simple ecology question

[Sarahisme]

Hi all,

Experimental observations in the lab revealed that the larvae of an insect species experiences a constant rate of mortality regardless of its age. That is qx is constant. However, the time from egg hatching to pupation increase in a linear manner with the density at which the larvae are raised. In this example is larval survival (number pupating/number hatched) density dependent?

i was just wondering you thoughts on this question... i think that larval survival is density dependent, but the answers say otherwise, would anyone be able to help explain this too me?

 

thanks

 

Sarah

[ecoli]

I've read your post several times, and I'm not 100% sure what you are asking. Could you be clearer?

[SkepticLance]

Taking the question at face value, my interpretation is :

Higher larval density means longer time to pupation.

Since mortality is constant (numbers of deaths per day), a longer time to pupation must been a higher percentage of larvae die before pupation.

[Sarahisme]

I've read your post several times, and I'm not 100% sure what you are asking. Could you be clearer?

 

i would love to be, but thats the question from the book

 

sorry ;p

[Sarahisme]

Taking the question at face value' date=' my interpretation is :

Higher larval density means longer time to pupation.

Since mortality is constant (numbers of deaths per day), a longer time to pupation must been a higher percentage of larvae die before pupation.[/quote']

 

so your saying it is density dependent?

[SkepticLance]

This question is extraordinarily badly worded. The examiner should have his/her arse kicked very solidly.

 

If the question is taken as worded, then yes. it is density dependent.

[CharonY]

I think it is only badly worded if you think about it in a, say, biological context. If you take it as it is that is, as a question of stochastics, it appears rather straightforward. Unless of course, I misunderstood it.

[Sarahisme]

I think it is only badly worded if you think about it in a, say, biological context. If you take it as it is that is, as a question of stochastics, it appears rather straightforward. Unless of course, I misunderstood it.

 

and so , do you understand it to be density dependent or independent?

[CharonY]

Sorry, just noticed that my post was quite meaningless. Posted in a hurry.

Basically it should not be population dependent since a longer pupation time correlates with a higher starting density,

Actually, upon reflecting I found one point a little bit unclear after all. Let me elaborate:

Assume that qx (or mortality rate) is 0.5. Furthermore the mortality rate is calculated in arbitrary steps (could be days, weeks, whatever).

Now assume a starting density of 10 and assume that one time step is needed till pupation for this density. This would result in 5 survivors and 5 deads (5/10=0.5). Or in other words, 5 larvae enter pupation.

Now double the density to 20.

Step 1: 10 survivors 10 dead. 10 survivors can enter step 2 (due to increased time it takes to enter pupation because of higher density)

Step 2: 5 survivors, 5 dead. So you end up again with 5 survivors.

 

And so on. But this would only apply if the start of the pupation phase is dependent on the current population size so that starting with a population size of 80 it does not take 8 steps but 4 (80 ->40 ->20 -> 10 -> 5-> pupation). At least that’s how I understood it.

[Sarahisme]

hmmm ok, so here is what i think is going on....

 

i think it is a 'trick' kind of question. where the linear manner part just trying to throw you off. the first thing it states is that the larvae have a constant mortality rate which implies that the survival of the larvae is not density dependent.

 

hmm....