1. Physics 414: Introduction to Biophysics Professor Henry Greenside November 30, 2017

2. Class discussion: How should we define a “pattern”?
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. Role of biological patterns can be difficult to understand
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
Many sea shells spend most of their life under sand and so are not visible, and they themselves have no vision. So why the spatial patterns?

4. Stripe formation in Drosophila embryo: how to get precise reproducible stripes?
1934 drawing by E. Wallace, Male on left, female on right
14 segments: 3 for head, 3 for thorax, 8 for abdomen
Pattern of gene expression at 13th generation. Stained protein products of genes Bicoid in blue, Even-skipped (EVE) in green, Caudal in red.
There are ~6,000 < 213 = 8192 nuclei in one large cell membrane. Radii of nuclei about 6 microns, L ~ 500 microns, about 130 nuclei per stripe.
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. Light-sheet microscopy to visualize all surface cells of fly embryo as it evolves into larva
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. Spatially varying concentrations of transcription factors in Drosophila embryo determine segments
Figure 19.2
p 804 of PBOC2
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
Stripe 2 of EVE gene

7. Different mechanism lays out somites (precursors to ribs) of verterbrates (somitogenesis)
Snake skeleton
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
Unlike stripes in fly embryo that form all at once, somites form sequentially in time. Synchronized oscillations of cells in presomitic mesoderm are stopped at different points in cycle by traveling wave from anterior to posterior. See Fig in Section , “clock-and wavefront” model of somitogenesis

8. 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

9. Missing morphogen gradients explains why adult human brain cannot recover from stroke: axonal growth cones don’t know where to go
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: 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

11. 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.

12. One-minute End-of-class Question