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A level Biology: Post your doubts here!

knowitall10

knowitall10

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Vounn Rose said:
So for part (ii) of the section I just have to write about this?
Click to expand...

Yup. Look at the mark allocation in the marking scheme: 2 bold A 's are written for the first part and the total mark allocation is 3 marks..so apparently, 2 marks for the first part, and one for the second part.
 
knowitall10

knowitall10

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Vounn Rose said:
May I know what it means by 9:3:3:1?
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I was wrong up there :/ I apologize. You're right, it's a 3:1 ratio because we are only crossing one trait.
However, if we were crossing two heterozygous traits (for eg: with tail x without tail + brown fur x white fur) then it would have been 9:3:3:1.
 
knowitall10

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Hey, the answer is pretty simple. Since there is no set ratio of the number of males and females born without tail, we cannot assume that the trait is sex-linked! Look at the figures again, brother. See what i'm saying?
 
Vounn Rose

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knowitall10 said:
Are you sure? Because I did this question months ago for M/J2013 examinations.. :unsure: I doubt I helped you rightly..
Click to expand...
Yeah. Totally. I am merely confused with the mark allocation. And you solved that. And I also refer to your explanation about the "male without tail won't pass his phenotype to his male offspring if it is sex-linked gene" it's perfect.
 
knowitall10

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Vounn Rose said:
Yeah. Totally. I am merely confused with the mark allocation. And you solved that. And I also refer to your explanation about the "male without tail won't pass his phenotype to his male offspring if it is sex-linked gene" it's perfect.
Click to expand...

I'm glad I could be of help :)
 
Namehere

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Jinosupreme said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Biology (9700)/9700_s03_qp_1.pdf

Need help in Q1, 20, 22, 36.
1D
20C
22B
36A (why other answer is not correct?)
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Q1 - D, because the RER contains ribosomes which synthesise proteins, which are made of amino acids, hence radioactivity is first detected here.
Q20 - C, because humans don´t use RNA, so reverse transcriptase is the enzyme responsible for converting the RNA into complementary DNA.
Q22- B, well... arteries carry oxygenated blood away from the heart but the pulmonary artery carries DEOXYGENATED blood away from the heart to the lungs. Therefore D would be somewhere like in the lungs, C would be something like a capillary near the lungs, A would be something like a respiring tissue so B would be the correct answer as it has a "rather low" pp of oxygen.
Q36- A, not quite sure of this one but I´d say B no because Memory cells would be present all the time from day 5 onwards, C no because I´d say second exposure occured at day 25 and D I´d say T helper cells are activated on day 5. (Really not sure on this one)
 
J

Jinosupreme

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Namehere said:
Q1 - D, because the RER contains ribosomes which synthesise proteins, which are made of amino acids, hence radioactivity is first detected here.
Q20 - C, because humans don´t use RNA, so reverse transcriptase is the enzyme responsible for converting the RNA into complementary DNA.
Q22- B, well... arteries carry oxygenated blood away from the heart but the pulmonary artery carries DEOXYGENATED blood away from the heart to the lungs. Therefore D would be somewhere like in the lungs, C would be something like a capillary near the lungs, A would be something like a respiring tissue so B would be the correct answer as it has a "rather low" pp of oxygen.
Q36- A, not quite sure of this one but I´d say B no because Memory cells would be present all the time from day 5 onwards, C no because I´d say second exposure occured at day 25 and D I´d say T helper cells are activated on day 5. (Really not sure on this one)
Click to expand...
Namehere said:
Q1 - D, because the RER contains ribosomes which synthesise proteins, which are made of amino acids, hence radioactivity is first detected here.
Q20 - C, because humans don´t use RNA, so reverse transcriptase is the enzyme responsible for converting the RNA into complementary DNA.
Q22- B, well... arteries carry oxygenated blood away from the heart but the pulmonary artery carries DEOXYGENATED blood away from the heart to the lungs. Therefore D would be somewhere like in the lungs, C would be something like a capillary near the lungs, A would be something like a respiring tissue so B would be the correct answer as it has a "rather low" pp of oxygen.
Q36- A, not quite sure of this one but I´d say B no because Memory cells would be present all the time from day 5 onwards, C no because I´d say second exposure occured at day 25 and D I´d say T helper cells are activated on day 5. (Really not sure on this one)
Click to expand...

Omg, thanks a lot! I just realized I made a mistake q36 is B, not A.

THANK YOU VERY MUCH STILL :D
 
Namehere

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Jinosupreme said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Biology (9700)/9700_s07_qp_1.pdf

Need help in Q15, 24
15C. 24C.

Need help pls!!! :eek: thanks!!!
Click to expand...

Q15 - C -> You can see that the graph for the 3-Carbon sugar is the same in both graphs, however on the 6-carbon sugar the graph is a straight line when nitrogen is used. Air contains oxygen, which is used for many life processes. Since the 3-carbon sugar graph is the same when oxygen and nitrogen is used you can infer that the 3-carbon sugar doesnt need oxygen to take up the sugar, meaning its diffusion. Since the 6-carbon sugar uses oxygen (rate of uptake increases) and no nitrogen (rate of uptake is constant) you can infer there is oxygen used, which means it is active transport.

Q24- C -> Hormones detected... in the cell surface membrane you have glycoproteins etc which are important in chemical recognitions, so the greater the surface area (the more glycoproteins per unit area in the cell surface membrane) the more hormones are detected. Carbon dioxide produced... the more volume a cell has the greater their gas production is.

Hope it helps.
 
J

Jinosupreme

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Namehere said:
Q15 - C -> You can see that the graph for the 3-Carbon sugar is the same in both graphs, however on the 6-carbon sugar the graph is a straight line when nitrogen is used. Air contains oxygen, which is used for many life processes. Since the 3-carbon sugar graph is the same when oxygen and nitrogen is used you can infer that the 3-carbon sugar doesnt need oxygen to take up the sugar, meaning its diffusion. Since the 6-carbon sugar uses oxygen (rate of uptake increases) and no nitrogen (rate of uptake is constant) you can infer there is oxygen used, which means it is active transport.

Q24- C -> Hormones detected... in the cell surface membrane you have glycoproteins etc which are important in chemical recognitions, so the greater the surface area (the more glycoproteins per unit area in the cell surface membrane) the more hormones are detected. Carbon dioxide produced... the more volume a cell has the greater their gas production is.

Hope it helps.
Click to expand...

Thanks!!!!! :D help a lot love u!!!
 
mintchocolate4567

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what are the pressure values for the afferent and efferent capillaries in the kidney? and the net pressure in the glomerulus?
(need a quick response plz)
 
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