1. Physics 414: Introduction to Biophysics Professor Henry Greenside November 21, 2017

2. What is rate equation for [E] for two coupled substrate-enzyme reactions with same enyzme?
Answer is The green dot is about 2/3 of the way between 0.01 and 0.1 so the 0.01 value is multiplied by 10^(2/3) = 100^(1/3). Since 4^3 = 64 and 5^3 = 125, 100^(1/3) is less than but closer to 125, so estimated value is 5 * 0.01 = 0.05.
To two significant digits, 100^(1/3) = 0.47.
Where some tickmarks would lie Log[10,2] = 0.3 or about 1/3 from the left
Log[10,3] = 0.48 or about 0.5
Log[10,4] = 0.60
Log[10,5] = 0.70
Log[10,6] = 0.78
Log[10,7] = 0.85
Log[10,8] = 0.90
Log[10,9] = 0.95

3. Demo of great power of enzymes catalase and hydrogen peroxide
H2O2 → H2O + (1/2)O2 is highly exothermic (Delta{G} about -41 kT) but quite slow in pure solution, 1M H2O2 at room temperature converts at 10^-8 M/s. Catalase speeds up reaction by 10^12
Catalase converts H2O2 to O2 and H2O at high rate, reaction velocity 10^4 M/s (this is at the high end of reaction velocities).

4. Converting rate equations into dimensionless form to reduce number of parameters
Answer is The green dot is about 2/3 of the way between 0.01 and 0.1 so the 0.01 value is multiplied by 10^(2/3) = 100^(1/3). Since 4^3 = 64 and 5^3 = 125, 100^(1/3) is less than but closer to 125, so estimated value is 5 * 0.01 = 0.05.
To two significant digits, 100^(1/3) = 0.47.
Where some tickmarks would lie Log[10,2] = 0.3 or about 1/3 from the left
Log[10,3] = 0.48 or about 0.5
Log[10,4] = 0.60
Log[10,5] = 0.70
Log[10,6] = 0.78
Log[10,7] = 0.85
Log[10,8] = 0.90
Log[10,9] = 0.95

5. Use Mathematica function NDSolve to find accurate approximate solutions to rate equations
Numerical algorithm requires same input as the mathematical problem (odes and ics) and requires further parameters that determine error of answer, e.g., a constant time step deltaT, or an upper bound to the relative error.
See NDSolve documentation to see the large number of parameters that can be adjusted.
Answer is The green dot is about 2/3 of the way between 0.01 and 0.1 so the 0.01 value is multiplied by 10^(2/3) = 100^(1/3). Since 4^3 = 64 and 5^3 = 125, 100^(1/3) is less than but closer to 125, so estimated value is 5 * 0.01 = 0.05.
To two significant digits, 100^(1/3) = 0.47.
Where some tickmarks would lie Log[10,2] = 0.3 or about 1/3 from the left
Log[10,3] = 0.48 or about 0.5
Log[10,4] = 0.60
Log[10,5] = 0.70
Log[10,6] = 0.78
Log[10,7] = 0.85
Log[10,8] = 0.90
Log[10,9] = 0.95

6. Numerical integration methods approximate the first few terms of a Taylor series in time
Answer is The green dot is about 2/3 of the way between 0.01 and 0.1 so the 0.01 value is multiplied by 10^(2/3) = 100^(1/3). Since 4^3 = 64 and 5^3 = 125, 100^(1/3) is less than but closer to 125, so estimated value is 5 * 0.01 = 0.05.
To two significant digits, 100^(1/3) = 0.47.
Where some tickmarks would lie Log[10,2] = 0.3 or about 1/3 from the left
Log[10,3] = 0.48 or about 0.5
Log[10,4] = 0.60
Log[10,5] = 0.70
Log[10,6] = 0.78
Log[10,7] = 0.85
Log[10,8] = 0.90
Log[10,9] = 0.95

7. Chemical equilibrium attained when production of C is slow compared to A ↔ B reactions
Answer is The green dot is about 2/3 of the way between 0.01 and 0.1 so the 0.01 value is multiplied by 10^(2/3) = 100^(1/3). Since 4^3 = 64 and 5^3 = 125, 100^(1/3) is less than but closer to 125, so estimated value is 5 * 0.01 = 0.05.
To two significant digits, 100^(1/3) = 0.47.
Where some tickmarks would lie Log[10,2] = 0.3 or about 1/3 from the left
Log[10,3] = 0.48 or about 0.5
Log[10,4] = 0.60
Log[10,5] = 0.70
Log[10,6] = 0.78
Log[10,7] = 0.85
Log[10,8] = 0.90
Log[10,9] = 0.95

8. Can solve constant-coefficient linear rate equations analytically using eigenvector representation
Answer is The green dot is about 2/3 of the way between 0.01 and 0.1 so the 0.01 value is multiplied by 10^(2/3) = 100^(1/3). Since 4^3 = 64 and 5^3 = 125, 100^(1/3) is less than but closer to 125, so estimated value is 5 * 0.01 = 0.05.
To two significant digits, 100^(1/3) = 0.47.
Where some tickmarks would lie Log[10,2] = 0.3 or about 1/3 from the left
Log[10,3] = 0.48 or about 0.5
Log[10,4] = 0.60
Log[10,5] = 0.70
Log[10,6] = 0.78
Log[10,7] = 0.85
Log[10,8] = 0.90
Log[10,9] = 0.95

9. Assume eps << 1, production of C is slow compared to forward and backward rates between A and B
Answer is The green dot is about 2/3 of the way between 0.01 and 0.1 so the 0.01 value is multiplied by 10^(2/3) = 100^(1/3). Since 4^3 = 64 and 5^3 = 125, 100^(1/3) is less than but closer to 125, so estimated value is 5 * 0.01 = 0.05.
To two significant digits, 100^(1/3) = 0.47.
Where some tickmarks would lie Log[10,2] = 0.3 or about 1/3 from the left
Log[10,3] = 0.48 or about 0.5
Log[10,4] = 0.60
Log[10,5] = 0.70
Log[10,6] = 0.78
Log[10,7] = 0.85
Log[10,8] = 0.90
Log[10,9] = 0.95

10. Eigenvector representation simplifies when parameter eps = r/k- is small
After time tau of order one (time of 1/(k+1)), second term has decayed away leaving slow decay of concentrations A and B, and slow increase of C. This describes exact solution well for times tau >~ 5
Answer is The green dot is about 2/3 of the way between 0.01 and 0.1 so the 0.01 value is multiplied by 10^(2/3) = 100^(1/3). Since 4^3 = 64 and 5^3 = 125, 100^(1/3) is less than but closer to 125, so estimated value is 5 * 0.01 = 0.05.
To two significant digits, 100^(1/3) = 0.47.
Where some tickmarks would lie Log[10,2] = 0.3 or about 1/3 from the left
Log[10,3] = 0.48 or about 0.5
Log[10,4] = 0.60
Log[10,5] = 0.70
Log[10,6] = 0.78
Log[10,7] = 0.85
Log[10,8] = 0.90
Log[10,9] = 0.95

11. Some nonlinear dynamics of rate equations: Poincare-Bendixson theorem
Answer is The green dot is about 2/3 of the way between 0.01 and 0.1 so the 0.01 value is multiplied by 10^(2/3) = 100^(1/3). Since 4^3 = 64 and 5^3 = 125, 100^(1/3) is less than but closer to 125, so estimated value is 5 * 0.01 = 0.05.
To two significant digits, 100^(1/3) = 0.47.
Where some tickmarks would lie Log[10,2] = 0.3 or about 1/3 from the left
Log[10,3] = 0.48 or about 0.5
Log[10,4] = 0.60
Log[10,5] = 0.70
Log[10,6] = 0.78
Log[10,7] = 0.85
Log[10,8] = 0.90
Log[10,9] = 0.95
For any two-variable set of differential equations with smooth functions f and g, only non-transient bounded solutions are fixed points (constant solutions) or limit cycles (periodic solutions)

12. Three-variable rate equations can have quasiperiodic and chaotic dynamics
Lorenz equations arise from three-mode Fourier truncation of Boussinesq fluid equations describing buoyancy-driven convection, look like rate equations.
By just varying parameters, same equations can produce different dynamics: constant, periodic, intermittent, and chaotic. As some parameter like rho is varied, dynamics can change type, this is called a bifurcation.
In 1963, Lorenz brilliantly used these equations to show why accurate weather forecasting would never be possible beyond a time window of about 10 days.
Answer is The green dot is about 2/3 of the way between 0.01 and 0.1 so the 0.01 value is multiplied by 10^(2/3) = 100^(1/3). Since 4^3 = 64 and 5^3 = 125, 100^(1/3) is less than but closer to 125, so estimated value is 5 * 0.01 = 0.05.
To two significant digits, 100^(1/3) = 0.47.
Where some tickmarks would lie Log[10,2] = 0.3 or about 1/3 from the left
Log[10,3] = 0.48 or about 0.5
Log[10,4] = 0.60
Log[10,5] = 0.70
Log[10,6] = 0.78
Log[10,7] = 0.85
Log[10,8] = 0.90
Log[10,9] = 0.95

13. Numerical integration of the Lorenz equations
Answer is The green dot is about 2/3 of the way between 0.01 and 0.1 so the 0.01 value is multiplied by 10^(2/3) = 100^(1/3). Since 4^3 = 64 and 5^3 = 125, 100^(1/3) is less than but closer to 125, so estimated value is 5 * 0.01 = 0.05.
To two significant digits, 100^(1/3) = 0.47.
Where some tickmarks would lie Log[10,2] = 0.3 or about 1/3 from the left
Log[10,3] = 0.48 or about 0.5
Log[10,4] = 0.60
Log[10,5] = 0.70
Log[10,6] = 0.78
Log[10,7] = 0.85
Log[10,8] = 0.90
Log[10,9] = 0.95

14. Enzyme-substrate dynamics 1
Answer is The green dot is about 2/3 of the way between 0.01 and 0.1 so the 0.01 value is multiplied by 10^(2/3) = 100^(1/3). Since 4^3 = 64 and 5^3 = 125, 100^(1/3) is less than but closer to 125, so estimated value is 5 * 0.01 = 0.05.
To two significant digits, 100^(1/3) = 0.47.
Where some tickmarks would lie Log[10,2] = 0.3 or about 1/3 from the left
Log[10,3] = 0.48 or about 0.5
Log[10,4] = 0.60
Log[10,5] = 0.70
Log[10,6] = 0.78
Log[10,7] = 0.85
Log[10,8] = 0.90
Log[10,9] = 0.95

15. Enzyme-substrate dynamics 2
Understand results qualitatively, e.g, why does [E] decrease and then increase with time for [E]=[S]=c0 at time t=0? What are long-time values of concentrations?
Answer is The green dot is about 2/3 of the way between 0.01 and 0.1 so the 0.01 value is multiplied by 10^(2/3) = 100^(1/3). Since 4^3 = 64 and 5^3 = 125, 100^(1/3) is less than but closer to 125, so estimated value is 5 * 0.01 = 0.05.
To two significant digits, 100^(1/3) = 0.47.
Where some tickmarks would lie Log[10,2] = 0.3 or about 1/3 from the left
Log[10,3] = 0.48 or about 0.5
Log[10,4] = 0.60
Log[10,5] = 0.70
Log[10,6] = 0.78
Log[10,7] = 0.85
Log[10,8] = 0.90
Log[10,9] = 0.95

16. Michaelis-Menten approximation of local equilibrium for enzyme-substrate reactions
Answer is The green dot is about 2/3 of the way between 0.01 and 0.1 so the 0.01 value is multiplied by 10^(2/3) = 100^(1/3). Since 4^3 = 64 and 5^3 = 125, 100^(1/3) is less than but closer to 125, so estimated value is 5 * 0.01 = 0.05.
To two significant digits, 100^(1/3) = 0.47.
Where some tickmarks would lie Log[10,2] = 0.3 or about 1/3 from the left
Log[10,3] = 0.48 or about 0.5
Log[10,4] = 0.60
Log[10,5] = 0.70
Log[10,6] = 0.78
Log[10,7] = 0.85
Log[10,8] = 0.90
Log[10,9] = 0.95

17. Michaelis-Menten describes many but not all enzymatic reactions well
Approximation can be poor when there are multiple binding sites per enzyme, when ES complex undergoes further transitions before producing product, when enzyme behavior modified by allosteric regulators.
Answer is The green dot is about 2/3 of the way between 0.01 and 0.1 so the 0.01 value is multiplied by 10^(2/3) = 100^(1/3). Since 4^3 = 64 and 5^3 = 125, 100^(1/3) is less than but closer to 125, so estimated value is 5 * 0.01 = 0.05.
To two significant digits, 100^(1/3) = 0.47.
Where some tickmarks would lie Log[10,2] = 0.3 or about 1/3 from the left
Log[10,3] = 0.48 or about 0.5
Log[10,4] = 0.60
Log[10,5] = 0.70
Log[10,6] = 0.78
Log[10,7] = 0.85
Log[10,8] = 0.90
Log[10,9] = 0.95
Experimental data for ATP hydrolysis by rabbit skeletal muscle myosin as function of ATP concentration. > Ouelette et al Arch Biochem. Biophy. 1952

18. One-minute End-of-class Question