- Messages
- 415
- Reaction score
- 252
- Points
- 53
Irfan1995 ??
-
We need your support!
We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
Click here to Donate Now (View Announcement)
A level Biology: Post your doubts here!
- Thread starter XPFMember
- Start date
- Messages
- 134
- Reaction score
- 117
- Points
- 28
I need help with this one too please..and if possible can anyone explain this HORRIBLE graph to me?
Thanks in advance
Things you have to know first:
- The atrio-ventricular (AV) valves open when the pressure of the blood in the atria is greater than that in the ventricles. We all know that blood flows from high pressure to low pressure. So when the pressure in the atria is high, blood will flow to the ventricles. The opposite is also true (i.e. the AV valves close when the pressure in the atria is less than the ventricles). If they did not close, then we would have some back-leak of the blood from the ventricles to the atria, but since there are AV valves, they close to prevent this from happening.
- The semi-lunar valves open when the pressure in the ventricles is greater than that in the aorta (this is to ensure that blood gets forced out of the ventricles to the aorta. The opposite is also true.
(1) The AV valves close (point where the blood pressure of atria becomes less than pressure in the ventricles)
(2) The semilunar valves open (point where the blood pressure of the ventricles is greater than that of the aorta)
(3) The semilunar valves close (point where the blood pressure of the ventricles is less than that in the aorta)
(4) The AV valves open (point where the blood pressure of the atria is greater than that in the ventricles)
If you want to find the time that both valves are closed, that would be between (1) and (2) and between (3) and (4)
The time between (1) and (2) is about 0.03 s
The time between (3) and (4) is about 0.04 s
The total time would be 0.07 s which corresponds to answer C.
Could you please confirm the answer with the mark scheme please? If my answer is different, please say so and I'll review my work!
- Messages
- 233
- Reaction score
- 90
- Points
- 38
0.5 has a higher resolution , because objects of that size can be seen get it?Can anyone explain to me about the resolution of the light/electron microscope? 0.5 or 200 nm which of this value has a higher resolution?
- Messages
- 233
- Reaction score
- 90
- Points
- 38
why
WHY the time between 1 and 2 and 3 and 4 is the time both the valves are closed??
- Messages
- 415
- Reaction score
- 252
- Points
- 53
absloutely right !
thanks
isnt at 2 S.V O .. why to find difference at this point ..
http://papers.xtremepapers.com/CIE/...nd AS Level/Biology (9700)/9700_w10_qp_12.pdf
Question 13. I have chosen D. So the concentrations of Na inside cell is low while outside is high. So Na will move against the concentration gradient through protein 2. The concentrations of K inside cell is high and outside cell is low. So K will move against the concentration gradient through active transport through protein 3. Diffusion will occur down the concentration gradient for each, so 4 for Na and 1 for K. Answer is A. I do not get it.
Question 37. How can there be 3 fragments if the heavy polypeptide chains are hydrolysed? I thought there would be 2, each having 1 heavy chain and 1 light chain.
Question 13. I have chosen D. So the concentrations of Na inside cell is low while outside is high. So Na will move against the concentration gradient through protein 2. The concentrations of K inside cell is high and outside cell is low. So K will move against the concentration gradient through active transport through protein 3. Diffusion will occur down the concentration gradient for each, so 4 for Na and 1 for K. Answer is A. I do not get it.
Question 37. How can there be 3 fragments if the heavy polypeptide chains are hydrolysed? I thought there would be 2, each having 1 heavy chain and 1 light chain.
- Messages
- 233
- Reaction score
- 90
- Points
- 38
yeah I don't get why to add the differences too :/absloutely right !
thanks
isnt at 2 S.V O .. why to find difference at this point ..
- Messages
- 375
- Reaction score
- 205
- Points
- 53
- Messages
- 375
- Reaction score
- 205
- Points
- 53
How do u tag someone?
Oct/Nov '09:
14: The initial rate of reaction is equivalent to the initial slope. Take up to 30 seconds (as the slope is nearly constant). The slope would be 3/30 = 0.1
20: They tell you that the code for valine is CAT. The mRNA codon would be the complement of CAT which is GUA (remember that RNA has U instead of T). The anticodon of GUA would be the complement of that, which is CAU.
22: The complete explanation is a bit long. I've typed it somewhere, around 20 pages ago. In short, semi-conservative replication means that the new DNA helix has one strand from the old DNA and one strand from the new DNA. So the N15 DNA (old strands) would halve every time. That is, it starts from 100%, then goes to 50%, 25%, 12.5%, 6.25% etc... This corresponds to option C.
As for the question at Oct/Nov '08, I made an image, but I don't really know how to upload it. If you can tell me how, I'd love to share it with you.
Q13:
Active transports pumps ions against a concentration gradient (from a region of low concentration to a region of high concentration). So the cell has a high concentration of potassium, but rather than going from high to low, it will go from low to high. So the low concentration of potassium outside will all be forced inside.
With sodium, the exact opposite happens.
Diffusion goes down a concentration gradient (from a region of high concentration to a region of low concentration). There is a very low concentration of sodium inside and a high concentration of sodium outside. So sodium will travel from outside the cell to inside the cell through diffusion. The opposite is true for potassium.
Q37:
We know that an antibody looks like a "Y". When it fragments, it does so in three parts. The top two are called the "fragment antigen-binding" (Fab). These are shaped like "\" and "/". Antigens can bind to those two parts. The bottom part is the constant region, or fragment crystallizable (Fc). It's the straight/vertical line shaped like "|"
I'll try to explain this as simply as possible.
At (1), the AV valves close. At (2), the semilunar valves open. Before point (2), the semilunar valves are still closed. So between (1) and (2), both valves are closed.
At (3), the semilunar valves close, and at (4), the AV valve starts to open. Before point (4), the AV valves are still closed (they were closed since point (1)).
- Messages
- 306
- Reaction score
- 278
- Points
- 38
.
Q13: That's exactly what I said. Now look at the answer and you'll see that it should be D and it's A.
The answer is A. How is it D?
Diffusion evens out the concentrations; it tries to attain equilibrium. For example, if solution A is 1M sucrose, and solution B is 3M sucrose, when the particles diffuse, both solutions would be of 2M sucrose.
Active transport does the opposite: it tries to widen the concentration gradient. For example, if solution A is 1M sucrose, and solution B is 3M sucrose, when a pump is placed, solution A would be 0.5M sucrose, and solution B would be 3.5M sucrose after a while.
Active transport makes something concentrated even more concentrated. Inside the cell, there is a high concentration of potassium. Active transport tries to make it even more concentrated, so it transports potassium inside the cell (that is, process number 2).
The same holds true for the other three processes.
EDIT: Your problem may be that you didn't differentiate between the outside and inside parts of the cell. From the diagram, you can clearly see that the inside of the cell is up and the outside of the cell is down. So processes 1 and 2 point towards the inside of the cell and processes 3 and 4 point towards the outside of the cell.
A primary structure of a protein is held by peptide bonds only. So there are no hydrostatic forces of attraction at all. This makes option A wrong.
There are no ionic bonds in a secondary structure, only hydrogen bonds (responsible for the alpha-helix and beta-plated sheet). This makes option B wrong.
Everything is correct with option C.
There are definitely hydrogen bonds holding a quaternary structure together. There are also no peptide bonds linking two different polypeptide chains.
One more point I'd like to clarify, When I say that a quaternary structure doesn't have any peptide bonds, I don't mean that there are no peptide bonds at all. All I'm saying is that it's not responsible for forming and maintaining a quaternary structure.
- Messages
- 233
- Reaction score
- 90
- Points
- 38
Now I get it thanks a lot , JAZAK ILALLAH 5AYRAN