B) Suppose the 100 ml of cellular extract is treated with 50 ml of 0.5 mM HCL. W
ID: 862921 • Letter: B
Question
B) Suppose the 100 ml of cellular extract is treated with 50 ml of 0.5 mM HCL. What is the final pH? [Assume that the HCI reacts with the phosphate buffer in the cell extract.] Which form of phosphate is most abundant now? 3. Thermodynamics of protein folding. The native and denatured forms of a protein are generally in equilibrium . For a certain solution of the protein ribonuclease A. in which the total protein concentration is 2.0 x 10^-3 M, the concentrations of the denatured and native proteins at 50 and 100 degree C are listed below. Temperature 50 degree C 100 degree C Protein (denatured) 5.1 x 10^-6 M 2.8 x 10^-4 Protein (native) 2.0 x 10^-3 1.7 X 10^-3 A) Determine triangle S degree and triangle H degree for the folding reaction. [Assume these quantities are independent of temperature.) B) Calculate triangle G degree for ribonuclease A folding at 25 degree C. Is this reaction Spontaneous at this temperature? C) What is the denaturation temperature of ribonuclease A?Explanation / Answer
3.
Protein denatured (PD) = protein native (PN).
50C
T1=50+273.15=323.15K
PD = 5.1*10^(-6)M
PN = 2*10^(-3)M
Keq1 = PN/PD = 392
100C
T2=100+273.15=373.15K
PD = 0.28*10^(-3)M
PN = 1.7*10^(-3)M
Keq2 = PN/PD = 6.1
Van't hoff equation
Log(Keq2/Keq1) =
DH0/(2.303*R)*(1/T1-1/T2)
R = 8.314 j K-1 mol-1
(a)
DH0 = ((2.303*R)/(1/T1-1/T2))*Log(Keq2/Keq1) = (2.303*8.314*Log(6.1/392))/(1/323.15-1/373.15) = -83415j/mole = -83.415kj/mole
DS0 = (DH0 - DG0)/T1 =
(DH0 -(-)R*T1*2.303*Log(Keq1))/T1 = DH0/T1 + 2.303*R*Log(Keq1) =
-83415/323.15 + 2.303*8.314*Log(392) = -258.13 + 49.65 = -208.5j/K mole
(B.)
At 25C T= 25+273.15=298.15K
DG0 = DH0-T*DS0= -83415-298.15*(-208.5) = -21256j/mole
DG0 < 0
So reaction is spontaneous at 25C
C.
At denaturation temperature T,
DH0 = T*DS0
T = DH0/DS0 = -83415/(-208.5) = 400.1K =400.1-273.15= 127C
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