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http://www.xtremepapers.com/papers/...and AS Level/Biology (9700)/9700_s07_qp_1.pdf
Question # 4 anyone?
Question # 4 anyone?
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A level Biology: Post your doubts here!
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s05: 14th) you can see that at B the graph is not steepest. the gradient is low. its not the optimum temperature. k.e of enzyme and substrate will get low at this point. whereas A its not far from the start of reaction so at this pooint bonds will be formed for enzyme substrate complex to be build upon...
6)for the enzyme ATPase to act on. as it needs to convert ATP to ADP and a phosphate which releases energy so this has to be a enzymatic reaction which takes place inside the cristae.
w05 , 30th ) my theory: haemoglobin is least saturated in muscles as the volume of blood which travels is very low, muscle cells are active and heamglobin releases oxygen easily at this comparitively low partial pressure. this is not a myoglobin oxygen dissociation curve ...
38th) yes A and C are the most closer options. we need to identify the most important factor which is C as you know that when plasmodium becomes a part of blood its hard to identify and build antibodies against it as it cultures inside red blood cells. the vaccines and drugs become ineffective. A is also a factor in not developing effective vaccine but not as much as C
1) after calibration 1 e.p.g = 2.5um , in the plant cell given below there are chloroplasts at the lower portion or the line, its about 6 divisions on average. 6 into 2.5 = 15 um , so closer to B.
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for question 14... k.e increases with increase in temperature. As temperature is still increasing k.e is also increasing and it is not highest at Q
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mean of the dilutions would mean that u have to calculate how much that particular solution is diluted , ie 50% , 25%...
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Anyone taking finals for Travel and Tourism either A or AS level??? Please reply
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Thankyou miss. :]
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Ohkaay!
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the question says *the fact that the palisade mesophyll is twice as thick as the spongy mesophyll* therefore C, cuz in A spongy mesophyll layer( lower one) is twice the palisade-upper one.
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Why is it important to determine initial rate of reaction when investigating the effect of a competitive inhibitor?
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initially enzyme is active all its active sites are empty, the reaction is fast but later competitive inhibitor will bind to the enzyme slowing down the reaction.reaction is slower with inhibitor
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lets see how much does one eye piece graticule division represent: 0.1/50 = 2.5 into 10^-3
in diagram 1, length is 5 divisions, 5 into upper reading = 0.01, diagram two length is 15 e.p.g divisions which equals 0.03. now 0.03-0.01 = 0.02 this is the change in length in mm, change in micrometer then divide by 4 hours, u get A
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in emphysema lung tissue gets destroyed. airways block. alveoli don't expand as they will in a normal lung, in chronic bronchitis there is a lot of mucus production and bronchus inflammation. so this gonna effect less then the emphysema. C is correct. x looks a normal healthy lung with vital capacity at 5 dm^3
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i don't get it.
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The question says *the fact that the palisade mesophyll is twice as thick as the spongy mesophyll* therefore C, cuz in A spongy mesophyll layer( lower one) is twice the palisade-upper one.i don't get it.
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They're asking the typ of food test which is NOT to be used. If u see (A) this works as we need heat in order to process benedicts test. In (C) heat again is required to process benedict test and boiling wid dilute acid for presence of non reducing sugarsm in (D) biuret test alone also works.whereas in (B) heat is not required to undergo biuret test therefore B is the right choice here.
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Role of capillaries in forming tissue fluid?