Essay on bio lab

Submitted By Clark-Huang
Words: 710
Pages: 3
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December 17 2013

Procedure
1. Mark and Recapture method would be the most useful for estimating the population of eagles since eagles are constantly moving so quadrat method would not be ideal here. Also, within a set boundary marking the eagles and recapturing can give us a good ratio of how many eagles are expected in the set area. Mark and recapture are effective for moving organisms.
2. Distribution Patter 1= Clumped as populations are mainly grouped together
Distribution Pattern 2- Random- No specific pattern to where populations are located
Distribution Pattern 3- Uniform- Population seems to be evenly dispersed where the distance between each is the same.
3. The first cause of clumped may be because the eagles want to stay together to hunt or the resources available are located in clumps so in order for eagles to get food, they must be dispersed in clumps as well. Another possibility is that useable land is located in specific areas so eagles reside there.
Random dispersion of the eagles can be that the eagles have no interaction with other eagles and fly about randomly where even their homes or resources are located. Resources available may also be in random areas so eagles follow the same pattern.
Uniform dispersion may occur because resources/food is spread out evenly across the land and each eagle is marking their own territory of where their hunting grounds are and each eagle wants to cover as much land as possible.
4. 1cm by 7cm= 1km by 7 km= 7km2 is area of each transect
5. Transect 1 (distribution pattern 1)= 18 eagles/7km2 Average for pop. 1 = 12 eagles/transect
Transect 2 (distribution pattern 1)= 6 eagles/ 7km2
Transect 1 (distribution pattern 2) =4 eagles / 7km2 Average for pop.2 = 6 eagles/ transect
Transect 2 (distribution pattern 2)= 8 eagles/7km2
Transect 1 (distribution pattern 3) = 4 eagles/7km2 Average for pop. 3= 3 eagles/ transect
Transect 2 (distribution pattern 3) = 2 eagles/7km2
6. Total area= 6.8cm by 7.3cm
= 6.8km by 7.3km
=49.64 km2

12eagles/ 7km2 = x eagles / 49.64km2
X= 85.1 eagles or 85 eagles for population distribution 1

6 eagles/ 7km2 = x eagles / 49.64km2
X= 42.5 eagles or around 43 eagles for population distribution 2

3 eagles/ 7km2 =x eagles/ 49.64km2
X= 21.2 or 21 eagles for population distribution 3
7. 35 pairs= 70 eagles for population distribution 1
26 pairs=52 eagles for population distribution 2
11 pairs=22 eagles for population distribution 3
Questions
1. 70-85/70 x 100% = 21.4 % error for population distribution 1 so it was off by quite a bit. Calculated population was off by 15 eagles.
52-43/52 x 100%=17.3 % error for population distribution 2 so it was still off by quite a bit but less than population distribution 1.Calculated number was off by around 9 eagles.
22-21/22 x 100%= 4.5% error for population so it was really close to actual population. Calculated population was only off by 1 eagle.

2. The reason why the estimate for population dispersion 1 was so off was
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