 |
|
Answer to calculation exercise - larvae resources
Both small changes in mortality and development
time are of great importance for how many fully developed larvae establish
themselves. This denotes that what happens to larvae in the water can greatly
effect the size of populations from year to year. That a certain species
is uncommon certain years, but common during other periods does not necessarily
denote the survival rate amongst adults of the species in the same area.
Number of offspring that survive to the end
of the day that the larvae are in the water.
R
(mortality rate) |
Spreading period
(t dagar) |
| 18 | 20 | 22 | | 0,62 | 14 000 | 4 000 | 1 200 | | 0,69 | 4 000 | 1 000 | 260 | | 0,76 | 1 200 | 260 | 56 |
This is what the calculation looks like when
1 000 females are replaced by 1 000 newly developed individuals:
The 103 females together produce 103x106
= 109 larvae. This is N0. Nt
= 109x2,718(-0,69x20)
= 109x2,718-13,8 = 109x1,017x10-6
= 1,017x103.
The values for R and t have been rounded off to two decimal
points, therefore Nt is rounded of to 1 000.
 |
Calculation
exercise
Page 2 of 2 |
|
|
