1. In class we calculated the value of the pressure for which diamond is in equilibrium with graphite. Suppose that you are working for a company that wants to make diamond and they determine that they need a chemical potential of -80 kJ/mol for the reaction. What pressure does this correspond to?

Solution: The difference here is that we must use the full relationship including DG (since DG is not zero here!). We do the same thing we did in class. We solve for the pressure.



 



 

2. Calculate a.) the activation energy and prefactor for the Arrhenius rate constant for the unfolding of the protein "foldase" given the data in the table. b.) Using the fraction folded determine the folding rate constant at each temperature. c.) Finally, use the data to determine DH^o and DS^o for the unfolding reaction.

1. We determine the activation energy and prefactor of the Arrhenius rate constant (k = Ae^-Ea/RT) by plotting ln(k) vs 1/T. On such a plot the data should be approximately linear with slope -E_a/R and intercept ln(A).



The slope is 4205.0 and the intercept is 22.3.

These values give E_a = 34.8 kJ/mol and A = 4.8 x 10⁹.

2. The equilibrium constant for the unfolding process f ß à u is K_u = [u]/[f] = 1-ff/ff where ff is the fraction folded. Once we have determined K_u we can determine the folding rate constant from microscopic reversibility (K_u = k_u/k_f or k_f = k_u/K_u).

 

 

 

 

 

T (K)	kobs (unfolding)	Fraction folded	Ku	kf
280	1450	0.84	0.19	7630
290	2420	0.70	0.43	5630
300	3936	0.50	1.00	3936
310	6180	0.31	2.22	2780
320	9470	0.18	4.56	2076

3. To determine the thermodynamic parameters we first use a van’t Hoff plot to determine the enthalpy.

The enthalpy is obtained from the slope; Slope = -7261.1 K, DH^o = (Slope in K)R = (7261.1)8.31 J/mol-K = 60,339.0 J/mol or approximately 60 kJ/mol.

To obtain the entropy use the fact that DG^o = 0 at 300 K (because the equilibrium constant is K = 1 at that temperature). Using the fact that DG^o = DH^o - TDS^o we can further state that DS^o = DH^o/300 K = 60,000 J/mol/ 300 K = 200 J/mol-K.

1. A pharmaceutical company is investigating a new lead for a drug known as Cureall. The drug binds to the active site of enzyme Blahase, which tends to make sick people feel lousy if it is not inhibited. The binding equilibrium is known to be:

1. Determine the binding (association) enthalpy for Cureall and Yuckose, respectively, with Blahase.

Solution: Use a van’t Hoff plot.

1/T (K-1)	ln(Ka) for (1)	ln(Ka) for (1)
0.00357	22.34	21.49
0.00345	21.16	20.74
0.00333	20.05	20.05
0.00322	19.01	19.41
0.00312	18.05	18.83



For (1) the slope is 9640 K.

Thus, DH^o = -R(9640 K) = -8.31 J/mol-K(9640 K) = - 80.1 kJ/mol

For (2) the slope is 5968 K.

Thus, DH^o = -R(5968 K) = -8.31 J/mol-K(6958 K) = - 49.5 kJ/mol

Notice that the slope of the van’t Hoff plot is positive because H^o < 0.

DH^o (1) = _________________. DH^o (2) = _________________.

2. Determine the binding (association) entropy for Cureall and Yuckose, respectively, with Blahase.

Solution: Calculate DG^o at 300 K. Note that it is the same for both the substrate and the inhibitor at this temperature.

DG^o = -RT ln K_a = -(8.31 J/mol-K)(300 K) ln (0.51 x 10⁹)

= - 49.98 kJ/mol ~ - 50 kJ/mol

DS^o (1) = (DH^o (1) - DG^o (1))/T ~ (- 80 kJ/mol + 50 kJ/mol)/300 K = -100 J/mol-K

DS^o (2) = (DH^o (2) - DG^o (2))/T ~ (- 50 kJ/mol + 50 kJ/mol)/300 K = -0 J/mol-K

DS^o (1) = _________________. DS^o (2) = _________________.

3. Cureall is a mimic for a carbohydrate, but it is a floppy molecule (i.e. it has many ether linkages and there any many possible conformations of the drug in solution). Your boss asks you to explain the sign of the entropy of binding of the drug.

The entropy change upon binding is large and negative. The reason for this is that the drug loses conformational entropy when it interacts with the protein. There is more freedom for the drug molecule to assume different conformations (and hence greater conformational entropy) when the drug is in solution.

 

 

4. Ascertain whether the drug binds more tightly than the native substrate Yuckose at body temperature.

e.) The pharmaceutical company states that the concentration of the drug Cureall can be as high as 10^-8 M while the native substrate has a concentration of 10^-6 M (1 micromolar). Assuming these concentrations and an enzyme concentration of 10^-6 M determine the effect of inhibitor on the enzyme kinetics at 290 K. The enzyme turnover number (i.e. the rate constant k_b) is 1000 s^-1 at 290 K. The binding half-life for the substrate Yuckose is 69.3 milliseconds. NOTE: The binding constants above are association constants. K_I for inhibition is usually reported as a dissociation constant (K_I = 1/K_a for 1).

K_M = ___________________. V_max = ___________________________.

V = ______________________________ for [S] = 10^-6 M.

V_I = ______________________________ for [S] = 10^-6 M and for [I] = 10^-8 M.

The rate slows by a factor of:

V/ V_I = ________________________________.

Using microscopic reversibility k_a' = k_a/K_a = 1.44 x 10⁷ M^-1s^-1/1.02 x 10⁹ M^-1 ~ 0.014 s^-1.

V_max = k_b[E]₀ = (1000 s^-1)(10^-6 M) = 10^-3 Ms^-1.

f.) Assuming the same concentrations as above determine the effect of inhibitor on the enzyme kinetics at 310 K. At this temperature the enzyme turnover number (i.e. the rate constant k_b) is unchanged at 10³ s^-1 and the binding half-life for the substrate Cureall is 6.93 milliseconds.

K_M = ___________________. V_max = ___________________________.

V = ______________________________ for [S] = 10^-6 M.

V_I = ______________________________ for [S] = 10^-6 M and for [I] = 10^-8 M.

With inhibitor

Experts suggest that an inhibitor should lower the enzyme rate by at least a factor 100 to be an effective drug. Should Smartmoney, Inc. invest in Cureall?

1. The enzyme convertase is inhibited by a non-competitive inhibitor that binds to an allosteric binding site with an association constant of K_a = 10⁴ M^-1. Convertase is observed to have a maximal velocity of 10.0 mmol/min for the conversion of sugar to carbohydrate. At a sugar concentration of 10^-6 M the rate is 5.0 mmol/min.

1. Determine K_M for convertase.
2. Determine effect on the maximal rate if the inhibitor concentration is 10^-2 M.

K_M = ____10^-6 M_______________.

V_max,I = ____100 mmol/min__________________________

In the presence of an inhibitor we have:

1. According to the chemiosmotic theory, an electrochemical proton gradient is used to synthesize ATP in mitochondria. The enzyme that does this is located on the inside of the mitochondrial membrane. The oxidation of carbohydrates and fats is used to pump protons outside the mitochondrial membrane until the steady state membrane potential is –140 mV and the pH gradient is DpH = 1.5 units. Inside the mitochondrion, pH = 7.0, [ATP] = 1 mM, [P_i] = 2.5 mM, [ADP] = 1mM, and T = 310 K and the standard free energy change is (DG^o = 31 kJ/mol).

1. A. How much free energy is required to synthesize ATP inside the microchondria?
2. B. How much free energy is made available by moving one mole of protons from the outside to the inside? Is this enough to drive ATP synthesis?
3. C. How many protons must be translocated per ATP synthesized?

1. Calculate the surface tension of a liquid if the capillary rise in a tube of radius 100 m is 0.1 m and the density is 0.8 gm/cm³.

Solution: rgh = 2g/R à g = rghR/2 = (800 kg/m³)(9.8 m/s²)(0.1 m)(10^-4 m)/2 = 0.039 J/m².

2. Myoglobin is a skeletal protein that binds oxygen. The standard free energy for the reaction

Myoglobin + O₂ à Oxymyoglobin

Is DG^o = -30.0 kJ/mol at 298 K and pH 7. The standard state of O₂ is the dilute solution molarity scale; therefore the concentration of O₂ must by in units of molarity (M). What is the ratio (oxymyoglobin)/(total myoglobin) in an aqueous solution at equilibrium with a partial pressure of oxygen P(O₂) = 30 Torr? Assume ideal behavior of O₂ gas and that Henry’s law holds for O₂ dissolved in water (The Henry’s Law Constant is K(O₂) = 43 x 10³ atm).

Solution: The concentration of O2 in solution is given by Henry’s law.

P(O₂) = K(O₂) x(O₂) or

x(O₂) = P(O₂)/K(O₂) = 30 Torr/43000 atm(760 Torr/atm) = 9.1 x 10^-7 ~ 10^-6.

Since the molarity of water is 55.5 molar the molarity of oxygen is approximately

[O₂] = 9.1 x 10^-7(55.5 M) ~ 5.0 x 10^-5 M.

K = exp{-DG^o/RT} = exp{30000/8.31/310} = 1.14 x 10⁵

q = K[O₂]/{1 + K[O₂]} = (1.14 x 10⁵)(5.0 x 10^-5)/{1 + (1.14 x 10⁵)(5.0 x 10^-5)} = 0.85

 

3. A. Determine the solution concentration of oxymyoglobin given that the extinction coefficient is 180,000 M^-1cm^-1 at the peak of the Soret band (l_max = 418 nm). The absorbance of the sample is 0.3 in a 1 millimeter pathlength cell.

Solution: A = ecd à c = A/ed = 0.3/(180,000 M^-1cm^-1)(0.1 cm) = 1.67 x 10^-5 M = 16.7 mM

B. Assuming that the photolysis yield for MbO₂ à Mb + O₂ is F_photolysis = 0.75 and that the absorbance of Mb is 0.1 at 418 nm, calculate the largest possible change in absorbance that can occur for laser photolysis of oxymyoglobin.

Solution: If all of the molecules are photoexcited then 25% will remain as MbO₂ (A =

0.3 x 0.25) and will be unchanged and 75% will be converted to Mb (A = 0.1 x 0.75).

The change in absorbance is thus DA = (0.3 – 0.1)0.75 = 0.15.

 

4. How many photons of light strike a molecule on average if the wavelength of the radiation is 500 nm and the total energy per unit area is 1 nanoJoule/mm² and the concentration of molecules is 1 mM in a 1 mm cuvette?