Chapter 1.10, Problem 31E

a

To determine

To calculate: To find the number of wild-type bacteria that mutates using the given information

Answer to Problem 31E

The number of wild-type bacteria that mutate is $0.1b_t$

Explanation of Solution

Given information: Suppose that a fraction 0.1 of wild-type mutate each generation, but that each wild-type individual produces 2.0 offspring while each mutant produces only 1.5 offspring.

Calculation:

Initially, let there be $b_t$ , wild-type and $m_t$ , mutants. Suppose that a fraction 0.1 of wild-type mutate each generation, but that each wild-type individual produces 2.0 offspring while each mutant produces only 1.5 offspring.

Now a fraction 0.1 of wild-type mutate each generation.

Therefore, the number of wild-type bacteria that mutate is $0.1b_t$

b

To determine

To calculate: To find the number of wild-type bacteria and the number of mutants after mutation using the given information

Answer to Problem 31E

The number of mutants after mutation is given by $m_t+0.1b_t$

Explanation of Solution

Given information: Suppose that a fraction 0.1 of wild-type mutate each generation, but that each wild-type individual produces 2.0 offspring while each mutant produces only 1.5 offspring.

Calculation:

Now the number of wild type bacteria after mutation is given by

  $b_t-0.1b_t=0.9b_t$

Also, the number of mutants after mutation is given by

  $m_t+0.1b_t$

c

To determine

To calculate: To find the number of wild-type bacteria and the number of mutants after reproduction using the given information

Answer to Problem 31E

The number of mutants after reproduction shall be

  $\begin{array}{l}{m}_{t+1}=\left(m_t+0.1b_t\right)\times 1.5\\ m_{t+1}=1.5m_t+0.15b_t\end{array}$

Explanation of Solution

Given information: Suppose that a fraction 0.1 of wild-type mutate each generation, but that each wild-type individual produces 2.0 offspring while each mutant produces only 1.5 offspring.

Calculation:

Now, each wild-type individual produces 2.0 offspring.

Therefore, the number of wild-type bacteria after reproduction is given by

  $\begin{array}{l}{b}_{t+1}=\left(0.9b_t\right)\times 2\\ b_{t+1}=1.8b_t\end{array}$

Also, each mutant produces only 1.5 offspring.

Therefore, the number of mutants after reproduction shall be

  $\begin{array}{l}{m}_{t+1}=\left(m_t+0.1b_t\right)\times 1.5\\ m_{t+1}=1.5m_t+0.15b_t\end{array}$

d

To determine

To calculate: To find the total number of bacteria after mutation and reproduction using the given information

Answer to Problem 31E

The total number of bacteria after mutation and reproduction is given by

  $1.8b_t+\left(1.5m_t+0.15b_t\right)=1.95b_t+1.5m_t$

Explanation of Solution

Given information: Suppose that a fraction 0.1 of wild-type mutate each generation, but that each wild-type individual produces 2.0 offspring while each mutant produces only 1.5 offspring.

Calculation:

The total number of bacteria after mutation and reproduction is given by

  $1.8b_t+\left(1.5m_t+0.15b_t\right)=1.95b_t+1.5m_t$

e

To determine

To calculate: To find the fraction of mutants after mutation and reproduction using the given information

Answer to Problem 31E

  $p_{t+1}=\frac{1.5p_t+0.15\left(1-p_t\right)}{1.5p_t+0.95\left(1-p_t\right)}$

Explanation of Solution

Given information: Suppose that a fraction 0.1 of wild-type mutate each generation, but that each wild-type individual produces 2.0 offspring while each mutant produces only 1.5 offspring.

Calculation:

To find the fraction of mutants after mutation and reproduction, divide the discrete-time dynamical system for $m_{t+1}$ by $b_{t+1}+m_{t+1}$

Therefore, we have

  $\begin{array}{l}{p}_{t+1}=\frac{m_{t+1}}{b_{t+1}+m_{t+1}}\\ =\frac{1.5m_t+0.15b_t}{1.8b_t+\left(1.5m_t+0.15b_t\right)}\\ =\frac{1.5m_t+0.15b_t}{1.95b_t+1.5m_t}\end{array}$

  $\begin{array}{l}\frac{\frac{1.5m_t}{m_t+b_t}+\frac{0.15b_t}{m_t+b_t}}{\frac{1.95b_t}{m_t+b_t}+\frac{1.5m_t}{m_t+b_t}}\\ p_{t+1}=\frac{1.5p_t+0.15\left(1-p_t\right)}{1.5p_t+0.95\left(1-p_t\right)}\end{array}$

f

To determine

To calculate: To find the equilibrium using the obtained values

Answer to Problem 31E

  $p^{•}=1/3\text{or}{p}^{•}=1.0$

Explanation of Solution

Given information: Suppose that a fraction 0.1 of wild-type mutate each generation, but that each wild-type individual produces 2.0 offspring while each mutant produces only 1.5 offspring.

Calculation:

Let $p^{•}$ the equilibrium

To find the equilibrium consider,

  $p^{•}=\frac{1.5p^{•}+0.15\left(1-p^{•}\right)}{1.5p^{•}+0.95\left(1-p^{•}\right)}$

Solving the above equation gives,

  $\begin{array}{l}{p}^{•}\left[1.5p^{•}+1.95\left(1-p^{•}\right)\right]=1.5p^{•}+0.15\left(1-p^{•}\right)\\ 1.5{\left(p^{•}\right)}^2+1.95p^{•}-1.95{\left(p^{•}\right)}^2-1.5p^{•}-0.15+0.15p^{•}=0\\ 0.45{\left(p^{•}\right)}^2-0.6p^{•}+0.15=0\\ 0.45\left(p^{•}-1/3\right)\left(p^{•}-1.0\right)=0\end{array}$

Therefore,

  $p^{•}=1/3\text{or}{p}^{•}=1.0$

g

To determine

To calculate: To determine and to find whether the equilibrium is stable or not using the obtained information

Answer to Problem 31E

The equilibrium at l/ 3 seems to be stable. The fraction of mutants will end up at about 33.33 percent.

Explanation of Solution

Given information: Suppose that a fraction 0.1 of wild-type mutate each generation, but that each wild-type individual produces 2.0 offspring while each mutant produces only 1.5 offspring.

Calculation:

The cobweb starting from the initial condition is as shown below:

  

The equilibrium at l/ 3 seems to be stable. The fraction of mutants will end up at about 33.33 percent.