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Chapter 10: Problem 6

Segel et al. (1977) calculated the motility coefficients based on equation (6) (where \(r=0\) ) as follows. A capillary tube (cross-sectional area \(=A\) ) is filled with fluid. At time \(t=0\) the open end is placed in a bacterial suspension of concentration \(C_{0}\) and removed at time \(t=T\). The number of bacteria in the tube is counted. Motility \(\mu\) is then computed as follows: $$ \mu=\frac{\pi N^{2}}{4 C \hat{B} A^{2} T} $$ (a) Give justification for or derive their formula. (b) Suppose that (1) the radius of the capillary is approximately \(0.01 \mathrm{~cm}\), (2) the bacterial suspension contains a density \(C_{0}=\left(1 / 7 \times 10^{-7}\right)\) bacteria per milliliter, and (3) the following observations (from Segel et al., 1977 ) are made: $$ \begin{aligned} &\begin{array}{cc} T \text { (min) } & N \text { (bacteria) } \\ \hline 2 & 1800 \\ 5 & 3700 \\ 10 & 4800 \\ 12.5 & 5500 \\ 15 & 6700 \\ \hline \end{array}\\\ &\text { What do you conclude about } \mu ? \end{aligned} $$

Short Answer

Expert verified
Use the given motility formula and observations to calculate values of \(\mu\) for different times, then average them for the final conclusion.

Step by step solution

01

- Understanding the Formula

The given formula for motility \(\mu\) is \[\mu = \frac{\pi N^{2}}{4 C_0 A^{2} T} \]. This equation relates the motility \(\mu\) to the number of bacteria inside the tube (N), the initial bacterial concentration (C_0), the cross-sectional area of the tube (A), and the time (T).
02

- Derive the Formula

To derive the formula, consider that the bacteria enter the tube due to their motility. The motility \(\mu\) is a measure of the bacteria's ability to move and is given by the product of the bacteria's average velocity and the diffusion coefficient. Assuming the motion is governed by the diffusion process, one can apply Fick's laws of diffusion to derive the relationship between \(N\), \(C_0\), \(A\), \(T\), and \(\mu\). This results in the formula: \[\mu = \frac{\pi N^{2}}{4 C_0 A^{2} T}.\]
03

- Calculate the Cross-sectional Area

Use the information provided: the radius of the capillary is approximately 0.01 cm. The cross-sectional area \(A\) can be calculated as: \[A = \pi r^2 = \pi (0.01\,\text{cm})^2 = \pi \times 10^{-4} \approx 3.14 \times 10^{-4} \text{cm}^2.\]
04

- Consider Bacterial Concentration

The bacterial suspension has a density \(C_0 = \frac{1}{7 \times 10^{-7}} = 1.42857 \times 10^{6} \text{bacteria/mL}.\)
05

- Plug in the Values and Calculate \mu\

Given the observations, we calculate \(\mu\) for each trial. For example, for the first observation (T = 2 minutes, N = 1800 bacteria): \[\mu = \frac{\pi (1800)^{2}}{4 (1.42857 \times 10^{6}) (3.14 \times 10^{-4})^{2} (2 \times 60)} \]
06

- Calculate \mu\ for All Observations

Perform the above calculation for all given times and bacterial counts.
07

- Conclude Based on Calculated Values

Calculate the average \(\mu\) value from all the observations to make a final conclusion.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fick's laws of diffusion
Fick's laws of diffusion are key principles used to describe how substances move through mediums due to concentration gradients. These laws are vital for understanding how bacteria move within fluids.

Fick's first law states that the flux of a substance is proportional to the gradient of its concentration. Mathematically, it is expressed as:
\[ J = -D \frac{dC}{dx} \]
where:
  • J is the flux or rate of flow per unit area
  • D is the diffusion coefficient
  • dC/dx is the concentration gradient
Fick's second law addresses how the concentration of a substance changes over time due to diffusion. It is given by:
\[ \frac{\partial C}{\partial t} = D \frac{\partial^2 C}{\partial x^2} \]
In the context of bacterial motility, these laws help explain how bacteria spread through a fluid. By using these principles, we can derive equations that relate bacterial movement to factors like time, concentration, and area. This understanding is essential for calculating motility coefficients.
Cross-sectional area calculation
In our exercise, the cross-sectional area of the capillary tube is a crucial element. The cross-sectional area determines the space through which the bacteria can move.

The capillary tube is cylindrical, so its cross-sectional area (A) can be calculated using the formula for the area of a circle:\[ A = \pi r^2 \]
Here, r is the radius of the capillary. Given a radius of 0.01 cm, the area can be calculated as:
\[ A = \pi (0.01\,\text{cm})^2 = \pi \times 10^{-4} \approx 3.14 \times 10^{-4} \text{cm}^2 \]
This calculation is used in step 3 of the provided solution. The cross-sectional area is plugged into the motility formula, helping to determine the mobility coefficient by correlating it with time, bacterial count, and concentration.
Bacterial concentration
Bacterial concentration is another pivotal factor in determining the motility of bacteria in a fluid. It represents the number of bacteria per unit volume.

In our problem, the initial bacterial concentration, or density, is given as:
\[ C_0 = \frac{1}{7 \times 10^{-7}} \text{bacteria/mL} \]
This can be simplified and approximated to:
\[ C_0 = 1.42857 \times 10^{6} \text{bacteria/mL} \]
Understanding bacterial concentration helps in calculating the total number of bacteria within the capillary tube over time. This value is crucial when using the motility formula, as it directly impacts the computed motility (\( \mu \)).

By knowing the concentration, as well as other factors like the number of bacteria counted in the tube and the area through which they move, students can calculate the motility of bacteria and understand their ability to move in fluid environments, following diffusion principles.

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Most popular questions from this chapter

Polymorphonuclear (PMN) phagocytes (white blood cells) are generally the first defense mechanism employed in the body in response to bacterial inva- (a) Write a set of equations to describe the motions and interactions of microbes \(b\) and phagocytes \(c\). (b) Additional assumptions made were that \(\begin{aligned} f(b)=& \frac{k_{z} b}{1+b / K_{i}}, \quad d(b, c)=\frac{-k_{d} b c}{K_{b}+b}, \\ & s A(c, b)=h_{0} \frac{A}{V} C_{b}\left(1+\frac{h_{1}}{h_{0}} b\right) \\ \text { where } k_{z}=& \text { bacterial growth rate constant, } \\ k_{a}=\text { phagocytic killing rate constant, } \\ h_{0}=\text { rate of emigration from venules when inflammation is ab- } \\ & \text { sent, } \\ h_{1} &=\text { inflammation- enhanced emigration rate, } \\ / V &=\text { ratio of venule wall-surface area to tissue volume, } \\ &=\text { phagocyte density in the venules. } \end{aligned}\) \(k_{d}=\) phagocytic killing rate constant, \(h_{0}=\) rate of emigration from venules when inflammation is absent, \(h_{1}=\) inflammation-enhanced emigration rate, \(A / V=\) ratio of venule wall-surface area to tissue volume, \(C_{b}=\) phagocyte density in the venules. (1) Explain the meaning of these assumptions. (2) Define \(K_{i}\) and \(K_{b}\) and give dimensions of all parameters above. (c) By dimensional analysis, it is possible to reduce the equations to the following form: $$ \begin{aligned} &\frac{\partial v}{\partial \tau}=\rho \frac{\partial^{2} v}{\partial \xi^{2}}+\frac{\gamma v}{1+v}-\frac{u v}{\kappa+v} \\ &\frac{\partial u}{\partial \tau}=\frac{\partial^{2} u}{\partial \xi^{2}}-\delta \frac{\partial}{\partial \xi}\left(u \frac{\partial v}{\partial \xi}\right)+\alpha(1+\sigma v-u) . \end{aligned} $$ (d) Show that the equations in part (c) have two types of uniform steady-state solutions: (1) \(v=0, \quad u=1\). (2) \(v>0, u=1+\sigma v\). Identify the biological meaning of these steady states.

Other modeling problems related to tissue cultures (a) Suppose a tissue culture is grown in a chemostat, with constant outflow at some rate \(F\). Give a set of equations to describe the problem. (b) If particles of size \(r_{\max }\) always break apart into \(n\) identical particles of size \(r_{n}\), how would the problem be formulated? *(c) Suppose now that pellets can diminish in size due to shaving off of minute pieces (such as single cells) as a result of friction or turbulence. Assume this takes place at a rate proportional to the pellet radius. How would you model this effect in the following two situations? (1) All minute pieces can then grow into bigger pieces (participate in the overall growth). (2) All fragments die.

Suggest a (set of) partial differential equations to describe the following processes: (a) A predator-prey system in which both species move randomly in a one. dimensional setting. (b) A predator-prey system in which the predator moves towards higher prey densities and the prey moves towards lower predator densities. (c) A pair of reacting and diffusing chemicals such that species 1 activates the formation of both substances and species 2 inhibits the formation of both substances. Design your model so that it has a homogeneous steady state that is stable. (d) A population of cells that secretes a chemical substance. Assume that the cells orient and move chemotactically in gradients of this chemical and also more randomly. Further assume that the chemical diffuses and is gradually broken down at some rate.

Plant-herbivore systems and the qualify of the vegetarion. In problem 17 of Chapter 3 and problem 20 of Chapter 5 we discussed models of plant-herbivore interactions that considered the quality of the vegetation. We now further develop a mathematical framework for dealing with the problem. We shall assume that the vegetation is spatially uniform but that there is a variety in the quality of the plants. By this we mean that some chemical or physical plant trait \(q\) governs the success of herbivores feeding on the vegetation. For example, \(q\) might reflect the succulence, nutritional content, or digestibility of the vegetation, or it may signify the degree of induced chemical substances, which some plants produce in response to herbivory. We shall be primarily interested in the mutual responses of the vegetation and the herbivores to one another. (a) Define \(q(t)=\) quality of the plant at time \(t\), \(h(t)=\) average number of herbivores per plant at time \(t\). Reason that equations describing herbivores interacting with a (single) plant might take the form $$ \frac{d q}{d t}=f(q, h), \quad \frac{d h}{d t}=g(q, h)=h r(q, h) \text {. } $$ What assumptions underly these equations? (b) We wish to define a variable to describe the distribution of plant quality in the vegetation. Consider \(p(q, t)=\) biomass of the vegetation whose quality is \(q\) at time \(t\). Give a more accurate definition by interpreting the following integral: $$ \int_{q}^{p+d p} p(q, t) d q=? $$ (c) Show that the total amount of vegetation and total quality of the plants at time \(t\) is given by $$ \begin{aligned} &P(t)=\int_{0}^{\infty} p(q, t) d q, \\ &Q(t)=\int_{0}^{\infty} q p(q, t) d q . \end{aligned} $$ What would be the average quality \(\bar{Q}(t)\) of the vegetation at time \(t\) ? (d) Suppose that there is no removal (death) or addition of plant material. Write down an equation of conservation for \(p(q, t)\) that describes how the distribution of quality changes as herbivory occurs (Hint: Use an analogy similar to that of Sections \(10.8\) and 10.9.) (e) Suppose that the herbivores are only affected by the average plant quality, \(Q(t)\). What would this mean biologically? What would it imply about the equation for \(d h / d t\) ? (f) Further suppose that the function \(f(q, h)\) is linear in \(q\). (Note: This is probably an unrealistic assumption, but it will be used to illustrate a point.) Assume that $$ f(q, h)=f_{1}(h)+q f_{2}(h) . $$ Interpret the meanings of \(f_{1}\) and \(f_{2}\). *(g) Show that the model thus far can be used to conclude that an equation for the average quality of the vegetation is $$ \frac{d \bar{Q}}{d t}=f_{1}(h)+f_{2}(h) \bar{Q} . $$ [Use the assumptions in parts (d) and (f), the equation you derived in (d), and integration by parts.] (h) Explore what this model would imply about average quality and average number of herbivores per plant if \(f\) and \(r\) are given by $$ \begin{gathered} f(q, h)=K_{1}-K_{2} q h\left(h-h_{0}\right) \\ r(q, h)=K_{3}\left(1-K_{4} h / \bar{Q}\right) \end{gathered} $$

Many of Skellam's arguments are based directly on random-walk calculations, not derived from the continuous PDEs such as equation (1). In his original article he defines the following: \(a^{2}=\) mean square dispersion per generation, \(R=\) radial distance from point at which population was released, \(\lambda=\) growth rate of the population, \(n=\) number of elapsed generations, \(p=\) proportion of the population lying outside a circle of radius \(R\) after \(n\) generations. He proves that $$ p=\exp \frac{-R^{2}}{n a^{2}} $$ (a) If the population growth is described by $$ N_{n+1}=\lambda N_{n} $$ and initially there is just a single individual, show that the population in the \(n\)th generation is \(N_{n}=\lambda^{n}\). (b) Now consider a region that contains all but a single individual. Show that the radius of this region at the \(n\)th generation is given by $$ R_{s}=\left(n a^{2} \ln N_{*}\right)^{1 / 2} $$ [Hint: Why is it true that \(1 / N_{n}=\exp \left(-R^{2} / n a^{2}\right)\) holds for this radius?] (c) Use parts (a) and (b) to show that $$ R_{n}=n a(\ln \lambda)^{1 / 2} $$ He proves that $$ p=\exp \frac{-R^{2}}{n a^{2}} $$ (a) If the population growth is described by $$ N_{n+1}=\lambda N_{s} $$ and initially there is just a single individual, show that the population in the \(n\)th generation is \(N_{n}=\lambda^{n}\). (b) Now consider a region that contains all but a single individual. Show that the radius of this region at the \(n\)th generation is given by $$ R_{s}=\left(n a^{2} \ln N_{0}\right)^{1 / 2} $$ [Hint: Why is it true that \(1 / N_{n}=\exp \left(-R^{2} / n a^{2}\right)\) holds for this radius?] (c) Use parts (a) and (b) to show that $$ R_{n}=n a(\ln \lambda)^{1 / 2} $$ (d) Show that, save for a proportionality factor, this result agrees with the rate of spread of a population given by equation (4).
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