3 Single-domain proteins under constant loading

Let $x$ be the end to end distance of the protein, $t$ be the time since loading began, $F$ be tension applied to the protein, $P$ be the surviving population of folded proteins. Make the definitions

$$v \equiv \deriv{t}{x}$$ the pulling velocity (1)

$$k \equiv \deriv{x}{F}$$ the loading spring constant (2)

$$P_0 \equiv P(t=0)$$ the initial number of folded proteins (3)

$$D \equiv P_0 - P$$ the number of dead (unfolded) proteins (4)

$$\kappa \equiv -\frac{1}{P} \deriv{t}{P}$$ the unfolding rate (5)

The proteins are under constant loading because

$$\deriv{t}{F} = \deriv{x}{F}\deriv{t}{x} = kv\;,$$ (6)

a constant, since both $k$ and $v$ are constant (Evans and Ritchie (1997) in the text on the first page, Dudko et al. (2006) in the text just before Eqn. 4).

The instantaneous likelyhood of a protein unfolding is given by $\deriv{F}{D}$, and the unfolding histogram is merely this function discretized over a bin of width $W$(This is similar to Dudko et al. (2006) Eqn. 2, remembering that $\dot{F}=kv$, that their probability density is not a histogram ($W=1$), and that their pdf is normalized to $N=1$).

$$h(F) \equiv \deriv{\text{bin}}{F} = \deriv{F}{D} \cdot \deriv{\te... ...v{F}{D} = -W \deriv{F}{P} = -W \deriv{t}{P} \deriv{F}{t} = \frac{W}{vk} P\kappa$$ (7)

Solving for theoretical histograms is merely a question of taking your chosen $\kappa$, solving for $P(f)$, and plugging into Eqn. 7. We can also make a bit of progress solving for $P$ in terms of $\kappa$ as follows:

$$\kappa \equiv -\frac{1}{P} \deriv{t}{P}$$ (8)

$$-\kappa \dd t \cdot \deriv{t}{F} = \frac{\dd P}{P}$$ (9)

$$\frac{-1}{kv} \int \kappa \dd F = \ln(P) + c$$ (10)

$$P = C\exp{\p({\frac{-1}{kv}\ensuremath{\int {\kappa} \dd {F}}})} \;,$$ (11)

where $c \equiv \ln(C)$ is a constant of integration scaling $P$.

3.1 Constant unfolding rate

In the extremely weak tension regime, the proteins' unfolding rate is independent of tension, we have

$$P = C\exp{\p({\frac{-1}{kv}\ensuremath{\int {\kappa} \dd {F}}})} = C\exp{\p({\frac{-1}{kv}\kappa F})} = C\exp{\p({\frac{-\kappa F}{kv}})}$$ (12)

$$P(0) \equiv P_0 = C\exp(0) = C$$ (13)

$$h(F) = \frac{W}{vk} P \kappa = \frac{W\kappa P_0}{vk} \exp{\p({\frac{-\kappa F}{kv}})}$$ (14)

Suprise! A constant unfolding-rate/hazard-function gives exponential decay. Not the most earth shattering result, but it's a comforting first step, and it does show explicitly the dependence in terms of the various unfolding-specific parameters.

3.2 Bell model

Stepping up the intensity a bit, we come to Bell's model for unfolding (Hummer and Szabo (2003) Eqn. 1 and the first paragraph of Dudko et al. (2006) and Dudko et al. (2007)).

$$\kappa = \kappa_0 \cdot \exp\p({\frac{F \dd x}{k_B T}}) = \kappa_0 \cdot \exp(a F) \;,$$ (15)

where we've defined $a \equiv \dd x/k_B T$ to bundle some constants together. The unfolding histogram is then given by

$$P = C\exp\p({\frac{-1}{kv}\ensuremath{\int {\kappa} \dd {F}}}) = C\... ...{kv} \frac{\kappa_0}{a} \exp(a F)}] = C\exp\p[{\frac{-\kappa_0}{akv}\exp(a F)}]$$ (16)

$$P(0) \equiv P_0 = C\exp\p({\frac{-\kappa_0}{akv}})$$ (17)

$$C = P_0 \exp\p({\frac{\kappa_0}{akv}})$$ (18)

$$P = P_0 \exp\p\{{\frac{\kappa_0}{akv}[1-\exp(a F)]}\}$$ (19)

$$h(F) = \frac{W}{vk} P \kappa = \frac{W}{vk} P_0 \exp\p\{{\frac{\kappa_... ...\frac{W\kappa_0 P_0}{vk} \exp\p\{{a F + \frac{\kappa_0}{akv}[1-\exp(a F)]}\}\;.$$ (20)

The $F$ dependent behavior reduces to

$$h(F) \propto \exp\p[{a F - b\exp(a F)}] \;,$$ (21)

where $b \equiv \kappa_0/akv \equiv \kappa_0 k_B T / k v \dd x$ is another constant rephrasing.

This looks an awful lot like the the Gompertz/Gumbel/Fisher-Tippett distribution, where

$$p(x) \propto z\exp(-z)$$ (22)

$$z \equiv \exp\p({-\frac{x-\mu}{\beta}}) \;,$$ (23)

but we have

$$p(x) \propto z\exp(-bz) \;.$$ (24)

Strangely, the Gumbel distribution is supposed to derive from an exponentially increasing hazard function, which is where we started for our derivation. I haven't been able to find a good explaination of this discrepancy yet, but I have found a source that echos my result (Wu et al. (2004) Eqn. 1).

Oh wait, we can do this:

$$p(x) \propto z\exp(-bz) = \frac{1}{b} z'\exp(-z')\propto z'\exp(-z') \;,$$ (25)

with $z'\equiv bz$. I feel silly... From Wolframhttp://mathworld.wolfram.com/GumbelDistribution.html, the mean of the Gumbel probability density

$$P(x) = \frac{1}{\beta} \exp\p[{\frac{x-\alpha}{\beta} -\exp\p({\frac{x-\alpha}{\beta}}) }]$$ (26)

is given by $\mu=\alpha-\gamma\beta$, and the variance is $\sigma^2=\frac{1}{6}\pi^2\beta^2$, where $\gamma=0.57721566\ldots$ is the Euler-Mascheroni constant. Selecting $\beta=1/a=k_BT/\dd x$, $\alpha=-\beta\ln(\kappa\beta/kv)$, and $F=x$ we have

$$P(F) = \frac{1}{\beta} \exp\p[{\frac{F+\beta\ln(\kappa\beta/kv)}{\beta} -\exp\p({\frac{F+\beta\ln(\kappa\beta/kv)} {\beta}}) }]$$ (27)

$$= \frac{1}{\beta} \exp(F/\beta)\exp[\ln(\kappa\beta/kv)] \exp\p\{{-\exp(F/\beta)\exp[\ln(\kappa\beta/kv)]}\}$$ (28)

$$= \frac{1}{\beta} \frac{\kappa\beta}{kv} \exp(F/\beta) \exp\p[{-\kappa\beta/kv\exp(F/\beta)}]$$ (29)

$$= \frac{\kappa}{kv} \exp(F/\beta)\exp[-\kappa\beta/kv\exp(F/\beta)]$$ (30)

$$= \frac{\kappa}{kv} \exp(F/\beta - \kappa\beta/kv\exp(F/\beta)]$$ (31)

$$= \frac{\kappa}{kv} \exp(aF - \kappa/akv\exp(aF)]$$ (32)

$$= \frac{\kappa}{kv} \exp(aF - b\exp(aF)] \propto h(F) \;.$$ (33)

So our unfolding force histogram for a single Bell domain under constant loading does indeed follow the Gumbel distribution.

3.3 Saddle-point Kramers model

For the saddle-point approximation for Kramers' model for unfolding (Evans and Ritchie (1997) Eqn. 3, () Eqn. 4.56c, van Kampen (2007) Eqn. XIII.2.2).

$$\kappa = \frac{D}{l_b l_{ts}} \cdot \exp\p({\frac{-E_b(F)}{k_B T}}) \;,$$ (34)

where $E_b(F)$ is the barrier height under an external force $F$, $D$ is the diffusion constant of the protein conformation along the reaction coordinate, $l_b$ is the characteristic length of the bound state $l_b \equiv 1/\rho_b$, $\rho_b$ is the density of states in the bound state, and $l_{ts}$ is the characteristic length of the transition state

$$l_{ts} = TODO$$ (35)

Evans and Ritchie (1997) solved this unfolding rate for both inverse power law potentials and cusp potentials.

3.3.1 Inverse power law potentials

$$E(x) = \frac{-A}{x^n}$$ (36)

(e.g. $n=6$ for a van der Waals interaction, see Evans and Ritchie (1997) in the text on page 1544, in the first paragraph of the section Dissociation under force from an inverse power law attraction). Evans then gets funky with diffusion constants that depend on the protein's end to end distance, and I haven't worked out the math yet...

3.3.2 Cusp potentials

$$E(x) = \frac{1}{2}\kappa_a \p({\frac{x}{x_a}})^2$$ (37)

(see Evans and Ritchie (1997) in the text on page 1545, in the first paragraph of the section Dissociation under force from a deep harmonic well). Download unfolding_distributions.pdf
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3 Single-domain proteins under constant loading
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