Monday, December 14, 2009

#12 ER lumen protein

GRE-ST question number 54 is about ER proteins and functions.
The KDEL sequence, found on luminal proteins of the ER, is responsible for
(A) translocation of proteins into the ER lumen
(B) insertion of proteins into the membrane of the ER
(C) quality control in the ER
(D) recognition by signal peptidase of the signal sequence
(E) retrieval of ER luminal proteins from the Golgi

Show answer
This question is just seems to be hard. Just google and you'll probably find this wikipedia article on ER, where KDEL function is described clearly.
Point is that there are two types of proteins in ER lumen: those which should be transported further (to Goldgi) and those which work inside ER (e.g. linking oligosaccharides to proteins). So there should be some anchor for that "stable" proteins and it links to KDEL sequence on the end of those proteins. And because they cannot be fixed any way (or it would be very, very energetically unfavourable) they are transported back from Golgi to ER by specific KDEL-recetors.
So right answer is E.

Sunday, December 6, 2009

#14 Protein secretion

GRE-ST question #98.
All of the following processes occur in the pathway leading to regulated protein secretion in animal cells EXCEPT
(A) formation of transport vesicles from the rough endoplasmic reticulum
(B) an increase in the concentration of cytosolic calcium ions prior to secretion
(C) synthesis of an amino-terminal signal sequence
(D) phosphorylation of a mannose residue in a glycoprotein
(E) trimming of N-linked oligosaccharides
Show answer
That is pretty simple. Question can be solved by just following common protein secretion pathway. That starts by transcription DNA's gene into RNA. Then that RNA — called matrix RNA, mRNA — caught by ribosome — which is actually RNA too, but ribosomal thus rRNA — and with assistance of different translation factors protein production begins (that is called translation mRNA into protein). Why secretion proteins are selected ones? That is because they have special signal at the beginning of polypeptide chain on N-terminal (which is actually some sequence at the beginning of mRNA, of course) which is caught by special cytoplasmic protein — Signal Recognition Particle or SRP. When SRP bind to that signal sequence translation stops and ribosome with mRNA and bound SRP flows to endoplasmic reticulum, where SRP-receptors are located. They bind SRP and passes signal through the lipid bilayer of ER. Then translation continues and protein get into ER lumen. After that happened, protein undergo different saccharides modifications. Pathway of Asn modification clearly described in Alberts book, includes trimming of N-linked oligosaccharides. Main steps of protein movement are: secretion in vesicle from ER to Goldgi apparatus and same movement through different compartments of Goldgi apparatus. After that vesicle with protein inside fuse into cell's membrane thus secreting protein into extracellular space.
So, I suppose that no calcium used in this pathway except signal transduction to muscle cells. So the answer is B.

Monday, November 9, 2009

Great Lodish's MCB dedicated web service

I have found recently very interesting site, hosted by publisher of Lodish et al. "Molecular Cell Biology" 6th edition.
For this blog's purpose most interesting part of that service are Online Quizzes. Nice feature there is that it is possible to subscribe your instructor to your results.
Other services are lots of animations, tutorials, review and more-more stuff. I would like to mention here "Classic Experiments" section of the site.
What is even more exciting is that there are similar sites for 5th and 4th editions users! And there is GRE kind test on site dedicated to the 5th edition.

I really admire those people, who maintain biological education services online and on such high level. I think it is dangerous lack of opportunities for online reviewing, testing and upgrading knowledge in russian bio education.

Friday, November 6, 2009

Questionnaire #2

Several GRE-ST questions linked together.
Questions 118-121 refer to the following cellular processes.
(A) Phagocytosis
(B) Exocytosis
(C) Endocytosis
(D) Transcytosis
(E) Apoptosis

118. Movement of plasma membrane receptors from the basolateral surface to the apical surface of polarized epithelial cells
119. Up-regulation of glucose transporters at the plasma membrane
120. Selective retrieval of cell-surface proteins for recycling or degradation
121. Neurotransmitter release

Questions 124-126 refer to the following protein-modifying reagents.
(A) Chymotrypsin
(B) Cyanogen bromide
(C) Iodoacetamide
(D) Phenylglyoxal
(E) Pyridoxal 5′-phosphate

124. Forms a Schiff-base linkage with the epsilon-amino group of lysine residues
125. Specifically cleaves polypeptides on the carboxyl side of methionine residues
126. Generally used as a sulfhydryl-modifying reagent

Questions 127-129 refer to the following cell structures.
(A) Thylakoid membrane
(B) Nuclear lamina
(C) Eubacterial cell wall
(D) Plant cell wall
(E) Endoplasmic reticulum

127. Its assembly is inhibited by penicillin.
128. It is formed from polymeric fibrils composed of cellulose cross-linked by pectin and hemicellulose.
129. It is the site of dolichol phosphate function.
Show answers
It is important here to notice that one answer can be right for several questions.
Now lets try to describe each process announced. So phagocytosis is process of "eating" external organells by the cell. It is quite close to endocytosis, when – in contrast – molecules are being absorbed. Exocytosis is opposite process – secretion of molecules into external space. It can be hormones, antibodies and so on. Transcytosis is interesting process of transporting molecules from apical (absorbing) space of polarized cells to basolateral space, which happens, for example, when antibodies are transported through baby rat's gut. Main idea is that molecules are absorbed into internal vesicles (as during endocytosis) and then after several steps containers' content is exposed as during exocytosis. No target molecules are left in cell's plasma. Interesting is that cell can regulate exposure of some proteins in the cell membrane using transcytosis. During latent phase (e.g. while there is no great need in glucose) transporters are stored incorporated in vesicles inside the cell. But in case of emergency (or if cell is stimulated, by insulin in our case) that vesicles fuse into plasma membrane and transporters expose into extracellular space and begin to work. Last one is apoptosis which is self-destroying of the cell. That happens if crucial malfunction appear or cell's mission was fulfilled. So the answer are like that. I suppose, answer D is suitable for both two first questions.
118 D. 119 D. 120 C. 121 B.

Chymotripsin is protein, enzyme actually, of so-called class of proteases. It cleaves specific polypeptide regions near serine, so it is called serine protease.
Cyanogen bromide is simple small molecule CNBr and, as written in wikipedia, is used to immobilize proteins by gluing them to agarose gel during chromatography. But more important is that CNBr cleaves polypeptide chains at C-terminus of methionine.
Iodoacetamide is substance that can bind covalently with cysteine to prevent formation of disulfide bonds. So it can be called sulfhydryl modifying reagent.
Phenylglyoxal modifies arginine.
Pyridoxal 5′-phosphate (or PLP) is active form of vitamin B6. As written in wikipedia (again, yeah) it forms Schiff-base linkage the ε-amino group of a specific lysine group of the aminotransferase enzyme.
124 E. 125 B. 126 C.

This questions uncovers very specific properties of very important and well-known objects. So I will not discuss all of them, but just give right answers and some interesting details if I find any.
All options are common in one: they are membranes or structures, that acts like membranes – support shape (cell walls and nuclear lamina) and protect internal content. But they have different structure, functions and behavior.
127 C. Notice that eubacteria is synonym for bacteria.
128 D. 129 E. Dolichol phosphate acts in formation of membrane-associated glycoproteins, so I suppose its site is on the ER membrane.

#13 Hydrogen bonds

Which of the following pairs of molecules could NOT hydrogen bond with each other?

Show answer
Answer can be easily found in MBoC in few first chapters (page 57 :). Main idea about hydrogen bonds is that it is an interaction between two partial charges: d+ and d- or partial charge d+ with net charge on atom. So if there is shift of electrical density on some atom from hydrogen and same situation happened with atom on another molecule (actually this can be one molecule and hydrogen bond will occur because of its flexibility) hydrogen bond will establish. Note that its strength is about 20 times less than covalent bond's.
Right answer thus is D because there is no shift of electron density to the ring. In other cases there is possibility exists for hydrogen bonds to occur: between H-O- and O=H2 (A), H-OH and O= (B), -NH and O- (C), -OH and O= (E).

Sunday, November 1, 2009

#10 Membrane is mosaic two-dimensional fluid (isn't it?)

GRE-ST question #109 – I reaaaally looove membranes, yeah?
All of the following statements about the fluid mosaic model of biological membranes are true EXCEPT:
(A) Lipid molecules in the membrane readily undergo lateral diffusion.
(B) Lipid molecules in the membrane readily undergo transverse (flip-flop) diffusion.
(C) Integral membrane proteins can undergo lateral diffusion.
(D) The saturated hydrocarbon chains of lipid molecules in the membrane undergo carbon-carbon bond rotation.
(E) The transition temperature of a membrane is sensitive to the composition of the lipid molecules in the membrane.
Show answer
That is rather easy question. To gt right answer, lets consider options one by one.
First, word "readily" was quite unclear, before I looked it up in the dictionary. It means "without hesitation or reluctance", i.e. without doubt, but in our case – rapidly, fast. So A is obviously true, because later (in plane) diffusion is primary characteristic of bilayer.
B is first candidate for right answer, because it is known, that one phohpolipid flip-flop by itself just once a month. But there are special proteins called phospholipid translocators. As wrote in MBoC, they "catalyze the rapid flip-flop of phospholipids from one monolayer to the other".
C is right too, because lots of proteins, e.g. channels and receptors, migrate in plane of bilayer rapidly.
E is true too. There two factors can be selected: saturation and length of phospholipids' tails and cholesterol amount in the bilayer. First two properties shift viscosity curve of membrane, while second smooth it as drawn on the picture below. Transition temperature is – in common sense – temperature when material undergoes phase transition. If tails are saturated (i.e. all carbon-carbon bonds are not double) and long, then it becomes easier to pack such bilayer tightly, thus temperature of freezing shifts to higher value. Contrary, if chains are unsaturated and short, their mobility is lower, thus freezing point will be shifted to lower temperatures. But cholesterol is thing that improve mobility of phospholipids at lower temperatures, as well as keep rigid structure at higher values.

Now, answer D. It is known, that tails of saturated hydrocarbons are very, vary mobile stuff. But they can't get far away from phospholipid core, hence they rotate around some links. Unsaturated phospholipids have double bonds in their tails. That double bonds are actually like two sticks, that connect atoms, so rotation around double bonds is impossible.
So right answer, as we excluded any other, is B and rapid flip-flop, even in presence of phospholipid translocators, doesn't happen.

#11 Amino acids identification

GRE-ST questions #162 and 163 are about amino acids: Arg, His, Cys, Pro and Trp.
A student was given a task of identifying the contents of five bottles of amino acids from which the labels had fallen off. Each of the original bottles contained one of the following: arginine, histidine, cysteine, proline, and tryptophan.

162. Which of the following methods could be most
readily employed to identify tryptophan?
(A) Electrophoresis
(B) Ultraviolet spectroscopy
(C) Gel filtration
(D) Analytical ultracentrifugation
(E) Optical rotation

163. Each amino acid was subjected to paper electrophoresis in a pH 9.5 buffer, and then the amino acids were visualized by spraying the paper with ninhydrin. Which amino acid is at the point labeled (1) in the figure above?
(A) Arginine
(B) Histidine
(C) Cysteine
(D) Proline
(E) Tryptophan
Show answer
First one question is rather easy. Tryptophan have one very useful feature that called fluorescence. That is emission of light after illumination. Trp is one of 20 amino acids and widely occur in proteins and its fluorescence provide useful data about protein's folding and structure. Trp's fluorescence is one of the most intense among all fluorescent amino acids (others are Phe and Tyr). Trp excitation wave length is 280 nm and emitted light has wave length 300-350 nm. Ultraviolet is part of spectrum from 10 to 400 nm. So right answer is B, because other amino acids from the list are not fluorescent.

Next part of question refers mainly to amino acids' behavior in different pH. So it is necessary to investigate polarity of substances in solutions with different acidity.
But what does pH actually means? It is defined as pH=-Log[H+], i.e. minus decimal logarithm of H+ (protons) concentration in solution. Neutral pH is 7 because in water there are 10^-7 M/liter of H+ ions due to H20 decay into H+ and OH-.
So, briefly, high pH means very low concentration of H+, and low pH's mean high concentration of H+, or low concentration of OH-. Actually, pH and pOH (defined in the same manner) are closely connected with each other. For pure water pH=pOH, of course.
Now lets go further. Practically all amino acids can be distinguished using pKa – pH level at which molecule is not charged. That very simple experiment described in the question. If there is any pKa and if acid (actually, amino acid residue) have NH2 or OH groups, then at pH higher than pKa protons will be taken away from acid so -OH will become -O- (oxygen with negative charge) and at lower that pKa pH acid's NH2 will become NH3+ (one positive charge). Alberts provide several examples, that make everything clear:

Using pKa values is quite hard to solve this problem, really. Best picture can be granted by pI value. That is isoelectric point, pH level, at which whole molecule of acid is electroneutral. There is a nice table of pI values on the web. If pH higher than pI, then protons will be taken away and charge will become negative. Otherwise at pH lower than pI molecule will be protonated. If you check linked page, you will find, that all molecules from questions have pI lower than 9.5 and only arginine have higher value. Hence, Pro, His, Cys and Trp will have negative charge and only Arg – positive. That is why four molecules will move on one side from point of sample application, and only one – on the other. So it will be Arg, the answer is A. Note, that picture is electrophoresis results, so + and - are signs of CAthode and ANode, which attracts CAtions and ANions respectively.

Additional information
OK, we figured out – using something called pI – right answer. But why does it happens to be right? And why – what is more important – why argynine differs so much from other amino acids?
First, what is pI actually? It is mean of pK(amino group) and pK(carboxyl group) for molecule, which have only one of those. pI=(pK(NH2)+pK(-OH))/2. When pH is rising from pK of -OH more and more groups undergo deprotonating. Similarly, when pH decrease from pK of NH2, more and more groups undergo protonating, so there is point of equilibrium, where number of protonated groups equal number of deprotonated groups, thus net charge is zero. We assume that this equilibrium point is average of pK's.
But there are several amino acid residues that have their own -NH2 or -OH groups and even not one as Arg (2 -NH2), His (1) and so on. In this case pI is counted as sum of pK-s of all -OH and -NH2 groups divided by two. That is why Arg has pI much greater than any amino acid's that lack -NH2 groups at all.

Saturday, October 31, 2009

MCB Q&A and wonderful Wordle service

Wordle: my blog in wordle

New version:
Wordle: MBC QnA-2

V4 (Nov 8)
Wordle: MCB:Q&A - 4

#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!