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Chapter 6: Problem 16

The following problem arises in molecular biology. The surface of a bacterium is supposed to consist of several sites at which foreign molecules - some acceptable and some not-become attached. We consider a particular site and assume that molecules arrive at the site according to a Poisson process with parameter \(\lambda\). Among these molecules a proportion \(\alpha\) is acceptable. Unacceptable molecules stay at the site for a length of time which is exponentially distributed with parameter \(\mu_{1}\), whereas an acceptable molecule remains at the site for an exponential time with rate \(\mu_{2}\). An arriving molecule will become attached only if the site is free of other molecules. What percentage of time is the site occupied with an acceptable (unacceptable) molecule?

Short Answer

Expert verified
The percentage of time the site is occupied by acceptable (\( \% \, Occupied \, by \, Acceptable\)) and unacceptable (\( \% \, Occupied \, by \, Unacceptable\)) molecules can be calculated using the given parameters 位, 伪, 渭鈧�, and 渭鈧� with the following formulas: 1. Calculate the service rates of acceptable (\(\mu_A\)) and unacceptable (\(\mu_U\)) molecules: \[ \mu_A = \alpha \mu_2 \] \[ \mu_U = (1 - \alpha) \mu_1 \] 2. Calculate the overall service rate (\(\mu\)) and probability of the site being occupied (\(P_{occupied}\)): \[ \mu = \mu_A + \mu_U \] \[ P_{occupied} = \frac{\lambda}{\lambda + \mu} \] 3. Calculate the probabilities of being occupied by an acceptable (\(P_A\)) and unacceptable (\(P_U\)) molecule: \[ P_A = P_{occupied} \cdot \frac{\mu_A}{\mu} \] \[ P_U = P_{occupied} \cdot \frac{\mu_U}{\mu} \] 4. Calculate the percentage of occupation time: \[ \% \, Occupied \, by \, Acceptable = P_A \cdot 100 \] \[ \% \, Occupied \, by \, Unacceptable = P_U \cdot 100 \]

Step by step solution

01

Given parameters from the problem: 位 is the arrival rate (Poisson), 伪 is the proportion of acceptable molecules, 渭鈧� is the service rate (exponential) of unacceptable molecules, and 渭鈧� is the service rate (exponential) of acceptable molecules. The variables we need to determine for our solution are: \(P_A\): The probability of the site being occupied by an acceptable molecule. \(P_U\): The probability of the site being occupied by an unacceptable molecule. #step 2: Find probability of the site being occupied#

As the molecules arrivals follow a Poisson process, we know that the probability of the site being occupied can be given by the ratio of the arrival rate to the sum of the arrival and service rates. \[ P_{occupied} = \frac{\lambda}{\lambda + \mu} \] where \[ \mu = \mu_A + \mu_U \] Given the proportion of acceptable molecules, 伪, we can find the service rates of acceptable (\(\mu_A\)) and unacceptable (\(\mu_U\)) molecules: \[ \mu_A = \alpha \mu_2 \] \[ \mu_U = (1 - \alpha) \mu_1 \] #step 3: Calculate the probabilities of being occupied by an acceptable and unacceptable molecule#
02

To find the probabilities of \(P_A\) and \(P_U\), we will multiply the probability of the site being occupied (\(P_{occupied}\)) by the proportions of acceptable and unacceptable molecules' service rates. \[ P_A = P_{occupied} \cdot \frac{\mu_A}{\mu} \] \[ P_U = P_{occupied} \cdot \frac{\mu_U}{\mu} \] #step 4: Calculate the percentage of time the site is occupied by acceptable and unacceptable molecules#

Now we have the probabilities of the site being occupied by acceptable and unacceptable molecules. To find the percentage of occupation time, we simply multiply these probabilities by 100. \[ \% \, Occupied \, by \, Acceptable = P_A \cdot 100 \] \[ \% \, Occupied \, by \, Unacceptable = P_U \cdot 100 \] By applying the formulas in steps 2-4, we can find the required percentage of occupation times for acceptable and unacceptable molecules in terms of given parameters 位, 伪, 渭鈧�, and 渭鈧�.

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Key Concepts

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

Poisson Process
The Poisson process is a stochastic process that models random events occurring over a continuous space or time. It's particularly useful in the context of molecular biology, where events such as molecular attachments to cell surfaces happen at random yet constant average rates.

In a Poisson process, the number of events occurring in a fixed interval of time or space follows a Poisson distribution, which is characterized by the parameter \( \lambda \), known as the rate parameter. This rate remains constant over time, meaning that events occur independently of each other.

Key Characteristics of a Poisson Process

  • Events occur one at a time.
  • The average rate (\(\lambda\)) at which events occur is constant.
  • The number of events occurring in one interval is independent of the number of events occurring in another, non-overlapping interval.
To apply this to the molecular biology problem, we identify the arrival of molecules at a specific site on a bacterium as the 'events' in the Poisson process. The rate at which these molecules arrive is given by \(\lambda\), allowing us to calculate the arrival and occupancy probabilities of these molecules on the bacterial surface.
Exponential Distribution
In probability theory and statistics, the exponential distribution is a continuous probability distribution that is often associated with the time between events in a Poisson process. It is used to model the time taken before an event occurs, given a constant average rate of occurrence.

The exponential distribution is parameterized by the rate \(\mu\), which is the reciprocal of the mean time between occurrences of events in the Poisson process. This distribution has a memoryless property, meaning that the probability of an event occurring in the next instant is always the same, regardless of how much time has already elapsed.

Applying the Exponential Distribution

In molecular biology, when a molecule attaches to a bacterium, the time it remains attached is exponentially distributed with a rate dependent on whether the molecule is acceptable (\(\mu_2\)) or unacceptable (\(\mu_1\)). These rates give us the average occupancy time for each type of molecule, with which we can calculate the probability that the site will be free for the next arriving molecule.
Occupancy Probability
Occupancy probability is the likelihood that a given site, like a receptor on a bacterium's surface, is occupied by a particle or molecule at any given time. This is an essential concept when considering biochemical reactions or molecular binding events on cellular surfaces.

Owing to the randomness of molecular arrivals and binding durations, we can calculate an occupancy probability using the rate of arrival and the rate at which molecules disassociate from the site. These calculations help predict how often the site will be occupied by either acceptable or unacceptable molecules.

Calculating Occupancy Probabilities

Using the rates from our Poisson process (\(\lambda\)) and exponential distributions for acceptable (\(\mu_A\)) and unacceptable (\(\mu_U\)) molecules, the proportion of time a site is occupied by any molecule is derived from the total occupancy probability. It's this probability, in combination with the proportion of acceptable and unacceptable molecules (\(\alpha\)), that gives us the overall occupancy likelihood for each molecule type.

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Most popular questions from this chapter

Four workers share an office that contains four telephones. At any time, cach worker is either "working" or "on the phone." Each "working" period of worker \(i\) lasts for an exponentially distributed time with rate \(\lambda_{i}\). and each "on the phone" period lasts for an exponentially distributed time with rate \(\mu_{i}, i=1,2,3,4\), (a) What proportion of time are all workers "working"? Let \(X_{i}(t)\) equal 1 if worker \(i\) is working at time \(t\), and let it be 0 otherwise. Let \(\mathbf{X}(t)=\left(X_{1}(t), X_{2}(t), X_{3}(t), X_{4}(t)\right)\). (b) Argue that \(\\{\mathbf{X}(t), t \geq 0]\) is a continuous-time Markov chain and give its infinitesimal rates. (c) Is \(\\{\mathbf{X}(t) \mid\) time reversible? Why or why not? Suppose now that one of the phones has broken down. Suppose that a worker who is about to use a phone but finds thern all being used begins a new "working'" period. (d) What proportion of time are all workers "working'r?

Consider a system of \(n\) components such that the working times of component \(i, i=1, \ldots, n\), are exponentially distributed with rate \(\lambda_{i} .\) When failed, however, the repair rate of component \(i\) depends on how many other components are down. Specifically, suppose that the instantaneous repair rate of component \(i, i=1, \ldots, n\), when there are a total of \(k\) failed components, is \(\alpha^{k} \mu_{i}\). (a) Explain how we can analyze the preceding as a continuous-time Markov chain. Define the states and give the parameters of the chain. (b) Show that, in steady state, the chain is time reversible and compute the limiting probabilities.

A total of \(N\) customers move about among \(r\) servers in the following manner. When a customer is served by server \(i\), he then goes over to server \(j, j \neq i\), with probability \(1 /(r-1) .\) If the server he goes to is free, then the customer enters service; otherwise he joins the queue. The service times are all independent, with the service times at server \(i\) being exponential with rate \(\mu_{i}, i=1, \ldots, r\), Let the state at any time be the vector \(\left(n_{1}, \ldots, n_{0}\right)\), where \(n_{i}\) is the number of customers presently at server \(l, i=1, \ldots, r, \Sigma_{i} n_{i}=N\). (a) Argue that if \(X(t)\) is the state at time \(t\), then \(\mid X(t), t \geq 0\\}\) is a continuous-time Markov chain. (b) Give the infinitesimal rates of this chain. (c) Show that this chain is time reversible, and find the limiting probabilities.

There are two machines, one of which is used as a spare. A working machine will function for an exponential time with rate \(\lambda\) and will then fail. Upon failure, it is immediately replaced by the other machine if that one is in working order, and it goes to the repair facility. The repair facility consists of a single person who takes an exponential time with rate \(\mu\) to repair a failed machine. At the repair facility, the newly failed machine enters service if the repairperson is free. If the repairperson is busy, it waits until the other machine is fixed. At that time, the newly repaired machine is put in service and repair begins on the other one. Starting with both machines in working condition, find (a) the expected value and (b) the variance of the time until both are in the repair facility. (c) In the long run, what proportion of time is there a working machine?

Fach individual in a biological population is assumed to give birth at an exponential rate \(\lambda\), and to die at an exponential rate \(\mu\). In addition, there is an exponential rate of increase \(\theta\) due to immigration. However, immigration is not allowed when the population size is \(N\) or larger. (a) Set this up as a birth and death model. (b) If \(N=3,1=\theta=\lambda, \mu=2\), determine the proportion of time that immigration is restricted.
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