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.