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Exp. 4 Isotopes

3723 F02

Biochemistry 3723Lecture 4b: RadioisotopesAugust 28, 2002

V. Scintillation Counting of - emissions: 3H, 14C, 35S, and 32P

A. Process of scintillation: use3H 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 aromaticcompounds-- toluene, xylene, alkylbenzenes, alkylnapthalenes

c.Fluor--various multicyclic compounds (aromatic) -delocalized electrons

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

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)


4.Coincidence circuit

a.Light flash must be seen by both PMTs at same time to be counted.

iTheoretical 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

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)

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) orSpectral 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

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 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 /  / % 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 = =

c.Example: 1500 cpm,

i.count 1 min:

means there is a 95% probability that 1500 is within ± 5.2% of the true mean dpm for this sample.

ii. count for 5 min:

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.

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.

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
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 = 1%

Substituting:; square both sides and cross multiply to get

40,000 = (4450 cpm) (x min); x = 8.99 minutes.