BIOCHEMISTRY 440
ANSWER KEY for Problem Set #3
Page 1
ANSWER (1a):
With 8 Cys, there are 7 combinations to make the 1st S-S; once that’s formed there are 5 possible
ways to make the 2nd S-S; then there are 3 ways to make the 3rd S-S; and 1 left for the final.
NOTE: there are other ways to think about the combinatorials that will give you the same
conclusion.
So: # possible combinations = 7x5x3 = 105
ANSWER (1b):
By the same logic, there are 9 different Cys residues that can be the reduced one and then for
each of those, there are 105 possible S-S pairings.
So: # possible combinations = 9 x 105 = 945
SO, only ONE additional Cys residue really increases the number of incorrect ways the protein
can fold! This progression illustrates why it is often so challenging to purify proteins that
contain many Cys residues in their native form…there are so many ways they can go wrong.
ANSWER (2):
In expt. (b), Anfinsen allowed the disulfides to reform randomly. They were not guided by the
3D structure of the folded protein because they were formed under denaturing conditions. In that
case, only 1/105 combinations is the same as in the native state. Only the proteins w/ that
combination could then fold to the correct 3D structure, so only 1/105 = 0.95%
!
1% will have
activity in expt. (b).
ANSWER (3a):
HoloMb is much more stable than ApoMb. It takes more than double the concentration of
GuHCl to unfold 50% of holoMb.
ANSWER (3b):
The heme group (which is non-polar) binds into a cleft formed by the 3D structure of Mb.
Without the heme, this cleft has its hydrophobic surface exposed to solvent, which is unfavorable
(destabilizing the apo-form). Also, binding of the heme provides additional vdW’s interactions
between the heme and the cleft (stabilizing the holo-form).
ANSWER (3c):
We know that ΔGfolding <0 because apoMb is folded under normal conditions. However, it is not
terribly stable so we can conclude that ΔGfolding is not very negative. We can write the overall
folding process for holoMb as
ΔGoverall
Mbunfolded + heme holoMb
where ΔGoverall = ΔGfolding + ΔGheme binding
We know from the expt. that ΔGoverall < ΔGfolding so, ΔGheme binding < 0 also.
Thus, Mb couples 2 exergonic processes together to produce a more highly favored process.
