 
RISK ASSESSMENT AND COMMUNICATION [Chronic
risk: part 2]
The
following are a collection of 17 revised examples of the calculations of intakes, chronic risks, and
hazard indices using the EPA fourstep model. They are displayed with the problem, the assumption
used, & the solutions. 
Reminder of the INTAKE calculation......
AS A HELP IN ENCOURAGING YOU TO WORK
OUT THE SOLUTIONS YOURSELF AND NOT JUST TO RELY ON THE ILLUSTRATED EXAMPLES, I HAVE TEMPORARILY MASKED SOME OF THE
SOLUTIONS IN THE FOLLOWING
17 CHALLENGES.
AFTER YOU TRY THE
CALCULATIONS AGAIN, YOU CAN REVEAL THE SOLUTION BY PULLING YOUR CURSOR OVER
THE 'EMPTY' CELL.
NEW!..... Solutions to questions on p.44
# 49
I_{c} =
[C mg/M^{3}](0.0037573 M^{3}/Kg.d)
# 50
I_{N} =
[C mg/M^{3}](0.418853
M^{3}/Kg.d)
Although it is not clear in the question, this is AIRBORNE
# 51
As carc: I_{c}
= 0.0040816 mg/Kg.d
As noncarc: I_{N}
= 0.0285714 mg/Kg.d
(Because there is evidence that
chloroform is both C and N, you
should complete assessments for
both effects (unless you are
asked to limit the assessment to
one of these) 

1. Dichlorvos
exposure in a teacher
What is your estimated
lifetime cancer risk for an adult individual who is exposed to drinking
water while at home containing 0.001 mg/L dichlorvos. The exposure is for 2
years. She is employed as a grade 3 teacher during that time at a school
not affected by the contaminated water. 
From the question, you have these values
C =
0.001mg/L
ED = 2 y
Other values and assumptions :
CR =
2L/d (standard tables)
BW =
70 kg ( "
)
AT =
(carc.) 25550d ( " )
EF =
daysequivalents exposed per year
=
full year MINUS work time
= 365d MINUS
(9
months x 20d/m x 8h/24h)
=
365d — 60d
=
305d
RR
=
1 (default)
ABS
=
1 ( " ) 
SOLUTION:
Ic =
( 0.001 mg/L ) (2 L/d) (302 d/y) (2y)
70kg 25,550d
=
6.75 x 10^{7} mg / Kg.d
Risk = 6.75 x 10^{7} mg /
Kg.d x 0.29 (mg / Kg.d)^{1}
=
1.95 x 10^{7 }
This is less than then the de minimis level of 1 in a million and is
probably acceptable by the community (although special considerations and
precautions are appropriate for children, infants and the unborn). 
2. Mineworker / nickel dust
You are
assisting in the assessment of a worker in a Nickel mine. The mean
concentration of nickel refinery dust in air is 2.0 microgram/M³. The workers are at work about 200 days a year
in single 8 hr
shifts. They have been at this work for 8 years.
Has the typical worker been exposed to excessive nickel? Calculate the risk
of additional risk of death from cancer. What recommendations, if any, would
you make to the company to reduce the risk for future workers? 
Assumptions and values
used
(NOTE: These values would normally be given to you)
[C] = 2 microgram/M³.
or 0.002 mg/M³.
CR
= 1.2 M³ /Hr (note this is in hours and must show days so
x(24h/d)
EF = 8 hr shift per day, x (8h/24h) x (200 d/yr)
ED = 8 y
BW = 70 Kg
AT = 25,550 d
ABS:
1
RR:
1

SOLUTION
Ic =
(0.002 mg/M³) x (1.2 M³ /h) x (24h/d) x
(8h/24h) x (200 d/y) x (8 y)
70 Kg x 25,550 d
Ic = 0.0000172 mg/Kg.d
RISK = 0.0000172
mg/Kg.d x 0.84 = 0.00001443
which is more than 14 times the de minimis level.
Recommendation: reduce the air
concentration to 1/14 the current [Conc]
(i.e. to 0.00014 mg/M³ or 0.14
micrograms/M³)

3. Carrots and Copper Cyanide (different
calculation)
(A) Find intake in mg/Kg.d for copper cyanide in carrots
detected at rate of 4 mg/Kg carrots. Assess for 9 y/o child, with daily
consumption of 100 g carrots per day, for her lifetime to date. If
these carrots are to be consumed in future, what would you recommend (B) as
the maximum intake of carrots a day assuming the concentration of copper
cyanide remains the same? (C) as the maximum concentration of copper
cyanide in the carrots assuming the consumption remains at 100 g per day? 
Assumptions and values used (NOTE: These values would normally be given to you)
C: 4 mg /Kg(food)
9 yr old
child
CR: 100g
carrots/d
EF:
daily
ED: 9
yr
AT:
(365x9) d
BW:
29 kg
ABS :
assume 1.0
RR:
assume 1.0 
Calculation (A)
I_{N
= }(4mg/Kgf) x (0.1Kgf/d) x (365d/y) x (9y) x (1) x (1)
29 Kg bw x (365 x
9) d
This
can be simplified because the EFxED cancel the AT
I_{N
= }(4mg/Kgf) x (0.1Kgf/d)
29 Kgbw
= 1.38 x 10^{2}
mg/Kg.d
H.Index : I_{N
}
mg/Kg.d / RfD
1.38E02 mg/Kg.d / 5.00E03 mg/Kg.d =
2.76
 which is nearly 3 times higher than the "action level" generally
considered for a noncarcinogen.
(B)
Assuming [C] in carrots remains
constant at 4mg/Kg,
the daily
intake would need to be reduced to 100g / 2.76 or 36 g a day
(C)
Assuming she must continue to eat
100g carrots a day,
the [C] in
carrots would need to be reduced to 4mg / 2.76 = 1.45 mg/Kg

4. Water contamination: Mining town
The water in a mining town has been found to be
contaminated with metal salts and chemicals associated with assaying and extraction of
ores. Carry out an assessment of any carcinogenic and noncarcinogenic effects arising
from an adult resident's lifetime of water consumption. The following are
the concentrations recorded.
zinc cyanide
0.04 mg/L
zinc phosphide
0.02 mg/L
nickel (soluble) salts
0.23 mg/L
manganese
0.19 mg/L

Assumptions and values used
(NOTE: These values would normally be given to you)
EF: 365d/y
ED: 70y
BW: 70Kg
Water consumpt: 2 L/d
Also assume that as the assessment is over a lifetime,
the AT rates and EDxEF rates are equivalent. 
Solution:
The intake calculation:
In
= [Cmg/L] x 2L/d x 1/70Kg =
[Cmg/L] x
0.0286
Solution: Creating the
assessment. .... There are no CPF data shown for any of these
chemicals.
Therefore there is no carcinogenic assessment possible.
The following is the noncarc. assessment
leading to
Hax.Index.


1 
2 
3 
4 
5 
6 
7 

CONC 
RfD 
TS 
TS% 
TS%ACCUM 
INTAKE 
H.Index 

MG/L 




Cx(0.0286) 

zinc phosphide 
0.02 
3.00E04 
66.67 
68.05% 
68.05% 
0.000572 
1.907 
manganese 
0.19 
1.00E02 
19.00 
19.39% 
87.44% 
0.005434 
0.543 
nickel (sol) salts 
0.23 
2.00E02 
11.50 
11.74% 
99.18% 
0.006578 
0.329 
zinc cyanide 
0.04 
5.00E02 
0.80 
0.82% 
100.00% 
0.001144 
0.023 



9.80E+01 
100.00% 


2.802 
5. Ethyl Ether in water
Estimate (A) the intake for 0.02 mg/L ethyl ether in
drinking water and (B) the expected health outcomes for sedentary male at residence for 1 year 
Assumptions and values used.
(NOTE: These values would normally be given to you)
[C] 0.02
mg/L (a noncarcinogen)
Assume 365
d/y,
2L/d water
cons,
ABS and RR:
100%
adult bw:
70 Kg.
AT: 365 d
ED: 1 y
EF: 365d/y 
Solution
(A)
In = (0.02
mg/L) (2L/d) (365d/y) (1y) (1) (1)
(70 Kg) (365 d)
Because EDxED are the same
as AT, we can simplify to
In =
(0.02 mg/L) (2L/d) = 5.714E04
mg/Kg.d
(70 Kg)
(B)
The HI is calculated as In /
RfD
5.714E04 mg/Kg.d / 2E01 mg/Kg.d
= 0.0028 (much below the level at which concern might be expected or
expressed) 
6. Airborne Solvent
Calculate the intake for an adult's 5year
occupational exposure to an airborne carcinogenic solvent. 
Assumptions and values used (NOTE: These
values would normally be given to you)
An 8hr
working day for 200 working days a year. ED: 5 y
The work is
light, so respiration of 1.2 m3/h, 70% RR and 100% Abs 
Solution
Ic = [C] x [CR] x
EF x ED x RR x ABS
BW x AT
= (1.2 m3/h) (8h/d) (200 d/y) (5 y) (0.7)
(1)
(70 Kg) (25550d)
= [C] x 0.00376 m3/Kg.d

7 Child: local airborne nonCarc.
Calculate the intake for a 20 Kg child of a nonc
continuously emitted from a chemical plant near the residence for one year. 
Assumptions and values used: (NOTE: These
values would normally be given to you). 20Kgbw (see table on p37)
suggests 6yr old, with resp. rate about 0.35 cu M per day. EFxED: 9 months x
20 d/mo x 6.5 h/d.
100% abs,
and RR. AT: 10 months at 30.5 d/mo 
Solutions
I_{N} = [C]
(CR) (EF) (ED) (RR) (ABS)
(BW) (AT)
= [C] (0.35 m3/h) (6.5h/d) (180d/y) (1 y)
(1) (1)
(20 Kg) (305d)
=
[C] x 0.0671 m3/Kg.d
. Note that any slight
variation of this type would be virtually INVISIBLE in the final outcome.
when we are concerned with orders of magnitude ( 10x, 100x, 10^{3},
10^{4},
etc.), something like the difference between 0.32 and 0.34 is of no
importance at all. 
8. gammaHexachlorocyclohexane
(γHCH) in water
Concentration of HCH in drinking water at 0.7 mg/L
Estimate intake for adult resident (exposed during hours of 5pm8am each
day. ED: 18 months. γHCH
is nonCarc. only. Also calc H.I. If H.I >1, calculate
water concentration which would maintain HI<1 
Assumptions and values used
(NOTE: These values would normally be given to you)
EF: (15h/24h) (365d/y)
ED:
1.5 y
AT: 1.5x365
d
CR: 2
L/d 
Solution
Intake (N)
= 0.7 mg.L x 2 L/d x 15h/24h x 365 d/y x 1.5 y
70 Kg (1.5 x 365)d
simplified to 0.7 mg.L x 2 L/d x 15/24
x (365 d/y x 1.5 y)
70 Kg x (1.5y x 365d/y)
= 0.0125 mg/Kg.d
H.I. =
( 0.0125 mg/Kg.d) / (3.00E04) =
41.67
To maintain H.I. <1, water concentration must be
(0.7 mg/L) x 1/42 = 0.0167 mg/L

9. 12 Dichloroethane in air
(adjusted April 5: the calculation should be done as a carcinogen.
Therefore the AT value should have been 70x365d)
Calculate intake for adult (assumed sedentary) daily
(8 hr shiftwork) for 2 years

Assumptions and values used (NOTE: These values would
normally be given to you)
[C] : 0.002 mg/m3
EF: (8h/24h) x (200d/y)
ED: 2 y
AT
(70x365)d
air
breathed: assume sedentary : 0.85 m3/h

Solution
Ic = [0.002mg/m3] x
(0.85 m3/h x 24 h/d) x (8h/24h) x (200d/y) x
2y x 1 x 1
70 Kg x (365 x
70)d
Ic =
3.042 x 10^{6}
mg/kg.d
risk=
Ic x SF
=
3.042 x 10^{6}
mg/kg.d x
9.10E2 (mg/kg.d)^{1} = 2.768E07 which is
close to, but less than, the de minimis level.

10
PCP in lumber mill workers (previously #7) Workers in a
lumber mill have an estimated daily intake (per person) of
pentachlorophenol (PCP) through food and water pathways combined of 0.11 mg.
Inhalation of PCP through airborne vapour adds another 0.04 mg per person
per day. What is the incremental cancer risk for these workers from these
combined exposures?

Assumptions and values used.
(Note that normally you would be given these values)
ED: working life of 50 y
EF: 200 d/y
Adult bw:
70 Kg
AT (carc):
25,550 d
Notes:
(1) that the INTAKE has already been calculated BUT as "per person, per
day".
and (2)
that whereas normally the different pathways (ingestion and inhalation)
would be separate calculations because they are subject to different SF/CPF,
in this case we only have a single SF/CPF, so we can combine to a single
intake and single calculation. The problem is that an inhalation CPF
is not given and instead the oral CPF is being used. This can be
contentious and may not always be acceptable. We would not give
you a situation such as this on an exam.

Solution:
Intake per person = 0.15 mg /d
We can assume the time factors (ED, EF, AT)
have similarly already been considered iin calculating the intake But we
need to calculate the intake as mg/Kg.d
= _0.15 mg / d_
= 0.002143 mg / Kg.d
70 Kg
The increased risk is Intake x SF
= (2.143E03 mg/Kg.d) x 1.20E01 (mg/Kg.d)^{1}
= 0.0002572, which is 257
times the de minimis level

11
Heptachlor and potatoes (previously #8) A group of
subsistence farmers grow potatoes on their land. They and their families eat
up to 0.7 Kg potatoes The soil has been found to contain a residual
contamination with heptachlor (at 0.05 mg/Kg soil). Water from the site's
shallow wells is also found to contain heptachlor (0.004 mg/L).
Calculate the incremental cancer risk and the hazard index for one year's
exposure.

Assumptions and values used:
CR: 0.7Kg
potatoes/person/day
EF: 365 d/y
EF: 1 y
Bioconcentration factor potatoes/soil : 2.5
The intake calculation is performed without
the EDxEF or AT. Of course this would be normal where the exposure is a
"24/7" lifetime exposure and the numerator cancels the denominator.
But in this case, the exposure is ONE year so EDxEF should be 1y x 365d/y, but
the AT should still be 25,550d

Solution:
we calculate the the two pathways separately and
combine the intakes (equivalent to column 6)
(FOOD) Ic = 0.05mg/Kgf x 2.5 x
0.7 Kgfd x 365d x 1 =
1.79E05
70Kgbw x 25,550d
(WATER) Ic = 0.004 mg/L x 2L/d x 365 d
= 1.63E06
70Kg x 25,550d
(COMBINED)
(1.79E05
+ 1.63E06) mg/Kg.d = 1.95E05 mg/Kg.d
The risk is calculated at I x SF
1.95E05 mg/Kg.d x
4.5E+00 = 8.79E05 which is 88 times higher than de minimis

(The equivalent calculation will yield the H.I.) 
12 Hydrazine Sulfate:
If the "de minimis" level is accepted as the "background" level for the
population of 150,000 persons, how many additional cases of cancer would you
expect if each person were to be exposed to 0.002 mg hydrazine per day on a
continuing basis? 
Assumptions and values
used: Note
that the intake has already been calculated! Also note that the intake
is given as mg/person per day. In other words we still need to get it
into the form we use, which is mg/Kg.d
Ic = per
person intake divided by average bw of 70 Kg: (0.002 mg/d) /
(70kg) = 0.0000285 mg/kg.d
It remains only to multiply
the Ic by the SF of 3.0 (having confirmed first that hydrazine
is a carcinogen)
Assume every day + all day
for lifetime, so EDxEF cancel with AT

NOTE:THIS ITEM IS UNDER REVIEW; PLEASE CHECK
BACK IN 24 HRS
Solution:
Risk = 2.85 x 10^{4}
mg/kg.d x 3.0 (mg/kg.d)^{1}
= 8.55 x 10^{5}
This is of course MORE than
de minimis (by 85.5 times)
The question of how many
additional cases would ensue from the population's exposure can be answered
by recognizing that this risk (8.55 x 10^{5} ) is for any ONE
person. Therefore we multiply ( 8.55 x 10^{5} ) by
150,000 to obtain 12.8 (or 13) additional deaths from cancers.

13 Any airborne carcinogenic
exposure Calculate the intake for an adult's
5 year occupational exposure to an airborne carcinogenic solvent during an 8
hr working day for 200 days per year. The work is fairly light, so
assume respiration at 1.2 m3/hr, with 70% retention and 100% absorption.
(Note that this calls for the intake calculation to be determined without
the concentration being included. In this way the value can be used
for exposure to ANY airborne carcinogens materials which are in the same
path and have the same characteristics) 
Assumptions and values used.
C: (not
used)
CR: (1.2
m3/hr) (given)
EF: 200 d/y
(given)
ED: 5 y
(given)
ABS: 1.0
(given)
RR: 0.7
(given)
BW: 70kg
(from standard values)
AT: 25,550
d (from standard values) 
Solutions:
Ic = [C mg/m3]
x (1.2 m3/h) (24h/d) (8h/24h) (200d/y) (5y) (1.0) (0.7)
( 70kg ) ( 25 550d )
= [C mg/m3] x (0.000376
m^{3}/kg.d)
This value would be used in
column 6 to calculate the intake values for a list of substances

14. Any noncarcinogen intake
value for a 20 Kg child Calculate the intake for a
20 Kg child for a noncarcinogen continuously emmitted from a chemical plant
near the residence for six months.
Assume all living and recreation, school, etc. is
in ther asame area. Assume 100 abs and 100 retention. 
Assumptions and values used
C: not used
CR:
0.35m3/h (20 kg child is probably about 6 yrs old (from standard values
table)
EF:
365 d/y (given)
ED:
0.5 yr (given)
Abs = RR =
1.0 (given)
BW= 20 kg
(given)
AT = days
in 0.5 y = 183d 
In =
[C] x (0.35 m3/h) (24h/d) (365d/y) (0.5 y) (1) (1)
( 20 kg ) ( 183 d )
Note that the EDxED cancel the
AT because it is full time exposure. So we can simplify using
In = [C
mg/m^{3}] x (0.35 m3/h) (24h/d)
( 20 kg )
giving
[Cmg/m^{3}]
x 0.42 m^{3}/K.d

15: Longterm exposure to
noncarcinogen: population of all ages. A town's
drinking water has been found to have been contaminated by a chemical
dump that was buried and forgotten in 1930. The first chemical of interest
is phenol, a noncarcinogen The water contains 230 ppb phenol and the
assessment will assume a constant level of exposure at this level for the
past 78 years. Assess the health status to this exposure up to the
present date. The the ground water has this year been replaced by a
municipal supply. 
Assumptions and Values used.
Some people may have been exposed for their whole life if they were present
continually since 1930. Others may have been born in the community
only this year. The best approach is to take an "average" exposure
duration therefore of 30 years (roughly half a lifetime, and allowing for
the community to have grown and other people not to have been present the
whole time). The AT would cancel out the EDxEF.
C:
0.23 mg/L (note the original C was 230 ppb or 230 micrograms/L
CR: 2 L/d
EF: 365d/y
ED: 30 y
70 kg
AT: 30 x
365 d

Solution
Ic =
( 0.23 mg/L) (2 L/d) (365 d/y) (30 y)
( 70 kg ) ( 365 d/y x 30 y)
simplifies to
Ic = ( 0.023 mg/L) (2
L/d) = 0.00657
mg/Kg.d
( 70 kg )
The H.I. becomes:
In / RfD = (0.00657) / (0.60) =
0.011
This is far below any action
level (1.0), and should probably be ignored, Note that the elimination
of the ED, EF and AT reduces the calculation to a constant hazard value for
any period of time.
Compare this with the
following (carcinogen) example.

16
Longterm exposure to
carcinogen: population of all ages.
The same water supply (see situation #16) contained
benzyl chloride at a concentration of 100 ppb. Use the same parameters to
estimate the carcinogenic risk

Assumptions and values used: In this case, the midrange of the ED (30
years) can be used but the AT must remain 70x365d.
C: 0.100 mg/L , a carcinogen. (note
the original C was quoted as 100 ppb or 100 micrograms/L
CR: 2 L/d
EF: 365d/y
ED: 30 y
70 kg
AT: 70 x
365 d

Solution:
Ic = ( 0.100 mg/L) (2 L/d) (365 d/y) (30 y)
( 70 kg ) ( 25,550d)
= 0.0012 mg/kg.d
The risk is therefore 0.0012 mg/kg.d
x 0.17 (mg/kg.d)^{1 = }2.03
E^{4}
This exceeds the de minimis level by two hundred
times.

17
1,2Diphenylhydrazine is being released during the daytime from a
refinery in the concentration of 12 ppb air. The exposure may have been
constant for a year and a half. Calculate the health effects of such
exposure for a sedentary typical adult who may have been permanently
resident in this area for that time.

Assumptions and values used
[C}: 12 ppb = 12 micro g / m3 =
0.012 mg/m3 1,2Diphenylhydrazine vapour in air
CR:
(0.85 m3/h) air
EF: 365d/y
ED:
1.5y
BW: 70kg
AT: 25550 d
(carcinogen) 
Solution:
Ic
= (0.012
mg/m3) ( 0.85m3/h) (24h/d) ( 365 d/y) (1/5y)
(70kg) (25550 d)
= 7.49E05 mg/kg.d
The risk estimate is
therefore Ic x SF or 7.49E05 x 0.8
= 5.99E05
This is 60 times greater than
the de minimis level, and should probably receive some prompt corrective
action

Another example of the composite calculation.
