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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.

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an amino acid
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
The TwentyAmino Acids

SMALL: hydrogen or methyl R group.

GLYCINE

Alanine

Types of AAs

NONPOLAR/HYDROPHOBIC:

R groups contain largely C, H atoms.

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

CHARGED

ALIPHATIC: no aromatic rings.

VALINE

Leucine

Isoleucine

Methonine

*Proline

AROMATIC: contains aromatic rings.

PHENYLALANINE

Tyrosine

Tryptophan

UNCHARGED: no ionizable group in R group.

ASPARAGINE

Glutamine

Serine

Threonine

*Cysteine

ACIDIC: acid in R group.

ASPARTATE

Glutamate

BASIC: base in R group.

LYSINE

Arginine

Histidine

slide4

SpecialAmino Acids

SMALL: hydrogen or methyl R group.

GLYCINE

Alanine

Types of AAs

NONPOLAR/HYDROPHOBIC:

R groups contain largely C, H atoms.

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

CHARGED

ALIPHATIC: no aromatic rings.

VALINE

Leucine

Isoleucine

Methonine

*Proline

UNCHARGED: no ionizable group in R group.

ASPARAGINE

Glutamine

Serine

Threonine

*Cysteine

an imino acid, with R group bound to amino group

thiol group can participate in disulfide bonding

peptide bond formation
Peptide Bond Formation

1

2

carboxyl terminal residue

“C terminal”

amino terminal residue

“N terminal”

acid base behavior of aas
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 aas1
Acid-Base Behavior of AAs

Equilibrium constant:Ka = [A-][H+]/[HA]

pH = -log[H+]…convenient shorthand for writing widely

variable [H+] concentrations

pKa = -log(Ka)…similar shorthand for writing variable Ka values

the henderson hasselbalch equation
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

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

isoelectric point
Isoelectric Point

Isoelectric point: the pH at which an AA or polypeptide has no net charge.

  • For a dibasic amino acid:
  • For a tribasic amino acid:

Isoelectric point = average of amino and carboxyl pka values

pka = 2.4

pka = 9.8

GLYCINE

= (2.4 + 9.8)/2 = 6.1

pka = 2.2

Isoelectric point = average of the two numerically closest pka values

= (2.2 + 4.3)/2 = 3.25

pka = 9.7

pka = 4.3

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

GLUTAMATE

titration curves
Titration Curves

This example shows the curve for a dibasic AA

pka of amino group

pkavalues always occur at the flattest parts of the curve (around 0.5, 1.5, 2.5 baseequivs.)

buffering works best at pka

Isoelectric point

pka of carboxyl group

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
Practice Problem #1: Titration of Aspartate

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.)

2

3

1

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

practice problem 1 titration of aspartate1
Practice Problem #1: Titration of Aspartate

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 aspartate2
Practice Problem #1: Titration of Aspartate

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-COOH

100%-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-COOH

100%-11.18% = 11.18% R-COO-

Average charge on R carboxyl:

(0)*(0.8882) + (-1)*(0.1118) = -0.1118

practice problem 1 titration of aspartate3
Practice Problem #1: Titration of Aspartate

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
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
Primary Structure
  • Genetically determined: specified by sequence of gene that encodes protein
primary structure1
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 structure1
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
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 structure1
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 structure2
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 structure2
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
  • 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 structure3
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
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 1 o structure
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
Ion Exchange Chromatography
  • Tells you what AAs a protein contains and the relative abundance of each
ion exchange chromatography1
Ion Exchange Chromatography
  • Digest protein in strong acid
  • Load onto column of negatively charged (sulfonate coated) beads; AAs bind because positively charged
  • Wash column with solutions of increasing pH, collecting eluate in small fractions
  • For each fraction, determine the presence/absence and abundance of the corresponding AA
ion exchange chromatography2
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.

Peak height indicates relative abundance

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

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

edman degradation
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
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
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 agents1
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
Electrophoresis
  • Used to separate proteins into bands according to their net charge.
      • Load protein samples into wells at one end of a gel.
      • Apply current; proteins will move through gel matrix towards the pole opposite their net charge.

(+) pole

wells

(-) pole

electrophoresis1
Electrophoresis
  • Used to separate proteins into bands according to their net charge.
      • Load protein samples into wells at one end of a gel.
      • Apply current; proteins will move through gel matrix towards the pole opposite their net charge.

(+) pole

Peptide with large net negative charge

Peptide with smaller net negative charge

wells

Peptide with net positive charge

(-) pole

methods to analyze higher order structure
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
X-Ray Diffraction
  • Crystallize protein of interest
  • Expose crystallized protein to X-ray source (wavelength 1.5 angstroms)
  • Record diffraction pattern
  • 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

Homologous proteins are proteins from different organisms that are very similar in structure and function.

Ex: insulin, cytochrome C

homologous proteins1
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

Insulin

8

9

homologous proteins2
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
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 clock1
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: 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 changes1
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

-

+

Glu to Val substitution results in a less negative net charge on HbS and slower migration towards the + pole.

Electrophoresis of HbA and HbS

hemoglobin kinetics regulation
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 weakly

binds O2 tightly

hemoglobin kinetics regulation1
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 regulation2
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 regulation3
Hemoglobin: Kinetics & Regulation

O2 Transport

Low O2

Low pH

High CO2

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
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 greenglobins1
Practice Problem 3: Greenglobins

There are four different greenglobins, A-D. You conduct binding experiments to learn more about them:

% of progreen binding sites occupied

practice problem 3 greenglobins2
Practice Problem 3: Greenglobins
  • For each greenglobin, determine whether it is oligomeric or monomeric or if you can’t tell:

Greenglobin A: oligomeric monomeric can’t tell

Greenglobin B: oligomeric monomeric can’t tell

Greenglobin C: oligomeric monomeric can’t tell

Greenglobin D: oligomeric monomeric can’t tell

% of progreen binding sites occupied

practice problem 3 greenglobins3
Practice Problem 3: Greenglobins
  • 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

% of progreen binding sites occupied

practice problem 3 greenglobins4
Practice Problem 3: Greenglobins
  • Rank the different greenglobins in their ability to transport progreen from the intestine to the eye.

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

% of progreen binding sites occupied

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