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The Hilton
Baltimore, Maryland
November 13, 2013
ACMT American College of Medical Toxicology
Seminars in Forensic Toxicology Consultation in the Civil & Criminal Arenas
RETROGRADE EXTRAPOLATION And Other Ethanol Calculations
Robert. B. Forney, Jr. Ph.D., DABFT.
Exigent circumstances Court found that blood alcohol evidence is fleeting as the drug is being eliminated from the body.
(Schmerber v. California, 348U.S. 757, 86 S. Ct. 1826, 16 L.Ed.2d 908 (1966))
Widmark Equation
One compartment model rho = modern Volume of Distribution
The amount (dose) of alcohol in the body is proportional to a BAC measured by a proportionality constant, “r”
A = c p r and c = • • A
p • r
where A = Alcohol in the body in grams
(rho) r = Reduction Factor (“reduced body mass”)
c p
= BAC in grams/kg = Body weight in kgs
Widmark’s rho • Erik Widmark, Sweden
• Earliest use of the distribution of a drug for forensic purposes
• Widmark’s rho = Whole body alcohol conc.. Blood alcohol conc.
• Now use volume of distribution
Calculate the AMOUNT CONSUMED
Volume Ingested
3 ozs.
Conversion to mLs
X 29.6 mL oz.
Conc. of beverage
X 40 mL EtOH 100 mL
Specific Gravity
X 0.789 g . mL
= Dose
= 28 g
• Assume a 150 lb. male ingests 3 ozs. of 80 Proof whiskey
• How much alcohol has he consumed, expressed in grams?
Widmark’s rho or
Volume of Distribution
k a Absorption rate constant
k el Elimination rate constant
Who can hold more liquor? Men or Women
• MEN Vd Ethanol = 0.68 L/kg
• WOMEN Vd Ethanol = 0.55 L/kg
Combining differences in weight and distribution,
a dose in a woman may be 1/2 the dose required in a man
for the same BAC
EXAMPLE: (180 x 2.2 lb/kg) x 0.68 L/kg = 56 L
EXAMPLE: (120 x 2.2 lb/kg) x 0.55 L/kg = 30 L
Elimination Rate • Zero-order (constant rate) process > 0.02
• Normal range (healthy adults) – Men 14.94 (+ 4.5 mg/dl/hr)
– Women 18.30 (+ 3.23 mg/dl/hr) KM Dubowski 1976 Alcohol Tech. Rep. 5:55-63
• Alcoholics = 23 mg/dl/hr (+ 5.8 mg/dl/hr) (13 - 36)
• Eskimos, Amer. Indians & Asians <<slower
AW Jones & B Sternebring 1992 Alcohol & Alcoholism 27(6):641-47
5 hours
Time (hours)
“Zero Order”
5 * 0.018% = 0.09%
means a STRAIGHT line on a Linear Plot
B A
C
0.10
0.12
0.14
0.08
0.06
0.04
0.02
B A
C
Time (hours)
“First Order” means a CURVED line on a Linear Plot
0.10
0.12
0.14
0.08
0.06
0.04
0.02
1. Straight line with "arithmentic" plot
2. Rate of elimination (or amount eliminated per unit of time) is CONSTANT and INDEPENDENT of the concentration.
3. Halflife (t1/2) and Elimination rate constant DO NOT APPLY!
4. Ethanol elimination is zero order between 0.30 and 0.02%
Characteristics of "Zero-Order" Elimination
► extrapolated from the elimination phase to the ordinate Distribution is determined from a point C 0
► represents conc. after complete absorption + no elimination
C 0
NOTE on the diagram, the relative plateau between the early rapidly rising BAC and the steadily declining BAC of the post-absorptive state.
Two blood samples
First BAC < Second BAC
12/199 6% Jones (1993)
0/432 0% Lund (1979)
47/2354 2% Neuteboom and Jones (1990)
Absorp/on Status of Drinking Drivers Most are post-absorptive
Gullberg, R.G., and McElroy, A.J., “Comparing Roadside with Subsequent Breath Alcohol Analyses and Their Relevance to the Issue of Retrograde Extrapolation.” Forensic Science International, 57:193-201,1992
PBT 1
PBT 2
2 Evidential Tests
161 Arrested Drivers
Time between PBTs was 25 to 120 minutes (mean 64 minutes)
BrACs ranged from 0.06 to 0.310 g/210 L
Absorp/on Status of Drinking Drivers
No evidence that any of the arrested drivers were in the ascending portion (in excess of breath sampling variation)
Using 0.015 g/210L/h there was a small but significant overestimation of PBT1 (mean 0.005 g/2lO L)
CONCLUSIONS
Absorp/on Status of Drinking Drivers
Gullberg, R.G., and McElroy, A.J., “Comparing Roadside with Subsequent Breath Alcohol Analyses and Their Relevance to the Issue of Retrograde Extrapolation.” Forensic Science International, 57:193-201,1992
Absorp/on Status of Drinking Drivers
Reference Year % Not Increasing Biasotti 1985 98% Jones 1987 97% Neuteboom 1990 98% Jones 1993 94% Lund 1993 100% Levine 2000 91% Jones 2002 88%
Six male four female subjects Consumed liquor ad libitum for three hours (range 160 to 181 minutes)
Laboratory Drinking Study
Increase in BrAC from end of drinking to highest BrAC was 0.005 g/210L (range 0.0 to 0.02 g/210L)
Ganert, P.M. and Bowthorpe, W.D., “Evaluation of Breath Alcohol Profiles Following a Period of Social Drinking.” Canadian Society of Forensic Science Journal, 33:137-43, 2000
Peak BrACs ranged from 0.085 to 0.190 g/210L
BrACs were measured with an Intoxilyzer 5000 every 5-15 mins.
Time to maximum BrAC was 12 minutes (range 4 to 22 mins.)
The peak BrACs were all obtained in first test conducted 15 minutes after the end of drinking
Laboratory Drinking Study
Cowan, J.M., Dennis, M.E.,III, and Smith, L.F., “A Comparison of Equal Alcohol Doses of Beer and Whiskey on Eleven Human Test Subjects. Canadian Society of Forensic Science Journal, 37:137-45, 2004 ,
Figure 6 Plot of alcohol concentration against time for the mean result for all subjects
Seven male and four female subjects consumed beer or whiskey (mixed) over two hours and forty-five minutes Mean peak BrAC was 0.113 g/210L (ranged 0.100 to 0.129 g/210L)
In some cases there is no clear or distinct peak. Instead there is a plateau, in which the rate of absorption is equal to the rate of elimination and the BAC is relatively constant.
Watts and Simonick (1989) fifteen drinking subjects with food, six subjects had a plateau (ranged 47 to 89 minutes)
BrA
C
Time (mins)
Plateau
BAC Plateau
Duration of Plateau
46 to 89 minutes Watts and Simonick (1989)
16 to 106 minutes Hodgson and Taylor (1992)
0 to 124 minutes Ganert and Bowthorpe (2000) Ganert, P.M. and Bowthorpe, W.D., “Evaluation of Breath Alcohol Profiles Following a Period of Social Drinking.” Canadian Society of Forensic Science Journal, 33:137-43, 2000
Loomis (1974) "If the incident-test interval is in excess of 2 hr, then the BAC could not have been lower at the time of the incident than it was at the time of the test, but it could have been higher by an amount equal to a range of rates of disappearance of alcohol from the blood of from 0.01 to 0.02% per hr...”
Plateau/Presumption
Lund (1979) deduct two hours then add 0.01 g/lOOmL/h and "possibility of obtaining a too high BAC value by backward calculation is virtually eliminated”
Loomis, T.A., “Blood Alcohol in Automobile Drivers: Measurement and Interpretation for Medicolegal Purposes. I. Effect of Time Interval Between Incident and Sample Acquisition.” Clinical Quarterly of Journal Studies on Alcohol, 35:458-72, 1974
Lund, A., “The Rate of Disappearance of Blood Alcohol in Drunken Drivers.” Blutalkahol, 16:395-98, 1979
Accumula/on when absorpHon > eliminaHon
Forney, R.B., and Hughes, F.W., “Alcohol Accumulation in Humans after Prolonged Drinking.” Clinical Pharmacology and Therapeutics, 5: 619-21, 1963
22 female and 35 male subjects who consumed either 1 or 2 fl. oz. of 50% v/v alcohol per hour per 70 kg (150 lb) of body weight. Drinking began after a breakfast of toast and fruit juice. Lunch consisted of meat sandwiches.
• extrapolating BACKWARD from a BAC or BrAC
Forward extrapolation
• often used with refusals
Retrograde extrapolation
• extrapolating FORWARD from a drinking history
Estimation of a BAC or BrAC:
• at the time of a crash or observed driving
• often used when BAC was determined on a sample collected later than statutory limits
• at the time of a crash or observed driving
* * *
Forward extrapolation
Widmark formula:
where: A = dose of ethanol in grams r = Widmark’s rho, a constant
A = r ρ (C + (ß t) t
C = BAC at time “t” in hours. t ß = Widmark’s beta, a constant
a zero order elim. rate constant = 0.015%/hr.
Ratio of amt. of alcohol in the body to that in the blood. Now called Vd, the volume of distribution
ρ = body weight in kilograms
= 0.68 L/kg in men; 0.55 L/kg in women
Retrograde Extrapolation
• Extrapolating BAC at the time of driving from an analysis performed some time later.
• Cannot simply be done by adding back expected elimination! Absorption may still be occurring.
• Exigent circumstances Schmerber v California, 1966
Necessary assumptions: Quantity of unabsorbed alcohol remaining in the GI tract at the time of driving (time of the stop, time of the accident, etc.) Quantity of alcohol consumed after the time of driving
OVER ESTIMATION of a prior BAC will result if either quantity is under estimated.
Appropriate values for:
• Volume of distribution • Rate of absorption
• Rate of elimination
Retrograde extrapolation of a BAC CANNOT simply be done by adding back for expected elimination using an average rate.
Continuing significant absorption means that the BAC will decrease less than expected per unit time until the post-absorptive state is reached.
Use CONSERVATIVE assumptions
• time of drinking
• time of driving
allowing for individual variations in pharmacokinetic parameters
• alcohol consumed AFTER the time of driving
• UNABSORBED alcohol remaining in the GI tract at the time of driving
SUBTRACT from the calculation:
Identify the sufficiency of evidence for
USE retrograde extrapolation to:
• evaluate/impeach as appropriate, testimony regarding drinking history (eg. "I only had two beers.")
• estimate whether BAC was above, below or close to a given threshold for the assertion of driving "under the influence", "impaired driving", a per se violation or other statute.
Bre
ath
Alc
ohol
Con
c.
Time (Hrs.)
X
O
Measured BrAC
Observed driving or crash
Bre
ath
Alc
ohol
Con
c.
Time (Hrs.)
X
O
Observed driving or crash
X
2 Measured BrACs
Drinking Pa<erns of Drinking Drivers
Cooper, P.J., and Rothe, J.P., “Drinking Establishment, Driving Risk and Ethno-‐Pharmacology” Proceedings of 35th Interna3onal Congress on Alcohol and Drug Dependence (Vol. 2), Oslo, Norway, 1988.
Extensive interviews with bartenders, servers, drinkers, accident involved drivers in BC
Drinking Percent Time Waited Location Time (hrs) Drink Beer minutes
Pub 2.0 73% 24 Hotel Bar 2.0 86% 29 Restaurant 2.2 36% 37
N = 435
Social Event 2.3 39% 42 Own Home 1.8 39% 36
Retrograde Extrapolation Problems
Subject weighs 180 lbs. and Drinks five, 12 oz. beers (3.5% v/v) Over a four hour period from 8:00 p.m. to 12:00 a.m. He is arrested at 1:00 a.m. and Tested at 2:00 a.m.
Problem #1
1. What would his BAC have been when he was TESTED?
Retrograde Extrapolation Problems Subject weighs 180 lbs. and Drinks five, 12 oz. beers (3.5% v/v) Over a four hour period from 8:00 p.m. to 12:00 a.m. He is arrested at 1:00 a.m. and Tested at 2:00 a.m.
1. What would his BAC have been when he was TESTED?
a. Calculate the dose ingested:
12 oz. 29.6 ml 3.5 ml 0.789 g beer oz. 100 ml ml
5 beers x x x x = 49 g
b. Calculate Co: = 81.8 kg 1.0 kg 2.2 lbs.
180 lbs. x
Dose (g) Vd x Weight x 10 L kg dL kg L
= 49 g 0.68 x 81.8 x 10
L kg dL kg L
= 0.089 g dL
Retrograde Extrapolation Problems Subject weighs 180 lbs. and Drinks five, 12 oz. beers (3.5% v/v) Over a four hour period from 8:00 p.m. to 12:00 a.m. He is arrested at 1:00 a.m. and Tested at 2:00 a.m.
1. What would his BAC have been when he was TESTED?
C. Calculate Conc. Eliminated:
= 0.090 %
Time for elimination = 8:00 pm to 2:00 am = 6 hours
Conc. Eliminated = 6 hrs. 0.015 % hr
x
Retrograde Extrapolation Problems Subject weighs 180 lbs. and Drinks five, 12 oz. beers (3.5% v/v) Over a four hour period from 8:00 p.m. to 12:00 a.m. He is arrested at 1:00 a.m. and Tested at 2:00 a.m.
1. What would his BAC have been when he was TESTED?
d. SUBTRACT Conc. Eliminated from Co:
= 0.00 % Co = 0.089%
Conc. Eliminated = 0.090%
0.089% - 0.090% BAC when tested =
Retrograde Extrapolation Problems Subject weighs 180 lbs. and Drinks five, 12 oz. beers (3.5% v/v) Over a four hour period from 8:00 p.m. to 12:00 a.m. He is arrested at 1:00 a.m. and Tested at 2:00 a.m.
1. What would his BAC have been when he was TESTED?
Retrograde Extrapolation Problems
Subject weighs 180 lbs. and Drinks five, 12 oz. beers (3.5% v/v) Over a four hour period from 8:00 p.m. to 12:00 a.m. He is
arrested at 1:00 a.m. and Tested at 2:00 a.m.
Problem #2
2. What would his BAC have been when ARRESTED?
Retrograde Extrapolation Problems Subject weighs 180 lbs. and Drinks five, 12 oz. beers (3.5% v/v) Over a four hour period from 8:00 p.m. to 12:00 a.m. He is arrested at 1:00 a.m. and Tested at 2:00 a.m.
2. What would his BAC have been when he was ARRESTED?
a. Calculate the dose ingested:
12 oz. 29.6 ml 3.5 ml 0.789 g beer oz. 100 ml ml
5 beers x x x x = 49 g
b. Calculate Co: 1.0 kg 2.2 lbs. = 81.8 kg 180 lbs. x
Dose (g) Vd x Weight x 10 L kg dL kg L
= 49 g 0.68 x 81.8 x 10
L kg dL kg L
= 0.089 g dL
Retrograde Extrapolation Problems Subject weighs 180 lbs. and Drinks five, 12 oz. beers (3.5% v/v) Over a four hour period from 8:00 p.m. to 12:00 a.m. He is arrested at 1:00 a.m. and Tested at 2:00 a.m.
2. What would his BAC have been when he was ARRESTED?
C. Calculate Conc. Eliminated:
= 0.075 %
Time for elimination = 8:00 pm to 1:00 am = 5 hours
Conc. Eliminated = 5 hrs. 0.015 % hr
x
Retrograde Extrapolation Problems Subject weighs 180 lbs. and Drinks five, 12 oz. beers (3.5% v/v) Over a four hour period from 8:00 p.m. to 12:00 a.m. He is arrested at 1:00 a.m. and Tested at 2:00 a.m.
2. What would his BAC have been when he was ARRESTED?
d. SUBTRACT Conc. Eliminated from Co:
= 0.014 % Co = 0.089%
Conc. Eliminated = 0.075%
0.089% - 0.075% BAC when tested =
Retrograde Extrapolation Problems Subject weighs 180 lbs. and Drinks five, 12 oz. beers (3.5% v/v) Over a four hour period from 8:00 p.m. to 12:00 a.m. He is arrested at 1:00 a.m. and Tested at 2:00 a.m.
2. What would his BAC have been when he was ARRESTED?
Retrograde Extrapolation Problems
Subject weighs 180 lbs. and Drinks five, 12 oz. beers (3.5% v/v) Over a four hour period from 8:00 p.m. to 12:00 a.m. He is arrested at 1:00 a.m. and Tested at 2:00 a.m.
Problem #3
3. HOW MANY BEERS (3.5% v/v) would he have to drink between 8:00 pm and 12:00 am in order to test
0.10% w/v when arrested?
Retrograde Extrapolation Problems
a. Calculate Dose in grams: Dose (g)
Vd x Weight x 10 L kg dL
kg L
= 54.8 g
BAC =
And, Dose = BAC x Vd x Weight x 10 L kg dL kg L
g dL
0.10% x 0.68 x 81.8 x 10 L kg dL
kg L g dL
b. Calculate the Conc. Eliminated:
= 0.090 %
Time for elimination = 8:00 pm to 2:00 am = 6 hours
Conc. Eliminated = 6 hrs. 0.015 % hr
x
Retrograde Extrapolation Problems Subject weighs 180 lbs. and Drinks five, 12 oz. beers (3.5% v/v) Over a four hour period from 8:00 p.m. to 12:00 a.m. He is arrested at 1:00 a.m. and Tested at 2:00 a.m.
3. HOW MANY BEERS (3.5% v/v) would he have to drink between 8:00 pm and 12:00 am in order to test 0.10% w/v when arrested?
= 49.3 g
Dose = BAC x Vd x Weight x 10 L kg dL
kg L g dL
0.09% x 0.68 x 81.8 x 10 L kg dL kg L
g dL
c. Calculate the dose to replace what has been eliminated:
Retrograde Extrapolation Problems Subject weighs 180 lbs. and Drinks five, 12 oz. beers (3.5% v/v) Over a four hour period from 8:00 p.m. to 12:00 a.m. He is arrested at 1:00 a.m. and Tested at 2:00 a.m.
3. HOW MANY BEERS (3.5% v/v) would he have to drink between 8:00 pm and 12:00 am in order to test 0.10% w/v when arrested?
Retrograde Extrapolation Problems Subject weighs 180 lbs. and Drinks five, 12 oz. beers (3.5% v/v) Over a four hour period from 8:00 p.m. to 12:00 a.m. He is arrested at 1:00 a.m. and Tested at 2:00 a.m.
3. HOW MANY BEERS (3.5% v/v) would he have to drink between 8:00 pm and 12:00 am in order to test 0.10% w/v when arrested?
= 104.1 g Dose = Dose for 0.10% w/v + Dose to replace elimination
54.8 g + 49.3 g
d. Add the dose to produce a 0.10% with the dose to replace what has been eliminated:
Retrograde Extrapolation Problems Subject weighs 180 lbs. and Drinks five, 12 oz. beers (3.5% v/v) Over a four hour period from 8:00 p.m. to 12:00 a.m. He is arrested at 1:00 a.m. and Tested at 2:00 a.m.
3. HOW MANY BEERS (3.5% v/v) would he have to drink between 8:00 pm and 12:00 am in order to test 0.10% w/v when arrested?
= 127.35 ozs.
e. Calculate the volume of beer required to deliver a dose of 104.1 g:
Vol. Ingested = conversion x beer conc. x spec. grav.
Dose (g)
29.6 ml 3.5 ml 0.789 g oz. 100 ml ml
x x 104.1 g
= 10.6, 12 oz Beers
Subject Drinking Examples
Time Betw Drinks Subject
1 2 3 4 5 6 7 8 9
10 11 12 13 14 15
34-35 min. 33-49 “ 25-48 “ 35-54 “ 28-32 “ 23-32 “ 32-33 “ 16-38 “ 21-30 “ 25-40 “ 35-45 “ 25-46 “ 10-28 “ 26-56 “ 16-38 “
Means: Time = 32.5 min. Drinks = 5.7
No. of Drinks
3 5 6 4 5 7 3 9 9 6 5 5 7 5 7
Peak Time Interval* Subject
1 2 3 4 5 6 7 8 9
10 11 12 13 14 15
12 min. 44 “ 15 “ 26 “ 21 “ 35 “ 14 “ 16 “ 61 “
55 “ 22 “ 23 “ 26 “ 12 “
Mean time = 27 min.
*Time FROM when drinking stopped TO peak
1.0 hr of Zero order elim: 0.015 g x 10 dL x 55.6 L = 8.34 g L dL
mL drink
Subject #1 Sex: Weight:
Beverage:
Food: Duration of Drinking:
Calculated Vd:
Male 180 lbs x 1.0 kg = 81.8 kg
2.2 lbs 0.68 L x 81.8 kg = 55.6 L
kg
Chinese dinner, 1.0 hr. at start
5.7 x 360 mL of 4% v/v Beer
204 mins.
= 360 mL x 4 mL x 0.789 g = 11.4 g 100 mL
One drink
= x = 0.020 g/dL 55.6 L 11.4 g 1.0 L
10 dL
0.04
C
0.02
Time (mins)
Subject #1, male 1.0 hr of zero order elim. = 0.015 g/dL
40 80 120 160 200 240 280 320
0.08
C
0.06 stop
= 1.0 drink = 0.020 g/dL
1.0 hr of Zero order elim: 0.015 g x 10 dL x 51 L = 7.65 g L dL
mL drink
Subject #2 Sex: Weight:
Beverage:
Food: Duration of Drinking:
Calculated Vd:
Male 165 lbs x 1.0 kg = 75 kg
2.2 lbs 0.68 L x 75 kg = 51 L
kg
NIDA dinner, 1.0 hr. at start
5 x 30 mL of 80 Proof Bourbon
193 mins.
= 30 mL x 40 mL x 0.789 g = 9.4 g 100 mL
One drink
= x = 0.018 g/dL 51 L 9.4 g 1.0 L
10 dL
0.04
C
0.02
Time (mins)
Subject #2, male 1.0 hr of zero order elim. = 0.015 g/dL
40 80 120 160 200 240 280 320
0.08
C
0.06 stop
= 1.0 drink = 0.018 g/dL
1.0 hr of Zero order elim: 0.018 g x 10 dL x 37.5 L = 6.75 g L dL
mL drink
Subject #3 Sex: Weight:
Beverage:
Food: Duration of Drinking:
Calculated Vd:
Female 150 lbs x 1.0 kg = 68.2 kg
2.2 lbs 0.55 L x 68.2 kg = 37.5 L
kg
NIDA dinner, 0.5 hr. at start
3 x 50 mL of 80 Proof Bourbon
85 mins.
= 50 mL x 40 mL x 0.789 g = 15.8 g 100 mL
One drink
= x = 0.042 g/dL 37.5 L 15.7 g 1.0 L
10 dL
0.04
C
0.02
Time (mins)
Subject #3, female 1.0 hr of zero order elim. = 0.018 g/dL = 1.0 drink = 0.042 g/dL
40 80 120 160 200 240 280 320
0.08
C
0.06 stop
1.0 hr of Zero order elim: 0.015 g x 10 dL x 72.6 L = 10.9 g L dL
mL drink
Subject #4 Sex: Weight:
Beverage:
Food: Duration of Drinking:
Calculated Vd:
Male 235 lbs x 1.0 kg = 106.8 kg
2.2 lbs 0.68 L x 106.8 kg = 72.6 L
kg
Large meal, 1.0 hr. at start
6.66 x 60 mL of 80 Proof Gin
191 mins.
= 60 mL x 40 mL x 0.789 g = 18.9 g 100 mL
One drink
= x = 0.026 g/dL 72.6 L 18.9 g 1.0 L
10 dL
0.04
C 0.02
Time (mins)
Subject #4, male 1.0 hr of zero order elim. = 0.015 g/dL
40 80 120 160 200 240 280 320
008
C 0.06
stop
= 1.0 drink = 0.026 g/dL
012
C
0.10
Recommended