AS Biology P1 MCQs Preparation Thread

[geek101]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Biology (9700)/9700_w06_qp_1.pdf
MCQ 15 PLS
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Biology (9700)/9700_s06_qp_1.pdf
MCQ 6,20,29 ,30

nov 06
15) as the concentration of sucrose decreases the water potential outside decreases as compared to inside so water moves in.
At 0.1 the water potential outside is lesser than inside so water moves it and the %change in length increases.

june 06
6) the actual length of the layer is 8nm >> 8 x 10^-9 m
measure with ur ruler to get the length of the image >> 0.02 m
magnification = image / actual
= 0.02 / (8 x 10^-9) = 2.5 x 10^6

20) radioactive nucleotides will be incorporated when there is an exposed template or something, this happens only during replication which occurs during the interpahse.

29) the curve shifts to the right when CO2 conc is high. when CO2 levels rise the conc of H+ ions also increases. at a low pH the H+ conc is very high. so D

30) this question i think was removed from the exam but anyway...
at 1 the av valves close and at 2 semilunar open....which means between 1 and 2 they are both closed and the time between 1 and 2 is 0.04. the same is for 3 and 4 but at 3 semilunar cose and at 4 av open, but between them they are both closed, and the time is 0.04. so the total time is 0.04 + 0.04 = 0.08, the closest answer is C

 

[fathimaash]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Biology (9700)/9700_s07_qp_1.pdf
mcq 1,15 and 26
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Biology (9700)/9700_w07_qp_1.pdf
MCQ 1,6,25,,31,40

q1- frm da scale bar u knw dat 2nm=1cm den u measure da width of da membrane which is 3.75cm. if 1cm=2nm den 3.75cm=7.5nm (3.75*2)
so 7.5/10^9= 7.5*10^-9 => C
q25- onli wen da water potential is higher in da soil water moves to da xylem due to water potential gradient.
onli in =>B da water potental in soil is more dan da water ( less negative)
q31-X shows da graph for a person with healthy lung look at Z da graph is v.low dis shows its emphysema patent bcoz da ari brathed out is v.low due to brakdown of elastin and Y of a person of chronic bronchits => C
q40- its 0.75/1.75v bcoz u knw NP=GPP-R so in producers NP=4-2.25=1.75 nd da % passd is o.75 so answr is =>C

 

[fathimaash]

may 2005 mcq 14 pls

k will go through da options first
B=> At Q, the kinetic energy of enzyme and substrate is highest.- no its not
C=> At R, peptide bonds in the enzyme begin to break-

At S, the substrate is completely denatured- no its da enzyme dats dentaured so option left is



============>A
hope u get it

 

[fathimaash]

hay guys i kinda need help with dis quiz http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Biology (9700)/9700_s04_qp_1.pdf
Q36 its regarding blood groups and some antibody/antigen thingy kinda stuck dr mind helping poor ol me pretty plzzzzzzzzzz

thanx

i think its no longer in da AS syllabus....

 

[Haniya Khan]

q1- frm da scale bar u knw dat 2nm=1cm den u measure da width of da membrane which is 3.75cm. if 1cm=2nm den 3.75cm=7.5nm (3.75*2)
so 7.5/10^9= 7.5*10^-9 => C
q25- onli wen da water potential is higher in da soil water moves to da xylem due to water potential gradient.
onli in =>B da water potental in soil is more dan da water ( less negative)
q31-X shows da graph for a person with healthy lung look at Z da graph is v.low dis shows its emphysema patent bcoz da ari brathed out is v.low due to brakdown of elastin and Y of a person of chronic bronchits => C
q40- its 0.75/1.75v bcoz u knw NP=GPP-R so in producers NP=4-2.25=1.75 nd da % passd is o.75 so answr is =>C

thanku

 

[Haniya Khan]

nov 06
15) as the concentration of sucrose decreases the water potential outside decreases as compared to inside so water moves in.
At 0.1 the water potential outside is lesser than inside so water moves it and the %change in length increases.

june 06
6) the actual length of the layer is 8nm >> 8 x 10^-9 m
measure with ur ruler to get the length of the image >> 0.02 m
magnification = image / actual
= 0.02 / (8 x 10^-9) = 2.5 x 10^6

20) radioactive nucleotides will be incorporated when there is an exposed template or something, this happens only during replication which occurs during the interpahse.

29) the curve shifts to the right when CO2 conc is high. when CO2 levels rise the conc of H+ ions also increases. at a low pH the H+ conc is very high. so D

30) this question i think was removed from the exam but anyway...
at 1 the av valves close and at 2 semilunar open....which means between 1 and 2 they are both closed and the time between 1 and 2 is 0.04. the same is for 3 and 4 but at 3 semilunar cose and at 4 av open, but between them they are both closed, and the time is 0.04. so the total time is 0.04 + 0.04 = 0.08, the closest answer is C

THANKU

 

[geek101]

n

THANKU

no problemo! Good luck

 

[raamish]

do the processes exocytosis and endocytosis require energy or not? and do proteins travel by exocytosis out of cells and endocytosis into cells?

 

[hmlahori]

For Q17, red blood cells are animal cells and therefore, do not have a cell wall. Cell wall is present in plant cells and withstands the turgor pressure to make the plant cell turgid. In red blood cells, there is no cell wall to withstand the pressure due to the water that moves in, so the cell bursts (known as lysis/haemolysis/cytolysis). Remember turgidity is only for plant cells, not animals cells. Also, haemoglobin is a large protein so it won't be able to pass through the transport/channel proteins. Only small molecules such as water soluble vitamins, glucose and amino acids can pass through these integral proteins.
For Q1, you calibrated the gracticule correctly (1 division is equal to 2.5 um), but you measured the chloroplast incorrectly! You need to align the ends of one of the chloroplasts with the scale given below the picture of the cell. One chloroplast is approximately equal to 4 divisions. (since 1 division is equal to 2.5 um so 4 divisions are equal to 2.5 x 4 = 10um). Also 1.8 cm is too large for a chloroplast (although you dont need to measure it, perhaps you are measuring another structure; chloroplasts are the small circular structures within the cell that are densely stained and most in number.)
For Q18, The telomeres are reduced in the replicated DNA since they are not completely replaced. With substance X the telomere quantity in replicated DNA remains the same as that in the DNA which was replicated (telomeres are completely replaced), so promotion of DNA replication (in interphase of cell cycle) is not reduced and the cell divides continually . Without substance X telomeres are not completely replaced so their amount in replicated DNA decreases, which in turn reduces promotion so the finally no DNA is replicated and cell division stops as a result (If there is no DNA replication, the cell cannot divide and reproduce). Don't confuse replacement of telomeres. Substance X does not take the place of the telormers, it simply causes them to be completely replaced in replicated DNA. The Answer is A.

Thanks! that helped me loads especially the explanation on the telomeres question.

 

[Pinklady]

PLEASE HELP.
MAY JUNE 06
Q #13 and 17

 

[raamish]

do the processes exocytosis and endocytosis require energy or not? and do proteins travel by exocytosis out of cells and endocytosis into cells?

 

[Pinklady]

PLEASE HELP.
MAY JUNE 06
Q #13 and 17

 

[JulyMei]

I'm confuse with this... the answer is A.. Why not B?

 

[JulyMei]

PLEASE HELP.
MAY JUNE 06
Q #13 and 17

link, please?

 

[daisy1234]

Bio on02 no.30,32
mjo6 no.30
on10/11 no.5,38,39.
URGENT!PLZ HELP!!!!

 

[Mashroor]

Its B i think. if it its right then i can make some solid explanation for that. in short i think this practice will not replenish the soil fertility as plant seeds are grown on the same soil. A option will provide the alternative to fertilizer as the previous crop waste material will act as fertiliser.

Heres an opinion.... I think leguminous plants can provide both ammonium ion and nitrate ion which synthetic urea or animal dung cannot provide!!And option A is not suitable since cereal crops cannot replenish the soil with any nitrogen compounds, which is obvious!

 

[Pinklady]

PLEASE HELP.
MAY JUNE 06
Q #13 and 17

Please help with thisssssssss!!!!

 

[Mashroor]

but how do i know the answer? the answer is B though.. about active transport..

Heyy I think a possible explanation is that in phloem the sucrose concentration is always greater than the sucrose concentration in the leaves so it has to be actively transported, right??

 

[thunderingthunder]

it is B cz they are not talkin abt da DNA frm which da mRNA ws made bt da non trancribed one therefore da ans is B
HOPE ITS HELPFUL!!!!!!!!!!!!!

thank you

 

[JulyMei]

Heyy I think a possible explanation is that in phloem the sucrose concentration is always greater than the sucrose concentration in the leaves so it has to be actively transported, right??

woopss.. which page was this? I forget which question I was asking.. LOL..