Optional Test: Calculations of PMI using ADH

Transcription

1 Optional Test: Calculations of PMI using ADH FS 14 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 o C. 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 o C require five days to reach the 3 rd instar, the ADD for a 3 rd instar would be (5) x 22 o C = 110. The ADH would be (5 x 24) x 22 o C= 2,640. If we raise the temperature, the time for development should decrease. At 26.7 o C, the ADH is still 2,640. To reach the 3 rd 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 22 o C. If the oldest maggots collected at the crime scene are newly hatched, 3 rd Average ADH Needed for Completion of Developmental instars, Stages Phormia regina at 22 o C the ADH for the maggots is ADH is calculated using the ADH for the end of the Stage Hours ADH previous stage if it is an early instar and the Egg ADH for the end of the current stage if it is a later instar. If the temperature at the crime 1 st instar scene has been an average of 22 o C, the body 2 nd instar was first exposed to fly activity 70 hours earlier ( ). If the insects on the Feeding3 rd instar body were collected at 10:00 am on Sunday, Postfeeding 3 rd instar based on the calculations, the body was exposed at approximately 12:00 pm the Pupa previous Thursday (70 hours earlier!). The PMI is 70 hours or approximately 3 days. 1

2 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 3 rd instars of Phormia regina. The ADH for early 3 rd instars (late 2 nd instars) should be 1540 regardless of the temperature at the crime scene. Weather data from the scene averaged 20 o C on Tuesday, 18 o C on the previous Monday (July 14 th ), 19 o C on Sunday and 23 o C 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 Average Daily Temperature from NOAA Tuesday, 15 h July 15 (12am-3pm) Monday, July 14 Sunday July 13 Saturday July 12 Daily ADH Σ ADH 20 o C 300 (15 x 20) h 18 o C ( ) 24h 19 o C 456 1,188 (19 x 24) 24h 23 o C (1, ) 1540 ADH Based on the chart above and the early 3 rd 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 the Saturday am ADH, convert to hours and add to Saturday 12:00 am = 200 ADH needed on Saturday for growth to the early 3 rd instar. If 23 o C x hours = 200 ADH, then 200 ADH / 23 o C = 8.7 h Add 8.7 h + 12 am Saturday= 8.7 am Saturday Determine the difference between the 1540 ADH and the Sunday am ADH, convert to hours and subtract from Sunday 12:00 am = 352 ADH not needed on Saturday for growth to the early 3 rd instar. If 23 o C x hours = 352 ADH, then 352 ADH / 23 o C = 15.3 h 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 12 th. 2

3 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 o C or 54 o F ( ), eggs will not hatch and larvae will not move, feed or grow in temperatures lower than 12 o C. A freezer would certainly have a temperature lower than the minimum threshold temperature for most insects. (For more information: ) 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 # Jane Smith s body was found in a deserted building in Jersey City, NJ (Essex County) on March 13 th, 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 3

4 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 0.6 to 4.0 mm to 8.0mm to 15mm 3 Table 2. Range in length (mm) for instars of Sarcophaga bullata * : Length 0.5 to 5.0mm to 10.0mm to 20mm 3 * The minimum threshold temperature for both species is 12 o C. Table 4. Time for Phormia regina larval and pupal development at 22 o C. Stage Cumulative Time to Cumulative ADH at END of Stage Egg 20 hrs st instar 25 hrs nd instar 25 hrs rd instar 25 hrs 2090 Prepupa 125 hrs 4840 end of 1 st instar, beginning of 2 nd instar. Pupa hrs 7403 Total 7 days Table 3. Time for Sarchophaga bullata larval and pupal development at 27 o C. Note that data for S. bullata was not available at 22 o C. Stage Cumulative Time to Egg 20 hrs (20 x 27)= 1st 26 hrs Cumulative ADH at END of stage 2nd 18 hrs rd Prepupae Pupae 50 hrs 112 hrs 12 days Total 17 days You must fill in the ADH for S. bullata. Some work has been done for you. 4

5 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 Hours Mean Temp. from NOAA ( o C) Daily ADH ADH 03/13/ /12/ /11/ /10/ /09/ /08/ /07/ /06/ /05/ /04/ /03/ ) 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 Larval measurements (mm) Approx. Minimum ADH for Development Phormia regina 3 1 mid =550/2= =715 P. regina 5 P. regina 5 2 mid 1265 P. regina 8 2 late/ 3 early P. regina 8 2 late/ 3 early Sarcophaga bullata S. bullata 7 2 mid =486/2= =1485 S. bullata 10 2 late 1728 S. bullata 15 3 mid S. bullata 15 3 mid 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. 5

6 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. 6

7 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 53 o C (127 o F). 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! 7