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1
Mathematical Modelling
Worksheet 5
Some problems in Heredity
1. Genetic Selection.
Hereditary traits are determined by genes, which occur on every cell of an
organism, grouped together on the chromosomes. Except in the reproductive
cells genes occur in pairs and appear on paired chromosomes. A particular
gene with two alleles R and r. The genes of an offspring result from the
pairing of two genes, one from each parent. There are three possible genotypes
of the organism relative to this gene:
RR 

rr 
known as homozygotes.
Rr
known as the heterozygote.
One of the two alleles R of a particular gene is said to be dominant if
genotypes RR and Rr are indistinguishable from one another. Provided that the
genotype rr is observably different, r is said to be recessive.
The gene frequency p of gene R in a population is:
PR 
(a)
number of R genes in the population
number of R genes  number of r genes in the population
If the genotypes RR, Rr, rr occur in the population in the proportions l: m: n,
where l + m + n =1, we can determine the gene frequencies PR and Pr .
2lP  mP
2P
1
l M
2
PR 
 Pr  n 
1
M
2
2
(b)
Suppose that in a large parent population genes R and r are present in
proportion p : q, where p + q =1. Assuming these proportions are the same for
male and female and that mating is random, we can determine the proportions
of the three genotypes RR, Rr and rr.
p
p
R
RR
r
Rr
p
R
Rr
q
r
R
q
S
t
qa
r
t
r
l  RR  p 2 ,
rr
m  Rr  2 pq,
n  rr  q 2
l
1
m  p 2  pq  p p  q   p  R
2
n
1
m  q 2  pq  q p  q   q  r
2
Therefore proportions of the three genotypes RR, Rr and rr in future generations are:
RR : p 2
(c)
Rr : 2 pq
rr : q 2
If a fraction s of the RR and Rr individuals survive to adulthood, while only a
fraction (1 – k)s (0<k<1) of rr individuals survive, then zygote rr has selective
disadvantage k.
Let pn , qn  1  pn  be the gene frequencies of genes R and r in the n’th
generation adults. We can find the proportions of the genotypes born and
surviving to adulthood in the next generation, assuming selective disadvantage
p
k of the rr individuals. Also by denoting n by y n , we can obtain a recurrence
qn
relation for yn .
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Born p 2 : 2 pq : q 2
Adults sp 2 n : 2 sp n q n : 1  k sq 2 n
PR 
2 sp 2 n  2 sp n q n

2 Sp 2 n  2 Sp n q n  1  k Sq 2 n
pn  pn qn
2


PR 
p
2
p
2
 2 p n q n 1  k q 2 n
n
n
pn  pn  qn 
 2 p n q n  q 2 n  kq 2 n
pn
1  kqn
2(1  k ) Sq n2  2Sp n q n
Pr 
2 Sp n2  2Sp n q n  1  k Sq n2



q n  p n  1  k q n 
1  kqn2


pn
1  kqn2
pn
PR



Pr 1  kqn2 q n  kqn2 q n  kqn2

pn
q n 1  kqn 

4
Now
pn
 y n so we get
qn
yn
1 - kq n
pn  qn  1
pn  yn qn
1  qn  yn qn
1  q n  y n  1
qn 
PR

Pr
1
and substitute back in. Now we get
yn  1
yn
 k 

1  
y

1
 n

y n 1 
(d)

y n  y n 1
yn  1  k
y n  y n  1
yn  1  k
If a recessive gene has initial frequency 0.99 and selective disadvantage k
=0.05, how many generations will it take for the gene frequency to drop to;
(i)
(ii)
0.1
0.01
y n 1 
y n  y n  1
y n  1  k 
dy n y n 1  y n

dn
n 1 n
dy n
 y n 1  y n
dn
5
dy n
y  yn
 n
 yn
dn
y n  1  k 
2
yn  yn
y  y  1  k 
 n n
y n  1  k 
y n  1  k 
2

dy n
kyn

dn
y n  1  k 

yn  1  k
dy n   d n
kyn
yn
 ky
n
dy n  
1 k
dn  n
kyn
1
0.95
 0.05 dy   0.05 y
putting k  0.05
n
n
n
n  20 y n  19 ln y n  c
n0
y0
p 0 0.01 1


q 0 0.99 99
20
1
 19 ln
c
99
99
20
1
c    19 ln
 87.105
99
99
 0
 the number of generation ' s it will take for the gene frequency to drop to 0.1 is
p n 0.99

 9  yn
qn
0.1
p n  0.1 , q n  0.9
n  209   19 ln 9  87.105  308.953
 n  309
and the number of generation ' s it will take for the gene frequency to drop to 0.01 is
p n 0.99

 yn
q n 0.01
n  2099   19 ln 99  87.105  2154.413
 n  2155
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(e)
We can consider the effect on the situation in (c) and (d) if the selective
disadvantage is associated with the dominant gene.
Sp 2 n : 2 Sp n q n : 1  k Sq 2 n
S 1  k  p 2 n : 21  k Sp n q n : Sq 2 n
 1  2
Sp 2 n : 2 Sp n q n : S 
q n
1 k 
dy n
y 2 n  yn

 yn
dn
y n  1  k 


dy n
y 2 n  y n 1  k  y n  y n 1  k   1


dn
y n 1  k   1
y n 1  k   1


 kyn
y n 1  k   1
y n 1  k   1
dy n   d n
 kyn
kyn
yn
1
dy n  
n
kyn
kyn
 ky
dy n  
yn 
1
y n  k ln y n  n
k
n
 1
 y n 1    k ln y n  0
 k
0  k  1
2. Sex Linked Characteristics.
Sex linked characteristics are not transmitted according to the previously
stated laws. One pair of chromosomes determines sex - XX for females and
XY for males. If a particular gene occurs only on the X chromosome a female
can be classified as RR, Rr or rr, but a male having only one X chromosome
can only be classified R or r.
(a)
Denoting the gene proportions in the female parent population by p and q, and
those in the male population by p’ and q’, (p + q = 1 = p’ + q’), we can show
that the corresponding proportions in the next female generation are the
arithmetic means of the proportions of male and female genes in the parent
population.
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Proportion R in Female = p
Proportion r in Female = q
Proportion R in Male = p’
Proportion r in Male = r’
l:m:n
p + q = 1 = p’+ q’
RR : Rr : rr
pp’: p’q + q’p : qq’
PR 
(b)
2l  m
2 pp'   p' q  pq 

2l  m  n  2 pp'   p' q  pq'  qq'
p2p'q'  p' q

2 p p' q'  q p' q'
pp'1  p' q

2 p  q 
p' p  q   p

2
p  p'

2
Using the results of 2(a), we can derive and sole a difference equation relating
the gene proportion of allele R in the female population in successive
generations.
From 2(a) we have:
p1 
p0  p0'
2
pn1 
pn  pn'
2
pn 1 
pn  pn 1
2
where pn'  pn1
This equation can be rearranged into a recurrence relation, which can be solved.
pn  2 pn1  pn1  0
If p n  s n
s  0
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s n  2s n 1  s n 1  0
 2s 2  s  1  0
Then
s  0.5s  1  0
s  
1
or s  1
2
n
 1
 p n  A    B
 2
Using the boundary conditions p0  p1' , p1  p2' we can find the values of A and B.
p1'  A  B
1
p 2'   A  B
2
Solving these two simultaneous equations we get:


2 '
p1  p 2'
3
21

B   p1'  p 2' 
32

A
n
2 '
1 1 '

' 
 p n   p1  p 2      p1  p 2' 
3 
 2 2


