Test 2

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

motif (+ examples)

AKA fold or supersecondary structure recognizable folding pattern involving 2 or more elements of secondary structure and the connections between them examples include β α β loop and β barrel

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

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

relationship of Kd to θ and [L]: binding of a ligand to a protein

Kd is the concentration of ligand at which half of the ligand binding sites on a protein are occupied ([L] when θ=1/2)

difference in structures b/t Hb T and R states and what occurs during transition

T state is stabilized by more ionic interactions, esp along the α1β1 (α2β2) axes compared to the R state T-->R: when O2 binds to heme (in a subunit of Hb) in the T state, it pulls on the alpha helices of heme (shifting position of proximal His), causing it to assume a slightly more planar conformation; these changes lead to adjustments in the ion pairs at the α1β2/(α2β1) interface, causing these subunits to slide past each other and rotate

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 contaminating phenylthiohydantoin (PTH) de- rivatives. This error grows with the number of cycles run until the point that two amino acids are released in equal amounts, and you cannot tell which one was supposed to be the correct one

basis of rigid and brittle CT during aging

as you age, there is an increase in the amount of cross-linking in collagen fibrils, leading to rigid and brittle CT

protein aggregates

partially unfolded proteins and protein folding intermediates that escape the quality control activities of chaperones and degradative pathways may become sticky (due to exposed hydrophobic surfaces), forming both disordered aggregates and amyloid-like aggregates that contribute to diseases and aging

proteome

the entire set of proteins expressed by a given cell or group of cells; maintained by proteostasis

11. Recall What is the basis for the separation of proteins by the follow- ing 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.

prion diseases (+ examples)

(proteinaceous infection) diseases that infect animal populations and spread through exposure to misfolded proteins; once exposed, the misfolded proteins recruit other proteins to become misfolded, forming plaques in the brain and eventually death ex: scrapie (in sheep), mad cow disease

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*

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

Ka and Kd: equations and what do they tell you?

*low Kd/high Ka means the ligand has a high affinity for/binds well to the protein* P+L⇌PL: Ka=[PL]/[P][L], is the association constant and Kd=1/Ka, is the dissociation constant these are measures of the affinity of a ligand for a protein/how well a ligand binds to a protein

why is Mb a good storer of O2 but a poor transporter of O2

-Kd is very small for O2: O2 has a high affinity for Mb; it rapidly binds and quickly saturates Mb binding sites -concentration of O2 in lungs is 13 kPa and in tissues it is 4 kPa; using equation, we find that the saturation of Mb in the lungs would be 98% and in the tissues it would be 94% -the ability to transfer O2 from lungs to tissues is proportional to the difference in binding b/t 2 locations, meaning O2 would be a poor transporter of O2 b/c it binds too tightly -tissues would be starved for O2

why does one need vitamin C (ascorbic acid) for collagen?

-Prolyl 4-hydroxylase catalyzes 2 separate reactions, one of which hydroxylates Pro to form hyPro (Y of GlyXY; forces Pro into exo conformation to form tight turns of collagen triple helices) (this rxn occurs via oxidative decarboxylation of alpha ketoglutarate to form CO2 and succinate) -Fe must be in its 2+ oxidation state on prolyl 4-hydroxylase in order for it to function properly -prolyl 4-hydroxylase also catalyses a separate oxidative decarboxylation of alpha ketogluterate (not coupled to pro hydroxlation); this rxn requires ascorbate to keep Fe in prolyl 4-hydroxylase in Fe2+ oxidation state to restore the enzyme's activity -without ascorbate, Fe becomes oxidized to Fe3+, therefore prolyl 4-hydroxylase cannot catalyze the hydroxylation of proline, and colllagen is therefore unable to form tight turns

keratin: -type of protein -structure -stabilization -2 other characteristices

-fibrous protein -Amino acids form a helix, 2 alpha helices intertwine forming a coiled-coil. Then 2 coiled coils twist together to form a protofilament, and protofilaments combine to form protofibrils -stabilized via disulfide bonds (high degree of cysteine residues) -composed of many hydrophobic residues -high tensile strength

O2 binding curve for Mb

-fraction of O2 bound to myoglobin as a function of concentration of O2 (partial pressure of O2) -is a hyperbolic function: O2 binds very rapidly and reaches saturation quickly -Kd is extremely small, therefore O2 has a very high affinity for Mb

globular proteins: folding

-globular proteins contain segments of polypeptide chain folded back on each other, making globular proteins compact and complex -this folding leads to the structural diversity necessary for proteins' wide array of functions

forces that provide protein stability & how they contribute to free energy

-hydrophobic effect (primary source of free energy and thus stability) -ionic interactions -H bonds: important for conformation of protein (secondary structure) but do not contribute much free energy b/c an H bond formed within the protein is equivalent to one formed in water -Van der Waals (difficult to quantify free energy difference b/t folded and unfolded state) -disulfide bridges

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*

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 -*

Salting out

1) Soluble proteins in the cytosol are separated out via ammonium sulfate (NH4)2SO4 2) Proteins remain soluble because of their interactions with water 3) Water interacts w/ the ammonium sulfate to form ion-dipole bonds, thus leaving less water to interact w/ the proteins 4) The proteins can then interact more with each other via hydrophobic interactions (hydrophobic interactions among proteins increases w/ less water available for hydration) 5) Different aliquots taken as function of salt concentration to get closer to desired protein sample of interest (30, 40, 50, 75% increments) 6) One fraction has protein of interest (generally @ 60-70% saturation)

proteostasis processes

1. proteins synthesized on the ribosome and must fold into their native conformations 2. multiple pathways contribute to protein folding, many of which require chaperones --chaperones also contribute to refolding of proteins that are partially/transiently unfolded 3. proteins that are irreversibly unfolded are subject to sequesteration and degradation of several other pathways, with the aid of proteasomes

Anfinsen's experiment

=denaturation/renaturation of ribonuclease -didn't know structure of ribonuclease, but knew that it cleaves RNA -purified and measured its activity: if the protein cut RNA he deduced that it was folded -added mercaptoethanol to reduce disulfide bonds and urea to break hydrophobic interactions, thus unfolding ribonuclease; then measured its activity and it no longer diced RNA -used dialysis (to remove mercaptoethanol and urea), and the ribonuclease worked again

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

myoglobin

A globular protein that has the ability to bind oxygen and functions as oxygen storage. Myoglobin is found in the muscle tissues and helps to store oxygen in the muscle for use in aerobic respiration

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.

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

binding of O2 vs CO to heme

CO binds better to free heme b/c it binds with the CO axis perpendicular to the plane of the porphyrin ring while O2 binds to heme with the O2 axis at a 120 deg angle; however, this angle is favored by heme bound to myoglobin b/c of the distal His when CO binds to heme in myoglobin it is forced to adopt a slight angle b/c of sterics; this weakens the binding of CO to heme myoglobin CO still binds better, but b/c there is such little concentration of CO in our bodies compared to O2, this is not an issue

how are high [CO2] and low pH related

CO2 and H+ are byproducts of cellular respiration, which occurs at the tissues CO2 is hydrated to form bicarbonate, which H+ is a byproduct of (CO2 + H2O⇌H2CO3⇌H+ + HCO3-), resulting in an increase in H+ concentration these stabilize the T state of Hb, decreasing O2 saturation

Cleavage by CnBr

Cleaves @ C-terminal of INTERNAL methionines

proteostasis

Continual maintenance of the necessary set of active cellular proteins (proteome). Requires regulation of synthesis, folding, refolding, and degradation (of irreversibly unfolded proteins) (proteasome)

Which of the chemicals or enzymes normally used for cutting proteins into fragments would be the least useful to you? Which enzymes or chemicals would you choose to use to cut the protein? Why?

Cyanogen bromide would be useless, because there is no methionine. Trypsin would be little better, because the protein is 35% basic residues. Trypsin would shred the protein into more than 30 pieces, which would be very hard to analyze Chymotrypsin would be a good choice. There are more than four residues of aromatic amino acids. The protein, containing 100 amino acids, would be cut four times, possibly yielding nice fragments roughly 20-30 amino acids long, which can be sequenced effectively by the Edman degradation.

heme bound to myoglobin (or Hb)

Fe2+ of heme is bound to the proximal His on myoglobin (or Hb); also bound to O2, which is bound to the distal His on myoglobin (or Hb) O2 is bound at a 120 degree angle to heme and is stabilized via a H bond to the distal His heme is found deep w/i a pocket of Mb or Hb

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

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.

Hb O2 binding curve: how does it make Hb a good transporter of O2?

Hb exhibits allostery in which there is communication b/t the 4 subunits of the heterotetramer: as soon as 1 O2 molecule binds to a subunit of Hb, it begins transition to the R state, thus increasing Hb's affinity for O2 (each molecule of O2 binds with more affinity) this results in the sigmoidal curve, as Hb transitions from T state to R state in the presence of increasing [O2], allowing for Hb to release 30% O2 in the tissues and pick up O2 in the lungs

how does sickle cell anemia work

HbSxHbS causes a point mutation in which the 6th a.a. on the beta subunit of Hb is changed from glu to val, thus changing it from negative a.a. to neutral a.a causes hydrophobic effect in RBCs, crystallizing Hb in RBCs

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

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 would happen during an amino acid sequencing experiment using the Edman degradation if you acci- dentally added twice as much Edman reagent (on a per-mole basis) as the peptide you were sequencing?

In the first cycle, the first and second amino acids from the N-terminal end would be reacted and released as PTH deriva- tives. You would get a double signal and not know which one was the true N-terminus.

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

A sample of a peptide of unknown sequence was treated with trypsin; another sample of the same peptide was treated with chymotrypsin. The sequences (N-terminal to C-terminal) of the smaller peptides produced by trypsin digestion were as follows:

Met!Val!Ser!Thr!Lys!Leu!Phe!Asn!Glu!Ser! Arg!Val!Ile!Trp!Thr!Leu!Met!Ile

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*

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

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

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 re- duce 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.

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.

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

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.

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

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.

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.

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

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

You are in the process of determining the amino acid sequence of a peptide. After trypsin digestion fol- lowed by the Edman degradation, you see the following peptide fragments: Leu!Gly!Arg Gly!Ser!Phe!Tyr!Asn!His Ser!Glu!Asp!Met!Cys!Lys Thr!Tyr!Glu!Val!Cys!Met!His What is abnormal concerning these results? What might have been the problem that caused it?

There are two fragments that have C-termini that are not lysine or arginine, which is what trypsin is specific for. Normally there would be only one fragment ending with an amino acid that was not Arg or Lys, and we would immediately know that it was the C-terminus. Histidine is a basic amino acid, although it is usually neutral and therefore does not react with trypsin. It is possible that, in the pH environment of the reaction, the histidine was positively charged and was recognized by trypsin.

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

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.

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

A sample of an unknown peptide was divided into two aliquots. One aliquot was treated with trypsin; the other was treated with cyanogen bromide. Given the following sequences (N-terminal to C-terminal) of the resulting fragments, deduce the sequence of the original peptide.

Val!Leu!Gly!Met!Ser!Arg!Asn!Thr!Trp!Met! Ile!Lys!Gly!Tyr!Met!Gln!Phe

What could be an advantage of using an an- ion 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.

2 ways to determine protein structure: advantageous and disadvantages

X-ray crystallography: allows you to see the density of atoms, thus determining their spatial arrangement (limitation is the protein must be crystallized first, which can cause slight changes in their 3D shape) 2D NMR: protein does not have to be crystallized first, but this method does not have as good of resolution

fibrin structure

a fibrous protein that consists of layers of antiparallel beta sheets rich in Ala and Gly residues (which are small and hydrophobic) -does not stretch b/c phi and psi angles in beta sheets are already fully extended -flexible due to numerous weak interactions (such as hydrophobic and H bonds)

heme: definition

a prosthetic group that is covalently attached to myoglobin and hemoglobin; contains iron in order for myoglobin and hemoglobin to perform their oxygen binding functions

allosteric protein

a protein w/ multiple ligand binding sites in which the binding of a ligand to one site affects the binding properties of another site on the same protein -ex=Hb and O2

hair perms and straightening is...

a reduction of disulfide bonds that make up keratin

proteasome

a supramolecular assembly of enzymatic complexes that function in the degredation of damaged or unneeded cellular proteins

Anfinsen's conclusion, and why is this not completely true today

amino acid sequence determines 3D structure (a.a. sequence contains all the info required to fold the chain into its native, 3D structure) however, even though all proteins contain the info to fold into their native structure, most require more assistance (i.e. chaperone proteins)

3D/tertiary structure

assemblage of polypeptide segments in the alpha helical and beta conformations, linked by connecting segments: distinguish fibrous from globular proteins

Hsp70

bind to regions of unfolded polypeptides that are rich in hydrophobic residues, thus breaking up aggregates or preventing new ones from forming; bind to and release polypeptides in a cycle that uses energy from ATP hydrolysis and several other proteins

heme: structure

central Fe2+ atom which has 6 coordination bonds: 4 of which are bonded to the 4 N atoms that are part of the flat protoporphyrin IX ring, and 2 that are perpendicular to protoporphyrin IX, which bind the proximal His and O2

collagen molecules/Gly-X-Y

collagen molecules are made of 3 interwound alpha helices (each alpha helix has 3 residues per turn and is therefore tighter than average) the alpha helices are composed of repeating tripeptides of Gly-X-Y gly=glycine X=proline (proline contains kinks which provide stability and tightness to the helices) Y=hyPro (hydroxylated proline)

collagen fibril

composed of many (triple helical) collagen molecules that are covalently bound via cross links (providing strength and stability) via a post-translational modification, a cross link forms between 2 lysine residues of separate triple helical collagen molecules, one of which is hydroxylated

difference b/t globular and fibrous proteins

differ mainly in their 3D structures -fibrous proteins have simple repeating elements of secondary structure, are spacious, and serve mainly structural roles -globular proteins' 3D structure is more complicated, often containing several types of secondary structure in the same polypeptide chain, have folding and are thus more compact and structurally diverse, and serve a variety of functions

similarities between myoglobin and hemoglobin structures

each subunit of Hb is similar to Mb: similar 3D structures, similar alpha helices, similar binding for heme w/ the proximal and distal His

interactions stabilizing the 4 subunits of Hb

hydrophobic interactions stabilize everywhere, H bonds, and ion pairs (salt bridges) more ionic interactions stabilizing α1β1 (and α2β2) along the horizontal axis than α1β2 (α2β1) along the vertical axis if Hb is subjected to mild conditions, α1β1 (and α2β2) remain intact while α1β2 (α2β1) break: this led to the discovery of the T and R states of Hb

% saturation of Hb in arterial blood vs venous blood

in arterial blood, Hb is 96% saturated vs in venous blood it is 64% saturated, meaning Hb releases app 30% O2 in tissues

proximal His: what is it's purpose?

in order to keep heme's central iron in its Fe2+ state (and not Fe3+), one of its coordination bonds is bound to the N of a histidine residue on the protein heme is bound to

factors that stabilize the T conformation of Hb, thus reducing its affinity for O2: how do each of these stabilize the T state?

increase in CO2: combines w/ N terminus of Hb and other blood proteins to form carbamates (further increasing H+) decrease pH (increase in [H+]): H+ helps to stabilize ion pairs, stabilizing the T state increase BPG: binds to positively charged residues between beta subunits in the T state; the transition to R state narrows this cavity and eliminates BPG binding site

how does your body compensate for the lack of oxygen at high altitudes

increases BPG, shifting O2-Hb saturation curve to the right, so that your tissues can get the O2 they need

Bohr effect

low pH and high [CO2] (in the tissues) stabilize the T state of Hb, decreasing the affinity of Hb for O2; therefore in the tissues, H+ and CO2 (byproducts of cellular respiration) are bound to Hb and O2 is released (in lungs, CO2 is excreted and pH rises, affinity for O2 increases and Hb binds more O2) lower pH's shift O2 Hb saturation curve to the right

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

free energy in proteins: what is it? what is biggest contributor?

protein stability is the difference in free energy b/t the folded and unfolded state of proteins (more free energy in folded state than unfolded) -hydrophobic effect (burial of hydrophobic a.a.'s to exclude water) is the primary contributor to the free energy -protein folding is hard to predict since folded protein is only marginally stable

formation of amyloid fibrils

proteins have a concentration of hydrophobic amino acids in a core region of beta sheets; core folds into a β sheet before the rest of the protein folds correctly; sheets from 2 or more incompletely folded proteins begin forming an amyloid fibril, which grows in the extracellular space; other parts of the protein then misfold, remaining on the outside of the β sheet core in the growing fibril

chaperone proteins

proteins that interact w/ partially folded or improperly folded polypeptides, facilitating the correct folding pathways or providing microenvironments in which folding can properly occur 2 types=Hsp and chaperonins

amyloidosis

set of diseases caused by *(normally soluble)* proteins that have been secreted by the cell in a *misfolded* state and converted into an *insoluble extracellular amyloid fibrils* -examples: Huntington's, type 2 DM, Alz, some CA

Hill plot and Hill equation

tells you the degree of cooperativity of a protein Hill equation: log(θ/(1/θ))=nlog[L]-logKd Hill plot: log(θ/(1/θ)) vs log[L]

how does the difference in O2 binding curves b/t Hb and Mb result in Hb's better ability to transport O2

the allosteric property of Hb allows its affinity for O2 increase as O2 binds: the first molecule binds with the majority of Hb in the T state (w/ a low affinity for O2), which begins the transition to the R state; R state has higher affinity for O2, so as Hb transitions to R state, the affinity increases -the [O2] in the lungs is high, at which point Hb is almost all in the R state which has a very high affinity for O2, so Hb picks up O2 in the lungs -the [O2] in the tissues is low, where Hb population is mostly in the T state w/ a low affinity for O2, so it releases O2 in the tissues the hyperbolic curve of Mb demonstrates that Mb has a high affinity for O2 that does not change, so it would not release O2 in the tissues even though the [O2] at the tissues is low

Hill's coefficient

the slope of a Hill plot which is a measure of the degree of cooperativity exhibited by a protein -if Hill's coefficient (slope)=1 ligand binding is not cooperative if Hill's coefficient is greater than 1, the ligand binding to the protein exhibits positive cooperativity if Hill's coefficient is less than one, the protein has negative cooperativity in which the binding of one molecule impedes the binding of others

how are CO2 and H+ transported in Hb

they stabilize the T state, CO2 is mostly transported as HCO3-, but CO2 can also combine w/ the N terminus of Hb and other blood protein to form carbamates H+ is transported bound to several side chains on Hb whose pKa's are altered by the transition from T to R (high pKa in the T state b/c H+ stabilizes ion pairs)

what is θ? what is its relationship to [L]?

θ=a fraction of how much ligand is bound to the protein (value ranges from 0 to one)=binding sites on a protein occupied divided by total # of binding sites θ is a hyperbolic function of [L], meaning that as [L] increases, the fraction of ligand binding sites occupied approaches saturation