AS Biology P1 MCQs Preparation Thread

[HubbaBubba]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Biology (9700)/9700_w12_qp_11.pdf
Q5 why is c wrong?

q5 - C may seem slightly correct, but B seems even more correct. 220nm is 0.22micrometers so you won't be able to see anything that's 0.2 micrometers in width. You should take the one that is more true. Also if I recall, I don't recall being able to see two distinct membranes during practicals in light microscopes, so I guess you had to remember that?

 

[sweetjinnah]

From the figure you have, choose one of the chloroplasts present and see how many eye pieces graticule it occupies with your eyes.

owk i got it my mistake i was taking the whole cell...sorryyyyyyy to disturb u...

 

[Cesar_malik]

3

http://papers.xtremepapers.com/CIE/...nd AS Level/Biology (9700)/9700_w12_qp_11.pdf
Q5 why is c wrong?

300nm is larger then 600nm and since we half the wavlength to get the smallest difference which can be visable it is correct
B is correct ans (incorrect choice because 220 is smaller then .2 micros (200nm is .22 micrometer)
Just guessing iam not sure still
Anyone helping me with q7 paper 13 2011 oct novb

 

[HubbaBubba]

3
300nm is larger then 600nm and since we half the wavlength to get the smallest difference which can be visable it is correct
B is correct ans (incorrect choice because 220 is smaller then .2 micros (200nm is .22 micrometer)
Just guessing iam not sure still
Anyone helping me with q7 paper 13 2011 oct novb

Each division on the SM has 40 EPG units. Since each SM division is 0.1mm, that means each 0.1mm has 40 EPG units. Therefore 1 EPG unit is = 0.0025 mm (2.5micrometers)

Width begins at 61 units and ends at 70 units on EPG so width of the nucleus is 9 units. 9 * 2.5 = 22.5. The answer would be C

 

[Cesar_malik]

Each division on the SM has 40 EPG units. Since each SM division is 0.1mm, that means each 0.1mm has 40 EPG units. Therefore 1 EPG unit is = 0.0025 mm (2.5micrometers)

Width begins at 61 units and ends at 70 units on EPG so width of the nucleus is 9 units. 9 * 2.5 = 22.5. The answer would be C

He says the difference between two small divisons are .1 mm doest this make .1mm=80 divisons that us confusion
Thanks anyways

 

[HubbaBubba]

He says the difference between two small divisons are .1 mm doest this make .1mm=80 divisons that us confusion
Thanks anyways

Nope when they say each division, they mean each small division. Don't get confused with that! You could lose a lot of marks!

 

[SalmaXXXX]

The graph simply shows effect of increasing substrate concentration on the rate of reaction, non-competitive inhibitor doesn't let the rate of reaction reach the maximum limit, competitive inhibitor only initially decrease the rate of reaction, but the increase in substrate concentration diminish it's effect, so it will reach the maximum limit after time.

(Remember that there is no time here, I got confused by it in the beginning)

Okay so how will the graph look like if it was the amount of product formed and time? Does adding a small amount of non- competitive inhibitors stop the reaction so that the amount of product formed becomes constant (so no reaction is taking place) when the uninhibited reaction forms more product. In other words, does the non competitive inhibitor affects the amount of product formed or only the rate?

 

[Cesar_malik]

Nope when they say each division, they mean each small division. Don't get confused with that! You could lose a lot of marks!

Yea so each small divison has a large divison in the middle doesnt it ?
My method worked for other questsion like this but not for this

 

[kiran11]

All you have to do is to find the GGP of the tertiary consumer ( fourth trophic level) which is not shown here. First calculate the GGP for 2nd TL 23000- 8000 = 15000. Then calculate GGP for 3rd TL 15000 - 10500 = 4500. Then calculate GPP for 4rth TL 4500 - 4200 = 300. Now calculate the percentage 300/23000 x 100 = 1.3. Hope that was helpful

very helpful! i also had a problem with this question. thank u!!

 

[HubbaBubba]

Yea so each small divison has a large divison in the middle doesnt it ?
My method worked for other questsion like this but not for this

You can find a site which explains it very clearly if you google calibration, but when the question says each division, they literally mean 'every division'

 

[Pwincessajwa]

http://papers.xtremepapers.com/CIE/...nd AS Level/Biology (9700)/9700_s11_qp_12.pdf q1,610,15,16 plzzz help me ????

 

[HubbaBubba]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Biology (9700)/9700_s11_qp_12.pdf q1,610,15,16 plzzz help me ????

q1 - They pretty much calibrated it for you. If one EPG unit is 0.005mm, then 3.5 units are (0.005 * 3.5) 0.0175mm which is 17.5 micrometers. C is the answer.

q10 - Only peptide bonds are formed between amino acids in the primary structure. Any other bonds are formed later in the secondary, tertiary, and Quaternary structures. D is correct.

q15 - Everything except for the carbohydrate chain has hydrophobic and hydrophillic regions. The inside of the transport protein is hydrophillic, while the outer part is hydrophobic. Cholesterol and proteins also have it. Usually the hydrophobic bit, is inside the bi layer while the hydrophillic part faces outwards. C is right.

q16 - You need to know that saturated hydrocarbons in the hydrophobic tails of the phospholipid make the membrane more rigid, while unsaturated hydrocarbons make it more fluid. C is the right answer

 

[SalmaXXXX]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Biology (9700)/9700_s11_qp_12.pdf q1,610,15,16 plzzz help me ????

Q1 ) the diameter of cell = the measure in eye piece graticule x 0.005 = 0.0175 mm x 1000 = 17.5 micrometre. which is approximately 18 micromtre.
Q6 ) Amylopectin and glycogen are the only branched chains we have in the syllabus. No glycogen is here, so the branched is amylopectin B, and cellulose is not branched.
Q 10) Peptide bond only. Hydrogen bonds occur between the R groups containing -COOH and _NH2, disulphide bond occurs between amino acids whose R groups contain S. S here is not in the R group so the only bond is peptide
Q15) All the components have both hydrophilic region which is facing the medium and hydrophobic region facing each other except glycoproteins and glycolipids since they only only face the water so are only hydrophilic.
Q16) double bonds in fatty acid chains makes "kinks" which increases the gap between phospholipids thus increasing the fluidity.
Hope I helped

 

[Alicee]

http://papers.xtremepapers.com/CIE/...nd AS Level/Biology (9700)/9700_w12_qp_12.pdf
12, 23, 33,34 and 39! pleeease explain

 

[1357911]

O/N/12 varient 11 q-33 and q-36can anybody help?

 

[1357911]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Biology (9700)/9700_w12_qp_12.pdf
12, 23, 33,34 and 39! pleeease explain

q-12
Its something that u have to learn, Tertiary and quartanary structures only contain hudrogen,disulphide,ionic bonds and hydrophobic interactions.
q-23
Look at the amino acid sequence. There are two amino acids of the same type. (leu).However if u look at the other options they are also correct, but they are only asking about the triplet code.

 

[Alicee]

q-12
Its something that u have to learn, Tertiary and quartanary structures only contain hudrogen,disulphide,ionic bonds and hydrophobic interactions.
q-23
Look at the amino acid sequence. There are two amino acids of the same type. (leu).However if u look at the other options they are also correct, but they are only asking about the triplet code.

no peptide bond? i thought all protien had them!!
what about the oders?
n thanks!!

 

[1357911]

no peptide bond? i thought all protien had them!!
what about the oders?
n thanks!!

And no tertiary and quarternary structures do not contain peptide bonds

Im sorry but which orders are u talking about

 

[Pwincessajwa]

Q1 ) the diameter of cell = the measure in eye piece graticule x 0.005 = 0.0175 mm x 1000 = 17.5 micrometre. which is approximately 18 micromtre.
Q6 ) Amylopectin and glycogen are the only branched chains we have in the syllabus. No glycogen is here, so the branched is amylopectin B, and cellulose is not branched.
Q 10) Peptide bond only. Hydrogen bonds occur between the R groups containing -COOH and _NH2, disulphide bond occurs between amino acids whose R groups contain S. S here is not in the R group so the only bond is peptide
Q15) All the components have both hydrophilic region which is facing the medium and hydrophobic region facing each other except glycoproteins and glycolipids since they only only face the water so are only hydrophilic.
Q16) double bonds in fatty acid chains makes "kinks" which increases the gap between phospholipids thus increasing the fluidity.
Hope I helped

thank u soooo muchhhhhhhhh i am gonna ask so many doubts if there is no prob with u ????

 

[SalmaXXXX]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Biology (9700)/9700_w12_qp_12.pdf
12, 23, 33,34 and 39! pleeease explain

12) A primary structure doesn't contain hydrostatic bonds so option A is out. A secondary structure doesn't contain ionic bond so B is out. A quatenary structure does contain hydrogen bond so D is out. B is the correct answer, by the type of bonds found, he means the types of bond which maintain that specific structure. The peptide bond doesn't maintain the tertiary structure.

23) None of B, C , D is supported by the information given. However, it's clear that Leu amino acid has more than one triplet genetic code that codes it so option A is correcr answer.

33) Deposition of fatty materials in arteries surely can result a heart failure. But he stated in arteries AND veins. Deposition in veins plays no role thus excluding 2. The other three are correct so option D is correct answer.

39) that's easy, nitrification takes place whether from ammonium to nitirites or from nitrites to nitrates so option C is corect.