1. Quantum Physical Phenomena in Life (and Medical) Sciences
III. Absorption and fluorescence spectroscopy
Péter Maróti
Professor of Biophysics, University of Szeged, Hungary
Suggested texts:
P. Maróti, G. Laczkó and L. Szalay: Medical Physics I-II, University of Szeged, 1988.
P. Maróti, L. Berkes and F. Tölgyesi: Biophysics Problems, A Textbook with Answers, Akadémiai Kiadó, Bp
S. Damjanovich, J. Fidy and J. Szőlősi: Medical Biophysics, Semmelweis, Budapest 2006
P.W. Atkins: Physical Chemistry, Oxford University Press, Oxford, 5th Edition, 1994

2. What determines the extinction coefficient
What determines the extinction coefficient? Orbital symmetry, overlap and spin multiplicity. (from chemistry to spectroscopy of molecules)
Brush up your mind....
First, let’s review how atomic orbitals are combined to form molecular orbitals. As an example, consider the molecule H2C=O, formaldehyde. The s and p atomic orbitals are combined to form sigma, π, n, and anti-bonding π* orbitals (see the schemes later).
Note that the non-bonding molecular orbitals, n, contain the “lone pair” electrons on the oxygen atom, which are rather exposed to solvent and susceptible to influence by solvent effects. The two pz orbitals on the O and C atoms combine to form two atomic orbitals, of which the lower energy orbital is the bonding π orbital and the higher energy orbital is the anti-bonding π* orbital. The absorption of a photon can promote an electron from the n orbital to the π* or from the lower energy π orbital to the π* orbital. These transitions are known as n-π* and π- π* transitions. These are the types of transitions that we deal with in organic molecules that comprise the components of biological macromolecules such as proteins and nucleic acids.

3. Molecular orbitals, energy levels, and transitional energy levels
EXAMPLE
Molecular orbitals, energy levels, and transitional energy levels
Formaldehyde
Antibonding orbitals of excited states
Energy levels of the orbitals

4. MOLECULAR ORBITALS of formaldehyde molecule
H 1s 1 electron
C 1s2 (2sp2)32pz 6 electrons
O 1 s2 2s2 2p2x 2py 2pz 8 electrons
C--H bonds are formed between the s electron of H and the carbon hybrid 2sp2. The third sp2 orbital on carbon forms a σ orbital with the 2px orbital of oxygen. The remaining orbitals are a 2pz for C and 2pz and 2py for O.
A molecular orbital can be formed by overlap of the single occupied 2pz of O with 2pz of C.
The remaining 2 electrons of O must be in the 2py atomic orbital (non-bonding n orbitals).
The two highest unoccupied states in formaldehyde are the binding π and the atomic n (2py) orbital of oxygen.
Considering only the π* level, two transitions are possible from the n state and from the-π state. To estimate absorption intensities, the matrix element of the transition dipole moment can be evaluated (not here).

5. , n, and * orbitals *  *  - * n - *  Formaldehyde 1e- H: 1s
6e- C: 1s22s22px2py s2(2sp2)32pz1
8e- O: 1s22s22py22px s2(2sp2)52pz1 *  *
2pz(C) + 2pz(O) n
 - *
n - * 
Excited state ground state excited state
Carey, Organic Chemistry, 1987

7. Electronic, vibrational and rotational energy components of the molecules
Anharmonic oscillator
Electronic excited state
Vibration and rotational energy levels
wavefunction
transition
Ground state
at room temperature

8. Franck-Condon principle
Electronic transitions in diatomic molecule
Franck-Condon principle
Electronic transitions are much faster, ~1015 s-1, than nuclear vibrations (1013 s-1).
Consequences:
1) Electronic transitions represent vertical displacement in diagrams, the nuclear position (as indicated on the r scale) does not change appreciably while the electron moves. This is the Franck-Condon principle. Similar diagrams could be drawn to represent transitions in more complicated molecules like chlorophylls.
2) The kinetic energy of the nuclei does not change during the electronic transition. This usually necessitates transitions between points where the vibrational energy levels meet the parabola (A, A’; B, B’; D, D’), when the nuclei are at rest. Transitions like C, C’ can only occur if the nuclei have the same kinetic energy, indicated by the hatched lines.
electronic transition
nucleaus „at rest”
nuclear vibration
nuclear coordinate
3) all transitions are quantized. If an excitation populates a higher vibrational level, the molecule can relax within the excited state to lower vibrational levels by emission of IR quanta.

9. Franck-Condon principle: Origin of Fluorescence
Where energy gaps are large, the downward transition is less probable, or “forbidden”. The life-time of the excited state is prolonged, and the energy is released in a single quantum of fluorescence. Because some energy can be lost rapidly through vibrational relaxation in the excited state, fluorescence is always shifted to the red.

10. Fluorescence + Phosphorescence
Thermal relaxation
Phosphorescence occurs at longer wavelengths with respect to fluorescence. Because the triplet to singlet transition is forbidden, the lifetime of the triplet state is very long (milliseconds to seconds)
Luminescence:
Fluorescence + Phosphorescence

11. Absorption spectrophotometry

13. OD = OD(λ) or ε = ε(λ) is the absorption spectrum.
The area under the absorption spectrum is the oscillator strength (f) of the dipole.
ε: molar absorption, c: velocity of light (in cm/s), me: mass (g) of the electron, e: charge (esu) of the electron, n: refractive index of the medium and N0: Avogadro’s number.
The oscillator strength characterizes the number of electrons per molecule that oscillate with the characteristic frequency.

14. Absorption photometry of the skin
The light of different colors penetrate to different depths of the skin. The penetration is strongest in the near infrared spectral region.
Absorption bands of key molecules in different layers of the skin:
Nucleic acids and amino acids: UV-C,B (present mainly in epidermal cells)
Haemoglobin, melanin, the carotenes and bilirubin: visible range
burn injury
pigment formation
photo-carcinogenesis
erythema
epidermis
dermis
subcutaneous layer

15. Absorption spectroscopy as a probe of the chromophore environment
Tuning the absorption spectrum of a chromophore bound to a protein by different electrostatic and steric interactions with the ground and excited states
Visual Pigments: the rhodopsin family
tuning by interactions
excited state
red shift if excited state has lower energy
due to interaction with an amino acid side chain
blue shift if excited state has higher energy
due to interaction with an amino acid side chain
ground state

16. rods: black/white vision cones: color vision

17. Absorption spectra of rhodopsins
Chemically different retinal chromophores or
the same chromophore but in different interactions with the environment?
Rod cell
red sensitive
Cone cell
green sensitive
Disks containing Rhodopsin
Nucleus
blue sensitive
absorption
Synaptic Ending
wavelength (nm)
Absorption spectra of rhodopsins

18. visual pigment in the eye rod cells responsible for dim light vision
Rhodopsin
visual pigment in the eye rod cells
responsible for dim light vision
max = 500 nm
color vision is due to similar pigment
proteins in cone cells:
max = 414 nm (blue)
max = 533 nm (green)
max = 560 nm (red)
Same chromophore: 11-cis retinal
“spectral tuning” by interaction with
amino acid residues nearby
Biochem (2001) 40,

19. Absorption spectra of rhodopsin in the retinal membrane of the visual cells of cones and rod

20. Light-induced conformational change of the retinal7 9 11 6 12 8 10
11-cis retinal h
all trans retinal

21. Residues in rhodopsin near the retinal chromophore
Biochem (2001) 40,

22. Residues mutated to shift the spectrum of rhodopsin
membrane spanning helixes
Biochem (2001) 40,

23. Blue shift of the rhodopsin spectrum due to altering
the residues in the protein binding site
WT: 500 nm
G90S: 487 nm
T118A: 484 nm
E122D: 477 nm
A292S: 489 nm
A295S: 498 nm
T/E/A triple mutant
453 nm
blue shift
T/E/A
Notation: G90S
The native amino acid glycine (G) at position 90 is mutated for serine (S).
Different electrostatic and steric interactions with the
ground state and excited state of retinal alter
their relative energy difference
Biochem (2001) 40,

24. Excitons and Energy Transfer
Förster resonance energy transfer
A B
kt = kf·(R0/R)6
at R = R0 (Förster’s distance)
kt = kf
excitedstate h kf
ground
A B
excited state
migration
A B
deexcite A
either by fluorescence, kf
or by energy transfer, kt
excite A
If excitation hops very fast, it cannot be localized in a single molecule
excited state covers both A and B (or more)
called exciton band
can view excitation as diffusing from one molecule to next.
Slow hopping (108 sec-1) is measurable
This leads to energy transfer, which provides a way to experimentally measure the distance (R) between A and B (assuming Förster’s mechanism of energy transfer): FRET (Förster’s Resonance Energy Transfer)

25. Three general situations to consider geometrically the interacting dipoles of the molecules
Representation of the dipoles by springs
I Card stack geometry
II Head to tail geometry
III Herringbone geometry mA mB
harmonic oscillators mA mB mA mB

26. +   Case I: Card stack geometry E blue shift seen with A AA poly A
monomer dimer polymer S1 S0 f 2f E

27. Case 2: Head-to-tail geometry+
and + - + - + - - +
oscillator strength f f
2·f
in phase
out of phase
allowed, lower energy (red shift)
forbidden (blue shift)
monomer dimer f 2f E
Red shift  e
monomer
dimer

28. Case 3: Herringbone geometry+ - - - + + + - f f
Oscillator strength
in phase
non-zero f
high energy
allowed
out of phase
non-zero f
low energy
allowed
monomer dimer E
This is called Davydov splitting  
monomer
dimer

29. An example of shifts in the absorption spectrum
due to molecular complex formation
anthocyanin complexes responsible for
the colors of flowers and fruits
cover the entire visible spectrum

30. Exiton couplings are responsible for many of the colors
of flowers and fruits
Due to non-covalent hydrogen bonded complexes of
anthocyanins
Commelinin is composed of
6 anthocyanin (A/blue) and
6 flavocommelins (yellow, F)
PNAS (2001)98,

31. Model of the synthetic complex between a dimelamine and barbituate
PNAS (2001)98,

32. Dilution causes the complex to dissociate
and results in color change: wavelength change of
the absorption spectrum
No Complex
(10-6 M)
NOTE: ISOSBESTIC POINT
all the spectra pass through a common
point
Complex
(10-3 M)
The complex behaves like card stack geometry:
blue shift
upon complex formation
PNAS (2001)98,

33. An isosbestic point implies that there are only two species
in equilibrium being observed
Main point: if there were three or more
species, the chances they all have the same
absorbance at any given wavelength is
very small.
The presence of isosbestics is used as evidence that there are no intermediate species of significant concentration between the reactants and products. A
[A] [B]
Ctot = [A] + [B]
Absorbance = [A]Al + [B]Bl
Absorbance = (fracACtot) Al + (fracBCtot) Bl
when A = B (at the isobestic point)
then
Absorbance = isoCtot(fracA + fracB)l
but (fracA + fracB) = 1.0
so Absorbance = isoCtotl
does not change as the ratio of
A and B change
constant absorbance
at the isosbestic point B

34. Bacteriochlorophyll Dimer –
Davydov splitting observed in long wavelength band
Red band
Blue (Soret) band
in 1/(M·cm)
Green window
Monomer and dimer spectra for solutions of bacteriochlorophyll. A pronounced splitting of the longest-wavelength band in the dimer is visible.
JACS 88:2681 (1996)

35. Hyperchromism A B A B A B
Due to interactions of neighboring molecules where a transition in molecule A interacts with different transitions (higher or lower energy) of molecule B
A B A B A B
Mixing of wavefunctions of higher excited states results in intensity borrowing
-Oscillator strengths of different transitions can increase or decrease
(hyperchromism)
(hypochromism)

36. Kuhn-Thomas Sum Rule:  +
Area of total Absorption spectrum is constant.
Molecular interactions can increase or decrease particular bands -
but the net area under the spectrum is not changed by the
molecular interactions.
e.g. Hypochromism in nucleic acids:
lower absorption in one region of the spectrum means there
must be an increased absorption elsewhere.  
Stacking results in decrease in the
intensity of the UV band, but an increase
in a far UV band. +

37. An Example of Hypochromisim
Lower absorbance at 260 nm due to stacking in DNA and RNA
Mononucleotides
Single-strand
unstacked DNA
Double-strand stacked DNA

38. Monitoring “melting” of double-strand DNA (from E. coli)
by the absorbance change at 260 nm
an example of hypochromism:
lower 260 for double-strand vs single-strand DNA
single-strand DNA
(high temp)
double-strand DNA
(low temp)

39. Fluorescence spectroscopy

40. 10-11 sec: internal conversion heat loss to solvent
Within about sec, all excited state molecules lose heat to decay to the lowest vibronic level of the first excited state S2
2’’
1’’
10-11 sec: internal conversion
heat loss to solvent
0’’
heat 0’ 1’ 2’ S1
heat
ABS
light
10-8 sec: larger energy gap results in slower
relaxation to ground electronic state 3 S0 2 1

41. Collisions with solvent molecules
The rate of internal conversion within the excited state manifold is due to loss of energy to the solvent via collisions and is dependent on the rate of collisions with the solvent
collision rate = kcoll [solvent]
~1010 M-1 sec-1
55M for water
In 55 M solvent, the rate of collisions of a single molecule is  sec-1 S2 S1 S0
Due to small energy gaps, collisions are very effective so that within sec all excited state molecules are in the lowest vibrational state of S1
However, the non-radiative loss of energy to the ground state, Internal Conversion to S0,is much slower due to the large energy gap

42. 10-11 sec: internal conversion
Phosphorescence results from emission of a photon from the lowest triplet state (electron spins aligned) S2
2’’
1’’
10-11 sec: internal conversion
0’’
heat 0’ 1’ 2’ S1
10-8sec: spin change,
singlet to triplet
intersystem crossing T1
ABS
fluorescence 3
phosphorescence
10-6 to 10-2 sec:
triplet to singlet S0 2 1

43. What can you monitor by fluorescence?
kI (all non-radiative modes) kf
Competition between rate of fluorescence and rates of non-radiative modes
Faster non-radiative de-excitation
less fluorescence
Many processes occur to a molecule during its excited state lifetime that influence the fluorescence:
(1) collisions (quenching) accessibility
(2) energy transfer distance
(3) solvent relaxation solvent polarity
(4) chromophore rotation molecular size/viscosity

44. Fluorescence Quenching
Rate of collisions = kQ ·[X*][Q]
between excited state X* and Q
The fastest collisional events are diffusion-limited
Dynamic quenching: collision with the excited state
1. X + h  X*
3. X*  Q + h’ kQ kf
2. X* + Q  X + Q + heat
absorption X*
kQ = second order rate constant
for collisional quenching
50Å
kQ is maximally about 1010 M-1sec-1
In the presence of [Q] = 1 M, each molecule of X* experiences 1010 collisions per second. Q So S1 kf
[Q]·kQ kI
Within the lifetime of the excited state (~ 10 ns), X* can collide with a quencher within a sphere of radius about 50 Å.

45. Dynamic Quenching So S1 kf [Q]·kQ kI No Quencher: +Quencher: Φf =
Φfo = kf
kf + kI
+Quencher: Φf =
kf + kI +kQ·[Q]
Ratio of the fluorescence yields:
Φfo Φf Fo F kf
kf + kI
kf + kI +kQ[Q]  = •
= 1 +
kQ[Q]
kf + kI
= 1 + kQo[Q] = 1 + K·[Q] Fo F
K = Stern-Volmer constant
kQ = quenching constant
Stern-Volmer Plot
Φf: fluorescence yield
F: intensity of the fluorescence 2
slope = K Fo F 1
[Q]

46. Experiment: Quenching of tryptophan in proteins-
Example:
Quenching of tryptophan in proteins-
An early indication of protein dynamics
-Observe proteins with single buried tryptophans (known from X-ray)
-Quench with (a) Iodide, (I-)
(b) O2 (use high pressures to increase [O2])
Experiment:
Result: (1) Tryptophans are inaccessible to I- but accessible to O2
(2) However, no room in X-ray structures for O2!
Lakowicz + Weber. Biochem (1973) 12,
Conclude: X-ray structure is only an average. The protein must open and close on a time scale sufficient to allow O2 inside

47. Most common methods of measuring fluorescence
Steady state measurement
source (continous)
monochrometer
polarizer
sample
detector for fluorescence
detector for measuring absorption
Dynamic methods
Pulse spectroscopy: Measure the intensity of emitted light after a very brief pulse of light (nsec duration)
-measure fluorescence lifetime
Also: Phase and Modulation Spectroscopy
See Ann. Rev. Biophys Bioeng. (1984) 13,

48. Fluorescence Parameters & Methods
1. Emission & Excitation Spectra
• local environment polarity, fluorophore concentration
2. Quantum yield of fluorescence
• how competitive other deactivation processes are with respect to fluorescence
3. Anisotropy & Polarization of Emission (not discussed here)
• rotational diffusion
4. Fluorescence Lifetime (not discussed here)
• dynamic processes (nanosecond timescale)
1. Quenching of fluorescence emission
• solvent accessibility
• character of the local environment
2. Resonance Energy Transfer (FRET)
• probe-to-probe distance measurements
3. Fluorescence microscopy, cell sorter (not discussed here)
• localization, separation
4. Fluorescence Correlation Spectroscopy
• translational & rotational diffusion
• concentration
• dynamics
5. Fluorescence Recovery after Photobleaching (FRAP)

51. e F 
0-0’

52. F (lex)  e (lex)
1-anilinonaphthalene-8-sulfonic acid (1,8-ANS) is a fluorescent probe

53. Fluorescence quantum yield
We define QUANTUM YIELD of the fluorescence by the ratio of
ΦF=kf / (kf+knr)
where kf and knr indicate the radiative (by fluorescence) non-radiative (dissipative) decay rates of the excited state, respectively. The maximum value of the quantum yield is 1. There are substances with quantum yield close to one (fluorescein, rhodamine).
excited state
emission
heat
absorption
knr kf
ground

54. Fluorescence Correlation Spectroscopy (FCS)
Special Techniques:
Fluorescence Correlation Spectroscopy (FCS)
By measuring fluctuations in fluorescence, the residence time
of a fluorescent molecule within a very small measuring volume
(1 femtoliter, L) is determined.
This is related to the Diffusion Coefficient
molecules moving into and out of the measuring volume:
fast slow
excitation
emitted photons

55. Fluorescence Correlation Spectroscopy: FCS
the time-dependence of the fluorescence is expressed as an autocorrelation function, G(),
the is the average value of the product of the fluorescence intensity at time t versus the intensity at a short time, , later. If the values fluctuate faster than time  then the product will be zero.
F(t)
F(t + τ)
δF(t) = F(t) - <F(t)> deviaton from the average intensity
δF(t)=F(t)–<F(t)>
average intensity: <F(t)> t
t + τ

56. simulated autocorrelation functions of
An example of FCS:
simulated autocorrelation functions of
a free fluorescence ligand and the same
ligand bound to a protein
1:1 mix of
free/bound ligand
bound ligand on
slow moving protein
G() goes to zero
at long times
free ligand (small, fast diffusion)

57. 1-photon
excitation
2-photon
excitation

58. Autocorrelation Adenylate Kinase –EGFP Chimeric Protein in HeLa Cells
Adenylate kinase (AK) is a ubiquitous enzyme that regulates the homeostasis of adenine nucleotides in the cell.. The cytosolic adenylate kinase (AK1) and its isoform (AK1beta) was fused with enhanced green fluorescence protein (EGFP) and expressed the chimera proteins in HeLa cells. Using two-photon excitation scanning fluorescence imaging, the localization of AK1-EGFP and AK1beta-EGFP in live cells were directly visualized. AK1beta-EGFP mainly localized on the plasma membrane, whereas AK1-EGFP distributed throughout the cell except for trace amounts in the nuclear membrane and some vesicles.
Fluorescence intensity
Examples of different Hela cells transfected with AK1-EGFP

59. GFP Spectral Tuning By Protein Environment
by mutagenesis
Protein (gene) is from a jellyfish:
Aequorea victoria
GFP
T203Y: -stacking of the tyrosine to the aromatic chromophore
stabilizes the excited state and results in a “red shift” Yellow Fluorescence
yellow emission maximum is 529 nm
Also there are blue (448 nm) and cyan (485 nm)
Autocatalytic reaction of three amino acids
in the protein: cyclization and Oxidation by O2

60. Autocorrelation functions
Autocorrelation of EGFP and Adenylate Kinase -EGFP
EGFP-AK1 in the cytosol
EGFPsolution
EGFP-AK1b in the cytosol
EGFPcell
A mixture of AK1b-EGFP in the cytoplasm and membrane of the cell.
Clearly more than one diffusion time

61. Autocorrelation Adenylate Kinaseb-EGFP in the HeLa cells
Cytosol
Plasma Membrane
For AK1-EGFP, only one diffusion component was observed in the cytoplasm. For AK1beta-EGFP, two distinct diffusion components were observed on the plasma membrane. One corresponded to the free diffusing protein, whereas the other represented the membrane-bound AK1beta-EGFP. The diffusion rate of AK1-EGFP was slowed by a factor of 1.8 with respect to that of EGFP, which was 50% more than what would be expected for a free diffusing AK1-EGFP.
The FCS is a powerful technique for quantitatively studying the mobility and interactions of the target protein and its function in live cells.
The color code represents diffusion constants in the range between μm2/s (blue, cytosol) and 0.12 – 0.18 μm2/s (yellow, membrane)

62. Special Techniques: Fluorescence Recovery After Photobleaching (FRAP)
Measuring the Diffusion of Proteins in the Cytoplasm of E. coli
Fluorescence Recovery After Photobleaching (FRAP)
Ready..
Aim...
Fire!
Diffusion of protein into the spot t1
E. coli cell t0 t2
1. Express a protein that is fluorescent: green fluorescent protein, GFP.
2. Use a laser to “photo-bleach” the fluorescent protein in part of a single
bacterial cell. This permanently destroys the fluorescence from
proteins in the target area.
3. Measure the intensity of fluorescence as the protein diffuses into
the region which was photo-bleached.

63. Diffusion of the Green Fluorescent Protein inside E. coli
Results:
D = 7.7 µm2/sec (7.7 x 10-8 m2/sec)
this is 11-fold less than the diffusion coefficient in water: 87 µm2/sec
Slow translational diffusion is due to the crowding resulting from the very high protein concentration in the bacterial cytoplasm ( mg/ml)
Single cell, expressing GFP
Bleach cell center with a laser, t0
t = 0.37 sec after flash
t = 1.8 sec after flash
one can observe the
molecules diffusing
back into the bleached
area
4 µm
J. Bacteriology (1999) 181,

64. Fluorescence Spectroscopy and SEX
Fluorescent Plumage of Parrots is a Sexual Signal!
white light
UV illumination
Science (2002) 295, 92

65. The blue pinwheel bioluminescence of an escaping deep-sea jellyfish
Mélytengeri medúza vészhelyzetben Biolumineszcenciával fénykereket működtet, hogy a ragadozó figyelmét lekösse.
The blue pinwheel bioluminescence of an escaping deep-sea jellyfish

66. Telepes medúza 2000 m-re a tenger felszíne alatt
Telepes medúza 2000 m-re a tenger felszíne alatt. Csaliként alkalmazott biolumineszcencia halak befogására (Science, július 8.)
Red bioluminescence to attract prey in depth of about 2000 m in the sea water.

67. Problems for Seminar
The 250 nm absorption band of nucleotides exhibits a 5 nm exciton splitting in polynucleotides due to the interaction of the nucleotides with their neighbors („Davydov splitting”). What will the interaction energy be in units of kJ/mol if only dimer formation is assumed?
The donor intrinsic fluorescence lifetime is τf = 2 ns (no acceptor is present). What will be the actual (observed) lifetime as a function of acceptor concentration? The Förster radius is R0 = 3 nm.
In an experiment of dynamic quenching of fluorescence, the following values of the fluorescence intensity (F) were measured at different concentrations of the quencher:
What is the Stern-Volmer constant?
4. Determine the isosbestic point of two interconverting species: A ↔ B. The equilibrium constant is Keq, the absorption spectra are Gaussian curves with peak positions of 500 nm and 520 nm, widths of 50 and 70 nm and maxima of 2·104 M-1cm-1 and 4·104 M-1cm-1 for the A and B species, respectively.
[Q] (mM) 1 2 4 8
F (arb. units)
10.5
9.1
8.5
7.1
5.3

68. Problems for Seminar
5. Solutions containing the amino acids tryptophan and tyrosine can be analyzed under alkaline conditions (0.1 M KOH) from their different UV spectra. The molar extinction coefficients (in M-1cm-1) under these conditions at 240 nm and 280 nm are
A 10 mg sample of the protein glucagon is hydrolyzed to its constituent amino acids and diluted to 100 mL in 0.1 M KOH. The absorbance (optical density) of this solution (in cuvette of 1 cm light path) was at 240 nm and at 280 nm. Estimate the content of tryptophan and tyrosine in μmol/(g protein).
6. The net reaction of CO2 fixation in photosynthesis is
CO2 + H2O → (CH2O) + O2
where (CH2O) represents 1/6 of a carbohydrate molecule such as glucose. The enthalpy change for this endothermic reaction is ΔH = +485 kJ/mol of CO2 fixed. The longest-wavelength absorption band of chlorophyll a peaks in vivo at a wavelength of about 680 nm. What is the minimum intensity of illumination (in unit of einstein, i.e. in number of mol photons) to be absorbed to provide the energy needed to fix 1 mol of CO2 via photosynthesis? What is the photochemical quantum yield of the process if 9 photons per CO2 molecule is the experimental value?
Tyr
Trp
240 nm
11,300
1,960
280 nm
1,500
5,380

69. Molar extinction, ε (1/(M·cm))
Problems for Seminar
7. The pH-indicator is a dye (D) whose spectrum changes with pH.
DH+ ↔ D + H+
Consider the following data for the absorption spectrum of an indicator (pK = 4) in its ionized (DH+) and non-ionized (D) form:
Wavelength (nm)
Molar extinction, ε (1/(M·cm))
DH+ D
400
10,000
420
15,000
2,000
440
8,000
460
12,000
480
3,000
The absorbance (optical density) of the indicator solution is measured in a 1 cm cuvette and found to be
λ (nm)
400
420
440
460
480
O.D.
0.250
0.425
0.400
0.300
0.075
Calculate the pH of the solution.
Calculate the absorbance of the solution at 440 nm and pH 6.37 for the same total indicator concentration.

70. Problems for Seminar
8. Consider the reaction of an absorbing reactant M to give an absorbing product P by the reaction
Show that no isosbestic points would be expected for this system, even if the intermediate N has no measureable absorbance in the same spectral region.
9. Compare the sensitivity of the radioactivity with that of fluorescence detection!
32P has a halftime of 14.3 days. How many disintegration per second will you obtain from 1,000 atoms of 32P? What is the maximum number of counts you can get?
Fluorescein has a molar absorption coefficient of 70,000 1/(M·cm) at 485 nm and a quantum yield for fluorescence of What is its molecular absorption coefficient (cross section) in cm2 per molecule?
1,000 molecules of fluorescein are irradiated with an argon laser which has an intensity of 2 mW/cm2 at 485 nm. How many fluorescence photons per second will be emitted?
Discuss the relative merits of detecting small number of molecules by radioactivity and by fluorescence. What other factors besides counting rate may be important?

71. Problems for Seminar
10. A swimmer enters a gloomier world on diving to greater depths. Given that the mean molar absorption coefficient of sea water in the visible region is 6.2·10-5 1/(M·cm), calculate the depth at which a divier will experience (a) half the surface intensity of light, (b) one-tenth the surface intensity.