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ERT 312
Lecture 3
Toxicology
What is toxicology?
•
•
•
•
Qualitative and quantitative study of adverse effects of
toxicants on biological organisms
Toxicant - A chemical or physical agent, including dusts,
fibers, noise and radiation
Toxicity – property of the toxicants describing its effect
on biological organisms
Toxic hazard – a likelihood of damage to biological
organisms based on exposure resulting from transport
and other physical factors of usage
Trivia
 Which
one can be reduced?
 Toxicity
 Toxic Hazard
Things that should be clarified
 Getting
the toxic into your body
 Ways to eliminate
 Harmful effects of toxicants
4 entry modes
 Ingestion
- mouth
 Inhalation – Respiratory
system
 Injection – skin cut
 Dermal absorption - skin
3 exit modes
 Excretion
– kidneys, liver,
lungs, skin
 Detoxification – downgrade
the toxicants into
something less harmful
 Storage – fatty tissue
Table 1: Various Responses to Toxicants
(Crowl & Louvar, 2002)
Irreversible
Effects
Reversible/Irreversible
Effects
Carcinogen
Mutagen
Reproductive hazard
Dermatoxic
Hemotoxic
Hepatoxic
Teratogen
Nephrotoxic
Neurotoxic
Pulmonotoxic
Individuals affected
Dose vs. Response
Low response
Average response
High response
Gaussian/Normal Distribution Curve
(Equa.1)
1 x 2
 (
)
1
2 
f ( x) 
e
 2
f(x)
x
σ
µ
the probability (or fraction) of individuals
experiencing a specific response
the response
the standard deviation
the mean
n
MEAN,  
 x f (x )
i 1
n
i
i
 f (x )
i
i 1
n
VARIANCE, 
Equa.3
2

Equa.2
 (x  ) f (x )
i 1
2
i
i
n
 f (x )
i 1
i
Example 1
A safety engineer of one leading fertilizer brand is very
concern on the irritancy effect of ammonia, a main raw
material used to produce the fertilizer. A toxicology study
has been conducted on 75 employees. The responses are
recorded on scale from 0 to 10, with 0 indicating no
response and 10 indicating a high response. Details of the
findings are presented in the table 2
Table 2
Response
Number of individuals affected
0
1
0
5
2
3
4
10
13
13
5
6
7
11
9
6
8
9
10
3
3
2
a.
b.
c.
Plot a histogram of the number of individuals affected
vs. the response
Determine the mean and the standard deviation
Plot the normal distribution on the histogram of the
original data
Answers
f ( x)  13.3e
0.100( x 4.51) 2



Mean, µ = 338/75 = 4.51
Variance, σ2 = 374.75/75 = 5
SD, σ = 2.24

Therefore; the normal distribution is,
f ( x)  0.178e
0.100( x 4.51)
2


To plot a normal distribution curve, you need to convert
a distribution equation to a function representing the
number of individuals affected.
In this case, total individuals affected = 75
Refer to table 2.3
(Crowl & Louvar, 2002)
x
f(x)
75f(x)
0
0.0232
1.74
1
0.0519
3.89
2
0.0948
7.11
3
10.6
4
13.0
4.51
13.3
5
13.0
6
10.7
7
7.18
8
3.95
9
1.78
10
0.655
Response – Log Dose Curve





For convenience, the response is plotted versus
the logarithm of the dose
If the response of the interest is death or lethality
= lethal dose curve, LD
LC = lethal concentration (gas)
If the response to the chemical or agent is minor
or irreversible = effective dose, ED
If the response to the agent is toxic (not lethal
but irreversible) = toxic dose, TD
Models for Dose and Response
Curves


The probit (probability unit) method is very common for
single exposure computational.
The probit variable Y is related to the probability P by
(Equa.4)
1
Y 5
2
u
P
)du
1  exp( 
2
2 
(2 )
Fig X: The probit transformation converts the sigmoidal response vs. log
dose curve into a straight line when plotted on a linear probit scale
Question 2.2 (Crowl & Louvar, 2002)

The effect of rotenone on macrosiphoniella sanborni sp. was investigated. Rotenone
was applied in a medium of 0.5% saponin, containing 5% alcohol. The insects were
examined and classified one day after spraying.The obtained date were:
Dose (mg/l)
Number of insects
Number affected
10.2
50
44
7.7
49
42
5.1
46
24
3.8
48
16
2.6
50
6
0
49
0

From the given data, plot the percentage of insects affected versus the natural
logarithm of dose

Convert the data to a probit variable, and plot the probit versus the natural
logarithm of the dose. If the results is linear, determine a straight line that fits the
data. Compare the probit and number of insects affected predicted by the straight
line fit to the actual data
Probit Variable Y

Equa.5
Y  k1  k2 lnV
k1, k2
V
Probit parameters
Causative factor represents the dose

OTOH, conversion from probits to percentage is given by
(Equa.6)
 Y 5
 Y  5 

P  501 
erf 
 2 
 Y 5
erf
the error function of Y