Biochemistry: Can somebody please help me answer the biochemistry questions abov

Question

Biochemistry:

Can somebody please help me answer the biochemistry questions above and show your work on how did you choose it so I can understand them better?

The 9 alphabets represent the 9 amino acids. For example, V stands for Valine, D stands for Aspartate, I stands for Isoleucine, S stands for Serine, etc.

You can look at the structure of the amino acids here:

The next 7 questions on this page pertain the following peptide sequence VDISVRAKF At pH-7, the peptide contains the following types of residues a. 6 nonpolar, 4 aromatic, 2 containing hydroxyls, 1 -branched b. 4 positively charged, 1 nonpolar, 2 polar, and 2 aromatic c. 1 sulfur containing, 1 aromatic, 2 hydroxyl containing, and 4 methyl containing d, 3 -branched, 4 nonpolar, 3 charged, 1 polar, and 1 aromatic e. none of the above 1) What is the pl of this peptide? a. 3.90 b. 6.05 C. 9.40 d. 10.05 e. none of the above 2) At the pH's of 3, 7, & 11, what are the respective charge states of this peptide? a. +3, +1, &1 3) 0 d. +1, -1, &2 e. +3, +1, &0 How many .p torsion angle pairs, total? a. 8 b. 9 C. 16 d. 18 e. none of the above 4) If this were an -helix, how long would it be? a. 1.5 A b. 13.5 A c. 15.0 A d. 32.4 e. none of the above 5) 6) If this were an a-helix, how many residues have only one of thelr main-chain hydrogen bonding groups (amino and carbonyl) hydrogen bonded? a. none of the below c. 3 d. 8 e. 9 The best reason that explains why this sequence cannot form an amphipathic parallel -sheet conformation is a. 7) possesses many helik forming residues Noe Ne has too few residues for such a conformation Nu* l- /a possesses many helix forming residues NY c. alternating nonpolar/polar residue pattern is not evident d, sequence can form parallel -sheet e. has the proper in register hydrogen bonding pattern.

Explanation / Answer

1. d - VVAI - Non polar

VVI - Beta Branched (Means there is a branch at the first carbon on the SIDE CHAIN)

DRK - charged

S - Polar

F- Aromatic

2. d. Pi = 10.05 - Reason is that two basic residues are there so one counters the acidic D residue and the other increases the Pi . Basic Residues have ~10 or higher Pi. If you are in doubt input this seq in the following

http://www.tulane.edu/~biochem/WW/PepDraw/

3. pH3= +3

pH7= +1 (One extra Positive charge on the second basic residue... others neutralized)

pH=-1 (after Pi = 10.05, net charge is to be negative)

so a is the answer

4. For 2 amino acid peptide there is 1 pair of phi psi angle... two for 3 aa peptides... n-1 for n aa peptides so answer is a. 8

5. b. 13.5Angstrom

Reason - Pitch of helix = 5.4 A

No.of Residues in a pitch = 3.6 A

Therefore, Pitch/residue = 5.4/3.6 = 1.5 A

Now peptides have 9 residues hence length of helix = 13.5 A

6. d. 8 ( This is tricky ... the CO of the Nth residue will bond with the NH of the N+4th residue . So, CO of 1st, 2,3,4,5 will be bonded to NH of 5,6,7,8 and 9 respectively. SO for 1,2,3,4 CO is bonded and NH is free. For 6,7,8,9 NH is free. Hence it is 8 residues in total.

7. b. Too many helix forming residues. Only Threonine/I/L and aromatic residues prefer the beta sheet. Rest (The majority) prefer alpha helix.

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