ca659 mathematical methods/ computational sciencemcrane/ca659/...2up.pdf · mathematical methods/...

Introduction & BasicsGrowth and Decay

Linear & Non-Linear Interaction Models

CA659Mathematical Methods/Computational Science

Dr. Martin Crane

February 8, 2011

Dr. Martin Crane CA659Mathematical Methods/Computational Science



Recommended Books for the Course

1 Essential Mathematical Biology by Nicholas F. Britton,Springer

2 Modelling with Differential & Difference Equations by G.Fulford, P. Forrester & Jones, Cambridge University Press

3 Mathematical Models in Biology by Leah Edelstein-Keshet,Random House press

4 Differential Equations, M.L. Spiegel, Schaums Outline Series,McGraw-Hill (Useful for D.E. Practice)

5 Linear Algebra & Its Applications, G, Strang, Academic Press

6 Mathematical Biology I, An Introduction by J.D. Murray,Springer




Assessment & Tutorials

Assessment

Exam in January

Three hours

Attempt Any Four from Six Questions

Tutorials

Tutorials every week

Two hours long

Some questions from tutorials will feature on the exam




Table of contents

1 Introduction & BasicsDiscrete V Continuous ModelsHigher Order Linear Difference EquationsNon-Linear ModelsPopulation Genetics

2 Growth and DecayIntroduction & Simple ModelsLogistic Growth ModelsLogistic Growth with Constant CullingLogistic Growth Revisited: The Chemostat

3 Linear & Non-Linear Interaction ModelsIntroductionLinear ModelsNon-Linear Models




Discrete V Continuous ModelsHigher Order Linear Difference EquationsNon-Linear ModelsPopulation Genetics

Chapter 1:

Introduction & Basics





First Order Linear Difference Equations

State at time t purely related to that at t − 1

Example in nature is cell division

Mn+1 = aMn (1.1)

a constant, n is the generation number

So number in nth generation related to that in first generationby:

Mn = aMn−1 = . . . = anM0 (1.2)

So if1 |a| > 1 the population will increase,2 |a| = 1 the population will be stable,3 |a| < 1 the population will decrease.





Higher Order Linear Difference Equations

Example 1: Rabbit Reproduction

Order of difference equation is number of terms determiningpresent state.

Examples of higher order difference eqns common in nature.

Leonardo of Pisa (Fibonacci) modelled rabbit reproduction.

Assumptions of Fibonacci model:1 Each pair of rabbits can reproduce from two months old2 Each reproduction produces only one pair of rabbits3 All rabbits survive.

Number of rabbit pairs at time n + 1, Mn+1 (for n months)given by:

Mn+1 = Mn + Mn−1. (1.3)

With M0 = 1, M1 = 1, (1 pair to start) number grows as1, 1, 2, 3, 5, 8, 13, . . .






Example 1: Rabbit Reproduction (cont’d)

Season: 0 1 2 3

Initial Pair : = =

:

Total Number of Pairs: 1 1

=

=

:

4

2 3

=

=

:

=

:

5

Figure 1.1: Fibonacci Number of Immature (:) & Mature Rabbits (=)






Example 1: Rabbit Reproduction (cont’d)

Rather than Eqn.(1.3), ‘one step’ eqn (like Eqn.(1.1)) isbetter.

Get this by writing Eqn.(1.3) in the form:

Mn+1 + Mn = Mn+2

Mn+1 = Mn+1(1.4)

which, by writing

un =

(Mn+1

Mn

)

takes the form

un+1 =

(1 11 0

)un. (1.5)





Digression: Matrix Basics

Matrices & Vectors

A matrix is an array of coefficients of the form:

A =

⎛⎜⎜⎜⎝

a11 a12 . . . a1n

a21 a22 . . . a2n...

.... . .

...am1 am2 . . . amn

⎞⎟⎟⎟⎠ (1.6)

This is called an m×n matrix as it has m rows and n columns.

A vector is an array of coefficients of the form:

A =

⎛⎜⎜⎜⎝

a11

a21...

am1

⎞⎟⎟⎟⎠





Digression: Matrix Basics cont’d

Matrix Systems

In the course we will see systems of equations of the form:

a11x1 + a12x2 = b1

a21x1 + a22x2 = b2(1.7)

for a system of two equations in two unknowns x1, x2 withconstant coefficients a11, a12, a21, a22 and a right-hand sideb1, b2.

With matrix multiplication, this can be written as:

Ax = b ≡(

a11 a12

a21 a22

)(x1

x2

)=

(b1

b2

)(1.8)






Matrix Inverse, Identity Matrix

It can be shown (c.f. Strang), that Eqn.(1.8) has a uniquesolution if the inverse of the matrix exists.

The inverse of the matrix A−1 has the property:

A × A−1 is the Identity Matrix I

The n × n identity matrix is given by:

I =

⎛⎜⎜⎜⎝

1 0 . . . 00 1 . . . 0...

.... . .

...0 0 . . . 1

⎞⎟⎟⎟⎠ (1.9)






Solutions to Matrix Systems: Matrix Determinant

To solve x = (x , y) in Eqn.(1.8) need to find A−1

For a 2× 2 A−1 is given by:

A−1 =1

det(A)

(a22 −a12

−a21 a11

)(1.10)

where det(A) is the determinant of the matrix A

The determinant of A is given by det(A) = a11a22 − a12a21.

Eqn.(1.10) holds for a 2× 2 matrix only.

The solution to the linear system in Eqn.(1.8)) will only existif the following condition is met:

det(A) ≡∣∣∣∣ a11 a12

a21 a22

∣∣∣∣ �= 0 (1.11)






Matrix Characteristic Equation, Trace

The characteristic equation for A is given by det(A− λI) = 0.

It arises in a number of circumstances, as we shall see later.

For a 2× 2 matrix, this expression becomes:

det(A− λI) = 0 ≡∣∣∣∣ a11 − λ a12

a21 a22 − λ

∣∣∣∣ = 0 (1.12)

which reduces to λ2 − (a11 + a22)λ + (a11a22 − a12a21) = 0which we rewrite as

λ2 − pλ + q = 0 (1.13)

where p = a11 + a22 is called the Trace of A and q = det(A).






Matrix Eigenvalues, Eigenvectors

The roots of the quadratic equation in Eqn.(1.13) are givenby:

λ1,2 =p

2±

√p2 − 4q

2(1.14)

are known as the eigenvalues of A.

It can be shown (see again Strang) that any matrix A can bedecomposed as follows:

A = SΛS−1 (1.15)

where S has eigenvectors of A, v1, v2 on the columns, & Λ isa matrix with eigenvalues as diagonals & zeros elsewhere:

A = S

(λ1 00 λ2

)S−1 (1.16)






Matrix Eigendecompostition, Singular Value Decomposition

Eqn.(1.15) comes in useful (amongst other things) for raisingmatrices to powers:

A3 =(SΛS−1

) (SΛS−1

) (SΛS−1

)=

(SΛ3S−1

)(1.17)

The eigenvectors v1, v2 are the solutions to the linear systemAx = λx for λ = λ1, λ2 respectively.

As with eigenvalues, these have important physical meaningsfor the system under consideration.

The process in Eqn.(1.15) is known as eigen decompositionfor a square matrix; where the matrix is not square, it isknown as singular value decomposition






Matrix Decomposition & Difference Equations

For difference equations the system at time step n is relatedto that at the previous step n − 1 through the system:

un = Aun−1 = Anu0 (1.18)

Using eigendecomposition A = SΛS−1 and setting(c1

c2

)= S−1u0 = S−1

(u1

u2

)n=0

we observe that

un =

(u1

u2

)= c1λ

n1v1 + c2λ

n2v2 (1.19)

where c1, c2 are constants.






Matrix Decomposition, Difference & Differential Equations

A similar result may be obtained for differential equationswhere the system of a second order equation (often) has asolution of the form:

x(t) =

(x(t)y(t)

)= c1v1e

λ1t + c2v2eλ2t . (1.20)

So the solutions of difference and differential equations can bebroken down into a linear combination of the eigenvalues andcorresponding eigenvectors of the original matrix system.

This is a very important result and, as we will see, comes invery useful for determining dominant or longterm solutions ofmatrix systems such as the Fibonacci series.





Back to Fibonacci Sequences

Eigenvalues and the Fibonacci Difference Equation

In order to find the long-term behaviour of the Fibonaccisystem in Eqn.(1.5), we can write (using Eqn.(1.17))

un = Anu0 = SΛnS−1u0 (1.21)

Given that

A =

(1 11 0

)from Eqn.(1.5), we find the characteristic equation to be

λ2 − λ− 1 = 0

(from Eqn.(1.12)).This gives the eigenvalues

λ1 =1 +

√5

2, λ2 =

1−√5

2.





Stability of Fibonacci Sequences

The full eigendecomposition for A can then be found to be

A =1

λ1 − λ2

(λ1 λ2

1 1

)(λ1 00 λ2

)(1 −λ2

−1 λ1

)(1.22)

Thus Eqn.(1.21) reduces to(Mn+1

Mn

)= S

(λn

1 00 λn

2

)S−1

(10

)(1.23)

The nth Fibonacci number is 2nd element of vector on lefthand side of Eqn.(1.23), Mn, can be shown to be:

Mn =λn

1 − λn2

λ1 − λ2=

1√5

⎡⎢⎣(

1 +√

5

2

)n

−��0(

1−√5

2

)n⎤⎥⎦

(1.24)





Stability of Fibonacci Sequences cont’d

The Golden Number & Fibonacci Sequences

φ = (1 +√

5)/2 is very important and was known to theAncient Greeks as the golden number because rectangles withsides in the ratio 1 : 1.618 were the most elegant.

The Golden Number occurs frequently in nature and persistsin the design of everyday items such as credit cards, ipods etc.

As λ2 > 1 & −1 < λ1 < 0, λ2 is the largest eigenvalue and itsmagnitude means the Fibonacci sequence is monotonicallyincreasing.

The fact that λ1 is negative and of magnitude less than 1means it contributes a slight oscillation that dies out as nincreases. This can be seen in Fig. 1.2.





Stability of Fibonacci Sequences cont’d

1 2 3 4 5 6 7 8 9 100

5

10

15

20

25

30

35

n

y n

Figure 1.2: Fibonacci Sequence to 10 Generations





Bonham Sequences

Example 1: Pig Reproduction

A pair of bonhams becomes a mature pair of pigs in the nextseason.

A mature pair produces six pairs of bonhams in the followingseason, and in every successive season thereafter

Each pair of bonhams produced takes one season to reachmaturity and a further season to start breeding (andproducing six young pairs) in every subsequent season.

This can be seen in the diagram (fig 1.3),

It is assumed that breeding is seasonal so that generations donot overlap and that pigs live a long time.





Bonham Sequences cont’d

Season: 0 1 2 3

Initial Pair : = =

: : :

: : :

Total Number of Pairs: 1 1 7

===

===

=

: : :

: : :

13

4

55

Figure 1.3: Number of Immature (:) & Mature Pigs (=)






As with eqn(1.5), we may derive an expression for number ofpairs of pigs in the n + 1th generation w.r.t. the nthgeneration:

un+1 =

(1 61 0

)un. (1.25)

which (from eqn(1.12)) leads to the eigenvalue problem:

det(A− λI) = 0 ≡∣∣∣∣ 1− λ 6

1 0− λ

∣∣∣∣ = 0 (1.26)

which reduces toλ2 − λ− 6 = 0 (1.27)

giving eigenvalues λ1 = 3 and λ2 = −2.






The full eigendecomposition can then be found to be:

A =1

5

(3 21 −1

)(3 00 −2

)(1 21 −3

)(1.28)

Thus, as in eqn(1.18) above for the Fibonacci example:

un = Aun−1 = Anu0 (1.29)

Which may be shown to be:

un =1

5

(3 21 −1

)(3n 00 −2n

)(1 21 −3

)u0 (1.30)

Which reduces to

un =1

5

[3(3n) + 2(−2)n

3n − (−2)n

](1.31)

for an initial population u0 = (1, 0)T (i.e. one breeding pair).





The Leslie Matrix

The Leslie matrix is a generalization of the above.

It is a matrix which describes the increases in numbers invarious age categories of a population year-on-year.

As above we write Pn+1 = APn where Pn, A are given by:

Pn =

⎛⎜⎜⎜⎝

p1n

p2n...

pmn

⎞⎟⎟⎟⎠ , A =

⎛⎜⎜⎜⎝

α1 α2 . . . αm−1 αm

σ1 0 . . . 0 0

0 σ2...

. . ....

0 0 . . . σm−1 0

⎞⎟⎟⎟⎠

(1.32)where the Leslie matrix A is made up of αi , σi , the number ofbirths in a given age class i in year n & the fraction/probabilitythat i year-olds survive to be i + 1 years old, respectively.





The Leslie Matrix cont’d

Long-term population demographics are found as withEqn.(1.21) using the eigenvalues of A in Eqn.(1.32) &det(A− λI) = 0 to give the general characteristic Leslieequation:

λn−α1λn−1−α2σ1λ

n−2−α3σ1σ2λn−3−· · ·−αn

n−1∏i=1

σi = 0

(1.33)where, again, the Leslie matrix A is made up of αi , σi , thenumber of births in a given age class i in year n & thefraction/probability that i year-olds survive to be i + 1 yearsold, respectively.






Eqn.(1.33) has one +ive eigenvalue λ∗ & correspondingeigenvector, v∗. for a general solution like Eqn.(1.19))

Pn = c1λn1v1 +��0

c2λn2v2 + · · ·+��0

cnλnmvm,

with dominant eigenvalue λ1 = λ∗ gives long-term solution:

Pn ≈ c1λn1v1 (1.34)

with stable age distribution v1 = v∗. The relative magnitudesof its elements give the proportions in the stable state.






Example 3: Leslie Matrix for a Salmon Population

A salmon population has three age classes & females in thesecond and third age classes produce 4 and 3 offspring,respectively after each iteration.

Suppose that 50% of the females in the first age class live oninto the second age class and that 25% of the females in thesecond age class live on into the third age class.

The Leslie Matrix (c.f. Eqn.(eqn:1.14) for this population is:

A =

⎛⎝ 0 4 3

0.5 0 00 0.25 0

⎞⎠ (1.35)

Fig. 1.4 shows the growth of age classes in the population.





Leslie Matrix cont’d


0 1 2 3 4 5 6 7 8 9 100

200

400

600

800

1000

1200

1400

Time

Pop

ulat

ion

First age classSecond age classThird age class

Figure 1.4: Growth of Salmon Age Classes







The eigenvalues of the Leslie matrix may be shown to be

Λ =

⎛⎝ λ1 0 0

0 λ2 00 0 λ3

⎞⎠ =

⎛⎝ 1.5 0 0

0 −1.309 00 0 −0.191

⎞⎠(1.36)

and the eigenvector matrix S to be given by

S =

⎛⎝ 0.9474 0.9320 0.2259

0.3158 −0.356 −0.5910.0526 0.0680 0.7741

⎞⎠ (1.37)

Dominant e-vector: (0.9474, 0.3158, 0.0526)T , can benormalized (divide by its sum), to give (0.72, 0.24, 0.04)T .






Example 3: Leslie Matrix for a Salmon Population cont’d

Thus, long-term, 72% of pop’n will be in first age class, 24%in second and 4% in third. As principal e-value λ1 = 1.5,population always increases.

Can verify answer by taking any initial distribution of ages &multiplying it by A. It always converges to the proportionsabove.





Stability in Difference Equations

If a system represented by a difference equation can bewritten in the form un = Aun−1 then the growth of thesequence as n →∞ will depend on the eigenvalues of A, inthe following way:

Whenever all eigenvalues satisfy |λi | < 1, the system is said tobe stable and un → 0 as n → ∞.Whenever all eigenvalues satisfy |λi | ≤ 1, the system is said tobe neutrally stable and un is bounded as n → ∞.Whenever at least one eigenvalue satisfies |λi | > 1, the systemis said to be unstable and un is unbounded as n → ∞.





Markov Processes

Often with difference equations don’t have certainties ofevents, but probabilities. So with Leslie Matrix Eqn.(1.32):

Pn =

⎛⎜⎜⎜⎝

p1n

p2n...

pmn

⎞⎟⎟⎟⎠ , A =

⎛⎜⎜⎜⎝

α1 α2 . . . αm−1 αm

σ1 0 . . . 0 0

0 σ2. . .

......

0 0 . . . σm−1 0

⎞⎟⎟⎟⎠(1.38)

σi is probability that i year-olds survive to i + 1 years old.Leslie model resembles a discrete-time Markov chain

Markov chain: discrete random process with Markov propertyMarkov property: state at tn+1 depends only on that at tn.

The difference between Leslie model & Markov model, is:In Markov αm + σm must = 1 for each m.Leslie model may have these sums <> 1.





Markov Processes cont’d

General form of discrete-time Markov chain is given by:

un+1 = Mun

where un, M are given by:

un =

⎛⎜⎜⎜⎝

u1n

u2n...

upn

⎞⎟⎟⎟⎠ , M =

⎛⎜⎜⎜⎝

m11 m12 . . . m1 p−1 m1p

m21 m22 . . . m2 p−1 m2p...

.... . .

...mp1 mp2 . . . mp p−1 mpp

⎞⎟⎟⎟⎠

(1.39)M is the p × p Transition matrix & its mij are calledTransition probabilities with property that

∑pi=1 mij = 1.

mij represents the probability that that item will go from statei at tn to state j at tn+1.






Example 4: Two Tree Forest Ecosystem

In a forest there are only two kinds of trees: oaks and cedars.

At any time n sample space of possible outcomes is (O, C)where O = % of tree population that is oak in a particularyear and C, = % that is cedar.

If life spans are similar & on death it is equally likely that anoak is replaced by an oak or a cedar but that cedars are morelikely (probability 0.74) to be replaced by an oak than anothercedar (probability 0.26).

How can the changes in the different types of trees be trackedover the generations?







This is a Markov Process as proportions of oak/cedar at timen + 1 etc are defined by those at time n.Transition Matrix (from Eqn.(1.39)) is Table 1.1:

FromOak Cedar

Oak 0.5 0.74To

Cedar 0.5 0.26

Table 1.1: Tree Transition Matrix

Table 1.1 in matrix form:

M =

(0.5 0.740.5 0.26

)(1.40)







To track time changes in the system, define un = (on, cn)T

describing probability of oak & cedar in the forest after ngenerations.

If forest is initially 50% oak and 50% cedar, thenu0 = (0.5, 0.5)T . Hence

un = Mun−1 = Mnu0 (1.41)

M can be shown to have one positive eigenvalue 1 &corresponding eigenvector (0.597, 0.403)T which is thedistribution of oaks and cedars in the nth generation.






Markov Models have a visible state and so the state transitionprobabilities and transition matrix can be noted by theobserver.

In another type of model, the Hidden Markov Model thisvisibility restriction is relaxed and these probabilities aregenerally not known.

These kind of models are very useful in Computational Biologyfor gene sequence analysis as well as many other applications.





Applications of Non-Linear Models: Logistic Growth

Linear difference equations are useful in that they permitclosed-form solutions to be easily obtained.

However, solutions often have the disadvantage that they donot agree with observation.

In many areas of biology & esp. population biology, non-linearmodels are better.

This section focuses on the development of some simplenon-linear models for the growth of populations over time.

The simplest model is the logistic equation

Introduce the logistic equation in discrete form in this chapter& study in detail in continuous form in the next.





Applications of Non-Linear Models: Logistic Growth cont’d

Simple linear model Mn+1 = aMn is generally unsuitable forreal populations due to unbounded growth as n increases.So if we express it in the form:

Mn+1 = Mn + r ×Mn (1.42)

r ≡ no. of births − no. of deaths per time period.and replace r by R(Mn) (non-constant growth rate affected bypopulation size) get the Logistic Growth Eqn:

Mn+1 = Mn + kMn

(1− Mn

K

). (1.43)

where

R(Mn) = k − k

KMn,

for growth rate k & carrying capacity of the population K (alarge number that limits the growth rate k).






In the Logistic Growth Eqn (1.43)

Mn+1 = Mn + kMn

(1− Mn

K

)

the term in brackets behaves as follows:

For small Mn,Mn

K is small & growth is unbounded.

For large Mn,Mn

K → 1 & Eqn (1.43) behaves like

Mn+1 = Mn + ε

for small ε.

In other words for large Mn overcrowding kicks in and thegrowth rate slows to zero.






By writing Eqn.(1.43) in the form:

Mn+1 −Mn = kMn

(1− Mn

K

). (1.44)

notice some important features; as LHS of eqn(1.44) ispopulation change between successive times.....

follows that if Mn < K then population in the next timeinterval will increase & decrease for Mn > K .expect a steady increase in population for small Mn but smalloscillations above & below K for large populations.

Model is realistic in this respect mirroring real populations.

Numerical problems (i.e. predicted -ve populations) happenfor certain k in Eqn.(1.44) & reduce applicability of thismodel.






In order to see how the Logistic Growth model performs inpractice examine plots of Mn V n for various k.

k corresponds to the average fertility of an individual in thepopulation.

k varies in the range 0 < k ≤ 3 for fixed carrying capacityK = 1000 & initial population M0 = 100.






0 5 10 15 20 25 30 35 40 45 50100

200

300

400

500

600

700

800

900

1000

n

y n

(a) k = 0.2

0 5 10 15 20 25 30 35 40 45 50100

200

300

400

500

600

700

800

900

1000

n

y n

(b) k = 0.8

Figure 1.5: Logistic Equation: Stable Growth

Populations in Fig 1.5 (a),(b) approach K as n increases.

Growth initially exp’l but K causes pop’n to level out.

Larger k causes steeper growth, more rapid overcrowding.






0 5 10 15 20 25 30 35 40 45 50100

200

300

400

500

600

700

800

900

1000

1100

n

y n

(a) k = 1.6

0 5 10 15 20 25 30 35 40 45 50100

200

300

400

500

600

700

800

900

1000

1100

n

y n

(b) k = 1.9

Figure 1.6: Logistic Equation: Damped Oscillations

Early on, pop’n growth is so rapid that it overshoots K beforeovercrowding is felt.Note: the higher the k value greater than 1, the longer foroscillations to die out.






0 5 10 15 20 25 30 35 40 45 500

200

400

600

800

1000

1200

n

y n

(a) k = 2.1

0 10 20 30 40 50 600

200

400

600

800

1000

1200

1400

n

y n

(b) k = 2.4

Figure 1.7: Logistic Equation: Cyclic Growth

In Fig 1.7(a,b) plot populations with 2 < k < 2.57.

Pop’n not damped but fluctuates above & below K , comesback every 2nd breeding season, a 2-cycle.






0 5 10 15 20 25 30 35 40 45 500

200

400

600

800

1000

1200

1400

n

y n

Figure 1.8: Logistic Equation: k = 2.5, a 4-cycle

As k → 2.57, pop’n repeats every 4th breeding season, a4-cycle.






Beyond k = 2.5 (& for k < 2.57), 4-cycles become 8-cyclesbecome 16-cycles - period doubling.

For 2.57 < k < 3 (Fig. 1.9 a,b), even though behaviour isstill predicted by simple difference eqn, pattern of growthappears to become more random for increasing k.

Discovery of chaotic behaviour was important in modelling asit questioned the fact that external events always causedcomplex population fluctuations.

Often difficult to tell whether they are chaotic or simplylong-term periodic.

One feature of chaotic behaviour is its sensitivity on the initialpopulation; if M0 is varied slightly, can cause wide variationsin pop’n in subsequent times.






0 5 10 15 20 25 30 35 40 45 500

200

400

600

800

1000

1200

1400

n

y n

(a) k = 2.6

0 5 10 15 20 25 30 35 40 45 500

200

400

600

800

1000

1200

1400

n

y n

(b) k = 3.0

Figure 1.9: Logistic Equation: Into the Chaotic






Why does pop’n oscillate at all? Answer is twofold:

pop’n is self-regulating through pop’n-dependent growth rate,regulating effect is felt in next time interval but determined incurrent time interval.

Gives rise to a natural time delay when pop’n is responding toovercrowding & a corresponding over-compensation whengrowth rate is sufficiently high. Over-compensation can leadto oscillatory behaviour & chaos.

Note: if pop’n is large enough (or time step small enough)that breeding can be assumed to be continuous, growth ratecan respond instant’ly to pop’n & oscillations do not normallyoccur, except due to external or seasonal factors.

Differential equations will be used to show this below.





Introduction

Previously difference equations used to model pop’n changesfrom one generation to next.

So, as chars of individuals change from one generation to nextthro heredity,seems natural to use diff eqns with laws ofprobability to predict dominant genes in a pop’n.

Basic laws that use are those based on Mendel’s ideas.

Main result of Mendel’s ideas is that certain traits of plants &animals determined by pairs of genes.

2 genes responsible for same char (eg eye colour) are calledalleles of each other & 2 genes are said to form pair of alleles.





Introduction cont’d

Different pairs of alleles, resp for diff characteristics, arelocated at diff points on chromosome. It is common to denotealleles by letters A and α respectively.

So in any individual in a species, the alleles can occur in justone form:

AA, Aα or αα

called genotypes. (Note that genotype Aα ≡ αA.)

A given genotype determines a physical char of individual.Which allele of A and α determines the particular char is saidto be dominant, (other is recessive).

If A is dominant & causes some char then all genotypes withthis allele will have that char (assuming 1 gene only isresponsible).






Next generation, assume (as per Mendel’s first law) that allelein egg/sperm is chosen at random from parent’s two alleles.

So if AA genotype mates with Aα, possible outcomes are AA,Aα, AA, αA.

Thus, on average, no. of AAs will be half total number ofoffspring, or, to generalize, proportion of AA genotypes givenby:

G (AA) =N (AA)

N(1.45)

where N (AA) is no. with genotype AA & N is total numberof offspring.






Can now introduce gene pool concept to which each memberof a pop’n contributes its genotype’s 2 alleles.

As gene pool contains all alleles in pop’n, can talk aboutproportion of alleles of one kind or another.

If P(A) & P(α) are proportions of A and α alleles in pop’nrespectively then can say that:

P(A) = 2N (AA)+N (Aα)2N ,

P(α) = 2N (αα)+N (Aα)2N

(1.46)






Note: can view eqns such as Eqn.(1.45, 1.46) as prob’s i.e.G (AA) is proportion of genotype AA in pop’n but also probthat a genotype selected at random is AA.

Similarly from Eqn.(1.46), P(A) gives prob that genotypeselected at random contains A allele.

As with all applications of probability must have:

G (AA) + G (Aα) + G (αα) = 1and P(A) + P(α) = 1

(1.47)





Random Mating with Equal Survival

Now progress to use these derivations to model changes ingene pool over time. Initially make a number of assumptionsto simplify the process:

1 Mating occurs at random & does not depend on a mate’sgenotype

2 Equal survival: each genotype has same chance of survivingfrom fertilized egg stage to mating stage (at end of generation)

3 Equal fertility: each couple produces on average same numberof viable sperm and eggs again irrespective of genotype.

Change of terminology:

Gk(AA) expected prop of genotype AA (say) at end of kthgeneration &G∗

k+1(AA) denotes that at start of next generation.

Similar sub-/ superscripts hold for proportions of alleles.





Random Mating with Equal Survival cont’d

Since prop of a particular genotype (say, again AA) in newgeneration will depend on prop’s of alleles making up thatgenotype in previous generation (meeting up and mating)then:

G ∗k+1(AA) = Pk(A)Pk(A)

= [Pk(A)]2

G ∗k+1(Aα) = Pk(α)Pk(A) + Pk(A)Pk(α)= 2Pk(A)Pk(α)

G ∗k+1(αα) = Pk(αα)Pk(αα)

= [Pk(α)]2

(1.48)






Recall that point of this was to model changes in gene poolover time.

This can be done by deriving difference equations for alleleproportions using a two step process:

1 calculate the probability/fraction of each genotype usingEqn.s(1.48) & constant survival fraction to maturity r ,

2 count up one of the allele proportions & eliminate the otherusing Pk(A) + Pk(α) = 1.






If N ∗k+1(AA) is no. of genotype AA at start of k + 1th gen’n

& N∗k+1 total no. of fertilized eggs which give rise to thatgen’n, then, from Eqn.(1.48)

N ∗k+1(AA) = [Pk(A)]2 N∗k+1

N ∗k+1(Aα) = 2Pk(A)Pk(α)N∗k+1

N ∗k+1(αα) = [Pk(α)]2 N∗k+1

(1.49)

&, assuming a survival fraction r of fertilized egg to end ofk + 1th gen’n & maturity:

Nk+1(AA) = r [Pk(A)]2 N∗k+1

Nk+1(Aα) = 2rPk(A)Pk(α)N∗k+1

Nk+1(αα) = r [Pk(α)]2 N∗k+1

(1.50)

which, for equal survivals, yields Pk+1(α) = Pk(α), ie const αallele pop’n over all gen’ns.





Random Mating with Lethal Recessives

Taking the case of lethal recessives, no assumption of EqualSurvival above rather that carriers of two copies of the allele α(say) do not survive to maturity & Nk+1(αα) = 0. Now, fromEqn.(1.45)

Pk+1(α) =no. of α alleles

total no. of alleles in gene pool,

therefore

Pk+1(α) =2Nk+1(αα) +Nk+1(αA)

2Nk+1(AA) + 2Nk+1(Aα) + 2Nk+1(αα)(1.51)

substituting from Eqn.(1.50), to get:

Pk+1(α) =Pk(A)Pk(α)

[Pk(A)]2 + 2Pk(A)Pk(α). (1.52)





Random Mating with Lethal Recessives cont’d

Which, given Pk(A) + Pk(α) = 1 gives

Xk+1 =Xk

1 + Xk(1.53)

where Xk = Pk(α). Eqn.(1.53) is a non-linear diff eqn which,has a closed-form solution. It can be seen that

X1 =X0

1 + X0

so

X2 =X1

1 + X1=

X01+X0

1 + X01+X0

=X0

1 + 2X0

and

X2 =X2

1 + X2=

X01+2X0

1 + X01+2X0

=X0

1 + 3X0





Random Mating with Lethal Recessives cont’d

So, from this, can write the general form for the pop’n ofα-alleles in kth generation:

Pk(α) = Xk =X0

1 + kX0(1.54)

where X0 is the initial population of α-alleles.

From eqn(1.54), if P0(A) = P0(α) = 0.5, will take 8 gen’nsfor recessive allele proportion to reduce to 0.1, 98 to reduce to0.01 and 998 to reduce to 0.001.

This slow elimination of a lethal recessive is due to thecarrying effect of hybrid genotype Aα.





Random Mating with Natural Selection

For a more true-to-life model than the above take NaturalSelection where a fraction r1 newborn AA & Aα populationssurvive to end of gen’n & a different fraction r2 of newbornαα survive. From Eqn.(1.51), have:

Pk+1(α) =2Nk+1(αα) +Nk+1(αA)

2Nk+1(AA) + 2Nk+1(Aα) + 2Nk+1(αα)(1.55)

&, subsitituting altered form of Eqn.(1.50):

Nk+1(AA) = r1 [Pk(A)]2 N∗k+1

Nk+1(Aα) = 2r1Pk(A)Pk(α)N∗k+1

Nk+1(αα) = r2 [Pk(α)]2 N∗k+1

(1.56)





Random Mating with Natural Selection

simplifying, to get:

Pk+1(α) =2r2X

2 + 2r1X (1− X )

2r1(1− X )2 + 4r1X (1− X ) + 2r2X 2(1.57)

where X = Pk+1(α) and 1− X = Pk+1(A). Hence

Pk+1(α) = r2X 2+r1X−r1X 2

r1(1−2X+X 2)+2r1X−2r1X 2+r2X 2

= X 2(r2−r1)+r1Xr1+X 2(r2−r1)

= X 2(β−1)+X1+X 2(β−1)

(1.58)

where β = r2/r1 is relative fitness of the genotype αα &measures the fitness to survive of the αα genotype relative tothe AA and Aα genotypes.

Note for β = 0 get lethal recessive model & for β = 1 getequal survival model.


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