Q1. You can call a  alpha-helix as 3.613 helix. What can you call a pi-helix in this kind of nomenclature?
     What are the n and r parameters for pi-helices?

Ans:
n - the number of residues per helical turn
r - the rise per helical residue

 4.416 Helix
 n-4.4 
 r-1.1 Angstroms
 
 
 Q.2

Ans:
 Helix boundary residues (the first and last helical residues) are called Ncap and Ccap at the N- and C-terminus, respectively (Richardson & Richardson, 1988). The proposed nomenclature uses the following definition:
...-N''-N'-Ncap-N1-N2-N3-............-C3-C2-C1-Ccap-C'-C''-...
But some amino acids have strong preferences, like this table shows.

Pro: For it's structure, I think the cis conformation plays an important role! And it has a hydrogen donor in -NH.
Gly: It is flexible, so it can be N-cap and C-cap.
Ser: The -NH in Ncap will form a hydrogen bond to residue bellow.
Thr: The R group has -OH, and the hydroxyl group can be a hydrogen bond aceptor.
Asn: Maybe the -NH in R group arm can be a hydrogen donor in ncap position!
Gln: Like r group of Asn, but Gln contribute it's C=O on R group arm to form the H-bond.
Asp & Glu: Appears in many places in N-terminal, for it has COOH in R group.Can form new H-bond with  previous a.a.'s  -NH .
Lys & Arg:The -NH on R group arm can be a hydrogen donor for synthsis Hydrogen-Bond.
 

Q3. How does a prolin residue lead to the kink of alpha-helices?
Ans: Pro (and hydroxyproline) acts as helix breaker(or helix terminator)  due to their unique structure, with fixes the value of the Calpha-N-C bond angle.
   
 Proline kinked helices are almost exclusively long helices (>4 turns) which might be a requirement to overcome the effect of the kink. Somewhat surprisingly, only one hydrogen bond (amide of the following residue, i+1, to the carbonyl oxygen of i-3) was found to be broken due to the kink in the helix.
 

Q4.
        A       AEAAAKEAAAKEAAAKA

        B       AKAAAEKAAAEKAAAEA

        C       AEAAKAEAAKAEAAKA Blue: positively charged residues

        D       AKAAEAKAAEAKAAEA red: negatively charged residues
 were compared (S. Marqusee and R. L. Baldwin, Proc. Natl. Acad. Sci. USA 84, 8898-8902,
     1987). The a -amino and a-carboxyl groups were blocked with acetyl and amide groups,
     respectively. How would the first and second peptides of each pair be expected to differ in their
     helicity? What's the reasons for these differences?

Ans: A & B are i+4 helix, and C & D are i+3 helix. The helicity of i+4 is better than i+3! So AB are better than CD! According  to the macrodipole, positive charges of N-terminal is more stable. So A is better than B, and C is better than D.  
 

 
 Q5. Ion pairs in proteins involving Arg residues have been observed to be energetically stronger
     than those involving Lys residues. If this were the case, in what ways might Arg and Lys residues
     be used differently in proteins? How could this hypothesis be tested? (Hint: D. B. Wigley et al.,
     Biochem. Biophys. Res. Commun. 149, 927-929, 1987)
Ans:Arg forms a tighter ion pair with a carboxylate group than dose Lys and is always for ion-pairs which are not broken during turnover in naturally-occuring enzymes. The ion pair with Arg was 20kJ/mol strong than the corresponding Lys-carboxylate ion pair, so the bond of Lys is weak. From the six proteins in the paper(Table 1). The bond between the protein and carboxylate is always with Arg when the protein carboxylate bond is not transformed during the enzyme reaction.
The enzyme involved Lys residues in this case, I think it may be a "transcarboxylase".   The carboxylate is transformes during the enzyme reaction.   The distinction can be well illustration with citrate synthase. Citrate contains three carbonxylates. Two of these remain attach to the protein durind catalysis and make bonds with Arg. The third, which is drived from the thioester in Acetyl-CoA is bound to the enzyme by catalytic His.
 

Q6.a) This is a schematically drawing of a four-stranded beta sheet. Please indicate the antiparallel
     and parallel strands.

Ans:

1  ------->
2  <-------
3  ------->
4  ------->

antiparallel strans: 1-2, 2-3
parallel strans:  3-4

 
 b) What is the diople moment of a beta-sheet? Is it plausible that parallel and antiparallel sheets
     could have substantially different dipole interactions? (Hint: W. G. Hol et al., Nature 294,
     532-536, 1981; P. T. van Duijnen et al., Biopolymers 24, 735-745, 1985)
 
Ans:In studying the diople moment of a beta-sheet phenomenon electrostatically,we examined the arrangement of peptide units in parallel beta-strands. From schematic drawings of hydrogen bondind patterns in parallel beta-sheet(Fig 2 on the paper), it is obvious that the N->H as well as the C->O diploes point backward with respect to the N to C-terminal direction of the strands. The suggests that the parallel beta-sheet has a significant overrall dipole moment, with the N-terminal end of the strands corresponding to the positive ends of the dipole(usually near C-terminal end).
Electrostatically, an energy  difference between two kinds of pleated sheet may be related to the fact that in an antiparallel sheet all peptide dipole moments seem to cancel each other out; there is no residual moment in either direction. Difference in electrostatic energy of ~0.8 kcal/mol between a parallel and antiparallel arrangement of three strands. And "Yes", it is significant ! they have substantically different dipole interactions.