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How many atoms compromise a repeating unit of a peptide backbone?
- 6 atoms:
- amide nitrogen & its H
- alpha carbon & its H
- carbonyl carbon & its O
(See slide #22)
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What is the minimum # of ionizable functional groups that can present itself in a peptide or a free amino acid? What are these functional groups?
2: amino (NH3+) & carboxylate (COO-)
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how to calculate the # of peptide bonds from the # of amino acids
# of peptide bonds = # of amino acids - 1
(Ex. 575 amino acids = 574 peptide bonds)
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Determine the total # of ionizable amino acids & total # of ionizable groups in the peptide sequence H3N+ - ELGSAHVK - COO-
- 4 ionizable amino acids: ESHK
- 6 ionizable groups: ESHK, NH3+, and COO-
(Ionizable amino acids: CDESTY HKR)
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Determine the # of stereoisomers of the peptide sequence H3N+ - QLTVGYARWIE - COO-
(11 total amino acids) - (1 for G) + (2 more chiral centers for I & T) = 12 total chiral centers
2^12 = 4096 stereoisomers
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A peptide has a total of 4,196,352 possible stereoisomers. How many chiral centers does it have?
(2)^n = # of stereoisomers
[log (2)^n]/log 2 = [log (4,196,352)]/log 2 = 22 chiral centers
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Determine the length (in amino acids) of a peptide that has 8192 stereoisomers. There are 2 Gs, 1 I, & 1 T residues.
Length of peptide = (total # of chiral centers + total # of G residues) - (total # I & T residues)
[log (8192)]/log 2 = 13 chiral centers
Length of peptide = (13+2) - (1+1) = 15 - 2 = 13
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Trypsin cleaves peptide bonds __
C-terminal (to the right of) Lys (K) & Arg (R) except when they are the last residue. (cannot be right next to COO- at the end)
Other exception: Pro (P) cannot be to the right of Lys & Arg. (Pro cannot be cleaved.)
See slide #30
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Chymotrypsin cleaves peptide bonds __
C-terminal (to the right of) Phe (F), Tyr (Y), Trp (W) except when they are the last residue. (cannot be right next to COO- at the end)
Other exception: Pro (P) cannot be to the right of Phe, Tyr, Trp. (Pro cannot be cleaved.)
See slide #31
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Cyanogen bromide turns __ into a homoserine (Hsr).
Met (M). After Met, the peptide would be cleaved off.
Ex. If using cyanogen bromide results in a tetrapeptide composed of Hsr & 3 other residues, Met would’ve been the last residue in that tetrapeptide.
(See slide #29 & problem set 1 #38)
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If one cycle of Edman degradation produces PTH-Tyr, where does Tyr go in the peptide sequence?
In the very beginning of the sequence next to H3N+ (N-terminus).
An Edman degradation technique identifies the amino terminal residue of a peptide.
(See slide #28 & problem set 1 #38)
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