final exam final 2013 for physical biology for the cell

8/10/2019 Final Exam Final 2013 for Physical Biology for the Cell

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Final Exam

Physical Biology of the Cell

Student Name: ..........................................................

Student ID #: ........................................

January 15, 2013


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PBC Course Final Exam 1

Physical Biology in Flow

Problem 1.1 (5 pts): Many cells in the body exist in an environment of flow. For example, the

endothelial cells in the 2D monolayer that lines the blood vessels exist in contact with flowingblood. Do cells that exist in 3D tissues exist in flow? Specifically, do cells in (a) cartilage or (b)

bone exist in flow? Answer yes or no to both, and write 2-3 sentences, and provide a sketch to

justify your answer.

Problem 1.2 (5 pts): Most cells have a specialized organelle that functions to sense flow. Name

this organelle, and provide 2-3 sentences and a sketch to describe this organelle.


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Adhesion

Problem 2.1 (5 pts): If an adhesive interaction (cell-matrix) between an adhesion receptor and a

matrix protein can withstand force F, it is often found that a force of N F is required to pull the cellaway from the matrix, where N is a number between, say, 10 and 50. Explain with a few sentences

and a diagram how this is possible.

Rate Expressions

Problem 3.1 (15 pts): Consider a microbial bioreactor in a continuous stirred tank mode. As

derived in class, the concentration of cells Ccis given as a function of dilution D by equation:

Cc= 1

YSc

CS0

D KS

max D

Where:

YSc is a yield coefficient

CS0 is the concentration of substrate in the in-flow

CSis the concentration (variable) in the well-mixed reactor and out-flow

Dis the dilution = v0V

, where v0is the volumetric in-flow rate and V is the bioreactor volume

KSand maxare the Monod constants

From this, derive an equation to give the maximal rate of cell production by the bioreactor, where

the rate of cell production is given by DCc. Show the derivation. (use other side of this page)


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Transcription Factor Structure

Problem 4.1 (3 pts): Which groove of the DNA do Transcription Factors most often bind to and

why?

Problem 4.2 (2 pts): Which type of protein secondary structure is most often used by transcription

factors to bind to DNA and why?

Problem 4.3 (3 pts): Name at least 3 different transcription factor families.


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Problem 4.4 (2 pts): Identify one specific and one non-specific TF - DNA interaction in the

following figure:


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Measurement Techniques

Problem 5.1 (2 pts): Indicate below which methods can be used to study protein - DNA interac-

tions:2 EMSA

2 SELEX

2 Phage Display

2 Yeast One-Hybrid (Y1H)

2 Yeast Two-Hybrid

2 PCA

2 PBMs

Problem 5.2 (3 pts): Explain the purpose of the control performed in lane 5 of the EMSA gel

below (the lane containing the Antibody). How is this type of control generally called?

, ,


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Microfluidics

Problem 6.1 (5 pts): What are the main advantages of microfluidics over current fluid handling

approaches used in biology? Explain why these advantages are important.

Problem 6.2 (5 pts): Explain why mixing is often challenging on a microfluidic device. Name 2

commonly used microfluidic mixers.


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Thermodynamics

Problem 7.1 (5 pts):Write down the ODE describing the change in [AB] as a function of time for

the following system:[A] + [B]

k1*)k2

[AB] + [C] k3*)k4

[ABC]

Problem 7.2 (3 pts): Solve the following ODE (show all the steps): d[AB]

dt = koff[AB]


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Problem 7.3 (10 pts): You are running an ELISA in the lab and you want to know how long you

need to incubate the Antigen with the Antibody. You assume that the observed association kinetics

follows the following equation:

[AB] = [A][B]0[A] + KD

1 e(kon[A]+koff)t

Your antigen is at a concentration of 1nM and you know the association and dissociation rates for

your particular antibody-antigen pair, which are: kon= 1x105M1s1 and koff= 3x10

4s1

How long will it take to reach 90% of the equilibrium concentration of [AB]?


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Transcription Factor Specificity

You obtained the following Matrix from a SELEX experiment for a particular Transcription Factor:

!" # $ % & ' ( ) * +

, !! " "# $% #" "& "! $ %'

- $! ($ ) $ )* ( ) (& &

. ($ )( ( ** ! ( $ )) "

/ " *& ( ! ) ( ( *% !

Problem 8.1 (2 pts): What is the consensus sequence for this Transcription Factor?

problem 8.2 (15 pts): Plot position 4 as a sequence logo with bits as the unit for the y-axis. you

can assume a 50% GC content. Show all intermediate steps!


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Statistical Mechanics

You have the following standard reaction between a receptor (R) and ligand (L):

[R] + [L] kon*)koff

[RL]

Problem 9.1 (15 pts): Setup the statistical mechanics model for this system.

Macrostates of the System:

Energy of the system:

Enumerate Microstates:

Statistical weight of each microstate:

Probability of theunboundstate:


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Promoter Architecture

Problem 10.1 (5 pts): In the following figure, what logic functions do the input-output functions

M1 and M2 most closely resemble? Write down the truth tables for these two logic functions.l l ll l l I I l

l l l l l

l I

l

l l

l

I l

l l I ll l

l l l l

l ll l

o

WT

M1

M2


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Extra Credit Questions

Extra Credit 1 (5 pts): Why is it important to measure individual cells rather than take an average

measurement of a large population of cells?

Extra Credit 2 (5 pts): Describe in detail how the NAPPA method works (include a schematic).


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Extra Credit 3 (5 pts):Explain in detail what is meant by base interdependence (or non-independence).

Extra Credit 4 (5 pts): Explain why transcriptional regulatory networks play an important role in

evolution. Explain why mutations in cis elements are more suitable for generating small adaptive

changes than mutations in trans elements.


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PBC Course Formula Sheet 1

Thermodynamics

[A] + [B] kon*)koff

[AB]

KD =[A][B]

[AB] =

koff

kon= K1A

YB = [A]

KD+ [A]

YB = [A]BmaxKD+ [A]

t1/2 = ln(2)

koff

[AB] = [AB]0ekofft

[AB] = [A][B]0[A] + KD

1 e(kon[A]+koff)t

G

=RT ln

(K

eq)

Keq = eGRT


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R is the gas constant:

R= 8.31451JK1

mol1

RT =kbT NA 2.4790kJmol1

0.592kcalmol1

Transcription Factor Specificity

Genome Lengths:

E. coli = 5 Mbp

S. cerevisiae = 12 Mbp

D. melanogaster = 120 Mbp

H. sapiens = 2.9 Gbp

P(s) =lY

i=1

P(si)

I(wn) =log 1

P(wn) = log(P(wn))

H(x) = nX

i=1

P(xi)logbP(xi)

H(i) = Xb

f(b, i)log2f(b, i)

Rseq(i) = 2 H(i)


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height(b, i) =f(b, i)Rseq(i)

Iseq(i) =Xb

f(b, i)log2

f(b, i)

P(b)

P(s) = 2

lP

i=1

I(si)

Gs=Xi

Gbsi ,i

P(s) = 1

eGs/RT + 1

P(s) = 1

eEs + 1

Es= Gs/RT

= ln

[TF]

KD,ref

Pocc = 1windowsY

s=1

(1 Ps)


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Pocc = 1windows

Ys=1

1 1

1 +eGs/RT

Statistical Mechanics

Sampling with Replacement:

nk

Sampling without replacement:

n(n 1)(n 2) . . . (n n+ 1) = n!

Permutations

n(n 1)(n 2) . . . (n k+ 1) = n!

(n k)!

P(n, k

) =

n!

(n

k)!

Combinations:

C(n, k) =

n

k

=

n!

k!(n k)!

final exam final 2013 for physical biology for the cell

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