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1. In Knoxville, Tennessee, there is a research facility, popularly known as 'the Body Farm', where research is conducted into the nature of human decomposition and the factors which affect the rate at which it occurs. Who was responsible for the creation of this facility? A. Ernest T Bass B. Gil Grissom C. Bill Clinton D. William Bass
2. What is studied in forensic palynology? A. Pollens and spores C. Soils B. Dust D. Fossilized micro-organisms
3. If you know what to look for, you can tell a male from a female skull. Which of the following statements is FALSE about a male skull? A. It is usually larger. C. It has a heavier jaw. B. It has a more prominent brow ridge. D. It has a more rounded chin.
4. In October 1974 part of a male torso was found floating in the River Thames in England. Several parts, including the head and hands, were missing so police could not use the usual methods of fingerprints, facial features and dental records to identify the corpse. How was it eventually identified? A. Presence of gallstones C. Skeletal characteristics on x-rays B. Blood type D. All the choices are correct.
1. In Knoxville, Tennessee, there is a research facility, popularly known as 'the Body Farm', where research is conducted into the nature of human decomposition and the factors which affect the rate at which it occurs. Who was responsible for the creation of this facility? A. Ernest T Bass B. Gil Grissom C. Bill Clinton D. William Bass
2. What is studied in forensic palynology? A. Pollens and spores C. Soils B. Dust D. Fossilized micro-organisms
3. If you know what to look for, you can tell a male from a female skull. Which of the following statements is FALSE about a male skull? A. It is usually larger. C. It has a heavier jaw. B. It has a more prominent brow ridge. D. It has a more rounded chin.
4. In October 1974 part of a male torso was found floating in the River Thames in England. Several parts, including the head and hands, were missing so police could not use the usual methods of fingerprints, facial features and dental records to identify the corpse. How was it eventually identified? A. Presence of gallstones C. Skeletal characteristics on x-rays B. Blood type D. All the choices are correct.
The official name of the body farm is the Tennessee Anthropological Research Facility (TARF) but some of Bass’s colleagues call it the Bass Anthropological Research Facility or BARF.
This case made legal history because it was the first time a body was identified without fingerprints or dental records. It belonged to a petty criminal called William Henry Moseley.
ObjectiveSWBAT determine the Post Mortem
Interval using data analysis.
Agenda1. Entomology Lab2. Data Analysis3. Exit Slip
Forensic Entomology Lab• Complete Examination of Taxa• Skip Separation of Taxa – there will be species
A and B samples at the front if you need to compare them
• Once you have completed the above get your sample from the front to complete the Analysis of Evidence section
• We will review the Data Analysis section together
• Answer Post Lab Questions when finished
Entomology Kit Climate Data Analysis Tutorial
Crosscutting Concepts
Updated 5/6/14
Degree-Hour DeterminationKnowns• Bodies discovered at 1:00PM on June 20• Insects collected at 3:00PM on June 20• Weather type (sunny, partly cloudy, overcast)• Weather events (rain, thunderstorms, snow)• Daily average temperature• Male and Female had the same species and lifecycles present
(Migrating 3rd Instar Species A, 2nd Instar Species B)Unknowns• Elapsed degree-hours for each day• Degree-hours for each life stage of both species• Cumulative degree-hours for each life stage of both species• Cumulative elapsed degree hours for each day• Which day the adult insect from both species laid its eggs• Earliest and latest time the insects began developing
Elapsed degree-hours for each day
• The collection time was 3:00PM on June 20, this tells us to multiply the first average temperature by 15 hours instead of 24
• Every other daily average temperature in the month will be multiplied by 24
Lab Procedure 2, Step 2: Determine the number of degree hours for each day using the weather
service data. To do this, multiply the average temperature times 24 hours for each day. This can be
performed in a spreadsheet.
Elapsed degree-hours for each dayDAY MAX MIN AVG DEPAR-
TURE FROM NORMAL
DEPAR-TURE FROM NORMAL
HEATING COOLING TOTAL WATER EQUIV
SNOW-FALL, ICE
PELLETS
SNOW, ICE PELLETS OR ICE ON GROUND
AVG SPEED (MPH)
AVG SPEED (KPH)
SKY COVER
SUNRISE-SUNSET
WEATHER OCCURENCES
PEAK WIND (KPH)
degree hours
1 18.3 10.6 16.1 -1.9 -1.0 2.2 0.0 0.00 0 0 6.3 10.08 6 0S 18.7
386.666667
2 19.4 12.8 15.0 -3.0 -1.7 3.3 0.0 0.09 0 0 13.8 22.08 6 0SE 26.2
360
3 17.2 9.4 12.8 -5.2 -2.9 5.6 0.0 0.19 0 0 17.2 27.52 8 5SW 33.3
306.666667
4 20.0 10.6 13.3 -4.7 -2.6 5.0 0.0 0.28 0 0 12.3 19.68 9 1W 38.6
320
5 21.1 12.8 19.4 1.4 0.8 0.0 1.1 0.00 0 0 11.1 17.76 7 1,2W 28.3
466.666667
6 25.6 16.7 20.6 2.6 1.4 0.0 2.2 0.00 0 0 8.1 12.96 6 0SW 24.3
493.333333
7 23.9 15.2 19.4 1.4 0.8 0.0 1.1 0.07 0 0 6.3 10.08 8 0S 16.7
466.666667
8 20.6 12.8 16.1 -1.9 -1.0 0.0 0.0 0.11 0 0 13.2 21.12 8 0S 27.9
386.666667
9 22.1 13.9 18.9 0.9 0.5 0.0 0.6 0.00 0 0 4.2 6.72 3 0SE 10.3
453.333333
10 25.0 12.6 16.1 -1.9 -1.0 0.0 0.0 0.00 0 0 8.88 14.208 6 0S 22.7
386.666667
11 22.1 8.9 15.3 -2.7 -1.5 0.0 0.0 0.67 0 0 19.6 31.36 6 3,5S 32.4
367.2
12 14.5 9.4 10.5 -7.5 -4.2 0.0 0.0 1.13 T 0 23.5 37.6 8 1,3,5SW 43.8
252
13 16.7 7.2 12.1 -5.9 -3.3 0.0 0.0 0.23 0 0 14.2 22.72 7 1,2S 29.8
290.4
14 19.3 9.3 15.0 -3.0 -1.7 0.0 0.0 0.02 0 0 10.5 16.8 8 1S 24.5
360
15 16.5 8.9 12.8 -5.2 -2.9 0.0 0.0 T 0 0 11.9 19.04 4 0SW 23
306.666667
16 18.9 10.7 13.3 -4.7 -2.6 0.0 0.0 0.00 0 0 6.4 10.24 3 0W 19.2
320
17 16.8 9.5 12.9 -5.1 -2.8 0.0 0.0 0.00 0 0 9.5 15.2 3 0S 27.2
309.6
18 19.4 10.7 16.4 -1.6 -0.9 0.0 0.0 0.00 0 0 11.1 17.76 2 0W 26.3
393.6
19 19.1 10.9 15.9 -2.1 -1.2 0.0 2.8 0.00 0 0 4.6 7.36 1 0SW 17.8
381.6
20 22.0 12.8 18.4 0.4 0.2 0.0 5.0 0.00 0 0 7.4 11.84 1 0W 23.1
276
Degree-hours for each life stage: Species A
Lab Procedure 2, Step 3: Determine the number of degree hours required for each life stage of both
species. To do this, multiply the number of hours by the degrees Celsius given in the table.
Temp °C Egg 1st Instar
2nd Instar Feeding 3rd Instar
Migrating 3rd Instar
Pupa
21 21 31 26 50 118 240
21*21 = 441
31*21 = 651
26*21 = 546
50*21 = 1050
118*21 = 2478
240*21 = 5040
Degree-hours for each life stage: Species B
Lab Procedure 2, Step 3: Determine the number of degree hours required for each life stage of both
species. To do this, multiply the number of hours by the degrees Celsius given in the table.
Temp °C Egg 1st Instar
2nd Instar Feeding 3rd Instar
Migrating 3rd Instar
Pupa
21 25 37 31 60 124 286
25*21 = 525
37*21 = 777
31*21 = 651
60*21 = 1260
124*21 = 2604
286*21 = 6006
Cumulative degree-hours for each life stage: Species A
Lab Procedure 2, Step 4: By adding all the degree hours for each of the six life stages together, you
calculate the cumulative degree hours required for an adult fly to develop at 21°C.
Temp °C Egg 1st Instar 2nd Instar Feeding 3rd Instar
Migrating 3rd Instar
Pupa
21 21 31 26 50 118 240
Deg Hrs 441 651 546 1050 2478 5040
Cum. Deg Hrs
441 651+441 = 1092
546+1092 = 1638
1050+1638 = 2688
2478+1638 = 5166
5040+5166 = 10206
Adult degree-hours = ∑ degree hours at each stage = cumulative degree hours = 10206
Cumulative degree-hours for each life stage: Species B
Lab Procedure 2, Step 4: By adding all the degree hours for each of the six life stages together, you
calculate the cumulative degree hours required for an adult fly to develop at 21°C.
Temp °C Egg 1st Instar 2nd Instar Feeding 3rd Instar
Migrating 3rd Instar
Pupa
21 25 37 31 60 124 286
Deg Hrs 525 777 651 1260 2604 6006
Cum. Deg Hrs
525 777+525 = 1302
651+1302 = 1953
1260+1953 = 3213
2604+3213 = 5817
6006+5817 = 11823
Adult degree-hours = ∑ degree hours at each stage = cumulative degree hours = 11823
Cumulative degree-hours for each dayLab Procedure 2, Step 5: Calculate elapsed degree hours for each of the days in the climatological
data provided. To do this, multiply the number of hours by the average temperature that day.
DAY MAX MIN AVG DEPAR-TURE FROM NORMAL
DEPAR-TURE FROM NORMAL
HEATING COOLING TOTAL WATER EQUIV
SNOW-FALL, ICE
PELLETS
SNOW, ICE PELLETS OR ICE ON GROUND
AVG SPEED (MPH)
AVG SPEED (KPH)
SKY COVER
SUNRISE-SUNSET
WEATHER OCCURENCES
PEAK WIND (KPH)
degree hours
cumulative degree hours
1 18.3 10.6 16.1 -1.9 -1.0 2.2 0.0 0.00 0 0 6.3 10.08 6 0S 18.7 386.66666
7 7283.73333
2 19.4 12.8 15.0 -3.0 -1.7 3.3 0.0 0.09 0 0 13.8 22.08 6 0SE 26.2
360 6897.06667
3 17.2 9.4 12.8 -5.2 -2.9 5.6 0.0 0.19 0 0 17.2 27.52 8 5SW 33.3 306.66666
7 6537.066674 20.0 10.6 13.3 -4.7 -2.6 5.0 0.0 0.28 0 0 12.3 19.68 9 1 W 38.6 320 6230.4
5 21.1 12.8 19.4 1.4 0.8 0.0 1.1 0.00 0 0 11.1 17.76 7 1,2W 28.3 466.66666
7 5910.4
6 25.6 16.7 20.6 2.6 1.4 0.0 2.2 0.00 0 0 8.1 12.96 6 0SW 24.3 493.33333
3 5443.73333
7 23.9 15.2 19.4 1.4 0.8 0.0 1.1 0.07 0 0 6.3 10.08 8 0S 16.7 466.66666
7 4950.4
8 20.6 12.8 16.1 -1.9 -1.0 0.0 0.0 0.11 0 0 13.2 21.12 8 0S 27.9 386.66666
7 4483.73333
9 22.1 13.9 18.9 0.9 0.5 0.0 0.6 0.00 0 0 4.2 6.72 3 0SE 10.3 453.33333
3 4097.06667
10 25.0 12.6 16.1 -1.9 -1.0 0.0 0.0 0.00 0 0 8.88 14.208 6 0S 22.7 386.66666
7 3643.73333
11 22.1 8.9 15.3 -2.7 -1.5 0.0 0.0 0.67 0 0 19.6 31.36 6 3,5S 32.4
367.2 3257.06667
12 14.5 9.4 10.5 -7.5 -4.2 0.0 0.0 1.13 T 0 23.5 37.6 8 1,3,5SW 43.8
252 2889.86667
13 16.7 7.2 12.1 -5.9 -3.3 0.0 0.0 0.23 0 0 14.2 22.72 7 1,2S 29.8
290.4 2637.86667
14 19.3 9.3 15.0 -3.0 -1.7 0.0 0.0 0.02 0 0 10.5 16.8 8 1S 24.5
360 2347.46667
15 16.5 8.9 12.8 -5.2 -2.9 0.0 0.0 T 0 0 11.9 19.04 4 0SW 23 306.66666
7 1987.4666716 18.9 10.7 13.3 -4.7 -2.6 0.0 0.0 0.00 0 0 6.4 10.24 3 0 W 19.2 320 1680.817 16.8 9.5 12.9 -5.1 -2.8 0.0 0.0 0.00 0 0 9.5 15.2 3 0 S 27.2 309.6 1360.818 19.4 10.7 16.4 -1.6 -0.9 0.0 0.0 0.00 0 0 11.1 17.76 2 0 W 26.3 393.6 1051.2
19 19.1 10.9 15.9 -2.1 -1.2 0.0 2.8 0.00 0 0 4.6 7.36 1 0SW 17.8
381.6 657.620 22.0 12.8 18.4 0.4 0.2 0.0 5.0 0.00 0 0 7.4 11.84 1 0 W 23.1 276 276
Which day the adult insect laid eggs on the body: Species A
Lab Procedure 2, Step 6a: Examine the species A life stages collected as evidence and identify the
oldest species A life stage collection for the adult male.DAY degree
hourscumulative degree hours
1386.66666
7 7283.73333
2 360 6897.06667
3306.66666
7 6537.066674 320 6230.4
5466.66666
7 5910.4
6493.33333
3 5443.73333
7466.66666
7 4950.4
8386.66666
7 4483.73333
9453.33333
3 4097.06667
10386.66666
7 3643.73333
11 367.2 3257.06667
12 252 2889.86667
13 290.4 2637.86667
14 360 2347.46667
15306.66666
7 1987.4666716 320 1680.817 309.6 1360.818 393.6 1051.2
19 381.6 657.620 276 276
On Day 11, the cumulative degree-hours were 3257. Species A takes 2688 degree-hours to complete development in the Feeding 3rd Instar and begin development in the Migration stage of the 3rd Instar. The temperature data alone suggests that the eggs were laid on the 12th, but there was a storm then, so we know that the eggs were laid before then since flies are not active during thunderstorms.
Feeding 3rd Instar
Migrating 3rd Instar
50 118
1050 2478
1050+1638 = 2688
2478+1638 = 5166
Which day the adult insect laid eggs on the body: Species B
Lab Procedure 2, Step 6a: Examine the species A life stages collected as evidence and identify the
oldest species B life stage collection for the adult male.DAY degree
hourscumulative degree hours
1386.66666
7 7283.73333
2 360 6897.06667
3306.66666
7 6537.066674 320 6230.4
5466.66666
7 5910.4
6493.33333
3 5443.73333
7466.66666
7 4950.4
8386.66666
7 4483.73333
9453.33333
3 4097.06667
10386.66666
7 3643.73333
11 367.2 3257.06667
12 252 2889.86667
13 290.4 2637.86667
14 360 2347.46667
15306.66666
7 1987.4666716 320 1680.817 309.6 1360.818 393.6 1051.2
19 381.6 657.620 276 276
On Day 17, the cumulative degree-hours were 1360. Species B takes 1302 degree-hours to complete development in the 1st Instar and begin development in the 2nd Instar.
1st Instar 2nd Instar
37 31
777 651
777+525 = 1302
651+1302 = 1953
Conclusion: Bodies have been dead for a minimum of 8 days, 16 hours
DAY degree hours
cumulative degree hours
9453.33333
3 4097.06667
10386.66666
7 3643.73333
11 367.2 3257.06667
12 252 2889.86667
13 290.4 2637.86667
14 360 2347.46667
15306.66666
7 1987.4666716 320 1680.817 309.6 1360.818 393.6 1051.2
19 381.6 657.620 276 276
THUNDERSTORM
Post Mortem Interval = ∑ hours(day) = h(20) + h(19) + h(18) … h(10) = 207 hours = 8 days, 15 hours
The post mortem interval was calculated to be 8 days, 15 hours, but we know that the storm occurred on the evening of the 11th, so we conjecture that at the flies were active at least an hour on the 11th, thus bringing our PMI to a minimum of 8 days, 16 hours. The actual PMI, which is unknown, may vary up to 12 hours more than this calculation due to weather. Students’ calculations may vary by up to a day later.
Homework
• Read Chapter 10, pages 356 – 367 and work on review questions #1-14 (they are due on Monday)
Exit SlipSeptember 18, 2014
• Go to m.socrative.com or the socrative app. • Room # is 417101 • QUESTION: What did you find difficult with
this lab (besides getting the species out of the vials)?