All worksheets and spreadsheets in this talk, including answers, can be found at http://www.georgetown.edu/faculty/ sandefur/genetics.htm
James Sandefur
Georgetown University

But first a request
Mathematical Lens
Ron Lancaster
Mathematics Teacher

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A and B alleles
One allele from each parent
Genotypes are AA, AB, BA, and BB
Allele from mother is independent of allele from father.
P(A) = a , P(B) = b

Using Basic Probability
Mom a=
Dad
A P(AA)=
A a= b= a=
B
A
P(AB)=
P(BA)=
B b= b=
B P(BB)=

Mom
P(A)=0.3
P(B)=0.7
Dad
P(A)=0.3 P(B)=0.7
0.09 0.21
AA AB
0.49 0.21
0.21 0.49
BA
BB
0.21 0.09

Basic Genetics Simulation
See first Worksheet
Understanding of Genetics

Fraction A this generation=0.3
P(AA)=0.09, P(“AB”)=0.42, P(BB)=0.49
Suppose 1000 children
90 AA, 420 “AB”, 490 BB
2(90)+420=600 A out of 2000
Fraction A next generation=0.3

Fraction A this generation= a
P(AA)= a 2 , P(“AB ”)=2a(1-a), P(BB)= (1-a) 2
Suppose 1000 children
2000
2
2000
[ a 2 + 2000
“AB”,
] a A-alleles
Fraction A
Total alleles

•
•
•
•
•
Proportion of alleles of each type remain constant from one generation to the next
Recessive traits remain constant over time
Assuming no additional effects such as
Selective advantage
Mutation

th
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Movie, Gattica
Frances Galton (positive eugenics) negative eugenics
Carrie Buck, 1927, Oliver Wendell Holmes http://www.eugenicsarchive.org/eugenics

Eugenics Simulation
See second Worksheet
Study of eugenics

Suppose in the current generation, 50% of the alleles are A, i.e.
Suppose 1000 children are born
Total=1000+500=1500
Fraction B = 500/1500= 1/3

P ( B )

1 x and P (


  1 x 


2
A )
1000 children

1

1 x
1

 
1 x
1 x

# AA

1000


Total alleles =
1 x 

 2
#
2000
"

1
AB "
1 x

2000

 
1
1 x

1 x

 
1
2000
[
1
1
2
+ x



1 x 


1 x
]
1 x

x
1

#B =

1
2000
2000







1 x 



 x 1 1 x x 



 


1 x 


1 x 


= fraction B

Given that currently,
P(B)=0.04, how many generations will it take until
P(B)=0.02? 25 generations (375 years)
P(B)=0.01?
Another 50 generations

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 parasite from Anopheles Mosquitoes
(CDC website) Forty-one percent of the world's population live in areas where malaria is transmitted (e.g., parts of Africa, Asia, the
Middle East, Central and South America,
Hispaniola, and Oceania)
(CDC) An estimated 700,000-2.7 million persons die of malaria each year, 75% of them African children

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Recessive Genetic Trait
Sickle shaped hemoglobin clogs small blood vessels—tissue damage
Sickle cell trait—mostly healthy http://www.sicklecelldisease.org

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Sickle cell trait gives partial immunity to malaria
Sickle cell allele is valuable in areas with high malaria risk

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A=normal, B=sickle cell
1/3 of AA children survive malaria
No BB children survive sickle cell
All “AB” children survive both diseases
3000 children born
How many children will reach adulthood?

Sickle Cell/Malaria
Survival Simulation
See 3 rd Worksheet
Study of Sickle Cell
Anemia/Malaria relationship

P(A)=1-x P(B)=x
P(AA)=(1-x) 2 , P(“AB”)=2x(1-x), P(BB)=x 2
#children
#AA=3000(1-x) 2
#“AB”=6000x(1-x)
#BB=3000 x 2
# adults
#AA=1000(1-x) 2
#“AB”=6000x(1-x)
#BB=0

x = fraction alleles B, sickle cell
1000(1-x) 2 6000x(1-x) x= adults=
0.1
0.3
1350 1750
0.6
1600
What fraction of A and B alleles maximizes number of adult survivors?

#AA=1000(1-x) 2 #“AB”=6000x(1-x)
Adults = f(x)= 1000(1-x) 2 +6000x(1-x)
=1000(1-x)[(1-x)+6x]
=1000(1-x)(5x+1]
=1000+4000x - 5000x 2
Maximum when x = 0.4

Sickle Cell/Malaria
Simulation
What happens over time?

#AA=1000(1-x) 2 #“AB”=6000x(1-x) x =
6000 x (1-x) total=
2
+
1

1
 u ( n )
1+5x=3
5

 x
1
1

3

3 u (
Now
5
5 u n
Now
(
 n
1

)
]

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How estimate mutation rate?
Lethal recessive trait, BB
Mutation from A to B

Lethal Trait
Mutation from normal allele
Simulation
See 4 th Worksheet
Mutation rates and lethal trait

Fraction A this generation= a
P(AA)= a 2 , P(“AB ”)=2a(1-a),
0
P(BB)= (1-a) 2
Suppose 1000 children
2000
2
[ a 2 + 2000 a (1- ) ] a
2
2000a(1-a) B-alleles
2000a [ 12 + 2000a ]
Total alleles

A-alleles
Before mutation
B-alleles
2000a(1-a)
9% mutation rate
2000a(2-a)
= fraction A 
0
2
.
91
 a

a

2

Equilibrium

0 .
91
2
 a
4
 a u
2
(
 n )
2

4 ( 0 .
91 ) a
2

0 .
0 91

0
2


2 u  ( n
2

0 .
1 )
=1.3
or 0.7
Fract. A=0.7
Frac. B=0.3 P(BB)=0.09

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Galactosemia used to be a lethal trait
Now easily diagnosed and treated
Recessive trait, BB
0.002%<Children born <0.01%
Mutation rate, 0.00002<m<0.0001

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Socially Relevant
Discuss with Biology Teachers
Evolution (Intelligent Design)