Using Genetics Applications James Sandefur Georgetown University All worksheets and spreadsheets in this talk, including answers, can be found at sandefur/genetics.htm

But first a request Mathematical Lens Ron Lancaster Mathematics Teacher

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

Using Basic Probability A B Mom a= b= 0.5 a= b= a= b= ABABABAB 0.5 P(AA)= P(AB)= P(BA)= P(BB)= (0.5)(0.5)=0.25 P(AB or BA) = P(AB) + P(BA) = = 0.5 Dad (0.3)(0.3)=0.09 (0.3)(0.7)=0.21 (0.7)(0.3)=0.21 (0.7)(0.7)= =0.42

Dad P(A)=0.3 P(B)=0.7 P(A)=0.3 Mom P(B)= AAAB BA BB

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

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

Eugenics Movement of early 20 th Century Movie, Gattica Movie, Gattica Frances Galton (positive eugenics) Frances Galton (positive eugenics) negative eugenics negative eugenics Carrie Buck, 1927, Oliver Wendell Holmes Carrie Buck, 1927, Oliver Wendell Holmes

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= =1500 Fraction B = 500/1500= 1/3 Suppose 1000 children are born 2500

P(AA)= P(“AB”)= 1000 children 2#AA = #”AB” =

=#B Total alleles = [ ] 1

Total =2000 #B = = fraction B

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 parasite from Anopheles Mosquitoes 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 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, million persons die of malaria each year, 75% of them African children (CDC) An estimated 700, million persons die of malaria each year, 75% of them African children

Sickle Cell Disease Recessive Genetic Trait Recessive Genetic Trait Sickle shaped hemoglobin Sickle shaped hemoglobin clogs small blood vessels—tissue damage clogs small blood vessels—tissue damage Sickle cell trait—mostly healthy Sickle cell trait—mostly healthy

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

Assumptions A=normal, B=sickle cell A=normal, B=sickle cell 1/3 of AA children survive malaria 1/3 of AA children survive malaria No BB children survive sickle cell No BB children survive sickle cell All “AB” children survive both diseases All “AB” children survive both diseases 3000 children born 3000 children born How many children will reach adulthood? 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-xP(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) x(1-x) x= adults= 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) x(1-x) =1000(1-x)[(1-x)+6x] =1000(1-x)(5x+1] = x x 2 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)12000 x(1-x) + 2 6[ ] 1 + 5x 6000 x(1-x) Fract. B = 3 x = 1+5x=3 x = 0.4

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

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

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

2000a(1-a) 2000a(2-a)Total alleles 2000a A-alleles Before mutation B-alleles 9% mutation rate 1820a+180a1820a = fraction A

Fraction A Equilibrium =1.3 or 0.7 Fract. A=0.7Frac. B=0.3P(BB)=0.09

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

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