Ch 2.8 Case Study: Dull, Dizzy or Dead?

Background

  • Alcohol requires no digestion and can be absorbed rapidly from stomach into bloodstream.
  • This is particularly the case if the stomach is empty.
  • Alcohol is absorbed even more rapidly from intestines.

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Background

  • Thus, unless alcohol is heavily diluted or taken with food, very little metabolism occurs in the GI-tract.
  • All of the alcohol is absorbed into the bloodstream.

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Background

  • Alcohol is distributed freely to all body fluids.
  • The concentration of alcohol in brain rapidly approaches that in the blood.

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Background

  • Most of the alcohol (90-98%) is oxidized through the liver and excreted.
  • The remainder leaves body through lungs, urine, saliva and sweat.
  • Liver can only metabolize alcohol at a constant rate if concentrations are not small.

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BAL

  • The blood alcohol level (BAL) is a measure of alcohol concentration in blood.
  • BAL is given by the total mass of alcohol in grams divided by the total fluid volume in the body.
  • Units for BAL are grams/100ml of blood.

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BAL Measurement

  • Thus 100 g/100ml yields BAL = 100.
  • Similarly, 100mg/100ml yields BAL = 0.1.
  • It is common to state BAL without units specified.

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Legal Limits for Driving

  • Australian law prohibits the driving of vehicles (including boats and horse- or camel-drawn vehicles) for those with a BAL above 0.05.
  • This then relates to 50mg/100ml alcohol in the bloodstream.

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Legal Limits for Driving

  • A person with a BAL of 0.15 is 25 times more likely to have a fatal accident than one with no alcohol.

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Limits & Deriving

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Assumptions

  • Alcohol intake into GI-tract is controlled by drinker.
  • Amount of alcohol subsequently absorbed into bloodstream depends on the following:
  • The concentration of alcohol and other liquids and food in GI-tract.
  • The weight and gender of individual.

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Assumptions

  • Alcohol is removed from bloodstream at constant rate by liver.
  • This is independent of body weight and gender of individual, as well as the concentration of alcohol in the bloodstream.
  • This assumes that the liver has not been damaged by large doses of alcohol.

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Assumptions

  • A small percentage leaves through sweat, saliva, breath and urine.
  • Ignoring this could mean the BAL estimate may be slightly above the true value.

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General compartmental model

Our compartmental model has two compartments:

  • GI-tract
  • Bloodstream.

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General Compartmental Model

  • The GI-tract compartment has a single input and output.
  • The bloodstream compartment has a single input and output.

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Word Equations

  • Balance law yields two word equations, one for each compartment.

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Variables

For the our variables, let

  • x(t) be the concentration of alcohol in the GI-tract at time t.
  • y(t) the concentration in the bloodstream at time t.

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Relate Variables

  • In GI-tract, input rate = \( I \) and output rate proportional to amount.
  • In bloodstream, initial amount of drug is zero, so \( y(0) = 0 \).
  • Bloodstream level increases as drug diffuses from GI-tract and decreases as kidneys and liver remove it.

\[ \begin{align*} \frac{dx}{dt} & = I - k_1x, \,\,\, x(0)= x_0 \\ \frac{dy}{dt} & = k_2x - \frac{k_3 y}{y+M}, \,\,\, y(0)= 0 \end{align*} \]

Relate Variables: Parameters

  • The parameters are:
  • \( k_1,\,k_2,\,k_3,\, M \) positive
  • \( M = 0.005 \) BAL
  • \( I = i/V_b \),
  • \( i \) = ingestion rate (BAL/hr)
  • \( V_b \) = volume of fluid in blood (100 ml)

\[ \begin{align*} \frac{dx}{dt} & = I - k_1x, \,\,\, x(0)= x_0 \\ \frac{dy}{dt} & = k_2x - \frac{k_3 y}{y+M}, \,\,\, y(0)= 0 \end{align*} \]

Relate Variables: Parameters

  • Model I: Empty stomach (\( k_1 = k_2 \))
  • Model II: Full stomach (\( k_1 > k_2 \))
  • From study: kidney removal rate = \( -(k_3 y)/(y+M) \)

\[ \begin{align*} \frac{dx}{dt} & = I - k_1x, \,\,\, x(0)= x_0 \\ \frac{dy}{dt} & = k_2x - \frac{k_3 y}{y+M}, \,\,\, y(0)= 0 \end{align*} \]

Kidney Removal Rate

\[ \frac{dy}{dt} \sim \frac{- k_3 y}{y+M} \propto \left\{ \begin{align*} -k_3, &\,\,\, y \, \gg M \\ - \frac{k_3}{M}y, &\,\,\, y \, \ll M \end{align*} \right. \]

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Known Constants for Initial Conditions

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Model I: Empty Stomach

  • Alcohol rapidly absorbed in first hour (\( k_1 = k_2 = 6 \)).
  • Number of initial drinks \( n \) consumed rapidly.
  • No further drinking after initial \( n \) amount, so \( I = 0 \).
  • Initial amount of alcohol in GI-tract is \( x_0 = nx_s \)
  • \( x_s \) = effective BAL produced by single drink.

\[ \begin{align*} \frac{dx}{dt} & = I - k_1x, \,\,\, x(0)= x_0 \\ \frac{dy}{dt} & = k_2x - \frac{k_3 y}{y+M}, \,\,\, y(0)= 0 \end{align*} \]

Model I: Determine Initial Concentration for x

  • One drink (\( n=1 \)) produces 14g effective alcohol.
  • Total volume of blood for a woman is (0.67 L/kg)(W kg)
  • Total volume of blood for a man is (0.82 L/kg)(W kg)
  • For 68 kg male with \( n=3 \) drinks:

\[ \begin{align*} x_0 = 3x_s & = \frac{3(14g)}{(0.82 L/kg)(68 kg))} = \frac{3(14g)}{(0.82L)(68)(100*10 ml/L)} \\ & = \frac{3(14) g}{(0.82)(68)(10)100 ml} \cong 0.0753 \, \mathrm{BAL} \end{align*} \]

Model I: Determine k3

  • Rate at which liver removes alcohol from blood \( \cong \) 8 g/hr.
  • Corresponding BAL reduction depends on total body fluids.
  • For 68 kg man, we have

\[ \begin{align*} k_3 & = \frac{8 g/hr}{(0.82 L/kg)(68 kg))} = \frac{8 g/hr}{(0.82L)(68)(100*10 ml/L)} \\ & = \frac{8 g/hr}{(0.82)(68)(10)100 ml} \cong 0.0143 \, \mathrm{BAL/hr} \end{align*} \]

Model I, Case 1: Three Drinks at Start on Empty Stomach

Ch28model11(6,500)

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Model I, Case 1: Three Drinks Discussion

Ch28model11(6,500)

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Model I, Case 1: R Script (Page 1 of 2) 

Ch28model11 <- function(T,N) {

#Chapter2.8 Model I; Empty Stomach, three drinks at start.
#Perform Rk4 for alcohol in GI-tract and bloodstream

#T is the time length for [0, T]
#N is the number of time steps
h = T/N  #This is the time step size

Model I, Case 1: R Script (Page 2 of 2)

#System Parameters
t0 <- 0
x0 <- 0.075     #initial value for GI-tract
y0 <- 0         #initial value for bloodstream 
k1 <- 6         #k1 value for GI-tract
k2 <- 6         #k2 value for bloodstream 
k3 <- 0.014     #k2 value for bloodstream 
M  <- 0.005     #Modeling constant from reading
I  <- 0         #Zero drinks after initial amount

#System of ODEs
f1 <- function(x) {I - k1*x}   
f2 <- function(x,y) {k2*x - k3*y/(y + M)}

Model I, Case 2: Continuous Drinking

  • We take \( I = (n/T)x_s \) BAL/hr and \( x_0=0 \).
  • Here, \( n \) is average number of drinks in \( T \) hours.

\[ \begin{align*} \frac{dx}{dt} & = I - k_1x, \,\,\, x(0)= 0 \\ \frac{dy}{dt} & = k_2x - \frac{k_3 y}{y+M}, \,\,\, y(0)= 0 \end{align*} \]

Model I, Case 2: Continuous Drinking 

Ch28model12(6,500,3)

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Model I, Case 2: Continuous Drinking 

Ch28model12(6,500,3)

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Model I, Case 2: R Script (Page 1 of 2)

Ch28model12 <- function(T,N,n) {

#Chapter2.8 Model I: Empty Stomach, continuous drinking.
#Perform Rk4 for alcohol in GI-tract and bloodstream

#T is the time length for [0, T]
#N is the number of time steps
#n is the number of drinks per hour
h = T/N  #This is the time step size

Model I, Case 2: R Script (Page 2 of 2)

#System Parameters
t0 <- 0
x0 <- 0         #initial value for GI-tract
y0 <- 0         #initial value for bloodstream 
k1 <- 6         #k1 value for GI-tract
k2 <- 6         #k2 value for bloodstream 
k3 <- 0.014     #k2 value for bloodstream 
M  <- 0.005     #Modeling constant from reading
I  <- 0.025*n   #BAL input from drinking continuously

#System of ODEs
f1 <- function(x) {I - k1*x}   
f2 <- function(x,y) {k2*x - k3*y/(y + M)}

Model II, Case 1: Three Drinks at Start on Full Stomach

  • Rate \( k_1 \) at which alcohol leaves GI-tract is greater than rate \( k_2 \) at which alcohol enters the bloodstream.
  • After substantial meal, rate of absorption into bloodstream is approximately halved.
  • Thus \( k_1 > k_2 = k_1/2 \).
  • Also, recall \( I = 0 \).

\[ \begin{align*} \frac{dx}{dt} & = I - k_1x, \,\,\, x(0)= x_0 \\ \frac{dy}{dt} & = k_2x - \frac{k_3 y}{y+M}, \,\,\, y(0)= 0 \end{align*} \]

Model II, Case 1: Three Drinks at Start on Full Stomach

Ch28model21(6,500)

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Model II, Case 1: Three Drinks at Start Discussion

Ch28model21(6,500)

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Model II, Case 1: R Script (Page 1 of 2)

Ch28model21 <- function(T,N) {

#Chapter2.8 Model II: Full Stomach, three drinks at start.
#Perform Rk4 for alcohol in GI-tract and bloodstream

#T is the time length for [0, T]
#N is the number of time steps
h = T/N  #This is the time step size

Model II, Case 1: R Script (Page 2 of 2)

#System Parameters
t0 <- 0
x0 <- 0.075     #initial value for GI-tract
y0 <- 0         #initial value for bloodstream 
k1 <- 6         #k1 value for GI-tract
k2 <- k1/2      #k2 value for bloodstream 
k3 <- 0.014     #k2 value for bloodstream 
M  <- 0.005     #Modeling constant from reading
I  <- 0         #Zero drinks after initial amount

#System of ODEs
f1 <- function(x) {I - k1*x}   
f2 <- function(x,y) {k2*x - k3*y/(y + M)}

Model II, Case 2: Continuous Drinking on Full Stomach

  • We take \( I = (n/T)x_s \) BAL/hr and \( x_0=0 \).
  • Here, \( n \) is average number of drinks in \( T \) hours.
  • As before, \( k_1 > k_2 = k_1/2 \).

\[ \begin{align*} \frac{dx}{dt} & = I - k_1x, \,\,\, x(0)= 0 \\ \frac{dy}{dt} & = k_2x - \frac{k_3 y}{y+M}, \,\,\, y(0)= 0 \end{align*} \]

Model II, Case 2: Continuous Drinking on Full Stomach

Ch28model22(6,500,3)

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Model II, Case 2: Continuous Drinking Discussion

Ch28model12(6,500,3)

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Model II, Case 2: R Script (Page 1 of 2)

Ch28model22 <- function(T,N,n) {

#Chapter2.8 Model II: Full Stomach, continuous drinking.
#Perform Rk4 for alcohol in GI-tract and bloodstream

#T is the time length for [0, T]
#N is the number of time steps
#n is the number of drinks per hour
h = T/N  #This is the time step size

Model II, Case 2: R Script (Page 2 of 2)

#System Parameters
t0 <- 0
x0 <- 0         #initial value for GI-tract
y0 <- 0         #initial value for bloodstream 
k1 <- 6         #k1 value for GI-tract
k2 <- k1/2      #k2 value for bloodstream 
k3 <- 0.014     #k2 value for bloodstream 
M  <- 0.005     #Modeling constant from reading
I  <- 0.025*n   #BAL input from drinking continuously

#System of ODEs
f1 <- function(x) {I - k1*x}   
f2 <- function(x,y) {k2*x - k3*y/(y + M)}