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A level Biology: Post your doubts here!

Student12

Student12

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http://papers.xtremepapers.com/CIE/...nd AS Level/Biology (9700)/9700_w09_qp_11.pdf
Q 2 D , Q 22 A , 17 B , 26 D , 29 B
http://papers.xtremepapers.com/CIE/...nd AS Level/Biology (9700)/9700_w09_qp_11.pdf
Q 14 B , 20 A, 22 C
http://papers.xtremepapers.com/CIE/...and AS Level/Biology (9700)/9700_w08_qp_1.pdf
Q 4 B, 24 D, 28 C
http://papers.xtremepapers.com/CIE/...and AS Level/Biology (9700)/9700_w07_qp_1.pdf
Q 6 A, 25 B,28 B (why not 4) ,40 C
http://papers.xtremepapers.com/CIE/...and AS Level/Biology (9700)/9700_s07_qp_1.pdf
Q 1B, 4 A, 15 C.
 
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Irfan1995

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magnesium said:
http://papers.xtremepapers.com/CIE/...nd AS Level/Biology (9700)/9700_w11_qp_53.pdf
CAN someone tell how to find expected values in Q2
Click to expand...

They tell you that there is a cross between two cats without a tail. Since we know that the 'no tail' allele is dominant, and that there are some offspring that have tails, then both parents are heterozygous for that gene (Aa). If one of the parents were homozygous (AA), then none of the offspring should have a tail.

If you take two heterozygous parents and cross them, you should get a phenotype ratio of 3:1 (dominant:recessive). So 75% of the cats don't have tails and 25% of them should have a tail.
The total number of offspring is 21 + 32 + 19 + 40 = 112
We expect that 3/4 of them should not have a tail. So 112 x (3/4) = 84 cats should not have a tail.
And 1/4 of them should have a tail, so 112 x (1/4) = 28 cats should have a tail.
These are only the data we expect to acquire.
 
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Irfan1995

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externityxzx said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Biology (9700)/9700_w12_qp_53.pdf
please help with 1.(b)(i)!! im not sure what to do with the given molar mass.... thanks!!
Click to expand...

I'm not sure if you took Chemistry before, but if you did, then this should be pretty simple.

We know that the number of moles = mass/molar mass

We want to prepare a 25 mmol/dm^3 solution. Simply add 25 mmol of sodium nitrate to 1 dm^3 of distilled water.
25 mmol = 25 x 10^-3 moles
So now we want to find the mass required.
25 x 10^-3 = mass/85
mass = 25 x 10^-3 x 85 = 2.125 g.

So you have to add 2.125 grams of sodium nitrate to 1 dm^3 of distilled water to get a 25 mmol/dm^3 solution of sodium nitrate.
 
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Irfan1995

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Student12 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Biology (9700)/9700_w09_qp_11.pdf
Q 2 D , Q 22 A , 17 B , 26 D , 29 B
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Biology (9700)/9700_w09_qp_11.pdf
Q 14 B , 20 A, 22 C
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Biology (9700)/9700_w08_qp_1.pdf
Q 4 B, 24 D, 28 C
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Biology (9700)/9700_w07_qp_1.pdf
Q 6 A, 25 B,28 B (why not 4) ,40 C
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Biology (9700)/9700_s07_qp_1.pdf
Q 1B, 4 A, 15 C.
Click to expand...

Oct/Nov '09:
2) 1 million nanometers = 1 millimeter. You can continue from there.
22) A tRNA molecule can bind to one type of amino acid. After it 'uses up' its amino acid, it goes back to the cytoplasm to bind with another one. After that, it returns to the ribosome to 'use it up' again.
The question tells you that there are only 4 types of amino acids, so you only need 4 different tRNA to translate the polypeptide.
17) Remember that active transport works against a concentration gradient (moving solutes from a region of low concentration to a region of high concentration). So potassium will move inside the cell and sodium moves out.
26) The solute potential is always negative, and the pressure potential is always positive. The water potential is the sum of the other two potentials, so it will be found somewhere in between them.
29) You are looking for the greatest CHANGE. So you're looking for the region where the slope is the highest. That's going to be region B
14) 20) 22) Are you sure you copied the right question paper?

Oct/Nov '08:
4) I think you just have to measure the distance with a ruler and calculate it from the magnification.
24) Actually, D is an adaptation. I'm pretty sure that option B is the right answer.
Water uptake is done through osmosis (a passive process which does not require energy). So mitochondria are not needed as energy is not required.
28) I mean, option C is correct while the others are wrong. Here are the corrections:
A: blood flows into the aorta when the LEFT ventricle contracts
B: blood flows into the left atrium through the pulmonary VEIN ...
D: blood flows into the left ventricle through the AV VALVE...

Oct/Nov '07:
6) Measure it by a ruler and calculate it using the magnification equation
25) Water flows from the roots to the xylem vessels. So the roots must have a higher water potential than the xylem vessels. That makes options C and D wrong. The soil water must contain some salts, so it's impossible to have a water potential of 0. It has to be less than that. So option A is wrong. Option B is the only one left.
28) They're asking for how many oxygen ATOMS are carried. Each oxygen molecule contains two atoms (O2). 4 molecules are carried by one haemoglobin molecule which is equivalent to 8 oxygen atoms.

May/June '07:
1) Simply divide the values to get a ratio of their sizes. Remember to convert 750 nm to 0.75 micrometers (or convert 15 micrometers to 15,000 nanometers)
So 15/0.75 = 20
Or, 15000/750 = 20
You get the same answer both ways.
4) The first figure tells you that 0.1 mm is equivalent to 50 divisions. So each division is 2 micrometers long.
By looking at the horizontal pollen tube, you can see that it grew from the 25th division to the 35th one. So it grew by 10 divisions which is equal to 20 micrometers. But, they also say that it grew 20 micrometers in 4 hours, so in 1 hour, it gre 20/4 = 5 micrometers. So the rate is 5 micrometers per hour.
15) Air contains oxygen, which is required for active transport (it's needed to produce ATP). Nitrogen cannot be used for active transport.
If you bubble air or nitrogen through the solution, the rate of the 3-carbon sugar doesn't change. This tells you that it doesn't involve active transport, which implies that it travels through diffusion.
The rate of the 6-carbon sugar was high when air was bubbled, but it completely stopped when nitrogen was bubbled. This tells you that air (hence oxygen) was necessary for uptake, so it must have been done through active transport.
 
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Irfan1995

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Student12 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Biology (9700)/9700_s06_qp_1.pdf
Q 6 C, 8 D, 22C, 23 B
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Biology (9700)/9700_w06_qp_1.pdf
Q 18 A
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Biology (9700)/9700_w05_qp_1.pdf
Q 23 B
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Biology (9700)/9700_w04_qp_1.pdf
Q 13 A
Click to expand...


May/June '06:
6) They tell you that the actual size was 8 nm. You have to measure the width of the bilayer using a ruler. Then use the magnification equation (magnification = measured size/actual size) to get the magnification.
8) Starch is broken down to maltose which is a reducing sugar. You use Benedict's test to confirm for that. If the solution turns from blue to brick red, then the solution contained a reducing sugar. So the correct answer is A. If you want to use iodine, then the colour change would be from blue-black to yellow.
22) You originally have two strands of N15 DNA. When the cell is dividing, the two strands split, and an N14 strand is added to each one. So you will have two pieces of DNA made up of N15-N14.
23) Only one strand is transcribed (not both), so only 3,000 bases will be transcribed.
Every 3 bases code for 1 amino acid, so you will have 1,000 amino acids in the end.

Oct/Nov '06:
18) When mitosis ends, you will have two cells with the exact same number of chromosomes as the original cell. That would correspond to option A.

Oct/Nov '05:
23) They give you two different anticodons which code for the same amino acid. So any one of those codons could code for a that base. For example, UCC codes for arginine, and GCG codes for arginine as well. Both are correct.
Now you just have to find an option which has any one of those anticodons. But remember that the original DNA template will replace all the uracil with thymine (so there will be T instead of U).

Oct/Nov '04:
13) Lets say you have 2 letters. How many 3-letter words can you make? Here are all the options: AAA, AAB, ABA, ABB, BAB, BAA, BBA, BBB. That's a total of 8 options. In other words, that's a total of 2^3 possibilities.
If you have a chain of length r amino acids (r empty slots), and you have n different amino acids (letters), then you can make n^r different polypeptides.
 
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raamish

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Irfan1995 said:
They tell you that there is a cross between two cats without a tail. Since we know that the 'no tail' allele is dominant, and that there are some offspring that have tails, then both parents are heterozygous for that gene (Aa). If one of the parents were homozygous (AA), then none of the offspring should have a tail.

If you take two heterozygous parents and cross them, you should get a phenotype ratio of 3:1 (dominant:recessive). So 75% of the cats don't have tails and 25% of them should have a tail.
The total number of offspring is 21 + 32 + 19 + 40 = 112
We expect that 3/4 of them should not have a tail. So 112 x (3/4) = 84 cats should not have a tail.
And 1/4 of them should have a tail, so 112 x (1/4) = 28 cats should have a tail.
These are only the data we expect to acquire.
Click to expand...

hey in n09.52 i dont understand the pts 2,3 and 4. can u plzz explain those pts to me so that i know what is their purpose and when to use them in my asnwers. By the way these pts should mostly be used in microscope right? i dont understand why the no. of grains are related with the area. what it also means by counting pollen in field of view. Thanx :)
 
magnesium

magnesium

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Mairaxo

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hela said:
ON 2004 Q 16B Q38 D Q29B Q30A
http://papers.xtremepapers.com/CIE/...and AS Level/Biology (9700)/9700_w04_qp_1.pdf
ON 2003 Q7D http://papers.xtremepapers.com/CIE/...and AS Level/Biology (9700)/9700_w03_qp_1.pdf
Click to expand...
Q16- phosphate and hydrophilic head occupies less space. Hydrocarbon chains are large and have a chain structure.
Q29- xylem is dead. heating will not affect it. only phloem will get affected.
Q30- xylem will be affected 1st. water moves through xylem and its the appoplast pathway. so if fungi grow there it gets affected.
Q 38- photosynthesis is needed for growth and respiration as it provides glucose. so 21500*500*1500=23500
Q7- water potential=solute potential+pressure potential. water and solute potential are equal so pressure is 0
 
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