Transcript Biochem Review, Part I: Protein Structure and Function

Biochem Review, Part I:
Protein Structure and Function
Lecture Notes pp. 1-38
An Amino Acid
different for
each AA
common to all
amino acids*
*except proline, in which the R group forms a
ring structure by binding to the amino group
The Twenty
Amino Acids
SMALL: hydrogen or methyl R group.
GLYCINE
Alanine
Types of AAs
NONPOLAR/HYDROPHOBIC:
R groups contain largely C, H atoms.
ALIPHATIC: no
aromatic rings.
AROMATIC: contains
aromatic rings.
VALINE
Leucine
Isoleucine
Methonine
*Proline
PHENYLALANINE
Tyrosine
Tryptophan
POLAR/HYDROPHILIC: R groups
typically contain O, N atoms.
CHARGED
ACIDIC: acid
in R group.
BASIC: base
in R group.
UNCHARGED: no
ionizable group in
R group.
LYSINE
Arginine
Histidine
ASPARAGINE
Glutamine
Serine
Threonine
*Cysteine
ASPARTATE
Glutamate
Special
Amino Acids
SMALL: hydrogen or methyl R group.
GLYCINE
Alanine
Types of AAs
NONPOLAR/HYDROPHOBIC:
R groups contain largely C, H atoms.
ALIPHATIC: no
aromatic rings.
VALINE
Leucine
Isoleucine
Methonine
*Proline
POLAR/HYDROPHILIC: R groups
typically contain O, N atoms.
CHARGED
UNCHARGED: no
ionizable group in
R group.
an imino acid, with R group
bound to amino group
thiol group can participate
in disulfide bonding
ASPARAGINE
Glutamine
Serine
Threonine
*Cysteine
Peptide Bond Formation
1
amino terminal residue
“N terminal”
2
carboxyl terminal residue
“C terminal”
Acid-Base Behavior of AAs
Each amino acid has at least TWO groups that display
acid-base behavior (gain or accept H+) – the carboxyl
group and amino group.
Acid-Base Behavior of AAs
Equilibrium constant: Ka = [A-][H+]/[HA]
pH = -log[H+]
pKa = -log(Ka)
…convenient shorthand for writing widely
variable [H+] concentrations
…similar shorthand for writing variable Ka values
The Henderson-Hasselbalch Equation
Relates three terms: pH, pKa, and [A-]/[HA]. If
you know two of these values, you can
determine the third.
pH = pKa + log([A-]/[HA])
When [A-] = [HA]:
pH = pKa + log(1)
pk is the pH at which a
functional group exists 50%
pH = pKa + 0
in its protonated form (HA)
and 50% in its deprotonated
pH = pKa
form (A-).
a
Isoelectric Point
Isoelectric point: the pH at which an AA or
polypeptide has no net charge.
• For a dibasic amino acid:
Isoelectric point = average of amino and carboxyl
pka values
= (2.4 + 9.8)/2 = 6.1
pka = 2.4
pka = 9.8
GLYCINE
• For a tribasic amino acid:
Isoelectric point = average of the two numerically
closest pka values
pka = 2.2
pka = 9.7
= (2.2 + 4.3)/2 = 3.25
Formulas can be used for polypeptides as long as
they have no more than one ionizable side chain.
pka = 4.3
GLUTAMATE
Titration Curves
This example shows the
curve for a dibasic AA
pka of amino group
pka values always
occur at the flattest
parts of the curve
(around 0.5, 1.5,
2.5 base equivs.)
pka of carboxyl group
buffering works best at pka
Isoelectric point
The isoelectric point will
always occur at at 1.0 or
2.0 base equivs.
buffering works best at pka
# of equivs. OH- tells whether AA is
dibasic (2 equivs) or tribasic (3 equivs.)
Practice Problem #1: Titration of Aspartate
2
Rule of thumb: if pH is >2 pH units
away from a group’s pka, that group
will effectively exist only in a single
form. (Below pka – protonated form;
above pka – deprotonated form.)
3
1
Posn
.
pH
1
~0
2
9.8
3
3.0
Major form(s)
present
Average charge on α
-carboxyl (pka = 2.1)
Average charge on R
carboxyl (pka = 3.9)
Average charge on amino
group (pka = 9.8)
Net
charge
Curve and structures from: http://www.cem.msu.edu/~reusch/VirtualText/proteins.htm
Practice Problem #1: Titration of Aspartate
Posn
pH
3
3.0
Major form(s)
present
Average charge on α
-carboxyl (pka = 2.1)
Average charge on R
carboxyl (pka = 3.9)
Average charge on amino
group (pka = 9.8)
Net
charge
For each relevant functional group , we will use the Henderson-Hasselbalch equation
to calculate [A-]/[HA] at pH 3.0.
Henderson-Hasselbalch equation: pH = pka + log([A-]/[HA])
Then, we will convert this ratio into % of groups protonated:
% groups protonated = [HA]/([HA]+[A-]) = 1/(1+[A-]/[HA]) x 100%
Practice Problem #1: Titration of Aspartate
Pos
n
pH
3
3.0
Major
form(s)
present
Average charge on
α -carboxyl (pka =
2.1)
For the alpha carboxyl:
Average charge on
R carboxyl (pka =
3.9)
Average charge on
amino group (pka = 9.8)
Net
charge
For the R group carboxyl
pH = pka + log([A-]/[HA])
pH - pka = log([A-]/[HA])
[A-]/[HA] = 10pH-pka
% protonated = 1/(1+10pH-pka) x 100%
% protonated = 1/(1+103.0-2.1) x 100%
% protonated = 11.18% alpha-COOH
100%-11.18% = 88.82% alpha-COO-
% protonated = 1/(1+103.0-3.9) x 100%
% protonated = 88.82% R-COOH
100%-11.18% = 11.18% R-COO-
Average charge on alpha carboxyl:
(0)*(0.1118) + (-1)*(0.8882) = -0.8882
Average charge on R carboxyl:
(0)*(0.8882) + (-1)*(0.1118) = -0.1118
Practice Problem #1: Titration of Aspartate
Posn
pH
3
3.0
Major
form(s)
present
Average charge on
α -carboxyl (pka =
2.1)
Average charge on
R carboxyl (pka =
3.9)
Average charge on
amino group (pka = 9.8)
Net
charge
-0.89
-0.11
+1
0
Finally, we calculate the fraction of molecules in each form by calculating the
probability of finding all three functional groups in the protonated/ deprotonated
states present in that form.
P(D) =
P(alpha-COO-) *
P(R-COOH) * P(NH3+) =
(0.8882)(0.8882)(1.0) =
0.7889
P(C) = P(alpha-COO-) * P(R-COO-) * P(NH3+) =
(0.8882)(0.1118)(1.0) = 0.0993
Orders of Protein Structure
Primary Structure: the sequence of a protein
(the AAs it contains and the order they are in.
Image from: http://www.umass.edu/molvis/workshop/imgs/protein-structure2.png
Primary Structure
• Genetically determined: specified by sequence
of gene that encodes protein
Primary Structure
• Contains all information necessary to specify
higher orders of structure (3D shape)
Denature
Renature
break
disulfide bonds
break
disulfide bonds
The 3D structure a protein
spontaneously assumes is
its most stable structure.
Orders of Protein Structure
Primary Structure: the sequence of a protein
(the AAs it contains and the order they are in.
Secondary Structure: regular 3D motifs formed
by H-bond interactions between N-H and C=O
groups of the peptide backbone.
Image from: http://www.umass.edu/molvis/workshop/imgs/protein-structure2.png
Secondary Structure
• Two major motifs: α helix and β sheet
• α helix is “default” structure for an AA chain
(energetically favorable)
The carbonyl of each AA Hbonds with the amino group
of an AA 4 residues down.
Secondary Structure
• Two major motifs: α helix and β sheet
• β sheet is composed of parallel strands of AAs
connected to one another by H-bonds
N terminus
turn in chain
C terminus
Like the α helix, the β sheet is formed by interactions WITHIN a single polypeptide chain.
Secondary Structure
Amino acid identity affects secondary structure.
Helix is destabilized by:
– Bulky R groups (e.g., Try)
– Adjacent R groups with like charges
– Proline, which cannot H-bond
Proline, in particular, tends to appear in
unstructured regions (turns).
Orders of Protein Structure
Primary Structure: the sequence of a protein
(the AAs it contains and the order they are in.
Secondary Structure: regular 3D motifs formed
by H-bond interactions between N-H and C=O
groups of the peptide backbone.
Tertiary Structure: the overall structure of a
single polypeptide chain.
Image from: http://www.umass.edu/molvis/workshop/imgs/protein-structure2.png
Tertiary Structure
• Tertiary structure can be complex and does
not typically consist of repeating units.
• A polypeptide will adopt the most stable
tertiary structure
In aqueous environment, this
occurs when hydropobic residues
are internal and hydrophilic
residues are external.
= charged
= hydrophobic
= neither
Orders of Protein Structure
Primary Structure: the sequence of a protein
(the AAs it contains and the order they are in.
Secondary Structure: regular 3D motifs formed
by H-bond interactions between N-H and C=O
groups of the peptide backbone.
Tertiary Structure: the overall structure of a
single polypeptide chain.
Quaternary Structure: interaction of multiple
polypeptides to form one functional protein.
Image from: http://www.umass.edu/molvis/workshop/imgs/protein-structure2.png
Quaternary Structure
• Proteins with quaternary structure have
multiple subunits and are oligomeric
• Subunits may be identical or different
• Oligomeric proteins have the potential for
cooperativity
All cooperative proteins must be
oligomeric, but NOT all oligomeric
proteins are cooperative.
Methods to Analyze 1o Structure
• Ion exchange chromatography: tells you
composition but NOT sequence
• Edman degradation: tells you sequence of a
short (<75 AA) fragment
• Proteolytic cleavage: generates short
fragments suitable for Edman degradation
• Electrophoresis: separates proteins according
to net charge
Ion Exchange Chromatography
• Tells you what AAs a protein contains and
the relative abundance of each
Ion Exchange Chromatography
1. Digest protein in strong acid
2. Load onto column of negatively
charged (sulfonate coated) beads;
AAs bind because positively
charged
3. Wash column with solutions of
increasing pH, collecting eluate in
small fractions
4. For each fraction, determine the
presence/absence and abundance
of the corresponding AA
Ion Exchange Chromatography
• An AA will elute approximately when the pH
of the solution equals its isoelectric point
Ion exchange
chromatography
can also be used
on small
polypeptides.
These too will
elute in an order
determined by
isoelectric point.
AAs with low isoelectric points
(such as acidic AAs) come off earlier
Peak
height
indicates
relative
abundance
AAs with high isoelectric points
(such as basic AAs) come off later
Edman Degradation
determine
identity
determine
identity
• Tells sequence of a short
peptide fragment (no
greater than 75 AAs)
• N-terminal AA is
labeled, cleaved, and its
identity determined
• Process is repeated in
successive rounds
Proteolytic Agents
• Used to cleave a protein into smaller
polypeptides (which can then be analyzed)
• Each agent has unique specificity:
–
–
–
Trypsin: cuts after Lys, Arg
Chymotrypsin: cuts after Phe, Tyr, Trp, Leu, Met
Cyanogen Bromide: cuts after Met
Practice Problem #2: Proteolytic Agents
You have digested a polypeptide with two
different agents and obtained these fragments:
Trypsin: [Met-Phe-Val-Arg] [Ala] [Glu-Lys]
Chymotrypsin: [Val-Arg-Ala] [Glu-Lys-Met-Phe]
What is the sequence of the polypeptide?
Practice Problem #2: Proteolytic Agents
You have digested a polypeptide with two
different agents and obtained these fragments:
Trypsin: [Met-Phe-Val-Arg] [Ala] [Glu-Lys]
Chymotrypsin: [Val-Arg-Ala] [Glu-Lys-Met-Phe]
What is the sequence of the polypeptide?
[Glu-Lys][Met-Phe-Val-Arg][Ala]
[Glu-Lys-Met-Phe] [Val-Arg-Ala]
Electrophoresis
• Used to separate proteins into bands
according to their net charge.
1. Load protein samples into wells at one end of a gel.
2. Apply current; proteins will move through gel matrix
towards the pole opposite their net charge.
(+) pole
wells
(-) pole
Electrophoresis
• Used to separate proteins into bands
according to their net charge.
1. Load protein samples into wells at one end of a gel.
2. Apply current; proteins will move through gel matrix
towards the pole opposite their net charge.
Peptide with large net
negative charge
Peptide with smaller
net negative charge
Peptide with net
positive charge
(+) pole
wells
(-) pole
Methods to Analyze Higher Order
Structure
• Nuclear Magnetic Resonance (NMR): can
be used for small proteins
• Electron Microscopy: gives overall shape
but not atomic resolution
• X-Ray Diffraction: the “gold standard,”
determines what atoms are in the protein
and the distances between them
X-Ray Diffraction
1. Crystallize protein of interest
2. Expose crystallized protein to Xray source (wavelength 1.5
angstroms)
3. Record diffraction pattern
4. Use intensities and positions of
spots to determine atom
identity and position
Larger atoms deflect X-rays
more than smaller atoms due to
greater electron density.
Homologous Proteins
Homologous proteins are proteins from
different organisms that are very similar in
structure and function.
Ex: insulin, cytochrome C
Homologous Proteins
Homologous proteins from different organisms
are similar but (usually) not identical.
•
•
•
•
Differences arise via mutation
Differences that survive must adequately preserve
function (natural selection)
Differences can only survive at certain residues
AAs at a given position tend to be chemically similar
Thr Gly Ile
8 9
Insulin
Homologous Proteins
Homologous proteins from different organisms
are similar but not identical.
•
•
•
•
•
Differences arise via mutation
Differences that survive must adequately preserve
function (natural selection)
Differences can only survive at certain residues
AAs at a given position tend to be chemically similar
Overall structure must be preserved
(structure function)
The “Molecular Clock”
Molecular clock theory: AA differences
accumulate over time, such that # of
differences between homologous proteins can
be used to calculate evolutionary distance
(time of divergence) for two species.
more AA differences  more distantly related
The “Molecular Clock”
Assume that differences accumulate at a
constant rate (# of differences is directly
proportional to time of divergence).
Example: Species A and Species B diverged 100 mya. Protein X
from Species A and B differs at 10 positions.
Protein X from species C differs from that of Species B at 20
positions. How long ago did Species B and C diverge?
The rate of difference accumulation
is unique to each protein (i.e.,
mutations accumulate at different
rates in insulin and CytC).
Hemoglobin: Intraspecific AA
Changes
• Hemoglobin (Hb): found in
RBCs, transports oxygen
from lungs to tissues
• Tetrameric (4 subunits: 2α,
2β)
• Each subunit has a heme
group where O2 binds
Hemoglobin: Intraspecific AA
Changes
low O2
-
+
Electrophoresis of HbA and HbS
• Sickle cell anemia: RBCs
sickle and form filaments
under low O2 conditions,
get stuck in capillaries
• Caused by a single Glu 
Val mutation at position 6
of the Hb β subunit
Glu to Val substitution results in a less
negative net charge on HbS and
slower migration towards the + pole.
Hemoglobin: Kinetics & Regulation
Hemoglobin can exist in one of two states:
binds O2 weakly
binds O2 tightly
These states exist in equilibrium.
In the presence of certain regulators, one state
will be preferentially stabilized, shifting the
equilibrium towards T or R.
Hemoglobin: Kinetics & Regulation
O2 is a homotropic
activator of Hb.
A homotropically
activated protein
displays cooperativity.
Binding of target molecule (O2) at
one active site enhances the
affinity of other active sites for
target molecule.
Hemoglobin: Kinetics & Regulation
H+, CO2, and BPG are
heterotropic inhibitors
of Hb.
They do not resemble
O2 and do not bind at
the active site.
Binding of heterotropic inhibitors
at non-active sites decreases
affinity for O2 at the active sites.
Hemoglobin: Kinetics & Regulation
Low O2
Low pH
High CO2
O2 Transport
High O2
High pH
Low CO2
Regulation of Hb
is optimized to
promote uptake
(high saturation)
of O2 in lungs and
deposition (low
saturation) of O2
in tissues.
Practice Problem 3:
Greenglobins
You are studying a strain of mice that have green eyes.
You believe that the green color is due to a molecule
called protogreen, which is produced in the intestine
and transported to the eye.
You hypothesize that a family of proteins called
greenglobins are the transporters, and that their
affinity for protogreen is affected by retinoin, a small
organic molecule present only in the eye.
Image from: http://thebluerepublic.com/Gallery/albums/album03/250px_Mus_Musculus_huismuis.jpg
Practice Problem 3: Greenglobins
% of progreen binding
sites occupied
There are four different greenglobins, A-D. You conduct
binding experiments to learn more about them:
% of progreen binding
sites occupied
Practice Problem 3: Greenglobins
1.
For each greenglobin, determine whether it is oligomeric or
monomeric or if you can’t tell:
Greenglobin A:
Greenglobin B:
Greenglobin C:
Greenglobin D:
oligomeric
oligomeric
oligomeric
oligomeric
monomeric
monomeric
monomeric
monomeric
can’t tell
can’t tell
can’t tell
can’t tell
% of progreen binding
sites occupied
Practice Problem 3: Greenglobins
2. For each greenglobin, which type(s) of regulation are illustrated in the graphs
above? (Circle all that apply)
Greenglobin A:
Greenglobin B:
Greenglobin C:
Greenglobin D:
homotropic activation
homotropic activation
homotropic activation
homotropic activation
heterotropic activation
heterotropic activation
heterotropic activation
heterotropic activation
heterotropic inhibition
heterotropic inhibition
heterotropic inhibition
heterotropic inhibition
% of progreen binding
sites occupied
Practice Problem 3: Greenglobins
3. Rank the different greenglobins in their ability to transport
progreen from the intestine to the eye.
1. _______ (best)
2. __________
3.__________
4.________ (worst)