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

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

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.

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

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?

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.

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.

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?

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.

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....