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

Class discussion: How should we define a “pattern”? 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

Role of biological patterns can be difficult to understand 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 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?

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 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

Light-sheet microscopy to visualize all surface cells of fly embryo as it evolves into larva 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 https://youtu.be/qiabYntNMC8

Spatially varying concentrations of transcription factors in Drosophila embryo determine segments Figure 19.2 p 804 of PBOC2 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 Stripe 2 of EVE gene

Different mechanism lays out somites (precursors to ribs) of verterbrates (somitogenesis) Snake skeleton 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 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 20.21 in Section 20.4.1, “clock-and wavefront” model of somitogenesis

French flag model of how fruit fly embryo sets up precise reproducible structures: a morphogen gradient 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

Missing morphogen gradients explains why adult human brain cannot recover from stroke: axonal growth cones don’t know where to go https://youtu.be/lU9QH7pB6nU 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 https://youtu.be/m1Y5ugEWF00

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 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

Compare exponential with lambda for different fly species D and lambda should be about same for all species 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 Decay constants lambda from fitted exponentials to Bicoid gradients. Lambda increases approximately linearly with embryo size.

One-minute End-of-class Question