Problem Set I

WITH CORRECTIONS

Problem Set I

Biochem 200

Fundamentals of Protein Structure and Function

1. (a) Explain the difference between DG°' and DG'.

DG°' is the standard free energy change at pH 7.0 and at 1 M reactants and products.

DG' = DG°' + RT log {[C] [D] / [A] [B]}. It is a measure of the thermodynamic drive for the reaction to occur at pH 7.0 given these starting concentrations of A,B,C, and D.

(b) One mole of compound X can be converted by phosphorylation to one mole of compound Y. This conversion of X to Y has a DG°' = + 4.5. At neutral pH and 25°C compound X is stable for many weeks.

(i) Is the reaction X to Y energetically favored?

(ii) What is the equilibrium constant at pH 7.0 and at standard conditions for the conversion of X to Y?

DG°' = -2.303 RT log K'eq

+ 4.5 kcal/mol = -2.303 RT log K'eq

log K'eq = -1.954/RT = (- 1.954 kcal/mol)/(1.98 x 10-3 kcal mol-1 °K-1)(298°K)

log K'eq = - 3.31

K'eq = 4.9 x 10-4

(iii) What can you deduce from the stability of compound X?

The simplest explanation is that the reaction is not favored.

(iv) If the conversion of X to Y had a DG°' = - 4.5, is it possible that the compound could still be stable for many weeks, and, if so, why?

Yes, because there could be a large energy of activation barrier to the conversion of X to Y.

2. You have discovered a new protein in leukocytes that is involved in chemotaxis of the cell. The protein is cytoplasmic and represents 0.1% of the total cytoplasmic protein. The amount of total cytoplasmic protein in a single leukocyte is 1.4 ng. The molecular weight of the protein is 50,000 Daltons, and it has one binding site for a ligand that regulates the activity of your protein. Consider the cell a sphere of 30 mm diameter.

(a) Calculate the Molar concentration of your protein in the cell.

Volume of the cell = (4/3)pi x r3 = (4/3)pi x (15 x 10-4 cm)3 = (1.33)(3.14)(3.375 x 10-9) = 1.4 x 10-8 cm3 = 1.4 x 10-8 ml

1.4 x 10-6 mg/1.4 x 10-8 ml = 100 mg/ml

100 mg/ml x 0.1% = 0.1 mg/ml

0.1 mg ml-1 / 50,000 mg mmole-1 = 2 x 10-6 M = 2 mM

(b) Given a ligand concentration in the cell of 2 mM, what would the Kd (the dissociation constant) for binding of ligand to the protein have to be for 50% of the protein to be in the form of the protein-ligand complex?

P + L = P.L

Kd = [P]x[L]/[P.L] = (2 x 10-3 M) [P]/[P.L] = 2 x 10-3 M

(c) Given a ligand concentration in the cell of 5 mM, what would the Kd (the dissociation constant) for binding of ligand to the protein have to be for 50% of the protein to be in the form of the protein-ligand complex?

P + L = P.L

Kd = [P]x[L]/[P.L] = (4 x 10-6 M) [P]/[P.L] = 4 x 10-6 M

(d) What fraction of the protein will be in the form of the protein-ligand complex if the Kd is 1 nM and the ligand concentration is 5 mM?

From what you learned in part (c), most of the P will be saturated with L with such a low Kd.

Therefore, the [P.L] is approximately 2 mM, and therefore the free ligand concentration will be about 3 mM. Thus, roughly,

Kd = [P]x[L]/[P.L] = 10-9 M = (~3 x 10-6 M) [P]/[P.L]

[P.L]/[P] = ~3 x 10+3

Strictly speaking, you should set up a quadratic equation, the solution of which, you will find, is within a factor of two of the value above.

3. (a) The ionization of a weak acid can be written as

HA = H+ + A-

Derive the Henderson-Hasselbalch equation that relates the pH of the solution to the ratio of the acid to the base.

answer: The apparent equilibrium constant is

K = [H+][A-]/[HA]

pK = pH - log {[A-]/[HA]}

pH = pK + log {[A-]/[HA]}

(b) The imidazole ring of histidine can carry a + charge. At the normal pH that exists in cells (pH 7.0), what is the ratio of uncharged histidine to charged histidine at 25°C?

answer: Use the Henderson-Hasselbalch equation (they need to look up the pKa for the side chain of his in Stryer on page 43 = 6.0) --

7.0 = 6.0 + log {[his]/[his+]}

log {[his]/[his+]} = 1.0

{[his]/[his+]} = 10

use the Henderson-Hasselbalch equation

4. (a) What would the apparent Km (Km,app) of an enzyme for a substrate be if a competitive inhibitor were present at a concentration equal to its KI (i.e., inhibitor concentration = [I] = KI)?

Answer: Twice the Km. The Km,app = Km (1 + [I]/KI)

(b) What would Km,app be in the above situation if the inhibitor were noncompetitive? Why?

Answer: Same as the Km. A noncompetitive inhibitor does not affect the binding of the substrate.

kcat/Km is a measure of the catalytic efficiency of the enzyme.

(d) What sets the upper limit of kcat/Km?

The upper limit (108 to 109 M-1s-1) is set by the rate of diffusion of the substrate in solution, which limits the rate at which it encounters the enzyme.

5. (a) When a native protein denatures it assumes a random coil, with many possible conformations.

(i) Will the contribution of DS to the free energy change be + or - ?

positive

(ii) What requirement does this impose on DH if proteins are to be stable structures?

since DG = DH - TDS, a positive DS yields a negative contribution to DG. Thus for proteins to be stable, which requires DG to be positive for the native to random coil transition, denaturation must involve a large positive DH and/or an additional negative contribution from DS.

(b) Given what you know about DH and DS, explain why it is reasonable that proteins denature as the temperature is increased.

DS is positive for denaturation since order is lost, and this contributes to a more negative DG. DH is positive, but as temperature is increased, the TDS terms predominates and DG becomes more negative and the denaturation is favored.

If many hydrophobic residues are present in a protein, DS could be negative for unfolding since order may be gained in the water structure; this would result in a positive contribution to DG. Perhaps more H-bonds can form with water upon denaturation than existed in the structure, resulting in a negative DH which might lead to an overall negative DG as the T is decreased because the -TDS term gets smaller.