retrograde extrapolation and other ethanol … · most are post-absorptive gullberg, r.g., and...

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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


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


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.


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



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



d. SUBTRACT Conc. Eliminated from Co:

= 0.00 % Co = 0.089%

Conc. Eliminated = 0.090%

0.089% - 0.090% BAC when tested =




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?


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



C. Calculate Conc. Eliminated:

= 0.075 %



x



d. SUBTRACT Conc. Eliminated from Co:

= 0.014 % Co = 0.089%

Conc. Eliminated = 0.075%

0.089% - 0.075% BAC when tested =




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?


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 %



x


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:





= 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:



= 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


Beverage:


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)


40 80 120 160 200 240 280 320

0.08

C

0.06 stop

= 1.0 drink = 0.018 g/dL


mL drink


Beverage:


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


mL drink


Beverage:


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)


40 80 120 160 200 240 280 320

008

C 0.06

stop

= 1.0 drink = 0.026 g/dL

012

C

0.10

retrograde extrapolation and other ethanol … · most are post-absorptive gullberg, r.g., and...

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