A level Biology: Post your doubts here!

[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

 

[Vounn Rose]

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?

So for part (ii) of the section I just have to write about this?

 

[knowitall10]

So for part (ii) of the section I just have to write about this?

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.

 

[Vounn Rose]

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

May I know what it means by 9:3:3:1?

 

[knowitall10]

May I know what it means by 9:3:3:1?

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]

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]

knowitall10

Thank you for your explanation help. I totally get it. =)

 

[knowitall10]

knowitall10

Thank you for your explanation help. I totally get it. =)

Are you sure? Because I did this question months ago for M/J2013 examinations.. I doubt I helped you rightly..

 

[Vounn Rose]

Are you sure? Because I did this question months ago for M/J2013 examinations.. I doubt I helped you rightly..

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]

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.

I'm glad I could be of help

 

[Jinosupreme]

http://papers.xtremepapers.com/CIE/...and AS Level/Biology (9700)/9700_w03_qp_1.pdf

Guys, need help in question 35, answer is C
Why is TB is the most successful prevention by vaccination?

 

[Jinosupreme]

http://papers.xtremepapers.com/CIE/...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?)

 

[Namehere]

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?)

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)

 

[Jinosupreme]

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)

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)

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

THANK YOU VERY MUCH STILL

 

[Namehere]

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

THANK YOU VERY MUCH STILL

np Hope u get Q36 anyways ^^

 

[Jinosupreme]

http://papers.xtremepapers.com/CIE/...and AS Level/Biology (9700)/9700_s07_qp_1.pdf

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

Need help pls!!! thanks!!!

 

[Namehere]

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!!! thanks!!!

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.

 

[Jinosupreme]

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.

Thanks!!!!! help a lot love u!!!

 

[Jinosupreme]

need some help in this
http://papers.xtremepapers.com/CIE/...and AS Level/Biology (9700)/9700_s08_qp_1.pdf
Q26, 34, 37
26A
34D
37B

Thanks

 

[mintchocolate4567]

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)