1. Physics 414: Introduction to Biophysics Professor Henry Greenside December 5, 2017

2. First 10 minutes to fill out online course evaluations
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. French flag model of how fruit fly embryo sets up precise reproducible structures: a morphogen gradient
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. At the blackboard: mathematics of a point source, diffusion, and uniform degradation leads to a puzzle: length scale does not depend on size of domain
Bicoid is first morphogen in temporal development that shows a pattern, starts off successive morphogen patterns. How to get what seems to be exponential spatial profile? Anterior source + diffusion + uniform degradation of Bicoid protein product gives exponential.
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. Compare exponential with lambda for different fly species D and lambda should be about same for all species
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
Decay constants lambda from fitted exponentials to Bicoid gradients. Lambda increases approximately linearly with embryo size contrary to prediction of diffusion-decay model.

6. Fitting exponential to data (xi, yi): least-squares fit a straight line to (xi, ln(yi))
Mathematica Fit function handles a general linear basis:
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. At the blackboard: solution to lambda(L) scaling puzzle, assume Bcd degradation occurs only in nuclei
Key biological insight: although fly embryos differ by factor of five in length, they have the same number of nuclei over that length. So if degradation occurs only inside nuclei, get decay length that varies with embryo length L.
Can use finite-difference numerical method to find approximate solutions to diffusion-degradation equation with given boundary conditions. Set degradation term -k b to zero in cytosol.
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. 1D simulation of diffusion and decay with source using finite differences on uniform space-time grid
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. Results of 1d simulations regarding establishment of bicoid gradient, scaling with embryo size
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. Spatial patterning of Bicoid and Hunchback
Bicoid is green
Hunchback is red
DNA is blue
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. Establishment of the Hunchback profile
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
Fig 20.7

12. Use statistical physics and Monod-Wyman-Changeux (MWC) model to deduce Hunchback profile from the Bicoid gradient (Fig 20.8)
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. Pattern formation in cyanobacterium Anabaena: how to get regularly spaced heterocysts
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. Solution: diffusive inhibitor RGSGR peptide
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. Turing’s insight: diffusion can induce instability of otherwise spatially uniform chemical reactions, leading to cellular patterns. Now called a “Turing instability:
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
Alan Turing

16. A famous paragraph about how best to create a theoretical model of some physical phenomenon
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. One-minute End-of-class Question