Determination of the Heat of Fusion of Insulin

Insulin exists as a hexamer, both in solution and as a crystal. Bergeron et al. $^{6}$ report the solubility of insulin in aqueous solution to be about 0.122 mg/mL at 10°C and 0.182 mg/mL at 25°C, and the molecular weight of the insulin hexamer to be 34 800. Use these data to estimate the heat of fusion of the insulin hexamer.

$^{6}$ L. Bergeron, L. F. Filobelo, O. Galkin, and P. G. Vekilov, Biophys. J., 85, 3935 (2003).

Question

Determination of the Heat of Fusion of Insulin

Insulin exists as a hexamer, both in solution and as a crystal. Bergeron et al.[latex]^{6}[/latex] report the solubility of insulin in aqueous solution to be about 0.122 mg/mL at 10°C and 0.182 mg/mL at 25°C, and the molecular weight of the insulin hexamer to be 34 800. Use these data to estimate the heat of fusion of the insulin hexamer.

[latex]^{6}[/latex]L. Bergeron, L. F. Filobelo, O. Galkin, and P. G. Vekilov, Biophys. J., 85, 3935 (2003).

Accepted Answer

Since the molecular weight of the insulin hexamer is so high, its mole fraction at both temperatures is very small. For example, at the higher concentration of 0.182 mg/mL,

[latex]x=\frac{\frac{0.000182}{34800}}{\frac{0.000182}{34800}+\frac{1}{18}}=9.414 imes 10^{-8}[/latex]

Therefore, we can assume that the activity coefficients that appear in Eq. 12.1-19 at both temperatures are the values at infinite dilution and (assuming the temperature dependence is not very large) will cancel in the ratio term in the equation. Also, the solubilities S in the problem statement are so low that they are linearly related to the mole fraction by

[latex]x=\frac{S imes 18}{34800}[/latex]

so that the reported solubilities can be used in the ratio in Eq. 12.1-19, and we have

[latex]\ln \left[\frac{x_{1}\left(T_{1} ight) \gamma_{1}\left(\underline{x}, T_{1} ight)}{x_{1}\left(T_{2} ight) \gamma_{1}\left(\underline{x}, T_{2} ight)} ight]=-\frac{\Delta_{ fus } \underline{H}_{1}}{R}\left[\frac{1}{T_{1}}-\frac{1}{T_{2}} ight]=-\frac{ riangle_{ fus } \underline{H}_{1}}{R}\left[\frac{T_{2}-T_{1}}{T_{1} T_{2}} ight][/latex] (12.1-19a)

[latex]\ln \left[\frac{x_{1}\left(T_{1} ight)}{x_{1}\left(T_{2} ight)} ight]=-\frac{\Delta_{ ext {fus }} \underline{H}_{1}}{R}\left[\frac{1}{T_{1}}-\frac{1}{T_{2}} ight]=-\frac{ riangle_{ ext {fus }} \underline{H}_{1}}{R}\left[\frac{T_{2}-T_{1}}{T_{1} T_{2}} ight][/latex] (12.1-19b)

[latex]\ln \left[\frac{S_{1}}{S_{2}} ight]=-\frac{\Delta_{ ext {fus }} \underline{H}}{R}\left[\frac{T_{2}-T_{1}}{T_{1} T_{2}} ight] \quad ext { or } \quad riangle_{ ext {fus }} \underline{H}=R\left[\frac{T_{1} T_{2}}{T_{1}-T_{2}} ight] \ln \left[\frac{S_{1}}{S_{2}} ight][/latex]

Consequently, for the crystallization of the insulin hexamer

[latex] riangle_{ ext {fus }} \underline{H}=8.314\left[\frac{283.13 imes 298.15}{283.15-298.15} ight] \ln \left[\frac{0.122}{0.182} ight]=18.7 \frac{ kJ }{ mol }[/latex]

Since the value of the heat of fusion (or crystallization) is positive, the solubility of insulin increases with increasing temperature.

Question 12.1.7: Determination of the Heat of Fusion of Insulin Insulin exist...

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