Question

PROJECT Modeling the Dynamics of Viral Infections A patient is infected with a virus that triples its numbers every hour. The immune system eventually kicks in and reduces the replication rate by a factor of in addition to kill ing 500,000 virus particles per hour, but this doesn't happen until the viral load reaches two million copies. To combat the infection, the infected person receives hourly doses of an antiviral drug. This drug further reduces the replication rate, to a value of 1.25, and the immune system and the drug together can kill 25,000,000 copies of the virus per hour. Let's model the phases of the infection using a discrete-time difference equation, 1. What is the recursion for the number of viral particles in the absence of treatment and before the immune response starts? What is the equation for the number of viral particles as a function of time? 2. How long does it take for the immune system to be activated after the infection arts? Derive an equation for this length of time for an arbitrary initial number of viral particles. 3. What is the recursion for the number of viral particles after the immune response has begun, but before the drug is used? 4. What is the condition for the viral population size to decrease over time solely because of the immune system? Is this possible? 5. What is the recursion for the number of viral particles in the presence of both the immune response and the drug? 6. What is the condition for the viral population size to decrease over time when the drug and immune system are both acting? Is this possible? 7. If an individual is infected by one virus, how much time do you have to start drug treatment in order to control the infection? 8. Often the outcome of an infection depends on the number of viral particles causing the infection. Derive an expression for the amount of time it takes from the initial infection for an individual to reach the critical viral loadbeyond which control of the infection is impossible as a function of the initial number of particles no and an arbitrary initial replication rate R.

Answer 1

You have asked a question with multiple sub-parts. As per Chegg Q&A Guidelines, I am answering the first 4 subparts.

1. Here, we have a virus that can replicate in the absence of any immune response or drug wherein in triples each hour. This is the replication number of the virus symbolized by R.

:.R=3

Now, the virus is exhibiting exponential growth. Let N(t) be the count of the virus at time t. The count of the virus at time t+1 will be given by N(t+1) = RN(t), where R is the replication number of the virus.

At time t = 0, the number of virus N(t = 0) = N0.

Then we have,

N4 = NR

In our case, R = 3. therefore, we have

N4 = No3

This equation specifies the number of particles that specifies the number of particles as a function of time.

2. Let the time spent by the virus replicating before the immune system kicks in be

t= T

Now, we can solve the equation derived in step 1 for time . Therefore, we have:

t= T

NT = No x 37

37:- NTE No

Taking natural logarithm on both side, we get

IN u = (£)u?

: Tiln(3) = n(

:.1; = In(3)

...T; = In(NT) - In(NO) In(3)

However, we are told that the immune system kicks in when the viral load reaches 2 million copies. Hence, we have:

NT, = 2 x 106

Substituting back in the equation for , we get:

t= T

T; = Inl In (2 x 10) - In(No) In(3) In (2) + In (106) – In (No) In(3)

:.T; = In(2) +6 x In (10) - In(No) In(3)

::T; = 0.693 + 6 x 2.303 – In(N) 1.099 0.693 + 13.818 – In (No) 1.099

14.511 - In(NO) :.T; = - 1.099 = 13.204 - 0.9099 x In (NO)

This is the equation for the time the immune system will take to get activated. As can be seen, the time is lower for the larger viral load at t=0.

3. In the second phase, we have two dynamic processes playing roles. Firstly, the virus is replicating at a new recursion number .

Ri= --R=0.5 x 3 = 1.5

Additionally, the virus is also being killed by the immune system. Let the viruses killed by the immune response in a unit time step be K.

Thus, we have a new recursion as:

(t+1) = R; (t) – K

4. For the viral population to decrease over time by the immune response, at each time step the change the viral count should be negative. That is to say:

AN = N(t+1) - N(t) <0

Now, from the recursion derived in step 3, we have:

AN = Rin(t) - K - N(t) = (Ri - 1)n(t) – K <0

:. (Ri - 1)n(t) <K

:N(t) <TRi - 1) NAK

Now, we have

R; = 1.5

and the virus kills 500,000 viral in one hour, so

K = 500000 = 0.5 X 106

The condition for the viral count to decrease by the immune response can be obtained by substituting the values of K and Ri in the above inequality.

N(t) 0.5 x 106 (1.5 – 1)

0.5 x 106 N(t) <= (0.5)

q0I XI> (+)N.

But as mentioned the immune response is generated only when the viral count crosses 2 million. Thus solely immune response will not be able to decrease the viral count.