Elastin like polypeptide (ELP) has interesting thermal behaviors. At room low te

Question

Elastin like polypeptide (ELP) has interesting thermal behaviors. At room low temperature, it is highly solubilized in the water, but it's precipitated out from the solution when it's heated up called inverse temperature transition (ITT). Here are the model ELP structures and their thermal behavior characterized by spectroscopy (Figure 2). In figure 2, n=50. Using UV-Vis spectrophotometer, we obtained the following observation related to IIT of the ELP (Figure 5).

(a) Calculate the entropy and enthalpy values related to this thermal behaviors of poly (Ile-25) at pH7 and poly (Lys-25) at pH13 and 7.

(b) Calculate the percent of the protonated amino side chains of Lys at pH 7 and pH 13 (5). Explain why poly (Lys-25) have pH dependent thermal behaviors and explain why this phenomenon is reflected in the thermodynamic parameters that you calculated.

Flexible Elastin Backbone H2N-I(Va-Pro-Gly-Val-Gly)(Val-Pro-Gly-Lys-Gly)ln CO2H Chemically reactive crosslinkable group NH3 Figure 2. Design of the repeat sequence of the cross-linkable elastin-mimetic protein polymer poly(Lys-25). In the repeat sequence of poly(Ile-25), an isoleucine residue replaces the lysine residue of poly(Lys-25)

Explanation / Answer

(a) We know, Keq = fun/(1-fun)

Where fun = fraction of protein unfolded

              Keq = equilibrium constant

Also we know, fun= (Aobs-Af)/(Aun-Af)

Where Aobs=oberved absorbace

             Af = absorbace of completely folded state

             Aun = absorbace of completely unfolded state

For lle-25 (pH 7)

Temparature(K)

1/T

fun

Keq

lnKeq

293

0.0034

0

0

1

308

0.00325

0.56

1.27

0.24

313

0.0032

1

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For Lys-25 (pH 13)

Temparature(K)

1/T

fun

Keq

lnKeq

293

0.0034

0

0

1

299

0.0033

0.478

0.916

-0.088

328

0.00305

1

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For Lys-25 (pH 7)

Temparature(K)

1/T

fun

Keq

lnKeq

293

0.0034

0

0

1

345

0.003

0.5

1

0

368

0.0027

1

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Now if we plt a linear fitted graph between lnKeq and 1/T then

Enthalpy= -R × slope

Entropy = R × intercept

Enthalpy(Joule.K-1)

Entropy (Joule.K-1)

Poly (lle-25)(pH 7)

-42118.7

-134.7

Poly (Lys-25)(pH 13)

-90456.3

-299.3

Poly (Lys-25)(pH 7)

-20785

-299.3

(b) We know

       pH=pKa + log (base/acid)   

pKa of lysine for the side chain is 2.18

So f= fraction of protonated side chain amino acid residue= acid/base

Hence for poly(Lys-25) %f= 0.0015(pH=7) and %f= 1.5 × 10-9

Poly (Lys-25) have polar protic amino residue side chain. Hence it degree of protonation or deprotonation depends on pH. Therefore the overall interaction in amino acid also depends on it. Hence the thermodynamic factors also depend on pH value.

Temparature(K)

1/T

fun

Keq

lnKeq

293

0.0034

0

0

1

308

0.00325

0.56

1.27

0.24

313

0.0032

1

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