1. Physics 414: Introduction to Biophysics Professor Henry Greenside November 14, 2017

2. How to read data off a log axis
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. On log axis, tickmarks lie fraction Log[10,i] between tickmarks for 10^k to 10^(k+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

4. Chapter 15: nonequilibrium chemical kinetics of cells and subcellular processes
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. Intricate cell cycle of Caulobacter crescentus
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. Gene expression dynamics of Caulobacter crescentus
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. Dynamics of cell reproduction: microtubule polymerization, molecular motors
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. Actin 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

9. Nonequilibrium growth and decay of cytoskeletal 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
Fish skin cell, a keratocyte

10. Crawling of an amoeba by actin 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

11. Surfing of Listeria monocytogenes inside a eukaryotic cell
Movie by Julie Theriot and collaborators
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
Bead driven by actin polymerization

12. Worthwhile online talks by Julie Theriot
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. Example 1 of chemical kinetics: Isomerization of retinal
all-trans
13-cis
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
1 kcal/mol ~ 1.7 kT for T ~ 300 K
1 green photon ~ 120 kT

14. Rate constant k is a time probability density, can be used to calculate distribution of lifetimes
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. Use conservation of mass to simplify mathematical analysis of rate 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

16. One-minute End-of-class Question