Using Genetics Applications James Sandefur Georgetown University

Using Genetics Applications

James Sandefur

Georgetown University

All worksheets and spreadsheets in this talk, including answers, can be found at

http://www.georgetown.edu/faculty/ sandefur/genetics.htm

But first a request

Mathematical Lens

Ron Lancaster

Mathematics Teacher

Basics of Simple Genetic Basics of Simple Genetic TraitTrait

A and B allelesA and B alleles One allele from each parentOne allele from each parent Genotypes are AA, AB, BA, and BBGenotypes are AA, AB, BA, and BB Allele from mother is independent of Allele from mother is independent of

allele from father.allele from father. P(A) = P(A) = aa, P(B) = , P(B) = bb

Using Basic Probability

A

B

Mom

a=

b=

0.5

0.5

a=

b=

a=

b=

A

B

A

B

0.5

0.5

0.5

0.5

P(AA)=

P(AB)=

P(BA)=

P(BB)=

(0.5)(0.5)=0.25

(0.5)(0.5)=0.25

(0.5)(0.5)=0.25

(0.5)(0.5)=0.25

P(AB or BA) = P(AB) + P(BA) =0.25+0.25 = 0.5

Dad

0.3

0.7

0.3

0.7

0.3

0.7

(0.3)(0.3)=0.09

(0.3)(0.7)=0.21

(0.7)(0.3)=0.21

(0.7)(0.7)=0.49

0.21+0.21=0.42

Dad

P(A)=0.3 P(B)=0.7

P(A)=0.3

Mom

P(B)=0.7

0.49 0.21

0.21 0.09

0.09 0.21

0.21 0.49

AA AB

BABB

Application

Basic Genetics Simulation

See first Worksheet

Understanding of Genetics

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=0.3

P(AA)=a 2, P(“AB”)=2a(1-a), P(BB)=(1-a) 2

Suppose 1000 children

1000a2 AA, 2000a(1-a) “AB”, 1000(1-a)2 BB

Fraction A this generation= a

2000 2000+a a (1- )2 A-alleles

[ ]aa

Total alleles

2000Fraction A

Hardy Weinberg LawHardy Weinberg Law

• Proportion of alleles of each type remain Proportion of alleles of each type remain constant from one generation to the nextconstant from one generation to the next

• Recessive traits remain constant over Recessive traits remain constant over timetime

• Assuming no additional effects such asAssuming no additional effects such as• Selective advantageSelective advantage• MutationMutation

Eugenics Movement of early Eugenics Movement of early 2020thth Century Century

Movie, Movie, GatticaGattica Frances Galton (positive eugenics)Frances Galton (positive eugenics) negative eugenicsnegative eugenics Carrie Buck, 1927, Oliver Wendell Carrie Buck, 1927, Oliver Wendell

HolmesHolmes http://http://www.eugenicsarchive.orgwww.eugenicsarchive.org

/eugenics/eugenics

Eugenics Simulation

See second Worksheet

Study of eugenics

Suppose in the current generation, 50% of the alleles are A, i.e.

P(A)=0.5 and P(B)=0.5

Then

P(AA)=0.25, P(“AB”)=0.5, P(BB)=0.25

#AA=250, #”AB”=500, #BB=

#A=2(250)+500=1000

#B=500

Total=1000+500=1500

Fraction B = 500/1500= 1/3

Suppose 1000 children are born 2500

xAP

xBP

11)( and

1)(

P(AA)= P(“AB”)=

1000 children

21

1

x

xx

11

12#AA = #”AB” =1000 2000

21

11000#

xAA

xxAB

11

12000""# =#B

Total alleles =

x

11

x

112000 4000

x

1+

22[ ]1

xx

11

112000

x

11

x

1

Total =

x

112000

2000#B =

x

11

= fraction B

x

x 1

1

1

x

Given that currently,

P(B)=0.04,

how many generations will it take until

P(B)=0.02?

P(B)=0.01?

25 generations (375 years)Another 50 generations

Malaria Malaria

parasite from Anopheles Mosquitoes parasite from Anopheles Mosquitoes (CDC website) Forty-one percent of the (CDC website) Forty-one percent of the

world's population live in areas where world's population live in areas where malaria is transmitted (e.g., parts of Africa, malaria is transmitted (e.g., parts of Africa, Asia, the Middle East, Central and South Asia, the Middle East, Central and South America, Hispaniola, and Oceania)America, Hispaniola, and Oceania)

(CDC) An estimated 700,000-2.7 million (CDC) An estimated 700,000-2.7 million persons die of malaria each year, 75% of persons die of malaria each year, 75% of them African childrenthem African children

Sickle Cell DiseaseSickle Cell Disease

Recessive Genetic TraitRecessive Genetic Trait Sickle shaped hemoglobinSickle shaped hemoglobin clogs small blood vessels—tissue clogs small blood vessels—tissue

damagedamage Sickle cell trait—mostly healthySickle cell trait—mostly healthy http://www.sicklecelldisease.orghttp://www.sicklecelldisease.org

Sickle Cell/Malaria Sickle Cell/Malaria RelationshipRelationship

Sickle cell trait gives partial Sickle cell trait gives partial immunity to malariaimmunity to malaria

Sickle cell allele is valuable in areas Sickle cell allele is valuable in areas with high malaria riskwith high malaria risk

AssumptionsAssumptions

A=normal, B=sickle cellA=normal, B=sickle cell 1/3 of AA children survive malaria1/3 of AA children survive malaria No BB children survive sickle cellNo BB children survive sickle cell All “AB” children survive both All “AB” children survive both

diseasesdiseases 3000 children born3000 children born How many children will reach How many children will reach

adulthood?adulthood?

Sickle Cell/Malaria Survival

Simulation

See 3rd 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

1000(1-x) 2 6000x(1-x)

x=

adults=

0.1 0.3 0.6

1350 1750 1600

What fraction of A and B alleles maximizes number of adult survivors?

#AA= #”AB”=Total= +

x = fraction alleles B, sickle cell

#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 - 5000x2

Maximum when x = 0.4

Sickle Cell/Malaria

Simulation

What happens over time?

#AA=1000(1-x) 2 #“AB”=6000x(1-x)#B =

total= 2000 (1-x) 12000x (1-x)+2 6[ ]1 + 5x

6000

x (1-x)Fract. B = 3x =

x51

31

1+5x=3 x = 0.4

Now

NowNext

51

3

)1(51

)1(3)(

nu

nunu

MutationMutation

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

Lethal Trait

Mutation from normal allele

Simulation

See 4th Worksheet

Mutation rates and lethal trait

P(AA)=a 2, P(“AB”)=2a(1-a), P(BB)= (1-a) 2

Suppose 1000 children

1000a2 AA, 2000a(1-a) “AB”, BB

Fraction A this generation= a

2000 2000+a a (1- )2 A-alleles

[ ]aa

Total alleles

1000(1-a) 2

2000a(1-a) B-alleles

2000a 2000a+ (1-a)[ ]12 a

0

—

2000a(1-a)

2000a(2-a)

Total alleles

2000a

A-alleles

Before mutation

B-alleles

9% mutation rate

1820a

+180a1820a

= fraction Aa

2

91.0

a

2

91.0Fraction A

Equilibrium

a

091.022 aa

2

)91.0(442 a

2

36.02 =1.3 or 0.7

Fract. A=0.7

Frac. B=0.3

P(BB)=0.09

)1(2

91.0)(

nunu

GalactasemiaGalactasemia

Galactosemia used to be a lethal traitGalactosemia used to be a lethal trait Now easily diagnosed and treatedNow easily diagnosed and treated Recessive trait, BBRecessive trait, BB 0.002%<Children born <0.01% 0.002%<Children born <0.01% Mutation rate, Mutation rate, 0.00002<m<0.00010.00002<m<0.0001

General Comments on General Comments on GeneticsGenetics

Socially Relevant Socially Relevant Discuss with Biology TeachersDiscuss with Biology Teachers Evolution (Intelligent Design)Evolution (Intelligent Design)