Alcohol calculations Flashcards
Standard drink
Beer [5% v/v] 12 oz
Wine [12% v/v] 5 oz
Whiskey [80 proof] 1.5 oz
How to calculate dose
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
Extrapolated alcohol conc.
starting alcohol conc given a drinking scenario and assumed subject parameters
Back-extraopolaate
final known conc given a drinking scenario and assumed subject parameters
What is required for retrograde
to be in post absorptive phase
Zig-Zag or Steepling Effect
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
Widmark Equation
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
Alcohol Standard equation
Dose=VdwtCp Cp(g/L) dose (g) Vd (L/kg) * W (kg)
Male drinks 6 shorts of whiskey
Calculation of maximum BAC – assuming instantaneous absorption and no elimination
Short: 1.5 oz
Wt = 185 pounds
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%
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
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%
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
= (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%
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
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
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
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
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
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%
