9.5. Thermal, chemical, and mechanical switching[[Student version, January 17, 2003]] 323
the true change of free energy upon making the first bond as−(2α− 4 γ)kBT.(Some authors refer
to the quantity e−^4 γas the “initiation parameter.”)
Your Turn 9i
Use the previous discussion to find a rough numerical estimate of the expected value ofγ.The above discussion assumed that the extra free-energy cost of initiating a tract of alpha helix
is purely entropic in character. As you found in Your Turn 9i, this assumption implies thatγis a
constant, independent of temperature. While reasonable, this assumption is just an approximation.
Wewill see, however, that it is quite successful in interpreting the experimental data.
9.5.3 Calculation of the helix-coil transition
Polypeptides, large N Having definedαandγ,wecan now proceed to evaluate〈σav〉≡
〈N−^1
∑N
i=1σi〉,which we know is related to the observable optical rotation. We characterize confor-
mations by listing{σ 1 ,...,σN},and give each such string a probability by the Boltzmann weight
formula. The probability contains a factor of eασifor each monomer, which changes by e^2 αwhen
σichanges from− 1 (unbonded) to +1 (H-bonded). In addition, we introduce a factor of eγσiσi+i
for each of theN− 1 links joining sites. Since introducing a single +1 into a string of−1’s creates
twomismatches, the total effect of initiating a stretch of alpha-helix is a factor of e−^4 γ,consistent
with the above definition ofγ.
WhenNis very large, the required partition function is once again given by Equation 9.18,
and〈σav〉=N−^1 ddαlnZ.Adapting your result from Your Turn 9h and recalling thatθis a linear
function of〈σav〉gives the predicted optical rotation as
θ=C 1 +
C 2 sinhα
√
sinh^2 α+e−^4 γ. (9.25)
In this expressionα(T)isthe function given by Equation 9.24, andC 1 ,C 1 are two constants.
Your Turn 9j
Derive Equation 9.25, then calculate the maximum slope of this curve. That is, find dθ/dTand
evaluate atTm.Comment on the role ofγ.The top curve of Figure 9.6 shows a fit of Equation 9.25 to Doty and Iso’s experimental data.
Standard curve-fitting software selected the values ∆Ebond =0. 78 kBTr,Tm= 285 K,γ =2.2,
C 1 =0.08, andC 2 =15. The fit value ofγhas the same general magnitude as your rough estimate
in Your Turn 9i.
The ability of our highly reduced model to fit the large-Ndata is encouraging, but we allowed
ourselves to adjust five phenomenological parameters to make Equation 9.25 fit the data! Moreover,
only four combinations of these parameters correspond to the main visual features of the S-shaped
(orsigmoid)curve in Figure 9.6, as follows:
- The overall vertical position and scale of the sigmoid fix the parametersC 1 andC 2.
- The horizontal position of the sigmoid fixes the midpoint temperatureTm.
- OnceC 2 andTmare fixed, the slope of the sigmoid at the origin fixes the combination
e^2 γ∆Ebond,according to your result in Your Turn 9j.
In fact, it is surprisingly difficult to determine the parametersγ and ∆Ebondseparately from
the data. We already saw this in Figure 9.4: There the top gray curve and dashed gray curve,