You would like to study some protein of red blood cell membrane. So you prepare ghosts of that cells and extract all proteins from the bilayer. Via SDS gel electrophoresis you select some protein with mass 106 kDa. Using some organism you get appropriate antibodies to that protein. They are polyclonal and monospecific.
After that you get new ghosts and reconstitute them into closed empty membranes. The experiment is designed in such way (it is its imperfection), that some ghosts will be right side out and the others will be inside out. Those different membranes are easily separated using chromatography because outer surface of natural red blood cells is covered with oligosaccharides and the inner is not (we will talk about it again in later posts about membrane proteins formation).
OK. You separate right out and inside out membranes into two different test tubes and add into them set of different proteases – enzymes that break down polypeptide chains (proteins) in different sites. Of course, some scraps will be washed into buffer and easily taken away.
To extract parts, that are still in the membrane, you use detergent (e.g. Triton X100) and sediment then your solution. Proteins will get to the bottom of the test tube after centrifugation.
After SDS gel electrophoresis of precipitate you got following sets of lines:
– right side out vesicles: 22.8, 19.2 and 38.4 (kDa)
– inside out vesicles: 14.4, 13.2, 10.8 (kDa)
which are visible because of reaction with antibodies you got earlier. You are able to see not all the protein's parts that were still in the membrane of course.
The question is what assumption can you make about that protein arrangement inside red cell membrane?
Show answer (its quite long, with images, black-jack and hookers)
That is really experimental exercise, an example of protein study. The idea here is that we know several types of protein-membrane interactions or how some protein can penetrate membrane.
As we know nothing about glycolipids links and another stuff, lets assume that protein snakes membrane in some way. Next, it is important to know what protein's composition is inside bilayer. It can be either alpha-helix (numbers 1, 2 and 4) or beta-barrel (number 3).
Alpha-helices that pass through the membrane are chains of 15 to 30 amino-acids residues. If one acid weight 100 to 200 Da, then one helix's molecular weight is 1.5 to 6 kDa.
If there is hydrophilic residue, it hides inside if the helix. Otherwise hydrophobic residues are exposed to the hydrophobic phospholipids' tails.
Barrels, made of several (8 to 22) beta-strands are something different. Strands form walls pore in the membrane often filled with water. As in helices, hydrophobic and hydrophilic residues' orientation persist: later are exposed to solute (water). But the difference is in linkage between strands and helices of multi-pass proteins. Barrel's "wall" is tightly fixed and rigid because of cross hydrogen bonds between residues when alpha-helices are much more mobile.
Barrel-type proteins are restricted to mitochondrial, chloroplast and bacterial outer membranes. vast majority of multipass transmembrane proteins in eucaryotic cells and in the bacterial plasma membrane are constructed from transmembrane alpha-helices, as Albers et al. wrote. This happens because of greater mobility of helices relative to each other that cause possibility of open-close switching of the channel.
But lets return to the problem. First, we will check if it is possible, that protein snakes through membrane as several helices.
Assume that antibodies will be produced to actually any helix and no parts will be unlabeled in the gel. We can do so because protein is rather large – 500 to 1000 amino acids residues. That means that it is impossible that protein snakes membrane, for example, 10 times, but we see only three chains (see image below as explanation; red are labeled antibodies and transparent regions are cleaved by protease so numbered regions are electrophoresis result).
Half-inserted helices as in picture 4 from first diagram are not possible because they are protected from washing away and can only act as additional parts that will not be shown in experiment with another side-out vesicles or there will be too much parts separated by protease.
So in "alpha-helices"-approach only possible answer is 5 transmembrane helices as on the picture. The question is only where N- and C-terminal amino acids are located. They cannot be on one side of the membrane, you can see in easily just drawing several snakes through bilayer. But it also cannot be determined during our data.
Also there is no need to distinguish exterior and interior any more. Lets just say, that from one side there is one stack of polypeptides and another from the opposite.
This is our model. Lets say, top is exterior and bottom is interior. Our model is essential for check if such composition ever possible. parts inside membrane are helices, and so, for example, part 1+2 can weight 10.8 kDa, because it is on the exterior of red blood cell and protected while protease cleave inside-out vesicle.
What's next? I suggest to write down hole system and solve it for each part (1, 2, 3 etc.). But there will be 6 equations (for number of conditions) and 11 unknown variables, that depends actually on helices weights. Vice versa weights of helices are functions of parts' weights (it is almost so).
My idea is to set different weights and check if our system will have positive and reliable solution on 1st, 3rd, 5th, 7th, 9th and 11th parts' weights.
Beta-barrels are not likely to be kind of answer because it is made of several, usually more that 5 strands, which ruins our model. But if we assume that there are 5 strands, everything will be OK if we just change even numbers in system for beta-barrel reliable. I hope it will fit, but check it up.
Phew! Get this tiny bonus from me.
As we know nothing about glycolipids links and another stuff, lets assume that protein snakes membrane in some way. Next, it is important to know what protein's composition is inside bilayer. It can be either alpha-helix (numbers 1, 2 and 4) or beta-barrel (number 3).
If there is hydrophilic residue, it hides inside if the helix. Otherwise hydrophobic residues are exposed to the hydrophobic phospholipids' tails.
Barrels, made of several (8 to 22) beta-strands are something different. Strands form walls pore in the membrane often filled with water. As in helices, hydrophobic and hydrophilic residues' orientation persist: later are exposed to solute (water). But the difference is in linkage between strands and helices of multi-pass proteins. Barrel's "wall" is tightly fixed and rigid because of cross hydrogen bonds between residues when alpha-helices are much more mobile.
Barrel-type proteins are restricted to mitochondrial, chloroplast and bacterial outer membranes. vast majority of multipass transmembrane proteins in eucaryotic cells and in the bacterial plasma membrane are constructed from transmembrane alpha-helices, as Albers et al. wrote. This happens because of greater mobility of helices relative to each other that cause possibility of open-close switching of the channel.
But lets return to the problem. First, we will check if it is possible, that protein snakes through membrane as several helices.
Assume that antibodies will be produced to actually any helix and no parts will be unlabeled in the gel. We can do so because protein is rather large – 500 to 1000 amino acids residues. That means that it is impossible that protein snakes membrane, for example, 10 times, but we see only three chains (see image below as explanation; red are labeled antibodies and transparent regions are cleaved by protease so numbered regions are electrophoresis result).
Half-inserted helices as in picture 4 from first diagram are not possible because they are protected from washing away and can only act as additional parts that will not be shown in experiment with another side-out vesicles or there will be too much parts separated by protease.
So in "alpha-helices"-approach only possible answer is 5 transmembrane helices as on the picture. The question is only where N- and C-terminal amino acids are located. They cannot be on one side of the membrane, you can see in easily just drawing several snakes through bilayer. But it also cannot be determined during our data.
Also there is no need to distinguish exterior and interior any more. Lets just say, that from one side there is one stack of polypeptides and another from the opposite.
This is our model. Lets say, top is exterior and bottom is interior. Our model is essential for check if such composition ever possible. parts inside membrane are helices, and so, for example, part 1+2 can weight 10.8 kDa, because it is on the exterior of red blood cell and protected while protease cleave inside-out vesicle.
My idea is to set different weights and check if our system will have positive and reliable solution on 1st, 3rd, 5th, 7th, 9th and 11th parts' weights.
1+2=10.8As you see, if we define all even weights (weights of helices actually) we will surprisingly fast achieve all pieces weights that are wanted just to be positive and reliable as polypeptides' weight in kDa. Thus if all helices weights (remember they are even numbers in system) are from 1.5 to 6 kDa, other pieces weights will vary from approximately 4 to 9 kDa (1st), 2 to 10 (5rd), 3 to 11 (9th), 26 to 35 (3rd), 10 to 19 (7th) and 13 to 17 (11th). Hence there will be from 10 to 175 amino acids residues average in each "tail" of protein that faces either exterior of the cell or cytoplasm. That is truly reliable (I loooove this word sooooo much, indeed!) because, for example, there is a protein called band 3 in the red cell membrane (no, question is not about it) with 12 alpha helices and about 930 amino acids in polypeptide. So if count a bit and draw picture of band-3 we'll see, that there are about 75 amino acids for each tail that is outside the bilayer. If we get average number of amino acids in "tails" from our exercise it will be (10+175)/2 = 92. What is quite close to band-3 result.
4+5+6=13.2
8+9+10=14.4
2+3+4=38.4
6+7+8=22.8
10+11=19.2
Beta-barrels are not likely to be kind of answer because it is made of several, usually more that 5 strands, which ruins our model. But if we assume that there are 5 strands, everything will be OK if we just change even numbers in system for beta-barrel reliable. I hope it will fit, but check it up.
Phew! Get this tiny bonus from me.
