Biology as level question!!!

Rate of entry increases, then plateaus, as concentration of X outside the cell increases.

If it could enter by diffusion through the phospholipid bilayer (pathway 1), then the rate wouldn't plateau. However as there will be a limited number of channel proteins, once they are all transporting substance X, increasing the concentration of X would have no effect on the rate.

Pathway 3 is not a transporter protein - it's a glycolipid/glycoprotein.

So, my guess would be that the answer is D. I may be wrong and misinterpreting the question though :s
EDIT: I can't tell from the photo - if the graph is still increasing slightly and not plateauing, then my answer would be A.

Do you have the markscheme?

Your answer A is correct. But it would be awesome If you could explain me how

It's because it is facilitated diffusion through that channel protein (and also simple through the bilayer). Take a look at this website, it'll give you a more clear explanation: http://www.biology.arizona.edu/cell_bio/problem_sets/membranes/07t.html

I just wanna know how concentration outside cell and rate of entry is related and how could i use that information to solve this?

I just wanna know how concentration outside cell and rate of entry is related and how could i use that information to solve this?

So,
The answer is A: X moves through the phospholipid bilayer and the transport proteins.
We can tell that diffusion happens, because the rate of movement in is directly proportional to the concentration of X. However, the graph goes flat.
Picture two cells. Cell A is in LOWER conc. and cell B is in HIGHER conc.. ok?
Now, for Cell A, X will diffuse in as there is a higher conc. outside the cell than in the cell. That is the staright part of the graph. this is simple diffusion.
In Cell B, X will also diffuse into the cell, BUT as there is a higher X conc., the cell contents will only be able to take X in via diffusion with a concentration gradient WHICH WILL DEPLETE OVER TIME. until there is no more gradient. There will still be movement but there will be NO NET MOVEMENT. this is the curve of the graph.
As for the membrane protein, logic says if it is small enough do diffuse through the phospholipid bilayer, then it will happily take an easier route through the channel protein, by FACILITATED DIFFUSION. faciltated diffusion still relies on conentration gradients though.
(The explanation for "cell B" may be a little lax? if it is, just ask for confirmation)

Yea as it says on that website:

If the particles can move through the lipid bilayer by simple diffusion, then there is no limit to the number that can fit through the membrane. The rate of diffusion increases linearly as we add more particles to one side of the membrane.

If the particles can only pass through protein channels, then the rate of diffusion is determined by the number of channels as well as the number of particles.

Once the channels operate at their maximal rate, a further increase in particle numbers no longer increases the apparent rate of diffusion. At this limited rate we describe the protein channel as being saturated.

Damnnnn buddy. Thats awesome explanation. You the hero man. Thanks a lot

So,
The answer is A: X moves through the phospholipid bilayer and the transport proteins.
We can tell that diffusion happens, because the rate of movement in is directly proportional to the concentration of X. However, the graph goes flat.
Picture two cells. Cell A is in LOWER conc. and cell B is in HIGHER conc.. ok?
Now, for Cell A, X will diffuse in as there is a higher conc. outside the cell than in the cell. That is the staright part of the graph. this is simple diffusion.
In Cell B, X will also diffuse into the cell, BUT as there is a higher X conc., the cell contents will only be able to take X in via diffusion with a concentration gradient WHICH WILL DEPLETE OVER TIME. until there is no more gradient. There will still be movement but there will be NO NET MOVEMENT. this is the curve of the graph.
As for the membrane protein, logic says if it is small enough do diffuse through the phospholipid bilayer, then it will happily take an easier route through the channel protein, by FACILITATED DIFFUSION. faciltated diffusion still relies on conentration gradients though.
(The explanation for "cell B" may be a little lax? if it is, just ask for confirmation)

So,
The answer is A: X moves through the phospholipid bilayer and the transport proteins.
We can tell that diffusion happens, because the rate of movement in is directly proportional to the concentration of X. However, the graph goes flat.
Picture two cells. Cell A is in LOWER conc. and cell B is in HIGHER conc.. ok?
Now, for Cell A, X will diffuse in as there is a higher conc. outside the cell than in the cell. That is the staright part of the graph. this is simple diffusion.
In Cell B, X will also diffuse into the cell, BUT as there is a higher X conc., the cell contents will only be able to take X in via diffusion with a concentration gradient WHICH WILL DEPLETE OVER TIME. until there is no more gradient. There will still be movement but there will be NO NET MOVEMENT. this is the curve of the graph.
As for the membrane protein, logic says if it is small enough do diffuse through the phospholipid bilayer, then it will happily take an easier route through the channel protein, by FACILITATED DIFFUSION. faciltated diffusion still relies on conentration gradients though.
(The explanation for "cell B" may be a little lax? if it is, just ask for confirmation)

What you said was right I understand it.But this graph shows us how increase in concentration of X affects rate of entry of X.So if the concentration of x inside and outside of cell is equal there would be no net movement ie diffusion,but if we increase concentration of x outside of the cell,then simple diffusion will occur,so there will be some rate of entry,its not 0 net movement, there is a net movement so graph can't plateau then,it must have some gradient.So then how can this be simple diffusion,is there some other reason.