Population Genetics
DNA Fingerprint
Reminders
• Genetic Disease Presentation - Friday
• Review - Monday
• Final Exam -Thursday, 1:30-3:30
Population Genetics Frequencies
Genotype Frequencies
P = freq. of AA
H = freq. of Aa
Q = freq. of aa
Allele Frequencies
p = freq. of A allele
q = freq. of a allele
p = P+1/2 H
q = Q+1/2 H
If Hardy-Weinberg Equilibrium
P = p2
H = 2pq
Q = q2
q=√Q
Hardy-Weinberg
Equilibrium
• Genotype and allele frequencies remain
constant from one generation to the next if
the following conditions are met:
• random mating
• no natural selection
• no migration
• no mutation
• large population size
Evolution
Population Genetics Equations
Natural Selection
fitness (from survival) w
inital P,H,Q x fitness = rel freq
rel freq/total = new P,H,Q
p’=P+1/2H, q’=Q+1/2H
Migration and Genetic Drift
N0=initial genotype numbers
Nt=new genotype numbers
Calculate P,H,Q
p’=P+1/2H , q’=Q+1/2H
Inbreeding
Random H*H*1/4
Siblings 2H * 1/16
Cousins 2H * 1/64
Mutation
p’ = p - µp
q’ = q + µp
DNA Fingerprint
• Shows alleles present in an individual
• Used in paternity, crime investigation, medical diagnosis
std. pers.1 pers.2
Tandem Repeats
• sequences of DNA that repeat
Short Tandem Repeats (STR) - 1,2,3 bases repeat 15-100
times
GACGACGACGACGAC
Variable Number of Tandem Repeats (VNTR) - 10-40
bases repeat 10’s to 1000’s
GACTTACAGCGACTTACAGCGACTTACAGC
Each Individual has 2 Alleles
Person A
Chromosome 1a
Chromosome 1b
GACGACGACGACGAC
GACGACGAC
3
Person B
Chromosome 1a
GACGACGACGAC
Chromosome 1b
GACGACGACGAC
4
4
5
PCR
• technique to amplify selected DNA region
GACGACGACGACGAC
GACGACGACGACGAC
GACGACGAC
GACGACGAC
Polymerase Chain Reaction (PCR)
• Heat DNA to open
3‘ATTGGTTCCGACCGACCGACCGACCGACCGAGGTTATT5’
5‘TAACCAAGGCTGGCTGGCTGGCTGGCTGGCTCCAATAA3’
Polymerase Chain Reaction (PCR)
• Primers attach
3‘ATTGGTTCCGACCGACCGACCGACCGACCGAGGTTATT5’
5’CCAA
GGTT5’
5‘TAACCAAGGCTGGCTGGCTGGCTGGCTGGCTCCAATAA3’
Polymerase Chain Reaction (PCR)
• Taq polymerase builds new DNA from primer to
end
3‘ATTGGTTCCGACCGACCGACCGACCGACCGAGGTTATT5’
5’CCAAGGCTGGCTGGCTGGCTGGCTGGCTCCAATAA3’
3‘ATTGGTTCCGACCGACCGACCGACCGACCGAGGTT5’
5‘TAACCAAGGCTGGCTGGCTGGCTGGCTGGCTCCAATAA3’
PCR Fingerprint
Gel
Electrophoresis
smallest
+
-
Stain
Ethidium
Bromide
D1S80
• Tandem Repeat
• Chromosome 1
• 16 nucleotides long
• repeats 15 - 40x
• multiple alleles (26)
D1S80
# Repeats
Caucasian
Hispanic
African Amer.
Asian
15
0
0.001
0
0
16
0.001
0.01
0.002
0.034
17
0.002
0.009
0.028
0.025
18
0.237
0.224
0.073
0.152
19
0.003
0.005
0.003
0.022
20
0.018
0.013
0.032
0.007
21
0.021
0.028
0.115
0.034
22
0.038
0.024
0.081
0.017
23
0.012
0.009
0.014
0.017
24
0.378
0.315
0.234
0.23
DNA Fingerprinting
19
0.003
20
0.018
21
0.021
22
0.038
23
0.012
• What is probability of a
Caucasian having a D1S80
repeat of 19 and 23?
• H=2pq
• H =2*0.003 * 0.012 =
(19,23)
0.000072
• 211.5 million Caucasians in US • 15,228 people
DNA Fingerprinting
• What is probability of a Caucasian having a D7S820
repeat of 8 and 14?
• H=2pq
H(8,14) = 2*.151 * 0.002 = 0.000604
• 211.5 million Caucasians in US • 127,746 people
DNA Fingerprinting
• D1S80 H = 0.000072
• D7S820 H = 0.000604 • Chance of both?
• 0.000072*0.000604 = 0.000000043 • 211.5 million caucasians * 0.000000043
• 9.2 people
(19,23)
(8,14)
13 Loci in DNA Fingerprints
DNA Fingerprint
• What if person has only 1 band?
• Homozygous for locus
• P=p
• What is the probability a
2
Hispanic person is homozygous
for D1S80 allele 20 repeats?
P(20,20)=0.013*0.013=0.000169
#
Caucasi Hispani
Repeats
an
c
18
0.237
0.224
19
0.003
0.005
20
0.018
0.013
21
0.021
0.028
22
0.038
0.024
Paternity Example
Activity 25