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

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 0.05. 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 0.02 Log[10,2] = 0.3 or about 1/3 from the left 0.03 Log[10,3] = 0.48 or about 0.5 0.04 Log[10,4] = 0.60 0.05 Log[10,5] = 0.70 0.06 Log[10,6] = 0.78 0.07 Log[10,7] = 0.85 0.08 Log[10,8] = 0.90 0.09 Log[10,9] = 0.95

Reminder of structure of actin polymers (Chapter 10) Answer is 0.05. 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 0.02 Log[10,2] = 0.3 or about 1/3 from the left 0.03 Log[10,3] = 0.48 or about 0.5 0.04 Log[10,4] = 0.60 0.05 Log[10,5] = 0.70 0.06 Log[10,6] = 0.78 0.07 Log[10,7] = 0.85 0.08 Log[10,8] = 0.90 0.09 Log[10,9] = 0.95 F-actin filament assembles from two protofilaments, diameter about 8 nm

Rate of actin polymerization, starting from monomers Answer is 0.05. 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 0.02 Log[10,2] = 0.3 or about 1/3 from the left 0.03 Log[10,3] = 0.48 or about 0.5 0.04 Log[10,4] = 0.60 0.05 Log[10,5] = 0.70 0.06 Log[10,6] = 0.78 0.07 Log[10,7] = 0.85 0.08 Log[10,8] = 0.90 0.09 Log[10,9] = 0.95

Rate of actin polymerization, starting from nucleation centers (triplets) Answer is 0.05. 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 0.02 Log[10,2] = 0.3 or about 1/3 from the left 0.03 Log[10,3] = 0.48 or about 0.5 0.04 Log[10,4] = 0.60 0.05 Log[10,5] = 0.70 0.06 Log[10,6] = 0.78 0.07 Log[10,7] = 0.85 0.08 Log[10,8] = 0.90 0.09 Log[10,9] = 0.95

Experimental equilibrium (steady state) approximately exponential distribution of actin filament lengths Answer is 0.05. 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 0.02 Log[10,2] = 0.3 or about 1/3 from the left 0.03 Log[10,3] = 0.48 or about 0.5 0.04 Log[10,4] = 0.60 0.05 Log[10,5] = 0.70 0.06 Log[10,6] = 0.78 0.07 Log[10,7] = 0.85 0.08 Log[10,8] = 0.90 0.09 Log[10,9] = 0.95

Rate models of increasing complexity for polymerization Answer is 0.05. 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 0.02 Log[10,2] = 0.3 or about 1/3 from the left 0.03 Log[10,3] = 0.48 or about 0.5 0.04 Log[10,4] = 0.60 0.05 Log[10,5] = 0.70 0.06 Log[10,6] = 0.78 0.07 Log[10,7] = 0.85 0.08 Log[10,8] = 0.90 0.09 Log[10,9] = 0.95

Microtubule “treadmilling”: one end grows at same rate that other end shrinks Answer is 0.05. 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 0.02 Log[10,2] = 0.3 or about 1/3 from the left 0.03 Log[10,3] = 0.48 or about 0.5 0.04 Log[10,4] = 0.60 0.05 Log[10,5] = 0.70 0.06 Log[10,6] = 0.78 0.07 Log[10,7] = 0.85 0.08 Log[10,8] = 0.90 0.09 Log[10,9] = 0.95

“Catastrophes” (dynamic instability) in microtubule growth, related to ATP hydrolysis Answer is 0.05. 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 0.02 Log[10,2] = 0.3 or about 1/3 from the left 0.03 Log[10,3] = 0.48 or about 0.5 0.04 Log[10,4] = 0.60 0.05 Log[10,5] = 0.70 0.06 Log[10,6] = 0.78 0.07 Log[10,7] = 0.85 0.08 Log[10,8] = 0.90 0.09 Log[10,9] = 0.95

At the blackboard: equilibrium and dynamic polymerization Sections 15.4.1 and 15.4.2 of Chapter 15, pages 602-612 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 0.05. 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 0.02 Log[10,2] = 0.3 or about 1/3 from the left 0.03 Log[10,3] = 0.48 or about 0.5 0.04 Log[10,4] = 0.60 0.05 Log[10,5] = 0.70 0.06 Log[10,6] = 0.78 0.07 Log[10,7] = 0.85 0.08 Log[10,8] = 0.90 0.09 Log[10,9] = 0.95

Average filament size based on simple kinetics Assume M nucleation centers so M filaments Answer is 0.05. 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 0.02 Log[10,2] = 0.3 or about 1/3 from the left 0.03 Log[10,3] = 0.48 or about 0.5 0.04 Log[10,4] = 0.60 0.05 Log[10,5] = 0.70 0.06 Log[10,6] = 0.78 0.07 Log[10,7] = 0.85 0.08 Log[10,8] = 0.90 0.09 Log[10,9] = 0.95 Length of filaments versus time for assembly of bacterial flagellin, Fig 15.27(B)

New and last chapter: Chapter 20 Biological Patterns: Order in Space and Time Answer is 0.05. 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 0.02 Log[10,2] = 0.3 or about 1/3 from the left 0.03 Log[10,3] = 0.48 or about 0.5 0.04 Log[10,4] = 0.60 0.05 Log[10,5] = 0.70 0.06 Log[10,6] = 0.78 0.07 Log[10,7] = 0.85 0.08 Log[10,8] = 0.90 0.09 Log[10,9] = 0.95

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 0.05. 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 0.02 Log[10,2] = 0.3 or about 1/3 from the left 0.03 Log[10,3] = 0.48 or about 0.5 0.04 Log[10,4] = 0.60 0.05 Log[10,5] = 0.70 0.06 Log[10,6] = 0.78 0.07 Log[10,7] = 0.85 0.08 Log[10,8] = 0.90 0.09 Log[10,9] = 0.95

One-minute End-of-class Question