Answers to home exercises
1 a) The N-terminal part of an -helix in a protein has partial + charge because of the macrodipole (No free –NH3+ in the N-terminal, since all amino groups are included in peptide bonds).
This positive charge should be neutralized, since it is located in the interior of the protein which is apolar. This can be done by an amino acid just outside the helix that is negatively charged.
The first helix turn has NH-groups that do not participate in H-bonds within the helix. These NH-groups should be stabilized with H-bonds from some side-chain group located on an amino acid in the vicinity of the N-terminal end.
All these protecting so called N-cap amino acids are not a part of the -helix, but are located just outside the helix structure in the protein.
Examples:
Asp, Glu are negative, lead to sinificant stabilization (Glu somewhat less strongly, since it has a longer side chain, which from entropic reasons are less favorable to have in a fixed state).
Thr, Ser: The OH-group stabilizes free NH-groups by forming H-bonds.
Asn, Gln: The amide side chain can do the same.
His is partially + charged, which will lead to repulsion. This will more or less cancel the positive effect by its H-bonding capacity
Val, Pro are hydrophobic and apolar and are therefore unfavorable amino acids in the polar environment at the helix end. (That Pro is a helix breaker is in this context not an explanation, since N-cap aimno acids are located just outside the helix).
b) The C-terminal has a negatively charged macrodipole: Arg, Lys are good candidates because they are positively charged.
Furthermore, amino acids that can provide H-bonds to the free CO-groups in the first turn of this end like Ser, Thr,Asn, Gln will likely play a stabilizing role.
2.a) Because of a smaller entropy loss. The reactants are already more or less in place with the correct orientation (=the effects: proximity and orientation).
b) If you compare the equilibrium constants between an intramolecular and an intermolecular reaction you will have:
Kintra/Kinter which will give the dimension M (here called the effective concentration = Ceff). This because Kintra is dimensionless and Kinter has the dimension M-1.
Here; Ceff = 3x103/10-2 = 3x105 M (for succinic acid anhydride),
3 a) = ΔΔ G#/ ΔΔ Go.
If = 1, then a removed interaction (by mutagenesis) has influenced the final state (native state) as much as the transistion state. Therefore, this interaction should have been developed already in the transition state. This requires that the structure is formed in the transistion state.
b) β-strand 3 to β-strand 6. These strands have -values above 0,5.
c)
(från = from)
4 a) Tm is approximately 34 ̊ C and 48 ̊ C, respectively.
b) It stabilizes the protein.
c) ΔH is obtained from the area under the Cp temperature curve. ΔH is determined by the non-covalent interactions that are broken during denaturation.