AS Biology P1 MCQs Preparation Thread

[Raweeha]

so,the general shape of the active site is also dependant on 3° only but the enzyme specificity is dependent on all 1,2,3,4°?

Questions of this type are loved by examiners so beware!
For example, in s12 p11:
10 Which level of protein structure maintains the globular shapes of enzymes?
A primary
B secondary
C tertiary
D quaternary

The answer is C, as the the globule is maintained by bonds between the alpha helices and beta sheets.
And the specifity of an enzyme is related to the active site which, in turn, is related to the globular structure, so the answer remains 3°.
If I'm mistaken please correct me

 

[Raweeha]

37 An enzyme hydrolyses the two heavy polypeptide chains of an antibody molecule. The hydrolysis
occurs at the hinge region and breaks the antibody into three fragments.
How many of these fragments are able to bind to antigens?
A 0
B 1
C 2
D 3


kindly explain dis ques

What paper is this from? Is the answer B?

 

[Bluejeans#]

What paper is this from? Is the answer B?

its C as far as i rmemembr

 

[Raweeha]

its C as far as i rmemembr

That kind of makes sense ... but then how would 3 fragments be formed?

 

[Bluejeans#]

That kind of makes sense ... but then how would 3 fragments be formed?

thats what i was asking

 

[Raweeha]

thats what i was asking

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

 

[polniks]

Can someone please explain these questions , I would really appreciate it thanks in advance:
http://papers.xtremepapers.com/CIE/...and AS Level/Biology (9700)/9700_s03_qp_1.pdf 4,6,14,19,20

http://papers.xtremepapers.com/CIE/...nd AS Level/Biology (9700)/9700_w11_qp_13.pdf2,19,32,33,35,38

 

[geek101]

Can someone please explain these questions , I would really appreciate it thanks in advance:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Biology (9700)/9700_s03_qp_1.pdf 4,6,14,19,20

4 - when the water potential of the sucrose is the same as the water potential of the cell sap, there will no change in the size of the strip due to no net exchange of water molecules. which means, that the length before (x1) is the same as the length after the 3 mins (x2), x1 = x2. therefore, if we calculate the ratio for this, x1/x2 = 1 . for a ratio of 1 from the graph the concentration is 0.45, which is the answer

6 - they say they need such vacuoles, to expel water, which would mean that the water potential in the animal cells would be lower compared to the surrounding water, so excess water would enter and it would have to be removed. why would plants not need it? A is the same case as the animal cells, and B is wrong. so eliminate those. and C is the one which makes sense, because if the water potential is the same inside and outside, there will be no net gain or loss of water from the cell vacuole, which is why fresh water plants would not require such vacuoles.

14 - if the enzyme is denatured at 50. it would take the most time for the reaction to be completed, so A and B are rejected. from C and D the graph is taking the longest time at 50 so thats the answer. It cannot be D, because 50 is the temp at which the enzyme activity is zero, the curve cannot go beyond it

19 - when 100% N15 DNA is mixed with N14, in each generation the number of pure N15 DNA becomes half. (curve C). this is because when the N15 DNA is mixed with N14. one strand of the N15 DNA will bind with N15, and one will bind with N14. so there will be 50% N15 DNA left, this process is repeated each time...

20 - during transcription DNA is made into mRNA using transcriptase enzyme, so when RNA is extracted from the beta cells, it has to be converted back to DNA, so reverse transcriptase is used...

hope that helps!

 

[Boo]

Tidal volume is the amount of air breathed in and out normally. As you can see in the first thirty seconds of the graph the tidal volume is at a constant 500 cm3. As for the Vital capacity - it is the maximum amount of air breathed out so if you count the number of squares for the large exhalation bit it would be 14.5 square. Since each 4 squares are 1000 cm3 that would equal to roughly (1000 + 1000 + 1000 + 500 + 250) which is 3750 cm3

Thank you!

 

[polniks]

4 - when the water potential of the sucrose is the same as the water potential of the cell sap, there will no change in the size of the strip due to no net exchange of water molecules. which means, that the length before (x1) is the same as the length after the 3 mins (x2), x1 = x2. therefore, if we calculate the ratio for this, x1/x2 = 1 . for a ratio of 1 from the graph the concentration is 0.45, which is the answer

6 - they say they need such vacuoles, to expel water, which would mean that the water potential in the animal cells would be lower compared to the surrounding water, so excess water would enter and it would have to be removed. why would plants not need it? A is the same case as the animal cells, and B is wrong. so eliminate those. and C is the one which makes sense, because if the water potential is the same inside and outside, there will be no net gain or loss of water from the cell vacuole, which is why fresh water plants would not require such vacuoles.

14 - if the enzyme is denatured at 50. it would take the most time for the reaction to be completed, so A and B are rejected. from C and D the graph is taking the longest time at 50 so thats the answer. It cannot be D, because 50 is the temp at which the enzyme activity is zero, the curve cannot go beyond it

19 - when 100% N15 DNA is mixed with N14, in each generation the number of pure N15 DNA becomes half. (curce C). this is because when the N15 DNA is mixed with N14. one strand of the N15 DNA will bind with N15, and one will bind with N14. so there will be 50% N15 DNA left, this process is repeated each time...

20 - during transcription DNA is made into mRNA using transcriptase enzyme, so when RNA is extracted from the beta cells, it has to be converted back to DNA, so reverse transcriptase is used...

hope that helps!

thanks

 

[Pwincessajwa]

Just post what you need help in, and I'll try to help

umm thanks alotttt
http://papers.xtremepapers.com/CIE/...nd AS Level/Biology (9700)/9700_s10_qp_12.pdf queshtion num 11,13,35,38,39 plx explain it to me

 

[HubbaBubba]

umm thanks alotttt
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Biology (9700)/9700_s10_qp_12.pdf queshtion num 11,13,35,38,39 plx explain it to me

q11 - It must be A, because as said in the question there is an efficiency of 10% of energy transfer. You begin with 1300 so for the zooplankton you have to do (1300 * 0.1 * 0.1 which is 13 or 1.3 * 10^1) To get the tuna you must multiply the 13 two more times (13 * 0.1 * 0.1 = 0.13 or 1.3 * 10^-1)

q13 - This one is a bit difficult, and you must know how to calibrate an EPG and SM. Each division is 0.1mm and there are 40 EPG units in each 0.1mm, which means each one EPG is equal to 0.0025 mm. The radius of the circle is 50 EPG units, so 50 * 0.0025 = 0.125mm. The question clearly needs it in micrometers so 0.125 mm = 125 micrometers. To find the area of a circle you do πr^2 so that is π * 125 * 125 which is C

q35 - A is clearly out, because although a primary and secondary response occurred for antigen X, there was NO secondary response for antigen Y because it was the first exposure to this antigen (It is a primary response). B is also wrong because memory B lymphocytes against antigen Y are different for antigen X. C is correct because the memory cells enabled a secondary response in the second exposure to X, and different lymphocyte cells were produced for Y. D is incorrect because the memory cells cannot produce antibodies that work on both antigens X and Y.

q38 - A is incorrect, because the semi lunar valve connects the aorta to the left ventricle not the left atrium. It shouldn't have affected the Bicuspid valve at all in the first place. B is also wrong because blood won't leak out of the ventricle during relaxation if the semi lunar valve is open a bit. C is wrong because it should have an opposite effect to what they state. Due to the hole, blood would leak backwards into the right ventricle during systole, so as a result the walls of the ventricle will thicken and work harder to keep blood flowing through the aorta, which makes D the correct answer.

q39 - Flow of current through atrium means Atrium systole, which means pressure will increase SLIGHTLY. Therefore it must mean P. recovery of the ventricular walls means diastole which means the ventricle would be filling up with blood which is during an atrial systole which is T. So the answer would be B - P & T.

I hope this made sense to you

 

[Pwincessajwa]

q11 - It must be A, because as said in the question there is an efficiency of 10% of energy transfer. You begin with 1300 so for the zooplankton you have to do (1300 * 0.1 * 0.1 which is 13 or 1.3 * 10^1) To get the tuna you must multiply the 13 two more times (13 * 0.1 * 0.1 = 0.13 or 1.3 * 10^-1)

q13 - This one is a bit difficult, and you must know how to calibrate an EPG and SM. Each division is 0.1mm and there are 40 EPG units in each 0.1mm, which means each one EPG is equal to 0.0025 mm. The radius of the circle is 50 EPG units, so 50 * 0.0025 = 0.125mm. The question clearly needs it in micrometers so 0.125 mm = 125 micrometers. To find the area of a circle you do πr^2 so that is π * 125 * 125 which is C

q35 - A is clearly out, because although a primary and secondary response occurred for antigen X, there was NO secondary response for antigen Y because it was the first exposure to this antigen (It is a primary response). B is also wrong because memory B lymphocytes against antigen Y are different for antigen X. C is correct because the memory cells enabled a secondary response in the second exposure to X, and different lymphocyte cells were produced for Y. D is incorrect because the memory cells cannot produce antibodies that work on both antigens X and Y.

q38 - A is incorrect, because the semi lunar valve connects the aorta to the left ventricle not the left atrium. It shouldn't have affected the Bicuspid valve at all in the first place. B is also wrong because blood won't leak out of the ventricle during relaxation if the semi lunar valve is open a bit. C is wrong because it should have an opposite effect to what they state. Due to the hole, blood would leak backwards into the right ventricle during systole, so as a result the walls of the ventricle will thicken and work harder to keep blood flowing through the aorta, which makes D the correct answer.

q39 - Flow of current through atrium means Atrium systole, which means pressure will increase SLIGHTLY. Therefore it must mean P. recovery of the ventricular walls means diastole which means the ventricle would be filling up with blood which is during an atrial systole which is T. So the answer would be B - P & T.

I hope this made sense to you

thankssssssssssssssssssssssssss alottt but i dnt khow why i am not able to understand question num 13 and 15 well thats my mistake thank u i will need ur help in so many of my pprs

 

[HubbaBubba]

thankssssssssssssssssssssssssss alottt but i dnt khow why i am not able to understand question num 13 and 15 well thats my mistake thank u i will need ur help in so many of my pprs

Try googling 'How to calibrate SM and EPG'. There was one website that explained it very well, but unfortunately I have forgotten the name :/ If you need any help, just ask me. I've solved all the papers from Nov 2002 - 2012.

 

[Pwincessajwa]

Try googling 'How to calibrate SM and EPG'. There was one website that explained it very well, bunfortunately I have forgotten the name :/ If you need any help, just ask me. I've solved all the papers from Nov 2002 - 2012.

uhttp://papers.xtremepapers.com/CIE/...nd AS Level/Biology (9700)/9700_s11_qp_12.pdf q num 21 q num7,32,30, plz ?t

 

[HubbaBubba]

uhttp://papers.xtremepapers.com/CIE/...nd AS Level/Biology (9700)/9700_s11_qp_12.pdf q num 21 q num7,32,30, plz ?t

q7 - Sucrose is made up of Fructose and alpha glucose. I cannot really explain this question, as you have to memorise the structures of those substances. Fructose is 3 and alpha glucose is 1 so the answer is B

q21 - These questions are usually confusing, so I remember it best as (tRNA is the same code as DNA but U is replaced with T) and (mRNA is the same as the sequence on the amino-acid itself) so since the sequence of tRNA is CUU-UUA-UGG-GGA, then the sequence of DNA would be CTT-TTA-TGG-GGA which is C

q30 - The answer is obviously B because as sucrose concentration decreases in the leaf, it decreases in the phloem and vise versa. While A is true, it is not what the question is asking for, as you cannot tell from the graph that water moves by osmosis into the phloem (water isn't mentioned at all!). C is the same as A, it is true, but you can't tell from the graph if sucrose moves in both directions. D - there is no mention of xylem on the graph so you can't tell if it's true or not.

q32 - You have to memorise these as well from the biology book. Trachea would be largest, then the bronchus, then bronchioles then the alveoli. Trachea is 18mm, Bronchus is 12mm, bronchioles are 0.50 mm, and finally alveoli are 0.25mm. Answer would be A

 

[Pwincessajwa]

q7 - Sucrose is made up of Fructose and alpha glucose. I cannot really explain this question, as you have to memorise the structures of those substances. Fructose is 3 and alpha glucose is 1 so the answer is B

q21 - These questions are usually confusing, so I remember it best as (tRNA is the same code as DNA but U is replaced with T) and (mRNA is the same as the sequence on the amino-acid itself) so since the sequence of tRNA is CUU-UUA-UGG-GGA, then the sequence of DNA would be CTT-TTA-TGG-GGA which is C

q30 - The answer is obviously B because as sucrose concentration decreases in the leaf, it decreases in the phloem and vise versa. While A is true, it is not what the question is asking for, as you cannot tell from the graph that water moves by osmosis into the phloem (water isn't mentioned at all!). C is the same as A, it is true, but you can't tell from the graph if sucrose moves in both directions. D - there is no mention of xylem on the graph so you can't tell if it's true or not.

q32 - You have to memorise these as well from the biology book. Trachea would be largest, then the bronchus, then bronchioles then the alveoli. Trachea is 18mm, Bronchus is 12mm, bronchioles are 0.50 mm, and finally alveoli are 0.25mm. Answer would be A

thyank u so much i will ask ma doubts from other papers tomorow bcz i have not solved them yet well thankss

 

[HubbaBubba]

thyank u so much i will ask ma doubts from other papers tomorow bcz i have not solved them yet well thankss

No problem, take your time

 

[lenin]

so who did variant 31 are u done with ur exams how was it ??????????

 

[ShootingStar]

q40 please.
http://papers.xtremepapers.com/CIE/...nd AS Level/Biology (9700)/9700_w12_qp_11.pdf