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 TwentyAmino Acids Types of AAs

SMALL: hydrogen or methyl R group.GLYCINEAlanine

NONPOLAR/HYDROPHOBIC: R groups contain largely C, H atoms.

POLAR/HYDROPHILIC: R groups typically contain O, N atoms.

ALIPHATIC: no aromatic rings.

VALINELeucineIsoleucineMethonine*Proline

AROMATIC: contains aromatic rings.

PHENYLALANINETyrosineTryptophan

ACIDIC: acid in R group.

ASPARTATEGlutamate

BASIC: base in R group.

LYSINEArginineHistidine

CHARGEDUNCHARGED: no ionizable group in R group.

ASPARAGINEGlutamineSerineThreonine*Cysteine

Types of AAs

SMALL: hydrogen or methyl R group.GLYCINEAlanine

NONPOLAR/HYDROPHOBIC: R groups contain largely C, H atoms.

POLAR/HYDROPHILIC: R groups typically contain O, N atoms.

ALIPHATIC: no aromatic rings.

VALINELeucineIsoleucineMethonine*Proline

CHARGEDUNCHARGED: no ionizable group in R group.

ASPARAGINEGlutamineSerineThreonine*Cysteine

an imino acid, with R group bound to amino group

thiol group can participate in disulfide bonding

SpecialAmino Acids

Peptide Bond Formation

amino terminal residue“N terminal”

carboxyl terminal residue“C terminal”

1 2

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+] …convenient shorthand for writing widelyvariable [H+] concentrations

pKa = -log(Ka) …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)

pH = pKa + 0

pH = pKa

pka is the pH at which a functional group exists 50% in its protonated form (HA) and 50% in its deprotonated form (A-).

Isoelectric PointIsoelectric point: the pH at which an AA or

polypeptide has no net charge.• For a dibasic amino acid:

• For a tribasic amino acid:

pka = 2.4

pka = 9.8

Isoelectric point = average of amino and carboxyl pka values

pka = 2.2

pka = 9.7

pka = 4.3

Isoelectric point = average of the two numerically closest pka values

= (2.2 + 4.3)/2 = 3.25

= (2.4 + 9.8)/2 = 6.1 GLYCINE

GLUTAMATEFormulas can be used for polypeptides as long as they have no more than one ionizable side chain.

Titration Curves This example shows the curve for a dibasic AA

pka of carboxyl group

pka of amino group

Isoelectric point

pka values always occur at the flattest parts of the curve (around 0.5, 1.5, 2.5 base equivs.)

The isoelectric point will always occur at at 1.0 or 2.0 base equivs.

# of equivs. OH- tells whether AA is dibasic (2 equivs) or tribasic (3 equivs.)

buffering works best at pka

buffering works best at pka

Practice Problem #1: Titration of Aspartate

1

3

2

Posn.

pH 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

1 ~ 0

2 9.8

3 3.0

Curve and structures from: http://www.cem.msu.edu/~reusch/VirtualText/proteins.htm

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.)

http://www.cem.msu.edu/~reusch/VirtualText/proteins.htm

http://www.cem.msu.edu/~reusch/VirtualText/proteins.htm

Practice Problem #1: Titration of AspartatePosn pH Major form(s)

presentAverage charge on α -carboxyl (pka = 2.1)



Net charge

3 3.0

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%


For the alpha 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-COOH100%-11.18% = 88.82% alpha-COO-

Average charge on alpha carboxyl:(0)*(0.1118) + (-1)*(0.8882) = -0.8882

For the R group carboxyl

% protonated = 1/(1+103.0-3.9) x 100%% protonated = 88.82% R-COOH100%-11.18% = 11.18% R-COO-

Average charge on R carboxyl:(0)*(0.8882) + (-1)*(0.1118) = -0.1118

Posn

pH Major form(s) present




Net charge

3 3.0


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.

Posn pH Major form(s) present




Net charge

3 3.0 -0.89 -0.11 +1 0

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

http://www.umass.edu/molvis/workshop/imgs/protein-structure2.png

Primary Structure

• Genetically determined: specified by sequence of gene that encodes protein

Primary Structure

Denature Renature

The 3D structure a protein spontaneously assumes is its most stable structure.

• Contains all information necessary to specify higher orders of structure (3D shape)

break disulfide bonds

break disulfide bonds



Secondary Structure: regular 3D motifs formed by H-bond interactions between N-H and C=O groups of the peptide backbone.



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 H-bonds 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

Like the α helix, the β sheet is formed by interactions WITHIN a single polypeptide chain.

N terminus

C terminus

turn in 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).




Tertiary Structure: the overall structure of a single polypeptide chain.



Tertiary Structure

• Tertiary structure can be complex and does not typically consist of repeating units.

• A polypeptide will adopt the most stable tertiary structure

= charged

= hydrophobic

= neither

In aqueous environment, this occurs when hydropobic residues are internal and hydrophilic residues are external.




Tertiary Structure: the overall structure of a single polypeptide chain.

Quaternary Structure: interaction of multiple polypeptides to form one functional protein.



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 Chromatography1. Digest protein in strong acid2. 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

AAs with low isoelectric points (such as acidic AAs) come off earlier

AAs with high isoelectric points (such as basic AAs) come off later

Peak height indicates relative abundance

Ion exchange chromatography can also be used on small polypeptides. These too will elute in an order determined by isoelectric point.

Edman Degradation

• 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

determine identity

determine identity

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.

wells

(+) pole

(-) 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.

wells

(+) pole

(-) pole

Peptide with large net negative charge

Peptide with smaller net negative charge

Peptide with net positive charge

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 interest2. Expose crystallized protein to X-

ray 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 ProteinsHomologous 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

8 9

Thr Gly Ile

Insulin

Homologous ProteinsHomologous 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

• 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

low O2

+-

Electrophoresis of HbA and HbS

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:

These states exist in equilibrium.In the presence of certain regulators, one state

will be preferentially stabilized, shifting the equilibrium towards T or R.

binds O2 tightlybinds O2 weakly


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.


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 & RegulationLow O2

Low pHHigh CO2

High O2

High pHLow CO2

O2 Transport

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

http://thebluerepublic.com/Gallery/albums/album03/250px_Mus_Musculus_huismuis.jpg

Practice Problem 3: GreenglobinsThere are four different greenglobins, A-D. You conduct

binding experiments to learn more about them:

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1. For each greenglobin, determine whether it is oligomeric or monomeric or if you can’t tell:

Greenglobin A: oligomeric monomeric can’t tellGreenglobin B: oligomeric monomeric can’t tellGreenglobin C: oligomeric monomeric can’t tellGreenglobin D: oligomeric monomeric can’t tell

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2. For each greenglobin, which type(s) of regulation are illustrated in the graphs above? (Circle all that apply)

Greenglobin A: homotropic activation heterotropic activation heterotropic inhibition Greenglobin B: homotropic activation heterotropic activation heterotropic inhibition Greenglobin C: homotropic activation heterotropic activation heterotropic inhibition Greenglobin D: homotropic activation heterotropic activation heterotropic inhibition

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3. Rank the different greenglobins in their ability to transport progreen from the intestine to the eye.

1. _______ (best) 2. __________ 3.__________ 4.________ (worst)

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biochem review, part i: protein structure and function

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