A level Biology: Post your doubts here!

[Brisingr]

So in that case B is correct? So why in the marking scheme did they chose seperation of centrioles not the formation of chromosomes (Condensation of chromosomes)?

chromosomes are seen or formed before nuclear division

 

[Mstudent]

chromosomes are seen or formed before nuclear division

Oh, ok, now I get it, thank you very much!

 

[SohaibButt]

Anyone?

 

[Mstudent]

Anyone?

For the first question,

first you have two DNA strands made of N15
Then you replicated them with N14 to make two MOLECULES , each with a new N14 strand and an old N15 strand
then you replicated one of them again with N14, so you will have, TWO MOLECULES, one with TWO N14 strands, the old one and a new N14 strand, and another MOLECULE with and old N15 strand and a new N14 strand,

so at the end you have 4 DNA strands, two OLD STANDS and two New strands, so 50%

 

[shahzaib ihsan]

View attachment 62446
Answer in ms is B.
But shouldn't the answer be A?
ER also says: T-lymphocytes can secrete cytokines. Cytotoxins are released by certain pathogens.
which means C is incorrect but A is correct, right? I never heard of T-lymphocytes leaving blood

i cant understand this too,
this looks the same like the question 36 october/november 2015 variant 12..

 

[shahzaib ihsan]

Anyone?

1. the question is about semiconservative replication.
the first offspring of n14 and n15 was all hybrid dna, i.e. 100% hybrid dna. when it was allowed to divide one more time, half of it was original and half was new, so that means 50% hybrid was left

2.
i) it says blood in human veins not the walls of human veins, so i don't think collagen will be present in blood.
ii) carbonic anhydrase is present in all red blood cells, and blood always contain red blood cells regardless of the vessel it is flowing through .
iii)the vessels which carries oxygenated blood from lungs to the heart are veins (pulmonary veins) which contain redblood cells containing oxyhaemoglobin.
so answer should be d

 

[SohaibButt]

1. the question is about semiconservative replication.
the first offspring of n14 and n15 was all hybrid dna, i.e. 100% hybrid dna. when it was allowed to divide one more time, half of it was original and half was new, so that means 50% hybrid was left

2.
i) it says blood in human veins not the walls of human veins, so i don't think collagen will be present in blood.
ii) carbonic anhydrase is present in all red blood cells, and blood always contain red blood cells regardless of the vessel it is flowing through .
iii)the vessels which carries oxygenated blood from lungs to the heart are veins (pulmonary veins) which contain redblood cells containing oxyhaemoglobin.
so answer should be d

For the first question,

first you have two DNA strands made of N15
Then you replicated them with N14 to make two MOLECULES , each with a new N14 strand and an old N15 strand
then you replicated one of them again with N14, so you will have, TWO MOLECULES, one with TWO N14 strands, the old one and a new N14 strand, and another MOLECULE with and old N15 strand and a new N14 strand,

so at the end you have 4 DNA strands, two OLD STANDS and two New strands, so 50%

Thank you!

 

[Mstudent]

Thank you!

You're welcome, anytime

 

[Metallic9896]

View attachment 62446
Answer in ms is B.
But shouldn't the answer be A?
ER also says: T-lymphocytes can secrete cytokines. Cytotoxins are released by certain pathogens.
which means C is incorrect but A is correct, right? I never heard of T-lymphocytes leaving blood

I think here you have to think from a different perspective. Yes we may not remember that T lymphcytes leave the blood or not. But we reject A because of the 'only found in blood' part. They AREN'T only found in blood because how can that be true when you know that they are produced in the bone marrow and mature in the thymus gland? Those parts aren't part of the blood so at any time when you scan a person for T lymphocytes you'll "find" them in 3 places at least: blood, bone marrow, and thymus gland. This is why I think A is rejected. The cytokines part is correct but your concern should be the 'only found in' part.

 

[darks]

View attachment 62446
Answer in ms is B.
But shouldn't the answer be A?
ER also says: T-lymphocytes can secrete cytokines. Cytotoxins are released by certain pathogens.
which means C is incorrect but A is correct, right? I never heard of T-lymphocytes leaving blood

I think B refers to killer T cells. The killer T cells DO leave the blood to reach infected cell (i mean how else can it reach an infected cell?) to attach to it and release toxins. A is definitely incorrect as it states the T-lymphocytes are only found in blood.

 

[darks]

I think here you have to think from a different perspective. Yes we may not remember that T lymphcytes leave the blood or not. But we reject A because of the 'only found in blood' part. They AREN'T only found in blood because how can that be true when you know that they are produced in the bone marrow and mature in the thymus gland? Those parts aren't part of the blood so at any time when you scan a person for T lymphocytes you'll "find" them in 3 places at least: blood, bone marrow, and thymus gland. This is why I think A is rejected. The cytokines part is correct but your concern should be the 'only found in' part.

Haha what a co incidence that we decided to ans. the same Q

 

[Metallic9896]

Haha what a co incidence that we decided to ans. the same Q

Hahah coincidence indeed. BTW I think they don't refer to killer T cells although what you said about killer T cells is correct they DO get at the site and release cytotoxins, I just checked. But that would make option C the most correct option as it says 'They can leave the blood and excrete cytotoxins when exposed to bacteria'. But answer is B cause they are asking about the T lymphocytes which haven't specialized into killer T cells nor formed themselves into helper T cells. They are plain ol' T lympoctyes which leave blood and accumulate at the site of infection, THEN divide to form killer T cells that release cytotoxines OR form helper T cells that release cytokines. So let's summarize the story as:

T lymphocytes are formed in the bone marrow, mature in the thymus gland, patrol in the blood, and can leave the blood to accumulate at the site of infection.
At the site of infection, they can differentiate to form killer T cells that release cytotoxins.
At the site of infection, they can differentiate to form helper T cells that release cytokines.

 

[darks]

Hahah coincidence indeed. BTW I think they don't refer to killer T cells although what you said about killer T cells is correct they DO get at the site and release cytotoxins, I just checked. But that would make option C the most correct option as it says 'They can leave the blood and excrete cytotoxins when exposed to bacteria'. But answer is B cause they are asking about the T lymphocytes which haven't specialized into killer T cells nor formed themselves into helper T cells. They are plain ol' T lympoctyes which leave blood and accumulate at the site of infection, THEN divide to form killer T cells that release cytotoxines OR form helper T cells that release cytokines. So let's summarize the story as:

T lymphocytes are formed in the bone marrow, mature in the thymus gland, patrol in the blood, and can leave the blood to accumulate at the site of infection.
At the site of infection, they can differentiate to form killer T cells that release cytotoxins.
At the site of infection, they can differentiate to form helper T cells that release cytokines.

That's a perfect answer Some information that i did not know.

 

[darks]

So in our book, There is section from page 168 to 171 (Mary Jones 4th edition, Haemoglobin, Bohr shift, CO2 Transport), no matter how much i learn and try to understand this, I basically always end up doing the 1-3 questions asked from this topic, wrong. I have attached example Qs.. If anyone can help me with this topic, Please do.

 

[shahzaib ihsan]

So in our book, There is section from page 168 to 171 (Mary Jones 4th edition, Haemoglobin, Bohr shift, CO2 Transport), no matter how much i learn and try to understand this, I basically always end up doing the 1-3 questions asked from this topic, wrong. I have attached example Qs.. If anyone can help me with this topic, Please do.
View attachment 62452 View attachment 62453

same case here, in almost every paper i pick the wrong answer to this topic's question.. i guess this topic in book is not good

 

[Anas Abbal]

how to solve this i am unable to get the answer

 

[Xaverian]

Please anyone who could assist me with the question no 2 of this link????? Please
Link: http://papers.gceguide.com/A Levels/Biology (9700)/9700_s16_qp_11.pdf

 

[shahzaib ihsan]

Please anyone who could assist me with the question no 2 of this link????? Please
Link: http://papers.gceguide.com/A Levels/Biology (9700)/9700_s16_qp_11.pdf

i have done this recently... you have to use the formula m=I/a (magnification= image size/actual size)

m=1500, image size of x-y = 3 cm acutal size: A=I/M = 3/1500= 0.002cm
so 1 cell is actually 0.002 cm long, how many cells will be 1 cm long? 1/0.002 (unitary method) = 500

 

[janedoe]

Question 4
The majority of candidates answered this incorrectly. A build-up of lipids in cells means that the excess lipids
are not being broken down. Excess lipids are normally broken down by hydrolytic enzymes found in the
lysosomes.

This is what the examiner report says. This is one of the reasons I really dislike Biology paper 1s. Some questions are simply too vague with overlapping correct options. Their emphasis is the 'build up' part and not the enzyme not being produced part. My first instinct was that Golgi apparatus is not part of 'producing' the enzyme, as we read and are expected to learn that it is involved in 'packing' the enzyme or adding non-protein elements to it, but that's something too trivial to be reason enough. More importantly, I guess what they mean to say is that in the production process, anything could be at fault, from the gene sequence of the enzyme to the ribosomes to the endoplasmic reticulum and the Golgi apparatus - all of these could be or one of these could be faulty. However, something that for CERTAIN won't function is the lysozomes because they are what the enzymes eventually should end up and if the enzymes aren't produced then for certain they won't end up there. These 2 arguments are what I could think of in favor of option A, but I do understand why anyone would have an issue with this question, including myself. They should have written something like, 'Which cell structure MUST not function correctly.." writing must in bold etc.

Ok, thank you.

 

[Metallic9896]

The Tale of Haemoglobin Affinity Curves and the Bohr Effect (dedicated to darks and some other folks on here who might have an issue with this topic)

You need to understand what that gorgeous curve actually represents. Take it out from your book as you read this. The x axis depicts the partial pressure of oxygen externally. This is a measure of how much oxygen is present around the red blood cell or haemoglobin. The y axis represents the degree to which the haemoglobin desires to bond with the oxygen, or formally known as the percentage saturation of haemoglobin with oxygen. As you know it's an S shaped curve. Look at the curve from right to left. What the curve implies is that if the partial pressure is high, then the haemoglobin would want to keep all that juicy oxygen to itself. It's chilling there with ample oxygen around itself and so wants to join to it as much as it can. As the partial pressure decreases though (you move towards the left), the percentage saturation of the haemoglobin also decreases. This means that for lower partial pressures, the haemoglobin doesn't want to keep the oxygen to itself, it wants to let it go, let it be free into the outside world. This in itself is simple enough, but complications arise when you compare two different S curves which is what I assume can be bothersome. http://www.mrothery.co.uk/images/scool2.gif This is a good looking simple Bohr effect picture. Open it and view it as you read this. I'll get to the reasons later, first just observe it. When there is a high carbon dioxide pressure, the entire curve shifts to the right. What does this mean? This means (and read this very slowly) that at a HIGHER carbon dioxide partial pressure outside, to get the SAME amount of haemoglobin to want to bind with oxygen as when the carbon dioxide partial pressure is lower, you need a HIGHER partial pressure of oxygen. Let me rephrase with examples. Let's say when carbon dioxide pressure is low (left curve), you want haemoglobin to be 40% saturated with oxygen. How much should the partial pressure outside be for that to happen? It should be around 3 kPa. When the carbon dioxide pressure is high though (right, shifted curve), what should the oxygen partial pressure be to get 40% haemoglobin saturation with oxygen? This time it shows up to be somewhere more than 4 kPa. You need a HIGHER O2 partial pressue for the SAME amount of haemoglobin saturation with O2. So what is the CO2 doing? It's maing the haemoglobin NOT-so-keen on binding with Oxygen. The presence of CO2 makes the haemoglobin say, 'Meh, this O2, I don't wanna be with it anymore'. (Again, I'm not getting into the reason of why this happens right now) Why does it benefit our body though? This is something I struggled with, but keep your mind clear. Our minds are programmed to think since childhood, 'Oxygen is good, blood needs oxygen, get max. oxygen, oxygen, oxygen, oxgen!' But NO! That's NOT what the situation should be. Oxygen in our blood is NOT always good. Why? Because in the tissues, you DON'T want the blood to keep and hog the oxygen. You want it to let the oxygen go, set it free, for the tissues that need it more. Imagine the entire route our blood takes. First it goes to the lungs. Here the carbon dioxide is low and oxygen is high. The curve will be at the left because of low Carbon Dioxide. So for even - relatively/comparatively - smaller amounts of external oxygen, our haemoglobin would happily take it in and be saturated with it. Now when it comes to the tissue, here the external pressure of oxygen is low. For a second, ignore the carbon dioxide and just imagine and look at the left curve. In the lungs let's say the partial pressure was 10, and saturation was around 90%. When it got to the tissue, oxygen pressure outside was lesser than the lungs, let's say around 6. At 6, as per the left curve, the saturation is 80%. It decreased by 10%, so in that decrease, the haemoglobin let go of some of the oxygen it was binded to, and this oxygen went over to the tissues. Everything seems good enough. But one can't help but wonder, only a 10% decrease? This doesn't seem to be too efficient. Now take the role of carbon dioxide into play. The tissues not only have a lower O2 conc., but also a higher CO2 conc. This makes the curve shift to the right. Now as per the right curve, when the oxygen pressure is 6, how muhc is the saturation? It's 60%. Now this looks yummy. It went from 90% to 60%, a 30% decrease. The amount of oxygen the haemoglobin would want to lose is now greater, and this means MORE oxygen for the tissues that need it. Effectively, our body gets the best of both worlds:

1. When we want our haemoglobin to keep the oxygen as much as it can, we use the left curve (by having lesser CO2 outside), to get as much saturation as possible compared to the right curve.
2. When we want our haemoglobin to lose the oxygen as much as it can, we use the right curve (by having more CO2 outside), to get as much a decrease in saturation as possible.

Now I'll tell you the reason for this. The reason is that a higher presence of CO2 automatically means a higher presence of H+ ion. You may wonder why. This is because CO2 normally likes to join with H2O and form H+ + HCO3-. The H+ then makes the entire medium more acidic, and our haemoglobin as you would know likes acting as a buffer the most. So the haemoglobin takes in the H+ and forms haemoglobinic acid. This entire process has an impact on the desire of haemoglobin to want to join with O2. It feels that binding with the H+ is MORE important, cause if it doesn't do that, the H+ are going to make EVERYTHING more acidic and destroy everything in our body. Thus, it does this at the cost of O2. But that cost of O2 is something we WANT. We WANT that to happen cause we don't want the haemoglobin to stay with O2 ALL the time. We only want haemoglobin to TRANSPORT the O2 to the tissues, but a good transport is not something that keeps its passengers, right? It should let go of its passengers when they reach the destination. So it's a win-win situation for our body. The decrease in the AFFINITY of haemoglobin for O2 caused by higher H+ (caused by higher CO2) is a benefit for us as this means it lets go of O2 more easily in the tissues. What happens in the lungs then? In the lungs, the O2 outside is high and CO2 is low. So CO2, normally there as HCO3-, reacts with the H+ again to form CO2 and H20, both of which leave our body through the air. No more CO2 means no more H+. No more H+ means no more reason for haemoglobin to hate O2. And this means we get back at the left curve and haemoglobin happily takes in O2 again.

*wipes sweat*

I really hope this made sense guys. Anynone else wants to add something or correct me feel free to do so.