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Using Genetics Applications James Sandefur Georgetown University
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Transcript of 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)