A level Biology: Post your doubts here!

[San_mk]

The anxiety levels creeping up as August approaches .............anyways just wanted to share a message I received from sir hasham yesterday. He will be starting A level bio tuitions in mid august. In case any of you (livin in karachi) are interested in taking A-level bio tuitions, dont miss this opportunity. Having studied frm him, I can vouch for him on any day .....u dont find many graduates from aga khan teaching a levels! And trust me hell go out lengths to help you out in any way he possibly can.....and the bonus thing is that if your batch irritates him enough to prepare you for Aga khan, hell squeeze n time to help you with it

Best of luck fr your results! nd pray for me too

 

[azhutosh]

can any one provide me chapter wise structured question of biology(AS level)?? I wud really feel grateful.

 

[shahroxe]

Bio
P1 35/40
P2 48/60
P3 25/40

Is this enough for an A..... I gave AS level only :/

 

[perky1]

can anyone explain these questions ?
23. In a genetic engineering experiment a piece of double-stranded DNA containing 6000 nucleotides
is transcribed and translated.
What is the total number of amino acids used?
A 500 B 1000 C 2000 D 3000




24 DNA from a chromosome is analysed and 20% of its bases are found to be cytosine.
Which percentage of uracil molecules will be found in mRNA transcribed from this DNA?
A 20 B 30 C 40 D 60

 

[perky1]

23 A polypeptide has the amino acid sequence glycine – arginine – lysine – serine.
The table gives possible tRNA anticodons for each amino acid.
amino acid tRNA anticodons
arginine UCC GCG
glycine CCA CCU
lysine UUC UUU
serine AGG UCG

Which sequence of bases on DNA would code for the polypeptide?
A CCACGCAAGAGC
B CCTTCCTTCTCG
C GGAAGGAAAAGC
D GGTTGGTTGTGC

 

[Kinki Tiffany 12]

http://papers.xtremepapers.com/CIE/...nd AS Level/Biology (9700)/9700_s11_qp_11.pdf
can anyone explain q14, 25, 33 and 39 for me please?

 

[asma tareen]

Do we need to study abt the classification and kindoms etc in bio cux our teacher havent taught it :/

 

[Vounn Rose]

Hi everyone!! I would like to ask one question from the past year Paper 5. Here's the link, http://papers.xtremepapers.com/CIE/...nd AS Level/Biology (9700)/9700_w11_qp_53.pdf

It is question no. 2, section (a), part (ii), about cross 3 and 4.

I attempted the question and referred to the mark scheme, here's the link for mark scheme, http://papers.xtremepapers.com/CIE/...nd AS Level/Biology (9700)/9700_w11_ms_53.pdf

I could understand the part about cross 4, but when it says about "the idea of 1:1 ratio in each sex", I got startled. I tried the genetic diagram for both if the gene is sex-linked or non sex-linked, and the ratio of 'with tail' to 'without tail' I got is the same for both conditions (sex-linked or not).

So I need you guys to help figure it out. Thank you.

 

[Vounn Rose]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Biology (9700)/9700_s11_qp_11.pdf
can anyone explain q14, 25, 33 and 39 for me please?

For q14, the answer scheme provided says "c", but I personally think 1,2 and 3, but no option having such combination there. An enzyme-substrate complex is dismantled when the active site changes shape and repels the substrate before a reaction can be performed in time.

For q25, the scheme says "b", the keyword lies in "reducing water potential gradient", in other words it can be that it reduces the rate at which water is evaporating from one stoma. Feature 1, 2 and 3 helps trapping water vapour and increasing the humidity near the stomata, hence reducing water potential gradient. Water is evaporating at much slower rate.

For q33, there is a total of 5 membranes between the air in alveolus and interior of red blood cells, that is the 2 plasma membranes of alveolar cell (alveolar cell has only single layer of plasma membrane, but here the "2 plasma membranes" simply means that an air molecule has to pass through the membrane twice, imagine that), 1 basal membrane between alveolus and blood capillary, the endothelium of blood capillary and then the plasma membrane of red blood cell. Oxygen has to pass through 5 layers because it is carried inside red blood cells, carbon dioxide can be either 4 or 5 because it is carried either in the red blood cell or blood plasma.

For q39, the answer scheme says "d", simply add all the values of energy used by the plant according to the given table, which is 21500 + 1500 + 500, all in kJ unit. One note that "respiratory heat losses" may not be so convincing as the energy used, but I personally think the heat loss maybe contributes to temperature increase in plants, which can be considered as a use too...

Hope they help, and please correct me if I am mistaken with any point. =)

 

[Vounn Rose]

23 A polypeptide has the amino acid sequence glycine – arginine – lysine – serine.
The table gives possible tRNA anticodons for each amino acid.
amino acid tRNA anticodons
arginine UCC GCG
glycine CCA CCU
lysine UUC UUU
serine AGG UCG

Which sequence of bases on DNA would code for the polypeptide?
A CCACGCAAGAGC
B CCTTCCTTCTCG
C GGAAGGAAAAGC
D GGTTGGTTGTGC

The sequence says "glycine - arginine - lysine - serine", so it maybe possible that the sequence of amino acids in the polypeptide is in the reverse order, so that it is "serine - lysine - arginine - glycine ", because it doesn't matter as the polypeptide can be flipped around in the three dimensions, the result is still the same. But one thing has to be made certain here is that, if the anticodon says UCC for arginine, it means the code must be read in the order of U, then C and then C again. It cannot be CCU, it will not code for arginine this way.

We know that mRNA is transcribed from the base template of DNA by complementary base pairing, so the sequence of amino acids in mRNA is complementary to that in DNA, for example, base G on mRNA means base C on DNA. On the other hand, tRNA carries anticodon which contains base sequence that is complementary to that of mRNA, so it makes sense that tRNA has exactly the same sequence of base with that of DNA, except that base U would be present on tRNA, instead of base T.

Ok sorry for the lengthy background information, now back to the question. Now, there are 2 information we have to work on. First, the base sequence on tRNA is same to that on DNA, with exception that base T would appear on DNA in place of base U. Second, each amino acid has 2 sets of anticodon, either of which can be used.

One way to solve the question is to look at the answer choices directly, to see if any of those matches the correct base sequence. Remember that the sequence can be read in either direction, from right to left or from left to right. My answer for this would be B, that is CCT(glycine) TCC(arginine) TTC(lysine) TCG(serine). Look that it matches with the information provided about the tRNA anticodon, which has the same base sequence with DNA, with the exception.

Hope they help, and correct me if I am mistaken with any point. =)

 

[abdo ibrahim]

Notes for biodiversity plzz

 

[Vounn Rose]

Hi everyone!! I would like to ask one question from the past year Paper 5. Here's the link, http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Biology (9700)/9700_w11_qp_53.pdf

It is question no. 2, section (a), part (ii), about cross 3 and 4.

I attempted the question and referred to the mark scheme, here's the link for mark scheme, http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Biology (9700)/9700_w11_ms_53.pdf

I could understand the part about cross 4, but when it says about "the idea of 1:1 ratio in each sex", I got startled. I tried the genetic diagram for both if the gene is sex-linked or non sex-linked, and the ratio of 'with tail' to 'without tail' I got is the same for both conditions (sex-linked or not).

So I need you guys to help figure it out. Thank you.

Can anyone help out for this past year question?

 

[knowitall10]

Can anyone help out for this past year question?

Look through the *biology p5 tips* thread.
In the last few posts I remember someone had answered thos doubt

 

[Vounn Rose]

Notes for biodiversity plzz

I have one website providing short notes for biodiversity, once you reach the link I am giving you, go to "Classification", http://www.s-cool.co.uk/a-level/biology

Otherwise, I don't seem to have a really detailed A-Level guide notes available in Internet. I would recommend you to read over the A-Level Biology Coursebook or the book published by Cambridge A-Level Press, which is "Biological Science" for highly detailed guide notes.

 

[Vounn Rose]

Look through the *biology p5 tips* thread.
In the last few posts I remember someone had answered thos doubt

Where is it exactly? Could you bother to send me the link please? Thank you.

 

[knowitall10]

Where is it exactly? Could you bother to send me the link please? Thank you.

Sure thing Lemme look for it..

 

[knowitall10]

I could understand the part about cross 4, but when it says about "the idea of 1:1 ratio in each sex", I got startled. I tried the genetic diagram for both if the gene is sex-linked or non sex-linked, and the ratio of 'with tail' to 'without tail' I got is the same for both conditions (sex-linked or not).

What ratio did you get?

 

[knowitall10]

Vounn Rose
The mark scheme says "a male without tail cannot pass the "without tail" allele to his male offspring if it's sex-linked". It's a one mark question and that's all you need to write down to get the mark..what don't you get?

 

[Vounn Rose]

Ok. I assumed if the gene is sex-linked, then the ratio of "with tail" to "without tail" in each sex is 1:3, if I have to take into account both types of allele combination for the "without tail" female, which is either X(H)X(N) or X(N)X(N), ( X(H) stands for X chromosome with allele coding for 'with tail' trait, while X(N) codes for 'without tail' trait ).

 

[knowitall10]

Ok. I assumed if the gene is sex-linked, then the ratio of "with tail" to "without tail" in each sex is 1:3, if I have to take into account both types of allele combination for the "without tail" female, which is either X(H)X(N) or X(N)X(N), ( X(H) stands for X chromosome with allele coding for 'with tail' trait, while X(N) codes for 'without tail' trait ).

If it were sex linked, the ratio would have been 9:3:3:1