Thursday, February 8, 2018

GRE Molecular & Cell BIology Practice test Q 76

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This question is concerned with organisms that can or cannot live without exogenous (added to the environment) chemical X. By definition auxotrophs are those that need some essential nutrient to be added, prototrophs can synthesize their own chemical X. Also, we need to know what is

"complementation group". By definition, that is pair of mutants that have same phenotype (say, they are auxotrophs for nutrient X), but when crossed will produce wild-type (prototroph) progeny. That means that mutations in these mutants occurred in different genes, and each parent "complemented" each other, supplied wild-type gene to correct for another parent's mutation. "Complementation group" is a group of genes that do not compliment each other. By definition, each gene can be in its own complementation group.

From data, mutants 2 and 4 do not complement each other, so are mutants 3 and 5. All what's left is mutant 1, which by definition doesn't complement itself, hence in its own group.

 Answer is C

 This is also a useful link:

Sunday, February 4, 2018

GRE Molecular & Cell Biology Questions 170, 171, 172

Today question is from GRE Biochemistry, Cell and Molecular Biology practice test.
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A1. From Fig. 1&2 we can see that protein synthesis continues for some time after inhibitor was added. That is caused by the fact that even when inhibitor is added and significant fraction of RNA Polymerase activity is halted, there is still a lot of "idle" or not yet used products of RNA Pol in the cell. This backup amount of RNAs is available for protein synthesis for short time. Answer is A.

A2. From Fig. 1 we can approximate rate of synthesis after inhibitor was added, and compare it to non-inhibited case. TO do that, let's just calculate derivative after inhibition, or slope of the lines. Take vertical distance (change in Counts Per minute, CPM), divide by change in time.

No inhibition: 6000-2000 / 20 hrs ~ 4000CPM/20min=200CPM/hr
With inhibition: 3000-2000 / 20 hrs ~ 50CPM/hr

After inhibition, rate goes down about 4x, thus we can estimate that inhibitor blocks about 75% of activity. Answer is B.

A3. In this question experiment was performed as following: proteins from nuclear extract were purified on a column. Peak were collected and used as a source for RNA polymerization. Since radioactive thymine was used, RNA polymerization activity of each fraction (peak) was measured afterwards.

In HeLa cells, there are 3 RNA polymerases:
RNA Pol I: synthesizes a pre-rRNA 45S, molecular weight 590 kDa
RNAPol II: synthesizes precursors of mRNAs and most snRNA and microRNAs, MW 550 kDa
RNA Pol III: synthesizes tRNAs, rRNA 5S and other small RNAs found in the nucleus and cytosol, MW 0.7MDa

Question is about type of the RNA Pol that is concentrated in first peak. Interestingly, Fig 3 gives much more information that asked in question, possibly setting up test-takers for a distraction. To answer, I had to Google RNA Pol chromatography, and it seems that peak 1 is mainly RNA Pol I.

 But even more widely known is that nucleolus is a site where ribosomal RNA (rRNA) is primarily produced. Answer is A.

Sunday, March 19, 2017

Bonus day: question 36 from Physics GRE

This question comes from practice Physics GRE exam (2011 edition)

The capacitor in the circuit above is charged. If switch S is closed at time t = 0, which of the following represents the magnetic energy, U, in the inductor as a function of time? (Assume that the capacitor and inductor are ideal.)

That's a relatively simple question with P+=50%. Now let's get to the answer: Show answer
Correct answer is A

To get to it we might use process of elimination: find all answers that are NOT correct and see what's left. First of all, question says "Assume that the capacitor and inductor are ideal". What that means right off the bat is that there should be no dissipation of energy. So answer D is incorrect right away. Also, we should remember that inductor's energy depends on current (contrast it to the fact that capacitor's energy depends on voltage across its plates). In initial conditions there is no current going thorugh the circuit because switch S is open. Hence, correct answer should have plot start at 0. This eliminates answers B, C, D (again). We are left with either E or A answers. You can pick one by again referring to ideal-ness or ideality of the circuit. Indefinite oscillations (answer A) are the only possibility in that case. By the way, the fact that this quite simple question has P+ of just 50% should tell you that it is not that hard to score high on Physics GRE. You can do it!

Sample GRE Q55

Turns out this blog is still up and running :-) so here is a new one: Sample GRE biochem test, question 55 with P+ of 31% (from 2016 edition) [*P+ is percentage of test takers who gave correct answer.]

An amino acid transporter protein is responsible for the transport of a specific amino acid across a membrane. The KI values of several competitive inhibitors of the amino acid transporter are shown above. Based on these data, which of the following is most likely the amino acid transported by this protein?

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Correct answer is E And here is why. First, let's unwrap some of the language. "The inhibitor constant, Ki, is an indication of how potent an inhibitor is; it is the concentration required to produce half maximum inhibition." (here) So, the higher the Ki, more inhibitor concentration is necessary to prevent our transporter from moving amino-acid X through the membrane. It is also mentioned in the question that inhibitors are competitive. That means, to some degree, that inhibitors are similar in structure to that of the amino-acid X. Most similar inhibitor will then have lowest Ki, is est will occupy transporter's "slot" very nicely preventing amino-acid X from binding. In other words, inhibitor with lowest Ki will have most similar structure. Lowest Ki belongs to competitive inhibitor #3 (calledethyl benzene, by the way). It might be useful to notice second-best inhibitor as well, #5 (isoamyl alcohol). Both of these compounds are relatively not charged. Now let's look at structures of all possible answers:

As you can see, only tyrosine (E) carries benzyl group and also has less charge than other compounds. hence the answer is E

Saturday, December 21, 2013

Question #97 of GRE Sample Subject test (MolBio)

Propagation of a regenerative action potential
along an axon can be accelerated by which of the
(A) A decrease in the transmembrane resistance
(B) A decrease in the axoplasmic resistance
(C) Reduced myelin wrapping
(D) Shortened internodal lengths
(E) Narrowing of the axon diameter

Show answer

First of all, what "regenerative" action potential means? Kandel's textbook defines just an action potential (AP) as a regenerative electrical signal.
Now we can unwrap answers and say what will happen in each case.
(A) A decrease in the transmembrane resistance = essentially that is increase in channels number. But not Hodgkin's channels, but so-called "leaky" channels that prevent absolute ion separation by Na/K-ATPase. If leaky channels were more abundant (or more active, you can imagine some drug that does that) it would be more difficult for membrane to stay polarized.

(C) Reduced myelin wrapping = As we all know, myelin facilitates rapid AP propagation by essentially increasing membrane resistance. Thinning of myelin will slow down impulse propagation.

(D) Shortened internodal lengths = shortening of distance between ranvier nodes. This will also slow down signal propagation. There are two ways for AP to travel along axon: as it happens in umyelinated axons and "jumping" from one node to another (saltatory conduction). In latter case, because of myelin isolation (high resistance of membrane), depolarization at one node is enough to trigger channels' opening in nearby nodes (on distance ~200-2000  um). "Inside" nodes signal is propagated in general fashion: channels opens one by one in orderly manner and go through open-closed/inactivate-close/active cycle.
Shortening of internodal length will slow down signal. However, increase in distance between nodes might decrease sensitivity of axon to signal propagation. That is, not every single AP or not every packets of APs will propagate from one node to distal one, because depolarization will essentially be lost on the road due to diffusion of ions.

(E) Narrowing of the axon diameter = Seemingly useful change for rapid ion diffusion, that will not speed up signals. Axon diameter is very important for AP since it is resistance that prevent signal from travelling. Lower resistance can be achieved by increasing diameter or ions concentration. However, I suppose, that ion concentration change will affect channels activity and other physiological properties before change in conductivity. Ions here act like free electrons in semiconductors.

(B) A decrease in the axoplasmic resistance. Now, that is one correct answer. As discussed on (E) resistance of cytoplasm in axon (axoplasm, duh) plays important role in AP propagation.
Now, what is bad with answer (A)? Honestly, it was on my correct-answers-list for that question. if you think about it, axon consists of two parts: conductivity and depolarization mechanisms. Cytoplasm resistance contribute to conductivity, however leaky channels and voltage-dependent channels influence depolarization and AP generation. AP propagation does not really change with membrane resistance, but rather act as filter for action potentials that are too low. Or other wise, permit easily AP generation by keeping cell membrane closer to threshold (you know what that is, right?)

Monday, January 31, 2011

#15 RNA stem-loop stability

104. Which of the following mRNA molecules would form the most stable stem-loop structure?
(A) 5'...GGCUU......UUCGG...3'
(B) 5'...GGCUU......AAGCC...3'
(C) 5'...GGCUU......GGCUU...3'
(D) 5'...GGCUU......CCGAA...3'
(E) 5'...AAGCC......AAGCC...3'

First of all, stem-loops are formed by so-called inverted repeats. Is est, one strand of loop is complement to the other:
5'-...GGGCCC... (small loop here)
If this sequence is linearized now, we will get
5'-...GGGCCC......GGGCCC...-3' strand
Stem is that paired part of looped RNA.
Any mismatches (when, e.g. G is opposed by A) are called bulges.

So, hairpin (synonym of stem-loop) stability is defined by number of mismatches and stability of each base pair in the stem. That one arise from the fact that A have 2 hydrogen bonds with U (or T in DNA) and G have 3 hydrogen bonds with C, so GC pair is much more stable.

From first fact we can eliminate answer choices A, C, D and E, since neither A binds to C, nor U to U or A to U (just compare two nucleotides on the inner edges of uncovered sequences). Lets check answer B to be sure.
Should be stable!

This question is from easiest ones, so should not take a lot of time, which we are short of.

Yes, and I am back, will try to complete this blog during preparation.

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.