Chapter 1.10, Problem 32E

a

To determine

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

Answer to Problem 32E

The number of wild-type bacteria that mutate is $0.2b_t$

Explanation of Solution

Given information: Suppose that a fraction 0.2 of wild-type mutate each generation, but that each wild-type individual produces 2.0 offspring while each mutant produces only 1.0 offspring.

Calculation:

Initially, let there be $b_t$ , wild-type and $m_t$ , mutants. Suppose that a fraction 0.2 of wild-type mutate each generation, but that each wild-type individual produces 2.0 offspring while each mutant produces only 1.0 offspring.

Now a fraction 0.2 of wild-type mutate each generation.

Therefore, the number of wild-type bacteria that mutate is $0.2b_t$

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 32E

The number of mutants after mutation is given by $m_t+0.2b_t$

Explanation of Solution

Given information: Suppose that a fraction 0.2 of wild-type mutate each generation, but that each wild-type individual produces 2.0 offspring while each mutant produces only 1.0 offspring.

Calculation:

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

  $b_t-0.2b_t=0.8b_t$

Also, the number of mutants after mutation is given by

  $m_t+0.2b_t$

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 32E

Therefore, the number of mutants after reproduction shall be

  $\begin{array}{l}{m}_{t+1}=\left(m_t+0.2b_t\right)\times 1.0\\ m_{t+1}=m_t+0.2b_t\end{array}$

Explanation of Solution

Given information: Suppose that a fraction 0.2 of wild-type mutate each generation, but that each wild-type individual produces 2.0 offspring while each mutant produces only 1.0 offspring.

Calculation:

Now, each wild-type individual produces 2.0 offspring.

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

  $\begin{array}{l}{b}_{t+1}=\left(0.8b_t\right)\times 2\\ b_{t+1}=1.6b_t\end{array}$

Also, each mutant produces only 1.5 offspring.

Therefore, the number of mutants after reproduction shall be

  $\begin{array}{l}{m}_{t+1}=\left(m_t+0.2b_t\right)\times 1.0\\ m_{t+1}=m_t+0.2b_t\end{array}$

d

To determine

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

Answer to Problem 32E

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

  $1.6b_t+\left(m_t+0.2b_t\right)=1.8b_t+m_t$

Explanation of Solution

Given information: Suppose that a fraction 0.2 of wild-type mutate each generation, but that each wild-type individual produces 2.0 offspring while each mutant produces only 1.0 offspring.

Calculation:

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

  $1.6b_t+\left(m_t+0.2b_t\right)=1.8b_t+m_t$

e

To determine

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

Answer to Problem 32E

  $p_{t+1}=\frac{p_t+0.2\left(1-p_t\right)}{p_t+1.8\left(1-p_t\right)}$

Explanation of Solution

Given information: Suppose that a fraction 0.2 of wild-type mutate each generation, but that each wild-type individual produces 2.0 offspring while each mutant produces only 1.0 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

  $\begin{array}{l}{p}_{t+1}=\frac{m_{t+1}}{b_{t+1}+m_{t+1}}\\ =\frac{m_t+0.2b_t}{1.6b_t+\left(m_t+0.2b_t\right)}\\ =\frac{m_t+0.2b_t}{1.8b_t+m_t}\end{array}$

  $\begin{array}{l}\frac{\frac{m_t}{m_t+b_t}+\frac{0.2b_t}{m_t+b_t}}{\frac{1.8b_t}{m_t+b_t}+\frac{m_t}{m_t+b_t}}\\ p_{t+1}=\frac{p_t+0.2\left(1-p_t\right)}{p_t+1.8\left(1-p_t\right)}\end{array}$

f

To determine

To calculate: To find the equilibrium using the obtained values

Answer to Problem 32E

  $p^{•}=1/4\text{or}{p}^{•}=1.0$

Explanation of Solution

Given information: Suppose that a fraction 0.2 of wild-type mutate each generation, but that each wild-type individual produces 2.0 offspring while each mutant produces only 1.0 offspring.

Calculation:

Let $p^{•}$ the equilibrium

To find the equilibrium consider,

  $p^{•}=\frac{p^{•}+0.2\left(1-p^{•}\right)}{p^{•}+1.8\left(1-p^{•}\right)}$

Solving the above equation gives,

  $\begin{array}{l}{p}^{•}\left[p^{•}+1.8\left(1-p^{•}\right)\right]=p^{•}+0.2\left(1-p^{•}\right)\\ {\left(p^{•}\right)}^2+1.8p^{•}-1.8{\left(p^{•}\right)}^2-p^{•}-0.2+0.2p^{•}=0\\ 0.8{\left(p^{•}\right)}^2-p^{•}+0.2=0\\ 0.8\left(p^{•}-1/4\right)\left(p^{•}-1.0\right)=0\end{array}$

Therefore,

  $p^{•}=1/4\text{or}{p}^{•}=1.0$

g

To determine

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

Answer to Problem 32E

The fraction of mutants will end up at about 33.33 percent.

Explanation of Solution

Given information: Suppose that a fraction 0.2 of wild-type mutate each generation, but that each wild-type individual produces 2.0 offspring while each mutant produces only 1.0 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.