NCBS Comprehension Exam, 2012 Dec

7/25/2019 NCBS Comprehension Exam, 2012 Dec

1/8

PART-1 Comprehensive Exam- December 2012Section-1 (Part A) 9AM 1PM

Instructions: Attempt 8 questions in this section. You have to pass 6 (six).Please spend

no more than 30 minutes on each question.

Begin the answer to each question on a fresh page

Q1

A gene that you work on encodes a transcription factor that expresses in developingmuscles of early embryos. Null mutants of this gene are embryonic lethal. While studying

its expression pattern you find it is also expressed in regions of the brain of adult animals.Describe, in some detail, at least two molecular genetic experimental plans that will help

in understanding its function in adult brain. You can choose any animal of your liking.

Ans: a) Tissue and stage specific RNAi knockdown followed by phenotypic

characterization and molecular changes in brain function e.g. altered transcriptionprofiles b) Weak mutants that survive as adults. This would require checking RNA and

protein levels of the TF in the brain first.

Q2

Mammalian erythrocytes (red blood cells) are created via a series of differentiation steps

from hematopoietic stem cells. In one of these steps, the cell extrudes its nucleus to givethe final enucleated (i.e. lacking a nucleus) RBC. Speculate on the molecular players and

pathways that might be involved in enucleation. (Hint: which well known cellular processdoes this enucleation process most closely resemble?) Describe in detail (molecules and

morphology) the series of events that must occur during enucleation.

ANS: Enucleation is related to cytokinesis, since you basically make two large blobs

from a single large blob. This should give clues about the rest of the answer: we requirethe contractile cytoskeleton (actin, myosin) as well as the machinery of vesicle traffic.The following steps must occur: 1. Asymmetry: move nucleus to one side (push with

actin). 2. Move relevant proteins to extrusion site (vesicle traffic). 3. Form a ring in thecortical cytoskeleton through which the nucleus is extruded (actin and regulators). 4.

Generate pressure that starts extrusion (contractile cytoskeleton). 5. Move membrane tothe ring to accommodate growing bulge (vesicle traffic). 6. Seal the gap (vesicle traffic).

Q3

How would you make use of DNA and RNA deep sequencing techniques to measure thefollowing on a genomic scale? (a) Chromosome topology. (b) Impact of topology on gene

expression. Your answer should include a step-wise description of the cellular ororganismal context, any treatments or perturbations, sample extraction and preparation,

and any analyses required in order to convert the sequencing reads you get intoinformative quantitative readouts.

Measuring DNA topology: There are many answers to this. Most students have based thison histone occupancy. In this case, one could measure this using either chromatin

immunoprecipitation (crosslinking - fragmentation - IP) using an anti-histone or a


2/8

histone-mark antibody (depending on what the student has chosen to study).

Other possibilities include crosslinking DNA fragments that are spatially close together,followed by restriction enzyme-mediated digestion and filling+ligation using biotinylated

oligos. This is followed by a pulldown again of biotinylated fragments and sequencing

from both ends.

One could similarly use methods that cross-link DNA in a manner that correlates withtopological features (e.g. ppsoralen).

Data analysis:

Irrespective of the experiment, the deep sequencing experiment will provide aquantitative readout across the genome. This can be obtained by mapping reads to

genome and counting number of reads that map to each base on the genome. By analying

the distributions so obtained one could come up with ways to identify reagions which are

statistically enriched for histone / histone-marks or cross-linkers. In this way one could

also get pairs of genomic regions which are close together spatially.

To analyse effect on gene expression, use appropriate perturbation method, extract RNA

and perform deep sequencing. Analyse data by mapping to genome, follwed bynormalisation and computation of differential expression. At a more advanced level, one

could design experiments that could account for the many pleiotropic effects that

inhibitors of DNA topology-modifying proteins might have.

Q4.

A. What are secondary structural propensities of amino acids?

B. How are these derived?C. Explain three contexts in bioinformatics where such propensity values can be

helpful.

Q5

Specialized DNA Polymerases belonging to the Y-family function to rescue stalledreplication by mediating synthesis across damaged nucleotides. Suggest a general

mechanism utilized by such DNA Polymerases to bypass damaged nucleotides to rescuereplication. Also, what are possible negative outcomes of the action of these enzymes?

ANS: Y-family DNA Polymerases possess specialized plastic active sites that allow themto accommodate damaged nucleotides and add nucleotides opposite these damagednucleotides onto the growing primer strand. A majority of these Polymerases also use

unusual modes of base pairing to select and stabilize the incoming nucleotide oppositethe damaged nucleotide. The major negative outcome of this strategy is that translesion

synthesis can be error-prone and lead to the appearance of mutations.


3/8

Q6

I have been given a small-molecule drug that the manufacturers claim will boost

performance in learning and memory tasks. The drug is administered by intravenousinjection, and is claimed to work in mice and man. I want to test the manufacturers claim

by measuring the performance of mice in a Morris water maze. In this navigation task,

mice are placed in a cylindrical pool of water, and are required to find a platform hiddenjust below the water surface. By making use of visual cues around the pool, mice are ableto find the platform relatively quickly after the first few attempts. I measure the time

taken to successfully complete the task. I run one trial with the vehicle injected (C), andanother with the drug injected (D). I first carry out all my C measurements. The platform

position and visual cues are then shuffled, and I carry out all my D measurements. After

my initial experiments on ! ! !individuals I have the following data (seconds), giving

the initial time taken (C0, D0), as well as the time taken after the mouse has learned over

several attempts (C1, D1).

Mouse 1 2 3 4 5 6 7 8 9 ! !

C0 !!" !#$ %& '( %) "$$ " "#% !#$ !'! $%

C1 '# %) '! "( (# ! !"( !#! )# $) #!

D0 '# ()& " %! !!% (&$ "'# !$) !'% !)# !&)

D1 !+ !#" ++ "( ($ !!! $$ )% #+ +% #(

C1/C0 &,#% &,+! &,+# &,#( &,() &,(% &,+& &,'$ &,+# &,'" &,!!

D1/D0 &,") &,(+ &,(" &,"% &,(! &,(+ &,(! &,'' &,") &,(# &,&%

D0/C0 &,#% ",+' ",'' !,'( !,(( !,!! !,"# &,$" !,&$ !,#! &,$#

D1/C1 &,") !,') !,") !,&& !,&% !,&$ &,+( &,+) &,#) &,)& &,#!

The t-test statistic for two equal-sized samples of equal variance is:

! !!!!!!

!!

!!!

!

!

!!!!!

where !gives sample means, !gives sample standard deviations, and !is the sample

size. Let us assume that a t-statistic value of 2.0 or greater is judged to be significant (thiswould be the 95% interval on a two-sided t-test with many degrees of freedom).

A.

What is the function !!!!that incorporates the effect of sample size?ANS: f(N)=!N

B.

All the measurements done here have large standard deviations. List 3 potential

causes of this variation.

ANS:

1. The mice could have different intrinsic abilities to navigate mazes.

2. The maze itself could set some level of difficulty, though the degree of difficulty is


4/8

mouse-specific since different mice might use different strategies.

3. Each run contains some random component, such as the initial position andorientation of the mouse.

C. Assuming that the drug acts on learning and memory pathways alone, but does

not otherwise affect spatial abilities, which pairs of numbers might I expect to be not

significantly different? Is this borne out by the data?ANS: The initial performance of the mice should be unaffected by the drug. So D0 and

C0 should be similar, and D0/C0 should be approximately 1. For the D0 to C0

comparison, we find t=0.96. For the D0/C0 to 1 comparison we find t=1.43. As expected,

neither of these is significant. Remember that the platform and cues are shuffled,producing a large variation around the mean of 1.

D. The manufacturers claim that the drug should reduce the final time to solve the

maze by up to 30%. The values of D1 and C1 seem consistent with this claim, showing a

drop of 15%, but the result is not very convincing. If the manufacturers are correct, andassuming that the measured trends persist, how many animals would you need to test in

order to establish this claim in a statistically significant manner using only a D1 to C1

comparison?

ANS: The value of the denominator of t-test statistic for the D1 to C1 comparison is 60.This is a measure of the amount of variation you expect by chance. Starting with a value

C1=79, a 15% drop gives C1-D1=12, while a 30% drop gives C1-D1=24. For the former,

significance requires!N"12/60>2 or N>100. Even assuming the manufacturers claim,significance requires!N"24/60>2, or N>25. Similar arguments can be made for usingthe ratio D1/C1

E. Suppose no further experiments are possible. Looking at the table above, are there any

other comparisons you could make that would establish that the drug is a success? Whydoes the nave comparison of D1 to C1 not show any result, whereas other comparisons

might?

ANS: Comparing D1/D0 to C1/C0, we obtain a value t=3.97, which is highly significant.

Assuming equal values of C0 and D0, we would expect D1/C1=0.34/0.52=0.65, an effectabout the size the manufacturer claims. The reason this comparison works is that the D1

to C1 comparison is confounded by large variations in the intrinsic abilities of mice,

which leads to the large standard deviations. This variation is eliminated if we first take

the ratio D1/C1, thereby normalizing out the ability of individual mice. This is oneversion of a paired-test experimental design.

Q7

From the gel below, draw the secondary structure of the RNA sequence.CACACGAUGACUGAACUACCGCAUGAAAGUGCGGAUCACAGUCGUCAAAAAAAA.

Lane 1: cleavage under denaturing conditions with RNAse A.Lane2: Cleavage under denaturing conditions using RNAse T1.

Lanes 3-5: Limited cleavage under folded conditions Lane 3: with RNAse ALane 4: RNAse VI

Lane 5: RNAse T1.


5/8

Q8The Lower, Middle and North Andaman Islands form a 3-island chain in the Bay of

Bengal. I am studying the movement patterns of a small bird species, which can flybetween neighboring islands. The year is split between a migrating season and a breeding

season. In a given year, I find that 1/2 the birds on any given island remain where they

are, while the other 1/2 tend to migrate. From the North and South islands all migrating

birds to go the Middle Island. From the Middle Island, the migrating birds split evenlybetween North and South Islands. During the subsequent breeding season, I find that

births and deaths are evenly balanced, so the population of offspring each year is the

same as the population of adults they replace.

A. Write down the matrix !relating the number of birds from year-to-year on each

island:!!"#$!

!!"##$%

!!"#$! !!!

! !

!!"#$!

!!"##$%

!!"#$! !

.

ANS: ! !

!!! !!! !

!!! !!! !!!

! !!! !!!

B. Find the eigen values of this matrix.

ANS: We have ! ! !"# !!

!" !!!!!

!!!!!!!

!!!so the eigenvalues are

! ! !! !!!! !.

C. Find the eigenvector corresponding to the largest eigenvalue.

ANS: For !! ! !, we have !! !!

!

!

.


6/8

D. Suppose I start with a population of 4000 birds on each of the three islands. What

values will the populations approach after a period of several years?

ANS: After several years the distribution will resemble the eigenvector with the largest

eigenvalue. The largest eigenvalue is 1 because the total population is conserved. We

therefore will have:

!!"#$!

!!"##$%

!!"#$! !!!

!

!"""

!"""

!"""

Q9

A complete answer to this question should contain the following elements:1. Draw an appropriately shaped EPSP fast rise time and slow decay, positive-

going as the connections are excitatory

2.

Show monosynaptic EPSPs reliably following the 40Hz train but the disynapticEPSPs do not follow the spike train in X (or A).

3. Partial credit for describing greater latency for disynaptic compared to

monosynaptic.4. When Calcium is replaced with Magnesium gradually, single APs in X cause

smaller and smaller EPSPs in Y if there is a monosynaptic connection. If there is adisynaptic connection, there will be failure of EPSPs in Y whenever the X to Z EPSP

amplitude falls below threshold.5. Partial credit for mentioning that calcium is important for vesicle release.

6. Additional test for testing monosynapticity: Synaptic latency variability (morethan its absolute value) is one test; Broaden the action potential in X pharmacologically

and see if EPSP amplitude in Y increases. If it does, then the connection is monosynaptic.If it doesnt, more synapses are involved. Also, in the presence of TTX, uncaging calcium

inside neuron X should cause EPSPs in Y if Y is monosynaptically connected to X.


7/8

Q10

Following antigen stimulation, mature B-cells undergo a transition to form antibody

producing plasma cells. This developmental transition involves a huge increase in thesecretory capacity of the cells to produce and secrete antibody molecules. Describe

signalling strategies that might be used by B-cells to detect and adapt to the increased

secretory load.

The answer should have the following elements:

1. The transition from B-cells to plasma cells should is associated with the functional

change namely the ability to produce and secrete large quantities of antibody.

2. This requires that the cell adjust its biology to be able to undertake enhanced levelsof protein synthesis.

3. To do this the major cell biological changes required are an increase in the cells

ability to increase the volume of its ER and also upregulate its secretory transport to dealwith increased secretory load.

4. They should then articulate a signaling mechanism where by following activation of

peripheral B-cell information is communicated to the cellular machinery to build roughER and readjust secretory transport.

5. In reality this involves the unfolded protein response and its multiple elements;

however if they don't know the specific molecular details of this it is OK. To pass thelogic of the signaling pathway required must come through in the answer as articulated

above.

Q11

Morphogens act over many cell-diameters to specify cell-fate in a concentrationdependent mechanism. In the Drosophila wing disc Hh is expressed in the posterior

compartment in En expressing cells. Yet, Dpp expression is activated, in a narrow row ofcells, by Hh signaling, only at the anterior-posterior (A-P) border on the anterior side of

the border. Write down a mechanism by which this narrow expression of Dpp is achievedby Hh and how you would use the tools of genetics available in flies to test the

mechanism proposed.

A. Hh is not active in En expressing cells because the activity of Ci, its target is regulatedto prevent activation of the pathway. A high concentration of Hh (close to the source at

the A-P border) turns on Dpp at the border. This model can be tested by ectopicexpression of Hh, En and Hh, and En.

Q12

(a) The isoelectric point of a substance (pI) is that pH at which the substance carries nonet charge. An amino acid like ALA has two ionizable groups (the amino group and the


8/8

acid group). Derive the value of pI in terms of the equilibrium constants of the twoionization reactions: K_1 and K_2.

(b) If a protein has a lot more ASP and GLU residues (their side chains contain -COOH)

than ARG, LYS or HIS residues (their side chains contain -NH_2 or -NH- groups) do you

expect its pI to be above or below a pH of 7? Explain your answer.

(c) A structured protein has a certain number of residues with ionizable side chains. The

ionization equilibrium constants (K_1, K_2 ... K_n) for these side chains have beenmeasured for the individual amino acids in solution. Do you expect the pI calculation for

the protein to be as simple as in (a)? Explain your answer.

ANS(a) Derive the relationship between K_1, K_2 and pI and not just state it.

(b) below pH = 7.0. more acidic groups means the pI will get closer and closer

to the pKa of the acidic groups.

(c) no, structure will change individual pKas and you can't now just addup the solution pKas.

NCBS Comprehension Exam, 2012 Dec

Documents

Transcript of NCBS Comprehension Exam, 2012 Dec