac{(84-53)^2}{53} + \frac{(131-196)^2}{196} + \frac{(216-182)^2}{182} \\
= & 46 \\
\end{align*}`

- Degrees of freedom (df) = n-k (#genotype - #alleles) =1
- Critical values of `$\chi^2$` with 1 df is `$3.84$` at `$\alpha=0.05$` and `$6.63$` at `$\alpha=0.01$`.

Our value of `$\chi^2 > 3.84$`, therefore, we reject H0 and conclude that the population is NOT in HWE.

---

# Chi-squared test of HWE

```r
nMM = 84; nMN = 131; nNN = 216
n = nMM + nMN + nNN
p = (2*nMM + nMN)/(2*n) # M allele freq
q = 1- p # N allele freq

obs = c(nMM, nMN, nNN)
exp = c(p^2, 2*p*q, q^2)*n
stat <- sum(((obs-exp)^2)/exp)
stat
```

```
## [1] 46.70601
```

```r
# pchisq gives the distribution function
1 - pchisq(stat, df=1)
```

```
## [1] 8.247514e-12
```

#### Note that chisq.test is not suitable for HWE as it computes the df as 2

---

# Test for HWE

### Chi-squared test
- Expected count should be `$≥5$` per category

- If allele frequencies are low and/or sample size is small results may be erroneous

### Exact Test for HWE

Using the multinomial probability equation:

`\begin{align*}
p(data) = \frac{n!}{n_{11}!n_{12}!n_{22}!}p^{2n_{11}}q^{2n_{22}}(2pq)^{n_{12}}
\end{align*}`

```r
exact.t <- function(n11, n12, n22){
  n <- n11 + n12 + n22
  p <- (2*n11 + n12)/(2*n)
  q = 1 - p
  prob1 <- factorial(n)/(factorial(n11) * factorial(n12) * factorial(n22))
  prob2 <- p^(2*n11)*q^(2*n22)*(2*p*q)^n12
  return(prob1*prob2)
}

exact.t(n11=2, n12=40, n22=2)
```

```
## [1] 2.893707e-09
```
---
# Wigginton et al, 2005

[Am J Hum Genet 76:887-883](http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1199378/)

.pull-left[
<div align="center">
<img src="extest.png" height=400>
</div>
]

.pull-right[
- Type1 error for Chi-squared (blue) and extact test (red) for HWE

- Both tests are better with higher minor allele freq and larger sample

- Chi-squared test has inflated type I error

- Exact test is always conservative while Chi-squared can be conservative or __anti-conservative__
]

---
# Work with your own data

```r
# This code implements an exact SNP test of Hardy-Weinberg Equilibrium as described in
# Wigginton, JE, Cutler, DJ, and Abecasis, GR (2005) A Note on Exact Tests of 
# Hardy-Weinberg Equilibrium. American Journal of Human Genetics. 76: 000 - 000

# NOTE: return code of -1.0 signals an error condition

source("http://csg.sph.umich.edu/abecasis/Exact/snp_hwe.r")

data <- data.frame(HET=c(100, 57, 31), 
                   HOM1=c(3, 14,32), 
                   HOM2=c(5, 50,51))

test_markers <- function(data){
  data$pval <- -9
  for (i in 1:nrow(data)){
    data$pval[i] <- SNPHWE(data[i, 1], data[i, 2], data[i, 3])
  }
  return(data)
}
test_markers(data)
```

```
##   HET HOM1 HOM2         pval
## 1 100    3    5 6.195094e-21
## 2  57   14   50 8.422798e-01
## 3  31   32   51 2.563773e-06
```