Biochemistry
Biochemistry
9th Edition
ISBN: 9781319114671
Author: Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Publisher: W. H. Freeman
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6. Would Lys-Glu-Ser have the same pl as Ser-Glu-Lys? Explain.
7. For a protein, how do you think you might estimate the pI?
8. Draw a dipeptide (use R, and R, for the side chain R groups) and the resonance structures of the peptide
bond.
9. Recall the geometry about atoms that participate in double bonds or partial double bonds. What atoms
form the rigid plane of the peptide bond (which atoms are coplanar)?
10. How do you expect the rigid plane of the peptide bond to impact folding?
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Transcribed Image Text:6. Would Lys-Glu-Ser have the same pl as Ser-Glu-Lys? Explain. 7. For a protein, how do you think you might estimate the pI? 8. Draw a dipeptide (use R, and R, for the side chain R groups) and the resonance structures of the peptide bond. 9. Recall the geometry about atoms that participate in double bonds or partial double bonds. What atoms form the rigid plane of the peptide bond (which atoms are coplanar)? 10. How do you expect the rigid plane of the peptide bond to impact folding?
The pl of a peptide is determined by examining the ionizable groups. The protonated and unprotonated
forms of each ionizable group are in equilibrium. Consider the peptide Lys-Glu-Ser shown below at pH
7.2. The complete structure is on on the left and a stylized structure with just the ionizable groups is on the
right. While the N-terminal is depicted as protonated, a sample of Lys-Glu-Ser is composed ofa population
of molecules and within that population some molecules may contain a non-protonated N-terminal group
at pH 7.2.
Model 2:
COO-
H,N*.
pK,= 3.5
N-CH-C N CH- -0-
H.
OH
:-
pK= 8.5
H,N*-CH-C-
H.
COO-
NH;*
pK= 10.0
CH2
pK=4.2
CH2
CH2
一
CH2
OH
CH2
CH2
C0
CH2
NH,*
In the stylized structures below, only one molecule is drawn. However each diagram represents a collection
of many molecules. Therefore "half protonated" implies that half the molecules present are protonated and
half are not.
HạN-
COOH
H3N*-
-COOH/
COO-
H3N-
COO-
OH
OH
COOH
NH3
COOH
NH3
СООН
NH3 COO-
At pH=1 all groups are protonated.
At pH=3.5 the C terminal is half proton-
ated and half deprotonated.
At pH=4.2 the R group carboxyl is half
protonated, half deprotonated.
NET charge = 0.5
NET charge = +2
NET charge = +1.5
H3N*
/H,N
COO-
H2N
COO-
H2N-
OH
COO-
CO0-
NH,
OH
COO-
OH
NH3 /NH2
COO-
At pH=8.5 the N terminal is half
protonated and half deprotonated.
NH2
At pH=10.0 the R group amino of lysine
is half protonated, half deprotonated.
NET charge -0.5
At pH=12 all groups are deprotonated.
%3D
NET charge = -1.5
NOTE: In peptides and proteins, the N-terminal and C-terminal groups have different pK,s from the parent
amino acid. The pK, of the N-terminal is about 8.5 whereas C-terminal pK, is about 3.5
NET charge =-2.0
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Transcribed Image Text:The pl of a peptide is determined by examining the ionizable groups. The protonated and unprotonated forms of each ionizable group are in equilibrium. Consider the peptide Lys-Glu-Ser shown below at pH 7.2. The complete structure is on on the left and a stylized structure with just the ionizable groups is on the right. While the N-terminal is depicted as protonated, a sample of Lys-Glu-Ser is composed ofa population of molecules and within that population some molecules may contain a non-protonated N-terminal group at pH 7.2. Model 2: COO- H,N*. pK,= 3.5 N-CH-C N CH- -0- H. OH :- pK= 8.5 H,N*-CH-C- H. COO- NH;* pK= 10.0 CH2 pK=4.2 CH2 CH2 一 CH2 OH CH2 CH2 C0 CH2 NH,* In the stylized structures below, only one molecule is drawn. However each diagram represents a collection of many molecules. Therefore "half protonated" implies that half the molecules present are protonated and half are not. HạN- COOH H3N*- -COOH/ COO- H3N- COO- OH OH COOH NH3 COOH NH3 СООН NH3 COO- At pH=1 all groups are protonated. At pH=3.5 the C terminal is half proton- ated and half deprotonated. At pH=4.2 the R group carboxyl is half protonated, half deprotonated. NET charge = 0.5 NET charge = +2 NET charge = +1.5 H3N* /H,N COO- H2N COO- H2N- OH COO- CO0- NH, OH COO- OH NH3 /NH2 COO- At pH=8.5 the N terminal is half protonated and half deprotonated. NH2 At pH=10.0 the R group amino of lysine is half protonated, half deprotonated. NET charge -0.5 At pH=12 all groups are deprotonated. %3D NET charge = -1.5 NOTE: In peptides and proteins, the N-terminal and C-terminal groups have different pK,s from the parent amino acid. The pK, of the N-terminal is about 8.5 whereas C-terminal pK, is about 3.5 NET charge =-2.0
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Is there a reason why you divided the pKa's by 2 when determining the pI?

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