Protocol for evaluating whether a test substance shows bioaccumulative properties
using Haber’s Law
Dr James E. Cresswell, University of Exeter (January 2017)
Overview
This protocol analyses the data describing mortality of honey bees during a chronic 10-day
oral exposure paradigm as specified in the OECD Test Guideline: Honey bee (Apis mellifera
L.), Chronic oral toxicity test,10-day feeding test in the laboratory (Kling & Schmitzer 2015*).
The protocol tests the relationship between the concentration of the test substance in the
honey bees’ diet, denoted C, and the duration of exposure that is required to produce a
specified level of injury (mortality) among the exposed bees, or the ‘time-to-effect’, denoted t.
Specifically, the protocol derives a t-vs.-C relationship from the data and determines its slope
when plotted on logarithmic axes, which indicates whether the data conform to Haber’s law.
The analysis can identify potentially bioaccumulative test substances, because they are not
expected to conform to Haber’s law.
*Kling, A. & Schmitzer, S. 2015: Proposal for a new OECD guideline for the testing of
chemicals on adult honey bees (Apis mellifera L.) in a 10 day chronic feeding test in the
laboratory and results of the recent ring test 2014. Julius-Kühn-Archiv 450: 69-74.
Principles of the analysis
Toxicants that conform to Haber’s law will produce the specified injury from any exposure
whose dose-duration combination conforms to a ‘constant product’ rule:
𝐶𝑡 𝑏 = 𝑘
Eq. 1
where the exponent takes the value b =1. Toxicological experiments on a system that
conforms to Haber’s Law will find that halving the dosage rate doubles the duration of the
exposure that is required to achieve a given level of injury.
By contrast, sustained exposure to bioaccumulative toxicants can produce ‘time-reinforced
toxicity’ (TRT), which occurs when the exponent in Eq 1 takes the value b >1. In a
toxicological system where b > 1, halving the dosage will require less than double the
duration of the exposure to achieve the given injury because time reinforces toxicity. When
TRT occurs, sustained exposures are disproportionately injurious relative to acute
exposures, which has implications for risk assessment if environmentally realistic exposures
are sustained.
When the t-vs.-C relationship is plotted on logarithmic axes, rearrangement of Eq. 1 reveals
that:
log(𝐶) = −𝑏 ∙ log(𝑡) + log(𝑘)
Eq. 2
Therefore, a constant-product relationship (e.g. Haber’s Law) appears as a straight line with
a slope of -1 (Fig. 1). In an idealized bioaccumulative toxicant that produces TRT, the slope
takes the value of -2 (Fig. 1; Appendix 1).
Haber’s Law test: 2
Fig. 1. The effects of reducing the dietary dose by a factor of 10 in two hypothetical
toxicological systems. In both systems, a continuous dose at dietary concentration C = 10
g L-1 produces a specified level of injury in t = 1 days; these data are indicated by the
square symbol. In the system that conforms to Haber’s Law, reducing dose by a factor of 10
to C = 1 g L-1 extends the exposure time required to produce the specified injury by a factor
of 10 to t = 10 days (indicated by the filled circle; the slope of the solid line is -1). In the
system that produces TRT, reducing dose by a factor of 10 to C = 1 g L-1 extends the
exposure time required to produce the specified injury only to t = 10 = 3.2 days (indicated
by the open circle; the slope of the dashed line is -2).
The protocol described below conducts a graphical test for conformity to Haber’s Law in data
comprising the observed time-to-effect, t, of a specified lethal endpoint (e.g. 50% mortality)
in batches of honey bees that were exposed to the test substance at various dietary
concentrations, C.
Implementation instructions
1. OBTAIN THE SOFTWARE
Open your EXCEL software and ensure that the Solver add-in package is installed.
If Solver is installed, it will appear at the right of the upper tool bar when the Data tab is
clicked (Fig. 2). If Solver is not visible, you will need to install the add-in; instructions are at:
https://support.office.com/en-us/article/Load-the-Solver-Add-in-612926fc-d53b-46b4-872ce24772f078ca
Fig. 2. Data tab tool bar in EXCEL showing Solver at right under Data Analysis
Haber’s Law test: 3
2. OPEN THE SOURCE AND ANALYSIS FILES
Open two* EXCEL files: the first should contain the raw data from a chronic 10-day oral
exposure of honey bees conducted according to the OECD draft protocol and it will be
organized in the spreadsheet using the conventional format of a professional testing
laboratory (here, an example file will be referred to as Test data example.xls); and the second
should be Haber’s Law test v1.xls, which provides a framework for implementing this data
analysis.
*If you intend to analyse data that is not organised in the conventional format, ignore step 3
COPY THE RAW DATA FROM THE SOURCE FILE and ignore step 4 PASTE THE RAW DATA INTO THE
ANALYSIS FILE and instead follow the instructions in APPENDIX 1: NON-STANDARD DATA before
proceeding to step 5 FIT A TIME-RESPONSE RELATIONSHIP.
Both files can be obtained from the University of Exeter’s online public-access archive (ORE)
at the using the following URL: to be confirmed
3. COPY THE RAW DATA FROM THE SOURCE FILE
Using Test data example.xls, select the worksheet Tab. Exact mortality data (Fig. 3) and then
highlight all the data in the blue-bordered box, which is cell range A1:S58.
Cell A1 is sometimes reduced in size, so place your cursor carefully.
Fig. 3. Table of mortality and
behavioural abnormalities of the
bees (raw data) in the worksheet
Tab. Exact mortality data. The
table extends below that shown.
Copy the data in the cell range A1:S58 using the Ctrl-C keystroke.
4. PASTE THE RAW DATA INTO THE ANALYSIS FILE
Using Haber’s Law test v1.xls, select the worksheet Tab exact mortality data, locate your cursor
in cell A1 and paste in the previously copied data (keystroke Ctrl-V), which will occupy the
cell range A1:S58.
In the same file, select the tab DATA EXTRACTION, which should now be populated (Fig. 4).
Haber’s Law test: 4
Fig. 4. After the copy-and-paste
operation, the raw data appear in a
worksheet under tab DATA
EXTRACTION.
Check that the operation has worked successfully by locating a distinctive pattern in Test data
example.xls#Tab. exact mortality data worksheet that recurs in the Haber’s Law test v1.xls#DATA
EXTRACTION worksheet (N.B. here the ‘#’ symbol denotes a worksheet in the preceding
file). For example, if the original numbers of dead bees at day 4 in the three replicates of the
highest dose are 10, 3 and 1 (Fig. 3, cell range N8:N10), the pattern recurs in DATA
EXTRACTION (Fig. 4, B6:D6) and as percentages (100, 30, 10) in the same worksheet (Fig.
4, B17:D17).
The subsequent analysis uses the mean levels of mortality in the replicates of each dose.
These means appear as a table in cells B27:F36 in the Haber’s Law test v1.xls#DATA
EXTRACTION worksheet. Specifically, cells B26:F36 are column headings that indicate dose
concentration, C, and cells A27:A36 are row headings that indicate the time (t days) at which
mortality measurements were recorded.
5. FIT A TIME-RESPONSE RELATIONSHIP FOR THE HIGHEST CONCENTRATION OF DOSE
Using Haber’s Law test v1.xls, select worksheet C1.
Fig. 5. Worksheet C1 showing
chart, data and the highlighted
cells before fitting.
The chart (Fig. 5) shows the mean levels of mortality (y-axis, %) in relation to exposure time
(x-axis, days).
The filled symbols are derived from the data in the DATA EXTRACTION worksheet.
The solid line is a fitted sigmoidal relationship that is controlled by values of the variables in
the yellow-highlighted cells (I6:I8): mortality = max / [1 = exp(-α(t – t50))]
Haber’s Law test: 5
The goodness of fit is indicated by the R-squared value that appears in the orangehighlighted cell (I2).
Select the Data tab from the upper toolbar and then click the Solver button at the far right of
the toolbar so that the Solver parameters dialogue box appears as in Fig 6.
Fig. 6. The Solver parameters dialogue box.
Our aim is to maximise the value of R-squared by changing the values of the yellowhighlighted variables, so we complete the fields of the dialogue box as follows.
Set objective: using the red-arrow button, return to the spreadsheet and click in the orangehigh-lighted cell, then use the red-arrow button to return to the Solver parameters dialogue
box; the Set Objective field should now contain the cell location $I$2.
To: click the radio button next to Max
By Changing the Variable Cells: using the red-arrow button, return to the spreadsheet and
use your mouse to select the cell range of the yellow-highlighted cells, then use the redarrow button to return to the Solver parameters dialogue box; the By Changing the Variable
Cells field should now contain the cell range $I$6:$I$7.
If necessary, you can type the cell range into the field of the dialogue box directly.
Click the Solve button with your mouse (or use the <carriage return> keystroke) to see the
Solver Results dialogue box (Fig. 7).
Fig. 7. The Solver Results dialogue box
Check that the Keep Solver Solution radio button is checked and click the Ok button with
your mouse (or use the <carriage return> keystroke).
Haber’s Law test: 6
Check the chart in the worksheet and ensure that the fitted sigmoidal relationship (solid line)
adequately describes the data points (i.e. R-squared >0.6).
6. FIT TIME-RESPONSE RELATIONSHIPS FOR THE OTHER CONCENTRATIONS OF DOSE (AND
TROUBLESHOOTING)
Repeat the Solver operations described in (5) above for each of the worksheets successively
under tabs C2, C3, C4, …, Cn (see examples in Fig. 8).
Fig. 8. Examples of time-dependent mortality (solid circles) and fitted sigmoidal models (solid
lines) produced by implementing Solver on each of five worksheets, C1 – C5; x-axis: days; yaxis: % mortality.
Do not perform the Solver operation if no bees died during the exposure (i.e. the entries are
all zero in the cell range B3-B12. It will do no harm if you implement Solver on a worksheet
with zero entries in this range, but Solver will report that it could not find a solution; if you
have implemented Solver, simply accept this outcome by clicking the Ok button with your
mouse (or by using the <carriage return> keystroke
Occasionally, Solver will return a solution where the fitted sigmoidal relationship clearly fails
to describe the data. In this case, you should adjust the values in the yellow-highlighted
cells (I6, I7) by hand until an approximate fit is obtained and then implement Solver again.
Initial values of cells I6 and I7 for an approximate fit can be obtained in either of two ways.
First, you can choose initial values by looking at cells I6 and I7 in the worksheet of the next
higher dose; i.e. if Solver failed to find a solution on worksheet C4, enter the values of the
Solver solution from worksheet C3 (values for cells I6 and I7) into cells I6 and I7 on
worksheet C4 and implement Solver again.
Second, you can guess the approximate solution by understanding the meaning of the
parameters. For example, spreadsheet cell I7, or parameter t50, is the value on the t-axis at
which the y-value of the sigmoidal function reaches half its maximum; spreadsheet cell I6, or
parameter α, controls the steepness of the sigmoidal curve: very shallow increases over the
Haber’s Law test: 7
relevant range in t might be described by α = 0.1. As a guide, the values corresponding to
the fitted curves in Fig. 8 are shown below in Table 1.
Dose
C1
C2
C3
C4
C5
alpha
0.957
0.732
0.583
0.225
0.149
t50
4.6
4.8
10.8
18.2
26.1
R-squared
0.98
0.95
0.97
0.90
0.50
Table 1. Parameter values for the least-squares best-fit sigmoidal relationships between
percentage mortality and duration of exposure (time in days), which are displayed in Fig. 8.
7. SAVE A COPY OF YOUR SPREADSHEET
Some of the manipulations that you will now make can cause irreparable changes to your
spreadsheet, but you may nevertheless wish to return to your original. Before proceeding to
step 8 below, save a copy of your spreadsheet as an archive and save another with a
different file name to be the working version.
8. CHOOSE AN ENDPOINT FOR TESTING HABER’S LAW
For statistical reliability, the conventional toxicological endpoint is L50, or 50% mortality
among the exposed bees. In a typical dataset, it is unlikely that exposures at the lower
doses will achieve L50, so that time-to-effect cannot be quantified if the specified effect is L50.
Other endpoints of lower mortality, such as L30 (i.e. 30% mortality) will allow more dose
levels to be included in your analysis, but they are also increasingly affected by stochastic
noise, so we must make a trade-off between including doses and incorporating less reliable
patterns.
Select the DATA EXTRACTION worksheet and inspect the means of the mortality-time data,
which are presented as a table in cells B26:F36 (Fig. 9)
Fig. 9. Mean mortality (%) for each dose is presented in
a column with the mortality achieved after 10 days in
row 36 (Dose 1 in B36, dose 2 in C36, etc.).
Aim to choose an endpoint that will enable at least four levels of dose to generate a time-toeffect.
Haber’s Law test: 8
In the example (Fig. 9), we could choose L10, but this would mean that the designated effect
(10% mortality) could be achieved in any cage of 10 bees by the death of a single individual,
which is unreliably stochastic.
In the example (Fig. 9), we could choose L30, but this would mean that the designated effect
(30% mortality) could be achieved only by three doses, even after 10 days of exposure.
A reasonable compromise is to choose L20 as the endpoint and to allow the time-to-effect in
dose C = 0.4 to be estimated by a small extrapolation (it will be slightly longer than 10 days
and so it has not been directly observed).
9. FINALISE THE DATASET FOR TESTING HABER’S LAW
In this step, all cell references apply to worksheet LOG-LOG plot.
Select worksheet LOG-LOG plot.
If your chosen endpoint is Lx, enter the value x in cell B4. For example, if choosing L20, enter
20 in B4.
Inspect the table in cells F23:J28 (Fig. 10)
TREATMENT
1
2
3
4
5
C
1.00
0.80
0.60
0.40
0.20
t
3.13
2.86
8.40
12.02
16.83
Status
INCLUDE
INCLUDE
INCLUDE
Fig. 10. Observed C-vs-t relationship
Column headed C lists the doses in the experiment.
Column headed t lists the estimated times to the specified effect, Lx.
Column headed Status lists whether the dose achieves the specified effect within the
allowed extrapolation.
If less than four dose treatments have Status = ‘INCLUDE’, then it may be possible to gain a
result from additional dose by making an extrapolation; you can include an extrapolation of
up to y% above observed mortality by entering the value y in cell B6.
This means that you will allow time-to-effect to be estimated for doses that have achieved (xy)% mortality after 10 days. In most cases, the sigmoidal relationships fitted to the timemortality mortality data are strong enough to justify extrapolation to an extra 10% of mortality
(i.e. one additional bee death), so you may reasonably enter the value 10 in cell B6.
If the extrapolation changes the Status entry to ‘INCLUDE’ for dose treatment Cn, you can
justify this assumption by checking the goodness of fit of the sigmoidal relationship, which is
given by its associated R2, or R-Squared. The R2 value is found in cell I2 of worksheet Cn,
and it should be greater than 0.8 if you intend to extrapolate safely.
Haber’s Law test: 9
For example, using the data displayed in Fig. 9, a time-to-effect of L20 for C = 0.4 will be
calculated by the spreadsheet by extrapolation even though the observed mean mortality
after 10 days was only 13.3%, but only if the permitted extrapolation was set at 10% in cell
B6. The result of allowing this extrapolation is shown in Fig. 11.
TREATMENT
1
2
3
4
5
C
t
1.00
0.80
0.60
0.40
0.20
Status
INCLUDE
INCLUDE
INCLUDE
INCLUDE
3.13
2.86
8.40
12.02
16.83
Fig. 11. Tabulated C-vs-t relationship
after allowing a 10% extrapolation.
Delete any entries in column t (H24:H28) that do not show ‘INCLUDE’ in the Status column
of the same row (J24:J28) – see Fig. 12.
TREATMENT
1
2
3
4
5
C
t
1.00
0.80
0.60
0.40
0.20
Status
INCLUDE
INCLUDE
INCLUDE
INCLUDE
3.13
2.86
8.40
12.02
Fig. 12. Tabulated C-vs-t relationship
after deleting an entry under t that did not
show INCLUDE in the corresponding
Status column. Value of 16.73 deleted –
compare with Fig 11.
10. TEST CONFORMITY TO HABER’S LAW
In worksheet LOG-LOG plot, inspect the chart (Fig. 13), which shows the best-fitting
constant-product relationship (solid line) in relation to the observed data (solid circles).
If necessary, adjust the maximum and minimum values of the axes so that all the relevant
data points are displayed. For example, if after all adjustments and extrapolations the table
shown in Fig. 12 has four data points designated by ‘INCLUDE’, then adjust the axes until
four data points appear on the graph.
The key result is the exponent, b, of the fitted relationship, y = ax-b.
Haber’s law is confirmed if the exponent of the fitted relationship (y = ax-b) approximates one,
i.e. b  1.
Log-log plot
10.0
y = 1.5912x-0.515
R² = 0.8696
C 1.0
0.1
1
10
t
100
Fig. 13. Chart showing the best-fitting
constant-product relationship (solid line) in
relation to the observed data (solid circles).
The inset legend gives the equation of the
best-fit relationship (y = 1.5912x-0.515) and the
corresponding value of R-squared (R2 =
0.8696).
Haber’s Law test: 10
In the example of Fig. 13, there is no evidence for time-reinforced mortality
(bioaccumulation), because b  - 0.5, which implies that the exposure time-to-effect at low
doses was disproportionately long when compared to the higher doses.
Haber’s Law test: 11
APPENDIX 1: PHARMACOLOGICAL FOUNDATIONS OF THE TEST
The exposition below draws on previously published ideas, and most notably those of: Bliss
(1941), who developed the logarithmic plots for analysing C-vs-t relationships; Rozman,
Doull & Hayes (2001), who proposed a geometric approach (rectangular vs. triangular toxic
loads) to link bioaccumulation of the toxicant to the accumulation of injury; and Tennekes
(2010), who first pointed out the utility of Haber’s Law for assessing the risk of pesticides to
bees.
When a toxicant binds reversibly to its target site and is susceptible to catabolic breakdown
and elimination, then during a sustained dietary exposure establishes a ‘steady state’
internal concentration inside the organism. If the daily rate of injury resulting from this
steady state is constant, a simple pharmacokinetic compartment model of toxic load (Fig.
A1.1) predicts that the accumulated total injury is proportional to the duration of the exposure
(Fig. A2.2).
Fig A1.1. Compartment model of pharmacokinetics during dietary exposure to a toxicant. Assume
that the pharmacokinetics of the toxicant in an animal’s body are governed by this simple compartment
model. The animal ingests the toxicant at a dose rate of d ng d-1 and assume that the animal’s
detoxification enzyme system has surplus capacity, which means that the rate of the detoxification is
proportional to the internal concentration of the toxicant, C. Hence, the toxicant is detoxified
metabolically (or otherwise eliminated from the animal’s body) with first order dynamics at a rate of
eC ng d-1. Let R denote the concentration of target receptors bound by the toxicant and assume that the
formation of the toxicant-receptor complex is governed by coefficients of association and dissociation,
denoted TA and TD respectively so that the rate at which the toxicant binds to receptors is R/TA., etc.
Assume that the animal incurs irreversible injury at a daily rate Ri. The total injury incurred by the
organism is denoted by circular box I (the circle is used to distinguish a box that accumulates an effect
from one that accumulates a mass) and the oblique arrow into the circular box indicates transfer of
influence, not mass.
Haber’s Law test: 12
Fig A1.2. Pharmacokinetics of an idealized non-bioaccumulative toxicant. When the toxicant binds
reversibly to its target site and is substantially susceptible to catabolic breakdown and elimination,
then during a sustained dietary exposure the continuous and opposing actions of ingestion and
elimination will establish a ‘steady state’ concentration inside the organism and C is constant. Since
R is proportional to C, R is also constant over time and injury accrues at a constant rate. Hence, I  t
and kI = Ct, where t denotes the duration of the exposure. In this hypothetical discrete-time example:
d = e = 1; TA = TD = 1.
This proportionality under steady-state conditions means that toxicological experiments on
such a system will find that halving the dosage rate doubles the duration of the exposure that
is required to achieve a given level of injury (Fig. A1.3).
Fig. A1.3. The rectangular geometry of toxic load in an idealized non-bioaccumulative toxicant.
The total injury across the exposure, or toxic load, is proportional to the area under the curve (AUC)
of the plot of C over time, which can be visualized as a rectangular geometry with area C× t (grey
fill). Consider two groups of animals that feed separately on diets whose toxicant concentrations
differ by a factor of α (i.e d1= d2 / α); in this hypothetical example, α = 2). If the feeding rates on the
diets are equal and the animals on the more toxic diet have an internal concentration of toxicant C1,
the internal concentration of toxicant of those that feed on the less toxic diet is C1/α. Assume that
animals feeding on the more toxic diet reach a given level of injury (toxic load) in t1 days and those in
the less toxic diet reach the same level in t2 days (in this hypothetical example, t1 = 2 days). Since the
AUCs have rectangular geometry, then for both groups to experience the same injury, those on the
Haber’s Law test: 13
less toxic diet must be exposed for t2 = αt1 days (i.e. t2 = 4 days). Generalisation for all conforming C
and t combinations yields Ct = k. Hence, subjects exposed to perfectly non-bioaccumulative toxicants
in appropriate ‘time-to-effect’ experiments will exhibit outcomes that conform to a constant-product
law of Cbt = k where b =1, which is Haber’s Law.
Toxicants with these properties will produce the specified injury from any exposure whose
dose-duration combination conforms to a ‘constant product’ rule known as Haber’s Law:
𝐶𝑡 𝑏 = 𝑘
Eq. 1
where C denotes the concentration of the toxicant in the diet, t denotes the duration of the
exposure and the exponent takes the value b =1, which reflects the proportionality
relationship.
If instead the organism’s internal concentration at the target site rises as intake proceeds
during the exposure because the toxicant bioaccumulates, the rate of injury increases with
time and so the accumulated total injury is not proportional to exposure time but instead
increases quasi-exponentially as a power function (Fig. A1.4).
Fig. A1.4. Pharmacokinetics of an ideal bioaccumulative toxicant. When the toxicant is not
susceptible to catabolic breakdown and elimination, then during a sustained dietary exposure
continuous ingestion will cause an accumulation of toxicant inside the organism and C increases over
time. Since R is proportional to C, R also increases over time and injury accrues at an increasing rate
as exposure progresses, which is ‘time reinforcement’. In this hypothetical discrete-time example: d
= 1; e = 0; TA = TD = 1.
Haber’s Law test: 14
Since the rate of injury accelerates over time towards the level required to produce a given
effect, it exhibits TRT. In this case, the exponent in Eq. 1 takes the value b > 1. In a
toxicological system where b > 1, halving the dosage will require less than double the
duration of the exposure to achieve the given injury because time reinforces toxicity. In the
pharmacokinetic model of an idealized bioaccumulative toxicant, the exponent in Eq. 1 takes
the value b = 2 (Fig. A1.5).
Fig. A1.5. Triangular geometry of toxic load in an idealized bioaccumulative toxicant. Given
constant ingestion of a bioaccumulative toxicant, let the internal concentration at time t be given by:
𝐶 = 𝛽𝑡
The total injury across the exposure, or toxic load, is proportional to the area under the curve (AUC) of
the plot of C over time, which can be visualized as a triangular geometry with area 0.5t × C (i.e. half
base × height). Consider two groups of animals that feed separately on diets whose toxicant
concentrations differ by a factor of α. If the feeding rates on the diets are equal, the animals on the more
toxic diet have an internal concentration of toxicant C1 = βt1 and those on the less toxic diet have C2 =
(β/α)t2. Since the AUC has triangular geometry, then for both groups to experience the same injury we
require:
0.5𝑡1 × 𝛽𝑡1 = 0.5𝑡2 ×
𝛽
𝑡
𝛼 2
Simplification yields:
𝑡12
𝑡22
=
𝛼
Multiplying both side by C1 yields:
𝐶1 𝑡12 =
𝐶1 2
𝑡
𝛼 2
Recall that the internal concentrations differ by a factor of α, so that:
Haber’s Law test: 15
𝐶1 𝑡12 = 𝐶2 𝑡22
Generalisation for all conforming C and t combinations yields C2t = k. Hence, subjects exposed to
perfectly bioaccumulative toxicants in appropriate ‘time-to-effect’ experiments will exhibit outcomes
that conform to a constant-product law of Ctb = k where b =2.
Taking logarithms of both sides of Ct2 = k and rearranging yields:
log(𝐶) = −2 log(𝑡) + log(𝑘)
Therefore, an ideal bioaccumulative toxicant delivered in a time-to-effect experiment will produce a Cvs.-t relationship with a slope of -2 on log-log axes.
References
Bliss, C. I. 1941. The relation between exposure time, concentration and toxicity in experiments on
insecticides. Annals Of The Entomological Society Of America 33:721-766.
Rozman, K. K., J. Doull, and W. J. Hayes. 2001. Dose, time, and other factors influencing, toxicity.
Pages 1-95 in R. I. Krieger and W. C. Krieger, editors. Handbook of Pesticide Toxicology (2nd
ed.) Academic Press, London.
Tennekes, H. 2010. The significance of the Druckrey–Küpfmüller equation for risk assessment - The
toxicity of neonicotinoid insecticides to arthropods is reinforced by exposure time. Toxicology
276:1-4.
Haber’s Law test: 16
APPENDIX 2: NON-STANDARD DATA
It is straightforward to use the spreadsheet Haber’s Law test v1.xls to analyse a t-vs.-C
relationship without extracting the raw data as described above in steps 3 and 4. Select the
DATA EXTRACTION worksheet and type your concentrations in cells B27:F26 and the
mortality-time data (mean % mortality) directly into cells B26:F36 (Fig. 9). Then proceed from
step 5.