SET 4 - Vishal Birudakota PDF

Preview

Summary

Vishal Birudakota...

Description

SET 4 Affinity chromatography

1) How does it work? 2) When to use this? 3) What are some problems? Form of column chromatography

Advantages: 1) Uses specific binding properties of molecules/proteins 2) Produces very pure proteins

How it works: 1) Stationary phase has a polymer that can be covalently linked to a compound called a ligand that specifically binds to the desired protein 2) After the unwanted proteins are eluted, more ligand or salt is added 3) Salt weakens the binding of protein to the ligand and extra free ligand competes with the column ligand in protein binding -a change of pH or ionic strength will disrupt the ligand-protein interaction-

Use this after getting partially pure protein to get pure protein

Problems: 1) ligand doesn't bind properly to polymer 2) ligand binds too strongly; can't remove 3) expensive

WORKED EXAMPLE:

Ion Exchange of Peptides A biochemist wants to separate two peptides by ion-exchange chromatography.

At the pH of the mobile phase to be used on the column:

Peptide A has a net charge of −3, due to the presence of more Glu and Asp residues than Arg, Lys, and His residues.

Peptide B has a net charge of +1. Which peptide would elute first from a cation-exchange resin? Which would elute first from an anion-exchange resin? Solution: A cation-exchange resin has negative charges and binds positively charged molecules, retarding their progress through the column. Peptide B, (+1) with its net positive charge, will interact more strongly with the cation-exchange resin than peptide A, (-3) and thus peptide A will elute first.

On the anion-exchange resin, peptide B (+1) will elute first. Peptide A, (-3) being negatively charged, will be retarded by its interaction with the positively charged resin.

Note: Separation can be optimized by changing pH or salt concentration.

WORKED EXAMPLE: What happens when the following amino acids are added to a column packed with a cation exchange resin?

Aspartate Serine Lysine lysine is eluted last due to being the most positively charge amino acid

Electrophoresis

1) basis? 2) what are the materials? 3) process? Basis: 1) Charged particles migrate in electric field toward opposite charge 2) Proteins have different mobility: - Charge - Size - Shape

Materials: 1) Agarose used as matrix for nucleic acids 2) Polyacrylamide (continuous cross-linked matrix that can manipulate pore sizes) - used mostly for proteins - has more resistance towards larger molecules than smaller

Steps: 1) Protein is treated with detergent (SDS) sodium dodecyl sulfate - this gives all proteins a uniform NEGATIVE charge - breaks all noncovalent interactions of the tertiary and quaternary levels, therefore component polypeptide chains can be analyzed 2) SDS-PAGE gives molecular weight of protein by comparing the sample w/ standard samples 3) Smaller proteins move through faster (charge and shape usually similar)

shape and charge are approximately the same for the proteins in the sample, thus the size of the protein becomes the determining factor in separation

Isoelectric Focusing Based on differing isoelectric points (pI) of proteins

PI: pH at which a protein (aa or peptide) has no net charge

Isolectric focusing is a variation of gel electrophoresis

How it works: 1) Gel is prepared with pH gradient that parallels electric-field 2) As protein moves through the gel, it encounters regions of different pH, so the charge on the protein changes, but protein stops at the region of its PI - Charge on the protein changes as it migrates - When it gets to pI, has no charge and stops

Isoelectric Focusing 2 This technique separates proteins according to their isoelectric points

A stable pH gradient is established in the gel by the addition of appropriate ampholytes

A protein mixture is placed in a well on the gel. With an applied electric field, proteins enter the gel and migrate until each reaches a pH equivalent to its pI

Remember that when pH = pI, the net charge of a protein is zero

Incubate longer to make it look pure

Primary structure determination Hydrolysis of proteins (HPLC analysis): 1) high resolution columns that can be run under high pressures are used 2) separation that might take hours on standard column can be done in minutes w/ HPLC 3) Reverse phase HPLC is used for the separation of nonpolar molecules

Sequence-specific cleavage: trypsin chymotrypsin CNBr

How is 1˚ structure determined? 1) Determine which amino acids are present (amino acid analysis by HPLC) 2) Determine the N- and C- termini of the sequence (a.a sequencing; check to see if there are more than one polypeptide chain) 3) Determine the sequence of smaller peptide fragments (most proteins > 100 a.a) - Edman degradation 4) Some type of cleavage into smaller units necessary - Edman degradation -

Primary Structure Determination Protein Cleavage Protein cleaved at specific sites by:

Enzymes: - Trypsin - Chymotrypsin

Chemical reagents: - Cyanogen bromide

Enzymes: 1) Trypsin: cleaves @ C-terminal of (+) charged side chains (e.g., lysine, arginine) 2) Chymotrypsin: cleaves @ C-terminal of aromatics (e.g., tyrosine, tryptophan, and phenylalanine)

Chemical reagents: 1) CNBr: cleaves internal methionine residues - sulfur of the methionine reactions w/ carbon of the CNBr to produce homoserine lactone @ the C-terminal end of fragment

Sequence of a set of peptides produce by one reagent overlap the sequences produced by another reagent; peptides can be arranged in the proper order after their own sequences have been determined

Primary Structure Determination Peptide Digestion Cyanogen bromide hydrolyzes peptide bonds at the carboxyl terminus of methionine residues

This reaction is used to reduce the size of polypeptide segments for identification and sequencing

Cleavage by CnBr Cleaves @ C-terminal of INTERNAL methionines

Determining Protein Sequence After cleavage, mixture of peptide fragments produced.

1) Can be separated by HPLC or other chromatographic techniques 2) Use different cleavage reagents to help in 1˚ determination

Peptide Sequencing 1) Can be accomplished by Edman Degradation 2) Relatively short sequences (30-40 amino acids) can be determined quickly 3) So efficient, today N-/C-terminal residues usually not done by enzymatic/chemical cleavage

method takes off an amino acid one at @ a time from N terminus

Cell fractionation To begin the process of purification, proteins are released from cells using homogenization using a variety of physical techniques such as blender processing or a tissue sonicator

This is followed by differential centrifugation. As the homogenate is subjected to increasing g forces in a series of steps, smaller and smaller cell components end up in the pellet

What comes out during differential centrifugation: 1) @ 600 x g: nuclei and any unbroken cells 2) @ 15,000 x g: mitochondria, lysosomes and microbes 3) @ 100,000 x g: ribosomes, microsomes, ER, golgi and plasma membrane fragments

Salting out? how does it work a process where by a highly ionic salt is used to reduce the solubility of a protein until it comes out of solution and can be centrifuge. the salt forms ion-dipole bonds with the water in the solution , which leaves less water available to hydrate the protein. non polar side cans begin to interact b/t protein molecules and the become insoluble

In protein purification, what are the methods highlighted that are based on column chromatography? Ion-exchange chromatography -Cation-exchange chromatography -Anion-exchange chromatography

Size-exclusion chromatography Affinity chromatography

Ion-Exchange Chromatography

1) basis? 2) how does it work? Protein separation based on NET charge; much like affinity chromatography, but is less specific

How it works: 1) Stationary phases (resins) have a ligand with different charges at a defined pH - positive or negative

Cation-exchange chromatography: 2) negatively charged resin (negative charges on the surface); therefore, there is an exchange of cations - generally bound to Na+ or K+

Anion-exchange chromatography: 3) positively charged resin (positive charges on the surface); therefore, there is an exchange of anions - generally bound to Cl-

4) Exchange resin is bound to counterions 5) Mixture of proteins is loaded on the column and allowed to flow through 6) Proteins that have a net charge opposite to that of the exchanger stick to the column, exchanging places w/ the bound counterions 7) Proteins that have no net charge or same charge as the exchanger elute 8) After unwanted proteins are eluted, eluent is changed either to a buffer that has a pH that removes the charge on the bound proteins or to one w/ a higher salt concentration 9) Higher salt concentration will outcompete the bound proteins for the limited binding space on the column 10) Desired molecules are eluted

Column Chromatography

1) What is the basis of separation? 2) What are the two phases? 3) What are the steps? Stationary vs. mobile phases

Basis of Chromatography: Different mobilities of the components of the sample are the basis of separation

Two phases: 1) Stationary: samples interacts with this phase 2) Mobile (eluent): flows over the stationary phase and carries along with it the sample to be separated

Steps: A) material of the stationary phase is packed in a column B) sample is a small volume of concentrated solution that is applied to the top of the column

C) eluent is passed through the column D) sample is diluted by eluent 5) separation process increases the volume occupied by the sample

Size-exclusion chromatography

1) Advantages? 2) How does it work? 2) What is eluted first? Size-exclusion (or gel filtration) chromatography is a form of column chromatography

Advantages: 1) proteins are separated by size 2) estimate molecular weight by comparing the sample w/ a set of standards

Stationary phase composed of cross-linked gel particles: 1) particles are in bead form 2) consists of one of two kinds of polymers; a carbohydrate polymer (dextran or agarose) or polyacrylamide

How it works: 1) Cross-linked structure of these polymers produce pores in the material 2) Extent of cross-linking can be controlled to produced desired pore size 3) Smaller molecules enter the pores and are delayed in elution time (only after escaping the pores) 4) Larger molecules do not enter and elute from column before smaller ones

Column Chromatography (Picture)

Ion-Exchange chromatography (picture)

Ion-Exchange chromatography (picture 2)

3. Recall What is meant by "salting out"? How does it work? Salting out is a process whereby a highly ionic salt is used to reduce the solubility of a protein until it comes out of solution and can be centrifuged. The salt forms ion-dipole bonds with the water in the solution, which leaves less water available to hydrate the protein. Nonpolar side chains begin to interact between pro- tein molecules, and they become insoluble.

4. Recall What differences between proteins are responsible for their differential solubility in ammonium sulfate? Their amino acid content and arrangements make some proteins more soluble than others. A protein with more highly polar amino acids on the surface is more soluble than one with more hydrophobic ones on the surface

11. Recall What is the basis for the separation of proteins by the following techniques? (a) gel-filtration chromatography (b) affinity chromatography (c) ion-exchange chromatography (d) reverse phase HPLC

(a) Size. (b) Specific ligand-binding ability (c) Net charge (d) Polarity.

12. Recall What is the order of elution of proteins on a gel-filtration column? Why is this so? The largest proteins elute first; the smallest elute last. Larger proteins are excluded from the interior of the gel bead so they have less available column space to travel. Essentially, they travel a shorter distance and elute first

13. Recall What are two ways that a compound can be eluted from an affinity column? What could be the advantages and disadvantages of each? A compound can be eluted by raising the salt concentration or by adding a mobile ligand that has a higher affinity for the bound protein than the stationary resin ligand does. Salt is cheaper but less specific. A specific ligand may be more specific, but it is likely to be expensive

14. Recall What are two ways that a compound can be eluted from an ion-exchange column? What could be the advantages or disadvantages of each? A compound can be eluted by raising the salt concentration or by changing the pH. Salt is cheap, but it might not be as specific for a particular protein. Changing the pH may be more specific for a tight pI range, but extremes of pH may also denature the protein

Why do most people elute bound proteins from an ion- exchange column by raising the salt concentration instead of chang- ing the pH? Raising the salt concentration is relatively safe. Most proteins will elute this way, and, if the protein is an enzyme, it will still be active. If necessary, the salt can be removed later via dialysis. Changing the pH enough to remove the charge can cause the proteins to become denatured. Many proteins are not soluble at the isoelectric points

Draw an example of a compound that would serve as a cation exchanger. Draw one for an anion exchanger See Figure 5.7

Design an experiment to purify protein X on an anion-exchange column. Protein X has an isoelectric point of 7.0 Set up an anion-exchange column, such as Q-Sepharose (qua- ternary amine). Run the column at pH 8.5, a pH at which the protein X has a net negative charge. Put a homogenate contain- ing protein X on the column and wash with the starting buffer. Protein X will bind to the column. Then elute by running a salt gradient.

What is the main difference between reverse phase HPLC and standard ion-exchange or gel filtration chromatography?

how would you purify protein X using ion-exchange chromatography if it turns out the protein is only stable at a pH between 6 and 6.5? In most chromatography systems, the ligands and solvents are polar. In reverse phase HPLC, a solution of nonpolar compounds is put through a column that has a nonpolar liquid immobilized on an inert matrix. A more polar liquid serves as the mobile phase and is passed over the matrix. The solute molecules are eluted in proportion to their solubility in the more polar liquid

Use a cation-exchange column, such as CM-Sepharose, and run it at pH 6. Protein X will have a positive charge and will stick to the column.

What could be an advantage of using an anion exchange column based on a quaternary amine [i.e., resin- N1(CH2CH3)3] as opposed to a tertiary amine [resin-NH1 (CH2CH3)2]? With a quaternary amine, the column resin always has a net positive charge, and you don't have to worry about the pH of your buffer altering the form of the column. With a tertiary amine, there is a dissociable hydrogen, and the resin may be positive or neutrally charged, depending on the buffer pH.

An amino acid mixture consisting of lysine, leu- cine, and glutamic acid is to be separated by ion-exchange chroma- tography, using a cation-exchange resin at pH 3.5, with the eluting buffer at the same pH. Which of these amino acids will be eluted from the column first? Will any other treatment be needed to elute one of these amino acids from the column? Glutamic acid will be eluted first because the column pH is close to its pI. Leucine and lysine will be positively charged and will stick to the column. To elute leucine, raise the pH to around 6. To elute lysine, raise the pH to around 11.

An amino acid mixture consisting of phenylala- nine, glycine, and glutamic acid is to be separated by HPLC. The stationary phase is aqueous and the mobile phase is a solvent less polar than water. Which of these amino acids will move the fastest? Which one will move the slowest?

In reverse-phase HPLC, the stationary phase is nonpolar and the mobile phase is a polar solvent at neutral pH. Which of the three amino acids will move fast- est on a reverse-phase HPLC column? Which one will move the slowest? A nonpolar mobile solvent will move the nonpolar amino acids fastest, so phenylalanine will be the first to elute, followed by gly- cine and then glutamic acid

The nonpolar amino acids will stick the most to the stationary phase, so glutamic acid will move the fastest, followed by glycine and then phenylalanine.

Gel-filtration chromatography is a useful method for removing salts, such as ammonium sulfate, from protein solu- tions. Describe how such a separation is accomplished. A protein solution from an ammonium sulfate preparation is passed over a gel-filtration column where the proteins of interest will elute in the void volume. The salt, being very small, will move through the column slowly. In this way, the proteins will leave the salt behind and exit the column without it.

If you had a mixture of proteins with different sizes, shapes, and charges and you separated them with electrophoresis, which proteins would move fastest toward the anode (positive electrode)? Those with the highest charge/mass ratio would move the fast- est. There are three variables to consider, and most electro- phoreses are done in a way to eliminate two of the variables so that the separation is by size or by charge, but not by both.

Why is the order of separation based on size opposite for gel filtration and gel electrophoresis, even though they often use the same compound to form the matrix? In a polyacrylamide gel used for gel-filtration chromatography, the larger proteins can travel around the beads, thereby having a shorter path to travel and therefore eluting faster. With electro- phoresis, the proteins are forced to go through the matrix, so the larger ones travel more slowly because there is more friction

The accompanying figure is from an electrophoresis experiment using SDS-PAGE. The left lane has the following standards: bovine serum albumin (MW 66,000), ovalbumin (MW 45,000), glyceraldehyde 3-phosphate dehydrogenase (MW 36,000), carbonic anhydrase (MW 24,000), and trypsinogen (MW 20,000). The right lane is an unknown. Calculate the MW of the unknown. The MW is 37,000 Da.

Why is it no longer considered necessary to determine the N-terminal amino acid of a protein as a separate step?

What useful information might you get if you did determine the N-terminal amino acid as a separate step? The Edman degradation will give the identity of the N-terminal amino acid in its first cycle, so doing a separate experiment is not necessary

It might tell you if the protein were pure or if there were subunits.

Show by a series of equations (with structures) the first stage of the Edman method applied to a peptide that has leucine as its N-terminal residue.

Why can the Edman degradation not be used effectively with very long peptides? Hint: Think about the stoichi- ometry of the peptides and the Edman reagent and the percent yield of the organic reactions involving them. The amount of Edman reagent must exactly match the amount of N-termini in the first reaction. If there is too little Edman reagent, some of the N-termini will not react. If there is too much, some of the second amino acid will react. In either case, there will be a small amount of contaminati...