RETROGRADE EXTRAPOLATION And Other Ethanol Calculations

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.
The Hilton
Baltimore, Maryland
November 13, 2013
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
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
c = BAC in grams/kg
p = Body weight in kgs
(rho) r = Reduction Factor (“reduced body mass”)
rho
=
modern Volume of Distribution
One compartment model
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
• Assume a 150 lb. male ingests 3 ozs. of 80 Proof whiskey
• How much alcohol has he consumed, expressed in grams?
Volume
Ingested
Conversion
to mLs
Conc. of
beverage
Specific
Gravity
=
3 ozs. X 29.6 mL X 40 mL EtOH X 0.789 g . =
oz.
100 mL
mL
Dose
28 g
ka
Absorption rate constant
Widmark’s rho
or
Volume of Distribution
k el
Elimination rate constant
Who can hold more liquor?
Men or Women
• MEN
Vd Ethanol = 0.68 L/kg EXAMPLE: (180 x 2.2 lb/kg) x 0.68 L/kg =
• WOMEN
56 L
Vd Ethanol = 0.55 L/kg
EXAMPLE: (120 x 2.2 lb/kg) x 0.55 L/kg =
30 L
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
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)
AW Jones & B Sternebring 1992 Alcohol & Alcoholism 27(6):641-47
•  Eskimos, Amer. Indians & Asians <<slower
“Zero Order” means a STRAIGHT line on a Linear Plot
0.14
B
A
0.12
0.10
0.08
5 * 0.018% =
0.09%
0.06
C
0.04
0.02
5 hours
Time (hours)
“First Order” means a CURVED line on a Linear Plot
0.14
B
A
0.12
0.10
0.08
0.06
C
0.04
0.02
Time (hours)
Characteristics of "Zero-Order" Elimination
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%
Distribution is determined from a point C 0
► extrapolated from the elimination phase to the ordinate
► represents conc. after complete absorption + no elimination
C0
NOTE on the diagram, the relative plateau between the early rapidly
rising BAC and the steadily declining BAC of the post-absorptive state.
Absorp/on Status of Drinking Drivers Most are post-absorptive
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 PBT 1
2 Evidential Tests
PBT 2
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
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 CONCLUSIONS
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)
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
Biasotti
Jones
Neuteboom
Jones
Lund
Levine
Jones
Year
% Not Increasing
1985
1987
1990
1993
1993
2000
2002
98%
97%
98%
94%
100%
91%
88%
Laboratory Drinking Study
Six male four female subjects
Consumed liquor ad libitum for three hours
(range 160 to 181 minutes)
BrACs were measured with an Intoxilyzer 5000 every 5-15 mins.
Peak BrACs ranged from 0.085 to 0.190 g/210L
Time to maximum BrAC was 12 minutes (range 4 to 22 mins.)
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
Laboratory Drinking Study
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)
The peak BrACs were all obtained in first test conducted 15 minutes
after the end of drinking
Figure 6 Plot of alcohol concentration against time for the mean result for all subjects
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 ,
BAC Plateau
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)
BrAC
Plateau
Time (mins)
Duration of Plateau
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
16 to 106 minutes
46 to 89 minutes
Hodgson and Taylor (1992)
Watts and Simonick (1989)
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”
Lund, A., “The Rate of Disappearance of Blood Alcohol in Drunken Drivers.” Blutalkahol,
16:395-98, 1979
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...”
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
Accumula/on when absorpHon > eliminaHon
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.
Forney, R.B., and Hughes, F.W., “Alcohol Accumulation in Humans after Prolonged
Drinking.” Clinical Pharmacology and Therapeutics, 5: 619-21, 1963
Estimation of a BAC or BrAC:
Forward extrapolation
• at the time of a crash or observed driving
• extrapolating FORWARD from a drinking history
• often used with refusals
Retrograde extrapolation
• at the time of a crash or observed driving
• extrapolating BACKWARD from a BAC or BrAC
• often used when BAC was determined on a sample
collected later than statutory limits
Forward extrapolation
Widmark formula:
A = r * ρ * (C t + (ß * t)
where: A = dose of ethanol in grams
r = Widmark’s rho, a constant
= 0.68 L/kg in men; 0.55 L/kg in women
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
Ct = BAC at time “t” in hours.
ß = Widmark’s beta, a constant
a zero order elim. rate constant
= 0.015%/hr.
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:
• Rate of absorption
• Volume of distribution
• 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 allowing for individual variations in pharmacokinetic
parameters
Identify the sufficiency of evidence for
• time of drinking
• time of driving
SUBTRACT from the calculation:
• alcohol consumed AFTER the time of driving
• UNABSORBED alcohol remaining in the GI tract at the time of driving
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.
Breath Alcohol Conc.
Observed driving or crash
O
Measured BrAC
X
Time (Hrs.)
Breath Alcohol Conc.
Observed driving or crash
O
2 Measured BrACs
X
X
Time (Hrs.)
Drinking Pa<erns of Drinking Drivers Extensive interviews with bartenders, servers, drinkers, accident involved drivers in BC Location
Drinking
Time (hrs)
Percent
Drink Beer
Time Waited
minutes
Pub
2.0
73%
24
Hotel Bar
2.0
86%
29
Restaurant
2.2
36%
37
Social Event
2.3
39%
42
Own Home
1.8
39%
36
N = 435
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. Retrograde Extrapolation Problems
Problem #1
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. 1. What would his BAC have been when he was TESTED?
a. Calculate the dose ingested:
5 beers
x
12 oz.
29.6 ml
3.5 ml
0.789 g
x
x
x
=
beer
oz.
100 ml
ml
49 g
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?
b. Calculate Co: 180 lbs. x
Vd
L
kg
Dose (g)
x Weight x 10
kg
dL
L
=
1.0 kg
2.2 lbs.
= 81.8 kg
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:
Time for elimination = 8:00 pm to 2:00 am = 6 hours
Conc. Eliminated = 6 hrs. x 0.015 %
hr
= 0.090 %
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:
Conc. Eliminated = 0.090%
Co = 0.089%
BAC when tested = 0.089% - 0.090%
= 0.00 %
Retrograde Extrapolation Problems
Problem #2
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 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:
5 beers
x
12 oz.
29.6 ml
3.5 ml
0.789 g
x
x
x
=
beer
oz.
100 ml
ml
49 g
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?
b. Calculate Co: 180 lbs. x
Vd
L
kg
Dose (g)
x Weight x 10
kg
dL
L
=
1.0 kg
2.2 lbs.
= 81.8 kg
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:
Time for elimination = 8:00 pm to 1:00 am = 5 hours
Conc. Eliminated = 5 hrs. x
0.015 %
hr
= 0.075 %
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:
Conc. Eliminated = 0.075%
Co = 0.089%
BAC when tested = 0.089% - 0.075%
= 0.014 %
Retrograde Extrapolation Problems
Problem #3
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
a. Calculate Dose in
grams:
BAC =
Vd
L
kg
Dose (g)
x Weight x 10
kg
dL
L
And, Dose = BAC x
g
dL
Vd
L
kg
x Weight
kg
0.10% x 0.68 x
g
L
dL
kg
81.8
kg
x
x
10
dL
L
10
dL
L
= 54.8 g
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?
b. Calculate the Conc.
Eliminated:
Time for elimination = 8:00 pm to 2:00 am = 6 hours
Conc. Eliminated = 6 hrs. x 0.015 %
hr
= 0.090 %
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?
c. Calculate the dose to replace what has been eliminated:
Dose = BAC x
g
dL
Vd
L
kg
x Weight
kg
0.09% x 0.68 x
g
L
dL
kg
x
10
dL
L
81.8 x 10
kg
dL
L
= 49.3 g
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?
d.  Add the dose to produce a 0.10%
with the dose to replace what has been eliminated:
Dose = Dose for 0.10% w/v + Dose to replace elimination
54.8 g
+
49.3 g
= 104.1 g
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?
e. Calculate the volume of beer required to deliver a dose of 104.1
Dose (g)
Vol. Ingested =
conversion x beer conc. x spec. grav.
104.1 g
29.6 ml x 3.5 ml x 0.789 g
oz.
100 ml
ml
= 127.35 ozs. = 10.6, 12 oz Beers
Subject Drinking Examples
Subject
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Time Betw Drinks
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.
No. of Drinks
3
5
6
4
5
7
3
9
9
6
5
5
7
5
7
Drinks = 5.7
Subject
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Peak Time Interval*
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
Subject #1
Sex:
Weight:
Calculated Vd:
1.0 hr of Zero order elim:
Beverage:
Male
180 lbs x 1.0 kg = 81.8 kg
2.2 lbs
0.68 L x 81.8 kg = 55.6 L
kg
0.015 g x 10 dL x 55.6 L = 8.34 g
dL
L
5.7 x 360 mL of 4% v/v Beer
One drink = 360 mL x 4 mL x 0.789 g = 11.4 g
100 mL
mL drink
11.4 g
1.0 L
=
x 10 dL = 0.020 g/dL
55.6 L
Food:
Duration of Drinking:
Chinese dinner, 1.0 hr. at start
204 mins.
1.0 hr of zero order elim. = 0.015 g/dL
= 1.0 drink = 0.020 g/dL
Subject #1, male
0.08
C
stop
0.06
0.04
C
0.02
40
80
120 160 200
Time (mins)
240 280 320
Subject #2
Sex:
Weight:
Calculated Vd:
1.0 hr of Zero order elim:
Beverage:
Male
165 lbs x 1.0 kg = 75 kg
2.2 lbs
0.68 L x 75 kg = 51 L
kg
0.015 g x 10 dL x 51 L = 7.65 g
dL
L
5 x 30 mL of 80 Proof Bourbon
One drink = 30 mL x 40 mL x 0.789 g = 9.4 g
100 mL
mL drink
9.4 g
1.0 L
=
x 10 dL = 0.018 g/dL
51 L
Food:
Duration of Drinking:
NIDA dinner, 1.0 hr. at start
193 mins.
1.0 hr of zero order elim. = 0.015 g/dL
= 1.0 drink = 0.018 g/dL
Subject #2, male
0.08
C
0.06
stop
0.04
C
0.02
40
80
120 160 200
Time (mins)
240 280 320
Subject #3
Sex:
Weight:
Calculated Vd:
1.0 hr of Zero order elim:
Beverage:
Female
150 lbs x 1.0 kg = 68.2 kg
2.2 lbs
0.55 L x 68.2 kg = 37.5 L
kg
0.018 g x 10 dL x 37.5 L = 6.75 g
dL
L
3 x 50 mL of 80 Proof Bourbon
One drink = 50 mL x 40 mL x 0.789 g = 15.8 g
100 mL
mL
drink
15.7 g 1.0 L
=
x 10 dL = 0.042 g/dL
37.5 L
Food:
Duration of Drinking:
NIDA dinner, 0.5 hr. at start
85 mins.
1.0 hr of zero order elim. = 0.018 g/dL
= 1.0 drink = 0.042 g/dL
Subject #3, female
0.08
C
0.06
stop
0.04
C
0.02
40
80
120 160 200
Time (mins)
240 280 320
Subject #4
Sex:
Weight:
Calculated Vd:
1.0 hr of Zero order elim:
Beverage:
Male
235 lbs x 1.0 kg = 106.8 kg
2.2 lbs
0.68 L x 106.8 kg = 72.6 L
kg
0.015 g x 10 dL x 72.6 L = 10.9 g
dL
L
6.66 x 60 mL of 80 Proof Gin
One drink = 60 mL x 40 mL x 0.789 g = 18.9 g
100 mL
mL drink
18.9 g 1.0 L
=
x 10 dL = 0.026 g/dL
72.6 L
Food:
Duration of Drinking:
Large meal, 1.0 hr. at start
191 mins.
1.0 hr of zero order elim. = 0.015 g/dL
= 1.0 drink = 0.026 g/dL
Subject #4, male
012
C 0.10
stop
008
C 0.06
0.04
C
0.02
40
80
120 160 200
Time (mins)
240 280 320