MCB137L/237L: Physical Biology of the CellSpring 2020

Homework 6: Biological Dynamics - Solution

Yang Joon Kim, Jiaxi (Jake) Zhao, Hernan G. Garcia

1 Solving Ligand-Receptor Multiple Ways

(a) In class, and then in the homework, we solved for the dynamics of mRNA production anddegradation using the dynamics protocol. In this problem, we are going to use that analysisas a jumping off point for thinking about ligand-receptor binding problems. Imagine asituation in which we have a receptor fixed at some point in space as shown in the top rightpanel of Figure 1. Write a rate equation for the concentration of ligand-receptor pairs interms of the concentration of ligands and receptors. Now, assume steady state and, giventhat equation, derive an expression for the dissociation constant

Kd =[L][R]

[LR](1)

in terms of the on and off rates. Make sure you explain the dimensions of your on and offrates and hence, the dimensions of Kd.

(b) A second route to considering ligand-receptor interactions is to think of binding proba-bilistically with the probability that the receptor is occupied given by

pbound =[LR]

[R] + [LR]. (2)

Given the definition of the dissociation constant introduced in the previous part of theproblem, find a simple expression for pbound([L]) that is only a function of the concen-tration of ligand. (NOTE: for now, we are ignoring the subtlety that the amount oftotal ligand and free ligand are not actually the same, though in the case consideredhere with a single receptor we have somewhat finessed that point.) Make a plot ofpbound([L]) as a function of [L] and comment on where Kd belongs on the axes. Lateron in the course, we will solve this problem in yet another way, by using statistical mechanics.

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(a) The rate equation for the concentration of ligand–receptor pairs is given by

d[LR]

dt= kon[L][R]− koff[LR]. (3)

Before proceeding to the steady state solution, let us explore the dimensionality of thedifferent terms. All concentrations have units of (1/volume). Therefore, the units of theleft-hand side are 1/(volume × time). The two terms on the right-hand side should havematching units. From this, we can deduce the dimensions of kon and koff, namely,

dimension of kon =volume−1time−1

(volume−1)2=

volume

time=

1

time× concentration, (4)

dimension of koff =volume−1time−1

volume−1 =1

time. (5)

At steady state, the time derivative of [LR] must vanish. Equating the right-hand side ofEq. 3 to zero, we obtain

koff[LR] = kon[L][R],

koff

kon

=[L][R]

[LR]. (6)

From the result above, we identify the dissociation constant as the ratio of the off- andon-rates, i.e. Kd = koff/kon, which has units of 1/volume or concentration.

(b) Using the expression for Kd obtained above, we can use Eq. 6 to express the concen-tration of the ligand-receptor complex as [LR] = [L][R]/Kd. Substituting this form into thedefinition of pbound provided in the problem, we find

pbound =[LR]

[R] + [LR]

=[L][R]/Kd

[R] + [L][R]/Kd

=[L]/Kd

1 + [L]/Kd

. (7)

As can be seen from Figure 1, Kd is the ligand concentration for which half of the receptorsare bound. This result also follows directly by evaluating Eq. 7 at [L] = Kd.

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Figure 1: Probability of the receptor being bound by as a function of the ligand concentration(scaled by Kd).

2 Cytoskeletal polymerization

In this problem, we explore how the same master equation approach presented in class todescribe the distribution of mRNA molecules can be used to understand how biologicalpolymers such as actin or tubulin are assembled. Specifically, we think of a polymer that canincorporate or lose monomers at one end, as shown in Figure 4. The rate of incorporationis r and the rate with which monomers are lost is γ. Note that, unlike the mRNA case, thenumber of monomers that leave the filament in a time interval ∆t is given by γ∆t and isindependent of the filament length.

(a) The master equation of the probability of a filament having length n at time t, p(n, t),is given by

p(n, t+ ∆t) = p(n, t) + r∆tp(n− 1, t)− r∆tp(n, t)− γ∆tp(n, t) + γ∆tp(n+ 1, t). (8)

Justify why this is the master equation describing the length of a filament by carefullyexplaining the significance of each term in the master equation.

(b) Using Python, solve the master equation assuming r = 1/s and γ = 1.1/s and plotthe resulting distribution for different time points. Assume that at time t = 0 thereare no filaments by setting p(n = 0, t = 0) = 1 and p(n > 0, t = 0) = 0. Be carefulabout the boundary conditions! In particular, remember that you cannot have filamentswith a negative number of monomers. Further, you cannot have Python calculate theprobability of an arbitrarily long polymer. Hence, you will have to choose a maximum

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length of the polymer for your calculation. In doing so, make sure that the probability ofhaving a filament of length n near your chosen maximum polymer length is very close to zero.

(c) Plot the length distribution in steady state. To make sure you’re in steady state,demonstrate that the distribution does not change as you extend the total simulation time.

(d) In class we solved the mRNA distribution of the constitutive promoter in steady state.Follow the same approach to now solve the steady-state distribution of filament lengths. Tomake this possible, explore the master equation for the specific cases of n = 0, n = 1, n = 2

in steady state and show that the distribution can be written as p(n) = p(0)(rγ

)n, which is

a geometric distribution.

(e) If you do the math, you can infer that p(0) = 1−r/γ (you don’t need to show this). Nowthat you have the full distribution, plot it together with your simulation results to show thatboth approaches are consistent. If you want to compare your results to real data (and figureout more realistic values of r/γ, you can look at Figure 15.23 of PBoC2. Note that an actinmonomer is about 2.4 nm in length.

(a) The probability of finding a filament of length n at time t + ∆t (this is, the fraction oflength n filaments / total number of filaments) is given by p(n, t + ∆t). This probability isequal to the probability at the previous time point p(n, t) adjusted by processes that occurduring ∆t that change the number of filaments with n monomers.

• r∆tp(n−1, t) represents filaments that in the previous time point had n−1 monomersand gained a monomer during ∆t, thus adding to p(n, t)

• γ∆tp(n+1, t) represents filaments that in the previous time point had n+1 monomersand lost a monomer during ∆t, thus adding to p(n, t)

• r∆tp(n, t) represents filaments with n monomers in the previous time point that gaineda monomer during ∆t, thus decreasing p(n, t)

• γ∆tp(n, t) represents filaments with n monomers in the previous time point that losta monomer during ∆t, thus decreasing p(n, t)

(b)(c) The plot of the the filament length distribution at different time points is shownin Figure 2. Since the distribution does not change significantly for three consequent timepoints, we can assume that it has reached the steady state. The Python code used to generatethis plot is posted on the course website.

(d) We begin by evaluating the master equation at n = 0.

p(0, t+ ∆t) = p(0, t) + r∆tp(0− 1, t)− r∆tp(0, t)− γ∆tp(0, t) + γ∆tp(0 + 1, t) (9)

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Figure 2: FInding Steady State Distribution

We now note that at steady state, by definition p(0, t + ∆t) = p(0, t). Furthermore, p(0 −1, t) = 0 (there are no −1 monomer length filaments) and γ∆tp(0, t) = 0 (because thereis no degradation from a nonexistent polymer). Additionally, because at steady state thefunction does not change with time, we keep n as the only variable. With all of this in mind,the expression becomes:

0 = −rp(0) + γp(1) (10)

and we can solve for p(1)

p(1) =r

γp(0) (11)

Let’s continue with n = 1. Invoking the same arguments as before, we get:

0 = rp(0)− rp(1)− γp(1) + γp(2) (12)

By invoking our results that γp(1) = rp(0), obtained above, we can reduce the expressionto:

0 = −rp(1) + γp(2) (13)

p(2) =r

γp(1) =

(r

γ

)2

p(0) (14)

For p(2), the master equation at steady state leads to:

0 = −rp(2) + γp(3) (15)

p(3) =r

γp(2) =

(r

γ

)3

p(0) (16)

Thus, the probability distribution associated with the master equation is p(n) = p(0)(rγ

)n.

(e) This analytical solutions is plotted along with the numerical solution (part c) in Figure3.

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Figure 3: Filament length distribution in steady state

r

n filaments

γ

Figure 4: Simple model of filament polymerization. Monomers can only be added at oneend of the filament.

3 Protein-mRNA ratio

In this problem we go beyond the calculation on mRNA production we did in class, andthink about how transcription and translation shape the protein-to-mRNA ratio inside cells.

(a) In class, we described the temporal evolution of the number of mRNA molecules usingthe equation

m(t+ ∆t) = m(t) + rm∆t− γmm(t)∆t. (17)

Here, m(t) is the number of mRNA at time t, rm is the rate of mRNA production, andγm is the mRNA decay rate. Write the corresponding equation for the number of proteinmolecules given a rate of protein production per mRNA of rp and a protein decay rateγp. Make sure to incorporate the fact that the number of mRNA molecules present willdetermine how many proteins are produced in a time interval ∆t.

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(b) Calculate the ratio of protein to mRNA in steady state, pSS/mSS and show that it isgiven by rp/γp. Find typical values for the various model parameters in E. coli and estimatethe ratio of proteins to mRNA molecules. How do your numbers compare to those measuredin Figure 3C of Taniguchi et al., which is provided on the course website?

We can also obtain this protein-mRNA ratio in the context of fruit flies.

(c) Using flies with different dosages of Bicoid-GFP, Petkova et al. measured the relationbetween the number of bicoid mRNA molecules deposited by the mother, and the resultingnumber of Bicoid proteins. Read their paper (available on the course website) and write ashort paragraph about how their Figure 3 is generated.

(d) The paper by Drocco et al. (available on the course website) uses a photoactivatablefluorescent protein to measure the lifetime of the Bicoid protein. Read the paper (availableon the course website) and explain the technique in one paragraph. You might find it usefulto draw a schematic plot such as shown in Figure 1f of the paper.

(e) Assuming that Bicoid-GFP is in steady state, use what you learned about rp and γp forthe Bicoid protein in order to calculate its protein-mRNA ratio rp/γp.

(a) The temporal evolution of the number of protein molecules is described by the equation

p(t+ ∆t) = p(t) + rpm(t)∆t− γpp(t)∆t .

Here, p(t) is the number of protein molecules at time t, rp is the rate of protein productionper mRNA molecule, and γp is the protein degradation rate. Note that the amount of proteinproduced in time ∆t depends on the number of mRNA molecules present, while the amountof protein molecules that are degraded in time ∆t depends on the amount of proteins present.

We can manipulate the discrete version of the equation describing the temporal evolutionof the number of proteins to arrive at a continuous time version as follows:

p(t+ ∆t) = p(t) + rpm(t)∆t− γpp(t)∆t=⇒ p(t+ ∆t)− p(t) = rpm(t)∆t− γpp(t)∆t=⇒ p(t+∆t)−p(t)

∆t= rpm(t)− γpp(t)

=⇒ dpdt

= lim∆t→0

p(t+ ∆t)− p(t)∆t

= rpm(t)− γpp(t) ,

(b) In steady state, p(t) = pss is constant and dpdt

= 0 such that

dp

dt= rpmss − γppss = 0

pssmss

=rpγp.

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Note that we have used the fact that m(t) = mss must also be constant for the number ofproteins to be constant.

The average translation initiation rate in E. coli is 5 min−1 mRNA−1 (BNID 112001)which corresponds to an average production rate of 5 proteins/min/mRNA. To find theaverage protein degradation rate for E.coli, we can use the fact that 1% of the bulk proteinpool is degraded in an hour (BNID 109924). Because the rate of protein degradation issmall, we can infer that γp ≈ 1/100 = .01 h−1 from the fact that 1 in 100 proteins in thebulk protein pool is degraded in an hour.

Note: For a more general approach with γp of arbitrary size, we convert this to a degra-dation rate as follows: Assuming all of the bulk protein pool decays via simple first-orderdegradation and that no protein is produced, the amount of protein at time t is given by

p(t) = p0e−γpt .

We can mathematically express the fact that 1% of the bulk protein pool is degraded in 1hour as

p(t = 1 h) = .99× p0 .

Plugging into our expression for p(t) in the case where there is no production and there arep0 proteins at t = 0, we find

p(t = 1 h) = .99× p0 = p0e−γp·1 h

=⇒ .99 = e−γp·1 h

=⇒ ln (.99) = −γp · 1 h

=⇒ γp = − ln (.99)

1 h≈ .01005 hour−1 ≈ .01 hour−1

Note that both approaches gave us the same value for γp in this case.

The average protein half-life in E. coli is very long such that the effect of protein degra-dation is dominated by the dilution rate resulting from cell division. When a cell divides theconcentrations of its proteins are cut in half. The effect of protein degradation is thereforenegligible when compared to dilution. Assuming a doubling time of ∼ 30 minutes (BNID109057), we can find γp:

1

2= e−γp·30 min =⇒ ln

(1

2

)= −γp · 30 min =⇒ γp =

1

30 minln 2 ≈ .02 min−1 = 1.4 hour−1 ,

where we have used the same approach as described in detail in the box above. Pluggingthese values for rp and γp into our expression for the ratio of proteins to mRNA moleculesgives:

pss/mss = rp/γp

=5 proteins ·min−1mRNA−1

.02 min−1

= 250 proteins/mRNA .

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Note that if we had used γp = .01 hour−1 ≈ 2× 10−4 min−1, we would have found

pss/mss = rp/γp

=5 proteins ·min−1mRNA−1

2× 10−4 min−1

= 25000 proteins/mRNA ,

which is 100-fold larger than our estimate of the ratio using the dilution rate.Figure 3C from Taniguchi et al. shows a scatter plot of mRNA copy number against

protein copy number. Notice that each dot is a gene and that the measurements were donein bulk, meaning they represent the average in a population of cells. There is a clear trendin the data that is consistent with pss/mss = rp/γp ≈ 103, which is about 4 times higherthan our estimate.

(c) To get the relationship between protein and mRNA level, Petkova, et al. have engineeredtransgenic flies that express Bcd-GFP at different levels. Then, they quantified differencesin protein concentration by comparing average nuclear Bcd-GFP intensities across AP po-sitions between a given transgenic test line and a reference line. Meanwhile, the maternallydeposited bicoid-gfp mRNA expression is measured by qPCR.

(d) Bcd-Dronpa has bistable bright and dark states, conversions between which are rapidand inducible in living and fixed material. As shown in Figure 5(A), Bcd-Dronpa can beconverted between dark and bright states by illuminating the embryo with either 496nm lightor 405nm light. Using a method that employs repeated photoswitching of the Bcd-Dronpafusion (Figure 5(C)), they can measure and modify the lifetime of Dronpa-Bcd in livingDrosophila embryos.

(e) We know that the translation rate of Bcd protein is about 2 proteins/mRNA·min(Petkova et al.) and the degradation rate is approximately 0.028 protein/min (Drocco etal.). Based on the equation we derived, we can calculate the protein-mRNA ratio of Bcdprotein

pss/mss = rp/γp

=2 proteins ·min−1mRNA−1

0.028 min−1

≈ 70 proteins/mRNA .

This is similar to the measured ratio from Petkova et al. (pss/mss ≈ 100).

4 Phase diagram for the logistic equation

In Homework 3, we solved the logistic equation numerically. This equation can describe thesaturation of a bacterial culture by accounting for a limited food supply

dN

dt= rN

(1− N

K

), (18)

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time

totalBicoid-Dronpa

brightBicoid-Dronpa

darkBicoid-Dronpa

time

time

turnDronpa

off

turnDronpaon

inactivate(496 nm)

activate(405 nm)

(A) (C)

(B)

Figure 5: Description of Dronpa-Bcd construct and degradation measurement. (A) Confo-cal images of in vivo Dronpa-Bcd-expressing embryos with Dronpa-Bcd predominantly inthe bright or dark state. (B) Confocal image of a fixed Dronpa-Bcd embryo with Dronpapredominantly converted to the dark state except an inscription produced by a targeted405-nm reactivation pulse. (C) Schematic of degradation measurement, beginning with aDronpa-Bcd population converted to the dark state

where N is the number of cells, r is the growth rate, and is the carrying capacity or maximumpopulation size. Note that this equation can also be written as

dN

dt= rN − rN2

K. (19)

Here, we can identify a “cell production” term and “cell destruction” term.

Like we did for the case of the constitutive promoter, plot a phase diagram where bothproduction and destruction terms are plotted (either by hand or in Python). Use this plotto graphically show that there are two stable points at which the production and destructionare balanced out.

Here we have two rates. Both “cell production” and “cell destruction” rates are functionsof the number of cells, N . The production term f(N) = rN should be a straight line witha slope of r and the destruction term, g(N) = rN2

K, should be a quadratic function of N .

Note that when N = 0 and N = K, the production and destruction terms are equal whichindicates that these are steady state solutions of the logistic growth equation.

Assuming r = 10 cells/min and K = 1000, we can plot production and destruction ratesas shown in Figure 4. We can see that there are two intersections between production anddestruction term which indicates that there are two stable points. The Python code used togenerate the figure is posted on the course website.

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5 Random walks and biological polymers

Physicists know how to solve just a handful of problems. Fortunately, many dissimilarphenomena in physics and biology alike can be mapped onto such problems for which weknow a solution. Here, we explore the mathematical connection between diffusion and thespatial arrangement of polymers such as DNA, actin, and microtubules.

(a) Read the introduction to Section 8.2 of PBoC (“Random Walk Models of MacromoleculesView Them as Rigid Segments Connected by Hinges”) to learn more about how polymerscan be thought of as chains of connected rigid segments. Pay close attention to Figures 8.1and 8.2. Here, the Kuhn length a is defined as the length of the segments. Look up theKuhn length for DNA, actin, and microtubules in order to get a feeling for these polymers.Note that you might find reference to the persistence length ξp = a/2 instead of the Kuhnlength.

(b) Now, think of a polymer chain of N segments in 1D. As shown in Figure 8.3 of PBoCeach segment can either be pointing to the right of to the left. Given nR and nL segmentspointing to the right and left, respectively, the position of the end of the chain is given byL = (nR − nL) a. Map this problem onto the diffusion problem we solved in class where wecalculated the 〈x〉 and 〈x2〉 of a random walker that start at the origin shown in Figure 3.To make this possible, note that each segment can be randomly pointing to the left or right.In particular, calculate 〈nR− nL〉 and 〈(nR− nL)2〉 and show that the size of the polymer isgiven by

size ≈√〈L2〉 = a

√N (20)

by repeating graphical the derivation we did in class.

(c) Think of the size of the polymer you derived in (b) as the linear dimension of the blob thepolymer will make on a surface such as shown in the figures below. Use the derived formula

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to estimate the genome length (in µm and bp) of the bacteriophage T2 shown in Figure 1.16of PBoC and of the E. coli in Figure 8.5 of PBoC. How well did your estimate do?

All relevant figures from PBoC can also be found in Figures 4 and 5 below.

(a) DNA: ξp ≈ 50 nm =⇒ a = 100 nm (BNID 103112)actin: ξp ≈ 10µm =⇒ a = 20µm (BNID 106830)microtubules: ξp ≈ 1.4 mm =⇒ a = 2.8 mm (BNID 105534)

(b) Here we use the same way of thinking we employed to derive a formula for the diffusionof particles in 1D. By following the same logic, we hope to be able to find a similar law forthe size of a polymer chain. We start by imagining the growth of a polymer. Each time asegment is added it “flips a coin” to decide whether it grows to the left or to the right.If polymer growth can be captured by the same law as diffusion, i.e. if the growing polymerbehaves like a random walk, then we expect:

〈nR − nL〉 = 0

〈(nR − nL)2〉 = N

We can check all possible combinations of “coin flips” (N), for cases where we have N = 1, 2,and 3 to confirm that the expected expressions for 〈nR−nL〉 and 〈(nR−nL)2〉 are correct. Allpossible individual paths are enumerated for N = 1, 2, 3 in Tables 1, 2, and 3, respectively:

L = a(nR − nL) nR − nL nR nL paths # of pathsprobabilityper path

totalprobability

a 1 1 0 right 1 12

12

−a -1 0 1 left 1 12

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Table 1: Probabilities for all possible values of ∆n given N = nR + nL = 1.

L = a(nR − nL) nR − nL nR nL paths # of pathsprobabilityper path

totalprobability

2a 2 2 0 right-right 1(

12

)2= 1

414

0 0 1 1right-leftleft-right

2(

12

)2= 1

412

-2a -2 0 2 left-left 1(

12

)2= 1

414

Table 2: Probabilities for all possible values of ∆n given N = nR + nL = 2.

A summary of this information is given in the first three rows of Table 4, from which we candeduce a general formulas for 〈nR − nL〉 and 〈(nR − nL)2〉 for a polymer with N segments(row 4 of Table 4).

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L = a(nR − nL) (nR − nL) nR nL paths # of pathsprobabilityper path

totalprobability

3a 3 3 0 right-right-right 1(

12

)3= 1

818

a 1 2 1right-right-leftright-left-rightleft-right-right

3(

12

)3= 1

838

−a -1 1 2right-left-leftleft-right-leftleft-left-right

3(

12

)3= 1

838

−3a -3 0 3 left-left-left 1(

12

)3= 1

818

Table 3: Probabilities for all possible values of ∆n given N = nR + nL = 3.

N 〈nR − nL〉 =N∑

nR=0

(nR − nL) · PN (nR) 〈(nR − nL)2〉 =

N∑nR=0

(nR − nL)2 · PN (nR)

1 a · 12

+ (−a) · 12

= 0 a2 · 12

+ (−a)2 · 12

= a2

2 2a · 14

+ 0 · 12

+ (−2a) · 14

= 0 (2a)2 · 14

+ (0)2 · 12

+ (−2a)2 · 14

= 2a2

3 3a · 18

+ a · 38

+ (−a) · 38

+ (−3a) · 18

= 0 (3a)2 · 18

+ a2 · 38

+ (−a)2 · 38

+ (−3a)2 · 18

= 3a2

N 0 Na2

Table 4: Summary of information found by considering all paths for each number of segmentsN .

Note: Proof of these relationships is given at the end of the solutions for Q4. It is notnecessary to go through this calculation for the purposes of your HW solutions, but we’veincluded it here in case you are interested in seeing a more formal proof.

Recall that the length of the polymer is defined as L = (nR − nL)a because the distanceyou travel in a walk is equal to the number of steps forward minus the number of stepsbackward multiplied by the length of each of these steps. Therefore:

L

a= (nR − nL) ,

and since 〈(nR − nL)2〉 = N we can replace

〈(L/a)2〉 = 〈(nR − nL)2〉 = N . (21)

Because a is a constant we can bring it out of the expectation value:

N =1

a2〈L2〉 , (22)

which leads to the final formula:

〈L2〉 = a2N (23)

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Thus the size of the polymer (or how far is its end from its beginning in 1D) can be approx-imated by:

L ≈√〈L2〉 = a

√N (24)

(c) From figures, we can estimate that Lphage ≈ 1µm. The length of the E. coli genome is

LE. coli ≈ 10µm. The Kuhn length of DNA is aDNA = 100 nm.Using the formula we derived above we can solve for the number of segments N in the

T2 bacteriophage genome:

Lphage ≈√〈L2

phage〉 = aDNA

√Nphage

Nphage =

(Lphage

aDNA

)2

=

(1µm

100 nm

)2

=

(103 nm

100 nm

)2

= 100 segments .

The length of the fully-stretched genome is simply the number of segments multiplied by thelength of each segment, which we can get from the Kuhn length of DNA:

Length of phage genome (µm) ≈ 100 segments× 100nm

segment= 104 nm = 10µm .

To convert this to base pairs, we simply divide this number by the average height for a basepair, which is .34 nm/bp = 3.4× 10−4 µm:

Length of phage genome (bp) = 10µm× 1bp

3.4× 10−4µm≈ 3× 104 bp .

This is in good agreement with the size of the bacteriophage T7 genome, which is ≈ 4×104 bp(BNID 106939), and the size of the bacteriophage lambda genome, which is ≈ 5 × 104 bp(BNID 105770). We can assume these genomes are fairly similar in size to that of the T2bacteriophage.

Taking a similar approach for E.coli :

LE. coli ≈√〈L2

E. coli〉 = aDNA

√NE. coli

NE. coli =

(LE. coli

aDNA

)2

=

(10µm

100 nm

)2

=

(104 nm

100 nm

)2

= 104 segments .

Length of E. coli genome (µm) ≈ 104 segments× 100nm

segment= 106 nm = 103 µm .

Length of E. coli genome (bp) = 103 µm× 1bp

3.4× 10−4µm≈ 3× 106 bp .

This is good agreement with the measured size of the E. coli genome, which is ≈ 5× 106 bp(BNID 100269).

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Returning to 4(b): Formal proofs showing that 〈nR − nL〉 = 0 and = 〈(nR − nL)2〉 = N

Note that nR + nL = N as every one of the N segments must either point to the right or tothe left. We can define a probability distribution P (nR) for nR = 0, . . . , N :

P (nR) =

(1

2

)nR(

1

2

)nL N !

nR! nL!

=

(1

2

)nR(

1

2

)N−nR N !

nR! (N − nR)!(because nL = N − nR)

=⇒ P (nR) =

(1

2

)NN !

nR! (N − nR)!

The sum of the probabilities P (nR) for nR = 0, . . . , N must be equal to 1:

N∑nR=0

P (nR) =N∑

nR=0

(1

2

)NN !

nR! (N − nR)!=

(1

2

)N N∑nR=0

N !

nR! (N − nR)!= 1 ,

=⇒N∑

nR=0

N !

nR! (N − nR)!= 2N .

Since nL = N − nR, then

〈nR − nL〉 = 〈nR − (N − nR)〉 = 〈2nR −N〉 = 2〈nR〉 −N .

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We now calculate 〈nR〉:

〈nR〉 =N∑

nR=0

nRP (nR)

=N∑

nR=0

nR

(1

2

)NN !

nR! (N − nR)!

=

(1

2

)N N∑nR=1

nRN !

nR(nR − 1)! (N − nR)!

=

(1

2

)N N∑nR=1

N !

(nR − 1)! (N − nR)!

=

(1

2

)N N−1∑m=0

N !

m! (N − (m+ 1))!(where m = nR − 1)

=

(1

2

)N N−1∑m=0

N · (N − 1)!

m! ((N − 1)−m)!

= N

(1

2

)N N−1∑m=0

(N − 1)!

m! ((N − 1)−m)!

= N

(1

2

)N2N−1

=N

2

Plugging this into our expression for 〈nR − nL〉:

〈nR − nL〉 = 2〈nR〉 −N = 2N

2−N = N −N = 0

Note that we could have arrived at this relationship without doing any calculations bynoticing that it must be the case that 〈nR〉 = 〈nL〉. The probability distribution P (nR)could be written equivalently as both a function of nR and nL as follows:

P (nR) = P′(nR, nL) =

(1

2

)nR(

1

2

)nL N !

nR! nL!

∣∣∣∣∣nL=N−nR

=

(1

2

)nR+nL N !

nR! nL!

∣∣∣∣∣nL=N−nR

This probability function P′(nR, nL) is symmetric, which means that it is unchanged by

swapping nR and nL, i.e. P′(nR, nL) = P

′(nL, nR). It follows that, for any function f(n),

〈f(nR)〉 = 〈f(nL)〉 ,

for the probability distribution P′(nR, nL) = P (nR). Now we can move on to calculating

〈(nR − nL)2〉. First we can manipulate this expression so it is written entirely as a functionof nR and constants by plugging in nL = N − nR and expanding:

〈(nR − nL)2〉 = 〈(nR − (N − nR))2〉 = 〈(2nR −N)2〉 = 4〈n2R〉 − 4N〈nR〉+N2 = 4〈n2

R〉 −N2 .

16

We now calculate 〈n2R〉:

〈n2R〉 =

N∑nR=0

n2RP (nR) =

N∑nR=0

n2R

(1

2

)NN !

nR! (N − nR)!

=

(1

2

)N N∑nR=0

nR(nR − 1 + 1)N !

nR! (N − nR)!

=

(1

2

)N N∑nR=0

[nR(nR − 1) + nR]N !

nR! (N − nR)!

=

[(1

2

)N N∑nR=0

nR(nR − 1)N !

nR! (N − nR)!

]+

[(1

2

)N N∑nR=0

nRN !

nR! (N − nR)!

]

=

[(1

2

)N N∑nR=0

nR(nR − 1)N !

nR! (N − nR)!

]+ 〈nR〉

=

[(1

2

)N N∑nR=2

nR(nR − 1)N !

nR(nR − 1)(nR − 2)! (N − nR)!

]+N

2

=

[(1

2

)N N∑nR=2

N !

(nR − 2)! (N − nR)!

]+N

2

=

[(1

2

)N N−2∑m=0

N !

m! (N − (m+ 2))!

]+N

2(where m = nR − 2)

=

[(1

2

)N N−2∑m=0

N(N − 1)(N − 2)!

m! ((N − 2)−m)!

]+N

2

=

[N(N − 1)

(1

2

)N N−2∑m=0

(N − 2)!

m! ((N − 2)−m)!

]+N

2

=

[N(N − 1)

(1

2

)N2N−2] +

N

2

=N(N − 1)

4+N

2

=N2

4+N

4.

Now we plug into our expression for 〈(nR − nL)2〉:

〈(nR − nL)2〉 = 4〈n2R〉 −N2 = 4

(N2

4+N

4

)−N2 = N2 +N −N2 = N

17