crystallization laboratory

Putting crystals in perspective. Why do I need them? How do I grow them?

M230D, January 2012

Crystallization Laboratory

sawaya

You might find it strange that during this 5 week course on crystallography, today is the only day we discuss how crystals are obtained.It is true that the most fun and excitement begins after the crystals are obtained.But, it is important to know that the bulk of time an effort of solving a crystal structure is usually spent in obtaining the crystal.

Crystallization is one step in the structure determination pipeline

1) chose gene product, source organism, full length, fragment, or fusion

2) chose vector, tag, location of tag (N or C?)

3) Chose host organism, temperature, media, purification scheme

4) Screen 1000 conditionsScreen for crystal quality

5) collect diffraction datamake heavy atom derivativedetermine heavy atom sitescalculate mapinterpret maprefine coordinates

6) publish

select protein target

clone

express

crystallize

solve

deposit in PDB

Mike Sawaya

Let's begin by looking at a road map illustrating the process of structure determination by X-ray crystallography.There are institutions that seek to solve the structures of hundreds of proteins in-parallel. They call this map a pipeline.The pipe has a large diameter for all the proteins that go through it. In academia, the pipeline is more narrow.We break this pipeline into 6 segments (steps). Each step can be broken into numerous sub-steps. Beginning with selection of the protein target. At this step a choice of gene product is made. One must chose which organism, and whether to express the full length, or a fragment, or a fusion, or a complex.The second step is to clone the chosen gene into a vector. One must chose which vector, whether to use a tag, where to put the tag.The third step is to express the protein. One must chose a host organism, temperature, media.The fourth step is to crystallize the protein. One usually screens 1000s of conditions. Once a crystal is obtained, it must be tested in the X-ray beam for diffraction quality. If it is not of sufficient quality, one must optimize the crystal or look for other conditions.The fifth step is to solve the structure. This includes collecting data, producing a heavy atom derivative, locating the positions of the heavy atom, calculating an electron density map, interpretting the map, building a model, refining it and validating its quality. This lab deals mostly with these steps.The last step is to deposit the model in the protein data bank where it becomes freely accessible to all.Which of these steps do you expect is the most time-consuming or difficult to achieve?

• Joint Center for Structural Genomics established. 2000.• Statistics reported http://www.jcsg.org/ on Jan 4, 2010.

Deposited in PDB: 968

Crystallized: 2128

Solved: 1045

Expressed: 27640

Cloned: 27959

Selected Targets: 33209

8% success

84% success

99% success

49% success

93% success

Target protein sequences

http://www.jcsg.org/

http://www.jcsg.org/prod/scripts/structure_gallery/gallery.shtml

http://www.jcsg.org/prod/scripts/public_targets/mounted.html

http://www.jcsg.org/prod/scripts/public_targets/structure.html

http://www.jcsg.org/prod/scripts/public_targets/expressed.html

http://www.jcsg.org/prod/scripts/public_targets/cloned.html

http://www.jcsg.org/prod/scripts/public_targets/selected.html

Mike Sawaya

We get some indication from the success rate of structural genomics consortiums.Here are illustrated the same six segments of the pipeline as shown on the previous slide.Next to each step is a number representing how many proteins successfully progressed to the corresponding step in the pipeline at JCSG since their beginning in late 2000.Out of 33209 targets chosen, 84% of these were succesfully cloned.Out of these cloned genes, 99% were successfully expressed.But only 8% of the expressed genes produced crystals. Here is the true bottleneck of the pipeline.If we continue down the pipeline, we see about 50% of the proteins that crystallized could be solved. And 93% of these structures have been deposited in the PDB.

Why is it necessary to grow crystals?

Growing a suitable crystal is such a hurdle!

In a crystal, the diffraction signal is amplified by the large number of repeating units (molecules).

Diffraction intensity is proportional to the number of unit cells in the crystal (Darwin’s formula, 1914).

Diffraction from a single molecule is not currently measurable.

A 100 m3 crystal contains 1012 unit cells

Michael Sawaya

Michael Sawaya1/4/2011No one has yet measured the X-ray diffraction pattern from a single molecule.3 barriers1) Beam diameter: One would have to be able to collimate an X-ray beam to the diameter of a single protein molecule, then train that beam on that single protein molecule (represented by this knight). 2) Beam intensity: X-rays have a relatively weak interaction with matter (compared to diffraction from an electron beam). The intensity of the beam would have to be 10-12 orders of magnitude brighter than what is currently acheivable at synchrotrons in order to generate a recordable number of scattered photons. 3)Beam induced damage: The intensity would obliterate the molecule, so the x-ray pulse would have to be short, calculations suggest 10 fs.Its a very short amount of time. In this time, light can travel only 3 microns.Scientists are working on femtosecond X-ray laser beam technology to enable single molecule diffraction studies to be performed. All our current progress in x-ray diffraction of macromolecules is made possible by amplification of the diffraction signal produced by the many repeating units of a crystal.This amplification is proportional to the number of repeating cells in the crystal (Darwin's formula 1914). So bigger crystal produce stronger diffraction patterns than smaller crystals.The bigger the crystal, the wider the incident beam can be, and the more diffracted photons can be recorded.Also, X-ray damage can be spread over a large number of molecules, allowing a longer lifetime of the sample. Standard crystals contain over 10^12 unit cells and so would produce a diffraction signal 10^12 more intense than a single crystal. The crystal shown here is only 27 unit cells and would not produce measurable diffraction with todays technology, butrecent experiments with crystals containing only 100 unit cells show significant diffraction. They achieve beams 8 orders of magnitude brighter than our best synchrotrons.Crystals also have the property of periodicity or 3D order.

In a crystal, the ordered, periodic arrangement of molecules produces constructive interference (i.e. amplification of diffraction signal).

a

b

c

Mike Sawaya

Periodicity in a crystal means that the molecules repeat at regular intervals.

Incident X-ray

crystal

Interference is constructive because path lengths differ by some integral multiple of the wavelength (n).

This situation is possible only because the diffracting objects are periodic.

12

34

56

12

34

56

7

12

34

5

In phase

67

78

89

detector

When a crystal is ordered, strong diffraction results from constructive interference of photons.

Michael Sawaya

Let's concentrate on the interaction of the x-rays with one part of the motif --the eyes.There will be a certain angle of diffracted rays in which the path lengths of the scattered rays differ by integer numbers of the the wavelength.At this angle (Bragg angle), the diffraction signal is strong...the scattered waves are all in phase...constructive interference.This constructive interaction is true for scattering from all parts of the motif (nose, ears, neck, etc.).

Irregularity in orientation or translation limits the order and usefulness of a crystal.

Rotational disorder Translational disorderPerfect order

Disorder destroys the periodicity leading toStreaky, weak, fuzzy, diffraction.

Irregularity in orientation or translation limits the order and usefulness of a crystal.

Rotational disorder Translational disorderPerfect order

Disorder destroys the periodicity leading toStreaky, weak, fuzzy, diffraction.

(CCML, Yeates Lab)(bacteriorhodopsin, Bowie Lab)

What makes crystallization such a difficult challenge?

Gcrystal=Hcrystal-T(Sprotein+Ssolvent)

protein crystalprotein in solution

Is Hcrystal favorable?

Enthalpic term Entropic term

Yes,Hcrystal is modestly favorable (0 to -17 kcal/mol)

lattice contacts

protein crystalprotein in solution• large area• specific• rigid

Is TSprotein favorable?


No, TSprotein is strongly unfavorable (+7 to +25 kcal/mol)

protein crystalprotein in solution3 degrees of freedom in orientation 3 degrees of freedom in translation

0 degrees of freedom in orientation 0 degrees of freedom in translation

Is TSsolvent favorable?


Yes,TSsolvent is favorable (-7.5 to -50 kcal/mol)

protein crystalprotein in solutionO

H H

OH

H

OH

H

OH H

OH H

OH H

OH H

OH H

OH H

OH H

OH H

OH H

OH H

OH

H

OH

H

OH H

OH H

OH H

OH H

OH H

OH H

OH H

OH H

OH H

OH H O

H H

OH H

OH H

OH H

OH H



Gcrystal=Hcrystal-T(Sprotein+Ssolvent)

Gcrystal= -small + large – large

Gcrystal= -small

Strategies to lessen the entropic penalty, TSprotein.

• Eliminate floppy, mobile termini (cleave His tags)

• Express individual domains separately and crystallize separately, or…

• Add a ligand (or protein binding partners) that bridges the domains and locks them together.

• Mutate high entropy residues (Glu, Lys) to Ala. (http://services.mbi.ucla.edu/SER/)

or

mutate

Increase [protein] to favor crystallization

G

N solublelysozymemolecules

nM→Mn

1 crystal(lysozyme)N

Increasing the monomer concentration [M] pushes the equilibrium toward the product.

nM→Mn

G=Go+RTln( [Mn]/[M]n )

Lesson: To crystallize a protein, you need to increase its concentration to exceed its solubility (by 3x). Force the monomer out of solution and into the crystal. Supersaturate!

Unstablenucleus

Three steps to achieve supersaturation.

1) Maximize concentration of purified protein

• Centricon-centrifugal force

• Amicon-pressure• Vacuum dialysis• Dialysis against high

molecular weight PEG• Ion exchange.• Slow! Avoid precipitation.

Co-solvent or low salt to maintain native state.

Concentrateprotein

Three steps to achieve supersaturation.2) Add a precipitating

agent• Polyethylene glycol

• PEG 8000• PEG 4000

• High salt concentration• (NH4)2SO4

• NaH2PO4/Na2HPO4Polyethylene glycol

• Small organics• ethanol• Methylpentanediol

(MPD)

PEGPolymer of ethylene glycol

Precipitating agents monopolize water molecules, driving proteins to neutralize their surface charges by interacting with one another. It can lead to (1) amorphous precipitate or (2) crystals.

Three steps to achieve supersaturation.

3) Allow vapor diffusion to dehydrate the protein solution

• Hanging drop vapor diffusion

• Sitting drop vapor diffusion• Dialysis• Liquid-liquid interface

diffusion Note: Ammonium sulfate concentration is 2M in reservoir and only 1M in the drop.

With time, water will vaporize from the drop and condense in the reservoir in order to balance the salt concentration.—SUPERSATURATION is achieved!

2M ammonium sulfate

Drop =½ protein + ½ reservoir

Michael Sawaya

There is osmotic pressure to equalize the ionic strength of the drop and the reservoir.How does the ionic strength become equalized?Either the salt has to migrate from reservoir to drop, or water has to migrate from drop to reservoir.Which is more likely?Salt isnt going to vaporize and go into drop.

Naomi E Chayen & Emmanuel Saridakis Nature Methods - 5, 147 - 153 (2008)Published online: 30 January 2008; |

doi:10.1038/nmeth.f.203

Precipitating agent concentration

• Shotgun- for finding initial conditions, samples different preciptating agents, pHs, salts.

• Systematic-for optimizing crystallization conditions.

First commerciallyAvailable crystallizationScreening kit.Hampton Crystal Screen 1

Conventionally, try shotgun screening first, then systematic screening

Heavy Atom Gel Shift Assay.Why?

Why are heavy atoms used to solve the phase problem?

• Phase problem was first solved in 1960. Kendrew & Perutz soaked heavy atoms into a hemoglobin crystal, just as we are doing today. (isomorphous replacement).

• Heavy atoms are useful because they are electron dense. Bottom of periodic table.

• High electron density is useful because X-rays are diffracted from electrons.

• When the heavy atom is bound to discrete sites in a protein crystal (a derivative), it alters the X-ray diffraction pattern slightly.

• Comparing diffraction patterns from native and derivative data sets gives phase information.

Why do heavy atoms have to be screened?

• To affect the diffraction pattern, heavy atom binding must be specific– Must bind the same site (e.g. Cys

134) on every protein molecule throughout the crystal.

– Non specific binding does not help.

• Specific binding often requires specific side chains (e.g. Cys, His, Asp, Glu) and geometry.– It is not possible to determine

whether a heavy atom will bind to a protein given only its amino acid composition.

Before 2000, trial & error was the primary method of heavy atom

screening• Pick a heavy atom compound

– hundreds to chose from• Soak a crystal

– Most of the time the heavy atom will crack the crystal.

– If crystal cracks, try lower concentration or soak for less time.

– Surviving crystal is sent for data collection.

• Collect a data set• Compare diffraction intensities

between native and potential derivative.

• Enormously wasteful of time and resources. Crystals are expensive to make. How many crystallization plates

does it take to find a decent heavy atom derivative?

Heavy Atom Gel Shift Assay• Specific binding affects

mobility in native gel.• Compare mobility of

protein in presence and absence of heavy atom.

• Heavy atoms which produce a gel shift are good candidates for crystal soaking

• Collect data on soaked crystals and compare with native.

• Assay performed on soluble protein, not crystal.

None Hg Au Pt Pb Sm

Procedures• Just incubate protein with

heavy atom for a minute. – Pipet 3 uL of protein on

parafilm covered plate.– Pipet 1 uL of heavy atom

(100 mM) as specified.– Give plate to me to load on

gel.• Run on a native gel• We use PhastSystem• Reverse Polarity

electrode• Room BH269 (Yeates

Lab)

The details of the experiment.

Goal: crystallize Proteinase K and its complex with PMSF

• Number of amino acids: 280

• Molecular weight: 29038.0

• Theoretical pI: 8.20• Non-specific serine

protease frequently used as a tool in molecular biology.

• PMSF is a suicide inhibitor. Toxic!

MAAQTNAPWGLARISSTSPGTSTYYYDESAGQGSCVYVIDTGIEASHPEFEGRAQMVKTYYYSSRDGNGHGTHCAGTVGSRTYGVAKKTQLFGVKVLDDNGSGQYSTIIAGMDFVASDKNNRNCPKGVVASLSLGGGYSSSVNSAAARLQSSGVMVAVAAGNNNADARNYSPASEPSVCTVGASDRYDRRSSFSNYGSVLDIFGPGTSILSTWIGGSTRSISGTSMATPHVAGLAAYLMTLGKTTAASACRYIADTANKGDLSNIPFGTVNLLAYNNYQA

Ala (A) 33 11.8%Arg (R) 12 4.3%Asn (N) 17 6.1%Asp (D) 13 4.6%Cys (C) 5 1.8%Gln (Q) 7 2.5%Glu (E) 5 1.8%Gly (G) 33 11.8%His (H) 4 1.4%Ile (I) 11 3.9%Leu (L) 14 5.0%Lys (K) 8 2.9%Met (M) 6 2.1%Phe (F) 6 2.1%Pro (P) 9 3.2%Ser (S) 37 13.2%Thr (T) 22 7.9%Trp (W) 2 0.7%Tyr (Y) 17 6.1%Val (V) 19 6.8%

Reservoir Solutions1) We are optimizing two types

of crystals. 1) ProK (rows AB)2) ProK+PMSF (rows CD).

2) There are three components to each reservoir: (NH4)2SO4, Tris buffer, and water.

3) We are screening six concentrations of ammonium sulfate and 2 buffer pHs.

4) Pipet one chemical to all reservoirs before pipeting next chemical—it saves tips.

Linbro or VDX plate

ProK

ProK+ PMSF

((

Practical Considerations

When reservoirs are ready, lay 6 coverslips on the tray lid,

Then pipet protein and corresponding reservoir on slips

Invert slips over reservoir.

Only 6 at a time, or else dry out.

tray containing reservoir solutions

Gently swirl tray to mix reservoir solutions.

tray lid tray

025

P20

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Proper use of the pipetor.

Which pipetor would you use for delivering 320 uL of liquid?

P1000 P200 P20

Each pipetor has a different range of accuracy

P1000 P200 P20

200-1000uL 20-200uL 1-20uL

Michael Sawaya

The number on the pipetor is the maximum volume that pipetor can deliver.

Which pipetor would you use for delivering 170 uL of ammonium

sulfate?

P1000 P200 P20

How much volume will this pipetor deliver?

027

P200

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170

P20

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027

P1000

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What is wrong with this picture?

027

P1000

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----

50 mL

Dip tip in stock solution, just under the surface.

027

P1000

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50 mL

Withdrawing and Dispensing Liquid.3 different positions

027

P1000

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027

P1000

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027

P1000

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First stopStart position

Second stop

Withdrawing solution: set volume, then push plunger to first stop to push air out of the tip.

027

P1000

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Second stop

----

50 mL

Dip tip below surface of solution. Then release plunger gently to withdraw solution

027

P1000

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Second stop

To expel solution, push to second stop.

027

P1000

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Second stop

When dispensing protein, just push to first stop.Bubbles mean troubles.

027

P1000

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Second stop

Hanging drop vapor diffusionstep two

Pipet 1.5 uL of concentrated protein (50 mg/mL) onto a siliconized glass coverslip.

Pipet 1.5 uL of the reservoir solution onto the protein drop

2M ammonium sulfate0.1M buffer

BUBBLES MEAN TROUBLESExpel to 1st stop, not 2nd stop!

Hanging drop vapor diffusionstep three

•Invert cover slip over reservoir quickly & deliberately.

•Don’t hesitate when coverslip on its side or else drop will roll off cover slip.•Don’t get fingerprints on coverslip –they obscure your view of the crystal under the microscope.

Preparing Proteinase K -PMSF complex

• Mix gently– Pipet up and down

<5 times– Stir with pipet tip

gently– Excessive mixing

leads to xtal showers

• No bubbles36 L of50 mg/mL ProK

Remove 18 uL of proteinase K and add to PMSF

5 L of 100 mM PMSF

Proteinase K time lapse photography

• illustrates crystal growth in 20 minute increments

• film ends after 5 hours

500 m

crystallization laboratory

Documents

corresponding step

crystal structure

heavy atom sitescalculate

heavy atom derivative

crystal quality5

host organism

segments steps

protein data bank