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AS Biology P1 MCQs Preparation Thread

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Ohh, I get it! You see on this diagram where the arrow points to 'hinge region'? Just draw a lines thru the solid bonds nest to the dotted hinge region. The molecule is now broken down to three fragments, and there are still two variable regions where the antigen can bind. I know my explanation's childish, but I hope it helps :)
ab3.gif

thankyouuu bt where r dose 2 variable regions?
 
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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 :)
 
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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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