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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.

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Using Genetics Applications James Sandefur Georgetown University

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  1. 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

  2. But first a request Mathematical Lens Ron Lancaster Mathematics Teacher

  3. Basics of Simple Genetic Trait • 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

  4. Using Basic Probability Mom Dad a= b= a= b= 0.3 0.7 0.3 0.7 0.5 0.5 0.5 0.5 A B A B 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 (0.3)(0.3)=0.09 (0.3)(0.7)=0.21 (0.7)(0.3)=0.21 (0.7)(0.7)=0.49 A a= 0.3 0.7 0.5 B 0.5 b= P(AB or BA) = P(AB) + P(BA) = 0.25+0.25 = 0.5 0.21+0.21=0.42

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

  6. Basic Genetics Simulation See first Worksheet Understanding of Genetics

  7. 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

  8. Fraction A this generation= a P(AA)=a 2, P(“AB”)=2a(1-a), P(BB)=(1-a) 2 Suppose 1000 children 2 [ a + a (1- ) a ] a A-alleles 2000 2000 1000a2 AA, 2000a(1-a) “AB”, 1000(1-a)2 BB Fraction A Total alleles 2000

  9. Hardy Weinberg Law • 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

  10. Eugenics Movement of early 20th Century • Movie, Gattica • Frances Galton (positive eugenics) • negative eugenics • Carrie Buck, 1927, Oliver Wendell Holmes • http://www.eugenicsarchive.org/eugenics

  11. Eugenics Simulation See second Worksheet Study of eugenics

  12. Suppose in the current generation, 50% of the alleles are A, i.e. P(A)=0.5 and P(B)=0.5 Then Suppose 1000 children are born #AA=250, #”AB”=500, #BB= 250 0 #A=2(250)+500=1000 #B=500 P(AA)=0.25, P(“AB”)=0.5, P(BB)=0.25 Total=1000+500=1500 Fraction B = 500/1500= 1/3

  13. P(AA)= P(“AB”)= #AA = #”AB” = 2 1000 2000 1000 children

  14. =#B Total alleles = ] 2 [ 2 + 4000 2000 1

  15. #B = 2000 = fraction B Total = 2000

  16. 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

  17. Malaria • 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

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

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

  20. Assumptions • 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?

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

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

  23. x = fraction alleles B, sickle cell 1000(1-x) 2 6000x(1-x) Total= + #AA= #”AB”= 0.6 x= 0.1 0.3 adults= 1350 1750 1600 What fraction of A and B alleles maximizes number of adult survivors?

  24. #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

  25. Sickle Cell/Malaria Simulation What happens over time?

  26. #AA=1000(1-x) 2 #“AB”=6000x(1-x) #B = x (1-x) 6000 3 Fract. B = x = 2 ] x (1-x) [ (1-x) 1 + 5x + 12000 6 total= 2000 x = 0.4 1+5x=3

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

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

  29. Fraction A this generation= a 0 P(AA)=a 2, P(“AB”)=2a(1-a), P(BB)= (1-a) 2 Suppose 1000 children 2 [ a + a (1- ) a ] a A-alleles 2000 2000 1000(1-a) 2 1000a2 AA, 2000a(1-a) “AB”, BB 2000a(1-a) B-alleles 2000a [ 2 1 + 2000a — a (1-a) ] Total alleles

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

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

  32. Galactasemia • 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

  33. General Comments on Genetics • Socially Relevant • Discuss with Biology Teachers • Evolution (Intelligent Design)

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