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Genetic Analysis II (385) –
Population Genetics Lecture 02
Date: 03/10/11
Instructor: Professor Derek Gordon
Office: Life Sciences Building, Room 128
Phone: 4453386
Phone: 445 3386
Email: [email protected]
Office Hours: Tues, Thurs: 12:00pm – 1:15pm,
and by appt.
HOW TO RECOGNIZE A POPULATION IN HWE
In Genetics you have to deal with real populations.
These
may or may not be in or near HWE.
may or may not be in or near HWE.
Consider a population with three genotypes
A
1
A
1,
A
1
A
2,
&
A
2
A
2
and the genotype array of
X + Y + Z.
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1.
First calculate the values of p
and q
.
p = X + 1/2Y
q = Z + 1/2Y
2.
Next calculate the e
xpected genotype
frequencies:
p
2
(A
1
A
1
) + 2pq(A
1
A
2
) + q
2
(A
2
A
2
).
3.
Calculate the
expected number
of each genotype by
multiplying the expected genotype frequencies by the total
number
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number.
4.
Compare the
expected genotype array
under HWE to
the
observed genotype array X + Y + Z
.
•In natural populations you would use a
Chisquare test
for goodnessoffit.
•If you are given a frequency distribution assume an
infinitely large population.
(Must have an exact fit)
5.
If the observed and expected distributions
are statistically
the same
then you can assume the population is in
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, then you can assume the population is in
HWE.
*
3
Are the following in HWE?
Observed
Distribution or Freq.
pq
Expected
Distribution or
Freq.
2
HWE
A
1
A
1
A
1
A
2
A
2
A
2
a.
.60
.20
.20
b.
533
734
233
c.
1000
6000
3000
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d.
2050
3900
2050
2
With 1 df, Critical value is 3.84 at .05 level of significance.
I'll step through the first one for you
X+ Y + Z = .6 + .2 + .2 = 1
p = X + 1/2(Y) = .6 + 1/2(.2) = .7
q = Z + 1/2(Y) = .2 + 1/2(.2) = .3
In HWE, X + Y + Z = (p+q)
2
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(p+q)
2 =
(.7+ .3)
2 =
.49 + .42 + .09
Evaluate: .49 + .42 + .09 does not equal .6 + .2 + .2. Not in
HWE
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EXAMPLES:
Observed
Distribution or Freq.
pq
Expected
Distribution or
Freq.
2
HWE
A
1
A
1
A
1
A
2
A
2
A
2
A
1
A
1
A
1
A
2
A
2
A
2
a.
.60
.20
.20
.70
.30
.49
.42
.09
NO
b.
533
734
233
c
1000
6000
3000
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c.
d.
2050
3900
2050
2
With 1 df, Critical value is 3.84 at .05 level of significance.
Example 2
533 + 734 + 233 = 1500
p =
[533 + 1/2(734)]/1500 = 900/1500 = .6
q =
[233 + 1/2(734)]/1500 = 600/1500 = .4
At HWE
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[.6
2
+ (2*.6*.4)
+ .4
2
] * 1500 =
[.36 + .48 +. 16] * 1500 = 540 + 720 + 240
Expected distribution
5
Example 2
533
734
233
Observed distribution
Expected distribution
ObsExp
(ObsExp)
2
(ObsExp)
2
/Exp
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Example 2
533
734
233
Observed distribution
540
720
240
Expected distribution
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Example 2
533
734
233
Observed distribution
540
720
240
Expected distribution
7
14
7
ObsExp
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Example 2
533
734
233
Observed distribution
540
720
240
Expected distribution
7
14
7
ObsExp
49
196
49
(ObsExp)
2
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7
Example 2
533
734
233
Observed distribution
540
720
240
Expected distribution
7
14
7
ObsExp
49
196
49
(ObsExp)
2
.09
.27
.20
(ObsExp)
2
/Exp
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Example 2
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This note was uploaded on 04/04/2011 for the course GENETICS 385 taught by Professor Brennemanandgordon during the Spring '11 term at Rutgers.
 Spring '11
 brennemanandgordon
 Genetics

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