Understanding Fluorescence: Excited States, Einstein Coefficients, and Quantum Yield, Study notes of Biotechnology

Download Understanding Fluorescence: Excited States, Einstein Coefficients, and Quantum Yield and more Study notes Biotechnology in PDF only on Docsity! IV-1 IV FLUORESCENCE So far we have focused on the absorption of light by chromophores. Now we turn to the fate of the absorbed light. If that chromophore subsequently re–emits the light it may be called a fluorophore. As we have already seen both G and S have a family of sub-states that arise from the various vibrational levels present. The splitting of these vibrational levels will be different in the two cases because G and S have different geometries; typically the molecule in S is expanded with respect to that in G. G S Fig. IV-1 G = Ground State S = First Excited Singlet ∆E (G ⇒ S) is of the order 50 kcal. As RT ~ 0.6 kcal the number of molecules (N) in S and G is determined by N s= exp(-(∆E/RT)), Ns = 0 (for all practical purposes). Ng Typical vibrational energies are 2-4 kcal so the lowest level of G is the level most heavily occupied. For the moment we simplify Fig 1(left) by ignoring all the vibrational levels to get Fig. 1, right. In lecture II we saw that the rate constant for the transition from G → S is symbolized as Bgs and the rate constant for the reverse transition (S → G), Bsg = Bgs. These are the Einstein IV-2 coefficients for stimulated absorption and emission. B is proportional to D (and thus approximately proportional to Am) In a light beam of intensity Io the number of molecules promoted = NgIoBgs and the number of molecules demoted = NsIoBsg For a two state system at equilibrium NgIoBgs = NsIoBsg OR Ng = Ns for any value of Io > 0. Calamity? the net absorption of light should disappear! Is this the end of spectrophotometry? In resolving this dilemma Einstein proposed that the S → G transition could occur spontaneously. This process he denoted as Asg (or kr), the rate constant for spontaneous emission. Einstein showed that Asg = 8 π νo3 Bgs (IV-1) c3 At very low frequencies (109 Hz, nmr, epr) stimulated emission (Bsg) dominates. At high frequencies (1015 Hz, Uv-Vis) spontaneous emission (Asg) dominates. Rearranging and simplifying equation (IV-1) (Bgs α D α [area under the absorption band]): Asg = 5.8 x 10-9νo2 [√πAM∆] (sec-1) .............(IV-2) (with spectral parameters in cm-1). Example : A very common fluorophore is ANS (1-anilinonapthalene-8-sulfonic acid. For structure of ANS and other common fluorophores see Fig IV-2. The spectral parameters of ANS are νo = 26500 cm-1; AM = 4900 M-1cm-1; ∆ = 1900 cm-1 So Asg = (5.8 x 10-9)(26500)2 (1.77)(4900)(1900) = 7.33 x 107 sec-1 A large B (= large D = large AM) will have a rapid spontaneous emission of light. We say that the state S has a characteristic radiative lifetime denoted as τr. τr = 1/Asg (IV-3) For ANS τr = 13 nsec. If we manipulate our sample such that a significant fraction of the molecules are in the excited stae at time=0 τr is the time for that fraction to decay to 1/e of the initial value. In general, the larger the value of AM the more rapidly a fluorophore emits fluorescence and the shorter τr. IV-5 F = IE φ %A G(θ) (IV-7) IE = intensity of the exciting light at a convenient wavelength φ = quantum yield %A is %absorption of the solution (= (100 - %T) = (100 - 102-A)) G(θ) is a geometric factor characteristic of the instrument (if standard and unknown are measured in the same machine it can be ignored). By arranging that the absorbance of the unknown and the standard are the same at the wavelength used for exciting the fluorescence then the %A factor can also be ignored and we get F u = φ u (IV-8) Fs φs. where u and s mean unknown and standard. The F's are the areas under the fluorescence emission spectrum . The only complication is that the original data has to be been corrected for the spectral response of the detection system (optical components and photon detectors do not respond equally well at all wavelengths and the spectral response of the instrument must be compensated for when comparing fluorescence emission in two different spectral ranges). The most commonly used standard is quinine sulfate in ca 0.1M sulfuric acid; its quantum yield is 0.55 (absorption maximum at 340 nm and emission maximum at 445 nm). IV-6 The fluorescence instrument Light Source Detector M1 M2 Sample 90' Geometry of Standard Fluorescence Instrument Fig IV-3 The modern fluorescence spectrometer (Fig IV-3) contains two monochromators: M1 selects the desired exciting wavelength. M2 is used to analyze the wavelength dependence of the fluorescence emission. The emission is normally analyzed at 90° to the incident radiation so that the detector is not overwhelmed by the much more intense exciting beam and produce spurious results. For typical samples this 90° geometry, in which a small amount of fluorescent light is detected over a background of no light, gives the fluorescence measurement about 1000x sensitivity over absorption measurements on the same solution. The instrumental geometry makes it clear that two kinds of spectra can be obtained. Excitation Spectra. M1 is varied at a fixed value of M2. We obtain fluorescence intensity as a function of the wavelength of the excitation. Because all excited states decay to the lowest vibrational level of S1 in a time short compared to τf, the excitation spectrum will have the same shape as the absorption spectrum. This is because the fluorescence intensity measures the number of molecules that have been excited and this number is directly proportional to the absorbance (provided the data has been suitably corrected). Typically IV-7 the excitation spectrum is used to verify that the fluorophore is the species you believe it to be; because the technique is so sensitive trace impurities of a highly fluorescent material might confer "fluorescence" to a material that, in reality, is non-fluorescent. Emission Spectra. M2 is varied at fixed M1; we obtain the wavelength distribution of the emission. The emission is from a single level, the lowest level of S1, regardless of where the electron is originally promoted. Consequently: 1) The shape of the fluorescence emission is independent of exciting wavelength. However the intensity will be proportional to the value of the absorbance at the wavelength selected for excitation. 2) Because promotion is from the lowest level of G to all the vibrational excited levels of S while the demotion is from the lowest level of S to all the levels of G, the fluorescence emission is red-shifted with respect to the last absorption band (G⇒S1) and has only small overlap with this band. The emission spectrum is approximately the mirror image of the lowest energy absorption band. This occurs because on promotion the vibrational levels (of S1) are on the high-energy side of the absorption curve while on demotion the vibrational levels (of G) are on the low-energy side. (The relative probability of reaching a given vibrational level decreases in order no, n1, n2 ). Routine Applications of Fluorescence Essentially the same as spectrophotometry, routine characterization and quantification of biomolecules though with a much higher sensitivity. Examples: Quantitative analysis of flavins Steady-state kinetics of NADH utilizing enzymes. regrettably most biomolecules are non-fluorescent. Important exceptions are: tryptophan in proteins (although tyrosine should make a weak contribution it is usually quenched by energy transfer). In nucleic acids only the Y-base of t-RNA has significant fluorescence. Consequently a large proportion of fluorescence research relies on biosystems that have been "labeled" with fluorescent markers, usually directed to a particular site such as a particular amino acid via covalent reaction (e.g. ANS), or by intercalation in the case of nucleic acids e.g. acridine (for examples see Fig IV-2) Environmental Effects. Absorption and CD are events that are complete in 10-15 sec; during this short time the environment has no chance to change and consequently these spectroscopic processes are relatively insensitive to environmental factors. However, the fluorescence lifetime is very long and all kinds of processes have an opportunity to occur (protonation, deprotonation, solvent reorientation, local conformational changes, and bulk movement of biological molecules, principally rotation). Thus fluorescence is a much more sensitive probe of changes in the environment than is absorbance. Fluorescence Quenching. The quenching (i.e. suppression) of fluorescence by added reagents called quenchers (Q). Examples of common quenchers are oxygen, bromide ion, cesium and acrylamide. IV-10 This is (singlet-singlet) energy or RESONANCE TRANSFER. It does NOT occur by the photon leaving D and then being re-absorbed by A (this is the so-called trivial process); it is direct. (A mechanical analogy is provided by two pendulums of identical frequency suspended from the same beam). Note that there is no requirement that A and D be in the same molecule (though this is often the case). The only requirements are: 1) There must be some overlap between the emission of D and absorption of A (matching of energy levels). 2) They must be within a reasonable distance; some estimates say that 100 Å is the upper maximum. 3) They must have a suitable geometric relationship (V12 ≠ 0) In energy transfer these geometric terms (2 and 3 above) of the dipole-dipole eqn. are parameterized as κ/r3. A "golden rule" of quantum mechanics: The rate of a process = (interactions responsible for the process)2 Consequently the geometric term enters as κ2/r6. The sixth power dependence in r makes this phenomenon exquisitely sensitive to distance. The method: We first measure the extent of energy transfer. There are three ways to accomplish this (in order of increasing desirability): 1) Measure the extent to which the presence of a fixed concentration of A suppresses the intrinsic emission of D. 2) If A is fluorescent measure the emission of A while exciting D; this process is called sensitized fluorescence. 3) Measure the extent to which A reduces τf for D; i.e. A is treated as a quencher. (1) and (2) are less desirable because they do not distinguish between true energy transfer and the trivial process whereby D emits and A reabsorbs which can look like energy transfer. This trivial process will not affect (3). BASIC THEORY Let kf = 1/ τf be the rate constant for depopulating D* in the absence of energy transfer and let kt represent the rate constant for the energy transfer process. Then the fraction of excited molecules that react by energy transfer (ET) is ET = k t (IV-10) kf + kt The fraction that decays normally = 1- ET = k f = τ T (kf + kt) τf where τf is the normal fluorescence lifetime and τT is the lifetime observed in the presence of energy transfer. Recall the definition of quantum yield: Φ = τf /τr. Then ΦT / Φf = 1 - ET where ΦT and Φf are the quantum yields in the presence and absence of acceptor. The efficiency of this energy-transfer process is governed by the dipole-dipole interaction and Forster has shown that IV-11 ET = Ro6 / (Ro6 + R6) or R = Ro [(1 - ET) / ET]1/6 (IV-12) R is the distance between D and A and Ro is a "critical" distance at which the efficiency of transfer is 50%. It is calculated via the formula Ro = 9800(J (n)-4 κ2Φf)1/6 (IV-13) where n is the refractive index (1.33 for water), Φf is the donor quantum yield and J is the spectral overlap, the extent to which the energy levels match. J is calculated from J = ∫ F D ( λ )A M ( λ ) λ4 d λ (ΙV−14) ∫ FD(λ)dλ A recipe for calculating J is given at the end of this chapter. Example: Rhodopsin is the visual pigment present in the disc membrane of the rod cells of the retina. It is composed of the fluorescent compound retinal attached to the protein opsin. Rhodopsin is presumed to span the membrane so that light impacting on the outer surface can cause biochemical processes at the inner surface. Electron microscopy shows that this membrane is ca. 75Å thick. Is rhodopsin "long" enough to meet the need? 1) The quantum yield of rhodopsin = 0.75 (= Fs). 2) After labeling the membrane with the fluorophore 1,5-I-AEDANS the quantum yield of the retinal emission drops to 0.68 (= Ft). 3) The overlap integral (J) is 1.84 x 10-13 M-1 cm2. We first calculate Ro. With (n)-4 = 0.32 and κ2 = 2/3 Ro = (9800) [(1.84 x 10-13 )(0.32)(0.67)(0.75)]1/6 = 54 Å. 4) 1 - Et = Ft/Fs = 0.68/0.75 (= 0.91) so Et = 0.09 . So R = 54[(0.91)/(0.09]1/6 = 79Å. Apparently rhodopsin does have appropriate dimensions to span the membrane. (Note that if a biomolecule is labeled with several chromophores it is always the acceptor closest to the donor that is detected,, by virtue of r1/6.) See Background Note 2 for a comment on κ2. IV-12 Fluorescence Depolarization. This is fluorescence conducted with plane polarized light (ppl). Using polarized light not all molecules are excited from G → S. Our First Approximation: Only those molecules that have their transition dipole moment parallel to the plane of polarization of the light are promoted. Consequently only these molecules can fluoresce. (This is an example of a technique called photoselection, the preferential excitation of a subset of molecules. ) The simplest case: the emission dipole and absorption dipole are parallel (normally true for G ⇒ S1) . Possibility 1: the molecules have not moved during the ≈1 nsec of fluorescence lifetime. Consequence: the polarization of the emission will be parallel to that of the exciting light. Possibility 2: The molecules have rotated! Consequence: the plane of emission will be changed by an amount which is directly proportional to the amount of rotation. Practical value: Measure the change in plane of polarization and deduce the rotational correlation time τc = 1/3'rd the rotational relaxation time (ρ). The rotational relaxation time can be thought of as follows: At time zero let all molecules be oriented in the same way. With time the molecules rotate in a random way. At some time later the average orientation will be within 37% of being completely random. This time is the rotational relaxation time. ρ = 3ηV/RT (Original Einstein equation) where η is the viscosity (poise) and V the molecular volume (cm3). To simplify some later equations it is usual practice to use the rotational correlation time (τc = ρ/3.) A very rough rule: τc ≈ 1 nsec per 2.5 kDa; So 50 kDa protein will have a τc of 20 nsec. INSTRUMENT (see note three for a contemporary instrument) Three polarizers are present. P1 selects the plane of polarization of the excited light P2 selects that component of the emission (Iº or I2) parallel to the excitation. P3 selects the component (IÁ or I3) perpendicular to the exciting light.