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A population of bacteria grows exponentially and has mean time between divisions \(t_{b}=2\) hours. Assume that cell death can be ignored (that is, \(m=0\) ). (a) Sketch on the same axes (of \(N(t)\) against \(t\) ) the size of the population over the interval \(0

Short Answer

Expert verified
The wild type population grows faster than the mutant strain on both linear and semilogarithmic graphs.

Step by step solution

01

Understanding exponential growth

The population of bacteria grows exponentially, which can be described by the equation \( N(t) = N_0 \cdot 2^{t/t_b} \), where \( N_0 \) is the initial number of cells and \( t_b = 2 \) hours is the mean time between divisions.
02

Calculate Population at Different Times for Initial Conditions

Let's calculate the population \( N(t) \) at \( t = 0, 2, 4, 6 \) hours for both initial conditions:- For \( N_0 = 1000 \):\[ N(0) = 1000, \quad N(2) = 1000 \cdot 2^{2/2} = 2000, \quad N(4) = 1000 \cdot 2^{4/2} = 4000, \quad N(6) = 1000 \cdot 2^{6/2} = 8000. \]- For \( N_0 = 3000 \):\[ N(0) = 3000, \quad N(2) = 3000 \cdot 2^{2/2} = 6000, \quad N(4) = 3000 \cdot 2^{4/2} = 12000, \quad N(6) = 3000 \cdot 2^{6/2} = 24000. \]
03

Sketch the Exponential Growth on Linear Scale

On a coordinate plane with \( N(t) \) on the y-axis and \( t \) (hours) on the x-axis, plot the points and draw smooth curves for the two populations:- Population starting with \( N_0 = 1000 \) will have points at (0, 1000), (2, 2000), (4, 4000), and (6, 8000).- Population starting with \( N_0 = 3000 \) will have points at (0, 3000), (2, 6000), (4, 12000), and (6, 24000).
04

Sketch the Graph on Semilogarithmic Axes

In semilogarithmic axes, the y-axis has a logarithmic scale (log base 10). Re-plot the populations:- For \( N_0 = 1000 \), points become (0, log(1000)), (2, log(2000)), (4, log(4000)), and (6, log(8000)).- For \( N_0 = 3000 \), points become (0, log(3000)), (2, log(6000)), (4, log(12000)), and (6, log(24000)).On the graph, the lines will be straight, showing the exponential growth as a linear curve.
05

Adjusting to Mutant Bacteria Growth Rate

The mutant strain divides two times slower, so the mean time between divisions \( t_b = 4 \) hours. The exponential growth formula becomes \( N(t) = N_0 \cdot 2^{t/4} \).For \( N_0 = 1000 \):\[ N(0) = 1000, \quad N(2) = 1000 \cdot 2^{2/4} \approx 1414, \quad N(4) = 1000 \cdot 2^{4/4} = 2000, \quad N(6) = 1000 \cdot 2^{6/4} \approx 2828. \]
06

Sketch Wild Type vs. Mutant on Linear Scale

Plot on the same axes as before:- Wild type \( N_0 = 1000 \): points at (0, 1000), (2, 2000), (4, 4000), and (6, 8000).- Mutant \( N_0 = 1000 \): points at (0, 1000), (2, 1414), (4, 2000), and (6, 2828).Compare the steeper curve of the wild type with the flatter curve of the mutant.
07

Sketch on Semilogarithmic Axes for Mutant

Repeat Step 4 on semilogarithmic axes:- Wild type \( N_0 = 1000 \): points (0, log(1000)), (2, log(2000)), (4, log(4000)), and (6, log(8000)).- Mutant \( N_0 = 1000 \): points (0, log(1000)), (2, log(1414)), (4, log(2000)), and (6, log(2828)).Again, lines are straight with a gentler slope for mutants.

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Key Concepts

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

Bacterial Population
Bacteria are fascinating microorganisms that reproduce through a process known as binary fission. This means a single bacterium splits into two identical cells. In an ideal environment, free from constraints like nutrient shortage or competition, a bacterial population can grow exponentially. This is described by the equation \( N(t) = N_0 \cdot 2^{t/t_b} \), which calculates the number of bacteria \( N(t) \) at a given time \( t \). In this equation:
  • \( N_0 \) is the initial number of bacteria, provided as the starting point of the population.
  • \( t_b \) denotes the mean time between cell divisions—here it is 2 hours.
A shorter \( t_b \) means the population doubles more quickly, while a longer \( t_b \) would indicate slower growth. This exponential growth can result in explosive increases in population size over short periods, especially in ideal conditions where cell death (\( m=0 \)) is ignored.
Semilogarithmic Graph
When dealing with exponential growth, plotting on a semilogarithmic graph provides a clearer perspective. Normally, exponential functions curve upwards steeply on linear axes, making it hard to compare or predict values. However, using a semilogarithmic graph, where the y-axis (population size \( N(t) \)) is scaled logarithmically, exponential curves become straight lines. To sketch a semilogarithmic graph, you plot:
  • \( t \) on the x-axis, representing time.
  • The logarithm (log base 10) of the cell number on the y-axis. For example, for \( N(t) = 1000 \), you'd plot log(1000) about 3.
The relationship remains linear, allowing different populations to be compared easily regardless of their initial size. It helps better visualize growth and compare wild types and mutant strains by the slope of the lines. The steeper the slope, the faster the growth rate.
Mutation Rate
A mutation is a change in a bacterium's DNA, potentially affecting its division rate. In our scenario, the mutant bacterial strain divides two times slower than its wild type counterpart. The division rate slows down because the mutant's mean time between divisions \( t_b \) increases from 2 to 4 hours.This change can be reflected in the modified exponential growth formula:\[ N(t) = N_0 \cdot 2^{t/4} \]Where:
  • The wild type remains at \( t_b = 2 \) hours for quicker division.
  • The mutant bacterium grows slower, doubling every 4 hours.
Understanding mutation rates is key in biological studies, as they explain variations in growth rates. In practical terms, slower-growing strains may adapt differently to environments due to changes in selection pressures associated with mutations. Thus, mutation rates significantly influence ecological balances and evolutionary dynamics in microbial communities.

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