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Chapter 2: Problem 11

Write down a formula for the population size, \(N_{t}\), as a function of time, \(t\). 11\. A strain of bacteria reproduces asexually every hour. That is, every hour, each bacterial cell splits into two cells. If, initially, there is one bacterium, find the number of bacterial cells after 1 hour, 2 hours, 3 hours, 4 hours, and 5 hours.

Short Answer

Expert verified
The population size formula is \( N_t = 2^t \). After 1, 2, 3, 4, and 5 hours, the populations are 2, 4, 8, 16, and 32, respectively.

Step by step solution

01

Understand the Problem

The bacteria replicate by splitting into two every hour. Initially, there is one bacterium at time 0. We want to find the number of bacteria after each hour up to 5 hours. This describes an exponential growth pattern.
02

Define the Formula Structure

Since each bacterium splits into two every hour, the number of bacteria doubles each hour. The general formula for exponential growth can be written as: \[ N_t = N_0 imes (r)^t \]where \( N_t \) is the population size at time \( t \), \( N_0 \) is the initial population size, and \( r \) is the growth factor per time unit. In this case, \( r = 2 \).
03

Formulate the Specific Expression

Given \( N_0 = 1 \) (the initial number of bacteria) and \( r = 2 \) (the growth factor as each bacterium doubles every hour), we can specifically write:\[ N_t = 1 imes 2^t \]Hence, \( N_t = 2^t \).
04

Calculate Population Size at Each Time Point

Using the formula \( N_t = 2^t \), calculate for each hour:- After 1 hour: \( N_1 = 2^1 = 2 \)- After 2 hours: \( N_2 = 2^2 = 4 \)- After 3 hours: \( N_3 = 2^3 = 8 \)- After 4 hours: \( N_4 = 2^4 = 16 \)- After 5 hours: \( N_5 = 2^5 = 32 \)

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Key Concepts

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

Bacterial Reproduction
Bacterial reproduction, particularly in species that reproduce asexually, is a fascinating process that involves a single organism creating a copy of itself. In the context of our example, the bacteria replicate by splitting into two every hour, which is a method known as binary fission.
During binary fission:
  • A single bacterium, the parent cell, doubles its genetic material.
  • It divides into two genetically identical offspring cells.
  • These daughter cells grow independently and prepare to divide again in the next cycle.
This cycle can repeat very rapidly, which is why bacterial populations can explode into large numbers in a short period. This rapid increase is a classic example of exponential growth, characterized by a constant doubling period.
Population Dynamics
Population dynamics is the study of how and why the size of populations changes over time. In the exponential growth of bacteria, we observe a simple and direct model of population dynamics.
When studying bacterial growth, the initial population size and growth rate play crucial roles. With our bacteria example:
  • The initial number of bacteria: 1 bacterium
  • Time taken for each reproduction cycle: 1 hour
  • Reproduction rate: doubling every hour
These variables help us understand how the bacterial population multiplies. As the bacteria reproduce, the population size at each time point can be calculated using the formula for exponential growth. The larger the initial population or the faster the generation time (shorter cycles), the quicker the population size will expand.
Growth Factor
The growth factor is a key concept in understanding exponential growth. It describes how much the population increases during each time period. In our bacteria example, the growth factor is 2, meaning the population doubles every hour.
For bacterial growth:
  • The growth factor remains constant over time.
  • Each hour, the existing number of bacteria is multiplied by the growth factor of 2.
  • This consistent multiplication leads to a predictable exponential increase in population size.
Because of the constant growth factor, predicting bacterial population using the formula \( N_t = N_0 \times r^t \) becomes straightforward. Here, \( N_0 \) is the initial size, \( t \) is time, and \( r \) is the growth factor. This formula gives us a powerful tool to predict population sizes without needing to manually count each bacterial cell as time progresses.

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Most popular questions from this chapter

Investigate the behavior of the discrete logistic equation $$ x_{t+1}=R_{0} x_{t}\left(1-x_{t}\right) $$ Compute \(x_{t}\) for \(t=0,1,2, \ldots, 20\) for the given values of \(r\) and \(x_{0}\), and graph \(x_{t}\) as a function of \(t .\) \(R_{0}=3.8, x_{0}=0.1\)

A drug has zeroth order elimination kinetics. At time \(t=0\) an amount \(a_{0}=20 \mathrm{mg}\) is present in the blood. One hour later, at \(t=1\), an amount \(a_{1}=14 \mathrm{mg}\) is present. (a) Assuming that no drug is added to the blood between \(t=0\) and \(t=1\), calculate the amount of drug that is removed from the blood each hour. (b) Write a recursion relation for the amount of drug \(a_{t}\) that is present at time \(t\). Assume no extra drug is added to the blood. (c) Find an explicit formula for \(a_{t}\) as a function of \(t\). (d) When does the amount of drug present in the blood first drop \(\operatorname{to} 0 ?\)

Tylenol in the Body A patient is taking Tylenol (a painkiller that contains acetaminophen) to treat a fever. The data in this question is taken from Rawlins, Henderson, and Hijab (1977). At \(t=0\) the patient takes their first pill. One hour later the drug has been completely absorbed and the blood concentration, measured in \(\mu \mathrm{g} / \mathrm{ml}\), is 15 . Acetaminophen has first order elimination kinetics; in one hour, \(23 \%\) of the acetaminophen present in the blood is eliminated. (a) Write a recursion relation for the concentration \(c_{t}\) of drug in the patient's blood. For \(t \geq 1\) you may assume for now that no other pills are taken after the first one. (b) Find an explicit formula for \(c_{t}\) as a function of \(t\). (c) Suppose that the patient follows the directions on the pill box and takes another Tylenol pill 4 hours after the first (at time \(t=4\) ). What is the concentration at the time at which the second pill is taken? In others words, what is \(c_{4}\) ? (d) Over the next hour \(15 \mathrm{\mug} / \mathrm{ml}\) of drug enter the patient's bloodstream. So, \(c_{5}\) can be calculated from \(c_{4}\) using the word equation: $$ c_{5}=c_{4}+ $$ nt added \(\quad\) amount eliminated blo Given that the amount added is \(15 \mu \mathrm{g} / \mathrm{ml}\), and the amount eliminated is \(0.23 \cdot c_{4}\), calculate \(c_{5} .\) (e) For \(t=5,6,7,8\) the drug continues to be eliminated at a rate of \(23 \%\) per hour. No pills are taken and no extra drug enters the patient's blood. Compute \(c_{8}\). (f) At time \(t=8\), the patient takes another pill. Calculate \(c_{9} .\) Do not forget to include elimination of drug between \(t=8\) and \(t=9\). (g) We want to calculate the maximum concentration of drug in the patient's blood. We know that concentrations are highest in the hour after a pill is taken, namely at time \(t=1, t=5, t=\) \(9, \ldots\) Define a sequence \(C_{n}\) representing the concentration of the drug one hour after the \(n\) th pill is taken. (h) What terms of the original sequence \(\left\\{c_{r}: t=1,2, \ldots\right\\}\) are \(C_{1}\), \(C_{2}\), and \(C_{3} ?\) (i) Explain why $$ C_{n+1}=(0.77)^{4} \cdot C_{n}+15 $$ and \(c_{1}=15\) (j) From the recursion relation, assuming that the patient continues to take Tylenol pills at 4 -hour intervals, calculate \(C_{1}, C_{2}\), \(C_{3}, C_{4}, C_{5}\), and \(C_{6}\) (k) Does \(C_{n}\) increase indefinitely, or do you think that it converges? (1) By looking for fixing point of the recursion relation in (h), find the limit of \(C_{n}\) as \(n \rightarrow \infty\).

\(\lim _{n \rightarrow \infty} a_{n}=a\). Find the limit \(a\), and determine \(N\) so that \(\left|a_{n}-a\right|<\epsilon\) for all \(n>N\) for the given value of \(\epsilon\) $$ a_{n}=\ln \left(1+\frac{1}{n}\right), \epsilon=0.1 $$

In Problems \(73-78\), use the formal definition of limits to show that \(\lim _{n \rightarrow \infty} a_{n}=a ;\) that is, find \(N\) such that for every \(\epsilon>0\), there exists an \(N\) such that \(\left|a_{n}-a\right|<\epsilon\) whenever \(n>N\). $$ \lim _{n \rightarrow \infty} \frac{3}{n}=0 $$
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