3 | Deterministic reaction kinetics: the law of mass action Flashcards
The (volume-dependent) in vivo infection rate constant of HIV infecting a CD4+ T-cell
HIV + CD4+T-cel l⟶ ( βV invivo) infected cell
has been estimated to be βVinvivo = 8⋅10−12 per day.
Assume that in vivo infection takes place in the plasma and interstitial lymph with average volume 41 ml per kg body weight and 121 ml per kg body weight, respectively, and that measurements were taken in a 70 kg adult.
What is the infection rate constant per day in an in vitro experiment with volume Vinvitro = 200μL?
Note: give the answer in units 10−7 per day, round to one decimal digit and provide the decimal point as a comma, for example if the answer was 2.162⋅10−7 per day you would have to enter 2,2.
(ungraded quiz)
- We have a second-order reaction. Hence, βV = β / NAV
- We calculate: βVinvitro = β / NAVinvitro = βVinvivo * ( Vinvivo / Vinvitro )
- According to the instructions, Vinvivo= (0,041 + 0,121)L/kg⋅70kg = 11,34L.
- Doing the calculations, we obtain βVinvitro = 4,536⋅10−7 per day.
–> The correct answer is: 4,5
Consider the following reaction system:
R1: A+B→C | R2∗→B | R3: B+C→2A | R4: C→∗
What is the corresponding ODE system (in molar conc and ordered ABC)?
a.
x′1 = −k1x1x2 + 2k2x2x3
x′2 = −k1x1x2 − k2x2x3
x′3 = k1x1x2 − k2x2x3
b.
x′1 = −k1x1x2 + 2k3x2x3
x′2 = −k1x1x2 + k2 − k3x2x3
x′3 = k1x1x2 − k3x2x3 − k4x3
c.
x′1 = k1x1x2 − k3x2x3
x′2 = k1x1x2 + k2 + k3x2x3
x′3 = − k1x1x2 + k3x2x3 − k4x3
d.
x′1 = k1x1x2 + 2k3x2x3
x′2 = k1x1x2 + k2x2 − k3x2x3
x′3 = −k1x1x2 − k3x2x3 − k4
(ungraded quiz)
b.
x′1 = −k1x1x2 + 2k3x2x3
x′2 = −k1x1x2 + k2 − k3x2x3
x′3 = k1x1x2 − k3x2x3 − k4x3
Assume that a molecular species is present within a cell at concentration 5nM, and that the cell is growing exponentially with growth rate kgrowth = 0.1/h.
Assuming that the molecular species is neither produced nor degraded or involved in any reaction: what is its concentration after three hours?
(ungraded quiz)
Since no reactions take place, we obtain the ODE
x′=−kgrowthx,
which can be solved to yield
x(t)=x0e−kgrowtht
Using the stated numbers we get
x(3 h)=5 nMe−0.3=3.704 nM
The correct answer is: 3.7
Consider the reaction scheme for species (An, Ac), within reaction volumes Vn and Vc, respectively:
An → Ac
Ac → ∗
What is the ODE system corresponding to this scheme, stated in molar concentrations (and ordered as stated above)?
a.
x1’ = -k1 Vc/Vn x1
x2’ = k1 x1 - k2 x2
b.
x1’ = -k1 x1
x2’ = k1 Vn/Vc x1 - k2 x2
c.
x1’ = -k1 Vn/Vc x1
x2’ = k1 x1 - k2 x2
d.
x1’ = -k1 x1
x2’ = k1 x1 - k2 Vc/Vn x2
e.
x1’ = -k1 x1
x2’ = k1 x1 - k2 Vn/Vc x2
f.
x1’ = -k1 x1
x2’ = k1 Vc/Vn x1 - k2 x2
The correct answer is:
b.
x1’ = -k1 x1
x2’ = k1 Vn/Vc x1 - k2 x2
Explanation:
k2 Vc/Vn x2:
This accounts for the conversion of An to Ac considering the volume ratios.
What is the law of mass action according to Guldberg and Waage (1864-1879)?
The law of mass action states:
For an elementary reaction Rμ in a spatially homogeneous reaction system with constant volume:
The reaction rate αμ is proportional to the product of the molar concentrations of the reactant molecules involved.
Mathematically expressed as:
αμ(x(t)) = kμ · x1(t)rμ,1 · … · xN(t)rμ,N
Where:
kμ is the reaction rate constant
xi(t) are molar concentrations
rμ,i are stoichiometric coefficients
fundamental in chemical kinetics for describing reaction rates.