FS 14
Optional Test: Calculations of PMI using ADH
Objective: To model the activities of a forensic entomologist and use math and science skills to determine a
more accurate PMI.
The scenarios you worked on in the Forensic Entomology Activity were simplified. In reality, most
crime scenes involving insect data are much more complicated. Unless the victim is in an environment that is
constant, as in Case number 2, temperature fluctuations during the course of the day or week will affect instar
development and the PMI (time since exposure).
In order for a forensic entomologist to come up with charts like the one in the table below, they must
rear the species of fly they are interested in at a standard temperature and
determine the time of development for each instar. This can be done in
mesh cages (see Bio Quips’s Bug Dorm to right) with raw meat as a food
source. The cage is placed in a rearing chamber where light, temperature
and humidity can be carefully controlled. Unlike the chamber, the
temperature at a typical crime scene fluctuates over time. Temperatures
can be significantly different from one day to the next. These temperatures
can have a significant effect on the development of maggots and the PMI.
Higher temperatures generally speed development (reduce PMI) while
lower temperatures slow development (increase PMI). All insects have a
minimum threshold temperature below which they will not develop at
all.
A technique for determining development times is to use
accumulated degree hours (ADH), or accumulated degree days (ADD).
ADH is computed by multiplying the number of hours to development by the temperature in oC. Since the time
required for development decreases as the temperature increases, the total number of ADH required for a
maggot to develop to any given instar remains constant.
For example, if Musca domestica maggots reared at 22 oC require five days to reach the 3rd instar, the
ADD for a 3rd instar would be (5) x 22 oC = 110. The ADH would be (5 x 24) x 22 oC= 2,640. If we raise the
temperature, the time for development should decrease. At 26.7 oC, the ADH is still 2,640. To reach the 3rd
instar at this temperature would require 2,640/26.7= 99h or only 4.1 days.
How does this help the forensic entomologist? At the crime scene, samples of maggots are collected and
the instar number determined by length of the larvae or size and shape of the posterior spiracles on the larvae.
Data from laboratory rearings of the same species of maggot are used to determine the ADH for each stage.
The following chart shows the developmental times and ADH for the stages of the calliphorid fly, Phormia
regina reared at 22oC.
If the oldest maggots collected at the
Average ADH Needed for Completion of Developmental
crime scene are newly hatched, 3rd instars,
Stages Phormia regina at 22 oC
the ADH for the maggots is 1540. ADH is
Stage
Hours
ADH
calculated using the ADH for the end of the
previous stage if it is an early instar and the
Egg
20
440
ADH for the end of the current stage if it is a
1st instar
25
990
later instar. If the temperature at the crime
scene has been an average of 22 oC, the body
2nd instar
25
1540
was first exposed to fly activity 70 hours
rd
Feeding3 instar
25
2090
earlier (20+25+25). If the insects on the
rd
body were collected at 10:00 am on Sunday,
Postfeeding 3 instar
125
4840
based on the calculations, the body was
Pupa
116.5
7403
exposed at approximately 12:00 pm the
previous Thursday (70 hours earlier!). The PMI is 70 hours or approximately 3 days.
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Most maggots are collected in an environment that has a fluctuating temperature. The PMI will have to
be calculated differently to account for the fluctuations. Weather station data from the area closest to the crime
scene can be obtained from NOAA or other weather agencies. The average temperature for a number of days
before the body was discovered can be obtained and the ADH for each day calculated.
For example, a body of a victim is discovered in the afternoon on Tuesday, July 15. Insect samples were
collected from the body and placed in alcohol jars at 3:00 pm. These jars, along with weather data from the
previous days at the site were sent to the forensic entomologist. The oldest instars were determined to be early
3rd instars of Phormia regina. The ADH for early 3rd instars (late 2nd instars) should be 1540 regardless of the
temperature at the crime scene. Weather data from the scene averaged 20 oC on Tuesday, 18oC on the previous
Monday (July 14th), 19oC on Sunday and 23oC on Saturday.
Calculation for the PMI are as follows:
1) Determine the ADH for each day prior to the discovery of the crime scene:
Day
Hours
Tuesday,
July 15
Monday,
July 14
Sunday
July 13
Saturday
July 12
15 h
(12am-3pm)
24h
Average Daily
Temperature from
NOAA
20oC
18oC
24h
19oC
24h
23oC
Daily
ADH
Σ ADH
300
(15 x 20)
432
300
456
(19 x 24)
552
732
(300 + 432)
1,188
1540 ADH
1740
(1,188+552)
Based on the chart above and the early 3rd instar developmental ADH of 1540, the time that the body was
exposed to insect activity was somewhere between Sunday, July 13 th (12:00am) and Saturday, July 12
(12:00am). That still leaves a 24 hour period to deal with. Most alibi’s don’t cover an entire 24 hour period
even if the suspect is innocent. You need to correct for partial days to get a closer time interval.
2) Correct for partial days to determine when the body was first attacked by insects:
You have two ways to calculate the difference in development times. You know that the body was
probably left between 12 am Saturday or 12 am Sunday. You also know that the 1540 ADH needed for growth
of the oldest larvae is closer to the Saturday 12 am ADH (1740) than to the Sunday 12 am ADH (1,188). You
just need to subtract the 1540 ADH from either and determine how many hours to add or subtract from each
time. You have two options:
Determine the difference between the 1540 ADH and
Determine the difference between the 1540 ADH and
the Saturday am ADH, convert to hours and add to
the Sunday am ADH, convert to hours and subtract
Saturday 12:00 am.
from Sunday 12:00 am.
 1740 - 1540 = 200 ADH needed on Saturday
 1540-1188= 352 ADH not needed on Saturday for
rd
for growth to the early 3 instar.
growth to the early 3rd instar.
 If 23 oC x hours = 200 ADH, then
 If 23 oC x hours = 352 ADH, then
200 ADH / 23 oC = 8.7 h
352 ADH / 23 oC = 15.3 h
 Add 8.7 h + 12 am Saturday= 8.7 am Saturday
 Subtract 15.3 h from 12 am Sunday = 8.7 am Sat
Either way, the results indicate that the body was exposed to insect activity at around 8:30-9:00 am
on Saturday, July 12th.
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3) Calculate the PMI: You can now calculate the PMI or post mortem interval to identify the number
of days the body has been exposed to insect feeding.
The PMI is the number of hours or days since death. In this situation, it is, more correctly, the time that
the victim has been exposed to feeding insects. To calculate, just add the number of hours required to achieve
1540 ADH:
Tuesday 15 h + Monday 24 h + Sunday 24 h + Saturday 15.3 h = 78.3hours or 78.3/24= 3.3 days
(24h-8.7h)
In some cases, this calculation coincides with the true PMI but in other cases, it might not. If a body is
held in a freezer for a month before being dumped in a wooded area, the PMI will only determine the time the
body was exposed to flies. A freezer is not a very welcoming place for an insect to grow and all insects have a
minimum threshold temperature below which they will not develop at all. For example, if the minimum
threshold temperature for P. regina is 12 oC or 54 oF (http://www.onlineconversion.com/temperature.htm ),
eggs will not hatch and larvae will not move, feed or grow in temperatures lower than 12 oC. A freezer would
certainly have a temperature lower than the minimum threshold temperature for most insects.
(For more information:
http://www.key-net.net/users/swb/forensics/index.htm )
Directions:
You must have my permission to attempt this alternate assignment. You may work with ONE other student.
EACH student should: 1) Read the following case report and fill out a “Forensic Entomology Data Form”
(you can find this on my teacher site), 2) Write (do not type) your calculations into this document and 3) Type
up a Forensic Entomology Case Report similar to the cases we discussed in class. It will be imperative that you
explain EXACTLY what you did to calculate the TOD for the victim.
You and your partners Data Forms and calculations should be the same or very similar but you must have
unique reports. No copies are allowed. All materials should be hand written except for the Case Report.
Case # 890223
Jane Smith’s body was found in a deserted building in Jersey City, NJ (Essex County) on March 13 th, 2009.
The body was identified by her boyfriend who stated she had fled his apartment 2 days ago because she could
not overcome her addiction to drugs. The boyfriend stated her age at 23 years. The body was fully clothed, in
jeans and a long-sleeved flannel shirt. There was no evidence of wounds or animal scavenging. The building
did not have windows so the body was exposed to open air. The body was in an active state of decay. One
maggot mass was observed near the mouth of the decedent and 15 larvae were collected at 4:00 pm. Ten
maggots were immediately preserved for later measurement and five were secured in a Maggot Motel to
determine the species of flies. No pupae were present on the body. The temperature over the past 10 days was
obtained from NOAA and will be available for your use.
2) Five of the collected larvae were reared to adult and identified to be of two different species: Phormia
regina and Sarcophaga bullata:
Phormia regina
Sarcophaga bullata
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3) The following information will help you to determine the PMI in this case. You will need to fill in any
missing information in the charts below.
Table 1 . Range in length (mm) for instars of Phormia regina*:
Length
Instar
0.6 to 4.0 mm
1
3.0 to 8.0mm
2
7.5 to 15mm
3
Table 2 . Range in length (mm) for instars of Sarcophaga bullata*:
Length
Instar
0.5 to 5.0mm
1
4.0 to 10.0mm
2
9.5 to 20mm
3
*
The minimum threshold temperature for both species is 12 oC.
Table 4. Time for Phormia regina larval and pupal development at 22 oC.
Stage
Cumulative Time to
Instar
Cumulative ADH at END of Stage
Egg
20 hrs
440
1st instar
25 hrs
990
2 instar
25 hrs
1540
3rd instar
25 hrs
2090
Prepupa
125 hrs
4840
Pupa
116.5 hrs
7403
Total
7 days
11099
nd
 end of 1st instar, beginning
of 2nd instar.
Table 3. Time for Sarchophaga bullata larval and pupal development at 27 oC. Note that data for S.
bullata was not available at 22 oC.
Stage
Cumulative
Time to
Instar
Egg
20 hrs
1st Instar
26 hrs
2nd Instar
18 hrs
3rd Instar
50 hrs
Prepupae
112 hrs
Pupae
12 days
Total
17 days
Cumulative ADH at END of stage
(20 x 27)=
You must fill in the ADH
for S. bullata. Some work
has been done for you.
1728
24894
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Table 4. Temperature data for the last week days from NOAA. Calculate the Daily ADH and the sum of
the daily ADH. Some dates have been done for you.
Day
03/13/09
Hours
16
Mean Temp. from NOAA (oC)
20
03/12/09
24
18
03/11/09
24
15
03/10/09
24
15
03/09/09
24
20
03/08/09
24
25
03/07/09
24
24
03/06/09
24
23
03/05/09
24
22
03/04/09
24
22
03/03/09
24
20
Daily ADH
320
∑ ADH
320
528
4208
4) The length of the larvae have already been measured for you by the criminalist who collected the larvae at
the crime scene. Complete the following table so you have all the information you need to determine the
PMI. Some answers have been done for you.
# of
Larvae
Species
1
Phormia regina
Larval
measurements
(mm)
3
2
P. regina
5
3
P. regina
Approx.
Instar
Minimum ADH for Development
1 mid
990-440=550/2=275+440=715
5
2 mid
1265
P. regina
8
1540
5
P. regina
8
2 late/ 3
early
2 late/ 3
early
6
Sarcophaga bullata
5
S. bullata
7
2 mid
1728-1242=486/2=243+1242=1485
8
S. bullata
10
2 late
1728
9
S. bullata
15
3 mid
10
S. bullata
15
3 mid
4
7
1540
Note: For any maggot that is a ‘mid’ stage, you will need to subtract the ADH from the previous stage
ADH to determine the number of ADH for that stage alone. Then divide by 2 to determine half the
number of ADH needed for development to the middle of the stage. Add that number of ADH to the
previous stage ADH and you will have the number of ADH to the middle of the stage in question.
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5) Questions to answer at this stage:
a. What was the scientific name of the oldest larva found at the crime scene? __________________________
b. What was the instar stage of the oldest larva(e) above? ________________________
c. What was the minimum ADH needed for development of that larva?______________________
d. What is the range in days for the PMI? _____________________________________________________
6) Correct for partial days to determine the time and date when the body was first attacked by insects.
Use both methods and check that your hours for each add up to 24.
(Remember that 3.15h is not equal to 3:15min. .15 h is equal to 9 minutes)
7) Calculate the final PMI.
8) Consider the case report. Was the boyfriend truthful in saying that his girlfriend went missing 2
days ago? If not, explain why the difference might have happened.
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Anterior End
Posterior End
Mouthparts
Life cycle of the blow fly Chrysomya megacephala
Maggot Facts:
In most cases, the adult fly is needed for species identification. The shape of mouthparts and posterior spiracles
on a maggot can sometimes be used to identify the species. Spiracles are an external breathing pore.
Surprisingly, they are on the posterior end of the insect. Spiracle shape and size are also used to determine the
stage of the instar.
Maggots feed in masses. Eggs are laid and hatch about the same time. The maggots lacerate tissues with their
hook-like mouthparts and inject salivary enzymes that digest tissues. The tissues become semi-liquid and can
be fed on by the maggot. In this way, a maggot mass is more efficient “feeding machine” than a single larva.
The maggot mass in a dead body can produce a considerable amount of heat. Temperatures in the mass may
rise as high as 53oC (127oF). In order to survive, maggots constantly move from the inside of the corpse where
they feed, to the outside of the mass to cool off. Thus, the mass is constantly in motion.
When maggots reach the final instar, they begin searching for a dry place to pupate. Most crawl to a higher
place but the larva of the cheese skipper fly arches its body and flings or pops its body away from the corpse!
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