1. September 24, 2015 Objectives: – Students will be able to use entomological patterns and weather data found in a case study to determine Time of Death (TOD). Agenda: – Turn in homework – Warm Up: What in a crime scene can affect entomology data? – Observing the information provided by a climatological data sheet. – Determine TOD of a case study - together Calculating accumulated degree hours (ADH) of the life stages of a fly Calculating ADH from climatological data Comparing the calculations to determine TOD of the victim – Independent assignment

2. Think-Pair-Share https://www.youtube.com/watch?v=dntO3YANo18 – Found a family of three extremely decomposed bodies. Medical examiners predicted the bodies to be dead for months but could only find 1 st instar stage flies. They need pupa flies to confirm this prediction. Why? Background: All they could find was 1 st instar

3. Case Study Bodies discovered at 1:00PM on June 20 Insects collected at 3:00PM on June 20 Victims had the same species and lifecycles present (2nd Instar) Collection day hours: (june 20 th ) – Amount of am + amount of pm = hours – 12 + 3 = 15

4. DAYMAXMINAVGDEPAR- TURE FROM NORMAL HEATINGCOOLINGTOTAL 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) 118.310.616.1-1.92.20.00.00006.310.0860 S 18.7 219.412.815.0-3.0-1.73.30.00.090013.822.0860 SE 26.2 317.29.412.8-5.2-2.95.60.00.190017.227.5285 SW 33.3 420.010.613.3-4.7-2.65.00.00.280012.319.6891 W 38.6 521.112.819.41.40.80.01.10.000011.117.7671,2 W 28.3 625.616.720.62.61.40.02.20.00008.112.9660 SW 24.3 723.915.219.41.40.80.01.10.07006.310.0880 S 16.7 820.612.816.1-1.93.30.00.110013.221.1280 S 27.9 922.113.918.90.90.50.00.60.00004.26.7230 SE 10.3 1025.012.616.1-1.92.20.00.00008.8814.20860 S 22.7 1122.18.915.3-2.7-1.51.10.00.670019.631.3663,5 S 32.4 1214.59.410.5-7.5-4.22.20.01.13T023.537.681,3,5 SW 43.8 1316.77.212.1-5.9-3.38.30.00.230014.222.7271,2 S 29.8 1419.39.315.0-3.0-1.75.60.00.020010.516.881 S 24.5 1516.58.912.8-5.2-2.93.30.0T0011.919.0440 SW 23 1618.910.713.3-4.7-2.61.10.00.00006.410.2430 W 19.2 1716.89.512.9-5.1-2.80.60.00.00009.515.230 S 27.2 1819.410.716.4-1.6-0.91.10.00.000011.117.7620 W 26.3 1919.110.915.9-2.1-1.20.02.80.00004.67.3610 SW 17.8 2022.012.818.40.40.20.05.00.00007.411.8410 W 23.1 WEATHERWEATHER SYMBOLS CLEAR -SCALE (0-3) 1=FOG PARTLY CLOUDY (SCALE 4-7) 2=FOG W/VISIBILITY 1/4 MILE OR LESS CLOUDY (SCALE=8-10) 3=THUNDER 4=ICE PELLETS 5=HAIL 6=GLAZE OR RIME 7=DUSTSTORM OR SANDSTORM 8=SMOKE OR HAZE 9=BLOWING SNOW

5. Adult Eggs Pupa 3rd Instar Larva 2nd Instar Larva 1st Instar Larva 21 hours 31 hours 26 hours 50 hours Fly Life Cycle 118 hours 240 Hours

6. Create a simplified table for Fly and weather data Date (June) Av. Temp. HoursDaily ambient thermal energy (Hours x Ave. Temp Accumulative Degree Hours - ADH (Current Daily Ambient Energy + Previous) 20 18.4 15 276 19 15.9 24 381.6657.6 18 16.4 24 393.61051.2 17 12.9 24 309.61360.8 16 13.3 24 3201680.8 15 12.8 24 307.21987.5 14 15.0 24 3602347.5 13 12.124290.42637.9 12 10.5242522889.9 11 15.324367.23257.1 10 16.124386.43643.7 9 18.924453.34097.1 8 16.124386.44483.7

7. Case Study Bodies discovered at 1:00PM on June 20 Insects collected at 3:00PM on June 20 Victims had the same species and lifecycles present (2nd Instar) To Calculation # of hours for collection day 12 (am) +3 = 15 hours

8. 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 °CEgg1 st Instar2 nd Instar3 rd InstarPupaAdult 21 312650118240 21*21 = 441 31*21 = 651 26*21 = 546 50*21 = 1050 118*21 = 2478 240*21 = 5040

9. 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 °CEgg1 st Instar2 nd Instar3 rd Instarpupaadult 21 312650118240 Deg Hrs441651546105024785040 Cum. Deg Hrs 441651+441 = 1092 546+1092 = 1638 1050+16 38 = 2688 2478+1638 = 5166 5040+5166 = 10206 Adult degree-hours = ∑ degree hours at each stage = cumulative degree hours = 10206

10. Find the Cummunitative Degree Hours Develop to 2 nd Instar Temp °CEgg1 st Instar2 nd Instar3 rd InstarPupaAdult 21 312650118240 Deg Hrs441651546105024785040 Cum. Deg Hrs 441651+441 = 1092 546+1092 = 1638 1050+1638 = 2688 2478+1638 = 5166 5040+5166 = 10206

11. Find the 1 st Number that is Greater than 1638 Date (June) Av. Temp. HoursDaily ambient thermal energy (Hours x Ave. Temp ADH (Current Daily Ambient Energy + Previous) 20 18.4 15 276 19 15.9 24 381.6657.6 18 16.4 24 393.61051.2 17 12.9 24 309.61360.8 16 13.3 24 3201680.8 15 12.8 24 306.71987.5 14 15.0 24 3602347.5 13 12.124290.42637.9 12 10.5242522889.9 11 15.324367.23257.1 10 16.124386.73643.7 9 18.924453.34097.1 8 16.124386.74483.7 1638

12. What is TOD? June 16 th – eggs were laid Add all the hours from June 20 th to June 16 th June 20 th + June 19 th + June 18 th + June 17 th +June 16th 15 + 24 + 24 + 24 + 24= 111 hours/24 hours 4.625 Days = 4 days and 15 hours – 1 day/.625 hours = 1.6(ratio) – 24hours/1.6 = 15 hours

13. New Case – Independent Work Bodies discovered at 6:00PM on June 20 Insects collected at 8:00PM on June 20 Investigator found 3 rd Instar located on the body What is the TOD of the victim?