MODELING OF CRISPR-BASED GENETIC CIRCUITS AND
THEIR DYNAMICS
1 CRISPR-BASED REPRESSION CIRCUITS
Problem 1. Consider a CRISPR interference (CRISPRi) system targeting a gene of interest.
The dynamics of mRNA production and degradation can be modeled by the following differential
equation:
dm
dt = α⋅ 1
1+(c
K)n−δm
where:
• m is the concentration of mRNA
• c is the concentration of CRISPR-dCas9 complex
• α is the maximum transcription rate
• K is the repression coefficient
• n is the Hill coefficient
• δ is the mRNA degradation rate
Given the following parameters:
• α = 10 nM/min
• K = 5 nM
• n = 2
• δ = 0.1 min−1
• Initial mRNA concentration m0= 0 nM
• CRISPR-dCas9 complex concentration c = 10 nM
a) Calculate the steady-state mRNA concentration.
b) Determine the time it takes to reach 50% of the steady-state concentration.
c) If the CRISPR-dCas9 complex concentration is doubled to 20 nM, what is the new steady-
state mRNA concentration?
Solution 1. a) To find the steady-state concentration, set dm
dt = 0:
0 = α⋅ 1
1+(c
K)n−δmss
Solving for mss:
mss =α
δ⋅1
1+(c
K)n=10
0.1⋅1
1+(10
5)2=20 nM
b) The dynamics of the system follow the equation:
m(t)= mss(1−e−δt)
To find the time to reach 50% of steady-state:
0.5 = 1−e−δt
Solving for t:
t = −ln(0.5)
δ= −ln(0.5)
0.1 ≈ 6.93 minutes
c) With c = 20 nM, the new steady-state concentration is:
mss =10
0.1⋅1
1+(20
5)2= 5.56 nM
Therefore: a) The steady-state mRNA concentration is 20 nM. b) It takes approximately 6.93
minutes to reach 50% of the steady-state concentration. c) When the CRISPR-dCas9 complex
concentration is doubled, the new steady-state mRNA concentration is 5.56 nM.
2 CRISPR-BASED TOGGLE SWITCH
Problem 2. Consider a genetic toggle switch implemented using two CRISPR-based
repressors. The system can be described by the following coupled differential equations:
dx
dt = αx⋅1
1+( y
Ky)ny−δxx
dy
dt = αy⋅1
1+( x
Kx)nx−δyy
where x and y are the concentrations of the two repressors.
Given the following parameters:
• αx= αy=100 nM/min
• Kx= Ky=40 nM
• nx= ny= 4
• δx= δy= 1 min−1
a) Find the steady-state concentrations of x and y when the system is balanced.
b) If the initial conditions are x0=10 nM and y0=80 nM, which state will the system likely
settle into?
c) Calculate the switching threshold for x that would cause the system to flip states.
Solution 2. a) At steady-state and balanced conditions, x = y and dx
dt =dy
dt = 0. Solving:
x = y = αx
δx⋅1
1+( x
Kx)nx
Substituting the values:
x = 100⋅1
1+( x
40)4
Solving numerically, we get x = y ≈ 50 nM.
b) With initial conditions x0=10 nM and y0=80 nM, y is much higher than x. The system will
likely settle into the state where y is high and x is low.
To verify, we can calculate the steady-state values:
For x: xss =100⋅1
1+(80
40)4≈ 5.88 nM For y: yss =100⋅1
1+(10
40)4≈98.77 nM
c) The switching threshold occurs when dx
dt =dy
dt = 0 and x = y. We already calculated this in
part (a):
The switching threshold for x is approximately 50 nM.
Therefore: a) The balanced steady-state concentrations are x = y ≈ 50 nM. b) The system will
likely settle into the state where x ≈ 5.88 nM and y ≈ 98.77 nM. c) The switching threshold for x
is approximately 50 nM.
3 CRISPR-BASED OSCILLATOR
Problem 3. Consider a CRISPR-based genetic oscillator consisting of three repressors (X, Y,
and Z) in a cycle. The system can be described by the following differential equations:
dx
dt = αx⋅1
1+( z
Kz)nz−δxx
dy
dt = αy⋅1
1+( x
Kx)nx−δyy
dz
dt = αz⋅1
1+( y
Ky)ny−δzz
Given the following parameters:
• αx= αy= αz=200 nM/min
• Kx= Ky= Kz=40 nM
• nx= ny= nz= 2
• δx= δy= δz= 0.1 min−1
a) Estimate the period of oscillation using the delay between repressors.
b) Calculate the maximum concentration of each repressor during the oscillation.
c) If we increase all Hill coefficients to 4, how would this affect the oscillator’s behavior?
Solution 3. a) To estimate the period, we can consider the time for one repressor to reach its
maximum after another is at its minimum. This is approximately the time to reach steady-state:
tss ≈4
δ=4
0.1 =40 minutes
The full period involves three such delays, so:
Period ≈ 3⋅40 =120 minutes
b) The maximum concentration occurs when a repressor is fully de-repressed. At this point:
xmax =αx
δx=200
0.1 =2000 nM
This is the same for y and z due to symmetry.
c) Increasing the Hill coefficients to 4 would:
• Make the repression more switch-like (steeper response)
• Likely increase the amplitude of oscillations
• Potentially make the oscillations more robust and stable
• Slightly increase the period of oscillation
To quantify these effects precisely would require numerical simulation of the system.
Therefore: a) The estimated period of oscillation is approximately 120 minutes. b) The
maximum concentration of each repressor during oscillation is 2000 nM. c) Increasing Hill
coefficients to 4 would likely result in more robust, higher amplitude oscillations with a slightly
longer period.
Consider a CRISPR interference (CRISPRi) system targeting a gene of interest. The dynamics of
mRNA production and degradation can be modeled by the following differential equation:
dm
dt = α⋅ 1
1+(c
K)n−δm
where:
• m is the concentration of mRNA
• c is the concentration of CRISPR-dCas9 complex
• α is the maximum transcription rate
• K is the repression coefficient
• n is the Hill coefficient
• δ is the mRNA degradation rate
Given the following parameters:
• α = 10 nM/min
• K = 5 nM
• n = 2
• δ = 0.1 min−1
• Initial mRNA concentration m0= 0 nM
• CRISPR-dCas9 complex concentration c = 10 nM
a) Calculate the steady-state mRNA concentration.
b) Determine the time it takes to reach 50% of the steady-state concentration.
c) If the CRISPR-dCas9 complex concentration is doubled to 20 nM, what is the new steady-
state mRNA concentration?
Solution 1. a) To find the steady-state concentration, set dm
dt = 0:
0 = α⋅ 1
1+(c
K)n−δmss
Solving for mss:
mss =α
δ⋅1
1+(c
K)n=10
0.1⋅1
1+(10
5)2=20 nM
b) The dynamics of the system follow the equation:
m(t)= mss(1−e−δt)
To find the time to reach 50% of steady-state:
0.5 = 1−e−δt
Solving for t:
t = −ln(0.5)
δ= −ln(0.5)
0.1 ≈ 6.93 minutes
c) With c = 20 nM, the new steady-state concentration is:
mss =10
0.1⋅1
1+(20
5)2= 5.56 nM
Therefore: a) The steady-state mRNA concentration is 20 nM. b) It takes approximately 6.93
minutes to reach 50% of the steady-state concentration. c) When the CRISPR-dCas9 complex
concentration is doubled, the new steady-state mRNA concentration is 5.56 nM.
4 CRISPR-BASED TOGGLE SWITCH
Problem 2. Consider a genetic toggle switch implemented using two CRISPR-based
repressors. The system can be described by the following coupled differential equations:
dx
dt = αx⋅1
1+( y
Ky)ny−δxx
dy
dt = αy⋅1
1+( x
Kx)nx−δyy
where x and y are the concentrations of the two repressors.
Given the following parameters:
• αx= αy=100 nM/min
• Kx= Ky=40 nM
• nx= ny= 4
• δx= δy= 1 min−1
a) Find the steady-state concentrations of x and y when the system is balanced.
b) If the initial conditions are x0=10 nM and y0=80 nM, which state will the system likely
settle into?
c) Calculate the switching threshold for x that would cause the system to flip states.
Solution 2. a) At steady-state and balanced conditions, x = y and dx
dt =dy
dt = 0. Solving:
x = y = αx
δx⋅1
1+( x
Kx)nx
Substituting the values:
x = 100⋅1
1+( x
40)4
Solving numerically, we get x = y ≈ 50 nM.
b) With initial conditions x0=10 nM and y0=80 nM, y is much higher than x. The system will
likely settle into the state where y is high and x is low.
To verify, we can calculate the steady-state values:
For x: xss =100⋅1
1+(80
40)4≈ 5.88 nM For y: yss =100⋅1
1+(10
40)4≈98.77 nM
c) The switching threshold occurs when dx
dt =dy
dt = 0 and x = y. We already calculated this in
part (a):
The switching threshold for x is approximately 50 nM.
Therefore: a) The balanced steady-state concentrations are x = y ≈ 50 nM. b) The system will
likely settle into the state where x ≈ 5.88 nM and y ≈ 98.77 nM. c) The switching threshold for x
is approximately 50 nM.
5 CRISPR-BASED OSCILLATOR
Problem 3. Consider a CRISPR-based genetic oscillator consisting of three repressors (X, Y,
and Z) in a cycle. The system can be described by the following differential equations:
dx
dt = αx⋅1
1+( z
Kz)nz−δxx
dy
dt = αy⋅1
1+( x
Kx)nx−δyy
dz
dt = αz⋅1
1+( y
Ky)ny−δzz
Given the following parameters:
• αx= αy= αz=200 nM/min
• Kx= Ky= Kz=40 nM
• nx= ny= nz= 2
• δx= δy= δz= 0.1 min−1
a) Estimate the period of oscillation using the delay between repressors.
b) Calculate the maximum concentration of each repressor during the oscillation.
c) If we increase all Hill coefficients to 4, how would this affect the oscillator’s behavior?
Solution 3. a) To estimate the period, we can consider the time for one repressor to reach its
maximum after another is at its minimum. This is approximately the time to reach steady-state:
tss ≈4
δ=4
0.1 =40 minutes
The full period involves three such delays, so:
Period ≈ 3⋅40 =120 minutes
b) The maximum concentration occurs when a repressor is fully de-repressed. At this point:
xmax =αx
δx=200
0.1 =2000 nM
This is the same for y and z due to symmetry.
c) Increasing the Hill coefficients to 4 would:
• Make the repression more switch-like (steeper response)
• Likely increase the amplitude of oscillations
• Potentially make the oscillations more robust and stable
• Slightly increase the period of oscillation
To quantify these effects precisely would require numerical simulation of the system.
Therefore: a) The estimated period of oscillation is approximately 120 minutes. b) The
maximum concentration of each repressor during oscillation is 2000 nM. c) Increasing Hill
coefficients to 4 would likely result in more robust, higher amplitude oscillations with a slightly
longer period.
Consider a CRISPR interference (CRISPRi) system targeting a gene of interest. The dynamics of
mRNA production and degradation can be modeled by the following differential equation:
dm
dt = α⋅ 1
1+(c
K)n−δm
where:
• m is the concentration of mRNA
• c is the concentration of CRISPR-dCas9 complex
• α is the maximum transcription rate
• K is the repression coefficient
• n is the Hill coefficient
• δ is the mRNA degradation rate
Given the following parameters:
• α = 10 nM/min
• K = 5 nM
• n = 2
• δ = 0.1 min−1
• Initial mRNA concentration m0= 0 nM
• CRISPR-dCas9 complex concentration c = 10 nM
a) Calculate the steady-state mRNA concentration.
b) Determine the time it takes to reach 50% of the steady-state concentration.
c) If the CRISPR-dCas9 complex concentration is doubled to 20 nM, what is the new steady-
state mRNA concentration?
Solution 1. a) To find the steady-state concentration, set dm
dt = 0:
0 = α⋅ 1
1+(c
K)n−δmss
Solving for mss:
mss =α
δ⋅1
1+(c
K)n=10
0.1⋅1
1+(10
5)2=20 nM
b) The dynamics of the system follow the equation:
m(t)= mss(1−e−δt)
To find the time to reach 50% of steady-state:
0.5 = 1−e−δt
Solving for t:
t = −ln(0.5)
δ= −ln(0.5)
0.1 ≈ 6.93 minutes
c) With c = 20 nM, the new steady-state concentration is:
mss =10
0.1⋅1
1+(20
5)2= 5.56 nM
Therefore: a) The steady-state mRNA concentration is 20 nM. b) It takes approximately 6.93
minutes to reach 50% of the steady-state concentration. c) When the CRISPR-dCas9 complex
concentration is doubled, the new steady-state mRNA concentration is 5.56 nM.
6 CRISPR-BASED TOGGLE SWITCH
Problem 2. Consider a genetic toggle switch implemented using two CRISPR-based
repressors. The system can be described by the following coupled differential equations:
dx
dt = αx⋅1
1+( y
Ky)ny−δxx
dy
dt = αy⋅1
1+( x
Kx)nx−δyy
where x and y are the concentrations of the two repressors.
Given the following parameters:
• αx= αy=100 nM/min
• Kx= Ky=40 nM
• nx= ny= 4
• δx= δy= 1 min−1
a) Find the steady-state concentrations of x and y when the system is balanced.
b) If the initial conditions are x0=10 nM and y0=80 nM, which state will the system likely
settle into?
c) Calculate the switching threshold for x that would cause the system to flip states.
Solution 2. a) At steady-state and balanced conditions, x = y and dx
dt =dy
dt = 0. Solving:
x = y = αx
δx⋅1
1+( x
Kx)nx
Substituting the values:
x = 100⋅1
1+( x
40)4
Solving numerically, we get x = y ≈ 50 nM.
b) With initial conditions x0=10 nM and y0=80 nM, y is much higher than x. The system will
likely settle into the state where y is high and x is low.
To verify, we can calculate the steady-state values:
For x: xss =100⋅1
1+(80
40)4≈ 5.88 nM For y: yss =100⋅1
1+(10
40)4≈98.77 nM
c) The switching threshold occurs when dx
dt =dy
dt = 0 and x = y. We already calculated this in
part (a):
The switching threshold for x is approximately 50 nM.
Therefore: a) The balanced steady-state concentrations are x = y ≈ 50 nM. b) The system will
likely settle into the state where x ≈ 5.88 nM and y ≈ 98.77 nM. c) The switching threshold for x
is approximately 50 nM.
7 CRISPR-BASED OSCILLATOR
Problem 3. Consider a CRISPR-based genetic oscillator consisting of three repressors (X, Y,
and Z) in a cycle. The system can be described by the following differential equations:
dx
dt = αx⋅1
1+( z
Kz)nz−δxx
dy
dt = αy⋅1
1+( x
Kx)nx−δyy
dz
dt = αz⋅1
1+( y
Ky)ny−δzz
Given the following parameters:
• αx= αy= αz=200 nM/min
• Kx= Ky= Kz=40 nM
• nx= ny= nz= 2
• δx= δy= δz= 0.1 min−1
a) Estimate the period of oscillation using the delay between repressors.
b) Calculate the maximum concentration of each repressor during the oscillation.
c) If we increase all Hill coefficients to 4, how would this affect the oscillator’s behavior?
Solution 3. a) To estimate the period, we can consider the time for one repressor to reach its
maximum after another is at its minimum. This is approximately the time to reach steady-state:
tss ≈4
δ=4
0.1 =40 minutes
The full period involves three such delays, so:
Period ≈ 3⋅40 =120 minutes
b) The maximum concentration occurs when a repressor is fully de-repressed. At this point:
xmax =αx
δx=200
0.1 =2000 nM
This is the same for y and z due to symmetry.
c) Increasing the Hill coefficients to 4 would:
• Make the repression more switch-like (steeper response)
• Likely increase the amplitude of oscillations
• Potentially make the oscillations more robust and stable
• Slightly increase the period of oscillation
To quantify these effects precisely would require numerical simulation of the system.
Therefore: a) The estimated period of oscillation is approximately 120 minutes. b) The
maximum concentration of each repressor during oscillation is 2000 nM. c) Increasing Hill
coefficients to 4 would likely result in more robust, higher amplitude oscillations with a slightly
longer period.
Consider a CRISPR interference (CRISPRi) system targeting a gene of interest. The dynamics of
mRNA production and degradation can be modeled by the following differential equation:
dm
dt = α⋅ 1
1+(c
K)n−δm
where:
• m is the concentration of mRNA
• c is the concentration of CRISPR-dCas9 complex
• α is the maximum transcription rate
• K is the repression coefficient
• n is the Hill coefficient
• δ is the mRNA degradation rate
Given the following parameters:
• α = 10 nM/min
• K = 5 nM
• n = 2
• δ = 0.1 min−1
• Initial mRNA concentration m0= 0 nM
• CRISPR-dCas9 complex concentration c = 10 nM
a) Calculate the steady-state mRNA concentration.
b) Determine the time it takes to reach 50% of the steady-state concentration.
c) If the CRISPR-dCas9 complex concentration is doubled to 20 nM, what is the new steady-
state mRNA concentration?
Solution 1. a) To find the steady-state concentration, set dm
dt = 0:
0 = α⋅ 1
1+(c
K)n−δmss
Solving for mss:
mss =α
δ⋅1
1+(c
K)n=10
0.1⋅1
1+(10
5)2=20 nM
b) The dynamics of the system follow the equation:
m(t)= mss(1−e−δt)
To find the time to reach 50% of steady-state:
0.5 = 1−e−δt
Solving for t:
t = −ln(0.5)
δ= −ln(0.5)
0.1 ≈ 6.93 minutes
c) With c = 20 nM, the new steady-state concentration is:
mss =10
0.1⋅1
1+(20
5)2= 5.56 nM
Therefore: a) The steady-state mRNA concentration is 20 nM. b) It takes approximately 6.93
minutes to reach 50% of the steady-state concentration. c) When the CRISPR-dCas9 complex
concentration is doubled, the new steady-state mRNA concentration is 5.56 nM.
8 CRISPR-BASED TOGGLE SWITCH
Problem 2. Consider a genetic toggle switch implemented using two CRISPR-based
repressors. The system can be described by the following coupled differential equations:
dx
dt = αx⋅1
1+( y
Ky)ny−δxx
dy
dt = αy⋅1
1+( x
Kx)nx−δyy
where x and y are the concentrations of the two repressors.
Given the following parameters:
• αx= αy=100 nM/min
• Kx= Ky=40 nM
• nx= ny= 4
• δx= δy= 1 min−1
a) Find the steady-state concentrations of x and y when the system is balanced.
b) If the initial conditions are x0=10 nM and y0=80 nM, which state will the system likely
settle into?
c) Calculate the switching threshold for x that would cause the system to flip states.
Solution 2. a) At steady-state and balanced conditions, x = y and dx
dt =dy
dt = 0. Solving:
x = y = αx
δx⋅1
1+( x
Kx)nx
Substituting the values:
x = 100⋅1
1+( x
40)4
Solving numerically, we get x = y ≈ 50 nM.
b) With initial conditions x0=10 nM and y0=80 nM, y is much higher than x. The system will
likely settle into the state where y is high and x is low.
To verify, we can calculate the steady-state values:
For x: xss =100⋅1
1+(80
40)4≈ 5.88 nM For y: yss =100⋅1
1+(10
40)4≈98.77 nM
c) The switching threshold occurs when dx
dt =dy
dt = 0 and x = y. We already calculated this in
part (a):
The switching threshold for x is approximately 50 nM.
Therefore: a) The balanced steady-state concentrations are x = y ≈ 50 nM. b) The system will
likely settle into the state where x ≈ 5.88 nM and y ≈ 98.77 nM. c) The switching threshold for x
is approximately 50 nM.
9 CRISPR-BASED OSCILLATOR
Problem 3. Consider a CRISPR-based genetic oscillator consisting of three repressors (X, Y,
and Z) in a cycle. The system can be described by the following differential equations:
dx
dt = αx⋅1
1+( z
Kz)nz−δxx
dy
dt = αy⋅1
1+( x
Kx)nx−δyy
dz
dt = αz⋅1
1+( y
Ky)ny−δzz
Given the following parameters:
• αx= αy= αz=200 nM/min
• Kx= Ky= Kz=40 nM
• nx= ny= nz= 2
• δx= δy= δz= 0.1 min−1
a) Estimate the period of oscillation using the delay between repressors.
b) Calculate the maximum concentration of each repressor during the oscillation.
c) If we increase all Hill coefficients to 4, how would this affect the oscillator’s behavior?
Solution 3. a) To estimate the period, we can consider the time for one repressor to reach its
maximum after another is at its minimum. This is approximately the time to reach steady-state:
tss ≈4
δ=4
0.1 =40 minutes
The full period involves three such delays, so:
Period ≈ 3⋅40 =120 minutes
b) The maximum concentration occurs when a repressor is fully de-repressed. At this point:
xmax =αx
δx=200
0.1 =2000 nM
This is the same for y and z due to symmetry.
c) Increasing the Hill coefficients to 4 would:
• Make the repression more switch-like (steeper response)
• Likely increase the amplitude of oscillations
• Potentially make the oscillations more robust and stable
• Slightly increase the period of oscillation
To quantify these effects precisely would require numerical simulation of the system.
Therefore: a) The estimated period of oscillation is approximately 120 minutes. b) The
maximum concentration of each repressor during oscillation is 2000 nM. c) Increasing Hill
coefficients to 4 would likely result in more robust, higher amplitude oscillations with a slightly
longer period.
Consider a CRISPR interference (CRISPRi) system targeting a gene of interest. The dynamics of
mRNA production and degradation can be modeled by the following differential equation:
dm
dt = α⋅ 1
1+(c
K)n−δm
where:
• m is the concentration of mRNA
• c is the concentration of CRISPR-dCas9 complex
• α is the maximum transcription rate
• K is the repression coefficient
• n is the Hill coefficient
• δ is the mRNA degradation rate
Given the following parameters:
• α = 10 nM/min
• K = 5 nM
• n = 2
• δ = 0.1 min−1
• Initial mRNA concentration m0= 0 nM
• CRISPR-dCas9 complex concentration c = 10 nM
a) Calculate the steady-state mRNA concentration.
b) Determine the time it takes to reach 50% of the steady-state concentration.
c) If the CRISPR-dCas9 complex concentration is doubled to 20 nM, what is the new steady-
state mRNA concentration?
Solution 1. a) To find the steady-state concentration, set dm
dt = 0:
0 = α⋅ 1
1+(c
K)n−δmss
Solving for mss:
mss =α
δ⋅1
1+(c
K)n=10
0.1⋅1
1+(10
5)2=20 nM
b) The dynamics of the system follow the equation:
m(t)= mss(1−e−δt)
To find the time to reach 50% of steady-state:
0.5 = 1−e−δt
Solving for t:
t = −ln(0.5)
δ= −ln(0.5)
0.1 ≈ 6.93 minutes
c) With c = 20 nM, the new steady-state concentration is:
mss =10
0.1⋅1
1+(20
5)2= 5.56 nM
Therefore: a) The steady-state mRNA concentration is 20 nM. b) It takes approximately 6.93
minutes to reach 50% of the steady-state concentration. c) When the CRISPR-dCas9 complex
concentration is doubled, the new steady-state mRNA concentration is 5.56 nM.
10 CRISPR-BASED TOGGLE SWITCH
Problem 2. Consider a genetic toggle switch implemented using two CRISPR-based
repressors. The system can be described by the following coupled differential equations:
dx
dt = αx⋅1
1+( y
Ky)ny−δxx
dy
dt = αy⋅1
1+( x
Kx)nx−δyy
where x and y are the concentrations of the two repressors.
Given the following parameters:
• αx= αy=100 nM/min
• Kx= Ky=40 nM
• nx= ny= 4
• δx= δy= 1 min−1
a) Find the steady-state concentrations of x and y when the system is balanced.
b) If the initial conditions are x0=10 nM and y0=80 nM, which state will the system likely
settle into?
c) Calculate the switching threshold for x that would cause the system to flip states.
Solution 2. a) At steady-state and balanced conditions, x = y and dx
dt =dy
dt = 0. Solving:
x = y = αx
δx⋅1
1+( x
Kx)nx
Substituting the values:
x = 100⋅1
1+( x
40)4
Solving numerically, we get x = y ≈ 50 nM.
b) With initial conditions x0=10 nM and y0=80 nM, y is much higher than x. The system will
likely settle into the state where y is high and x is low.
To verify, we can calculate the steady-state values:
For x: xss =100⋅1
1+(80
40)4≈ 5.88 nM For y: yss =100⋅1
1+(10
40)4≈98.77 nM
c) The switching threshold occurs when dx
dt =dy
dt = 0 and x = y. We already calculated this in
part (a):
The switching threshold for x is approximately 50 nM.
Therefore: a) The balanced steady-state concentrations are x = y ≈ 50 nM. b) The system will
likely settle into the state where x ≈ 5.88 nM and y ≈ 98.77 nM. c) The switching threshold for x
is approximately 50 nM.
11 CRISPR-BASED OSCILLATOR
Problem 3. Consider a CRISPR-based genetic oscillator consisting of three repressors (X, Y,
and Z) in a cycle. The system can be described by the following differential equations:
dx
dt = αx⋅1
1+( z
Kz)nz−δxx
dy
dt = αy⋅1
1+( x
Kx)nx−δyy
dz
dt = αz⋅1
1+( y
Ky)ny−δzz
Given the following parameters:
• αx= αy= αz=200 nM/min
• Kx= Ky= Kz=40 nM
• nx= ny= nz= 2
• δx= δy= δz= 0.1 min−1
a) Estimate the period of oscillation using the delay between repressors.
b) Calculate the maximum concentration of each repressor during the oscillation.
c) If we increase all Hill coefficients to 4, how would this affect the oscillator’s behavior?
Solution 3. a) To estimate the period, we can consider the time for one repressor to reach its
maximum after another is at its minimum. This is approximately the time to reach steady-state:
tss ≈4
δ=4
0.1 =40 minutes
The full period involves three such delays, so:
Period ≈ 3⋅40 =120 minutes
b) The maximum concentration occurs when a repressor is fully de-repressed. At this point:
xmax =αx
δx=200
0.1 =2000 nM
This is the same for y and z due to symmetry.
c) Increasing the Hill coefficients to 4 would:
• Make the repression more switch-like (steeper response)
• Likely increase the amplitude of oscillations
• Potentially make the oscillations more robust and stable
• Slightly increase the period of oscillation
To quantify these effects precisely would require numerical simulation of the system.
Therefore: a) The estimated period of oscillation is approximately 120 minutes. b) The
maximum concentration of each repressor during oscillation is 2000 nM. c) Increasing Hill
coefficients to 4 would likely result in more robust, higher amplitude oscillations with a slightly
longer period.
Consider a CRISPR interference (CRISPRi) system targeting a gene of interest. The dynamics of
mRNA production and degradation can be modeled by the following differential equation:
dm
dt = α⋅ 1
1+(c
K)n−δm
where:
• m is the concentration of mRNA
• c is the concentration of CRISPR-dCas9 complex
• α is the maximum transcription rate
• K is the repression coefficient
• n is the Hill coefficient
• δ is the mRNA degradation rate
Given the following parameters:
• α = 10 nM/min
• K = 5 nM
• n = 2
• δ = 0.1 min−1
• Initial mRNA concentration m0= 0 nM
• CRISPR-dCas9 complex concentration c = 10 nM
a) Calculate the steady-state mRNA concentration.
b) Determine the time it takes to reach 50% of the steady-state concentration.
c) If the CRISPR-dCas9 complex concentration is doubled to 20 nM, what is the new steady-
state mRNA concentration?
Solution 1. a) To find the steady-state concentration, set dm
dt = 0:
0 = α⋅ 1
1+(c
K)n−δmss
Solving for mss:
mss =α
δ⋅1
1+(c
K)n=10
0.1⋅1
1+(10
5)2=20 nM
b) The dynamics of the system follow the equation:
m(t)= mss(1−e−δt)
To find the time to reach 50% of steady-state:
0.5 = 1−e−δt
Solving for t:
t = −ln(0.5)
δ= −ln(0.5)
0.1 ≈ 6.93 minutes
c) With c = 20 nM, the new steady-state concentration is:
mss =10
0.1⋅1
1+(20
5)2= 5.56 nM
Therefore: a) The steady-state mRNA concentration is 20 nM. b) It takes approximately 6.93
minutes to reach 50% of the steady-state concentration. c) When the CRISPR-dCas9 complex
concentration is doubled, the new steady-state mRNA concentration is 5.56 nM.
12 CRISPR-BASED TOGGLE SWITCH
Problem 2. Consider a genetic toggle switch implemented using two CRISPR-based
repressors. The system can be described by the following coupled differential equations:
dx
dt = αx⋅1
1+( y
Ky)ny−δxx
dy
dt = αy⋅1
1+( x
Kx)nx−δyy
where x and y are the concentrations of the two repressors.
Given the following parameters:
• αx= αy=100 nM/min
• Kx= Ky=40 nM
• nx= ny= 4
• δx= δy= 1 min−1
a) Find the steady-state concentrations of x and y when the system is balanced.
b) If the initial conditions are x0=10 nM and y0=80 nM, which state will the system likely
settle into?
c) Calculate the switching threshold for x that would cause the system to flip states.
Solution 2. a) At steady-state and balanced conditions, x = y and dx
dt =dy
dt = 0. Solving:
x = y = αx
δx⋅1
1+( x
Kx)nx
Substituting the values:
x = 100⋅1
1+( x
40)4
Solving numerically, we get x = y ≈ 50 nM.
b) With initial conditions x0=10 nM and y0=80 nM, y is much higher than x. The system will
likely settle into the state where y is high and x is low.
To verify, we can calculate the steady-state values:
For x: xss =100⋅1
1+(80
40)4≈ 5.88 nM For y: yss =100⋅1
1+(10
40)4≈98.77 nM
c) The switching threshold occurs when dx
dt =dy
dt = 0 and x = y. We already calculated this in
part (a):
The switching threshold for x is approximately 50 nM.
Therefore: a) The balanced steady-state concentrations are x = y ≈ 50 nM. b) The system will
likely settle into the state where x ≈ 5.88 nM and y ≈ 98.77 nM. c) The switching threshold for x
is approximately 50 nM.
13 CRISPR-BASED OSCILLATOR
Problem 3. Consider a CRISPR-based genetic oscillator consisting of three repressors (X, Y,
and Z) in a cycle. The system can be described by the following differential equations:
dx
dt = αx⋅1
1+( z
Kz)nz−δxx
dy
dt = αy⋅1
1+( x
Kx)nx−δyy
dz
dt = αz⋅1
1+( y
Ky)ny−δzz
Given the following parameters:
• αx= αy= αz=200 nM/min
• Kx= Ky= Kz=40 nM
• nx= ny= nz= 2
• δx= δy= δz= 0.1 min−1
a) Estimate the period of oscillation using the delay between repressors.
b) Calculate the maximum concentration of each repressor during the oscillation.
c) If we increase all Hill coefficients to 4, how would this affect the oscillator’s behavior?
Solution 3. a) To estimate the period, we can consider the time for one repressor to reach its
maximum after another is at its minimum. This is approximately the time to reach steady-state:
tss ≈4
δ=4
0.1 =40 minutes
The full period involves three such delays, so:
Period ≈ 3⋅40 =120 minutes
b) The maximum concentration occurs when a repressor is fully de-repressed. At this point:
xmax =αx
δx=200
0.1 =2000 nM
This is the same for y and z due to symmetry.
c) Increasing the Hill coefficients to 4 would:
• Make the repression more switch-like (steeper response)
• Likely increase the amplitude of oscillations
• Potentially make the oscillations more robust and stable
• Slightly increase the period of oscillation
To quantify these effects precisely would require numerical simulation of the system.
Therefore: a) The estimated period of oscillation is approximately 120 minutes. b) The
maximum concentration of each repressor during oscillation is 2000 nM. c) Increasing Hill
coefficients to 4 would likely result in more robust, higher amplitude oscillations with a slightly
longer period.
Consider a CRISPR interference (CRISPRi) system targeting a gene of interest. The dynamics of
mRNA production and degradation can be modeled by the following differential equation:
dm
dt = α⋅ 1
1+(c
K)n−δm
where:
• m is the concentration of mRNA
• c is the concentration of CRISPR-dCas9 complex
• α is the maximum transcription rate
• K is the repression coefficient
• n is the Hill coefficient
• δ is the mRNA degradation rate
Given the following parameters:
• α = 10 nM/min
• K = 5 nM
• n = 2
• δ = 0.1 min−1
• Initial mRNA concentration m0= 0 nM
• CRISPR-dCas9 complex concentration c = 10 nM
a) Calculate the steady-state mRNA concentration.
b) Determine the time it takes to reach 50% of the steady-state concentration.
c) If the CRISPR-dCas9 complex concentration is doubled to 20 nM, what is the new steady-
state mRNA concentration?
Solution 1. a) To find the steady-state concentration, set dm
dt = 0:
0 = α⋅ 1
1+(c
K)n−δmss
Solving for mss:
mss =α
δ⋅1
1+(c
K)n=10
0.1⋅1
1+(10
5)2=20 nM
b) The dynamics of the system follow the equation:
m(t)= mss(1−e−δt)
To find the time to reach 50% of steady-state:
0.5 = 1−e−δt
Solving for t:
t = −ln(0.5)
δ= −ln(0.5)
0.1 ≈ 6.93 minutes
c) With c = 20 nM, the new steady-state concentration is:
mss =10
0.1⋅1
1+(20
5)2= 5.56 nM
Therefore: a) The steady-state mRNA concentration is 20 nM. b) It takes approximately 6.93
minutes to reach 50% of the steady-state concentration. c) When the CRISPR-dCas9 complex
concentration is doubled, the new steady-state mRNA concentration is 5.56 nM.
14 CRISPR-BASED TOGGLE SWITCH
Problem 2. Consider a genetic toggle switch implemented using two CRISPR-based
repressors. The system can be described by the following coupled differential equations:
dx
dt = αx⋅1
1+( y
Ky)ny−δxx
dy
dt = αy⋅1
1+( x
Kx)nx−δyy
where x and y are the concentrations of the two repressors.
Given the following parameters:
• αx= αy=100 nM/min
• Kx= Ky=40 nM
• nx= ny= 4
• δx= δy= 1 min−1
a) Find the steady-state concentrations of x and y when the system is balanced.
b) If the initial conditions are x0=10 nM and y0=80 nM, which state will the system likely
settle into?
c) Calculate the switching threshold for x that would cause the system to flip states.
Solution 2. a) At steady-state and balanced conditions, x = y and dx
dt =dy
dt = 0. Solving:
x = y = αx
δx⋅1
1+( x
Kx)nx
Substituting the values:
x = 100⋅1
1+( x
40)4
Solving numerically, we get x = y ≈ 50 nM.
b) With initial conditions x0=10 nM and y0=80 nM, y is much higher than x. The system will
likely settle into the state where y is high and x is low.
To verify, we can calculate the steady-state values:
For x: xss =100⋅1
1+(80
40)4≈ 5.88 nM For y: yss =100⋅1
1+(10
40)4≈98.77 nM
c) The switching threshold occurs when dx
dt =dy
dt = 0 and x = y. We already calculated this in
part (a):
The switching threshold for x is approximately 50 nM.
Therefore: a) The balanced steady-state concentrations are x = y ≈ 50 nM. b) The system will
likely settle into the state where x ≈ 5.88 nM and y ≈ 98.77 nM. c) The switching threshold for x
is approximately 50 nM.
15 CRISPR-BASED OSCILLATOR
Problem 3. Consider a CRISPR-based genetic oscillator consisting of three repressors (X, Y,
and Z) in a cycle. The system can be described by the following differential equations:
dx
dt = αx⋅1
1+( z
Kz)nz−δxx
dy
dt = αy⋅1
1+( x
Kx)nx−δyy
dz
dt = αz⋅1
1+( y
Ky)ny−δzz
Given the following parameters:
• αx= αy= αz=200 nM/min
• Kx= Ky= Kz=40 nM
• nx= ny= nz= 2
• δx= δy= δz= 0.1 min−1
a) Estimate the period of oscillation using the delay between repressors.
b) Calculate the maximum concentration of each repressor during the oscillation.
c) If we increase all Hill coefficients to 4, how would this affect the oscillator’s behavior?
Solution 3. a) To estimate the period, we can consider the time for one repressor to reach its
maximum after another is at its minimum. This is approximately the time to reach steady-state:
tss ≈4
δ=4
0.1 =40 minutes
The full period involves three such delays, so:
Period ≈ 3⋅40 =120 minutes
b) The maximum concentration occurs when a repressor is fully de-repressed. At this point:
xmax =αx
δx=200
0.1 =2000 nM
This is the same for y and z due to symmetry.
c) Increasing the Hill coefficients to 4 would:
• Make the repression more switch-like (steeper response)
• Likely increase the amplitude of oscillations
• Potentially make the oscillations more robust and stable
• Slightly increase the period of oscillation
To quantify these effects precisely would require numerical simulation of the system.
Therefore: a) The estimated period of oscillation is approximately 120 minutes. b) The
maximum concentration of each repressor during oscillation is 2000 nM. c) Increasing Hill
coefficients to 4 would likely result in more robust, higher amplitude oscillations with a slightly
longer period.
Consider a CRISPR interference (CRISPRi) system targeting a gene of interest. The dynamics of
mRNA production and degradation can be modeled by the following differential equation:
dm
dt = α⋅ 1
1+(c
K)n−δm
where:
• m is the concentration of mRNA
• c is the concentration of CRISPR-dCas9 complex
• α is the maximum transcription rate
• K is the repression coefficient
• n is the Hill coefficient
• δ is the mRNA degradation rate
Given the following parameters:
• α = 10 nM/min
• K = 5 nM
• n = 2
• δ = 0.1 min−1
• Initial mRNA concentration m0= 0 nM
• CRISPR-dCas9 complex concentration c = 10 nM
a) Calculate the steady-state mRNA concentration.
b) Determine the time it takes to reach 50% of the steady-state concentration.
c) If the CRISPR-dCas9 complex concentration is doubled to 20 nM, what is the new steady-
state mRNA concentration?
Solution 1. a) To find the steady-state concentration, set dm
dt = 0:
0 = α⋅ 1
1+(c
K)n−δmss
Solving for mss:
mss =α
δ⋅1
1+(c
K)n=10
0.1⋅1
1+(10
5)2=20 nM
b) The dynamics of the system follow the equation:
m(t)= mss(1−e−δt)
To find the time to reach 50% of steady-state:
0.5 = 1−e−δt
Solving for t:
t = −ln(0.5)
δ= −ln(0.5)
0.1 ≈ 6.93 minutes
c) With c = 20 nM, the new steady-state concentration is:
mss =10
0.1⋅1
1+(20
5)2= 5.56 nM
Therefore: a) The steady-state mRNA concentration is 20 nM. b) It takes approximately 6.93
minutes to reach 50% of the steady-state concentration. c) When the CRISPR-dCas9 complex
concentration is doubled, the new steady-state mRNA concentration is 5.56 nM.
16 CRISPR-BASED TOGGLE SWITCH
Problem 2. Consider a genetic toggle switch implemented using two CRISPR-based
repressors. The system can be described by the following coupled differential equations:
dx
dt = αx⋅1
1+( y
Ky)ny−δxx
dy
dt = αy⋅1
1+( x
Kx)nx−δyy
where x and y are the concentrations of the two repressors.
Given the following parameters:
• αx= αy=100 nM/min
• Kx= Ky=40 nM
• nx= ny= 4
• δx= δy= 1 min−1
a) Find the steady-state concentrations of x and y when the system is balanced.
b) If the initial conditions are x0=10 nM and y0=80 nM, which state will the system likely
settle into?
c) Calculate the switching threshold for x that would cause the system to flip states.
Solution 2. a) At steady-state and balanced conditions, x = y and dx
dt =dy
dt = 0. Solving:
x = y = αx
δx⋅1
1+( x
Kx)nx
Substituting the values:
x = 100⋅1
1+( x
40)4
Solving numerically, we get x = y ≈ 50 nM.
b) With initial conditions x0=10 nM and y0=80 nM, y is much higher than x. The system will
likely settle into the state where y is high and x is low.
To verify, we can calculate the steady-state values:
For x: xss =100⋅1
1+(80
40)4≈ 5.88 nM For y: yss =100⋅1
1+(10
40)4≈98.77 nM
c) The switching threshold occurs when dx
dt =dy
dt = 0 and x = y. We already calculated this in
part (a):
The switching threshold for x is approximately 50 nM.
Therefore: a) The balanced steady-state concentrations are x = y ≈ 50 nM. b) The system will
likely settle into the state where x ≈ 5.88 nM and y ≈ 98.77 nM. c) The switching threshold for x
is approximately 50 nM.
17 CRISPR-BASED OSCILLATOR
Problem 3. Consider a CRISPR-based genetic oscillator consisting of three repressors (X, Y,
and Z) in a cycle. The system can be described by the following differential equations:
dx
dt = αx⋅1
1+( z
Kz)nz−δxx
dy
dt = αy⋅1
1+( x
Kx)nx−δyy
dz
dt = αz⋅1
1+( y
Ky)ny−δzz
Given the following parameters:
• αx= αy= αz=200 nM/min
• Kx= Ky= Kz=40 nM
• nx= ny= nz= 2
• δx= δy= δz= 0.1 min−1
a) Estimate the period of oscillation using the delay between repressors.
b) Calculate the maximum concentration of each repressor during the oscillation.
c) If we increase all Hill coefficients to 4, how would this affect the oscillator’s behavior?
Solution 3. a) To estimate the period, we can consider the time for one repressor to reach its
maximum after another is at its minimum. This is approximately the time to reach steady-state:
tss ≈4
δ=4
0.1 =40 minutes
The full period involves three such delays, so:
Period ≈ 3⋅40 =120 minutes
b) The maximum concentration occurs when a repressor is fully de-repressed. At this point:
xmax =αx
δx=200
0.1 =2000 nM
This is the same for y and z due to symmetry.
c) Increasing the Hill coefficients to 4 would:
• Make the repression more switch-like (steeper response)
• Likely increase the amplitude of oscillations
• Potentially make the oscillations more robust and stable
• Slightly increase the period of oscillation
To quantify these effects precisely would require numerical simulation of the system.
Therefore: a) The estimated period of oscillation is approximately 120 minutes. b) The
maximum concentration of each repressor during oscillation is 2000 nM. c) Increasing Hill
coefficients to 4 would likely result in more robust, higher amplitude oscillations with a slightly
longer period.
Consider a CRISPR interference (CRISPRi) system targeting a gene of interest. The dynamics of
mRNA production and degradation can be modeled by the following differential equation:
dm
dt = α⋅ 1
1+(c
K)n−δm
where:
• m is the concentration of mRNA
• c is the concentration of CRISPR-dCas9 complex
• α is the maximum transcription rate
• K is the repression coefficient
• n is the Hill coefficient
• δ is the mRNA degradation rate
Given the following parameters:
• α = 10 nM/min
• K = 5 nM
• n = 2
• δ = 0.1 min−1
• Initial mRNA concentration m0= 0 nM
• CRISPR-dCas9 complex concentration c = 10 nM
a) Calculate the steady-state mRNA concentration.
b) Determine the time it takes to reach 50% of the steady-state concentration.
c) If the CRISPR-dCas9 complex concentration is doubled to 20 nM, what is the new steady-
state mRNA concentration?
Solution 1. a) To find the steady-state concentration, set dm
dt = 0:
0 = α⋅ 1
1+(c
K)n−δmss
Solving for mss:
mss =α
δ⋅1
1+(c
K)n=10
0.1⋅1
1+(10
5)2=20 nM
b) The dynamics of the system follow the equation:
m(t)= mss(1−e−δt)
To find the time to reach 50% of steady-state:
0.5 = 1−e−δt
Solving for t:
t = −ln(0.5)
δ= −ln(0.5)
0.1 ≈ 6.93 minutes
c) With c = 20 nM, the new steady-state concentration is:
mss =10
0.1⋅1
1+(20
5)2= 5.56 nM
Therefore: a) The steady-state mRNA concentration is 20 nM. b) It takes approximately 6.93
minutes to reach 50% of the steady-state concentration. c) When the CRISPR-dCas9 complex
concentration is doubled, the new steady-state mRNA concentration is 5.56 nM.
18 CRISPR-BASED TOGGLE SWITCH
Problem 2. Consider a genetic toggle switch implemented using two CRISPR-based
repressors. The system can be described by the following coupled differential equations:
dx
dt = αx⋅1
1+( y
Ky)ny−δxx
dy
dt = αy⋅1
1+( x
Kx)nx−δyy
where x and y are the concentrations of the two repressors.
Given the following parameters:
• αx= αy=100 nM/min
• Kx= Ky=40 nM
• nx= ny= 4
• δx= δy= 1 min−1
a) Find the steady-state concentrations of x and y when the system is balanced.
b) If the initial conditions are x0=10 nM and y0=80 nM, which state will the system likely
settle into?
c) Calculate the switching threshold for x that would cause the system to flip states.
Solution 2. a) At steady-state and balanced conditions, x = y and dx
dt =dy
dt = 0. Solving:
x = y = αx
δx⋅1
1+( x
Kx)nx
Substituting the values:
x = 100⋅1
1+( x
40)4
Solving numerically, we get x = y ≈ 50 nM.
b) With initial conditions x0=10 nM and y0=80 nM, y is much higher than x. The system will
likely settle into the state where y is high and x is low.
To verify, we can calculate the steady-state values:
For x: xss =100⋅1
1+(80
40)4≈ 5.88 nM For y: yss =100⋅1
1+(10
40)4≈98.77 nM
c) The switching threshold occurs when dx
dt =dy
dt = 0 and x = y. We already calculated this in
part (a):
The switching threshold for x is approximately 50 nM.
Therefore: a) The balanced steady-state concentrations are x = y ≈ 50 nM. b) The system will
likely settle into the state where x ≈ 5.88 nM and y ≈ 98.77 nM. c) The switching threshold for x
is approximately 50 nM.
19 CRISPR-BASED OSCILLATOR
Problem 3. Consider a CRISPR-based genetic oscillator consisting of three repressors (X, Y,
and Z) in a cycle. The system can be described by the following differential equations:
dx
dt = αx⋅1
1+( z
Kz)nz−δxx
dy
dt = αy⋅1
1+( x
Kx)nx−δyy
dz
dt = αz⋅1
1+( y
Ky)ny−δzz
Given the following parameters:
• αx= αy= αz=200 nM/min
• Kx= Ky= Kz=40 nM
• nx= ny= nz= 2
• δx= δy= δz= 0.1 min−1
a) Estimate the period of oscillation using the delay between repressors.
b) Calculate the maximum concentration of each repressor during the oscillation.
c) If we increase all Hill coefficients to 4, how would this affect the oscillator’s behavior?
Solution 3. a) To estimate the period, we can consider the time for one repressor to reach its
maximum after another is at its minimum. This is approximately the time to reach steady-state:
tss ≈4
δ=4
0.1 =40 minutes
The full period involves three such delays, so:
Period ≈ 3⋅40 =120 minutes
b) The maximum concentration occurs when a repressor is fully de-repressed. At this point:
xmax =αx
δx=200
0.1 =2000 nM
This is the same for y and z due to symmetry.
c) Increasing the Hill coefficients to 4 would:
• Make the repression more switch-like (steeper response)
• Likely increase the amplitude of oscillations
• Potentially make the oscillations more robust and stable
• Slightly increase the period of oscillation
To quantify these effects precisely would require numerical simulation of the system.
Therefore: a) The estimated period of oscillation is approximately 120 minutes. b) The
maximum concentration of each repressor during oscillation is 2000 nM. c) Increasing Hill
coefficients to 4 would likely result in more robust, higher amplitude oscillations with a slightly
longer period.
Consider a CRISPR interference (CRISPRi) system targeting a gene of interest. The dynamics of
mRNA production and degradation can be modeled by the following differential equation:
dm
dt = α⋅ 1
1+(c
K)n−δm
where:
• m is the concentration of mRNA
• c is the concentration of CRISPR-dCas9 complex
• α is the maximum transcription rate
• K is the repression coefficient
• n is the Hill coefficient
• δ is the mRNA degradation rate
Given the following parameters:
• α = 10 nM/min
• K = 5 nM
• n = 2
• δ = 0.1 min−1
• Initial mRNA concentration m0= 0 nM
• CRISPR-dCas9 complex concentration c = 10 nM
a) Calculate the steady-state mRNA concentration.
b) Determine the time it takes to reach 50% of the steady-state concentration.
c) If the CRISPR-dCas9 complex concentration is doubled to 20 nM, what is the new steady-
state mRNA concentration?
Solution 1. a) To find the steady-state concentration, set dm
dt = 0:
0 = α⋅ 1
1+(c
K)n−δmss
Solving for mss:
mss =α
δ⋅1
1+(c
K)n=10
0.1⋅1
1+(10
5)2=20 nM
b) The dynamics of the system follow the equation:
m(t)= mss(1−e−δt)
To find the time to reach 50% of steady-state:
0.5 = 1−e−δt
Solving for t:
t = −ln(0.5)
δ= −ln(0.5)
0.1 ≈ 6.93 minutes
c) With c = 20 nM, the new steady-state concentration is:
mss =10
0.1⋅1
1+(20
5)2= 5.56 nM
Therefore: a) The steady-state mRNA concentration is 20 nM. b) It takes approximately 6.93
minutes to reach 50% of the steady-state concentration. c) When the CRISPR-dCas9 complex
concentration is doubled, the new steady-state mRNA concentration is 5.56 nM.
20 CRISPR-BASED TOGGLE SWITCH
Problem 2. Consider a genetic toggle switch implemented using two CRISPR-based
repressors. The system can be described by the following coupled differential equations:
dx
dt = αx⋅1
1+( y
Ky)ny−δxx
dy
dt = αy⋅1
1+( x
Kx)nx−δyy
where x and y are the concentrations of the two repressors.
Given the following parameters:
• αx= αy=100 nM/min
• Kx= Ky=40 nM
• nx= ny= 4
• δx= δy= 1 min−1
a) Find the steady-state concentrations of x and y when the system is balanced.
b) If the initial conditions are x0=10 nM and y0=80 nM, which state will the system likely
settle into?
c) Calculate the switching threshold for x that would cause the system to flip states.
Solution 2. a) At steady-state and balanced conditions, x = y and dx
dt =dy
dt = 0. Solving:
x = y = αx
δx⋅1
1+( x
Kx)nx
Substituting the values:
x = 100⋅1
1+( x
40)4
Solving numerically, we get x = y ≈ 50 nM.
b) With initial conditions x0=10 nM and y0=80 nM, y is much higher than x. The system will
likely settle into the state where y is high and x is low.
To verify, we can calculate the steady-state values:
For x: xss =100⋅1
1+(80
40)4≈ 5.88 nM For y: yss =100⋅1
1+(10
40)4≈98.77 nM
c) The switching threshold occurs when dx
dt =dy
dt = 0 and x = y. We already calculated this in
part (a):
The switching threshold for x is approximately 50 nM.
Therefore: a) The balanced steady-state concentrations are x = y ≈ 50 nM. b) The system will
likely settle into the state where x ≈ 5.88 nM and y ≈ 98.77 nM. c) The switching threshold for x
is approximately 50 nM.
21 CRISPR-BASED OSCILLATOR
Problem 3. Consider a CRISPR-based genetic oscillator consisting of three repressors (X, Y,
and Z) in a cycle. The system can be described by the following differential equations:
dx
dt = αx⋅1
1+( z
Kz)nz−δxx
dy
dt = αy⋅1
1+( x
Kx)nx−δyy
dz
dt = αz⋅1
1+( y
Ky)ny−δzz
Given the following parameters:
• αx= αy= αz=200 nM/min
• Kx= Ky= Kz=40 nM
• nx= ny= nz= 2
• δx= δy= δz= 0.1 min−1
a) Estimate the period of oscillation using the delay between repressors.
b) Calculate the maximum concentration of each repressor during the oscillation.
c) If we increase all Hill coefficients to 4, how would this affect the oscillator’s behavior?
Solution 3. a) To estimate the period, we can consider the time for one repressor to reach its
maximum after another is at its minimum. This is approximately the time to reach steady-state:
tss ≈4
δ=4
0.1 =40 minutes
The full period involves three such delays, so:
Period ≈ 3⋅40 =120 minutes
b) The maximum concentration occurs when a repressor is fully de-repressed. At this point:
xmax =αx
δx=200
0.1 =2000 nM
This is the same for y and z due to symmetry.
c) Increasing the Hill coefficients to 4 would:
• Make the repression more switch-like (steeper response)
• Likely increase the amplitude of oscillations
• Potentially make the oscillations more robust and stable
• Slightly increase the period of oscillation
To quantify these effects precisely would require numerical simulation of the system.
Therefore: a) The estimated period of oscillation is approximately 120 minutes. b) The
maximum concentration of each repressor during oscillation is 2000 nM. c) Increasing Hill
coefficients to 4 would likely result in more robust, higher amplitude oscillations with a slightly
longer period.
Consider a CRISPR interference (CRISPRi) system targeting a gene of interest. The dynamics of
mRNA production and degradation can be modeled by the following differential equation:
dm
dt = α⋅ 1
1+(c
K)n−δm
where:
• m is the concentration of mRNA
• c is the concentration of CRISPR-dCas9 complex
• α is the maximum transcription rate
• K is the repression coefficient
• n is the Hill coefficient
• δ is the mRNA degradation rate
Given the following parameters:
• α = 10 nM/min
• K = 5 nM
• n = 2
• δ = 0.1 min−1
• Initial mRNA concentration m0= 0 nM
• CRISPR-dCas9 complex concentration c = 10 nM
a) Calculate the steady-state mRNA concentration.
b) Determine the time it takes to reach 50% of the steady-state concentration.
c) If the CRISPR-dCas9 complex concentration is doubled to 20 nM, what is the new steady-
state mRNA concentration?
Solution 1. a) To find the steady-state concentration, set dm
dt = 0:
0 = α⋅ 1
1+(c
K)n−δmss
Solving for mss:
mss =α
δ⋅1
1+(c
K)n=10
0.1⋅1
1+(10
5)2=20 nM
b) The dynamics of the system follow the equation:
m(t)= mss(1−e−δt)
To find the time to reach 50% of steady-state:
0.5 = 1−e−δt
Solving for t:
t = −ln(0.5)
δ= −ln(0.5)
0.1 ≈ 6.93 minutes
c) With c = 20 nM, the new steady-state concentration is:
mss =10
0.1⋅1
1+(20
5)2= 5.56 nM
Therefore: a) The steady-state mRNA concentration is 20 nM. b) It takes approximately 6.93
minutes to reach 50% of the steady-state concentration. c) When the CRISPR-dCas9 complex
concentration is doubled, the new steady-state mRNA concentration is 5.56 nM.
22 CRISPR-BASED TOGGLE SWITCH
Problem 2. Consider a genetic toggle switch implemented using two CRISPR-based
repressors. The system can be described by the following coupled differential equations:
dx
dt = αx⋅1
1+( y
Ky)ny−δxx
dy
dt = αy⋅1
1+( x
Kx)nx−δyy
where x and y are the concentrations of the two repressors.
Given the following parameters:
• αx= αy=100 nM/min
• Kx= Ky=40 nM
• nx= ny= 4
• δx= δy= 1 min−1
a) Find the steady-state concentrations of x and y when the system is balanced.
b) If the initial conditions are x0=10 nM and y0=80 nM, which state will the system likely
settle into?
c) Calculate the switching threshold for x that would cause the system to flip states.
Solution 2. a) At steady-state and balanced conditions, x = y and dx
dt =dy
dt = 0. Solving:
x = y = αx
δx⋅1
1+( x
Kx)nx
Substituting the values:
x = 100⋅1
1+( x
40)4
Solving numerically, we get x = y ≈ 50 nM.
b) With initial conditions x0=10 nM and y0=80 nM, y is much higher than x. The system will
likely settle into the state where y is high and x is low.
To verify, we can calculate the steady-state values:
For x: xss =100⋅1
1+(80
40)4≈ 5.88 nM For y: yss =100⋅1
1+(10
40)4≈98.77 nM
c) The switching threshold occurs when dx
dt =dy
dt = 0 and x = y. We already calculated this in
part (a):
The switching threshold for x is approximately 50 nM.
Therefore: a) The balanced steady-state concentrations are x = y ≈ 50 nM. b) The system will
likely settle into the state where x ≈ 5.88 nM and y ≈ 98.77 nM. c) The switching threshold for x
is approximately 50 nM.
23 CRISPR-BASED OSCILLATOR
Problem 3. Consider a CRISPR-based genetic oscillator consisting of three repressors (X, Y,
and Z) in a cycle. The system can be described by the following differential equations:
dx
dt = αx⋅1
1+( z
Kz)nz−δxx
dy
dt = αy⋅1
1+( x
Kx)nx−δyy
dz
dt = αz⋅1
1+( y
Ky)ny−δzz
Given the following parameters:
• αx= αy= αz=200 nM/min
• Kx= Ky= Kz=40 nM
• nx= ny= nz= 2
• δx= δy= δz= 0.1 min−1
a) Estimate the period of oscillation using the delay between repressors.
b) Calculate the maximum concentration of each repressor during the oscillation.
c) If we increase all Hill coefficients to 4, how would this affect the oscillator’s behavior?
Solution 3. a) To estimate the period, we can consider the time for one repressor to reach its
maximum after another is at its minimum. This is approximately the time to reach steady-state:
tss ≈4
δ=4
0.1 =40 minutes
The full period involves three such delays, so:
Period ≈ 3⋅40 =120 minutes
b) The maximum concentration occurs when a repressor is fully de-repressed. At this point:
xmax =αx
δx=200
0.1 =2000 nM
This is the same for y and z due to symmetry.
c) Increasing the Hill coefficients to 4 would:
• Make the repression more switch-like (steeper response)
• Likely increase the amplitude of oscillations
• Potentially make the oscillations more robust and stable
• Slightly increase the period of oscillation
To quantify these effects precisely would require numerical simulation of the system.
Therefore: a) The estimated period of oscillation is approximately 120 minutes. b) The
maximum concentration of each repressor during oscillation is 2000 nM. c) Increasing Hill
coefficients to 4 would likely result in more robust, higher amplitude oscillations with a slightly
longer period.
Consider a CRISPR interference (CRISPRi) system targeting a gene of interest. The dynamics of
mRNA production and degradation can be modeled by the following differential equation:
dm
dt = α⋅ 1
1+(c
K)n−δm
where:
• m is the concentration of mRNA
• c is the concentration of CRISPR-dCas9 complex
• α is the maximum transcription rate
• K is the repression coefficient
• n is the Hill coefficient
• δ is the mRNA degradation rate
Given the following parameters:
• α = 10 nM/min
• K = 5 nM
• n = 2
• δ = 0.1 min−1
• Initial mRNA concentration m0= 0 nM
• CRISPR-dCas9 complex concentration c = 10 nM
a) Calculate the steady-state mRNA concentration.
b) Determine the time it takes to reach 50% of the steady-state concentration.
c) If the CRISPR-dCas9 complex concentration is doubled to 20 nM, what is the new steady-
state mRNA concentration?
Solution 1. a) To find the steady-state concentration, set dm
dt = 0:
0 = α⋅ 1
1+(c
K)n−δmss
Solving for mss:
mss =α
δ⋅1
1+(c
K)n=10
0.1⋅1
1+(10
5)2=20 nM
b) The dynamics of the system follow the equation:
m(t)= mss(1−e−δt)
To find the time to reach 50% of steady-state:
0.5 = 1−e−δt
Solving for t:
t = −ln(0.5)
δ= −ln(0.5)
0.1 ≈ 6.93 minutes
c) With c = 20 nM, the new steady-state concentration is:
mss =10
0.1⋅1
1+(20
5)2= 5.56 nM
Therefore: a) The steady-state mRNA concentration is 20 nM. b) It takes approximately 6.93
minutes to reach 50% of the steady-state concentration. c) When the CRISPR-dCas9 complex
concentration is doubled, the new steady-state mRNA concentration is 5.56 nM.
24 CRISPR-BASED TOGGLE SWITCH
Problem 2. Consider a genetic toggle switch implemented using two CRISPR-based
repressors. The system can be described by the following coupled differential equations:
dx
dt = αx⋅1
1+( y
Ky)ny−δxx
dy
dt = αy⋅1
1+( x
Kx)nx−δyy
where x and y are the concentrations of the two repressors.
Given the following parameters:
• αx= αy=100 nM/min
• Kx= Ky=40 nM
• nx= ny= 4
• δx= δy= 1 min−1
a) Find the steady-state concentrations of x and y when the system is balanced.
b) If the initial conditions are x0=10 nM and y0=80 nM, which state will the system likely
settle into?
c) Calculate the switching threshold for x that would cause the system to flip states.
Solution 2. a) At steady-state and balanced conditions, x = y and dx
dt =dy
dt = 0. Solving:
x = y = αx
δx⋅1
1+( x
Kx)nx
Substituting the values:
x = 100⋅1
1+( x
40)4
Solving numerically, we get x = y ≈ 50 nM.
b) With initial conditions x0=10 nM and y0=80 nM, y is much higher than x. The system will
likely settle into the state where y is high and x is low.
To verify, we can calculate the steady-state values:
For x: xss =100⋅1
1+(80
40)4≈ 5.88 nM For y: yss =100⋅1
1+(10
40)4≈98.77 nM
c) The switching threshold occurs when dx
dt =dy
dt = 0 and x = y. We already calculated this in
part (a):
The switching threshold for x is approximately 50 nM.
Therefore: a) The balanced steady-state concentrations are x = y ≈ 50 nM. b) The system will
likely settle into the state where x ≈ 5.88 nM and y ≈ 98.77 nM. c) The switching threshold for x
is approximately 50 nM.
25 CRISPR-BASED OSCILLATOR
Problem 3. Consider a CRISPR-based genetic oscillator consisting of three repressors (X, Y,
and Z) in a cycle. The system can be described by the following differential equations:
dx
dt = αx⋅1
1+( z
Kz)nz−δxx
dy
dt = αy⋅1
1+( x
Kx)nx−δyy
dz
dt = αz⋅1
1+( y
Ky)ny−δzz
Given the following parameters:
• αx= αy= αz=200 nM/min
• Kx= Ky= Kz=40 nM
• nx= ny= nz= 2
• δx= δy= δz= 0.1 min−1
a) Estimate the period of oscillation using the delay between repressors.
b) Calculate the maximum concentration of each repressor during the oscillation.
c) If we increase all Hill coefficients to 4, how would this affect the oscillator’s behavior?
Solution 3. a) To estimate the period, we can consider the time for one repressor to reach its
maximum after another is at its minimum. This is approximately the time to reach steady-state:
tss ≈4
δ=4
0.1 =40 minutes
The full period involves three such delays, so:
Period ≈ 3⋅40 =120 minutes
b) The maximum concentration occurs when a repressor is fully de-repressed. At this point:
xmax =αx
δx=200
0.1 =2000 nM
This is the same for y and z due to symmetry.
c) Increasing the Hill coefficients to 4 would:
• Make the repression more switch-like (steeper response)
• Likely increase the amplitude of oscillations
• Potentially make the oscillations more robust and stable
• Slightly increase the period of oscillation
To quantify these effects precisely would require numerical simulation of the system.
Therefore: a) The estimated period of oscillation is approximately 120 minutes. b) The
maximum concentration of each repressor during oscillation is 2000 nM. c) Increasing Hill
coefficients to 4 would likely result in more robust, higher amplitude oscillations with a slightly
longer period.
Consider a CRISPR interference (CRISPRi) system targeting a gene of interest. The dynamics of
mRNA production and degradation can be modeled by the following differential equation:
dm
dt = α⋅ 1
1+(c
K)n−δm
where:
• m is the concentration of mRNA
• c is the concentration of CRISPR-dCas9 complex
• α is the maximum transcription rate
• K is the repression coefficient
• n is the Hill coefficient
• δ is the mRNA degradation rate
Given the following parameters:
• α = 10 nM/min
• K = 5 nM
• n = 2
• δ = 0.1 min−1
• Initial mRNA concentration m0= 0 nM
• CRISPR-dCas9 complex concentration c = 10 nM
a) Calculate the steady-state mRNA concentration.
b) Determine the time it takes to reach 50% of the steady-state concentration.
c) If the CRISPR-dCas9 complex concentration is doubled to 20 nM, what is the new steady-
state mRNA concentration?
Solution 1. a) To find the steady-state concentration, set dm
dt = 0:
0 = α⋅ 1
1+(c
K)n−δmss
Solving for mss:
mss =α
δ⋅1
1+(c
K)n=10
0.1⋅1
1+(10
5)2=20 nM
b) The dynamics of the system follow the equation:
m(t)= mss(1−e−δt)
To find the time to reach 50% of steady-state:
0.5 = 1−e−δt
Solving for t:
t = −ln(0.5)
δ= −ln(0.5)
0.1 ≈ 6.93 minutes
c) With c = 20 nM, the new steady-state concentration is:
mss =10
0.1⋅1
1+(20
5)2= 5.56 nM
Therefore: a) The steady-state mRNA concentration is 20 nM. b) It takes approximately 6.93
minutes to reach 50% of the steady-state concentration. c) When the CRISPR-dCas9 complex
concentration is doubled, the new steady-state mRNA concentration is 5.56 nM.
26 CRISPR-BASED TOGGLE SWITCH
Problem 2. Consider a genetic toggle switch implemented using two CRISPR-based
repressors. The system can be described by the following coupled differential equations:
dx
dt = αx⋅1
1+( y
Ky)ny−δxx
dy
dt = αy⋅1
1+( x
Kx)nx−δyy
where x and y are the concentrations of the two repressors.
Given the following parameters:
• αx= αy=100 nM/min
• Kx= Ky=40 nM
• nx= ny= 4
• δx= δy= 1 min−1
a) Find the steady-state concentrations of x and y when the system is balanced.
b) If the initial conditions are x0=10 nM and y0=80 nM, which state will the system likely
settle into?
c) Calculate the switching threshold for x that would cause the system to flip states.
Solution 2. a) At steady-state and balanced conditions, x = y and dx
dt =dy
dt = 0. Solving:
x = y = αx
δx⋅1
1+( x
Kx)nx
Substituting the values:
x = 100⋅1
1+( x
40)4
Solving numerically, we get x = y ≈ 50 nM.
b) With initial conditions x0=10 nM and y0=80 nM, y is much higher than x. The system will
likely settle into the state where y is high and x is low.
To verify, we can calculate the steady-state values:
For x: xss =100⋅1
1+(80
40)4≈ 5.88 nM For y: yss =100⋅1
1+(10
40)4≈98.77 nM
c) The switching threshold occurs when dx
dt =dy
dt = 0 and x = y. We already calculated this in
part (a):
The switching threshold for x is approximately 50 nM.
Therefore: a) The balanced steady-state concentrations are x = y ≈ 50 nM. b) The system will
likely settle into the state where x ≈ 5.88 nM and y ≈ 98.77 nM. c) The switching threshold for x
is approximately 50 nM.
27 CRISPR-BASED OSCILLATOR
Problem 3. Consider a CRISPR-based genetic oscillator consisting of three repressors (X, Y,
and Z) in a cycle. The system can be described by the following differential equations:
dx
dt = αx⋅1
1+( z
Kz)nz−δxx
dy
dt = αy⋅1
1+( x
Kx)nx−δyy
dz
dt = αz⋅1
1+( y
Ky)ny−δzz
Given the following parameters:
• αx= αy= αz=200 nM/min
• Kx= Ky= Kz=40 nM
• nx= ny= nz= 2
• δx= δy= δz= 0.1 min−1
a) Estimate the period of oscillation using the delay between repressors.
b) Calculate the maximum concentration of each repressor during the oscillation.
c) If we increase all Hill coefficients to 4, how would this affect the oscillator’s behavior?
Solution 3. a) To estimate the period, we can consider the time for one repressor to reach its
maximum after another is at its minimum. This is approximately the time to reach steady-state:
tss ≈4
δ=4
0.1 =40 minutes
The full period involves three such delays, so:
Period ≈ 3⋅40 =120 minutes
b) The maximum concentration occurs when a repressor is fully de-repressed. At this point:
xmax =αx
δx=200
0.1 =2000 nM
This is the same for y and z due to symmetry.
c) Increasing the Hill coefficients to 4 would:
• Make the repression more switch-like (steeper response)
• Likely increase the amplitude of oscillations
• Potentially make the oscillations more robust and stable
• Slightly increase the period of oscillation
To quantify these effects precisely would require numerical simulation of the system.
Therefore: a) The estimated period of oscillation is approximately 120 minutes. b) The
maximum concentration of each repressor during oscillation is 2000 nM. c) Increasing Hill
coefficients to 4 would likely result in more robust, higher amplitude oscillations with a slightly
longer period.
Consider a CRISPR interference (CRISPRi) system targeting a gene of interest. The dynamics of
mRNA production and degradation can be modeled by the following differential equation:
dm
dt = α⋅ 1
1+(c
K)n−δm
where:
• m is the concentration of mRNA
• c is the concentration of CRISPR-dCas9 complex
• α is the maximum transcription rate
• K is the repression coefficient
• n is the Hill coefficient
• δ is the mRNA degradation rate
Given the following parameters:
• α = 10 nM/min
• K = 5 nM
• n = 2
• δ = 0.1 min−1
• Initial mRNA concentration m0= 0 nM
• CRISPR-dCas9 complex concentration c = 10 nM
a) Calculate the steady-state mRNA concentration.
b) Determine the time it takes to reach 50% of the steady-state concentration.
c) If the CRISPR-dCas9 complex concentration is doubled to 20 nM, what is the new steady-
state mRNA concentration?
Solution 1. a) To find the steady-state concentration, set dm
dt = 0:
0 = α⋅ 1
1+(c
K)n−δmss
Solving for mss:
mss =α
δ⋅1
1+(c
K)n=10
0.1⋅1
1+(10
5)2=20 nM
b) The dynamics of the system follow the equation:
m(t)= mss(1−e−δt)
To find the time to reach 50% of steady-state:
0.5 = 1−e−δt
Solving for t:
t = −ln(0.5)
δ= −ln(0.5)
0.1 ≈ 6.93 minutes
c) With c = 20 nM, the new steady-state concentration is:
mss =10
0.1⋅1
1+(20
5)2= 5.56 nM
Therefore: a) The steady-state mRNA concentration is 20 nM. b) It takes approximately 6.93
minutes to reach 50% of the steady-state concentration. c) When the CRISPR-dCas9 complex
concentration is doubled, the new steady-state mRNA concentration is 5.56 nM.
28 CRISPR-BASED TOGGLE SWITCH
Problem 2. Consider a genetic toggle switch implemented using two CRISPR-based
repressors. The system can be described by the following coupled differential equations:
dx
dt = αx⋅1
1+( y
Ky)ny−δxx
dy
dt = αy⋅1
1+( x
Kx)nx−δyy
where x and y are the concentrations of the two repressors.
Given the following parameters:
• αx= αy=100 nM/min
• Kx= Ky=40 nM
• nx= ny= 4
• δx= δy= 1 min−1
a) Find the steady-state concentrations of x and y when the system is balanced.
b) If the initial conditions are x0=10 nM and y0=80 nM, which state will the system likely
settle into?
c) Calculate the switching threshold for x that would cause the system to flip states.
Solution 2. a) At steady-state and balanced conditions, x = y and dx
dt =dy
dt = 0. Solving:
x = y = αx
δx⋅1
1+( x
Kx)nx
Substituting the values:
x = 100⋅1
1+( x
40)4
Solving numerically, we get x = y ≈ 50 nM.
b) With initial conditions x0=10 nM and y0=80 nM, y is much higher than x. The system will
likely settle into the state where y is high and x is low.
To verify, we can calculate the steady-state values:
For x: xss =100⋅1
1+(80
40)4≈ 5.88 nM For y: yss =100⋅1
1+(10
40)4≈98.77 nM
c) The switching threshold occurs when dx
dt =dy
dt = 0 and x = y. We already calculated this in
part (a):
The switching threshold for x is approximately 50 nM.
Therefore: a) The balanced steady-state concentrations are x = y ≈ 50 nM. b) The system will
likely settle into the state where x ≈ 5.88 nM and y ≈ 98.77 nM. c) The switching threshold for x
is approximately 50 nM.
29 CRISPR-BASED OSCILLATOR
Problem 3. Consider a CRISPR-based genetic oscillator consisting of three repressors (X, Y,
and Z) in a cycle. The system can be described by the following differential equations:
dx
dt = αx⋅1
1+( z
Kz)nz−δxx
dy
dt = αy⋅1
1+( x
Kx)nx−δyy
dz
dt = αz⋅1
1+( y
Ky)ny−δzz
Given the following parameters:
• αx= αy= αz=200 nM/min
• Kx= Ky= Kz=40 nM
• nx= ny= nz= 2
• δx= δy= δz= 0.1 min−1
a) Estimate the period of oscillation using the delay between repressors.
b) Calculate the maximum concentration of each repressor during the oscillation.
c) If we increase all Hill coefficients to 4, how would this affect the oscillator’s behavior?
Solution 3. a) To estimate the period, we can consider the time for one repressor to reach its
maximum after another is at its minimum. This is approximately the time to reach steady-state:
tss ≈4
δ=4
0.1 =40 minutes
The full period involves three such delays, so:
Period ≈ 3⋅40 =120 minutes
b) The maximum concentration occurs when a repressor is fully de-repressed. At this point:
xmax =αx
δx=200
0.1 =2000 nM
This is the same for y and z due to symmetry.
c) Increasing the Hill coefficients to 4 would:
• Make the repression more switch-like (steeper response)
• Likely increase the amplitude of oscillations
• Potentially make the oscillations more robust and stable
• Slightly increase the period of oscillation
To quantify these effects precisely would require numerical simulation of the system.
Therefore: a) The estimated period of oscillation is approximately 120 minutes. b) The
maximum concentration of each repressor during oscillation is 2000 nM. c) Increasing Hill
coefficients to 4 would likely result in more robust, higher amplitude oscillations with a slightly
longer period.
Consider a CRISPR interference (CRISPRi) system targeting a gene of interest. The dynamics of
mRNA production and degradation can be modeled by the following differential equation:
dm
dt = α⋅ 1
1+(c
K)n−δm
where:
• m is the concentration of mRNA
• c is the concentration of CRISPR-dCas9 complex
• α is the maximum transcription rate
• K is the repression coefficient
• n is the Hill coefficient
• δ is the mRNA degradation rate
Given the following parameters:
• α = 10 nM/min
• K = 5 nM
• n = 2
• δ = 0.1 min−1
• Initial mRNA concentration m0= 0 nM
• CRISPR-dCas9 complex concentration c = 10 nM
a) Calculate the steady-state mRNA concentration.
b) Determine the time it takes to reach 50% of the steady-state concentration.
c) If the CRISPR-dCas9 complex concentration is doubled to 20 nM, what is the new steady-
state mRNA concentration?
Solution 1. a) To find the steady-state concentration, set dm
dt = 0:
0 = α⋅ 1
1+(c
K)n−δmss
Solving for mss:
mss =α
δ⋅1
1+(c
K)n=10
0.1⋅1
1+(10
5)2=20 nM
b) The dynamics of the system follow the equation:
m(t)= mss(1−e−δt)
To find the time to reach 50% of steady-state:
0.5 = 1−e−δt
Solving for t:
t = −ln(0.5)
δ= −ln(0.5)
0.1 ≈ 6.93 minutes
c) With c = 20 nM, the new steady-state concentration is:
mss =10
0.1⋅1
1+(20
5)2= 5.56 nM
Therefore: a) The steady-state mRNA concentration is 20 nM. b) It takes approximately 6.93
minutes to reach 50% of the steady-state concentration. c) When the CRISPR-dCas9 complex
concentration is doubled, the new steady-state mRNA concentration is 5.56 nM.
30 CRISPR-BASED TOGGLE SWITCH
Problem 2. Consider a genetic toggle switch implemented using two CRISPR-based
repressors. The system can be described by the following coupled differential equations:
dx
dt = αx⋅1
1+( y
Ky)ny−δxx
dy
dt = αy⋅1
1+( x
Kx)nx−δyy
where x and y are the concentrations of the two repressors.
Given the following parameters:
• αx= αy=100 nM/min
• Kx= Ky=40 nM
• nx= ny= 4
• δx= δy= 1 min−1
a) Find the steady-state concentrations of x and y when the system is balanced.
b) If the initial conditions are x0=10 nM and y0=80 nM, which state will the system likely
settle into?
c) Calculate the switching threshold for x that would cause the system to flip states.
Solution 2. a) At steady-state and balanced conditions, x = y and dx
dt =dy
dt = 0. Solving:
x = y = αx
δx⋅1
1+( x
Kx)nx
Substituting the values:
x = 100⋅1
1+( x
40)4
Solving numerically, we get x = y ≈ 50 nM.
b) With initial conditions x0=10 nM and y0=80 nM, y is much higher than x. The system will
likely settle into the state where y is high and x is low.
To verify, we can calculate the steady-state values:
For x: xss =100⋅1
1+(80
40)4≈ 5.88 nM For y: yss =100⋅1
1+(10
40)4≈98.77 nM
c) The switching threshold occurs when dx
dt =dy
dt = 0 and x = y. We already calculated this in
part (a):
The switching threshold for x is approximately 50 nM.
Therefore: a) The balanced steady-state concentrations are x = y ≈ 50 nM. b) The system will
likely settle into the state where x ≈ 5.88 nM and y ≈ 98.77 nM. c) The switching threshold for x
is approximately 50 nM.
31 CRISPR-BASED OSCILLATOR
Problem 3. Consider a CRISPR-based genetic oscillator consisting of three repressors (X, Y,
and Z) in a cycle. The system can be described by the following differential equations:
dx
dt = αx⋅1
1+( z
Kz)nz−δxx
dy
dt = αy⋅1
1+( x
Kx)nx−δyy
dz
dt = αz⋅1
1+( y
Ky)ny−δzz
Given the following parameters:
• αx= αy= αz=200 nM/min
• Kx= Ky= Kz=40 nM
• nx= ny= nz= 2
• δx= δy= δz= 0.1 min−1
a) Estimate the period of oscillation using the delay between repressors.
b) Calculate the maximum concentration of each repressor during the oscillation.
c) If we increase all Hill coefficients to 4, how would this affect the oscillator’s behavior?
Solution 3. a) To estimate the period, we can consider the time for one repressor to reach its
maximum after another is at its minimum. This is approximately the time to reach steady-state:
tss ≈4
δ=4
0.1 =40 minutes
The full period involves three such delays, so:
Period ≈ 3⋅40 =120 minutes
b) The maximum concentration occurs when a repressor is fully de-repressed. At this point:
xmax =αx
δx=200
0.1 =2000 nM
This is the same for y and z due to symmetry.
c) Increasing the Hill coefficients to 4 would:
• Make the repression more switch-like (steeper response)
• Likely increase the amplitude of oscillations
• Potentially make the oscillations more robust and stable
• Slightly increase the period of oscillation
To quantify these effects precisely would require numerical simulation of the system.
Therefore: a) The estimated period of oscillation is approximately 120 minutes. b) The
maximum concentration of each repressor during oscillation is 2000 nM. c) Increasing Hill
coefficients to 4 would likely result in more robust, higher amplitude oscillations with a slightly
longer period.
Consider a CRISPR interference (CRISPRi) system targeting a gene of interest. The dynamics of
mRNA production and degradation can be modeled by the following differential equation:
dm
dt = α⋅ 1
1+(c
K)n−δm
where:
• m is the concentration of mRNA
• c is the concentration of CRISPR-dCas9 complex
• α is the maximum transcription rate
• K is the repression coefficient
• n is the Hill coefficient
• δ is the mRNA degradation rate
Given the following parameters:
• α = 10 nM/min
• K = 5 nM
• n = 2
• δ = 0.1 min−1
• Initial mRNA concentration m0= 0 nM
• CRISPR-dCas9 complex concentration c = 10 nM
a) Calculate the steady-state mRNA concentration.
b) Determine the time it takes to reach 50% of the steady-state concentration.
c) If the CRISPR-dCas9 complex concentration is doubled to 20 nM, what is the new steady-
state mRNA concentration?
Solution 1. a) To find the steady-state concentration, set dm
dt = 0:
0 = α⋅ 1
1+(c
K)n−δmss
Solving for mss:
mss =α
δ⋅1
1+(c
K)n=10
0.1⋅1
1+(10
5)2=20 nM
b) The dynamics of the system follow the equation:
m(t)= mss(1−e−δt)
To find the time to reach 50% of steady-state:
0.5 = 1−e−δt
Solving for t:
t = −ln(0.5)
δ= −ln(0.5)
0.1 ≈ 6.93 minutes
c) With c = 20 nM, the new steady-state concentration is:
mss =10
0.1⋅1
1+(20
5)2= 5.56 nM
Therefore: a) The steady-state mRNA concentration is 20 nM. b) It takes approximately 6.93
minutes to reach 50% of the steady-state concentration. c) When the CRISPR-dCas9 complex
concentration is doubled, the new steady-state mRNA concentration is 5.56 nM.
32 CRISPR-BASED TOGGLE SWITCH
Problem 2. Consider a genetic toggle switch implemented using two CRISPR-based
repressors. The system can be described by the following coupled differential equations:
dx
dt = αx⋅1
1+( y
Ky)ny−δxx
dy
dt = αy⋅1
1+( x
Kx)nx−δyy
where x and y are the concentrations of the two repressors.
Given the following parameters:
• αx= αy=100 nM/min
• Kx= Ky=40 nM
• nx= ny= 4
• δx= δy= 1 min−1
a) Find the steady-state concentrations of x and y when the system is balanced.
b) If the initial conditions are x0=10 nM and y0=80 nM, which state will the system likely
settle into?
c) Calculate the switching threshold for x that would cause the system to flip states.
Solution 2. a) At steady-state and balanced conditions, x = y and dx
dt =dy
dt = 0. Solving:
x = y = αx
δx⋅1
1+( x
Kx)nx
Substituting the values:
x = 100⋅1
1+( x
40)4
Solving numerically, we get x = y ≈ 50 nM.
b) With initial conditions x0=10 nM and y0=80 nM, y is much higher than x. The system will
likely settle into the state where y is high and x is low.
To verify, we can calculate the steady-state values:
For x: xss =100⋅1
1+(80
40)4≈ 5.88 nM For y: yss =100⋅1
1+(10
40)4≈98.77 nM
c) The switching threshold occurs when dx
dt =dy
dt = 0 and x = y. We already calculated this in
part (a):
The switching threshold for x is approximately 50 nM.
Therefore: a) The balanced steady-state concentrations are x = y ≈ 50 nM. b) The system will
likely settle into the state where x ≈ 5.88 nM and y ≈ 98.77 nM. c) The switching threshold for x
is approximately 50 nM.
33 CRISPR-BASED OSCILLATOR
Problem 3. Consider a CRISPR-based genetic oscillator consisting of three repressors (X, Y,
and Z) in a cycle. The system can be described by the following differential equations:
dx
dt = αx⋅1
1+( z
Kz)nz−δxx
dy
dt = αy⋅1
1+( x
Kx)nx−δyy
dz
dt = αz⋅1
1+( y
Ky)ny−δzz
Given the following parameters:
• αx= αy= αz=200 nM/min
• Kx= Ky= Kz=40 nM
• nx= ny= nz= 2
• δx= δy= δz= 0.1 min−1
a) Estimate the period of oscillation using the delay between repressors.
b) Calculate the maximum concentration of each repressor during the oscillation.
c) If we increase all Hill coefficients to 4, how would this affect the oscillator’s behavior?
Solution 3. a) To estimate the period, we can consider the time for one repressor to reach its
maximum after another is at its minimum. This is approximately the time to reach steady-state:
tss ≈4
δ=4
0.1 =40 minutes
The full period involves three such delays, so:
Period ≈ 3⋅40 =120 minutes
b) The maximum concentration occurs when a repressor is fully de-repressed. At this point:
xmax =αx
δx=200
0.1 =2000 nM
This is the same for y and z due to symmetry.
c) Increasing the Hill coefficients to 4 would:
• Make the repression more switch-like (steeper response)
• Likely increase the amplitude of oscillations
• Potentially make the oscillations more robust and stable
• Slightly increase the period of oscillation
To quantify these effects precisely would require numerical simulation of the system.
Therefore: a) The estimated period of oscillation is approximately 120 minutes. b) The
maximum concentration of each repressor during oscillation is 2000 nM. c) Increasing Hill
coefficients to 4 would likely result in more robust, higher amplitude oscillations with a slightly
longer period.
