Molecular Structure Techniques

Question 1

Brief explanation of Recombinant DNA

Answer 1

gene for protein of interest cloned, inserted into a plasmid and then used to control its expression in different cell types. These are grown in selective media, labelled (for NMR) or chemically modified AA added (X ray phasing). Cultures harvested by centrifugation where the desired material of supernatant is purified.

Question 2

Types of cells used for recombinant DNA techniques

Answer 2

E.coli, yeast, Chinese Hamster Ovary (CHO), Human embryonic kidney cells, baculovirus infected insect cells.

Question 3

Describe Purification methods

Answer 3

Purification helped by TAGS on the protein at either N/C terminus, forming fusion proteins. For example, histidine tag or glutathione S transferase fusion.

Tagged proteins passed over metal ion chelating column (Ni2+) where it sticks to column and rest passed through.

Imidazole wash elutes target.

Chromatography used to further purify. If protease cleave site incorporated the tag is cleaved, leaving just the target.

Question 4

What we can do now as a result of recombinant DNA, heterologous expression and purification with current instrumentation?

Answer 4

Wider range of samples accessible.

Less material needed (<milligrams)

• isolate and purify small quantities from tissues
• use PCR to find genes
• use genomic data to clone
• select domains
• make mutants
• find specific state/factors/form of target.

Question 5

Recombinant expression system in E.coli

Answer 5

restriction enzyme cut at cloning site, ligate gene of interest, transform in to cell culture, plate, select cells with plasmid, inoculate media with plasmid cells to give a purified culture.

Question 6

What role does IPTG have in gene activation?

Answer 6

It does this through binding to lac repressor. ie, genes switched off by the repressor can be expressed.

It switches off transcription of the lac operon and therefore causes gene expression of any gene regulated by the lac operator..

Question 7

What colour change effect does IPTG have?

Answer 7

When it is present there is a blue colour, when it is absent there is a white colour.

Question 8

How do you prepare an E.coli recombinant expression system?

Answer 8

LacI induces expression of lac Z to produce B-galactosidase. Insertion into clone site with gene of interest should disrupt Lac Z to either allow blue or white selection when grown on ampicillin media.

White colonies without IPTG, arent repressed and have to correctly assembled expression plasmid.

Question 9

Describe how the expression vectors are made different?

Answer 9

The vector can be altered at the cloning site leading to protein modifications at N/C terminus.

We can add 1)Histidine

2) glutathione-S-transferase
3) maltose binding protein
4) signal sequence (localisation)

These encourage the solubility of samples and helps purificaion.

Question 10

Decisions to be made for expression system using recombinant DNA.

Answer 10

1) find amino acid sequence
2) find heterologous expression system (e.coli, baculovirus)
3) check target gene doesnt have a restriction site (for enzymes that might be used)
4) check amino acids do not have protease recognition site (for tag removal)
5) decide expression vector, a tag, a purification method.

Question 11

Name Recombinant DNA purification methods.

Answer 11

the OVERPRODUCED target protein is either in cells or secreted out.

So cells opened by chemical (lysozymes) or mechanical (sonicator, homogenisation)

Centrifugation (High speed) and nucleases and filtration remove cell debris, decrease viscosity.

If protein secreted, media concentrated, possible use of affinity column with an antibody with transition metal ion (Ni2+), glutathione ect.

Question 12

What is Chromatography?

Answer 12

The separation between mobile and stationary phase.
Sample in buffer is mobile.
Sample in solid media is stationary where particles are packed.

Question 13

List Chromatographic methods

Answer 13

Stationary: differential separation by chemical attraction

1) Ion exchange (charge interactions) gives high resolution results
2) Size exclusion (using gel filtration) give low resolution.
3) Hydrophobic interactions give medium resolution.
4) Affinity chemical binding gives high resolution.

Question 14

Name two affinity tags.

Answer 14

Tag 1 is (GST) gluathione binds to an agarose bead.

Tag 2 is His tag binds to Nickel imidazole side chain.

Tag2 is

Question 15

Describe Affinity Chromatography

Answer 15

One type of molecule will attach to a molecule on the resin (in the column) Unattached molecules wash through, attached molecules however stay in a stationary phase.

Question 16

Describe Ion exchange chromatography.

Answer 16

Adsorption to charged stationary resin, then elution in buffer of greater ionic strength.

pH determines the net charge of the protein.

If the pH is ABOVE its isoelectric point (pI) protein bind to POSITIVELY charged anion exchanger

If BELOW pI, protein bind to NEGATIVELY charged cation exchanger.

Question 17

Describe Hydrophobic interaction

Chromatography.

Answer 17

Resin (with hydrophobic groups) binds to protein from aqueous solutions to differing degrees. This depends on structure of protein as well as salt conc, pH, temp, organic solvents

Question 18

Describe Size exclusion Chromatography.

Answer 18

A porous gel made of beads.
SLOW: Smaller molecules enter matrix pores and have to move through the matrix of column.

MEDIUM SPEED: intermediate size can enter matrix (stationary phase)

FAST: Larger molecules excluded from stationary matrix

Question 19

What techniques can be used to assess purification?

Answer 19

1) Enzyme/binding assay
2) SDS-PAGE and Mass Spec
3) Bradford assay or UV/VIS spec (concentration)

Question 20

Describe the theory of SDS-PAGE (sodium dodecylsulfate polyacrylamide gel electrophoresis)

Answer 20

acrylamide gel can separate proteins depending on mass.

SDS is detergent that gives a UNIFORM charge to sample.

Electrical current used to cause protein migration.

Here, the smallest proteins travel fastest (opposite from size exclusion) Heavier molecules slower as there is no solvent space.

Coomassie dye stains proteins and indicates what is present.

Question 21

The Bradford assay and Coomassie blue

Answer 21

Coomassie blue is a protein binding dye.

Acidic conditions means the dye is brown where max absorption is 465nm.

When protein is present, it is blue and can absorb up to 595nm.

Influenced by detergents.

Graph curve made with absorbance (y axis) and mass of protein (x axis)

Question 22

How can protein concentration be determined by spectroscopic methods?

Answer 22

Amino acid sequence allows a molar extinction co efficient to be given, and absorbance (280nm) and apply Beer lambert law.

Aromatic residues (F,Y,W) absorb UV light at 280nm.
A280=coefficient280 x conc x pathlength. 

At 280nm, chromatography flow can be measured.

Question 23

What happens after Comassie Blue binds and what data does it show?

Answer 23

After binding, protein-dye complex becomes negative.

Gives information on the QUATERNARY structure which is compared to gel filtration results.

Peaks from gel filtration can correspond to masses of protein components, eg, a dimer.

Both gel filtration and SDS-PAGE also show if proteins do not interact- if eluted in two separate peaks on the gel.

Question 24

Summarise Sample Acquisition

Answer 24

1) Plan approach- sequence, protease, tags required
2) Recombinant DNA methods and cell growth to form samples with affinity tags
3) Purify by using His, GST tags to target.
4) Affinity step
5) cleave tag, final polish
6) Assess level of purity and concentration

Question 25

Why are samples made pure, and made into crystals

Answer 25

Diffraction methods for structure determination need ordered crystals.

This means samples must be pure, homogeneous of 2-5ng.

(DNA, RNA USED- there are few proteins and ribosomes that can be purified in this way)

Question 26

Crystals: What is SUPERSATURATION?

Answer 26

Supersaturation is the formation of a crystal, or precipitate nuclei which will grow until equilibrium is reached.

A crystal is an ordered precipitate of an ordered array of atoms or molecules.

Samples are between 0.05-0.5mm

Question 27

What are the zones on a protein crystallisation phase diagram?

Answer 27

1) Nucleation: where molecules come come together and begin to precipitate
2) Supersaturation zone: greater amount of sample is in solution than be dissolved.
3) Metastable zone: is unstable but long lasting which results in crystal growth.

Question 28

How do you grow crystals?

Answer 28

Solution of the molecule is subjected to changing condition which will bring about the precipitation zone where nucleation is undergone. As molecules come out of solution they aggregate in an ordered fashion.

Question 29

What factors can influence crystal growth?

Answer 29

1) pH
2) Ionic strength
3) temp
4) conc and volume of sample
5) conc/type of precipitant
6) purity
7) sequence of protein
8) rate of equilibrium
9) dust, contaminants
10) additives, ligands, substrates
11) gravity, vibrations

Question 30

Macromolecular crystals are…

Answer 30

Fragile, unstable and have to be treated carefully and fresh. They can consist of between 40-80% bulk solvent (they are very hydrated.)

In order to crystalise the samples are more complex, labile, flexible and sensitive to their environment; they mus be kept hydrated, near their pH and temp.

Question 31

What are the methods used in the crystalisation of macromolecules?

Answer 31

1) BATCH method: solution mixed with precipitant until solution becomes cloudy, then add water.
2) DIALYSIS: Sample is in a dialysis tubing with porous membrane which small/solvents pass through
3) LIQUID-LIQUID DIFFUSION: sample in capillary and precipitant placed in contact allowing interface diffusion, promoting precipitation.
4) VAPOUR DIFFUSION

Question 32

Vapour Diffusion method crystalise macromolecules is most common. How does it work?

Answer 32

Small amount of sample is placed in sealed chamber with a precipitant such as polyethylene glycol (PEG) reservoir at the base of chamber at 50%.

or methylpentane diol or ammonium sulfate.

A little of PEG of 10% added to drop of sample. As a result, water diffuses from 10% to the 50% PEG until equilibrium is reached.

PEG concentration and macromolecular concentration increases until precipitation occurs.

Question 33

How do you crystallize small molecules?

Answer 33

By chance, cooling, slow evaporation, sublimation, vapour diffusion.

Question 34

Crystalisation: diffraction measurments

Answer 34

Seeding methods useful to get crystals to this stage.

They are picked up in a NYLON loop.

Crystals can be damaged by radiation but they are protected using polyethylene glycol and cooled to 100 K (cryo-cooled with nitrogen gas.)

Crystals then fixed in a gless within the loop and exposed to X RAYS.

The intensity of refraction is measured.

Question 35

List the methods used to investigate affinity, kinetics and thermodynamics

Answer 35

Isothermal Titration Calorimetry (ITC)

Fluorscence Thermal Shift Assay (FTSA) or DIFFERENTIAL scanning fluorimetry (DSF)

SURFACE PLASMON RESONCE (SPR)

Bio-layer Interferometry (BLI)

Question 36

Isothermal Titration Calorimetry (ITC)

Answer 36

iso; constant temperature;

titration; addition of standard reagent to a known volume of a molecule, in carefully measured amounts until reaction completed (color change, conductance etc). Can then calculate an unknown concentration or stoichiometry

Calorimetry - Measurement of the amount of heat evolved or absorbed in a chemical reaction/interaction or change of state.

This technique is label free and analyses the binding interactions through heat change.

Addition of ligand to protein in the sample cell produces a heat change. Power is altered to maintain the same temperature as the reference cell. The added or reduced power is what is measured for each addition/titration point.

Question 37

ITC

Answer 37

The typical experiment involves a series of injections of aliquots of ligand solution
(known volume and concentration) with mixing into a known volume and
concentration of protein. Each injection produces a heat change/pulse which after
a correction for a baseline and integrated over time gives the total heat released
or absorbed.

The change in temperature is very small and requires low instrument sensitivity of 20microjoules

Question 38

list the sources of heat in ITC

Answer 38

1) The protein-ligand interaction (objective)
2) Dilution of ligand on injection into larger volume of sample
3) Dilution of macromolecule by added ligand solution
4) Mechanical mixing effects

When ligand is added heat can be released (EXOthermic) or absorbed (ENDOthermic) due to the interactions.

Question 39

What is the c value.

Answer 39

a unitless parameter that determines the shape of the isotherm

c=Ka[M]n
c=association constant[macromolecule conscentration] stoichiometry

Question 40

ITC graph

Answer 40

Negative peak; less energy required to get temperature the same as heat dissipates- EXOTHERMIC

Positive peak; more energy needed to be equal to reference ENDOTHERMIC

The area of each peak is the heat in or out; as more ligand is injected, protein binging sites become saturated and so heat signal decreases, no change

Question 41

ITC GRAPH; what is the binding enthalpy?

Answer 41

The total integral of all signal peaks is the
binding enthalpy delta H. The shape of the curve
gives Ka, the association constant and
n, stoichiometry

Question 42

What is enthalpy (delta H)

Answer 42

change in energy as flow of heat at constant pressure.

Positive in endothermic reactions,

negative in exothermic (kcal mol-1 or kJ mol-1) This is enthalpically favoured binding which involves well aligned H-bonds

Indicates changes in hydrophobic/ van der Waals relative to the solvent; how strong is the ligand interaction.

Question 43

What is entropy (delta S)

Answer 43

disorder in a thermodynamic system.

The greater disorder, the more positive the ΔS. Less disorder results in more negative ΔS (J K-1 mol-1) changes in hydrophobic/conformation change interactions

positive, more disorder is favoured through hydrophobic binding. Solvent entropy is increased as hydrophobic groups added and release water
molecules from binding sites or that are associated with ligands.

Also expect minimal loss of conformational degrees of freedom with positive DS.

Question 44

What is Gibbs free energy (delta G)

Answer 44

spontaneity of a reaction; where spontaneous reactions are negative.

Related to binding affinity.

Question 45

What is shown through ITC

Answer 45

it measures Ka, Binding stoichiometry, change in enthalpy

This determines the dissociation constant (kd), change in entropy (delta S) and change in Gibbs free energy.

Question 46

Applications of ITC

Answer 46

drug discovery: it guides ligand optimisation for binding. It aims to find specific molecule interactions and which factors need to be altered to guide certain modifications to ligand scaffolds

Question 47

What are the advantages of ITC

Answer 47

Any soluble components can be studied (protein-ligand, inhibitor, protein, nucleic acid, lipid etc etc.)
No labeling 
No immobilization
Direct measurement 
Wide range of affinities:

Question 48

How does DSF

Answer 48

1) protein mixed with organic dye and heated
2) protein unfolds; dye bind to exposed hydrophobic surfaces;
3) this changes the fluorescence of dye. The wavelength used to detect fluorescence depends on the dye.
4) melting curve is measured; melting temperature determined.
5) molecule structure/stability will change melting temperature.

Question 49

Applications of DSF

Answer 49

1) screening buffers
2) screening different constructs
3) screening ligand binding

Question 50

What is SPR?

Answer 50

1) light passes through material of HIGH refractive index (prism) in to material with a LOW refractive index so it BENDS toward the interference plane.
2) incidence angle increased to critical angle when reflection where it is reflected out
3) beyond critiacl angle, light is refected back into the prism (total internal reflection)

This is an optical method to detect ligand interacting with a target

rate and equilibrium binding constants can be obtained

Question 51

Describe the instrumentation of SPR

Answer 51

prism coated with a thin layer of an inert metal, gold, on the reflection side.

Photon electric field transfers energy to outer shell gold electrons at a particular angle of incidence. free Electrons become surface plasmons and oscillate plasma waves, resulting in the dip in intensity.

energy taken up due to resonance effects and there is a particular angle that the LOSS IS MAXIMISED (angleSPR) determined by he refractive index of the medium.

SPR angle depends on the mass of a bound material at the sensor chip surface; increase mass, increase angle

Photodetector records change in the SPR angle; proportional to the change in absorbed mass on the sensor.

Question 52

What can be found form Surface Plasmon Resonance

Answer 52

association, dissociation, steady state.

This determines Kd (off rate)
ka (on rate)

KD (Dissociation constant) =kd/ka

Question 53

What is BLI?

Answer 53

Another optical method

interference pattern of white light reflected from immobilised protein on a biosensor.

changes in interference pattern and optical thickness caused by varying numbers of molecules on the biosensor.

Unbound molecules, change the refractive index of the surrounding medium, or changes in flow rate do not affect the interference pattern (only immobilised proteins)

Monitors change in wavelength to determine:

i) binding specificity
ii) rates of association and dissociation, or concentration

Question 54

What is interferometry

Answer 54

use of interference phenomena between a reference light wave and an experimental light wave or between two parts of an experimental wave to determine wavelengths and wave velocities, measure very small distances and thicknesses, and calculate indices of refraction.

Question 55

What is plotted on BLI graph?

Answer 55

Change in material; changes interference pattern; wavelength maxima changes as ligands attach or detach molecules.

wavelength (nm) vs. time