Question

Two intracellular molecules A and B, are normally synthesized at a constant rate of 1000 molecules per second per cell. The degradation rate of A is 0.01/s and B is 0.1/s

A}Write a differential equation to describe how the number of molecules of A and B change over time

B}What is the steady state value of A and B? How many molecules of A and B will there be after 1 second at steady state?

C}If the rate of synthesis of both A and B were suddenly increased 10-fold to 10,000 molecules per second—without any change in their degradation rate—how many molecules of A and B would there be after one second?

D}Which molecules would be preferred for responding rapidly changing environment? Explain your answer

Answer 1

Two intracellular molecules A and B, are normally synthesized at a constant rate of 1000 molecules per second per cell.
The degradation rate of A is 0.01/s and B is 0.1/s

A}Write a differential equation to describe how the number of molecules of A and B change over time

Rate of synthesis of A, rsA = 1000 molecules/sec
Rate of degradation of A, rdA = -k*A = -0.01*A molecules/s
Rate of reaction of A, rA, d[A]/dt = 1000 - 0.01*A molecules/s

Rate of synthesis of B, rsB = 1000 molecules/sec
Rate of degradation of B, rdB = -k*B = -0.1*B molecules/s
Rate of reaction of B, rB, d[B]/dt = 1000 -0.1*B molecules/s

B}
(i) What is the steady state value of A and B?
Steady state: d[A]/dt = 0 and d[B]/dt = 0

1000 - 0.01*A = 0
Ass = 1000/0.01 = 100,000 molecules

1000 -0.1*B = 0
Bss = 10,000 molecules

(ii) How many molecules of A and B will there be after 1 second at steady state?
After steady steady state, A and B values will be remain constant as d[A]/dt = 0 and d[B]/dt = 0
So, A = Ass = 100,000 molecules; B = Bss = 10,000 molecules

C}If the rate of synthesis of both A and B were suddenly increased 10-fold to 10,000 molecules per second—without any

change in their degradation rate—how many molecules of A and B would there be after one second?

Now, d[A]/dt = 10,000 - 0.01*[A ]
Integrating, A from steady state value Ass to A and time t=0 to t
t = -2.303/0.01*Log[(10,000 - 0.01*[A])/(10,000 - 0.01*[Ass])]

Substitute, t = 1s, Ass = 100,000
Log[(10,000 - 0.01*[A])/(10,000 - 0.01*[Ass])] = -0.01/2.303
(10,000 - 0.01*[A])/(10,000 - 0.01*[Ass]) = 10^(-0.01/2.303) = 0.99

(10,000 - 0.01*[A]) = 0.99*(10,000 - 0.01*100,000) = 8910
[A] = 100,900 molecules

Now, d[B]/dt = 10000 - 0.1*[B ]
Integrating, B from steady state value Bss to B and time t=0 to t
t = -2.303/0.1*Log[(10000 - 0.1*[B])/(10000 - 0.1*[Bss])]

Substitute, t = 1s, Bss = 10,000
Log[(10000 - 0.1*[B])/(10000 - 0.1*[Bss])] = -0.1/2.303
(10,000 - 0.1*[B])/(10,000 - 0.1*[Bss]) = 10^(-0.1/2.303) = 0.9

(10,000 - 0.1*[B]) = 0.9*(10,000 - 0.1*10,000) = 8100
[B] = 19,000 molecules

D}Which molecules would be preferred for responding rapidly changing environment? Explain your answer
Change in A molecules from steady state value = A - Ass = 100,900 -100,000 = 900 molecules
Change in B molecules from steady state value = B - Bss = 19,000 - 10,000 = 9,000 molecules

B responds rapidly due to higher degradation constant.