AS Biology P1 MCQs Preparation Thread

[perky1]

can anyone explain area of field of view ?? Q.3 M/J2010


http://papers.xtremepapers.com/CIE/...nd AS Level/Biology (9700)/9700_s10_qp_11.pdf

 

[6Astarstudent]

as it's a circle therefore the area is gonna be pi.r^2 so the radius is 55micrometer(found using eyepiece graticule) and they have already given you the value of pi i.e 3.14 so the area is found by multiplying 3.14*55*55=9498.5 and when written in standard form 9498.5 = 9.5*10^3

no answer is C
its 3.14*125*125=4.9*10^4
the radius is 50 small divisions and we know that 40 small divisions = 0.1mm or 100micrometers
each small division is therefore 100/40 = 2.5 micrometers
radius is thus 50x2.5= 125 micrometers

 

[Apuurv A Mehra]

no answer is C
its 3.14*125*125=4.9*10^4
the radius is 50 small divisions and we know that 40 small divisions = 0.1mm or 100micrometers
each small division is therefore 100/40 = 2.5 micrometers
radius is thus 50x2.5= 125 micrometers

shit sorry its jus slipped out my mind! :x

 

[panpan101]

Ans is C, cause only 9 bubbles can be produced, the mean number can't be in points? Imaging 9.2 bubbles that seems wrong, hence it's A, common sense, any idea about q19 same year and variant?

I think that the ans is A... cause if DNA replication stops, mitosis cannot occur any longer. Thus, the process of cell division will stop, but not immediately, because the telomeres are not completely replaced. So it eventually slows and stops until enough cells are produced.

If the telomeres are replaced, there will be DNA replication happening relentlessly. So the cell can keep on dividing.

Is the ans even A? lol

 

[sam1234]

I think that the ans is A... cause if DNA replication stops, mitosis cannot occur any longer. Thus, the process of cell division will stop, but not immediately, because the telomeres are not completely replaced. So it eventually slows and stops until enough cells are produced.

If the telomeres are replaced, there will be DNA replication happening relentlessly. So the cell can keep on dividing.

Is the ans even A? lol

Ans is A but your explanation doesn't seem to be right. Please explain more :/

 

[panpan101]

Basically, if interphase doesnt take place, mitosis will not happen.

With X,
Telomeres are replaced. So DNA replication will continue and mitosis will still continue

Without X,
Telomeres are still replaced... but NOT COMPLETELY. So once these telomeres gradually decrease in number, the process of cell division will eventually slow and stop. It does not stop immediately as there are still some telomeres but these telomeres are being used up every division. Once they are all used up, cell division stops.

 

[Trevor Jts]

Hi all How to do Nov 2008 Q8?

 

[Lujain M.]

The answer is C because basically sucrose is glucose + fructose. Fructose is an isomer of glucose so it has to be either C or D, C is the answer because it's alpha glucose as the OH group is belowCarbon . Hope it's clear.

 

[Trevor Jts]

The answer is C because basically sucrose is glucose + fructose. Fructose is an isomer of glucose so it has to be either C or D, C is the answer because it's alpha glucose as the OH group is belowCarbon . Hope it's clear.

Thank you!

 

[Kinki Tiffany 12]

http://papers.xtremepapers.com/CIE/...nd AS Level/Biology (9700)/9700_w12_qp_12.pdf

hey guys, how to do question 34?

 

[Trevor Jts]

Tha

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Biology (9700)/9700_w12_qp_12.pdf

hey guys, how to do question 34?

I've tried a few times and I finally figured out..
5dm3 = 5000cm3
100 cm3 of arterial blood carries 19 cm3 of oxygen. 100 cm3 of venous blood carries 12.5 cm3 of oxygen. Therefore oxygen retained in the body = 19-12.5 = 6.5
5000/100 = 50cm3
50cm3*6.5cm3 = 325cm3, hence the answer.
Hope this helps

 

[Trevor Jts]

Hey guys, this is a J/11/11 question 14
Some inhibitors of enzyme reactions bind to the enzyme/substrate complex.
which statements about this type of inhibtion are correct?
1 The active site changes shape
2 The inhibitor is non competitive
3 The initial rate of reaction is reduced
4. The maximum rate of reaction is increased
A. 1 and 2 only B. 1 and 3 only C 2 and 3 only D. 2, 3, 4
The answer is C, which includes 2 and 3. Why 1 is not included? Pls help

 

[6Astarstudent]

Hey guys, this is a J/11/11 question 14
Some inhibitors of enzyme reactions bind to the enzyme/substrate complex.
which statements about this type of inhibtion are correct?
1 The active site changes shape
2 The inhibitor is non competitive
3 The initial rate of reaction is reduced
4. The maximum rate of reaction is increased
A. 1 and 2 only B. 1 and 3 only C 2 and 3 only D. 2, 3, 4
The answer is C, which includes 2 and 3. Why 1 is not included? Pls help

there are 2 types of non competitive inhibition
one is by binding to the allosteric site (ie. anywhere not on active site) and change the structure of active site. this type of inhibition only occurs when the substrate is NOT bond to enzyme yet.

The question states that the inhibitor binded to the enzyme/substrate complex, this is the second type of non competitive inhibition.
This inhibition can occur either like a competitive inhibitor by binding on the enzyme active site but permanently, or by binding on to the enzyme substrate complex and stop formation of product (the complex can only revert back to reactants, or an enzyme inhibitor complex).
Active site shape is not changed in this way.

 

[Trevor Jts]

there are 2 types of non competitive inhibition
one is by binding to the allosteric site (ie. anywhere not on active site) and change the structure of active site. this type of inhibition only occurs when the substrate is NOT bond to enzyme yet.

The question states that the inhibitor binded to the enzyme/substrate complex, this is the second type of non competitive inhibition.
This inhibition can occur either like a competitive inhibitor by binding on the enzyme active site but permanently, or by binding on to the enzyme substrate complex and stop formation of product (the complex can only revert back to reactants, or an enzyme inhibitor complex).
Active site shape is not changed in this way.

Your explanation is really helpful. Thank you!

 

[Kinki Tiffany 12]

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

can anyone explain question 18 and 40 to me?

 

[6Astarstudent]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Biology (9700)/9700_w12_qp_11.pdf

can anyone explain question 18 and 40 to me?

#18 answer is A because mitosis is a form of asexual reproduction
characteristics: daughter cell and parent cells are identical ie. same jobs, DNA unchanged
so B,C,D is all correct
and semi conservative replication occurs before mitosis during interphase so A is wrong

#40
we find the energy transferred to tertiary, so we deduct all energy loss 23000 - 8000 - 10500- 4200=300
300/23000 x100% = 1.3%

 

[perky1]

can anyone explain area of field of view ?? Q.3 M/J2010
http://papers.xtremepapers.com/CIE/...nd AS Level/Biology (9700)/9700_s10_qp_11.pdf

 

[6Astarstudent]

can anyone explain area of field of view ?? Q.3 M/J2010
http://papers.xtremepapers.com/CIE/...nd AS Level/Biology (9700)/9700_s10_qp_11.pdf

its 3.14*125*125=4.9*10^4
the radius is 50 small divisions and we know that 40 small divisions = 0.1mm or 100micrometers
each small division is therefore 100/40 = 2.5 micrometers
radius is thus 50x2.5= 125 micrometers
area = pi x ( radius squared)

 

[Trevor Jts]

Hey anyone? For question 21 in June 2005 exam,
How could I identify metaphase?

 

[6Astarstudent]

Hey anyone? For question 21 in June 2005 exam,
How could I identify metaphase?

firstly question states its either anaphase or metaphase
metaphase is when the chromosomes line up at equator ie. not separated
anaphase is when its separated to chromatids

in the picture they're not separated so it must be metaphase