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AS Biology P1 MCQs Preparation Thread

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All you have to do is to find the GGP of the tertiary consumer ( fourth trophic level) which is not shown here. First calculate the GGP for 2nd TL 23000- 8000 = 15000. Then calculate GGP for 3rd TL 15000 - 10500 = 4500. Then calculate GPP for 4rth TL 4500 - 4200 = 300. Now calculate the percentage 300/23000 x 100 = 1.3. Hope that was helpful :)

plzzzzzzzzz ans my qts too...it is posted above....
 
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A polypeptide chain is coded by the Exon strand of DNA, not both strands. So first you'll divide 120/2 = 60. So 60 nucleotides are coding for this chain. Since every amino acid is coded by 3 nucleotides, you'll divide 60 by 3 = 20 amino acids.

thanks a ton!! :D
 
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Firstly, at some stages lysosomes are present in the plant cells, when it's newly formed as far as I remember before the cell wall is formed. And which questions do you exactly want in this test :)

nd ans mcq 1 of w o6 qp 1 plzzzzzzz
 
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Could someone please explain to me these graphs? Q 14
and also how it would look if the amount of product formed on Y-axis and time on X-axis. It was a question in June 2002 but I can't find it here.

http://papers.xtremepapers.com/CIE/...and AS Level/Biology (9700)/9700_w07_qp_1.pdf

The graph simply shows effect of increasing substrate concentration on the rate of reaction, non-competitive inhibitor doesn't let the rate of reaction reach the maximum limit, competitive inhibitor only initially decrease the rate of reaction, but the increase in substrate concentration diminish it's effect, so it will reach the maximum limit after time.

(Remember that there is no time here, I got confused by it in the beginning)
 
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this is a crazy ass question, help me out please? (A) answwer is C
 

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w11qp13 question 33 pls help


In question 33, you don't really need to read all of the question since he told you at the end that the DNA is all now hybrid so one strand of DNA is N15 and one is N14. Remember that DNA is replicated by semiconservative replication. Imagine now that we have a bacteria with this hybrid DNA, when it starts to divide, her first daughter would get the N15 strand and would have the other strand made of nucleotides of N14 ( the medium it's grown in) her other daughter would get her mother's N14 strand and her other strand N14 from the nucloutides in her medium. Thus 50% of her daughters is hypride and the other 50 % is N14.

Look at the semi-conservative replication in this picture.

http://www.visionlearning.com/library/modules/mid187/Image/VLObject-5567-110727100736.jpg
 
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w10 qp13 pls explai q24 and 38

q24 - Each division on the SM has 40 EPG units. Since each SM division is 0.1mm, that means each 0.1mm has 40 EPG units. Therefore 1 EPG unit is = 0.0025 mm (2.5micrometers) 10/2.5 = 4 so the answer is B.

q38 - Somebody explained this a few pages earlier, just browse through, and you'll find it!
 
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nd ans mcq 1 of w o6 qp 1 plzzzzzzz


Okay, first of all you have to calculate how many graticule divisions does one stage macrometre occupies. It's 40 graticule divisions here. Then measure how many eye piece graticule units does one chloroplast occupies in the 2nd picture, that's 4 units. Now, you have to calculate how many stage macrometer divisions is the chloroplast.

1 S.M = 40 eye P.G
X SM = 4 eye p.G

x = 1 x 4 / 40 = 0.1 SM. The 1 SM = 0.1 mm. Thus the cloroplast = 0.1 x 0.1 = 0.01 mm. 0.01 mm x 1000 = 10 micrometer. Answer B :)
 
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Okay, first of all you have to calculate how many graticule divisions does one stage macrometre occupies. It's 40 graticule divisions here. Then measure how many eye piece graticule units does one chloroplast occupies in the 2nd picture, that's 4 units. Now, you have to calculate how many stage macrometer divisions is the chloroplast.

1 S.M = 40 eye P.G
X SM = 4 eye p.G

x = 1 x 4 / 40 = 0.1 SM. The 1 SM = 0.1 mm. Thus the cloroplast = 0.1 x 0.1 = 0.01 mm. 0.01 mm x 1000 = 10 micrometer. Answer B :)

but how do v knw width is 4 units...
 
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