Saturday, October 31, 2009

MCB Q&A and wonderful Wordle service

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#8 Membrane carrier vs. channel proteins

GRE-ST #26 question about difference between carrier and channel proteins. I just discovered that I don't even kind of idea of what membrane carrier protein is.
Membrane carrier proteins differ from membrane channel proteins by which of the following characteristics?
(A) Carrier proteins are glycoproteins, while channel proteins are lipoproteins.
(B) Carrier proteins transport molecules down their electrochemical gradient, while channel proteins transport molecules against their electrochemical gradient.
(C) Carrier proteins can mediate active transport, while channel proteins cannot.
(D) Carrier proteins do not bind to the material transported, while channel proteins do.
(E) Carrier proteins are synthesized on free cytoplasmic ribosomes, while channel proteins are synthesized on ribosomes bound to the endoplasmic reticulum.

Show answer
The best answer is provided in MBoC, of course. It is pretty short and uncovered in chapter 11, paragraph "There are two main classes of membrane transport proteins: carriers and channels". Briefly, carriers catch molecule of interest and push it inside or outside of compartment or cell by change in conformation, while channels organize – actually – a channel (often, filled with water or solution) that creates favorable conditions for transporting molecules of interest in- or outside.

Problem of transport through membrane is in its impermeability for charged molecules. That is why transport proteins are crucially important for cell. Transport differ, in common, into two groups: active and passive. It is transport against or down electrochemical gradient of molecules.
What is gradient? This topic is widely covered in calculus, but can be described simply with analogy of hill. If you put piece of rock on side of hill, rock will ran down to decrease its potential energy. Thus rock will roll down potential energy gradient. Same thing happens in the cell. If there are lots of molecules out of the cell and little inside, they will attempt to penetrate membrane. This process is spontaneous and do not need any energy, because it is actually hidden in solution and molecules themself, as energy for rolling is actually potential energy.
Another story with active transport. There we, obviously, need to take energy to drag molecule against its concentration gradient as we need energy to roll rock back on the hill.
Electrochemical gradient referred to two types of gradients. First one is gradient of electric charge. Because particles discussed now are charged, electric forces are involved. It takes some energy to pull positive charge into positively charged solute. We discussed chemical, or concentration, gradient above, it is about forces that drive molecules to occupy the larges volume. When speaking about electrochemical gradient, we are speaking actually about some mixture of those forces.

Now lets continue our talk about transport through membrane.
Important point here is that channels can't participate in active transport, i.e. force molecules against their electrochemical gradient, it was made possible by carrier proteins. As I mentioned, channel proteins create pore with channel, that act as defense from hydrophobic forces from phospholipid tails.

But what is the answer? That is difference in mechanism of transport. Using our brief discussion it is easy to identify false answers and explain their fallibility. Some trouble can be caused by answer A, because it is not clear are those proteins glycosilated or bound by any part to lipids. All – carriers and channel – proteins are membrane proteins, thus are synthesized on ER ribosomes. The difference between carriers and channels are, actually, in their names. As road tunnels don't ever touch cars, but ferries do, carriers closely interact and fix molecules on their surface, while channels just opens free road for rolling down gradient, so the right answer is D.

Friday, October 30, 2009

#9 Protein translocation in the ER

GRE-ST #39 question continues membrane proteins-related series of posts.
The common pathway of entry into the endoplasmic reticulum (ER) of secretory, lysosomal, and plasma membrane proteins is best explained by which of the following?
(A) Binding of their mRNAs to a special class of ribosomes attached to the ER
(B) Addition of a common sorting signal to each type of protein after completion of synthesis
(C) Addition of oligosaccharides to all three types of proteins
(D) Presence of a signal sequence that targets each type of protein to the ER during synthesis
(E) Presence of a zinc finger-binding domain in these three types of proteins
Show answer
ER plays important role in proteins' conformation maturation, transportation and translocation, which means their distribution in membrane bilayer.
Proteins from ER move out of the cell, into cell membrane and into cell's organelles membranes (e.g. lysosomes). Some of those proteins are expressed into cytoplasm after maturation (which includes proper folding and binding of oligosaccharides).
There are no special ribosomes, that are bound constantly to ER membrane and wait for mRNA to translate into protein. After gene (DNA) transcription of gene, mRNA is expressed into cytoplasm where ribosomes catch them. Actually, there are several binding sites on one mRNA so several ribosomes produce proteins simultaneously on one strand.
Main idea of that process is called "signal hypothesis", which says that there should be some signal that provide binding ribosome to ER and producing protein into ER, but not into cytoplasm. That signal is in first few amino acids of newly synthesized protein. After being exposed, this signal is recognized by special molecule – Signal Recognition Particle (SRP) which move ribosome with mRNA to SRP binding receptor on membrane of ER. This is very excited process which are uncovered in the MBoC clearly. Also prof. Schekman's lectures are very usefull.
Right answer is D. Gain more information in sources mentioned, because this is very interesting process, indeed.

Tuesday, October 27, 2009

#7 Bacteria and mammalian protein in vitro synthesis

GRE-ST question #29.
When bacteria produce mammalian proteins,
cDNA is used rather than genomic DNA. Which of the following is the best explanation?

(A) It is easier to clone cDNA than genomic DNA of comparable size.
(B) It is easier to clone RNA than DNA.
(C) It is not possible to clone the entire coding region of the gene.
(D) Most eukaryotic genes have introns that cannot be removed in bacteria.
(E) Most eukaryotic gene promoters do not function in bacteria.
Show answer
cDNA – complementary DNA, synthesized from mature mRNA (matrix RNA, protein-coding RNA, which is translated by ribosomes).
Intron – non-coding DNA region, that is transcribed into precursor RNA, but removed during splicing while RNA maturing.
Promoters – DNA regions, that signals begin of protein-coding sequences.

This question is uncovered by Alberts et al. in a tiny piece of text about cDNA libraries. Shortly, they remind that bacteria or yeast can't delete non-sense sequences (cos' their DNA have none of those) which are removed during splicing in mammalian. Hence right answer is D.

Monday, October 26, 2009

#6 Antibodies' monoclonality

#17 GRE Subject Test question
A major advantage of monoclonal antibodies compared to polyclonal antibodies is that monoclonal antibodies
(A) have identical binding sites that recognize a specific epitope
(B) cross-link molecules that share antigenic sites
(C) are more easily coupled with probes such as fluorescent dyes
(D) have higher-affinity binding to antigens
(E) can be produced against proteins that are immunogenic in rabbits
Show answer
Here we again should define used words accurately to find right answer. Antibodies are polypeptides, consisted of four chains, that can bind to specific sites of specific molecules (antigens). Monoclonal antibodies are ones that can bind to unique site of exactly one antigen. Polyclonal antibodies bind to wide range of sites of the antigen. That sites are called antigenic determinants or epitopes. They are recognized by antibody's so-called paratope. Thus, right answer is A.
Antibodies are produced using wide range of animals (rabbits, rats etc), so E is also false answer. And all those antibodies are mainly polyclonal, but monoclonal antibodies can be extracted using special technique called postfusional selection.
Answer C is not good one, because third tail of antibodies (they are shaped as Y) is not depend on poly- or monoclonality.
Answer D is sure false, because polyclonal antibodies have higher affinity, obviously: they can hook molecules by different sites so probability of binding act is higher. B is false too, bacause of mentioned above: cross-links are formed with polyclonal antibodies more rapidly.

#5 SNARE and mitochondrial proteins

#24 GRE Subject test question
SNARE proteins are found in the membranes of all of the following compartments EXCEPT
(A) Mitochondria
(B) Golgi complex
(C) Early endosome
(D) Endoplasmic reticulum
(E) Synaptic plasma membrane
Show answer
This one is about such important process as membranes fusion. SNARE are twin membrane proteins (referred to as v-SNARE [vesicle] and t-SNARE [target]), that act as anchors when two vesicles decide to fuse into one. This process is so important because it participate in molecular transport in the cell. Membrane and secretory proteins pass through several membrane-bounded organelles. Simple pathway of protein can be described as (you will find much more info in Alberts' MBoC) nucleus –> endoplasmic reticulum (ER) –> Golgi apparatus (or complex) -> plasma membrane OR endosome (pro-lysosome) OR exterior. Origin of vesicles plays its role in sinaptic signal transduction. The reason why there are no SNARE proteins in the mitochondria membrane is described while talking about mitochondria itself.
But before speak about mitochondria, lets remember why vesicles are essential for proteins movement. The main idea – as I see – is so-called co-translational translocation of proteins. In plain english it means, that proteins are translated by ribosomes only when they are bound to ER membrane. Thus proteins are translated into the ER, because if they were translated into cytoplasm, they could fold unproperly (by the way, about 80% of all membrane proteins fold unproperly even inside ER). But mitochondrial proteins that are imported (because mitochondria have its own gene translation machinery and so produce many essential proteins by itself) are translocated post-translationally, after ribosomes actually detach from polypeptide. After that so-called precursor proteins are imported into mitochondrial matrix through TOM and TIM complexes led by signal polypeptide chain (note that there are two membranes! that complexes act like gateway, really fascinating structure) just like membrane proteins being synthesized by ribosome are dragged to ER by Signal Recognition Particle (SRP) bound to protein's signal chain.
Precursor protein does not fold in the cytoplasm because of interactions with other proteins, some of which are general chaperones like hsp70 family.
There is a lot of things to tell about mitochondrial proteins translocation (imagine that some proteins, produced by the cell, are membrane mitochondrial proteins and their translocation requires several different signal polypeptide chains). But the point is that there is no possibility for membrane fusion between vesicles and mitochondrial bi-membranes so no SNARE needed for protein translocation. And the answer is A.

#4 Membrane protein arrangement

Following problem is from my current MCB class at the Universuty.
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.
As 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.

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.

Questionnaire #1

That is first series of short (with short description as well as answer) questions from GRE Subject Test.
Most of the dry mass in the trunk of a tree was originally derived from
(A) the soil
(B) light energy
(C) amino acids
(D) CO2
(E) glucose
Show answer
As soon as dry mass of the trunk of a tree consists mainly of oligosaccharides its source should contain carbon and hydrogen. Soil is too much common answer for me. Light energy can't be right answer because energy is not converted into matter easily. Amino acids is strange idea also. So let's check two options: CO2 and glucose. As this question – obviously – touches photosynthesize process, which yields glucose and other carbohydrates, I pick up CO2 as an answer.

Which of the following types of molecules is always found in virions?
(A) Lipid
(B) Protein
(C) Carbohydrate
Show answer
First, we should define what virion is. Surprisingly, it is just one instance of object virus. So you talk about virus (e.g. fag T4) properties, but when you meet one in the shop, it is virion.
What do we know about virions? They are simple systems which are not really cells because of lack of membrane and other important compartments as well as functions and facilitates. They have got capsid – hard-breaking protein-lipid armor around internal stuff. Inside they have some proteins and nucleic acid: either DNA or RNA, either single- or double-strained molecules. So the answer is proteins as molecules, that are found in all viruses and thus virions.

The difference between the molecular weight of sucrose and that of the sum of the molecular weights of its components (glucose and fructose) is
(A) 0
(B) 1
(C) 16
(D) 18
(E) 180
Show answer
If you remember scheme structure of sucrose as O_O (:-) you get the idea of question. The reaction of such polymerization looks like that:
Glu-OH + OH-Fru = Glu-O-Fru + HOH
Thus, difference is in one water molecule which is 2 hydrogen atoms and 1 oxygen. The answer is D: 18 = 2*1 + 16.

An alpha-helical conformation of a globular protein in solution is best determined by which of the following?
(A) Ultraviolet-visible absorbance spectroscopy
(B) Fluorescence spectroscopy
(C) Electron microscopy
(D) Analytical ultracentrifugation
(E) Circular dichroism
Show answer
I really looooove fluorescence. So I hope it is the right answer, but lets be patient.
Electron microscopy is not good idea, because it can't show us structure of the protein as x-ray crystallography make it possible.
Ultracentrifugation, I suppose, will not help us because it will just sediment peptides and nothing more.
Circular dichroism is damn strange stuff I don't know much about. But google fetched me nice page about circular dichroism. I got the idea of that method: difference in absorbances of left- and right-polarized light is higher if peptide is structured. It is rather similar to fluorescence anisotropy. High anisotropy shows that fluorophore is fixed in space (and time :-) It can also predict stable structures as alpha-helices or beta-layers.

Sunday, October 25, 2009

#3 Membrane phospholipids

This one is MBoC:PB question 10-28.

A classic paper studied the behavior of lipids in the two monolayers of a membrane by labeling individual molecules with nitroxide groups, which are stable free radicals (Figure 10-3). These spin-labeled lipids can be detected by electron spin resonance (ESR) microscopy, a technique that does not harm living cells. Spin-labeled lipids are introduced into small lipid vesicles, which are then fused with cells, thereby transferring the labeled lipids into the plasma membrane.
The two spin-labeled phospholipids shown in figure 10-3 were incorporated into intact human red blood cell membranes in this way. To determine whether they were induced equally into the two monolayers of the bilayer, ascorbic acid (vitamin C), which is a water-soluble reducing agent that does not cross membranes, was added to the medium to destroy any nitroxide radicals exposed on the outside of the cell. The ESR signal was followed as a function of time in the presence and absence of ascorbic acid as indicated in Figure 10-4A and B.
A. Ignoring the moment of the difference in extent of loss of ESR signal, offer an explanation for why phospholipid 1 (10-4A) reacts faster with ascorbate than does phospholipid 2 (10-4B). Note that phospholipid 1 reaches plateau in about 15 minutes, whereas phospholipid 2 takes almost an hour.
B. To investigate the difference in extent of loss of ESR signal with the two phospholipids, the experiments were repeated using red cell ghosts that had been resealed to make them impermeable to ascorbate (Figure 10-4C and D). Resealed red cell ghosts are missing all of their cytoplasm, but have an intact plasma membrane. In these experiments the loss of ESR signal for both phospholipids was negligible in the absence of ascorbate and reached plateau at 50% in the presence of ascorbate. What do you suppose might account for the difference in extent of loss of ESR signal in experiments with red cell ghosts (10-4C and D), versus those with normal red cells (10-4A and B).
C. Were spin-labeled phospholipids introduced equally into the two monolayers of the red cell membrane?
Show answer
We will go number-by-number.
A. I think that the speed of ESR signal extinction is linked with probability of reaction between ascorbate and nitroxide group. Thus, speed of phospholipids 2 signal extinction should be significantly less than for phospholipids 1, because it is not as easy for ascorbate to hit side chain than head group.
B. Simple comparison of ghosts membranes and intact cells tells us that there should be kind of agent – reducing agent – in the red cells' membranes that can attack head groups but not side chains. Of course red cell ghosts lack that agent, so only ascorbate acts. That is why loss of ESR signal in 10-4A is 50% in abscence and 100% in presence of ascorbate. Also this idea explain ESR signal reaction in ghosts also.
C. Yes, they were introduced equally in the interior and exterior layers of membrane. This is right both for intact cells ("natural" 50% signal decrease) and for ghosts (50% decrease in presence of ascorbate).

Now lets check the right answer.
A. The difference in rate of loss of the ESR signals is due to the location of the nitroxide radical on the two phospholipids. The nitroxide radical in phospholipid 1 is on the head group and is therefore in direct contact with the external medium. Thus, it can react quickly with ascorbate. The nitroxide radical in phospholipid 2 is attached to a fatty acid chain and is therefore partially buried in the interior of the membrane. As a consequence, it is less accessible to ascorbate and is reduced more slowly.
B. The key observation is that the extent of loss of ESR signal in the presence and absence of ascorbate is the same in resealed red cell ghosts, but different in red cells. These results suggest that there is an undefined reducing
agent in the cytoplasm of red cells (which is absent from red cell ghosts). Like ascorbate, this cytoplasmic agent can reduce the more exposed phospholipid 1, but not the less exposed phospholipid 2. Thus, in red cells, phospholipid 2 is stable in the absence of ascorbate; in the presence of ascorbate, the spin-labeled phospholipids in the outer monolayer are reduced, causing loss of half the ESR signal. Phospholipid 1, on the other hand, is not stable in red cells in the absence of ascorbate because the phospholipids in the cytoplasmic monolayer are exposed to the cytoplasmic reducing agent, which destroys half the ESR signal. When ascorbate is added, labeled phospholipids in the outer monolayer are also reduced, causing loss of the remaining ESR signal.
C. The results in Figure 10–4 indicate that the labeled phospholipids were introduced equally into the two monolayers of the red cell plasma membrane. Phospholipid 2 was 50% sensitive to ascorbate, indicating that half the label was present in the outer monolayer, and 50% insensitive to ascorbate, indicating that half was present in the cytoplasmic monolayer. Similarly, phospholipid 1 was 50% sensitive to the cytoplasmic reducing agent and 50% sensitive to ascorbate, indicating an even distribution between the cytoplasmic and outer monolayers.

Saturday, October 24, 2009

#2 Sweetened salad dressing

Ok, this one is from GRE Subject test (question number 16) and rather "chemical".
If sucrose and monosodium glutamate (MSG) are added to vinegar and oil salad dressing and shaken, the mixture will eventually separate into two phases of different density and polarity. Where will most of the sucrose and the MSG de located following phase separation?
(A) Both will concentrate in the vinegar
(B) Both will concentrate in the oil
(C) Both will concentrate at the interface
(D) Sucrose will concentrate in the oil and MSG – in the vinegar
(E) Sucrose will concentrate in the vinegar and MSG – in the oil
Show answer
Lets think a bit. The key word in the question for me is "density" and also I can't understand what "polarity" means and what role does it play.
So we will use density approach.
If you google a bit, you'll find – as I did – that oil density is about 0.9 g/ml and sucrose density is 1.5 g/cm3 (=1.5 g/ml) when MSG density is about twice less (lets say 0.7g/ml). So oil will be on the top, above vinegar. Sucrose will  precipitate (so it will be in vinegar). The question is where MSG will get to. Its approximate density is less than oil's so it will be on the top of the mixture.
So my answer is E. I have no idea how to get another way to find answer, maybe it is just knowledge test.

#1 phospholipids and hydrophobicity

It is 10-14 question from MBoC:PB. It is a bit jokey, but raises different physical and biophysical problems about energy, forces and so on.
Five students in your class always sit together in the front row. This could be because (1) they really like each other or (2) nobody else in your class wants to sit next to them? Which explanation holds for the assembly of a lipid bilayer? Explain your answer. If the lipid bilayer assembled for the opposite reason, how would its properties differ?
Show answer
Main property of phospholipids is that they are amphiphatic molecules: they have hydrophobic tails and hydrophilic heads. Consequently, molecules – when crowded – try to hide tails away from water and the only way to do that is to get to the water surface (and swing tails into the air) or form vesicle (bubble), where tails are hidden inside. That is tails are forced to get away from water.
So, I suppose, the right answer is (2): every one in the class force them to get on the "surface" of class (in our case it is water).
If phospholipids liked each other, then, I think, no molecules (proteins) could stay inside membrane or among lipids.

Lets check the right answer from the CD, that supply the book.
Lipid bilayers assemble because the surrounding water molecules exclude
the component lipids; thus, analogy (2) is the correct one. If bilayers formed
because of attractive forces among the lipids—analogy (1)—the properties
of the bilayer would likely be quite different. Molecules ‘attract’ one another
by forming specific bonds that hold them together. Such bonding among
lipids would make the bilayer less fluid, perhaps even rigid, depending on
the strength of the interaction.
I was quite right, except forces approach. But yes, attracted molecules are more durable system, no molecules can intercept it or destroy. So my idea on proteins is also right.

#0 question

Question number 1 from GRE-Subject test.
Which of the following represents the most
reduced form of carbon?
(A) R-CH3
(E) CO2
Show answer
Let's go from the opposite. Antonym for reduced is oxidized, which means "with deleted electrons". So 5 is obviously not the right answer, because oxygen is highly electron-horny element, it pulls electrons from all around (only F – fluorine – is more powerful). So number 1 is our choose.
Test of blog ans script (answer toggler) is pretty well passed.

Blog's structure

Questions will be presented in no order. I will number them sequentially as write down, but some questions (as example look in MBoC: The Problems Book [MBoC:PB]) are linked tightly so they will have hyper links and will be posted in order.
You can browse problems using labels (it is blogspot's term; I prefer tags). If some question is about cell membrane, it will have tag cell membrane. But if question is about endoplasmic reticulum, then it will be labeled as ER, so common acronyms will be in use. If the question is about some common properties of membranes it will be labeled as just membrane. Uh.
Hope you will find what you are looking for. If not – please, leave comment or (whats is preferable) drop me an email (andrugatorandreev at

Friday, October 23, 2009


OK, here we start our journey into fascinating biochemistry, molecular and cell biology tests and general questions. There are several sources for them as:  
  • GRE subject test (180 questions that are in booklet on the website) on MCB and biochemistry
  •  Lodish's great Molecular Cell Biology book's questions at the end of each chapter (approx. 11/chapter*23 chapters=253)
  • maybe questions from Principles of Fluorescence Spectroscopy (by J.R. Lakowicz; there are a lot of questions, only specific or more common will be selected)
  • The Molecular Biology of The Cell: The Problems Book [MBoC:PB] (by Wilson and Hunt; a bunch of questions for each Alberts' MBoC chapter)
  • I have received midterm exam questions from prof. Schekman's great lectures (Berkely's MCB130 course that is available as webcasts) so they will be also included (5 to 9 multi-item thought questions per test in about 5 or 6 tests).

I wish to solve all that questions and record solutions. Probably, this blog can help someones' else training.
Good luck and good answers!