Part 8: Gene Expression Regulation

Step 1: Define the MWC two-state model

A protein with n identical subunits exists in two conformations: T (tense, low affinity) and R (relaxed, high affinity), in equilibrium characterized by L = [T₀]/[R₀]:

$$L = \frac{[T_0]}{[R_0]} \qquad c = \frac{K_R}{K_T} \ll 1 \qquad \alpha = \frac{[S]}{K_R}$$

Step 2: Write the MWC binding function

Each subunit binds ligand independently within its state. The exact MWC fractional saturation is:

$$\bar{Y} = \frac{Lc\alpha(1+c\alpha)^{n-1} + \alpha(1+\alpha)^{n-1}}{L(1+c\alpha)^n + (1+\alpha)^n}$$

Step 3: Take the extreme cooperativity limit (c → 0)

When the T state has negligible ligand affinity (c ≈ 0), the T terms in the numerator vanish:

$$\bar{Y} \approx \frac{\alpha(1+\alpha)^{n-1}}{L + (1+\alpha)^n}$$

Step 4: Further simplify for saturating conditions

For large L (strong T-state preference) and focusing on the transition region where α ∼ L^1/n, the binding curve becomes switch-like. When (1+α)ⁿ ≈ αⁿ for moderate α:

$$\bar{Y} \approx \frac{\alpha^n}{L + \alpha^n} = \frac{[S]^n / K_R^n}{L + [S]^n / K_R^n} = \frac{[S]^n}{L \cdot K_R^n + [S]^n}$$

Step 5: Identify the Hill equation

Defining K_d,eff = L^1/n · K_R as the effective half-saturation constant:

$$\bar{Y} \approx \frac{[S]^n}{K_{d,\text{eff}}^n + [S]^n} \qquad \text{(Hill equation with } n_H = n\text{)}$$

Step 6: Interpret the Hill coefficient

In the MWC model, the apparent Hill coefficient n_H depends on L and c. Maximum cooperativity (n_H → n) occurs when c → 0 and L is large. In practice, n_H < n because binding is not perfectly concerted. For hemoglobin (n = 4), n_H ≈ 2.8. For transcription factors, effective cooperativity can exceed the number of binding sites when combined with DNA looping or multimerization.

$$1 \leq n_H \leq n \qquad \text{(always between non-cooperative and maximum)}$$

Step 1: Enumerate all promoter states

The lac promoter can exist in four states: (1) empty, (2) RNAP bound, (3) Repressor bound, (4) both bound (mutually exclusive for overlapping binding sites). Each state has a Boltzmann weight:

$$w_{\text{empty}} = 1, \quad w_{\text{RNAP}} = \frac{[P]}{K_P}, \quad w_{\text{Rep}} = \frac{[R]}{K_R}, \quad w_{\text{P+CAP}} = \frac{[P]}{K_P}\cdot\frac{[A]}{K_A}\cdot\omega$$

Step 2: Construct the partition function

The partition function Z sums over all possible promoter configurations:

$$Z = 1 + \frac{[P]}{K_P} + \frac{[R]}{K_R} + \frac{[P]}{K_P}\frac{[A]}{K_A}\omega_{AP}$$

Step 3: Calculate RNAP occupancy probability

The probability that RNAP is bound (transcription occurs) is the sum of all RNAP-containing states divided by Z:

$$P_{\text{RNAP}} = \frac{\frac{[P]}{K_P}\left(1 + \frac{[A]}{K_A}\omega_{AP}\right)}{Z}$$

Step 4: Incorporate the repressor-inducer equilibrium

The effective repressor concentration depends on inducer (IPTG/allolactose). Inducer binding reduces repressor-DNA affinity by factor f:

$$[R]_{\text{eff}} = \frac{[R]_{\text{total}}}{1 + ([I]/K_I)^n} \qquad \text{(Hill-like inducer response)}$$

Step 5: Include CAP-cAMP activation (catabolite repression)

CAP-cAMP activates transcription when glucose is low (cAMP is high). The CAP activation factor depends on glucose through its effect on cAMP:

$$[\text{cAMP}] = \frac{[\text{cAMP}]_{\text{basal}}}{1 + ([\text{Glucose}]/K_{\text{glu}})^2} \qquad f_{\text{CAP}} = \frac{[\text{cAMP}]}{K_{\text{cAMP}} + [\text{cAMP}]}$$

Step 6: Final expression rate

The transcription rate is proportional to RNAP occupancy, integrating all regulatory inputs:

$$\text{Rate} = k_{\text{esc}} \cdot P_{\text{RNAP}} = k_{\text{esc}} \cdot \frac{\frac{[P]}{K_P}(1 + f_{\text{CAP}}\cdot\omega)}{1 + \frac{[P]}{K_P}(1 + f_{\text{CAP}}\cdot\omega) + \frac{[R]_{\text{eff}}}{K_R}}$$

This reproduces the known lac operon behavior: maximal expression requires both low glucose (high cAMP/CAP) and presence of inducer (low effective repressor). Neither condition alone is sufficient.

Step 1: Write the toggle switch ODEs

Two genes mutually repress each other with Hill-type repression (Gardner et al., 2000):

$$\frac{du}{dt} = \frac{\alpha_1}{1 + v^\beta} - u \qquad \frac{dv}{dt} = \frac{\alpha_2}{1 + u^\gamma} - v$$

Step 2: Find the nullclines (steady-state curves)

Set each derivative to zero to find the nullclines. The u-nullcline (du/dt = 0) and v-nullcline (dv/dt = 0) are:

$$u = \frac{\alpha_1}{1 + v^\beta} \qquad (\text{u-nullcline}) \qquad v = \frac{\alpha_2}{1 + u^\gamma} \qquad (\text{v-nullcline})$$

Step 3: Determine intersection conditions

Steady states occur where nullclines intersect. Substituting the v-nullcline into the u-nullcline gives a self-consistency equation:

$$u = \frac{\alpha_1}{1 + \left(\frac{\alpha_2}{1 + u^\gamma}\right)^\beta} \equiv F(u)$$

Step 4: Analyze the symmetric case (α₁ = α₂ = α, β = γ = n)

At the symmetric fixed point u* = v* = u_s, the self-consistency equation becomes:

$$u_s = \frac{\alpha}{1 + u_s^n} \implies u_s(1 + u_s^n) = \alpha$$

Step 5: Linearize and derive the bistability condition

The Jacobian at a fixed point (u*, v*) determines stability. Bistability requires that the symmetric fixed point be unstable (a saddle point). The condition for instability at the symmetric point is that the product of the nullcline slopes exceeds 1:

$$\left|\frac{du}{dv}\right|_{\text{u-null}} \times \left|\frac{dv}{du}\right|_{\text{v-null}} > 1 \implies \frac{\alpha_1 \beta\, v_s^{\beta-1}}{(1+v_s^\beta)^2} \cdot \frac{\alpha_2 \gamma\, u_s^{\gamma-1}}{(1+u_s^\gamma)^2} > 1$$

Step 6: Simplified bistability criterion

For the symmetric case (α₁ = α₂, β = γ = n), bistability requires the Hill coefficients to be sufficiently large and the production rates to be sufficiently high. The critical condition simplifies to:

$$n > 1 + \frac{1}{\log(\alpha/2)} \qquad \text{(approximate, for large } \alpha\text{)}$$

In practice: n > 2 almost always guarantees bistability for reasonable α values. When n = 1 (no cooperativity), the nullclines intersect only once and the system is monostable. Cooperativity (n > 1) is essential for creating the S-shaped nullclines that enable three intersections.

Step 1: Define the two-stage model

mRNAs are produced at rate k_m and degraded at rate δ_m. Each mRNA produces proteins at rate k_p, and proteins are degraded at rate δ_p:

$$\emptyset \xrightarrow{k_m} m \xrightarrow{\delta_m} \emptyset \qquad m \xrightarrow{k_p} m + P \qquad P \xrightarrow{\delta_p} \emptyset$$

Step 2: Solve for mean mRNA and protein levels

At steady state, the mean copy numbers are:

$$\langle m \rangle = \frac{k_m}{\delta_m} \qquad \langle P \rangle = \frac{k_m k_p}{\delta_m \delta_p}$$

Step 3: Define the burst size

Since mRNA is short-lived compared to protein (δ_m >> δ_p), each mRNA produces a “burst” of proteins before being degraded. The average burst size is:

$$b = \frac{k_p}{\delta_m} \qquad \text{(proteins per mRNA lifetime)}$$

Step 4: Compute the protein variance from the master equation

Using the generating function method on the chemical master equation (Thattai & van Oudenaarden, 2001), the protein variance has two components — intrinsic (Poisson) noise and extrinsic (burst) noise:

$$\sigma_P^2 = \langle P \rangle + \frac{k_p}{\delta_m} \cdot \langle P \rangle \cdot \frac{1}{1 + \delta_p/\delta_m}$$

Step 5: Simplify in the limit δ_m >> δ_p

When mRNA degrades much faster than protein (typical in bacteria: mRNA half-life ~2–5 min, protein half-life ~hours), δ_p/δ_m → 0:

$$\sigma_P^2 \approx \langle P \rangle + b \cdot \langle P \rangle = \langle P \rangle(1 + b)$$

Step 6: Extract the Fano factor

The Fano factor is the variance-to-mean ratio:

$$F = \frac{\sigma_P^2}{\langle P \rangle} = 1 + b = 1 + \frac{k_p}{\delta_m}$$

When b = 0 (no translation bursting), F = 1 (Poisson statistics). For typical E. coli genes with b ≈ 1–5, F ≈ 2–6, meaning protein fluctuations are 2–6× larger than Poisson. This “burstiness” enables phenotypic heterogeneity even in clonal populations, driving phenomena like antibiotic persistence and competence switching.