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Tidbits from the Sciences: Tidbits from the Sciences: Examples for Calculus and Examples for Calculus and Differential Equations Differential Equations
Tidbits from the Sciences: Tidbits from the Sciences: Examples for Calculus and Examples for Calculus and Differential Equations Differential Equations
Bruce E. ShapiroBruce E. Shapiro
California State University, California State University, NorthridgeNorthridge
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• Satellite navigation• Genomic variation• Cooking potatoes• Enzymatic reactions & switching• The dynamics of love• Measuring the human genome
Examples
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€
kt = M = E − esin E
t = time since perigee passagek = 2/periode<1 in an elliptical orbit
M is easy to calculate
E is easily converted to position in orbit
Problem: Find E as a function of time
Inversion of Kepler’s Equation
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• Solve using fixed point iteration
• Since e<1 for an elliptical orbit:
• Fixed point always converges
Inversion of Kepler’s Equation
€
kt = M = E − esin E
€
g(E) = M + esin E
€
| ′ g (t) |= e | cos E |≤ e <1
• Example: M= /4, e=1/4 to 3 digits:
€
g(E) = 0.785 + 0.25 *sin E
E0 = 0.785
E1 = g(E0 )
= 0.785 + 0.25sin(0.785)
= 0.962
E2 = g(E1)
= 0.785 + 0.25sin(.962)
= 0.990
E3 = 0.994; E4 = 0.995
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• Satellite navigation• Genomic variation• Cooking potatoes• Enzymatic reactions & switching• The dynamics of love• Measuring the human genome
Examples
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Imag
es: http://www.sciencemag.com, http://www.nature.com
Genomes are being sequenced at an exponential Rate
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• Individual genomic differences occur at every 1000 “base-pairs” of our DNA: – ≈1,000,000 significant points of difference between any
two individuals• Not all the same locations in everyone
– differences in drug metabolism, disease sensitivity, eye color, ...
• Yet we are 99.9% the same!
Genomic Variation
http://creative.gettyimages.com
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Micro Array Data
Time
Ge
ne
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Genetic Similarity• Samples have concentration vectors
(x1,x2,….,xn), (y1,y2,….,yn) – Can be two points on a time course or
samples from two different individual!
• Come up with different measures of similarity:– Dot product/angle– Euclidean distance– Various vector norms– Projections along principal components
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Data clusters in two dimensions
x
y
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Data clusters in two dimensions
x
y
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ABO BLOOD GROUPO A B AB
Armenians 31 50 13 6Basques 51 44 4 1Czechs 30 44 18 9Danes 41 44 11 4Dutch 45 43 9 3Hungarians 36 43 16 5Icelanders 56 32 10 3Latvians 32 37 24 7Lithuanians 40 34 20 6Slovaks 42 37 16 5Spanish 38 47 10 5Swedes 38 47 10 5Swiss 40 50 7 3
Source: http://www.bloodbook.com/world-abo.html
• Are the Slovaks and Czechs closer genealogically to the each other or to the Spanish? Use the following distance measurements:
€
d1(x,y) = cos−1 x ⋅y| x || y | ⎛ ⎝ ⎜
⎞ ⎠ ⎟
d2 (x,y) = (x − y) ⋅(x − y)
d3(x,y) = | x − y |ii∑
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ABO Blood Group
€
d1(Slovak,Spanish) = 69.4o
d1(Slovak,Czech) = 71.2o
d1(Czech,Spanish) = 69.9o
d2(Slovak,Spanish) = 12.32
d2(Slovak,Czech) = 14.59
d2(Czech,Spanish) = 12.37
d3(Slovak,Spanish) = 20
d3(Slovak,Czech) = 25
d3(Czech,Spanish) = 23
By all three methods the Czechs and the Slovaks are more closely related to the Spanish than they are to each other!
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• Satellite navigation• Genomic variation• Cooking potatoes• Enzymatic reactions & switching• The dynamics of love• Measuring the human genome
Examples
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The Potato Problem*• The rate of change of temperature T of a
potato in a pre-heated oven is proportional to the difference between the temperature of the oven and the potato*
• Preheat the oven to 420˚• Assume room temperature is 70˚• After 3 minutes the potato is 150˚. • When will it reach 300˚?
*Newton’s law of heating as formulated by a student
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The Potato Problem (Solution)
€
′ T (t) = k(420 − T ), T(0) = 70, T(3) = 150
Solve ODE :dT
T − 420∫ = − kdt∫ ⇒ T = 420 + Ce−kt
Solve for C : T(0) = 70⇒ C = -350⇒ T = 420 − 350e−kt
Solve for k :T(3) = 150⇒ 150 = 420 − 350e−3k ⇒ k = −13
ln270350
≈ 0.086
General Solution : T(t) = 420 − 350e−0.086t
T = 300 = 420 − 350e−0.086t ⇒ t = 12.44 minutes
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The Potato Problem• Can be treated as either IVP or
BVP– IVP plus “fitting” data to IVP to get
second constant, or as– BVP with two boundary conditions
• Linear Separable First Order ODE• Introduces idea of Canonical forms
in nature with something other than Capacitors
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• Satellite navigation• Genomic variation• Cooking potatoes• Enzymatic reactions & switching• The dynamics of love• Measuring the human genome
Examples
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• Model phenomona that appear in a wide variety of situations in nature:
• “Exponential Relaxation” of y to steady state with time constant
• One of the most common models in biology!!
Canonical Models
€
dydt
=y∞ − y
τ
€
y∞
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Law of Mass Action• The rate of a reaction is
proportional to the concentrations of the reactants
• Single Reactant:• Multiple Reactants:
•Multiple Reactions: add terms from each reaction
€
x → y ⇒ ′ x = −kx, ′ y = kx
€
w + x + y → z ⇒ ′ z = kxyz = − ′ x = − ′ y = − ′ z
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Application of Mass Action• Protein in Two States
– x=amount in “on” state – y=amount in “off” state
• Conservation of mass x+y=N=constant
• Chemical Equation:
€
xα
⏐ → ⏐ y
yβ
⏐ → ⏐ x
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Two-State Protein• Normalize variables (N=1)
• Solution:
€
xα
⏐ → ⏐ y
yβ
⏐ → ⏐ x
€
′ x = −αx + βy = −αx + β (1− x)
= β − (α + β )x
= (α + β )β
α + β− x
⎛ ⎝ ⎜
⎞ ⎠ ⎟
=x∞ − x
τ, x∞ =
βα + β
, τ =1
α + β
€
x + y = 1
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Enzymatic Cascades• Traditional Enzymatic Reaction:
• More common situation in nature:
€
Source + Enzyme → Product + Enzyme
€
Source + Enzyme → Complex
Complex → Source + Enzyme
Complex → Product + Enzyme
written as
Source ⇒ ProductEnzyme
or :
Source + EnzymeF Complex → Product + Enzyme
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MAPK Cascade ModelMAPK=Mitogen Activated Protein Kinase
10
100
1000
10000
100000
2 3 4 5 6# Slots
# ReactionsSingle PhosphorylationDouble Phosphorylation
€
KKK ⇒ KKKpsignal
KK ⇒KKKp
KKp ⇒KKKp
KKpp
K ⇒KKKpp
Kp ⇒KKKpp
Kpp
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€
KKK ⇒ KKKpsignal
KK ⇒KKKp
KKp ⇒KKKp
KKpp
K ⇒KKKpp
Kp ⇒KKKpp
KppAs chemical
reactions
As a cascade
As differential equations
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Differential equations
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• Satellite navigation• Genomic variation• Cooking potatoes• Enzymatic reactions & switching• The dynamics of love• Measuring the human genome
Examples
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Strogatz’s Romeo & Juliet• Juliet is strangely attracted to Romeo:
– The more Romeo loves Juliet, the more she wants to run away
– When Romeo gets discouraged, she finds him strangely attractive
• Romeo echo’s Juliet’s love:– he warms up when she loves him– he loses interest when she hates him
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Romeo and Juliet• R(t) = Romeo’s Love/Hate for Juliet• J(t)=Juliet’s Love/Hate for Romeo• Postive Values signify love, negative
values hate• Dynamical Model: • Outcome: a never-ending cycle of love
and hate with a center at (R,J)=(0,0); they manage to simultaneously love one another 25% of the time€
′ R = aJ , ′ J = −bR
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Romeo and Juliet• General Model:
– Can a cautious lover(a<0,b>0) find true love with an eager beaver (c>0,d>0)?
– Can two equally cautious lovers get together? (a=d<0, b=c>0)?
– What if Romeo and Juliet are both out of touch their own feelings (a=d=0)?
– Fire & Water: Do opposites attract (c=-a,d=-b)?
– How do Romantic Clones interact (a=d, b=c)?
€
′ R = aR + bJ , ′ J = cR + dJ
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• Satellite navigation• Genomic variation• Cooking potatoes• Enzymatic reactions & switching• The dynamics of love• Measuring the human genome
Examples
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Chromosomal Structure
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(1/
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Base Pairs = BarbellsAdenine
Thymine
GuanineCytosineA
T
CG
AT
GC
A-T T-AC-G G-C
“hydrogen bonds”
4 flavors
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Put one barbell on each spoke of a ladder
... then twist the ladder
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... and you get the “Double Helix”
Sugar-Phosphate "backbone"deoxyribose
phosphodiester bond
base
hydrogen bondsAdenine
Thymine
Guanine
Cytosine
BASES
DNA =deoxyribonucleic acid
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The SEQUENCE of the Human Genome
• 23 chromosome pairs• 2.91 Giga Base Pairs• 691 MB (at 2 bits/Base Pair)• 39,114 genes: functional units• 26,383 “known” function• Average gene ≈27 kBP• Genes ≈ 1/3 of genome
GATCTACCATGAAAGACTTGTGAATCCAGGAAGAGAGACTGACTGGGCAACATGTTATTCAGGTACAAAAAGATTTGGACTGTAACTTAAAAATGATCAAATTATGTTTCCCATGCATCAGGTGCAATGGGAAGCTCTTCTGGAGAGTGAGAGAAGCTTCCAGTTAAGGTGACATTGAAGCCAAGTCCTGAAAGATGAGGAAGAGTTGTATGAGAGTGGGGAGGGAAGGGGGAGGTGGAGGGATGGGGAATGGGCCGGGATGGGATAGCGCAAACTGCCC...
4592 miles 364,000 pages(12 point font)
(100x80 char/page)
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If you stretched out the DNA in your body it would be HOW long?
€
2.9 ×109 bases in the genome
× 1013 cells in the human body
× 2 copies in each cell
× 0.34 ×10-9 meters/base (0.34 nM)
× 1 kilometer /1000 meters
× 1 A.U./1.5 ×108 kilometers
= 131 A.U.
http://www.ornl.gov/hgmis/education/images.html
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Research examples can …• Awaken • Motivate• Consolidate
– Relate math to other disciplinesContact for more information:[email protected]
http://www.bruce-shapiro.com/presentations.html