Alcohol calculations Flashcards

1
Q

Standard drink

A

Beer [5% v/v] 12 oz
Wine [12% v/v] 5 oz
Whiskey [80 proof] 1.5 oz

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2
Q

How to calculate dose

A

Beer [5% v/v]
12 oz * 30 mL/oz * 5/100 * 0.79 g/mL = 14.22 g EtOH
Wine [12% v/v]
5 oz * 30 mL/oz * 12/100 * 0.79 g/mL = 14.22 g EtOH
Whiskey [80 proof]
Need to convert proof to % EtOH by volume
80/2 = 40% EtOH by volume
1.5 oz * 30 mL/oz * 40/100 * 0.79 g/mL = 14.22 g EtOH

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3
Q

Extrapolated alcohol conc.

A

starting alcohol conc given a drinking scenario and assumed subject parameters

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4
Q

Back-extraopolaate

A

final known conc given a drinking scenario and assumed subject parameters

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5
Q

What is required for retrograde

A

to be in post absorptive phase

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6
Q

Zig-Zag or Steepling Effect

A

Zig-Zag or irregular blood/breath alcohol time course instead of a smooth post-absorption elimination
Alleged impact on the back or forward calculations
One-two minute sampling vs 20-30 minute sampling [insert]
Subject received the same dose of alcohol [IV] on two consecutive days
Blood was taken every 5 minutes for 4 hours and then every 10-15 minutes for three more hours
Note lack of any marked irregularity or zig-zag pattern in these plots with a 5 minute interval between successive samples
Linear decay profile changes into a “hockey-stick” shape at low BAC

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7
Q

Widmark Equation

A

Earliest use of calculations for distribution of alcohol
Widmark’s rho factor
Reduced body mass
Whole body EtOH conc. Blood etOh conc.
Modern term VOlume of distribution
A=CPrho A= etOH in grams

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8
Q

Alcohol Standard equation

A

Dose=VdwtCp Cp(g/L) dose (g) Vd (L/kg) * W (kg)

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9
Q

Male drinks 6 shorts of whiskey
Calculation of maximum BAC – assuming instantaneous absorption and no elimination
Short: 1.5 oz

Wt = 185 pounds

A

Whiskey is 90 proof
90/2 = 45% EtOH by volume
(1.5 oz * 6 shots) * 30 mL/oz * 45/100 * 0.79 g/mL = 95.98 g EtOH = D
Cp (g/L) = D (g) / (Vd (L/kg) * Wt (kg))
Cp = (95.98 g EtOH) / (0.7 L/kg * 84.1 kg) = 1.63 g/L / 10 = 0.163 g%

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10
Q

Young female ESU student stopped for suspected DUI [over 21 years old]
Claims she had only two drinks [capt morgan 2 oz of 90 proof rum in each drink]
Is she over the legal limit – assuming instantaneous absorption and no elimination
Wt 125 lb

A

D = (2 drinks x 2 oz) * 30 mL/oz * 45/100 * 0.79 g/mL = 42.66 g EtOH
Cp (g/L) = 42.66 g EtOH / (0.6 L/kg * 56.8 kg) = 1.25 g/L / 10 = 0.125 g%

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11
Q

Frat brother is bragging at party, he said he just “polished off” a 12 pack of beer
12 oz cans with 6% v/v EtOH
Claims under the legal limit and takes a breath test to prove it BAC is 0.079 g%
155 lbs

A

= (12 cans * 12 oz) * 30 mL/oz * 6/100 * 0.79 g/mL = 204.77 g EtOH
Cp = (204.77 g EtOH) / (0.7 L/kg * 70.5 kg) = 4.15 g/L = 0.415 g%

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12
Q

Young man found dead in a car outside bar during winter
Car not running
Outside temp was -2 F
Autopsy BAC was 0.252 g%, Wt - 160 lbs
Bar claims they only served two mixed drinks
1.5 oz of 80-proof whiskey

A

D = 2.52 g/L * 72.7 kg * 0.7 L/kg = 128.29 g EtOH
Number of drinks = 128.29 g EtOH / (1.5 oz * 30 mL/oz * 40/100 * 0.79 g/mL) = 9 drinks

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13
Q

An English teacher is involved in MVC (motor vehicle collision)
She is given a blood test one hr after the collision and BAC was 0.070 g%
Evidence indicates she was drinking gin and tonics [1.5 oz gin per drink; gin was 80 proof]
Determine the minimal # of drinks she consumed
135 lb

A

D = 0.9 g/L * 61.4 kg * 0.6 L/kg = 33.1 g EtOH
Number of drinks = 33.1 g EtOH / (1.5 oz * 30 mL/oz * 40/100 * 0.79 g/mL) = 2.3

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14
Q

Professor went to local restaurant to have dinner after class
At dinner, he consumed three glasses of napa valley cab sav wine [5 oz 12% EtOH]
What is maximal BAC achieved? Assume instantiates absorption and no eliminate
Male Vd = 0.7 L/kg, Wt = 165 lbs = 75 kg

A

D = 3 drinks * 5 oz * 30 mL/oz * 12/100 * 0.79 g/mL = 42.66 g EtOH
Cp = (14.22 g EtOH) / (0.7 L/kg * 75 kg) = 0.81 g/L = 0.081 g%

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