Define the variables

pages/activators.md

@@ -35,4 +35,14 @@ That was a short section. Onward!
 
 <script>
     plot('#binding-curve', (kd, beta0) => [`${beta0}/(1 + x/(${kd}))`, `${beta0} * ((x/${kd})/(1 + (x/${kd})))`], ['#kd', '#beta0'], [[0, 10], [0, 10]])
+
+    defineVars([
+        ['P', 'The promoter of the DNA.'],
+        ['A', 'The activator.'],
+        ['p', 'A concentration of promoters, either unbound, bound (pbound), or total (ptot).'],
+        ['x', 'The gene in question.'],
+        ['K', 'Kd is the dissociation constant, a measure of the likelihood that the activator will unbind.'],
+        ['β', 'β(x) is the simple binding curve. β0 is the maximum expression level.'],
+        ['a', 'The concentration of activator.']
+    ])
 </script>

pages/hill-functions.md

@@ -59,4 +59,16 @@ Onwards!
 <script>
     plot('#hill-graph', (k, n) => [`((x/${k})^(${n}))/(1 + ((x/${k})^(${n})))`, `1/(1 + ((x/${k})^(${n})))`], ['#k', '#n'], [[0, 10], [0, 1]])
     plot('#hill-graph-prod', (k, n, b0) => [`(${b0}) * ((x/${k})^(${n}))/(1 + ((x/${k})^(${n})))`, `(${b0})/(1 + ((x/${k})^(${n})))`], ['#k-prod', '#n-prod', '#b0'], [[0, 10], [0, 10]])
+
+    defineVars([
+        ['f', 'The Hill function.'],
+        ['P', 'The promoter of the DNA.'],
+        ['A', 'The activator.'],
+        ['p', 'A concentration of promoters, either unbound, bound (pbound), or total (ptot).'],
+        ['k', 'The concentration at which the function attains half of its maximum value.'],
+        ['n', 'The Hill coefficient, a measurement of how ultrasensitive the response is.'],
+        ['β', 'β(x) is the simple binding curve. β0 is the maximum expression level.'],
+        ['r', 'The concentration of repressor.'],
+        ['a', 'The concentration of activator.']
+    ])
 </script>

pages/repressors.md

@@ -27,7 +27,7 @@ Specifically, if \\(p_\text{tot}\\) is the total concentration of promoters, bot
 
 $$\frac{p}{p_\text{tot}}=\frac1{(1+r/K_d)}$$
 
-Where \\(K_d\\) is the **dissociation constant** — a measurement of the likelyhood that a repressor will unbind from its binding site — calculated as \\(\frac{k_-}{k_+}\\). Becayse we have a separation of timescales and can simplify things, the rate of production of gene product should be proportional to the probability of the promoter being unbound:
+Where \\(K_d\\) is the **dissociation constant** — a measurement of the likelyhood that a repressor will unbind from its binding site — calculated as \\(\frac{k_-}{k_+}\\). Because we have a separation of timescales and can simplify things, the rate of production of gene product should be proportional to the probability of the promoter being unbound:
 
 $$\beta(r)=\beta_0\frac{p}{p_\text{tot}}=\frac{\beta_0}{1 + r/K_d}$$
 
@@ -53,7 +53,7 @@ For small values of \\(r\\), note that the slope is \\(-\frac{\beta_0}{K_d}\\).
 
 ## Leaky expression
 
-Unfortunately, repressors aren't perfect. Many genes can never get entirely repressed, and have a baseline, or "basal", level or expression. We can represent this by adding another term, $\alpha_0$, to $\beta(r)$:
+Unfortunately, repressors aren't perfect. Many genes can never be entirely repressed, and have a baseline, or "basal", level or expression. We can represent this by adding another term, $\alpha_0$, to $\beta(r)$:
 
 $$\beta(r)=\frac{\beta_0}{1 + r/K_d}+\alpha_0$$
 
@@ -88,4 +88,15 @@ Onwards!
     plot('#binding-curve', (kd, beta0) => [`${beta0}/(1 + x/(${kd}))`, `-(${beta0})x/(${kd}) + ${beta0}`], ['#kd', '#beta0'], [[0, 10], [0, 10]])
 
     plot('#binding-curve-leaky', (kd, beta0, alpha0) => [`${beta0}/(1 + x/(${kd})) + ${alpha0}`, `-(${beta0})x/(${kd}) + ${beta0} + ${alpha0}`], ['#leaky-kd', '#leaky-beta0', '#leaky-alpha0'], [[0, 10], [0, 10]])
+
+    defineVars([
+        ['P', 'The promoter of the DNA.'],
+        ['R', 'The repressor.'],
+        ['p', 'A concentration of promoters, either unbound, bound (pbound), or total (ptot).'],
+        ['x', 'The gene in question.'],
+        ['K', 'Kd is the dissociation constant, a measure of the likelihood that the repressor will unbind.'],
+        ['β', 'β(x) is the simple binding curve. β0 is the maximum expression level.'],
+        ['r', 'The concentration of repressor.'],
+        ['α', 'α0 is the baseline, or "basal", level of expression of the gene.']
+    ])
 </script>

pages/simplest-circuit.md

@@ -28,7 +28,7 @@ $$ \frac{dx}{dt} = \beta - \gamma x $$
 Since \\(\gamma\\) counts both degradation and dilution, we can say that:
 $$ \gamma = \gamma_\text{degradation} + \gamma_\text{dilution} $$
 
-Since we're getting into the math, a quick tip: if you (like me) forget what a variable does halfway through the page, just hover over it and it'll tell you what it does. Anyway, back to the show.
+Since we're getting into the math, a quick tip: if you (like me) forget what a variable does halfway through the page, just hover over it and it'll tell you what it does. Because of the way it's rendered, it might not work in some cases, but feel free to try it. Anyway, back to the show.
 
 To find the net production of the protein under steady state conditions, set the derivative to zero and solve for \\(x\\):