#  Limitations on concentration measurements and gradient discerning times in
cellular systems

Vaibhav Wasnik<EMAIL_ADDRESS>Indian Institute of Technology, Goa

###### Abstract

This work reports on two results. At first we revisit the Berg and Purcell
calculation that provides a lower bound to the error in concentration
measurement by cells, by considering the realistic case when the cell starts
measuring the moment it comes in contact with the chemoattractants, instead of
measuring after equilibrating with the chemotactic concentration as done in
the classic Berg and Purcell paper. We find that the error in concentration
measurement is still the same as evaluated by Berg and Purcell. We next derive
a lower bound on measurement time below which it is not possible for the cell
to discern extra-cellular chemotactic gradients through spatial sensing
mechanisms. This bound is independent of diffusion rate and concentration of
the chemoattracts and is instead set by detachment rate of ligands from the
cell receptors. The result could help explain experimental observations.

## Introduction

Cells have to make sense out of their surroundings. They have to go towards
sources of nutrition and away from sources of danger. Berg and Purcell berg ,
found that concentration measurement by cells such as _E. Coli_ approach one
of optimum design. The fluctuations in ligand receptor binding were later
incorporated by bialek  who found that cells perform concentration
measurements within limits set by these fluctuations. Since then, there have
been many theoretical works in understanding the limitations imposed on
concentration measurements by the cell. Some of these include szabo  that
considered effect of ligand diffusion on fluctuations in occupancy of
receptors. Understanding how increasing number of receptors could affect
reduction in measurement noise due to receptor noise were considered in rappel
. Corrections to contributions to diffusive arrival of ligands obtained by
bialek  were considered in ten wolde . Constraints placed by energy
consumption in concentration measurement of cells was considered in mehta
-energy2 . Limits to concentration sensing by a cell in an environment of
interfering ligands was considered in mora . Maximum likelihood estimation of
concentration of ligands by looking at history of attachment detachment of
ligands to the receptor was considered in wingreen endres 3 . Most
calculations that involve evaluation of limits to concentration detection
assume that the cell receptors have reached equilibrium with the surrounding
ligands after which the attachment/detachment dynamics is considered. However
realistic calculations to error measurements should consider measurement of
concentration beginning the moment exposure to chemoattractants occurs. We
label this scenario as the non-equilibrium case in comparison to the
equilibrium case which was considered in the Berg Purcell calculation where
the receptor is assumed to have been in equilibrium with the ligand
concentration throughout the measurement history. In the first part of the
paper we show that the bound on error in concentration measurement in the non-
equilibrium scenario is still given by what is expected from Berg-Purcell like
studies, in the limit of large measurement times. Berg and Purcell considered
the quantity getting measured to be the fraction of time the receptor is
occupied by the ligand.On considering a generic quantity defining a
measurement as the linear combination of powers of the fraction of time a
receptor is occupied, we show that the error in concentration measurement is
similar in the equilibrium and non-equilibrium cases for large measurement
times.

In addition to concentration measurement, cells also have to measure
extracellular gradients of chemoattractants. Cells have evolved to measure
gradients with great accuracy. Cells can detect gradients of $1-2\%$
difference across the cell gradient exp1 \- gradient exp3 . An optimal
response has been seen in the cells where the difference in receptor occupancy
between front and back of cells is only $10$ occupied receptors gradient exp4
. Understanding the limits to chemotactic gradient measurement by the cells
were studied in wingreen endres  which idealized the cell as a perfectly
absorbing sphere and a perfectly monitoring sphere . Fluctuation dissipation
theorem was used to consider limitations imposed on gradient sensing due to
ligand receptor kinetics in wingreen endres 2 . Modelling the surface of the
cell as an Ising spin chain showed improved ability to detect gradients if
receptor cooperativity was introduced levine . A signal transduction modelling
to understand gradient readouts was considered in levine rappel . If one
considers the works wingreen endres , wingreen endres 2 , it was shown that
the limits to gradient measurements by the cell went like
$\frac{1}{\sqrt{DacT}}$, upto multiplicative constants. This implies that one
could decrease the time of measurement $T$ to an arbitrary degree, by
increasing the concentration of chemoattractants and still be able to discern
the gradient. In the present work we represent the cell as sphere covered with
receptors with a point source of chemoattractants a particular distance away.
We show that there is a bound on the measurement time, below which it is not
possible to discern the concentration gradient set up by the source. This time
being independent of the concentration and diffusion rates of the
chemoattractant, implies one cannot simply increase these parameters to
decrease the measurement times to being as small as possible. For
_Dictyostelium_ we evaluate this time to be around $\sim 4s$. The time
duration of the pulses not being sufficient to discern gradients, could be a
reason as to why _Dictyostelium_ cells subjected to $5s$ pulses cyclic  do not
respond to the chemotactic gradients of varied concentrations, while being
placed at varied distances from the chemoattractant source.

In the next section we present the calculation which evaluates the error in
concentration measurement by cells, when the concentration detection starts as
soon as the cells are bought in contact with chemoattractants. The calculation
of the minimum time below which it is not possible to discern concentration
gradients is presented in the following section. We end with conclusions.

##  Concentration measurements commencing after immediate exposure.

Cells would in general start measuring chemotactic concentration the moment
they come in touch with the chemoattractants. Berg and Purcell berg  assumed
the receptor has already equilibrated with the surrounding chemotactic
concentration before concentration measurement are considered. Let us instead
consider the more realistic case when the cell has started measuring the
moment it got in touch with the chemotactic concentration. To understand this,
assume the cell represented by a sphere is dropped into a chemoattractant
concentration $c$ as shown in Fig. 1. The occupation probability of each cell
receptor obeys the relation

$\displaystyle\frac{dp(t)}{dt}=k_{+}c(1-p(t))-k_{-}p(t).$ (1)

With the initial condition $p(0)=0$, we get the solution

$\displaystyle p(t)=\frac{k_{+}c}{k_{+}c+k_{-}}(1-e^{-(k_{+}c+k_{-})t})$ (2)

Figure 1:  In (a) we have a cell in a constant concentration of
chemoattractants. The limitations to concentration measurement were studied in
the classic work of Berg Purcell, where the cell was idealised as a sphere in
a constant concentration of chemoattractants. In (b) The cell moves towards a
higher concentration of chemoattractants. We simulate this scenario with the
sphere dropped into a bottle of chemoattractants. In this paper we show that
the error in concentration measurement in (b) is similar to the error in in
(a) in limit of large measurement times. (Images created with BioRender.com)

Let us say that the receptor occupancy at time $t$ is given by $O(t)$, which
equals to zero if the receptor is not occupied and $1$ if its occupied. The
fraction of time a receptor is occupied in a time duration $T$ is

$\displaystyle m(T)=\frac{1}{T}\int_{0}^{T}O(t)dt$ (3)

This is something that the cell can measure. If we split the time $T$ in to
$\frac{T}{dt}$ intervals, the probability of having $n$ intervals in which
receptor was occupied is simply

$\displaystyle e^{-\int_{0}^{T}p(t)dt}\frac{[\int_{0}^{T}p(t)dt]^{n}}{n!}$ (4)

Hence an average estimate of $m(T)$ would be

$\displaystyle\langle m(T)\rangle$
$\displaystyle=\frac{1}{T}\sum_{n=0,\infty}ne^{-\int_{0}^{T}p(t)dt}\frac{[\int_{0}^{T}p(t)dt]^{n}}{n!}=\frac{1}{T}\int_{0}^{T}p(t)dt$
(5)
$\displaystyle=\frac{k_{+}c}{k_{+}c+k_{-}}-\frac{(1-e^{-(k_{+}c+k_{-})T})}{(k_{+}c+k_{-})T}\frac{k_{+}c}{k_{+}c+k_{-}}$

which in limit of large $T$ becomes

$\displaystyle\langle m(T)\rangle$
$\displaystyle=\frac{k_{+}c}{k_{+}c+k_{-}}\left[1-\frac{1}{(k_{+}c+k_{-})T}\right]$
(6) $\displaystyle\sim$ $\displaystyle\frac{k_{+}c}{k_{+}c+k_{-}},\quad
T\rightarrow\infty$

The average occupancy could be inverted by the cell to evaluate the
chemoattractant concentration $c$. To evaluate the noise in this estimate of
$c$ we need to evaluate $\langle m(T)^{2}\rangle$ for which we need to
evaluate $G(t,t^{\prime})=\langle O(t)O(t^{\prime})\rangle$. Assume the
receptor is occupied at time $t$. The receptor would be still found occupied
at time $t^{\prime}$, if the receptor remains attached for some time $\tau$
after time $t$, then has $n$ pairs of time intervals $t_{d}^{i},t_{a}^{i}$
with $i\in[0,n]$, where the receptor is in a detached/ attached state, such
that $\tau+\sum_{i=1,n}t_{d}^{i}+t_{a}^{i}=t-t^{\prime}$. For any particular
timing combination, the probability of realising the same is

$\displaystyle
e^{-\int_{0}^{\tau}k_{-}dx}[k_{+}ck_{-}]^{n}\Pi_{i=1,n}dt_{d}^{i}dt_{a}^{i}e^{-k_{+}c\int_{0}^{t_{d}^{i}}dx}e^{-k_{-}\int_{0}^{t_{a}^{i}}dx}$

One then sums over all possible timing combinations and all values of $n$ to
get the probability of still finding the receptor occupied at time
$t^{\prime}$. Let us call this probability $P(t,t^{\prime})$. Since the only
information getting into the evaluation of $P(t,t^{\prime})$ is that the
ligand is attached at time $t$ and since the ligand attachment detachment is
Markovian, the past history before attachment at time $t$ is irrelevant, so
whether the receptor has equilibrated with the ligands or not is irrelevant in
evaluation of $P(t,t^{\prime})$. It is obvious that
$G(t,t^{\prime})=p(t)P(t,t^{\prime})\quad t<t^{\prime}$. Since, we already
know from berg  that in the case where receptor has equilibrated with the
chemoattractant concentration

$\displaystyle
G_{eq}(t,t^{\prime})=\bar{p}^{2}+\bar{p}(1-\bar{p})e^{-\frac{k_{-}|t-t^{\prime}|}{1-\bar{p}}}$
(8)

with

$\displaystyle\bar{p}=\frac{k_{+}c}{k_{+}c+k_{-}}$ (9)

We get

$\displaystyle
P(t,t^{\prime})=\bar{p}+(1-\bar{p})e^{-\frac{k_{-}|t-t^{\prime}|}{1-\bar{p}}}$
(10)

So,

$\displaystyle
G(t,t^{\prime})=p(t)[\frac{k_{+}c}{k_{+}c+k_{-}}+\frac{k_{-}}{k_{+}c+k_{-}}e^{-(k_{+}c+k_{-})|t^{\prime}-t|}],\quad
t<t^{\prime}$ (11)

Hence,

$\displaystyle\langle m(T)^{2}\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{T^{2}}\int_{0}^{T}\int_{0}^{T}G(t,t^{\prime})dtdt^{\prime}$
$\displaystyle=$
$\displaystyle\frac{1}{T^{2}}\int_{0}^{T}dt\int_{0}^{t}dt^{\prime}p(t^{\prime})P(t^{\prime},t)$
$\displaystyle+$
$\displaystyle\frac{1}{T^{2}}\int_{0}^{T}dt^{\prime}\int_{0}^{t^{\prime}}dtp(t)P(t,t^{\prime})$
$\displaystyle=$
$\displaystyle\frac{2}{T^{2}}\int_{0}^{T}dt^{\prime}\int_{0}^{t^{\prime}}dtp(t)P(t,t^{\prime})$

The second last equation arises from the one above, because
$P(t,t^{\prime})=P(t^{\prime},t)$ and on interchanging
$t^{\prime}\leftrightarrow t$, we find the two integrals in the sum are equal.
Hence as shown in the Appendix, in the limit where $T$ is very large
($T^{-1}<<k_{+}c,k_{-}$).

$\displaystyle\langle m(T)^{2}\rangle-\langle m(T)\rangle^{2}$
$\displaystyle=$
$\displaystyle\left(\frac{k_{+}c}{k_{+}c+k_{-}}\right)^{2}\frac{2}{(k_{+}c+k_{-})T}\frac{k_{-}}{k_{+}c}$
(13)

and

$\displaystyle\frac{\delta c}{c}$ $\displaystyle=$
$\displaystyle\sqrt{\frac{2(k_{+}c+k_{-})}{T}\left[\frac{k_{-}}{k_{+}c}\right]}\frac{1}{k_{-}}.$

This is the same result as obtained by Berg and Purcell berg . What we hence
see is that the error in the non-equilibrium case is the same as the
equilibrium case when $(k_{+}c+k_{-})T>>1$. Berg and Purcell assumed that the
measurement made by the cell was the fraction of time the receptor was
occupied. Receptor activation starts a series of downstream reactions in the
cell that leads to a cellular response. There is no reason to expect that the
cellular response however it is quantified would simply be proportional to the
fraction of time the receptor was occupied. It is more likely that as a
general case the measurement could be proportional to a linear combinations of
powers of the fraction of time the receptor was occupied. Such a quantity
could be written as $Q(T)=\sum_{i>0}a_{i}m(T)^{i}$, where $a_{i}$ are
constants dependent on the nature of the system being analysed and the
corresponding readout. Now,

$\displaystyle\langle m(T)\rangle^{n}$
$\displaystyle=[\frac{1}{T}\int_{0}^{T}p(t)dt]^{n}=[\frac{k_{+}c}{k_{+}c+k_{-}}]^{n}[1-\frac{1}{T}\int_{0}^{T}e^{-(k_{+}c+k_{-})t}dt]^{n}$
(15)
$\displaystyle=[\frac{k_{+}c}{k_{+}c+k_{-}}]^{n}[1-\frac{n}{T}\int_{0}^{T}e^{-(k_{+}c+k_{-})t}dt+\frac{n(n-1)}{2!T^{2}}[\int_{0}^{T}e^{-(k_{+}c+k_{-})t}dt]^{2}]$
$\displaystyle=\langle
m(T)\rangle_{eq}^{n}-[\frac{k_{+}c}{k_{+}c+k_{-}}]^{n}\frac{n}{(k_{+}c+k_{-})T}+\mathcal{O}(\frac{1}{T^{2}}),\quad
if\;(k_{+}c+k_{-})T>>1$

Also,

$\displaystyle\langle
m(T)^{n}\rangle=\frac{1}{T^{n}}\int_{0}^{T}dt_{1}...\int_{0}^{T}dt_{n}\langle
O(t_{1})...O(t_{n})\rangle$ (16) $\displaystyle=$
$\displaystyle\frac{n!}{T^{n}}\int_{0}^{T}dt_{n-1}...\int_{0}^{t_{1}}dtp(t)P(t,t_{1})P(t_{1},t_{2})...P(t_{n-2},t_{n-1})$
$\displaystyle=$ $\displaystyle\langle
m(T)^{n}\rangle_{eq}-\frac{n!}{T^{n}}\int_{0}^{T}dt_{n-1}...\int_{0}^{t_{1}}dt\frac{k_{+}c}{k_{+}c+k_{-}}e^{-(k_{+}c+k_{-})t}[\frac{k_{+}c}{k_{+}c+k_{-}}+\frac{k_{-}}{k_{+}c+k_{-}}e^{-(k_{+}c+k_{-})|t_{1}-t|}]..$
$\displaystyle..[\frac{k_{+}c}{k_{+}c+k_{-}}+\frac{k_{-}}{k_{+}c+k_{-}}e^{-(k_{+}c+k_{-})|t_{n-1}-t_{n-2}|}]$
$\displaystyle=$ $\displaystyle\langle
m(T)^{n}\rangle_{eq}-[\frac{k_{+}c}{k_{+}c+k_{-}}]^{n}\frac{n!}{T^{n}}\int_{0}^{T}dt_{n-1}...\int_{0}^{t_{1}}dte^{-(k_{+}c+k_{-})t}$
$\displaystyle-\frac{k_{-}}{k_{+}c+k_{-}}[\frac{k_{+}c}{k_{+}c+k_{-}}]^{n-1}\frac{n!}{T^{n}}\int_{0}^{T}dt_{n-1}...\int_{0}^{t_{1}}dte^{-(k_{+}c+k_{-})t}[e^{-(k_{+}c+k_{-})|t_{1}-t|}+...+e^{-(k_{+}c+k_{-})|t_{n-1}-t_{n-2}|}]$
$\displaystyle+terms\;higher\;order\;in\;\frac{1}{T}$
$\displaystyle\sim\langle
m(T)^{n}\rangle_{eq}-[\frac{k_{+}c}{k_{+}c+k_{-}}]^{n}\frac{n}{(k_{+}c+k_{-})T}+\mathcal{O}(\frac{1}{T^{2}});\quad
if\;(k_{+}c+k_{-})>>1$

The factor of $n!$ in the second equation has a origin similar to the factor
of $2$ in Eq.LABEL:msquared. Now,

$\displaystyle\langle m(T)^{n}\rangle_{eq}-\langle m(T)\rangle_{eq}^{n}$
$\displaystyle=$
$\displaystyle\frac{n!}{T^{n}}\int_{0}^{T}dt_{n-1}...\int_{0}^{t_{1}}dt\frac{k_{+}c}{k_{+}c+k_{-}}[\frac{k_{+}c}{k_{+}c+k_{-}}+\frac{k_{-}}{k_{+}c+k_{-}}e^{-(k_{+}c+k_{-})|t_{1}-t|}]..$
$\displaystyle..[\frac{k_{+}c}{k_{+}c+k_{-}}+\frac{k_{-}}{k_{+}c+k_{-}}e^{-(k_{+}c+k_{-})|t_{n-1}-t_{n-2}|}]-[\frac{k_{+}c}{k_{+}c+k_{-}}]^{n}$
$\displaystyle=$
$\displaystyle\frac{k_{-}}{k_{+}c+k_{-}}[\frac{k_{+}c}{k_{+}c+k_{-}}]^{n-1}\frac{n!}{T^{n}}\int_{0}^{T}dt_{n-1}...\int_{0}^{t_{1}}dt[e^{-(k_{+}c+k_{-})|t_{1}-t|}+...+e^{-(k_{+}c+k_{-})|t_{n-1}-t_{n-2}|}]$
$\displaystyle+terms\;higher\;order\;in\;\frac{1}{T}$ $\displaystyle=$
$\displaystyle\frac{k_{-}}{k_{+}c+k_{-}}[\frac{k_{+}c}{k_{+}c+k_{-}}]^{n-1}\frac{n(n-1)}{(k_{+}c+k_{-})T}+\mathcal{O}(\frac{1}{T^{2}}),\quad
if\;(k_{+}c+k_{-})T>>1$

Hence,

$\displaystyle\langle Q(T)^{2}\rangle-\langle Q(T)\rangle^{2}$
$\displaystyle=$ $\displaystyle\sum_{i,j>0}a_{i}a_{j}[\langle
m(T)^{i+j}\rangle-\langle m(T)^{i}\rangle\langle m(T)^{j}\rangle]$
$\displaystyle=$ $\displaystyle\sum_{i,j>0}a_{i}a_{j}[\langle
m(T)^{i+j}\rangle_{eq}-[\frac{k_{+}c}{k_{+}c+k_{-}}]^{i+j}\frac{i+j}{(k_{+}c+k_{-})T}+\mathcal{O}(\frac{1}{T^{2}})$
$\displaystyle-[\langle
m(T)^{i}\rangle_{eq}-[\frac{k_{+}c}{k_{+}c+k_{-}}]^{i}\frac{i}{(k_{+}c+k_{-})T}+\mathcal{O}(\frac{1}{T^{2}})][\langle
m(T)^{j}\rangle_{eq}-[\frac{k_{+}c}{k_{+}c+k_{-}}]^{j}\frac{j}{(k_{+}c+k_{-})T}+\mathcal{O}(\frac{1}{T^{2}})]]$
$\displaystyle,\quad if\;(k_{+}c+k_{-})T>>1$ $\displaystyle=$
$\displaystyle\sum_{i,j>0}a_{i}a_{j}[\langle m(T)^{i+j}\rangle_{eq}-\langle
m(T)^{i}\rangle_{eq}\langle
m(T)^{j}\rangle_{eq}]+\mathcal{O}(\frac{1}{T^{2}}),\quad
if\;(k_{+}c+k_{-})T>>1$ $\displaystyle=$ $\displaystyle\langle
Q(T)^{2}\rangle_{eq}-\langle
Q(T)\rangle_{eq}^{2}+\mathcal{O}(\frac{1}{T^{2}}),\quad
if\;(k_{+}c+k_{-})T>>1$ $\displaystyle=$ $\displaystyle\langle
Q(T)^{2}\rangle_{eq}-\langle Q(T)\rangle_{eq}^{2},\quad
if\;(k_{+}c+k_{-})T>>1$

This is because $\sum_{i,j>0}a_{i}a_{j}[\langle m(T)^{i+j}\rangle_{eq}-\langle
m(T)^{i}\rangle_{eq}\langle m(T)^{j}\rangle_{eq}]$ is
$\mathcal{O}(\frac{1}{T})\quad if\;(k_{+}c+k_{-})T>>1$, as it contains terms
of the form $\langle m(T)^{n}\rangle_{eq}-\langle m(T)\rangle_{eq}^{n}$ which
are $\mathcal{O}(\frac{1}{T})$ in the same limit. For a specially chosen
combination of $\\{a_{i}\\}$’s, we could have that
$\sum_{i,j>0}a_{i}a_{j}[\langle m(T)^{i+j}\rangle_{eq}-\langle
m(T)^{i}\rangle_{eq}\langle m(T)^{j}\rangle_{eq}]$ may be of
$\mathcal{O}(\frac{1}{T^{2}})$, but for a generic combination of
$\\{a_{i}\\}$’s, this is not the case and hence $\langle
Q(T)^{2}\rangle-\langle Q(T)\rangle^{2}=\langle Q(T)^{2}\rangle_{eq}-\langle
Q(T)\rangle_{eq}^{2},\quad if\;(k_{+}c+k_{-})T>>1$. Also,

$\displaystyle\langle Q(T)\rangle$ $\displaystyle=$
$\displaystyle\sum_{i>0}a_{i}\langle m(T)^{i}\rangle$ (19) $\displaystyle=$
$\displaystyle\sum_{i>0}a_{i}[\frac{k_{+}c}{k_{+}c+k_{-}}]^{i}+\mathcal{O}(\frac{1}{T}),\quad
if\;(k_{+}c+k_{-})T>>1$ $\displaystyle=$ $\displaystyle\langle
Q(T)\rangle_{eq}+\mathcal{O}(\frac{1}{T}),\quad if\;(k_{+}c+k_{-})T>>1$
$\displaystyle=$ $\displaystyle\langle Q(T)\rangle_{eq},\quad
if\;(k_{+}c+k_{-})T>>1$

We hence see that for generic $\\{a_{i}\\}$’s, the evaluation of $\frac{\delta
c}{c}$ is the same in the limit $(k_{+}c+k_{-})T>>1$, whether we consider the
Berg Purcell like equilibrium calculations which assume that the receptor has
already equilibrated with the ligands before measurement starts or we consider
the realistic non equilibrium case where the measurement commences the moment
the receptor is bought in contact with the ligands.

## Lower bound on gradient measurement times in cells.

To consider the problem of time required to measure a chemotactic gradient by
the cell, consider the following setup illustrated in Fig.2. A sphere
representing the cell is covered with receptors, such that the probability of
the receptor at position $(R,\theta)$ being occupied is $p(\theta,t)$ which at
equilibrium becomes $p(\theta)$. The source of chemottractants is at a
distance $a$ from the center of the sphere. Let us say the concentration of
chemoattractants at a position $(r,\theta)$, is $c(r,\theta,t)$ which at
equilibrium becomes $c(r,\theta)$. We hence have that

Figure 2:  (i) A cell getting exposed to a point source of chemoattractants.
Experiments have shown cyclic , that for stimulation with $5s$ pulses for a
range of cAMP concentrations, no evidence of chemotaxis was observed at any
distance from the cAMP pipette. Calculations in this paper suggest a
measurement time bound of $4s$, below which it is not possible for cells to
discern gradients irrespective of concentration of chemoattractants involved.
(Images reated with BioRender.com.) (ii) The various quantities entering into
the calculation in the text. The cell is represented by a sphere of radius
$R$, and the source of chemoattractant is represented by in green colour at a
distance of $a$ from the center of the sphere.

$\displaystyle\frac{dp(\theta,t)}{dt}=k_{+}c(R,\theta,t)(1-p(\theta,t))-k_{-}p(\theta,t)$
(20)

and

$\displaystyle-D\frac{\partial c(r,\theta,t)}{\partial
r}|_{r=R}=k_{+}c(R,\theta,t)(1-p(\theta,t))-k_{-}p(\theta,t)$ (21)

At equilibrium we have

$\displaystyle k_{-}p(R,\theta)$
$\displaystyle=k_{+}c(R,\theta)(1-p(R,\theta))$ $\displaystyle D\frac{\partial
c(r,\theta)}{\partial r}|_{r=R}$ $\displaystyle=0$ (22)

The symmetry of the problem implies no $\phi$ dependence. Since the source of
chemoattractants is located at $r=a$, we have the equation

$\displaystyle D\nabla^{2}c(r,\theta,\phi)=C\delta(r-a)\delta(\theta)$ (23)

whose solution as described in the Appendix is

$\displaystyle c(R,\theta)$ $\displaystyle=$ $\displaystyle\frac{C}{4\pi
Da}+\frac{C}{4\pi Da}\sum_{n=1,\infty}\frac{R^{n}}{a^{n}}P_{n}(\cos\theta).$
(24)

Since

$\displaystyle p(R,\theta)$
$\displaystyle=\frac{k_{+}c(R,\theta)}{k_{+}c(R,\theta)+k_{-}},$

we have the error in concentration measurement berg  for large times

$\displaystyle\frac{\triangle c_{rms}(R,\theta)}{c(R,\theta)}$
$\displaystyle=\sqrt{\frac{2}{k_{-}Tp(R,\theta)}}=\sqrt{\frac{2(k_{+}c(R,\theta)+k_{-}))}{k_{-}Tk_{+}c}}$

However, including terms to all orders in $\frac{1}{T}$, we have.

$\displaystyle\frac{\triangle c_{rms}(R,\theta)}{c(R,\theta)}$
$\displaystyle=\sqrt{\frac{2[(k_{+}c(R,\theta)T+k_{-}T)-(1-e^{-(k_{+}c(R,\theta)+k_{-})T})]}{(k_{-}T)(k_{+}c(R,\theta)T)}}$

Hence, in order to be able to discern a gradient we should have that

$\displaystyle c(R,\theta=0)-c(R,\theta=\pi)$ $\displaystyle>$
$\displaystyle\triangle c_{rms}(R,\theta=0)+\triangle c_{rms}(R,\theta=\pi)$
$\displaystyle=$
$\displaystyle\sqrt{\frac{2c(R,\theta=0)[(k_{+}c(R,\theta=0)T+(k_{-}T-1)+e^{-(k_{+}c(R,\theta=0)+k_{-})T})]}{k_{-}k_{+}T^{2}}}$
$\displaystyle+$
$\displaystyle\sqrt{\frac{2c(R,\theta=0)[(k_{+}c(R,\theta=\pi)T+(k_{-}T-1)+e^{-(k_{+}c(R,\theta=\pi)+k_{-})T})]}{k_{-}k_{+}T^{2}}}$

Assume that the measurement time $T$ is such that $k_{-}T>1$, then

$\displaystyle c(R,\theta=0)-c(R,\theta=\pi)$ $\displaystyle>$
$\displaystyle\sqrt{\frac{2c(R,\theta=0)(k_{+}c(R,\theta=0))}{k_{-}k_{+}T}}$
$\displaystyle+$
$\displaystyle\sqrt{\frac{2c(R,\theta=\pi)(k_{+}c(R,\theta=\pi))}{k_{-}k_{+}T}}$
$\displaystyle\frac{C}{4\pi Da}\frac{2R/a}{1-R^{2}/a^{2}}$ $\displaystyle>$
$\displaystyle\sqrt{\frac{2}{k_{-}T}}[c(R,\theta=0)+c(R,\theta=\pi)]$
$\displaystyle\frac{C}{2\pi Da}\frac{R/a}{1-R^{2}/a^{2}}$ $\displaystyle>$
$\displaystyle\sqrt{\frac{2}{k_{-}T}}[\frac{C}{2\pi Da}+\frac{C}{2\pi
Da}\frac{R^{2}/a^{2}}{1-R^{2}/a^{2}}]$ $\displaystyle T$ $\displaystyle>$
$\displaystyle\frac{2}{k_{-}}[\frac{1+\frac{R^{2}/a^{2}}{1-R^{2}/a^{2}}}{\frac{R/a}{1-R^{2}/a^{2}}}]^{2}=\frac{2}{k_{-}}\frac{a^{2}}{R^{2}}$
(29)

We hence get a bound on the measurement time, independent of the nature of the
chemoattractant or its concentration, below which it is not possible for the
cell to decipher the concentration gradient. The smallest value this bound can
take is when $a=R$ or $T_{smallest}=\frac{2}{k_{-}}$. Since $k_{-}T\geq 2$,
the assumption $k_{-}T\geq 1$ used to derive above implies consistency of the
derivation above. Experiments rates  have evaluated the detachment rates of
Cy3-cAMP molecules from receptors on the surface of dictyostelium and found
that $k_{-}$ was in the range of $1.1-.39s^{-1}$ in the anterior pseudopod
region of the cell and $.39-.1s^{-1}$ in the posterior tail region. So
assuming an rate of $k_{-}\sim.5s^{-1}$, gives a bound on
$T_{smallest}=\frac{2}{k_{-}}=4s$. This is in line with cyclic  where it was
seen that the simulation of _Dictyostelium_ with $5s$ pulses using a wide
range of cAMP concentrations, shows no evidence of chemotaxis by cells at any
distance from the pipette. A possible reason for this could be the fact that
the time period for which the signal was active, was not sufficient for
discerning the gradient by the cell, leading to no response.

## Conclusion

Any calculation that evaluates the error in measurement of concentation should
consider the cell from the time when it first gets in touch with the ligand
concentration. In the first part of the paper we evaluated the error in such a
scenario. We showed that the error is the same, in the limit of large
measurement time, as evaluated in the Berg Purcell calculation which assumed
that the cell was already in equilibrium with the ligand concentration when
measurement commenced. In the non-equilibrium case and for large measurement
times $T>>(k_{+}c+k_{-})^{-1}$, the receptor will have equilibrated with the
ligand concentration for most of its measurement history and hence it is
expected that $\langle m(T)\rangle$ as well $\langle m(T)^{2}\rangle$ will
yield result similar to the case in which the receptor was already in
equilibrium with the surrounding concentration when the measurement commenced.
However, there is no reason apriori that the difference between the two
$\langle m(T)^{2}\rangle-\langle m(T)\rangle^{2}$ will yield similar values.
We however see an equality between the equilibrium and non-equilibrium
estimates. The cellular response would generally be proportional to some
linear combination of powers of the fraction of time the receptor is occupied.
By considering the measured quantity to go as $Q(T)=\sum_{i>0}a_{i}m(T)^{i}$,
where the $\\{a_{i}\\}$’s are generically chosen, we found that the error in
concentration measurement in the equilibrium as well as non equilibrium cases
is the same for large measurement times.

bialek  have utilised a fluctuation dissipation framework to talk about how
noisy ligand attachment/detachment events add limitations to concentration
measurements. Their work first considers the case without diffusion, producing
an rms error in the estimate of receptor occupancy $n$ given by

$\displaystyle\delta
n_{rms}=\sqrt{\frac{2\bar{n}(1-\bar{n})}{(k_{+}\bar{c}+k_{-})\tau}}$ (30)

including diffusion increases this error to

$\displaystyle\delta
n_{rms}=\sqrt{\frac{2\bar{n}(1-\bar{n})}{(k_{+}\bar{c}+k_{-})T}+\frac{(\bar{n}(1-\bar{n}))^{2}}{\pi
Da\bar{c}T}}$ (31)

where $\bar{c},\bar{n}$ are averages of measured concentration and receptor
occupancy, $T$ is the measurement time, $D$ is the diffusion coefficient and
$a$ is the receptor size. Hence, including the effect of diffusion only serves
to increase the error. Since bialek , considered the situation where the
ligands have equilibrated with the receptor, they could use the fluctuation
dissipation frameworks to do their calculations. Our work in part one of the
paper is considering the case when the receptor is suddenly got in touch with
the ligands implying a non-equilibrium framework, so fluctuation dissipation
frameworks to do calculations are not possible here. Hence, including
diffusion effects if possible would be more involved and we do not attempt the
same in our paper. However from the lessons of equilibrium calculations as
done in bialek , we can say that diffusion effects will only add to
measurement errors. How this addition will look like or whether it will yield
the same form as equation above, given the similarity in the equilibrium and
non-equilibrium case when diffusion is not considered, is an open question.

It is also quite interesting that the observation that the bound on
measurement times below which gradients cannot be detected, being independent
of the chemoattractant properties such as concentration and diffusion was not
noticed in theoretical works up until now. For example in wingreen endres 2
the authors consider the noise in receptor ligand binding/unbinding to
evaluate the limits to detection of gradients. The only information of the
chemotactic gradient that appears in their analysis is the ligand
concentration at points on the surface of the cell. For example the section
_Two Receptors_ evaluates the variance in the difference between concentration
between two points on the cell surface and relates the same to the
concentration at these two points. The work does not calculate any relation
between the ligand concentration of various points on the cell surface, but
assumes the same to be given. In our work we consider a point source that
produces the chemotactic gradient around the cell itself and calculate the
concentration on various points on the cell surface. Using this crucial
information we could evaluate a bound on the measurement time itself below
which no gradient could be discerned irrespective of concentrations involved,
an observation that was missed/not obtained in works such as wingreen endres 2
.

The FRET measurements for example in fret  point to various signalling
elements in the cell getting activated, however they do not imply or suggest
that any chemotaxis in direction of the source is attained for a pulse of
magnitude below 5 seconds, since as is known from cyclic  there is no
chemotactic response in direction of the gradient for pulses of time duration
5 sec. The part two of the paper provides a reason as to why for all possible
ligand concentrations there is no chemotactic response observed for the 5
second pulse, tracing this reason to the limitations imposed by the ligand
binding unbinding noise.

## Appendix

###  I

We elaborate on some calculations from the section on concentration
measurements. We know that

$\displaystyle\langle m(T)^{2}\rangle$ (32) $\displaystyle=$
$\displaystyle\frac{1}{T^{2}}\int_{0}^{T}\int_{0}^{T}G(t,t^{\prime})dtdt^{\prime}$
$\displaystyle=$
$\displaystyle\frac{1}{T^{2}}\int_{0}^{T}dt\int_{0}^{t}dt^{\prime}p(t^{\prime})[\frac{k_{+}c}{k_{+}c+k_{-}}+\frac{k_{-}}{k_{+}c+k_{-}}e^{-(k_{+}c+k_{-})(t-t^{\prime})}]^{\prime}$
$\displaystyle+$
$\displaystyle\frac{1}{T^{2}}\int_{0}^{T}dt^{\prime}\int_{0}^{t^{\prime}}dtp(t)[\frac{k_{+}c}{k_{+}c+k_{-}}+\frac{k_{-}}{k_{+}c+k_{-}}e^{-(k_{+}c+k_{-})(t^{\prime}-t)}]$
$\displaystyle=$
$\displaystyle\frac{2}{T^{2}}\int_{0}^{T}dt^{\prime}\int_{0}^{t^{\prime}}dtp(t)[\frac{k_{+}c}{k_{+}c+k_{-}}+\frac{k_{-}}{k_{+}c+k_{-}}e^{-(k_{+}c+k_{-})(t^{\prime}-t)}]$
$\displaystyle=$
$\displaystyle\frac{2}{T^{2}}\int_{0}^{T}dt^{\prime}\int_{0}^{t^{\prime}}dt[\frac{k_{+}c}{k_{+}c+k_{-}}](1-e^{-(k_{+}c+k_{-})t})[\frac{k_{+}c}{k_{+}c+k_{-}}+\frac{k_{-}}{k_{+}c+k_{-}}e^{-(k_{+}c+k_{-})(t^{\prime}-t)}]$
$\displaystyle=$
$\displaystyle\frac{1}{T^{2}}\left(-\frac{2ck_{-}k_{+}}{\left(ck_{+}+k_{-}\right){}^{4}}+\frac{2ck_{-}k_{+}e^{-T\left(ck_{+}+k_{-}\right)}}{\left(ck_{+}+k_{-}\right){}^{4}}+\frac{c^{2}k_{+}^{2}}{\left(ck_{+}+k_{-}\right){}^{4}}-\frac{c^{2}k_{+}^{2}e^{-T\left(ck_{+}+k_{-}\right)}}{\left(ck_{+}+k_{-}\right){}^{4}}\right)$
$\displaystyle+$
$\displaystyle\frac{1}{T}\left(-\frac{c^{3}k_{+}^{3}}{\left(ck_{+}+k_{-}\right){}^{4}}+\frac{c^{2}k_{-}k_{+}^{2}e^{-T\left(ck_{+}+k_{-}\right)}}{\left(ck_{+}+k_{-}\right){}^{4}}+\frac{ck_{-}^{2}k_{+}e^{-T\left(ck_{+}+k_{-}\right)}}{\left(ck_{+}+k_{-}\right){}^{4}}+\frac{ck_{-}^{2}k_{+}}{\left(ck_{+}+k_{-}\right){}^{4}}\right)$
$\displaystyle+$
$\displaystyle\left(\frac{c^{4}k_{+}^{4}}{2\left(ck_{+}+k_{-}\right){}^{4}}+\frac{c^{3}k_{-}k_{+}^{3}}{\left(ck_{+}+k_{-}\right){}^{4}}+\frac{c^{2}k_{-}^{2}k_{+}^{2}}{2\left(ck_{+}+k_{-}\right){}^{4}}\right)$

where we plugged in to the second last equation Eq.2

Also we have

$\displaystyle\langle m(T)\rangle^{2}$ (33) $\displaystyle=$
$\displaystyle[\int_{0}^{T}\frac{k_{+}c}{k_{+}c+k_{-}}](1-e^{-(k_{+}c+k_{-})t})]^{2}$
$\displaystyle=$
$\displaystyle\frac{1}{T^{2}}\left(\frac{c^{2}k_{+}^{2}}{\left(ck_{+}+k_{-}\right){}^{4}}+\frac{c^{2}k_{+}^{2}e^{-2T\left(ck_{+}+k_{-}\right)}}{\left(ck_{+}+k_{-}\right){}^{4}}-\frac{2c^{2}k_{+}^{2}e^{-T\left(ck_{+}+k_{-}\right)}}{\left(ck_{+}+k_{-}\right){}^{4}}\right)$
$\displaystyle+$
$\displaystyle\frac{1}{T}\left(\frac{2c^{3}k_{+}^{3}e^{-T\left(ck_{+}+k_{-}\right)}}{\left(ck_{+}+k_{-}\right){}^{4}}-\frac{2c^{3}k_{+}^{3}}{\left(ck_{+}+k_{-}\right){}^{4}}+\frac{2c^{2}k_{-}k_{+}^{2}e^{-T\left(ck_{+}+k_{-}\right)}}{\left(ck_{+}+k_{-}\right){}^{4}}-\frac{2c^{2}k_{-}k_{+}^{2}}{\left(ck_{+}+k_{-}\right){}^{4}}\right)$
$\displaystyle+$
$\displaystyle\left(\frac{c^{4}k_{+}^{4}}{\left(ck_{+}+k_{-}\right){}^{4}}+\frac{2c^{3}k_{-}k_{+}^{3}}{\left(ck_{+}+k_{-}\right){}^{4}}+\frac{c^{2}k_{-}^{2}k_{+}^{2}}{\left(ck_{+}+k_{-}\right){}^{4}}\right)$

Hence

$\displaystyle\langle m(T)^{2}\rangle-\langle m(T)\rangle^{2}$ (34)
$\displaystyle=$
$\displaystyle\frac{1}{T^{2}}\left(-\frac{c^{2}k_{+}^{2}e^{-2T\left(ck_{+}+k_{-}\right)}}{\left(ck_{+}+k_{-}\right){}^{4}}+\frac{c^{2}k_{+}^{2}}{\left(ck_{+}+k_{-}\right){}^{4}}+\frac{4ck_{-}k_{+}e^{-T\left(ck_{+}+k_{-}\right)}}{\left(ck_{+}+k_{-}\right){}^{4}}-\frac{4ck_{-}k_{+}}{\left(ck_{+}+k_{-}\right){}^{4}}\right)$
$\displaystyle+$
$\displaystyle\frac{1}{T}\left(-\frac{2c^{3}k_{+}^{3}e^{-T\left(ck_{+}+k_{-}\right)}}{\left(ck_{+}+k_{-}\right){}^{4}}+\frac{2c^{2}k_{-}k_{+}^{2}}{\left(ck_{+}+k_{-}\right){}^{4}}+\frac{2ck_{-}^{2}k_{+}e^{-T\left(ck_{+}+k_{-}\right)}}{\left(ck_{+}+k_{-}\right){}^{4}}+\frac{2ck_{-}^{2}k_{+}}{\left(ck_{+}+k_{-}\right){}^{4}}\right)$

which in large $T$ limit is

$\displaystyle\langle m(T)^{2}\rangle-\langle m(T)\rangle^{2}$
$\displaystyle=$
$\displaystyle\left(\frac{k_{+}c}{k_{+}c+k_{-}}\right)^{2}\frac{2}{(k_{+}c+k_{-})T}\frac{k_{-}}{k_{+}c}$
(35)

Since in limit of large $T$ ($T^{-1}<<k_{+}c,k_{-}$)

$\displaystyle 1$ $\displaystyle=$
$\displaystyle\frac{k_{-}}{k_{+}c}\frac{\langle m(T)\rangle}{1-\langle
m(T)\rangle}$ $\displaystyle\frac{\delta c}{c}$ $\displaystyle=$
$\displaystyle\frac{\delta\langle m(T)\rangle}{\langle m(T)\rangle(1-\langle
m(T)\rangle)}$ $\displaystyle=$
$\displaystyle\sqrt{\left(\frac{k_{+}c}{k_{+}c+k_{-}}\right)^{2}\frac{1}{(k_{+}c+k_{-})T}\left[2\frac{k_{-}}{k_{+}c}\right]}\frac{(k_{+}c+k_{-})^{2}}{k_{+}ck_{-}}$
$\displaystyle=$ $\displaystyle\sqrt{\frac{2}{k_{+}cT}+\frac{2}{k_{-}T}}$

## II

We next, consider calculations involved in evaluating the bound on gradient
measuring time. We have to solve

$\displaystyle D\nabla^{2}c(r,\theta,\phi)=C\delta(r-a)\delta(\theta)$ (37)

with the boundary condition

$\displaystyle D\frac{\partial c(r,\theta)}{\partial r}|_{r=R}$
$\displaystyle=0$ (38)

The symmetry of the problem implies no $\phi$ dependence. One solution to
above equation is

$\displaystyle c(r,\theta)$ $\displaystyle=$ $\displaystyle\frac{C}{4\pi
D\sqrt{r^{2}+a^{2}-2ar\cos\theta}}$

In the region close to $r=R$, the above becomes

$\displaystyle c(r,\theta)$ $\displaystyle=$ $\displaystyle\frac{C}{4\pi
Da}\sum_{n=0,\infty}\frac{r^{n}}{a^{n}}P_{n}(\cos\theta)$ (40)

where $P_{n}(\cos\theta)$ are the Legendre Polynomials.
$c(r,\theta)+b(r,\theta)$ is also a solution of Eq.37 if

$\displaystyle\nabla^{2}b(r,\theta)=0$ (41)

Solution of above equation for region $r>R$ is

$\displaystyle b(r,\theta)$ $\displaystyle=$
$\displaystyle\sum_{n=0,\infty}\frac{A_{n}}{r^{n+1}}P_{n}(\cos\theta)$ (42)

Hence for the region close to $r=R$, a solution is

$\displaystyle c(r,\theta)$ $\displaystyle=$ $\displaystyle\frac{C}{4\pi
Da}\sum_{n=0,\infty}\frac{r^{n}}{a^{n}}P_{n}(\cos\theta)+\sum_{n=0,\infty}\frac{A_{n}}{r^{n+1}}P_{n}(\cos\theta)$
(43)

The boundary condition

$\displaystyle D\frac{\partial c(r,\theta)}{\partial r}|_{r=R}$
$\displaystyle=0$ (44)

implies

$\displaystyle 0$ $\displaystyle=$ $\displaystyle\frac{C}{4\pi
Da}\sum_{n=1,\infty}\frac{nR^{n-1}}{a^{n}}P_{n}(\cos\theta)-\sum_{n=0,\infty}\frac{(n+1)A_{n}}{R^{n+2}}P_{n}(\cos\theta)$
(45)

Hence $A_{0}=0$ and for $n>0$

$\displaystyle A_{n}=\frac{C}{4\pi Da}\frac{nR^{2n+1}}{(n+1)a^{n}}$ (46)

Hence,

$\displaystyle c(R,\theta)$ $\displaystyle=$ $\displaystyle\frac{C}{4\pi
Da}\sum_{n=0,\infty}\frac{R^{n}}{a^{n}}P_{n}(\cos\theta)+\frac{C}{4\pi
Da}\sum_{n=1,\infty}\frac{nR^{2n+1}}{(n+1)a^{n}}\frac{1}{R^{n+1}}P_{n}(\cos\theta)$
(47) $\displaystyle=$ $\displaystyle\frac{C}{4\pi Da}+\frac{C}{4\pi
Da}\sum_{n=1,\infty}\frac{R^{n}}{a^{n}}P_{n}(\cos\theta).$

So

$\displaystyle c(R,\theta=0)$ $\displaystyle=$ $\displaystyle\frac{C}{4\pi
Da}+\frac{C}{4\pi Da}\sum_{n=1,\infty}\frac{R^{n}}{a^{n}}$ $\displaystyle=$
$\displaystyle\frac{C}{4\pi Da}+\frac{C}{4\pi Da}\frac{R/a}{1-R/a}$
$\displaystyle=$ $\displaystyle\frac{C}{4\pi Da}\frac{1}{1-R/a}$
$\displaystyle c(R,\theta=\pi)$ $\displaystyle=$ $\displaystyle\frac{C}{4\pi
Da}+\frac{C}{4\pi Da}\sum_{n=1,\infty}(-1)^{n}\frac{R^{n}}{a^{n}}$
$\displaystyle=$ $\displaystyle\frac{C}{4\pi Da}-\frac{C}{4\pi
Da}\frac{R/a}{1+R/a}$ $\displaystyle=$ $\displaystyle\frac{C}{4\pi
Da}\frac{1}{1+R/a}$

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