A level Biology: Post your doubts here!

[Anas Abbal]

can anyone help me with this i cant seem to get the correct answer

 

[Metallic9896]

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

I hope my previous post made sense. Now I'll directly address the question.

Q28: What the scientists basically found is that the curve looks similar to that of the Bohr effect, i.e., the curve is to the right the same way it is to the right when there is a higher carbon dioxide partial pressure. So ignore everything, just imagine that Bohr effect curve picture I linked in my previous post.

A can't be right because that applies to a curve that is on the left, not the right. The more shifted to the left it is, the more oxygen haemoglobin would want to carry for ANY partial pressure of oxygen and vice versa.
B Wrong, when the curve is at the right, this is what I've been trying to explain, that the oxygen will HAPPILY let go of the oxygen at lower oxygen pressures compared to when the curve is to the left.
C Wrong again for the same reason as above. It saturates with oxygen LESS easily. The right-er the curve is, the harder it becomes to saturate haemoglobin with oxygen, the more pressure of oxygen we need to get the same amount of saturation.
D Yes, this is correct. It UNloads it easily because it DOESN'T WANT the oxygen as much as it would want it if the curve was to the left. So this means it unloads it easily. The complication is the way we perceive the situation. Unloding easily sounds like a good thing and our minds are programmed to think as if good things can't happen with higher CO2 or a right-er curve as in the case here, but good things DO happen that's the entire purpose of the Bohr shift or a right-er curve, that the oxygen will be unloaded easily, and that means that more oxygen will be available to the tissues that need it. In the lungs, we want the opposite of what we want in the tissues. THERE you want haemoglobin to cling onto as much oxygen as it can and that's why a left-er curve will be better there. For loading, a left-er curve is good. For unloading, a right-er curve is good. It all boils down to what you want.

Q30: For this question you can actually get it right without even thinking about options 1 and 3. This is because option 2 says carbon dioxide forms CARBOXYhaemoglobin. That is not true. Carbon MONOXIDE forms carboxyhaemoglobin, while here they mentioned carbon DIOXIDE. On the other hand, carbon DIOXIDE, some of it, does directly join with haemoglobin but we call that carbaminohaemoglobin. The rest of the carbon dioxide as explained joins with H2O to form H2CO3 which breaks to form H+ and HCO3- (this reaction is controlled by carbon anhydrase just so you know), and some of the CO2 simply travels as CO2, but that's a comparatively smaller fraction. Anyway, statements 1 and 3 are correct. 1 is correct because when blood passes through respiring tissues, there is more CO2 present. More CO2 present means more of it joins with H2O to form H2CO3, and as I just told you, carbon anhydrase oversees this entire reaction in which it breaks into H+ and HCO3-, so yes, its activity increases here. 3 is correct for the reasons I have explained in this and my previous post. Hope I helped.

 

[Metallic9896]

can anyone help me with this i cant seem to get the correct answer

Know which ones are by viruses and which ones are by bacteria and which ones are by the protists.

HIV, smallpox, and measles are all by viruses.
TB and cholera are by bacteria.
Malaria is by a protoctist.

In your question, the HIV and measles are the ones by viruses. Since they are asking the number of deaths, you aren't concerned with the percentage column. All you have to do now is add up the deaths by HIV and by measles. 2.8 + 0.6 = 3.4 = B is the correct answer. Hope I helped.

 

[Anas Abbal]

Know which ones are by viruses and which ones are by bacteria and which ones are by the protists.

HIV, smallpox, and measles are all by viruses.
TB and cholera are by bacteria.
Malaria is by a protoctist.

In your question, the HIV and measles are the ones by viruses. Since they are asking the number of deaths, you aren't concerned with the percentage column. All you have to do now is add up the deaths by HIV and by measles. 2.8 + 0.6 = 3.4 = B is the correct answer. Hope I helped.

ohh yea got it i thought it has to ddo something with the percentages silly mistake from me thanks man

 

[darks]

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.

I just can't thank you enough for how much you have helped me with my exams!! You explained this better than the book! THANKS A LOT !!!!

 

[nigeller]

Hi guys , the answer to this question is D. Why cant it be A?

 

[Metallic9896]

I just can't thank you enough for how much you have helped me with my exams!! You explained this better than the book! THANKS A LOT !!!!

You're ever welcome man!! This truly means a lot and I do this happily because if it weren't for the internet I wouldn't have half the knowledge I have today. I stopped going to school in the last 6 months cause school didn't cut it. Self-studying FTW, and posts I read by other members on xtremepapers and other sites is what helped me, so I feel it's my obligation to pay back. Feel free to ask more I'll be available for another 6 hours before calling it a night. xD

 

[zellyman]

Can someone Explain Active Loading and unloading to me,including everything about the water potential?

 

[Metallic9896]

Hi guys , the answer to this question is D. Why cant it be A?View attachment 62458

Very cheeky question. I got it wrong the first time I did it. I did A too. But I think we should visualize what's happening before choosing the option. Think about it. Hungry enzymes, all vacant, free to get a hold on those delicious substrates. Then why would the conc. of ES complexes rise slowly? No, the very instant the reaction begins, ALL enzymes would become saturated immediately and we'd have a MAX. amount of possible ES substrates at any one given time. It stays this way as the substrate is in excess so as soon as even one enzyme becomes vacant, another substrate hits it and forms an ES complex. Then, there comes a time when the substrate is NO longer in excess. The enzymes are in excess and the substrate conc. begins decreasing. As the substrate conc. decreases, this isn't enough substrate to keep all enzymes busy, so the overall amount of ES complexes at any one given time decreases, and it continues decreasing until we run out of substrate and the amount of ES complexes becomes 0. So it's MAX at first, remains that way for a while as long as substrate is in excess, and then it decreases to 0 as substrate gets used up.

 

[darks]

can't understand statement 3..

 

[Metallic9896]

Can someone Explain Active Loading and unloading to me,including everything about the water potential?

I absolutely HATE the way CIE asks about this topic. I have some very ugly questions about this topic. Here it goes:

1. COMPANION cells' membranes have H+ pump carrier proteins that use ATP (active transport) to pump the H+ ions from within the companion cells to the cell walls/outside the cell.
2. This builds up positive charge as well as H+ ion concentration outside the cell and the H+ ions wants to get back into the companion cells down both its conc. gradient as well as its electrical gradient, known as the 'electrochemical' gradient. (This just means that there's an URGENT need for H+ to get back into companion cells, not just diffusion due to its higher conc. but diffusion due to the negative charge inside the cell as well.)
3. The only way back into the companion cells is through a channel protein that allows facillitated diffusion of H+ ions back into the companion cells.
4. This channel protein is really naughty, it not only has a site for H+ ions, but also room for sucrose.
5. The sucrose is present in the cells surrounding the companion cells where it is produced and would also normally want to get into companion cells from a place of higher conc. to a place of lower conc., and its only way in is also through that same channel protein which has room for both H+ and sucrose.
6. As the H+ joins the channel protein, sucrose also joins the channel protein, and thus, sucrose enters the cell through diffusion.
7. You may wonder, why doesn't sucrose just diffuse through that channel protein? The reason is that probably that channel protein needs BOTH the H+ AND sucrose to bind to it for functioning, and the only way H+ can bind to it is if H+ is present outside the cell to begin with, and to get H+ outside, the cell needs to actually pump it out from within it by active transport.
8. This channel protein is what we call a co-trnasporter protein as it has sites for two distinct molecules.
9. Once this co-transporter protein brings sucrose into the companion cells, the sucrose then DIFFUSES into the sieve tube elements through plasmodesmata.
10. Once in the sieve tube elements, the water potential within the sieve tube elements LOWERS, and thus, water moves in from surrounding cells into the sieve tube elements, INCREASING the volume of liquid at this end of the sieve tube elements.
11. This increase in quantity of solutes as well as the volume of liquid creates a high hydrostatis pressure at this end.
12. This high hydrostatis pressure drives mass flow of this liquid in the sieve tube elements from the source down to the sink.
13. At the sink, sucrose diffuses from the sieve tube elements back to the cells that need it.
14. This INCREASES the water potential in the sieve tube elements at the sink and causes water to move out from sieve tube elements into surrounding cells, and thus, DECREASES the volume of liquid at this end of the sieve tube elements.
15. This process can occur in reverse at the sink too if the plant decides to load sucrose from the sink into the sieve tube elements back to the source (thus making the sink, the source, and making what was previously the source, the sink )

I hope this makes sense. Anyone else correct me if I'm wrong.

EDIT: Even though sucrose diffuses from companion cells to sieve tube elements, in the grand scheme of things, we say that it is actively loaded into the sieve tube elements (at least as per a question in May/June 2012) because saying that it diffuses, although correct specifically, does not portray the entire picture and can be misleading.

 

[Metallic9896]

View attachment 62459
can't understand statement 3..

Ugly question again. I got this wrong too when I did it first. I googled it and turns out this is an actual thing. Apparently, it's just a fact that we are expected to know. -_-

Worst thing is, the options have '2 only' and '3 only' so we can't even eliminate the options. *facepalm*

I guess the logic is that fludity increases with temperature = more kinetic energy = phospholipids like moving more = more movement makes it more permeable..? XD I still don't get it but whatever. I guess CIE just can't tolerate anyone getting 40/40 and just want someone to screw up in one place or another. XD

 

[darks]

Ugly question again. I got this wrong too when I did it first. I googled it and turns out this is an actual thing. Apparently, it's just a fact that we are expected to know. -_-

Worst thing is, the options have '2 only' and '3 only' so we can't even eliminate the options. *facepalm*

I guess the logic is that fludity increases with temperature = more kinetic energy = phospholipids like moving more = more movement makes it more permeable..? XD I still don't get it but whatever. I guess CIE just can't tolerate anyone getting 40/40 and just want someone to screw up in one place or another. XD

Exactly! i selected "2 only". But yeah, i get the way it works now. Thanks

 

[darks]

so what are the sizes we are supposed to learn from chapter 1?
plant cell 40um
Animal 5 to 40um
cell membrane 7nm
ribosome 25nm
centriole 200nm
prokaryote 0.5-5um
virus 20-300nm
Thats all?

 

[Metallic9896]

so what are the sizes we are supposed to learn from chapter 1?
plant cell 40um
Animal 5 to 40um
cell membrane 7nm
ribosome 25nm
centriole 200nm
prokaryote 0.5-5um
virus 20-300nm
Thats all?

Beautifully compiled. I think this is the most. In fact directly I've only seen prokaryote, eukaryotic cell, and the cell membrane one, but yeah rest are good too. Also it helps to learn ribosomes in S units too. 70S in prokaryotic cytoplasms, as well as the stroma of chloroplasts and matrix of mitochondiron (they were actually prokaryotes that became part of eukaryotic cells as part of the endosymbiotic theory, beautiful work of Mother Nature), whereas eukaryotic cytoplasms contain 80S ribosomes. Umm, yeah that should be it.

 

[techgeek]

so what are the sizes we are supposed to learn from chapter 1?
plant cell 40um
Animal 5 to 40um
cell membrane 7nm
ribosome 25nm
centriole 200nm
prokaryote 0.5-5um
virus 20-300nm
Thats all?

Add to it:
mitochondria and bacteria: 1-5 um

 

[darks]

Unable to fully understand this... ans D

 

[Metallic9896]

Before mRNA leaves nucleus, the DNA should have replicated and an mRNA strand should have been synthesized, i.e. transcription. However, the DNA being replicated is not part of the question because the very first statement says that the events being discussed are from transcription. I made the mistake of ignoring this part (thanks for reminding me with this quesiton btw, tomorrow we have to read the stem of the questions VERY carefully, each word is relevant). First make a mental note of transcription:

- DNA breaks into two strands by bonds between complementary bases breaking, 1 occurs the first time.
- One strand acts as template strand and free RNA nucleotides pair with this template strand's complementary nucleotides, 4 occurs the first time.
- Bonds form between complementary bases, the free RNA nucleotides and the complementary nucleotides to hold the RNA nucleotides in place, 2 occurs the first time.
- RNA polymerase I believe then joins up the RNA nucleotide backbone by forming sugar-phosphate bonds while they are held in place by the previous step, thus, 3 occurs the first time.
- Once this RNA strand is complete (the mRNA has formed), complementary bonds between its nucleotides and the template DNA's nucleotides break, 2 occurs for the second time
- The DNA forms bonds with the complementary bases of the other strand which was broken off in the very first step to make the DNA molecule again, 1 occurs for the second time.
- DNA has been formed again, mRNA has formed, and thus, mRNA is ready to leave and it does so.

So only 1 and 2 occur twice. Hope I helped.

 

[zellyman]

View attachment 62461
Unable to fully understand this... ans D

DNA Unwinds---------->Bonds broken once
Bonds form between complementary bases of the DNA sense strand and mRNA strand being formed ----->Bonds formed once
Once mRNA is formed bonds are broken again --------> Bonds broken twice
DNA strands reform complimentary base pair bonding ------->Bonds formed Twice
3 and 4 happen more than twice ,at an undefined number of times depending on the length of RNA formed.

 

[darks]

Before mRNA leaves nucleus, the DNA should have replicated and an mRNA strand should have synthesized, i.e. transcription. However, the DNA being replicated is not part of the question because the very first statement says that the events being discussed are from transcription. I made the mistake of ignoring this part (thanks for reminding me with this quesiton btw, tomorrow we have to read the stem of the questions VERY carefully, each word is relevant). First make a mental note of transcription:

- DNA breaks into two strands by bonds between complementary bases breaking, 1 occurs the first time.
- One strand acts as template strand and free RNA nucleotides pair with this template strand's complementary nucleotides, 4 occurs the first time.
- Bonds form between complementary bases, the free RNA nucleotides and the complementary nucleotides to hold the RNA nucleotides in place, 2 occurs the first time.
- RNA polymerase I believe then joins up the RNA nucleotide backbone by forming sugar-phosphate bonds while they are held in place by the previous step, thus, 3 occurs the first time.
- Once this RNA strans is complete (the mRNA has formed), complementary bonds between its nucleotides and the template DNA's nucleotides break, 2 occurs for the second time
- The DNA forms bonds with the complementary bases of the other strand which was broken off in the very first step to make the DNA molecule again, 1 occurs for the second time.
- DNA has been formed again, mRNA has formed, and thus, mRNA is ready to leave and it does so.

So only 1 and 2 occur twice. Hope I helped.

Perfect!