Conceptual Problems

1) (10 points) Give two reasons why biological molecule absorbance spectra are typically very broad (i.e. they have a large spectral width)?

First, biological molecules tend to have large numbers of closely spaced vibrational levels on top of the electronic transitions. Second, biological molecules exist in many different conformations, resulting in large inhomogeneous broadening.

2) (10 points) Explain in a couple of sentences why molecules absorb light at certain wavelengths and not others.

 

The wavelength of a photon of light is inversely proportional to its energy. For absorption to occur, the energy of the photon must match the energy difference between two electronic levels. Absorption can only take place at wavelengths were these energies match.

 

3) (10 points) Write down an example of a reversible reaction which is first order in the forward direction and second order in the reverse direction.

4) (10 points) Consider a protein that binds two different small molecules. The binding of the two molecules occurs at two distinct and separate sites. If A binds first to the protein, then this decreases the dissociation constant for B by a factor of two, relative to the dissociation constant for B when there is no A bound. If B binds first to the protein, will this change the dissociation constant for A, relative to the dissociation constant for A when no B is bound? If so, how much and in what direction?

Must have the same equilibrium constant overall as

Therefore, the cooperativity coefficient for B binding to PA must be the same as for A binding to PB. Therefore A’s dissociation constant is decreased by a factor of two when B is bound just as B’s dissociation constant is decreased by a factor of two when A is bound.

Quantitative problems

5) (15 points) In my lab, we work with a molecule called thimazole orange (TO). This molecule has a quantum yield of fluorescence that is about 0.25 when it is bound to DNA, but only about 0.0001 when it is free in solution. If the lifetime of the excited state is about 1 nanosecond (10^-9 seconds) when the TO is bound to DNA, what is its natural radiative rate constant? What is the lifetime of the excited state TO in the unbound form? (Assume that the natural radiative rate constant for the molecule is the same in the bound and unbound forms.)

When the TO is bound to the DNA:



When TO is not bound to the DNA:



Therefore the lifetime of the state is 0.4 picoseconds.

 

6) (15 points) Consider the binding of a hormone (H) to a cell surface receptor (R). If the initial values of [H], [R] and [HR] are 0.0005, 0.2, and 0.1 mM respectively, what are the initial rates of change for each of these component concentrations assuming that the forward rate constant is 1 mM^-1 s^-1 and the reverse rate constant is 2 s^-1 in the reaction: H+R ß à HR? Make sure your answer includes the rate equations associated with each of the species in the reaction.

7) Consider two aberrant forms of hemoglobin. The first binds only one oxygen molecule. The second binds two oxygen molecules cooperatively.

a) (15 points) Determine for both forms of hemoglobin what N (the number of oxygen molecules bound per hemoglobin) is in terms of the oxygen concentration, the dissociation constant for binding of the first oxygen (K_O2) and the cooperativity coefficient (a).

 

Hb binding one oxygen:

 



 

Hb binding two oxygens:

 



 

b) (15 points) Next, determine for both forms of hemoglobin

which is the ratio of the oxygen concentration required go from 25% occupancy of the available sites to 75% occupancy. Note that for a hemoglobin that binds only one oxygen, 25% is N=0.25 and 75% is N = 0.75. For hemoglobin which binds two oxygens, 25% occupancy means N = 0.5 (since the max N is 2.0) and 75% means N = 1.5. Use a cooperativity coefficient of 0.5 for the form of hemoglobin that binds two oxygen molecules.

Since all we want are relative values, lets consider [O₂]/K instead of just [O₂]. We will call this x. We could do it in terms of [O₂] itself, but the math would be messier.

Single Oxygen bound:

Two oxygens bound with cooperativity

This is the only positive solution. You do exactly the same thing for N=0.5 and get 0.274. The ratio is now 6.65 which is significantly smaller than 9. Thus, the range of oxygen required to go from 25% occupancy to 75% occupancy is smaller in the case where two oxygens bind.