AS Biology P1 MCQs Preparation Thread

[Bluejeans#]

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

 

[fizzah96]

18 Which statement describes a cell that is capable of reproduction and belonging to a haploid
organism?
A It has chromosomes that contain one polynucleotide chain.
B It is capable of carrying out a reduction division to form gametes.
C It possesses two copies of each gene as a result of fertilisation.
D It will undergo cell division by mitosis during asexual reproduction.

ans is D .. but i am pretty sure its B ... from o/n 2009 va'11

as it is already haploid so it cannot undergo meiosis and to maintain the chromosome no. it will carry out mitosis and asexual reproduction so its D not B because chromosome no. cannot be further reduced.

 

[Boo]

Hey, can someone explain to me QNS 32 in o/n 04 how do we get the ans B? :s

 

[ahmed abdulla]

as it is already haploid so it cannot undergo meiosis and to maintain the chromosome no. it will carry out mitosis and asexual reproduction so its D not B because chromosome no. cannot be further reduced.

thanks ...
can u help with this

 

[fizzah96]

thanks ...
can u help with this

see at 1 atrio ventricular valves close and semilunar are also closed.semilunar open at 2,so the time period between 1 and 2 is 0.03 sec(both are closed)
at 3 semilunar open while atrioventricular are still closed that open at 4.the time period between these is 0.04 sec.so total time period becomes 0.07 sec.

 

[sweet life]

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Biology%20(9700)/9700_s12_qp_12.pdf

guys sorry can u plz just tell me Q1 and Q11
it will be a great help thank you!

 

[sweet life]

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Biology%20(9700)/9700_s12_qp_12.pdf

plz can any one help me with Q1 ,11 and 30 of this paper it will be really a great help thank you

 

[h4rriet]

http://papers.xtremepapers.com/CIE/...nd AS Level/Biology (9700)/9700_s12_qp_12.pdf

plz can any one help me with Q1 ,11 and 30 of this paper it will be really a great help thank you

1. That's just something you have to memorise.
11. A glycogen molecule has 1,4 and 1,6 bonds.
30. The stroke volume will increase when the ventricular end-diastolic volume increases, but not proportionally because not all the blood will leave the systole.

 

[sweet life]

1. That's just something you have to memorise.
11. A glycogen molecule has 1,4 and 1,6 bonds.
30. The stroke volume will increase when the ventricular end-diastolic volume increases, but not proportionally because not all the blood will leave the systole.

Thank you soo much

 

[Pwincessajwa]

i need help in so many papers will anyone help plz ?

 

[HubbaBubba]

Hey, can someone explain to me QNS 32 in o/n 04 how do we get the ans B? :s

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

 

[HubbaBubba]

i need help in so many papers will anyone help plz ?

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

 

[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!