Part 4: DNA Repair Mechanisms

Step 1: Replication error contribution

The per-base-pair mutation rate from replication errors is the product of three sequential escape probabilities (base selection, proofreading, MMR):

$$\mu_{\text{rep}} = \epsilon_{\text{ins}} \times (1 - f_{\text{proof}}) \times (1 - f_{\text{MMR}}) \approx 10^{-4} \times 10^{-2} \times 10^{-2} = 10^{-8}\text{--}10^{-10}$$

Step 2: Damage-induced mutation contribution

For each type of lesion $i$, the fraction that escapes repair before the replication fork arrives depends on the relative rates of damage and repair. The probability a lesion persists at replication is:

$$p_{\text{unrepaired},i} = \frac{k_{\text{dam},i}}{k_{\text{dam},i} + k_{\text{rep},i}}$$

This comes from the steady-state solution: at equilibrium, the fraction of time a site is damaged equals the damage rate divided by the sum of damage and repair rates.

Step 3: Combine with mutagenic potential

Not all unrepaired lesions cause mutations; some block replication (lethal) and some are bypassed accurately. Multiply by the probability $p_{\text{mut},i}$ that the lesion causes a mutation if unrepaired:

$$\mu_{\text{damage}} = \sum_{i} \frac{k_{\text{dam},i}}{k_{\text{dam},i} + k_{\text{rep},i}} \times p_{\text{mut},i}$$

Step 4: Total per-bp mutation rate

The total mutation rate per bp per division is the sum of replication and damage contributions:

$$\mu_{\text{total}} = \mu_{\text{rep}} + \mu_{\text{damage}}$$

Step 5: Scale to whole-genome mutations per generation

Multiply by genome size $G$ to get total mutations per cell division:

$$M = \mu_{\text{total}} \times G$$

For humans: $M \approx 10^{-9} \times 6.4 \times 10^9 \approx 0.5\text{--}1.0$ mutations per division. For E. coli: $M \approx 10^{-10} \times 4.6 \times 10^6 \approx 5 \times 10^{-4}$ mutations per generation (one mutation every ~2,000 generations).

Step 6: Drake's rule

John Drake (1991) observed that the per-genome mutation rate is remarkably constant across DNA-based microbes:

$$\mu \times G \approx 0.003\text{--}0.004 \text{ per genome per generation}$$

This $1/G$ scaling of per-bp rate suggests that the fidelity machinery has evolved to maintain a near-constant genomic mutation rate, balancing the cost of higher fidelity against the benefit of reduced deleterious mutations.

Step 1: Write the reaction scheme

A repair glycosylase E binds a damaged base (lesion L) in DNA to form an enzyme-substrate complex:

$$E + L \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} EL \xrightarrow{k_{\text{cat}}} E + P$$

where P is the abasic site product (for monofunctional glycosylases) or the nicked DNA (for bifunctional glycosylases with AP lyase activity).

Step 2: Apply the steady-state approximation

Assume $d[\text{EL}]/dt = 0$ (the ES complex concentration changes slowly compared to its formation and breakdown):

$$k_1[E][L] = (k_{-1} + k_{\text{cat}})[\text{EL}]$$

$$[\text{EL}] = \frac{[E][L]}{K_M} \quad \text{where } K_M = \frac{k_{-1} + k_{\text{cat}}}{k_1}$$

Step 3: Use the enzyme conservation equation

Total enzyme is partitioned between free and bound forms: $[E]_0 = [E] + [\text{EL}]$. Substituting:

$$[E] = [E]_0 - [\text{EL}] = [E]_0 - \frac{[E][L]}{K_M}$$

Solving for [E]: $[E] = \frac{[E]_0 \cdot K_M}{K_M + [L]}$

Step 4: Derive the rate equation

The repair rate is $v = k_{\text{cat}}[\text{EL}]$. Substituting the expressions:

$$v = \frac{k_{\text{cat}}[E]_0[L]}{K_M + [L]} = \frac{V_{\max}[L]}{K_M + [L]}$$

where $V_{\max} = k_{\text{cat}}[E]_0$ is the maximum repair rate at enzyme saturation.

Step 5: Solve for steady-state lesion level

At steady state, damage formation rate equals repair rate: $k_{\text{dam}} = v$:

$$k_{\text{dam}} = \frac{k_{\text{cat}}[E]_0[L]_{\text{ss}}}{K_M + [L]_{\text{ss}}}$$

Rearranging: $k_{\text{dam}}(K_M + [L]_{\text{ss}}) = k_{\text{cat}}[E]_0[L]_{\text{ss}}$

$$[L]_{\text{ss}} = \frac{k_{\text{dam}} \cdot K_M}{k_{\text{cat}}[E]_0 - k_{\text{dam}}} = \frac{k_{\text{dam}} \cdot K_M}{V_{\max} - k_{\text{dam}}}$$

Step 6: Biological interpretation

This result has critical implications: (1) A finite steady state exists only when $V_{\max} > k_{\text{dam}}$ (repair capacity exceeds damage rate). (2) As $k_{\text{dam}} \to V_{\max}$, $[L]_{\text{ss}} \to \infty$ — the system cannot cope. (3) Cells with reduced repair enzyme levels (e.g., haploinsufficiency in heterozygous carriers of repair gene mutations) have a higher steady-state lesion burden, explaining cancer predisposition.

Step 1: Linearize the M-M rate at low substrate

When $[L] \ll K_M$, the denominator $K_M + [L] \approx K_M$, so:

$$v_{\text{repair}} \approx \frac{V_{\max}}{K_M}[L] = k_{\text{eff}}[L]$$

where $k_{\text{eff}} = k_{\text{cat}}[E]_0/K_M$ is the effective first-order repair rate constant.

Step 2: Write the ODE for lesion dynamics

The rate of change of lesion number includes both constant damage production and first-order repair:

$$\frac{d[L]}{dt} = k_{\text{dam}} - k_{\text{eff}}[L]$$

Step 3: Solve the ODE (repair only, no new damage)

If damage production stops (e.g., UV source removed) and we track only repair of existing lesions starting from $N_0$ lesions:

$$\frac{dN}{dt} = -k_{\text{eff}} \cdot N \implies N(t) = N_0 \cdot e^{-k_{\text{eff}} t}$$

This exponential decay is observed experimentally: CPD removal by NER shows first-order kinetics with $t_{1/2} \approx 4\text{--}12$ hours in human cells.

Step 4: Solve the full ODE with constant damage

The general solution with initial condition $[L](0) = [L]_0$:

$$[L](t) = \frac{k_{\text{dam}}}{k_{\text{eff}}} + \left([L]_0 - \frac{k_{\text{dam}}}{k_{\text{eff}}}\right) e^{-k_{\text{eff}} t}$$

As $t \to \infty$: $[L] \to k_{\text{dam}}/k_{\text{eff}}$ (the steady state).

Step 5: Time constant and half-life

The system relaxes to steady state with time constant $\tau = 1/k_{\text{eff}}$ and half-life:

$$t_{1/2} = \frac{\ln 2}{k_{\text{eff}}} = \frac{K_M \ln 2}{k_{\text{cat}}[E]_0}$$

For 8-oxoG repair by OGG1 in human cells: $t_{1/2} \approx 30$ min, consistent with rapid turnover of oxidative lesions.

Step 1: Define the photon absorption rate

The rate of photon absorption by the photolyase-CPD complex depends on the light intensity $I$ (photons/cm$^2$/s) and the absorption cross-section $\sigma$ (cm$^2$):

$$k_{\text{abs}} = \sigma \cdot I$$

The antenna chromophore (MTHF or 8-HDF) has $\sigma \approx 10^{-16}$ cm$^2$ at 380 nm (peak absorption).

Step 2: Energy transfer efficiency

The antenna absorbs a photon and transfers energy to the catalytic FADH$^-$ cofactor via Förster resonance energy transfer (FRET). The transfer efficiency depends on the donor-acceptor distance $r$ and the Förster radius $R_0$:

$$\eta_{\text{FRET}} = \frac{R_0^6}{R_0^6 + r^6}$$

In photolyase, $r \approx 17$ Angstroms and $R_0 \approx 25$ Angstroms, giving $\eta_{\text{FRET}} \approx 0.9\text{--}0.97$.

Step 3: Electron transfer from FADH$^-$ to CPD

Excited FADH$^{-*}$ donates an electron to the CPD with efficiency $\eta_{\text{ET}}$. The electron injection competes with FADH$^{-*}$ fluorescence and non-radiative decay:

$$\eta_{\text{ET}} = \frac{k_{\text{ET}}}{k_{\text{ET}} + k_{\text{fl}} + k_{\text{nr}}}$$

Ultrafast spectroscopy shows $k_{\text{ET}} \approx 10^{12}$ s$^{-1}$ (picosecond timescale), far faster than fluorescence ($k_{\text{fl}} \sim 10^8$ s$^{-1}$), so $\eta_{\text{ET}} > 0.99$.

Step 4: CPD ring splitting probability

Once the CPD receives an electron, the cyclobutane ring cleaves via a radical anion mechanism with probability $\eta_{\text{split}}$. The alternative is back electron transfer to FADH before ring cleavage. Experimentally, $\eta_{\text{split}} \approx 0.85\text{--}0.95$.

Step 5: Overall quantum yield

The quantum yield $\Phi$ is the product of all step efficiencies:

$$\Phi = \eta_{\text{FRET}} \times \eta_{\text{ET}} \times \eta_{\text{split}}$$

$$\Phi \approx 0.95 \times 0.99 \times 0.90 \approx 0.85$$

This is remarkably efficient for a photochemical repair process. Measured quantum yields for E. coli CPD photolyase range from 0.7 to 0.9, consistent with this calculation.

Step 6: Repair rate under sunlight

The overall photoreactivation rate is:

$$k_{\text{repair}} = \sigma \cdot I \cdot \Phi \cdot f_{\text{bound}}$$

where $f_{\text{bound}}$ is the fraction of CPDs bound by photolyase ($= [\text{PL-CPD}] / [\text{CPD}]_{\text{total}}$). Under bright sunlight and saturating photolyase, CPDs can be repaired with a half-life of minutes, compared to hours for NER.