Circular Dichroism as a gauge of protein secondary structure: refolding of egg white proteins. CHE 554, 2014

Macroscopic observations: what happens when light traverses matter. It gets weaker (absorption). Rays bend (diffraction). Colours diverge (dispersion). 1

Interaction of light with matter: absorption If matter absorbs light, a lower intensity of light exits the material than entered it. This interaction is embodied by the extinction coefficient ε. εCl = log(Io/I) Io is the intensity of light entering, I is the intensity exiting. l is the thickness of the material and C is the concentration of chromophores in it. ε=γNAπr2/2.303 (γ is probability that a molecule will absorb a photon that ‘hits’ it and πr2 is the molecule’s cross-section (likelyhood of being ‘hit’ by a photon). Ι/Ιo = e-2.303εCl (review)

Demonstration: http://www.enzim.hu/~szia/cddemo/edemo10.htm

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Interaction of light with matter: refraction Matter can also retard light causing it to have a lower velocity in a material than in the vacuum. While the frequency is unchanged the wavelength shrinks in a material. This interaction is embodied by the refractive index n.

n=c/v

where c is the speed of light in vacuum and v is its speed in the material. Recall that c=λν. n=λo/λ λo is the wavelength in vacuum, and λ the wavelength in a material.

http://www.enzim.hu/~szia/cddemo/edemo12.htm

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Macroscopic observations: what happens when light traverses matter.

Contraction of the wavelength produces the effect of altering the direction of propagation k.

http://en.wikipedia.org/wiki/Refractive_index

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Macroscopic observations: what happens when light traverses matter. The phase velocity of light (v) is different in a medium than in vacuum (c). (v: phase velocity: how fast peaks travel.) The change is described by the refractive index ‘n’. n=c/v Snell’s law:

Refractive indices of some common materials (at 589 nm) Air Water Olive oil Glass Plexiglass Diamond

n1

n1 sinθ1 = n2 sinθ2

1.000 1.33 1.47 1.4-1.6 1.49 2.42

n1 normal

Total internal reflection: When a beam leaves a higher-n medium to enter a lower-n medium there is a maximum angle of attack beyond which light will be reflected (internal reflection).

refractive index

n2

θ2

normal

θ1

n2 θ2

θ1

http://en.wikipedia.org/wiki/Refractive_index 5

A little physics, getting complex An oscillating electric field (a photon) can be described as more general E(z,t) = cos(kz-ωt) = Re (Eoei(kz-ωt)) (eix = cos(x) + isin(x) ) oscillation in space k is the wave number = 2π/λ = 2πn/λo

oscillation in time

λn = λo/n

Warning: although IR spectroscopists call a wavenumber 1/λ = ν/c physicists prefer units of radians s-1 to Hz (=s-1) and set c=1 (fundamental).

Generalizing n to be a complex number too n = n + iκ Now we have that in a medium k = 2π(n + iκ)/λo E(z,t) = Re (Eoei(kz-ωt)) = Eo Re(ei(2πnz/λo + i2πκz/λo -ωt)) = Eo Re(e-2πκz/λo)(ei(2πnz/λo -ωt)) = Eo(e-2πκz/λo) Re (ei(2πnz/λo -ωt)) Amplitude decays exponentially with distance travelled: z Affects the phase of the light, when the peak comes. 6

Interaction of light with matter: absorption If matter absorbs light, a lower intensity of light exits the material than entered it. This interaction is embodied by the extinction coefficient ε. εCl = log(Io/I) Io is the intensity of light entering, I is the intensity exiting. l is the thickness of the material and C is the concentration of chromophores in it. ε=γNAπr2/2.303 (γ is probability that a molecule will absorb a photon that ‘hits’ it and πr2 is the molecule’s cross-section (likelyhood of being ‘hit’ by a photon). Ι/Ιo = e-2.303εCl (review)

Demonstration: http://www.enzim.hu/~szia/cddemo/edemo10.htm

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Circularly polarized light: plane-polarized light is the sum of left and right circularly polarized light The sum of right-circularly polarized and left-circularly polarized light of the same intensity is a plane-polarized beam. (Similarly, the sum of vertically-polarized light and horizontally-polarized light that is 90° out of phase in its oscillation yields circularly-polarized light. See http://www.enzim.hu/~szia/cddemo/edemo7.htm .

http://www.enzim.hu/~szia/cddemo/edemo8.htm

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Circular birefringence (= optical rotation) If RCP and LCP have different refractive indices, so that one is delayed more than the other, the axis that bisects them is rotated. bi refringence = two refractive indices.

θ=

πl (n - n ) λ - +

Re(n), ‘-’ for left-circ pol. light ‘l’ is path length. http://www.enzim.hu/~szia/cddemo/edemo15.htm

http://physics.unl.edu/~tgay/content/OA2.html

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Circular dichroism Circular dichroism means that RCP and LCP light have different extinction coefficients. One is absorbed more than the other. Di Chroism: two absorbances. Here the green vector has been absorbed upon passage through the material. The resulting light is elliptically polarized, and its major axis remains parallel to the plane of polarization that entered the medium.

http://www.enzim.hu/~szia/cddemo/edemo14.htm

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Circular dichroism is also related to ellipticity via the difference between the major and minor axes of the ellipse that results.

Previously plane-polarized light emerges from the sample elliptically polarized.

Ellipticity of ψ rotated by θ

http://www.photophysics.com/tutorials/circular-dichroism-cd-spectroscopy/2-chiral-molecules 11

Circular dichroism

Δε = ε L − ε R ε is a function of λ and so is Δε. Δε can be either negative or positive (vs. ε which is always positive). CD only occurs near an absorption band (i.e. ε≠0) It also only occurs for molecules that are ‘optically active’. Optical activity: a molecule cannot be superimposed on its mirror image. Opposing enantiomers have oppositely signed CD. 12

Absorbance vs. CD

Because CD can change signs, it can be easier to distinguish adjacent peaks in a CD spectrum than in an absorbance spectrum.

opposite enantiomer has negative CD but same absorbance.

http://www.photophysics.com/tutorials/circular-dichroism-cd-spectroscopy/2-chiral-molecules 13

Molecular origins of CD Covalent structure of the molecule is inherently asymmetric. Chiral centres of L-amino acids (vs. R-) Secondary structures that are helical: RH α helices of proteins and A-form and B-form double helix of DNA. Local asymmetry associated with tertiary structure. Tryptophan side chains possess inherent asymmetry. When these are immobilized in an asymmetric environment with asymmetric electric fields that constrain valence electrons’ freedom to move, the CD is strongly enhanced. 14

Proteins:

Not only are 19/20 amino acids chiral (optically active), but the major secondary structural elements are helical.

Molar CD is Δε Also used: Mean Residue molar CD ΔεMR = Δε/nr, nr is the # residues. Makes sense when observing backbone, with one peptide bond / residue. CD signal can be quite strong. 15

Typical signatures of common secondary structure Regular repeating structures

reinforce the CD of each residue resulting in large Δε.

note small CD amplitude in the absence of repeating structure.

http://www.photophysics.com/tutorials/circular-dichroism-cdspectroscopy/6-cd-signatures-of-structural-elements

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Resolubilization of protein from hard-boiled egg white The white solid is highly aggregated protein. It is denatured. We can use a chemical denaturant such as guanidinium hydrochloride to solubilize protein. It interacts so favorably with protein that it outcompetes the aggregation interactions and enables the protein molecules to release one-another and dissolve in solution with the guanidium hydrochloride. You should be able to see the protein aggregate dissolving. Question: Look up the structure of guanidiniumH+ (GuH). Draw a short segment of polypeptide and draw in a few molecules of GuH+ making favourable interactions with the protein backbone and at least one side chain. 17

Refolding the protein of egg white. Although the egg white protein is now back in solution, we can confirm that the protein in solution is still not structured, based on its lack of CD spectrum. However the capacity for structure has survived the hard-boiling and the GuHCl. It is inherent in the proteins’ amino acid sequence. We demonstrate this by observing recovery of the CD signature upon dilution of the GuHCl to low concentrations that are no longer denaturing. The initial demonstration of this principle earned Anfinsen the Nobel Prize in 1972.

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Anticipated results: the CD spectrum of ovalbumin The major protein in egg white is egg ovalbumin (it’s all in the name !) Question: The structure of ovalbumin is shown below. Based on a guesstimate of the % made up of α helix and the % made of β sheet sketch your guess of what the CD spectrum of fully folded ovalbumin should look like.

1OVA.pdb 19

Interpretation of your data Your ‘refolded’ egg white CD spectrum may look different from that of ovalbumin. For one thing, this could be because there are several other proteins besides ovalbumin in egg white. (In addition, some mutants of ovalbumin can adopt a slightly different structure. This is one of the hallmarks of serpin proteins: that they have more than one stable structure. Note that we do not think our eggs are from mutant chickens.)

1JTI.pdb After cleavage of the backbone, here is the other structure this mutant ovalbumin can adopt. Find the difference ! (just for fun, no grades.)

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Aside: fun facts on ovalbumin Ovalbumin is a member of the serpin family of proteins. Most serpins are protease inhibitors, but ovalbumin is not. The protease inhibitors permanently inhibit proteases by undergoing the first half of the protease reaction in which a covalent acyl enzyme intermediate is formed. However they do not detach from the protease (unlike normal substrates). Instead they undergo a striking conformational change that makes permanent their attachment to the protease, causing the protease to be permanently bound and therefore unable to do any more proteolysis.

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What you will do Sample of egg white prior to boiling. 1. Obtain a tube of 10 ml denaturing buffer and a tube of 10 ml native buffer, from your T.A. 2. Also take an eppendorf tube containing uncooked egg white. Record the weight of this after taring the balance on an empty centrifuge tube of the same kind (Note, due to its viscosity, raw egg white is difficult to deliver in predetermined volume. Eppendorf tubes are very uniform in weight. The error produced by assuming they weigh the same is not important in the context of this exercise.) 3. Convert the mass of raw egg white to volume using the density of water and add 8 volumes of native buffer. 4. Mix by swirling, DO NOT VORTEX.

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Sample of egg white after boiling. 1. Obtain a few large pieces of hard-boiled egg totalling ≈ 10 ml. 2. Rinse the hard-boiled egg twice by gently swirling in deionized water and decanting the rinse. 3. After the second rinse transfer the egg pieces to a mortar and grind to a paste using a pestle. 4. Weigh out 1.25 g of the ground egg white and transfer to a 15 ml conical tube containing 10 ml denaturing buffer. 5. Carefully watch and record your observations as you add the egg white to the buffer and swirl to mix. 6. Incubate several minutes, swirling periodically and recording your observations. DO NOT VORTEX. 7. When most of the egg has dissolved, transfer 1.0 ml of the solution to a 1.5 ml centrifuge tube. 8. Centrifuge the native and denatured samples for 1 min at 13,000 rpm. 9. Carefully remove 700 μl of each solution to a fresh centrifuge tube without disturbing the pellet. These are your clarified solutions. 10. Keep all solutions on ice.

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Branch A: estimating the concentrations of your proteins 1. Make three dilutions of each of your solutions to produce solutions that are 1%, 3% and 10% the concentration of your stock solutions (3 ml each). 2. Measure the A280 of each in quartz cuvettes with the appropriate buffer as the background. 3. Establish the region of linearity and if necessary make further dilutions to identify a linear regime. (It is likely that you will need to make at least one additional dilution.) 4. To determine the concentration of your two solutions you will assume that each is make predominantly of ovalbumin (we will test this) and use the known amino acid sequence to calculate an extinction coefficient for the denatured protein (as for the photometry lab, see http://www.ncbi.nlm.nih.gov/protein/223299). 5. Convert the mM concentration of your solution to mg/ml using the published molecular weight of ovalbumin. (Find this in the references provided for this lab.) 6. For the native protein we will use the published extinction coefficient. This is E1% = 7.3 and means that a 10 mg/ml solution has an absorbance of 7.0 for a 1 cm path length (Pace, 1995). 24

Branch B: preparing CD samples of your proteins, and obtaining CD spectra. Prepare CD samples as per the table below. Fill in the two cells for which information is not provided. Sample

Protein

8 M GuHCl

native buffer

Native egg white

30 μL of raw egg white solution

0

1.5 mL

Denatured egg 30 μL of denatured egg white white solution

1.125 mL

0.375

Refolded egg 30 μL of denatured egg white white solution

0.188

1.312 mL

resulting [protein]

resulting [GuHCl]

Centrifuge all samples for 1 min at 13,000 rpm, transfer clarified supernatant to fresh tubes. 25

Obtaining CD spectra 1. Centrifuge all samples for 1 min at 13,000 rpm, transfer clarified supernatant to fresh tubes. 2. Your T.A. will assist you in obtaining CD spectra. 3. Each will be the average of 7 scans collected from 250 nm to 210 nm using 30 sec/scan. 4. Use your determined protein concentrations to present your CD spectra with mean per-residue elipticities (MRE) as the Y-axis.

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Analysis and questions 1. Why is it important to maintain a flow of N2 gas during collection of CD data ? 2. Compare your CD spectra with a published spectrum representing pure α helix, as well as one representing pure β sheet. (Show the spectra and provide references to their sources.) 3. Comment on the secondary structure content of egg white protein before boiling, after denaturing and after ‘refolding’. 4. What was the state of the protein in the hardboiled egg, before you added denaturing buffer to it ? 5. What did the denaturing buffer do to the protein, in molecular terms ? 6. What was the state of the egg white protein at the end of the experiment after dilution into ≈0 M GdHCl ? 7. This lab adopts numerous simplifications but nonetheless succeeds remarkably well. What is an aspect of complexity or potential pitfall that has been neglected ? What might have been a consequence of neglecting it ? (Note that in this exercise we get away with numerous simplifications and short-cuts, you need only discuss one here, but feel free to list more). 27

References Batra PP, Sasa K, Ueki T, Takeda K. (1989) "Circular dichroic study of conformational changes in ovalbumin." J Protein Chem. Apr;8(2):221-9. Edelhoch H. (1967) "Spectroscopic determination of tryptophan and tyrosine in proteins." Biochemistry 6: 1948-1954. Nisbet,A.D., Saundry,R.H., Moir,A.J., Fothergill,L.A., Fothergill,J.E. (1981) "The complete amino-acid sequence of hen ovalbumin" Eur. J. Biochem. 115 (2), 335-345 Pace CN,Vajdos F, Fee L, Grimsley G, Gray T. (1995) "How to measure and predict the molar absorption coefficient of a protein." Protein Sci. Nov;4(11):2411-23. CD Sample Preparation. (2011). Accessed February 25, 2012, from: http://structbio.vanderbilt.edu/wetlab/cd.sample.prep.php ALSO the amino acid sequence of hen egg white ovalbumin is available in electronic form (one-letter amino acid codes) at http://www.ncbi.nlm.nih.gov/ protein/223299 The molecular weight can be predicted at http://web.expasy.org/compute_pi/

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*Denaturing Buffer 6 M Guanidinium hydrochloride (GdHCl), 15 mM dithiothreitol (DTT), 1mM disodium ethylenediaminetetraacetic acid (Na2-EDTA), and 10 mM potassium phosphate (K2HPO4), pH 7.4 *Guanidinium Stock 8 M Guanidinium hydrochloride (GdHCl) dissolved in 10 mM potassium phosphate (K2HPO4), pH 7.4 *Native buffer: Potassium Phosphate Buffer 10 mM potassium phosphate, pH 7.4

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