Question

A cell that is surrounded by chemoattractant molecules will sense or 'measure' the number of molecules, \(N\), in roughly its own volume \(a^{3}\), where \(a\) is the radius of the cell. Suppose the chemoattractant aspartate is in concentration \(c=1 \mu \mathrm{M}\) and has diffusion constant \(D=\) \(10^{-5} \mathrm{~cm}^{2} \mathrm{~s}^{-1}\). For E. coli, \(a \approx 10^{-6} \mathrm{~m}\). (a) Compute \(N\). (b) The error or noise in the measurement of a quantity like \(N\) will be approximately \(\sqrt{N}\). Compute the precision \(P\) of E. coli's measurement, which is the noise/signal ratio \(P=\sqrt{N} / N\). (c) Compute the 'clearance time' \(T_{c}\) that it takes for a chemoattractant molecule to diffuse roughly the distance \(a\). This is the time over which \(E\). coli makes its 'measurement.' (d) If the cell stays in place for \(m\) units of time \(T=m \tau_{c}\), to make essentially \(m\) independent measurements, the cell can improve its precision. Write an expression for the precision \(P_{m}\) of the cell's measurement after \(m\) time steps, as a function of \(a, D, c\), and \(T\). (e) Given \(\tau=1 \mathrm{~s}\), compute \(P_{m}\).

Short answer

N = 2.52 x 10^3 molecules. P = 0.0199. T_c = 1 s. P_m = 0.0199.

Step-by-step solution

Compute the Number of Molecules, N

First, calculate the volume of the cell, which is approximately given by the volume of a sphere: \[ V = \frac{4}{3} \pi a^3 \] Given the radius of the cell is \( a = 10^{-6} \text{ m} \), the volume is: \[ V = \frac{4}{3} \pi \left(10^{-6} \text{ m}\right)^3 \approx 4.19 \times 10^{-18} \text{ m}^3 \] Now, convert this volume to liters: \[ 1 \text{ m}^3 = 10^3 \text{ L} \] so, \[ V = 4.19 \times 10^{-18} \text{ m}^3 = 4.19 \times 10^{-15} \text{ L} \] Now, using the concentration \( c = 1 \mu M = 10^{-6} \text{ mol/L} \), we can find the number of molecules, \(N\), using Avogadro's number \(N_A = 6.022 \times 10^{23} \text{ molecules/mol} \): \[ N = c \times V \times N_A = 1 \times 10^{-6} \text{ mol/L} \times 4.19 \times 10^{-15} \text{ L} \times 6.022 \times 10^{23} \text{ molecules/mol} \] \[ N \approx 2.52 \times 10^3 \text{ molecules} \]

Calculate the Precision P

The precision \(P\) is given by the noise-to-signal ratio: \[ P = \frac{\sqrt{N}}{N} \] From Step 1, \(N = 2.52 \times 10^3\), so: \[ \sqrt{N} = \sqrt{2.52 \times 10^3} \approx 50.2 \] Thus, \[ P = \frac{50.2}{2.52 \times 10^3} \approx 0.0199 \]

Compute the Clearance Time \( T_c \)

The clearance time \(T_c\) is the time it takes for a molecule to diffuse a distance \(a\). This is given by: \[ T_c = \frac{a^2}{D} \] Given \(a = 10^{-6} \text{ m}\) and \(D = 10^{-5} \text{ cm}^2/\text{s} = 10^{-9} \text{ m}^2/\text{s}\): \[ T_c = \frac{(10^{-6} \text{ m})^2}{10^{-9} \text{ m}^2/\text{s}} = 1 \text{ s} \]

Expression for Precision After m Time Steps

To improve its precision, the cell can make \(m\) independent measurements over time \(T = mT_c\). The new precision \(P_m\) can be written as: \[ P_m = \frac{1}{\sqrt{m}} P \] Substituting the earlier derived expression for \(P\): \[ P = \frac{\sqrt{N}}{N} \] yields \[ P_m = \frac{1}{\sqrt{m}} \cdot \frac{\sqrt{N}}{N} \]

Compute the Precision for Given \( \tau = 1 \text{ s} \)

Given \( T = 1 \text{ s} \), we have \( m \tau_c = 1 \text{ s} \). Hence, \( m = 1 \). Substituting \( m = 1 \): \[ P_m = \frac{1}{\sqrt{1}} \cdot \frac{50.2}{2.52 \times 10^3} = 0.0199 \]

Key concepts

Diffusion Constant

The diffusion constant, often symbolized as \( D \), represents how quickly molecules spread out in a given medium. It plays a crucial role in determining how fast molecules like chemoattractants reach a cell. For our problem, we used the diffusion constant for aspartate, which is \( 10^{-5} \) \( \text{cm}^2/\text{s} \), converting to \( 10^{-9} \) \( \text{m}^2/\text{s} \). This value helps us calculate the clearance time (\( T_c \)), which is the time it takes for a molecule to travel roughly the radius of the cell. The formula used is:

\[ T_c = \frac{a^2}{D} \]
Understanding the diffusion constant is essential for comprehending how quickly the chemoattractants will surround E. coli.

E. coli

E. coli is a type of bacterium often utilized in scientific experiments because of its simple structure and ease of cultivation. For this exercise, it's essential to note that E. coli cells typically have a radius (\( a \)) of about \( 10^{-6} \) meters, meaning their sensing volume \( V \) can be approximated by:

\[ V = \frac{4}{3} \pi a^3 \]
With this volume, E. coli can measure the number of chemoattractant molecules within that space. As these bacteria navigate their environment, they use such measurements to decide where to move, highlighting the relevance of calculations involving \( N \) (number of sensed molecules).

Precision Calculation

Precision in these measurements showcases the accuracy with which E. coli can detect chemoattractants. It's expressed as the noise-to-signal ratio \( P \):

\[ P = \frac{\sqrt{N}}{N} \]
Here, \( N \) comes from computations involving the volume of the cell and the concentration of the chemoattractant. The number of molecules sensed (\( N \)) leads to a calculation where the square root of \( N \) defines the noise, yielding the precision error estimate. This ratio provides insight into how reliably E. coli can gauge its environment chemically.

Concentration

Concentration (\( c \)) in chemistry denotes the amount of a substance present in a mixture. For this problem, the concentration of aspartate was given as \( 1 \mu \text{M} \) (micromolar), equivalent to \( 10^{-6} \text{ mol/L} \). By combining this concentration with the cell’s volume converted to liters, we can derive the number of molecules (\( N \)):

\[ N = c \times V \times N_A \]
Where \( N_A \) is Avogadro's number. The concentration directly impacts the number of molecules available to the cell, hence influencing the precision and measurement noise.

Measurement Noise

Measurement noise is an indication of uncertainty or error inherent to the cell’s measurement process. For the calculated number of molecules \( N \), the noise is expressed as \( \sqrt{N} \). This arises from statistical fluctuations, a common phenomenon when dealing with small numbers of particles or molecules. The lower the noise-to-signal ratio, the better the measurement precision:

\[ P = \frac{\sqrt{N}}{N} \]
With more measurement intervals or larger \( N \), precision improves, and noise diminishes. This underpins the significance of repeated measurements or increased molecule counts to enhance the accuracy of the cell's chemical sensing capabilities.