1. Physics 414: Introduction to Biophysics Professor Henry Greenside November 28, 2017

2. Two talks this week and next on morphogenesis
Thursday, Nov 30, 12:30-1:30 pm, Nanaline Duke 147
Physicist Boris Shraiman of UCSB on
“Physics and Biology of Morphogenesis”
Wednesday, Dec 6, 4:30-5:30 pm, French 2231
Applied mathematician L. Mahadevan on
“Geometry, physics, and biology”
Especially intended for undergaduates
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. Reminder of structure of actin polymers (Chapter 10)
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
F-actin filament assembles from two protofilaments, diameter about 8 nm

4. Rate of actin polymerization, starting from monomers
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. Rate of actin polymerization, starting from nucleation centers (triplets)
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. Experimental equilibrium (steady state) approximately exponential distribution of actin filament lengths
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. Rate models of increasing complexity for polymerization
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. Microtubule “treadmilling”: one end grows at same rate that other end shrinks
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. “Catastrophes” (dynamic instability) in microtubule growth, related to ATP hydrolysis
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. At the blackboard: equilibrium and dynamic polymerization
Sections and of Chapter 15, pages
Key points:
Exponential distribution of actin filament lengths in equilibrium, Eq. (15.76), see also Problem 6.7 on page 277 (Problem 3 of Assignment 3)
Critical monomer concentration c*, below which filaments decay in length, above which filaments grow (according to simple model on page 605). c* = K_d, the dissociation constant for one monomer to add to a filament of any length.
The argument leading to Eq. (15.85): in equilibrium lengths fluctuate according to a random walk with a diffusion constant a^2 k_off that can be calculated. This leads to a wrong prediction, it takes too long for an equilibrium filament to increase its length by a fluctuation so there must be an energy-dependent non-equilibrium mechanism.
Eq. (15.88) for a simple rate model of how a filament grows over time, leading to Eq. (15.93) for an explicit solution and prediction.
The logic of Eq. (15.100): the probability to find a monomer of length n at time t can change over a short time interval dt because of four pathways: addition to a shorter n-1 polymer (increase), subtraction from a longer n+1 polymer (increase), subtraction for the existing n polymer (decrease), and addition to an existing n polymer (decrease). The resulting evolution equation can be used to find the mean polymer length as a function of time, Eq. (15.107)
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. Average filament size based on simple kinetics Assume M nucleation centers so M filaments
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
Length of filaments versus time for assembly of bacterial flagellin, Fig 15.27(B)

12. New and last chapter: Chapter 20 Biological Patterns: Order in Space and 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

13. Discuss three mechanisms for understanding biological pattern formation
“French flag” or “gradient-threshold” models: externally established chemical gradient (morphogen) causes cells to differentiate when they sense a concentration greater than some threshold.
Turing instability: intrinsic pattern formation from homogeneous state caused by at last two reacting and diffusing morphogens, slow-diffusing activator and fast-diffusing inhibitor.
Lateral inhibition or the “Notch-Delta” concept that create checkboard-like two-dimensional grids of cells.
Discussion is chance to integrate and practice what we have learned over the semester: diffusion, rate equations, statistical physics, biology
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. One-minute End-of-class Question