Replace eq tags with $$

selection/mutation-drift-selection-statistics.md

@@ -14,7 +14,7 @@ This is all assuming a Wright-Fisher population of discrete non-overlapping gene
 
 A new mutant appears in the population at an initial frequency *p* of $$1/N$$. It has a chance of fixing of
 
-{% eq \mathrm{Pr}(\mathrm{fix}) = \frac{ 1-e^{-2s} }{ 1-e^{-2Ns} } %}
+$$\mathrm{Pr}(\mathrm{fix}) = \frac{ 1-e^{-2s} }{ 1-e^{-2Ns} }$$
 
 In the limit as $$\lim_{N \to \infty}$$, this becomes $$\mathrm{Pr}(\mathrm{fix}) \approx 2s$$.
 
@@ -30,7 +30,7 @@ It's also clear that larger population sizes are more efficient at purging delet
 
 We also know the general probability of an allele fixing that is at current frequency *p*. This is
 
-{% eq \mathrm{Pr}(\mathrm{fix}) = \frac{ 1-e^{-2Nsp} }{ 1-e^{-2Ns} } %}.
+$$\mathrm{Pr}(\mathrm{fix}) = \frac{ 1-e^{-2Nsp} }{ 1-e^{-2Ns} }$$.
 
 This gives the following relationship between *Ns* and *p* and chance of fixation.
 

selection/mutation-drift-statistics.md

@@ -14,15 +14,15 @@ This is all assuming a Wright-Fisher population of discrete non-overlapping gene
 
 Genetic diversity is most commonly summarized with the statistic *π*, which is equal to the average number of mutations per site between two random individuals in the population. *π* is most commonly measured in terms of substitutions per site. The expectation of *π* follows
 
-{% eq E[\pi] = \theta = 2N\mu %}
+$$E[\pi] = \theta = 2N\mu$$
 
 *π* for *Drosophila* and *π* for flu is approximately 0.01, while *π* for humans is approximately 0.001. This means that for an average length gene of 1000 basepairs, two random fruit flies or two random flues will probably differ at ~10 sites, while two random humans will differ at ~1 site.
 
 ## Unique variants
 
 The number of unique haplotypes in a sample of *n* sequences can be estimated from [Ewen's sampling formula](https://en.wikipedia.org/wiki/Ewens's_sampling_formula). Ewen's sampling formula gives the probability of observing *a*<sub>1</sub> copies of haplotype 1, *a*<sub>2</sub> copies of haplotype 2, etc... in a sample of *n* sequences. The sole parameter of the sampling formula is *&theta;*. Thus *&theta;* is sufficient to predict the entire distribution of haplotype frequencies. The expectation of *k* unique haplotypes follows:
 
-{% eq E[k] = \sum^{n}_{i=1} \frac{\theta}{\theta + i - 1} %}
+$$E[k] = \sum^{n}_{i=1} \frac{\theta}{\theta + i - 1}$$
 
 With *&theta;* = 0.2, there is usually only a single dominant haplotype in the population.