# BIOL4 Biology Unit 4 Exam - 13th June 2011 watch

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1. (Original post by spottyzebra)
Also inorganic fertilizers tend to be very soluble and can be easily leached away by heavy rain and therefore need to be reapplied often.
didn't he say that?
2. Can anyone help me with January 2010 (past paper) question 2b?
its the calculation question with the graph about catterpillars.
aqa says the answer is 40 but i have no idea how to get there!!

3. (Original post by sms21)
Can anyone help me with January 2010 (past paper) question 2b?
its the calculation question with the graph about catterpillars.
aqa says the answer is 40 but i have no idea how to get there!!

The rate of growth iis the rate of change of mass with time. So that would be (change in mass)/(time). The rate of growth at any given moment is the gradient of the graph as this is change in mass over time. It follows that the fastest rate of growth is where the gradient is steepest, in this case the straight bit at the end of the graph. I read off the mass at two points, in my case 200 mg at day 20 and 400 mg at day 25. By then working out the change in mass over the five days we can work out the rate of mass change per day. In this case 200mg/5days so 40mg/day.
4. (Original post by jonnyboy1993)
The rate of growth iis the rate of change of mass with time. So that would be (change in mass)/(time). The rate of growth at any given moment is the gradient of the graph as this is change in mass over time. It follows that the fastest rate of growth is where the gradient is steepest, in this case the straight bit at the end of the graph. I read off the mass at two points, in my case 200 mg at day 20 and 400 mg at day 25. By then working out the change in mass over the five days we can work out the rate of mass change per day. In this case 200mg/5days so 40mg/day.
Thank you very much, very clearly explained too, I'd rep you, but i already have apparently
5. I'm going to fail this...again
6. you guys staying ol night revising bio
7. (Original post by al_habib)
you guys staying ol night revising bio
I need to, seriously dreading this exam AND unit 5! I hate it...argh
8. (Original post by Safx)
I need to, seriously dreading this exam AND unit 5! I hate it...argh
the same hate population esp the ecology bit. its long time i havent touched unit 5 for a while but i fink unit 5 makes alot more sense than unit 4.how many exams have you got
9. (Original post by al_habib)
the same hate population esp the ecology bit. its long time i havent touched unit 5 for a while but i fink unit 5 makes alot more sense than unit 4.how many exams have you got
It's frustrating cause bio is the only one letting me down, I just dont get hws or anything. I always think I've done well or alright but I end up failing. wtf! aaaaa. Sorry for the mini rant lol.
10. (Original post by Safx)
It's frustrating cause bio is the only one letting me down, I just dont get hws or anything. I always think I've done well or alright but I end up failing. wtf! aaaaa. Sorry for the mini rant lol.
4 exams left biol4 n 5 and chem4(wed) n chem5. hate this unit. do lots of past papers helps alot
11. So anyone hav any ideas/topics to come up on the paper on monday??
12. (Original post by sunil10)
So anyone hav any ideas/topics to come up on the paper on monday??
woop your name is sunil - so is mine... .... and topics wise not sure really. What about you?
13. Lol great name! I hav no idea! iv literally started me revision 2dai and is now worried!
14. (Original post by jonnyboy1993)
The rate of growth iis the rate of change of mass with time. So that would be (change in mass)/(time). The rate of growth at any given moment is the gradient of the graph as this is change in mass over time. It follows that the fastest rate of growth is where the gradient is steepest, in this case the straight bit at the end of the graph. I read off the mass at two points, in my case 200 mg at day 20 and 400 mg at day 25. By then working out the change in mass over the five days we can work out the rate of mass change per day. In this case 200mg/5days so 40mg/day.
Thanks what I did was 400 / (25-14.5) = 38.xx
Obviously its wrong ..
Dunno how u say 200 mg!!!

Also for 2d: dunno how they say caterpillar compensating to eat more plant increases the yield

Help me out
15. (Original post by arvin_infinity)
Thanks what I did was 400 / (25-14.5) = 38.xx
Obviously its wrong ..
Dunno how u say 200 mg!!!

Also for 2d: dunno how they say caterpillar compensating to eat more plant increases the yield

Help me out
What they are trying to say is that despite caterpillars growing slower on the plants grown in high CO2 concentrations, this doesn't mean that growing plants in these conditions will increase the yield as the caterpillars may just eat more plant to make up for the fact they gain less mass whilst eating the plants grown in these conditions.
16. (Original post by jonnyboy1993)
What they are trying to say is that despite caterpillars growing slower on the plants grown in high CO2 concentrations, this doesn't mean that growing plants in these conditions will increase the yield as the caterpillars may just eat more plant to make up for the fact they gain less mass whilst eating the plants grown in these conditions.
+rep
got it..
Didnt understand what ATP is to do with C02!
17. (Original post by arvin_infinity)
+rep
got it..
Didnt understand what ATP is to do with C02!
It's about realising they are testing your knowledge of aerobic and anaerobic respiration. Before being moved the apple is respiring almost entirely aerobically. During this, for every glucose molecule, 6 molecules of CO2 are produced. But equally a lot of ATP molecules are produced (36, 38 or 40 depending on how you work out your net production, assuming full efficiency, which is unrealistic, but not really to be worried about in this question). What matter is that we get around 6 ATP per CO2 molecule produced. After being moved to the nitrogen atmosphere, so no oxygen is present, the apple respires anaerobically so we get 2 CO2 and a net production of 2 ATP per glucose. This means we only get 1 ATP per CO2. Now if we assume the apple is using up ATP in metabolic processes etc at the same rate in both atmospheres then for every 6 CO2 molecules produced in anaerobic conditions only one is produced in aerobic conditions. As a result the rate of CO2 production is higher in the nitrogen atmosphere, despite the fact this might intuitively appear wrong.
18. (Original post by arvin_infinity)
+rep
got it..
Didnt understand what ATP is to do with C02!
which questions are you doing
19. (Original post by jonnyboy1993)
It's about realising they are testing your knowledge of aerobic and anaerobic respiration. Before being moved the apple is respiring almost entirely aerobically. During this, for every glucose molecule, 6 molecules of CO2 are produced. But equally a lot of ATP molecules are produced (36, 38 or 40 depending on how you work out your net production, assuming full efficiency, which is unrealistic, but not really to be worried about in this question). What matter is that we get around 6 ATP per CO2 molecule produced. After being moved to the nitrogen atmosphere, so no oxygen is present, the apple respires anaerobically so we get 2 CO2 and a net production of 2 ATP per glucose. This means we only get 1 ATP per CO2. Now if we assume the apple is using up ATP in metabolic processes etc at the same rate in both atmospheres then for every 6 CO2 molecules produced in anaerobic conditions only one is produced in aerobic conditions. As a result the rate of CO2 production is higher in the nitrogen atmosphere, despite the fact this might intuitively appear wrong.
Producing say 32 ATP anaerobically need 16 times respiration which produces 32 co2 molecules (which is higher)

Still not sure about 2b tho
20. it's tomorrow guys.. oh my god

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