Our first step is to determine the frequency of each allele in this data set.

The code below will retrieve the allelic frequencies for the “Canadian Rivers” Distinct Population Segment (DPS).

To make this analysis simpler for the sake of demonstration, we are selecting just two of the five alleles that Sturgeon in this DPS express at this loci.

alleles <- sturg %>% 
  filter(DPS == "Canadian Rivers" & (LS68 == 151 | LS68 == 155) & 
           (LS68.1 == 151 | LS68.1 == 155)) %>% 
  select(DPS, LS68, LS68.1) %>% 
  group_by(DPS, LS68, LS68.1) %>% 
  summarise(count=n())

allele_frequencies <- alleles %>% 
  ungroup() %>% 
  mutate(frequency = round(count / sum(count), 3))

Let’s assign each allele in the following way:

p = Allele 151, q = Allele 155

Then, the genotypes have the following frequencies:

\(p^2\) = 0.107, 2pq = 0.321, \(q^2\) = 0.571

From here, we can calculate the frequency of each allele:

\(\text{frequency of p} = p^2 + 0.5(2pq) = 0.268\)
\(\text{frequency of q} = 1 - p = 0.732\)

Now that we know the observed frequencies of each allele, we can use that to calculate the expected genotype frequencies.

\(\text{frequency of pp} = p^2 = 0.268^2 = 0.072\)
\(\text{frequency of pq} = 2pq = 2 * 0.268 * 0.732 = 0.392\)
\(\text{frequency of qq} = q^2 = 0.732^2 = 0.536\)

Using a \(\chi^2\) test, we can determine if the population of Sturgeon are in Hardy Weinberg equilibrium at the LS68 locus. Hardy Weinberg equilibrium is our null hypothesis, so if our result differs enough from equilibrium, we can reject the null hypothesis and state that the population is not in equilibrium. Otherwise, we fail to reject the null hypothesis and state that the population is in equilibrium.

\(\chi^2 = \Sigma \frac{(\text{O - E}^2)}{\text{E}}\)

Where O is the observed value and E is the expected value. In the \(\chi^2\) test, we use counts of individuals with each genotype, rather than frequencies.

Using the expected frequencies and a total population size of 28, we get that the expected values for each genotype are 2, 11, and 15, while the observed genotypes are 3, 9, and 16, respectively.

Thus,

\(\chi^2 = \frac{(\text{3-2})^2}{\text{2}} + \frac{(\text{9-11})^2}{\text{11}} + \frac{(\text{16-15})^2}{\text{15}} = 0.93\)

With 2 degrees of freedom, a \(\chi^2\) value of 0.93 corresponds to a p-value of roughly 0.63, meaning that we fail to reject the null hypothesis, and this population seems to be in Hardy Weinberg Equilibrium.

• Using a simple function shown on the next slide, we can see how the equilibrium will shift under different conditions. This can be used to test how real world events like migration may affect equilibrium.
• The function takes the frequency of the “p” and “q” alleles, as well as the population size, and returns the \(\chi^2\) value.

hw_simulator <- function(p, q, n) {
  stopifnot(p + q <= 1)
  pq = 1 - (p + q)
  
  p_freq = p + 0.5 * (pq)
  q_freq = 1 - p_freq
  
  pp_freq = p_freq^2
  pq_freq = 2 * p_freq * q_freq
  qq_freq = q_freq^2
  
  obs_p_count = round(p * n)
  obs_q_count = round(q * n)
  obs_pq_count = round(pq * n)
  exp_p_count = round(pp_freq * n)
  exp_q_count = round(qq_freq * n)
  exp_pq_count = round(pq_freq * n)
    
  chi_squared = ((obs_p_count - exp_p_count)^2) / exp_p_count + 
    ((obs_q_count - exp_q_count)^2) / exp_q_count +
    ((obs_pq_count - exp_pq_count)^2) / exp_pq_count
  
  return(chi_squared)
}

We can plot this to get a visual representation of how different factors may affect the Hardy Weinberg equilibrium of a population. On the x-axis is the frequency of allele “p” and on the y-axis is the frequency of the “q” allele. The size of each point corresponds to the \(\chi^2\) value. Larger \(\chi^2\) values correspond to populations further away from equilibrium.

We can also plot the frequencies of the three genotypes in 3D to see how all three values affect the chi-squared value.

## Warning: `line.width` does not currently support multiple values.