Human Exposure Pathways - Dealing with Contaminated Sites: From Theory Towards Practical Application - page 477

Table 11.3 Soil intake rates of animals [kg/d]

References Cattle Pigs Lambs Poultry

2.10 − 3 Veerkamp and Ten

Berge ( 1994 )
[kg/d]

0.72 0

IAEA ( 1994 ) [%] Grazing cattle: 6% of

dry weight of feed

1.5% of dry weight of
feed (if confined
feed)

10 % of dry
weight of
feed

10 − 2 GCNC ( 2002 ) [kg/d] 0,7 0 0.32

GRNC ( 2002 ) [kg/d] 0.1-0.5

10 − 2 to
3.10 − 2

US DOE ( 2003 )
[kg/d]

Beef cattle: 0.4-1.0
Dairy cows:
0.8-1.1

2.10 − 2 US EPA ( 2005 )

[kg/d]

Beef cattle: 0.5
Dairy cows: 0.4

0.37

11.3.3.2 Parameters for Estimating the Concentration in Animal Tissues

To calculate the concentration of a contaminant in animal tissues it is necessary to
know the oral absorption, elimination and degradation rates, or the biotransfer or
bioconcentration factors. The choice of values for these parameters is a very impor-
tant step in the modelling. These parameters may vary with the animal (type of ani-
mal, age, gender) and with the contaminant (type of contaminant, speciation, dose).
The determination of these parameters, which is done by performing experiments on
domestic animals, is relatively difficult, time-consuming and, therefore, expensive.
As a consequence, data for these input parameters are rare and uncertain. Values
collected from the literature may vary over several orders of magnitude. When
empirical data are lacking, regression models are often used to predict values of
bioconcentration or biotransfer coefficients. To reduce uncertainties on biotransfer
and bioconcentration coefficients, data and regression relationships collected from
literature have to be critically reviewed in order to make sure that the experimental
protocol used to generate the data was relevant and that the data are of good quality.

Method for Estimating Depuration and Absorption Rates

From the last day of animal exposure, the depuration rate can be estimated from the
slope of the curve where the concentration in tissue is plotted against time.
When I a = 0 and M a, 1 is constant, this yields:

dC a ,1
dt

=− × C a ,1 (11.36)

in which

depuration rate [d − 1 ]

= k a + λ a =