biology paper 1 doubts

[insomniac]

An oxygen molecule diffuses directly from the air in an alveolus to haemoglobin in a red blood cell.
What is the minimum number of cell surface membranes though which this molecule must pass?

A- 2
B- 3
C- 4
D- 5
ans is d:5 but howww ??????
DON'T FORGET TO MENTION THE REASON.
Thanks in advance =)

 

[Irfan1995]

According to my book, five:
1) Into a cell in the alveolar wall
2) Out of a cell in the alveolar wall
3) Into a cell in the capillary wall
4) Out of a cell in the capillary wall
5) Into a red blood cell.

 

[xtremeforums]

http://www.xtremepapers.com/papers/...and AS Level/Biology (9700)/9700_w07_qp_1.pdf
In nov 07 question 19...
It says that the centromeres divide... in which stage of mitosis does that occur? Does it mean that the chromatids are being pulled apart during anaphase, thus their centromere divides?

And please explain why the answer is D?

 

[Irfan1995]

Yes, you've just explained what happens to the centromere. Think of a centromere as the bond between two chromatids, so when the centromere breaks (or, divides in this case), you'll have separate chromatids.

Normally, after DNA replication, the two chromatids are held by a centromere to form a chromosome (5). During metaphase, the chromatids move to opposite poles of the cell (3). This is followed by anaphase, where the "bond" between the two chromatids is broken (1) and the spindle fibres pull the chromatids to opposite poles of the cell (2). Finally, in telophase, the nuclear envelope reforms and the chromosomes uncoil (4)*.

*Remember that during prophase, chromosomes condense/coil (i.e. they become shorter and thicker) so that they can become visible under a light microscope. So when nuclear division is over, the chromosomes have to return to their natural, uncoiled state.

 

[Kumkum]

Please help me with question number 5:
question paper : http://olevel.sourceforge.net/papers/9700/9700_s12_qp_12.pdf
marking scheme: http://olevel.sourceforge.net/papers/9700/9700_s12_ms_12.pdf
My answer was not correct....this is how i did it:
Magnification = Image size/Actual size
I measured the virus particles and i got the image size as 4mm hence
Actual size = (4*1000)/(24000) = 0.16
The answer is D
Please explain how it is D.

 

[Twinks05]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Biology (9700)/9700_w07_qp_1.pdf
In nov 07 question 19...
It says that the centromeres divide... in which stage of mitosis does that occur? Does it mean that the chromatids are being pulled apart during anaphase, thus their centromere divides?

And please explain why the answer is D?

Centromere is the region where two sister chromatids of a chromosome is joined together. During anaphase the microtubules of spindle pulls on the centromere first so that the sister chromatids can be pulled apart; this is what "centromere divides" refers to.
Stages of mitosis are: prophase, metaphase, anaphase and telophase.
According to the data given in the question:
Two chromatids are joined together by a centromere (prophase) - 5
Chromosome (containing two sister chromatids each, joined together at the centromere) line up along the equator with the help of microtubules (metaphase) - 3
Then the microtubules of spindle begin to to shorten, and since they're attached to centromere of the chromatids they pull on them, hence pulling the sister chromatids apart. This is referred by "centromere divides" (anaphase) - 1
Once the sister chromatids are pulled fully apart, they move to opposite poles by the shortening microtubules (late anaphase)-2
After reaching opposite ends, the chromatids are now chromosomes and begin to uncoil (telophase)-4
Hence, the order is: 5-3-1-2-4 (D)
Hope this helped =)

 

[Irfan1995]

Please help me with question number 5:
question paper : http://olevel.sourceforge.net/papers/9700/9700_s12_qp_12.pdf
marking scheme: http://olevel.sourceforge.net/papers/9700/9700_s12_ms_12.pdf
My answer was not correct....this is how i did it:
Magnification = Image size/Actual size
I measured the virus particles and i got the image size as 4mm hence
Actual size = (4*1000)/(24000) = 0.16
The answer is D
Please explain how it is D.

Okay, so you measured the size to be 4 mm, that's equal to 4x10^6 nm = 4,000,000 nm
This is the image size.
What you want to find is the actual size.
From your equation, magnification = image size/actual size
Rearranging that equation gives you actual size = image size/magnification.
Magnification is 24,000 as given in the question,
So 4000000/24000 = 166.67 nanometers which is pretty close to 150 nm = 1.5x10^2 nm (which is D).

Also, if you left the image size as 4 mm, you'd get the actual size to be 4/24000 = 1.66667x10^-4 millimeters. Which is equivalent to 167 nm. It doesn't really matter which units you're using as long as you're consistent. That is, if you measure the image size in millimeters, you'll get the actual size in millimeters.

Hope that helped!

 

[Kumkum]

Okay, so you measured the size to be 4 mm, that's equal to 4x10^6 nm = 4,000,000 nm
This is the image size.
What you want to find is the actual size.
From your equation, magnification = image size/actual size
Rearranging that equation gives you actual size = image size/magnification.
Magnification is 24,000 as given in the question,
So 4000000/24000 = 166.67 nanometers which is pretty close to 150 nm = 1.5x10^2 nm (which is D).

Also, if you left the image size as 4 mm, you'd get the actual size to be 4/24000 = 1.66667x10^-4 millimeters. Which is equivalent to 167 nm. It doesn't really matter which units you're using as long as you're consistent. That is, if you measure the image size in millimeters, you'll get the actual size in millimeters.

Hope that helped!

k thnx!

 

[SincerlyMe]

Why is answer D wrong ?

 

[ZohaibAsad]

1 is wrong. Haploid means having half diploid number of chromosomes but that doesn't mean that the chromosomes contained are all identical, they can vary hence there is genetic variation. If the chromosomes are also the same then that situation is called homoploid which is a different thing.

 

[Irfan1995]

View attachment 18105

Why is answer D wrong ?

I'm not really sure if you've studied A2 Biology yet, but as a result of independent assortment and crossing over, the gametes an organism produces would be all very different. To explain this simpler, we all have two sets of genes, one from our mother and the other from our father. During meiosis (reduction division), gametes are produced each with one set of chromosomes. But the set of chromosomes in the haploid is going to be a mix of the initial two sets (you get a few from the first set and a few from the other one). This is why you get completely different sets of genes. Look at our community today. Excluding twins, even brothers look very different.

 

[TCHS]

Can someone pls help with Question 24 from Oct/Nov 2005. I just dont understand it :|
http://www.xtremepapers.com/papers/...and AS Level/Biology (9700)/9700_w05_qp_1.pdf
The answer is B

 

[Irfan1995]

Can someone pls help with Question 24 from Oct/Nov 2005. I just dont understand it :|
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Biology (9700)/9700_w05_qp_1.pdf
The answer is B

A, B, C, and D are four different bases (adenine, guanine, cytosine, and thymine). They want to know which one is adenine.
By looking at sample 3, we can know this is a piece of RNA because it has uracil instead of thymine. This also proved to us that choice C is thymine because it is absent in RNA.
Now you just have to look at any other sample (1, 2, or 4) and find which choice is almost identical to C (since the amount of thymine is equal to that of adenine)
In sample 1, the percentage of thymine is 30%, and the closest one to that would be choice B (31%). Since B is almost identical to thymine, then B is adenine.
In sample 2, thymine is 24%, and the closest choice to that would be B (23%)
In sample 4, thymine is 33%, and the closest choice would be B (32%).

All three samples prove that base B is adenine since it is almost identical to thymine. The reason for the slight difference is experimental error.

 

[TCHS]

A, B, C, and D are four different bases (adenine, guanine, cytosine, and thymine). They want to know which one is adenine.
By looking at sample 3, we can know this is a piece of RNA because it has uracil instead of thymine. This also proved to us that choice C is thymine because it is absent in RNA.
Now you just have to look at any other sample (1, 2, or 4) and find which choice is almost identical to C (since the amount of thymine is equal to that of adenine)
In sample 1, the percentage of thymine is 30%, and the closest one to that would be choice B (31%). Since B is almost identical to thymine, then B is adenine.
In sample 2, thymine is 24%, and the closest choice to that would be B (23%)
In sample 4, thymine is 33%, and the closest choice would be B (32%).

All three samples prove that base B is adenine since it is almost identical to thymine. The reason for the slight difference is experimental error.

Thanks a lot! (Y)

 

[nelly trich]

I'm not really sure if you've studied A2 Biology yet, but as a result of independent assortment and crossing over, the gametes an organism produces would be all very different. To explain this simpler, we all have two sets of genes, one from our mother and the other from our father. During meiosis (reduction division), gametes are produced each with one set of chromosomes. But the set of chromosomes in the haploid is going to be a mix of the initial two sets (you get a few from the first set and a few from the other one). This is why you get completely different sets of genes. Look at our community today. Excluding twins, even brothers look very different.

But the Question says they develop from unfertilised eggs thus It's asexual reproduction ,no room for variation to occur

 

[Irfan1995]

But the Question says they develop from unfertilised eggs thus It's asexual reproduction ,no room for variation to occur

The process of PRODUCING the eggs creates variation. They produce eggs through meiosis which involves independent assortment and crossing over (which leads to variation).

 

[athaan]

Here's a question from MJ2003.

What is the main limiting factor on the amount of work that muscles can perform during aerobic exercise?
A) the percentage saturation of haemoglobin with oxygen in the lungs
B) he speed of dissociation of oxygen from haemoglobin in the muscles
C) the volume of blood flow through the lungs
D) the volume of blood flow through the muscles

correct answer is D. why not B?

 

[Irfan1995]

Here's a question from MJ2003.

What is the main limiting factor on the amount of work that muscles can perform during aerobic exercise?
A) the percentage saturation of haemoglobin with oxygen in the lungs
B) he speed of dissociation of oxygen from haemoglobin in the muscles
C) the volume of blood flow through the lungs
D) the volume of blood flow through the muscles

correct answer is D. why not B?

The Bohr shift already allows maximum rate of dissociation of oxygen from haemoglobin, so it can't be a limiting factor.

However, how quickly RBCs can deliver oxygen is a limiting factor. If the volume of blood flow through the muscles is low per unit time, not enough RBCs can be there to provide muscles with oxygen. If they don't deliver enough oxygen to the muscles, they have to go through anaerobic respiration which produces lactic acid (which is toxic). If lactic acid builds up, the cells will have to stop working or else they will die from the large quantities of toxins.

EDIT: the syllabus changed in early 2007, and the concept of anaerobic respiration is no longer included. So I don't think you're supposed to know the answer to this question. Anaerobic respiration is covered in greater depth in the A2 Biology course.

 

[athaan]

The Bohr shift already allows maximum rate of dissociation of oxygen from haemoglobin, so it can't be a limiting factor.

However, how quickly RBCs can deliver oxygen is a limiting factor. If the volume of blood flow through the muscles is low per unit time, not enough RBCs can be there to provide muscles with oxygen. If they don't deliver enough oxygen to the muscles, they have to go through anaerobic respiration which produces lactic acid (which is toxic). If lactic acid builds up, the cells will have to stop working or else they will die from the large quantities of toxins.

EDIT: the syllabus changed in early 2007, and the concept of anaerobic respiration is no longer included. So I don't think you're supposed to know the answer to this question. Anaerobic respiration is covered in greater depth in the A2 Biology course.

Ah, I see. Thank you for the information and the explanation. But I guess learning/knowing a little bit fact about it will be no harm.

I have a couple of questions too concerning MJ2012, link: http://olevel.sourceforge.net/papers/9700/9700_s12_qp_11.pdf
I have problem understanding no 16 (answer is C).

 

[Irfan1995]

Ah, I see. Thank you for the information and the explanation. But I guess learning/knowing a little bit fact about it will be no harm.

I have a couple of questions too concerning MJ2012, link: http://olevel.sourceforge.net/papers/9700/9700_s12_qp_11.pdf
I have problem understanding no 16 (answer is C)

First of all, I really like your desire to learn more

As for question 16, you need to know two (very obvious) things:
1) Oxygen is needed for active transport (to break down food for energy)
2) Air contains oxygen

With or without oxygen, the rate of uptake of 3-carbon sugar was the same, so you can be sure it is a passive process. So it could either be transported through diffusion or facilitated diffusion. Since the rate does not level off near the end, you can be sure that it moves by diffusion.
In the second graph, nitrogen gas is bubbled through, so there's no oxygen which is required for active transport. As shown by the graph, the rate of uptake for the 6-carbon sugar was zero, which means that oxygen was necessary to take in the 6-carbon sugar. This implies that it was taken in by active transport.
The only option which matches this would be C