Exp. 4 Isotopes
3723 F02
Biochemistry 3723
1
Lecture 4b: Radioisotopes
August 28, 2002
V. Scintillation Counting of - emissions: 3H, 14C, 35S, and 32P
A. Process of scintillation: use 3H as an example:
1. - decays with energy between 0 and 0.0186 MeV (= 18.6 KeV)
2. If radioactive solute dissolved in appropriate solvent, collision of - with solvent
transfers some of energy to solvent, “exciting” solvent molecule.
a. non-bonding π electrons excited to higher state without loss of energy
b. - continues to excite solvent until all energy of particle dissipated.
c. -**** + S  -*** + S*; -*** + S  -** + S* ; etc.(* = energy)
3. Excited solvent transfers energy to fluor, a compound that accepts energy easily from
S and decays to ground state with emission of light
S* + F  S + F*;
F* F + light (h)
4. Look at elements of overall process
a. - emission  photons takes < 5 nsec.
i. Yield ≈ 10 photons /KeV. Therefore most energetic 3H decay ≈ 186 photons
ii. # photons ≈ linear w/energy of -. Average 5.5 KeV ≈ 55 photons
b. Solvent--various aromatic compounds-- toluene, xylene, alkylbenzenes,
alkylnapthalenes
c. Fluor--various multicyclic compounds (aromatic) -delocalized electrons
CH3
H3 C
N
H
H
C C
C
C
O
H
H
PPO max =375 nm
bis-MSB  max = 425
d. Scintillation Cocktail: Fluor dissolved in Solvent
e. Additions to allow solution of aqueous samples-i.e. DMSO
f. Many proprietary formulations (companies won't tell exactly how they are
formulated, because then you wouldn't buy their expensive products.)
B. The Scintillation Counter: Necessary parts—in your book (p 177)
1. Sample chamber--where you put the scintillation vial containing sample dissolved in
scintillation coctail
Exp. 4 Isotopes
3723 F02
2
2. Photomultiplier tubes
a. Two of them "see" light flashes.
b. Resolution time ≈ 20 nsec. Means absolute limit ≈ 5 x 107 cps
3. Pulse summation (counts # events)
PMT
PMT
Pulse Summation
Coincidence Circuit
Analysis
Printer
4. Coincidence circuit
a. Light flash must be seen by both PMTs at same time to be counted.
i Theoretical minimum to be seen: 2 photons/- emission
ii. Practical minimum? ≈ 1 KeV (10 photons)
b. Greatly reduces electronic noise, which would excite only one PMT at a time.
5. Analysis: Two parallel technologies
a. Beckman instruments: Pulse height analyzer--Divide counts into 2 or 3 Channels
or “windows” by energy. Often can be set by operator--typically set channel A to
measure 3H, channel B for 14C/35S and channel C (if it exists) for 32P
Exp. 4 Isotopes
3723 F02
3
A
C
3
H
B
14
C
32
P
#
Energ
b. Packard Tricarb 2000CA is what we have
--4096 windows, each of 0.5 KeV (from 0-2000 KeV)
--each isotope gives characteristic spectrum--distribution in windows
--instrument analyzes spectrum to determine isotope/quenching
5. Printer: Pretty obvious--gives printout of results. usually time, cpm, dpm, etc.
C. Quenching: The problem with counting is that not every decay event is registered at its
proper energy. Must have a way of accounting for this.
1. Counting efficiency = (# voltage pulses)/(# decay events) = cpm/dpm <1
Often expressed as % efficiency (Efficiency  100)
2. Look at scheme of events: Quenching can occur at all steps
a. excitation of solvent by -: -**** + S  -*** + S*; etc.
b. transfer of energy from S* to Fluor: S* + F  S + F*
c. Relaxation of excited Fluor to ground state: F*  F + h
d. Counting of light by PMT  voltage pulse
energy of pulse ≈ photons counted ≈ energy of –
3. Define quenching as anything that reduces the apparent number or energy of decay
events.
a. POINT Quench: -*  – + heat (no S excitation)
Exp. 4 Isotopes
3723 F02
4
i. energy absorbed within sample, especially when sample insoluble in solvent
or is on a solid support (filter, TLC medium)
ii. Because of this not all – particles reach and excite solvent
iii. How deal with this: Change solvent to include emulsifyer: (usually
detergents--proprietary) to dissolve sample, so it is in direct contact with
solvent.
b. CHEMICAL Quench:
S* or F* + Chemical (Ch)  S or F + Ch* Ch + heat
i. energy transferred to contaminant that relaxes to ground state to generate heat
rather than light.
ii. Typically chlorinated hydrocarbons, organic solvents, H2O, O2
iii. How to deal with it? Often not possible to remove quencher from sample.
Must calculate quench correction
c. COLOR quench: h + Color Comp (CoC) CoC* CoC + heat
i. Colored compound absorbs light, gives off energy as heat (or light of longer
wavelength -lower energy).
ii. How deal with it? Purify sample or use quench correction
d. DILUTION quench
i. Insufficient S or F to absorb all - energy
ii. Solution? Use less sample, or different scintillation cocktail
D. Quench correction: Works for chemical and color quench only
1. Internal Standard on each sample is the best way, but it is costly and time consuming.
Seldom used and won’t discuss.
2. Quench curve determined from Sample Channels Ratio (SCR) or Spectral Index of
the Sample (SIS)
a. Method based on shift of energy spectrum
i. quench causes shift to lower energy
ii. quench reduces total # events because some at low energy do not generate
enough photons to overcome coincidence circuitry
b. Sample Channels Ratio (SCR) for Beckman instruments
i. Count series of standards of known dpm:  cpm
ii. Determine cpma/cpmb (SCR) for series of stds of known dpm
Exp. 4 Isotopes
3723 F02
5
iii. Plot eff (= cpm/dpm) vs SCR
iv. For each sample, determine counts, and SCR. Use graph to convert cpm to
dpm
c. Spectral Index of the Sample (SIS) for Packard Tricarb 2000CA
i. Functionally exactly same as SCR
ii. For standards of known dpm, measure cpm, SIS (determined by counter and
printed by printer)
iii. Plot eff vs. SIS for standards. This is a quench curve.
iv. For samples, determine counts and SIS. Go to quench curve to determine
efficiency and from that the dpm of sample. This is what you need to know
how to do.
3. Methods based on External Standard
a. Count -emitter stored (shielded) within counter 137Cs, 226Ra or 133Ba
b. In Beckman instruments, count in 2 windows, get ESCR (External Standard
Channels Ratio)
c. In Packard instrument, measure spectrum, get tSIE (transformed Spectral Index of
the External Standard
d. Functionally used just like SCR of SIS
4. Compare: SIS better for high sample counts, tSIE for low counts.
5. Different efficiencies of counting for common isotopes. Based on tSIE; SIS would be
similar. Note 14C and 35S have very similar energy profiles and therefore will have
very similar quench curves. Practically, you can use a 14C quench curve for a 35S
Exp. 4 Isotopes
3723 F02
6
sample.
6. Sample calculation:
You have a sample of 3H-leucine. You spot 0.100 mL onto a cellulose nitrate filter,
dissolve the filter in an appropriate scintillation fluid and count, getting 3578 cpm and
a tSIE of 375. Using the quench correction curve above, calculate the dpm/mL of that
sample
Let's assume that a tSIE of 375 corresponds to an efficiency of 0.450 (45.0%) for 3H.
To calculate dpm simply calculate:
3578 cpm/ x dpm = 0.450
dpm = 7951 = 7950 (3 sig figs)
This is the dpm in 0.100 mL, so dpm/ml = 7950 dpm/0.100 ml = 79,500 dpm/mL
V. Counting Statistics: BRIEFLY
A. For a single counting of randomly occurring event:
  total counts; generalize as # counts increase,  as % of counts decreases
Total counts
% total

4
50

400
5

40,000
0.5

B. Normal distribution about mean of single measurement
1. If repeat measurement, 68% ± 1; 95.5% ± 2, 99.7% ± 3
2. The longer you count, the narrower the distribution curve
3. Look at “95% (2) confidence limit”
a. from Exp. 1: % relative error (%) = ± (/average value) x 100%
b. % 2 =
2 total counts
2
x 100%
x 100% =
total counts
total counts
Exp. 4 Isotopes
3723 F02
7
=
200
total counts
c. Example: 1500 cpm,
200
200
=
=  5.17 %
1500
38.7
means there is a 95% probability that 1500 is within ± 5.2% of the true mean
dpm for this sample.
200
200
ii. count for 5 min: % 2  =
=
=  2.3%
7500 86.6
There is a 95% probability that 1500 cpm is within ± 2.3% of the true mean
dpm for this sample
iii. By counting longer, get greater accuracy (but only as √ of time).
4. Rule of thumb: Goal is % 2confidence limit) = ± 2%;
a. Note--this is 10,000 counts
b. Usually count to % 2 = ± 2% or to a set time (10 min. practical limit)
C. Background
1. Must always determine background--due to electronic noise of phototubes,
atmospheric radiation, etc.
2. Determine if background SIGNIFICANT compared to sample counts.
3. If significant, background should be subtracted before quench correction, because
background is resistant to quench.
D. Dual labeling: Counters set up so they can distinguish between two different labels used
in same experiment. If this is an issue for your research, you must learn how to handle,
calculate dpm of different isotopes.
i. count 1 min: % 2  =
VI. With regard to the laboratory exercise.
A. Scintillation counting experiment results returned
1. Standards. Note CPMA, SIS, for each standard
2. Sample-- unknown is the sample with your group number
3. Plot efficiency vs SIS using standards. You calculate current dpm of standards from
information given. This will involve correcting from date prepared to today's
date!(First order rate decay equation)
4. Determine efficiency of counting and dpm of your unknown.
B. Autoradiograms--Next week
1. Judy will develop
2. Take chromatogram to lab and spray with ninhydrin
a. Wrap in plastic wrap
b. Make exact tracing of ninhydrin results and autoradiogram results on same
tracing
3. Only tracing leaves laboratory. Chromatogram is radioactive waste.
Exp. 4 Isotopes
3723 F02
8
VII. Sample Problems
A. How much radioactivity in a sample.
1. Calculate the specific activity today (Feb. 20) of a 32P-ATP sample shipped on
February 1st with a specific activity of 550. Ci/mmole and a concentration of 0.150
µM.
Solution: First of all, you don't need the concentration, just the specific activity. You
also need the half-life of 32P and the time elapsed, and the first order decay equation.
ln Nt/No = –0.693 (t/t1/2)
Substituting: ln [x / (550 Ci/mmole)]= –0.693 (19.0 days/14.3 days)
ln x – ln 550 = - 0.921
ln x –6.321 = - 0.921
ln x
= 6.321 – 0.921 =5.400
x
= 221 Ci/mmole
Think about it—somewhat more than one half life has passed, therefore the
specific activity should be somewhat less than half what we started with. 221
Ci/mmole is somewhat less than 550 Ci/mmole, so the result makes sense.
2. Calculate (based on the radioactivity on Feb. 1) the volume of 32P-ATP solution
needed to get 3.5 x 106 dpm.
Solution: For this problem I need SA, concentration, and the conversion from Ci to
dpm.
550 Ci/mmole x 2.22x1012 dpm/Ci = 1.22 x 1015 dpm/mmole
(X mmole) (1.22 x 1015 dpm/mmole) = 3.5 x 106 dpm
X
= 2.87 x 10-9 mmole
(0.150 µmol/l) x (mmole/103µmol) x l /103ml x X ml = 2.87 x 10-9 mmole
X ml = 0.0191 ml = 19.1 µl
3. What volume of this sample would you need on Feb. 20 to get the same 3.5 x 106
dpm?
Solution: Since we know the ratio of radioactivity today is 221/550 what it was on
Feb. 1, we need 550/221 as much volume to get the same radioactivity.
550/221 x 19.1 µl = 47.5 µl
B. Scintillation counting
1. Using 3H standards containing 150,000 dpm each, you make a series of quenched
standards and count them for one minute each, with the following results:
CPM
SIS
75,000 19.5
60,000 16.1
Exp. 4 Isotopes
3723 F02
9
45,000 14.3
30,000 12.9
15,000 12.0
Determine the counting efficiency of each sample and include in the table above.
Draw the standard curve.
Solution: For counting efficiency of Standards, divide cpm/dpm. Thus:
75,000/150,000 = 0.500 (or 50.0%); 60,000/150,000= 0.400; 45,000/150,000=0.300;
30,000/150,000=0.200; 15,000/150,000=0.100
Your curve should look something like the one above (but since I did this on the
computer, and not in a graphing program, it is only an approximation).
2. Determine the dpm of a sample with 4450 cpm and an SIS of 15.0
Solution: To calculate the dpm in an unknown sample, first determine the %
Efficiency graphically by finding the SIS value on the x axis and determining the
corresponding % efficiency (y value) from the curve. For SIS = 15.0, Eff = 34.7%.
Then calculate:
4450 = (34.7 %) (X); X = 12,800 dpm
3. How long would you have to count the sample above to get a 2 confidence limit of
±1.0%?
Solution: The formula is %2 
Substituting:
200
4450 cpm x min 
200
200
= 1%

total counts

 1% ; square both sides and cross multiply to get
40,000 = (4450 cpm) (x min); x = 8.99 minutes.