PV92 PCR/Informatics Kit

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PV92 PCR/Informatics Kit

• Population Genetics and Informatics

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To estimate frequency of Alu within a
population

• Amplify Alu insert from representative sample
  population
• Calculate the expected allelic and genotypic
  frequencies
• Perform Chi-squared Test

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Calculating Allelic and Genotypic Frequencies

• Within your class, how unique is your particular
  combination of Alu alleles? By calculating an
  allele frequency, you can begin to answer this
  question. An allele frequency is the percentage
  of a particular allele within a population of
  alleles. It is expressed as a decimal. You can
  calculate an allele frequency for the Alu PV92
  insertion in your class by combining all your
  data. For example, imagine that there are 38
  students in your class and the genotype
  distribution within the class is as follows

Genotype / /- -/-
Total (N) of people 25
5 8 38
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Calculating Allelic Frequencies allele

• Total number of alleles 2N 2(38) 76
• Number of alleles
• / 25 with two alleles 50
  alleles
• /- 5 with one alleles 5
  alleles
• Total 55 alleles
• Frequency of number of alleles 55
  0.72
• total alleles
  76
• Calculation for the alleles would be similar,
  and the result would be .28
• Notice ( allele) (- allele) 1.0

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Calculating Genotypic Frequencies

• How does the distribution of Alu genotypes in
  your class compare with the distribution in other
  populations? For this analysis, you need to
  calculate a genotype frequency, the percentage of
  individuals within a population having a
  particular genotype. Remember that the term
  allele refers to one of several different forms
  of a particular genetic site whereas the term
  genotype refers to the specific alleles that an
  organism carries. You can calculate the frequency
  of each genotype in your class by counting how
  many students have a particular genotype and
  dividing that number by the total number of
  students.
• Given the ethnic makeup of your class, might you
  expect something different?
• How can you estimate what the expected frequency
  should be?

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Calculating Observed Genotypic Frequencies

• Genotype / (p2) /- (2pq)
  -/- (q2) Total (N)
• of people 25 5
  8 38
• Observed 0.66 0.13
  0.21 1.00
• frequency
• Calculation
• / genotypic frequency with genotype
• total number of people (N)
• 25/38
• .66

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Alu and Population Genetics

• If within an infinitely large population no
  mutations are acquired, no genotypes are lost or
  gained, mating is random, and all genotypes are
  equally viable, then that population is said to
  be in Hardy-Weinberg equilibrium. In such
  populations, the allele frequencies will remain
  constant generation after generation. Genotype
  frequencies within this population can then be
  calculated from allele frequencies by using the
  equation
• p2 2pq q2 1.0
• p and q are the allele frequencies for two
  alternate forms of a genetic site. The genotype
  frequency of the homozygous condition is either
  p2 or q2 (depending on which allele you assign to
  p and which to q). The heterozygous genotype
  frequency is 2pq.

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Alu and Population Genetics

• Hardy-Weinberg Equilibrium

p2 2pq q2 1
p
q
pp
pq
p
/ p2 /- 2pq -/- q2
q
pq
qq
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Using Hardy-Weinberg

• Determine p2, 2pq, and q2 values
• Expected genotypic frequencies
• p 0.72 , so q 0.28 since p q 1
  p2 2pq q2 1
• (0.72)2 2 (0.72)(0.28) (0.28)2
  1
• 0.52 0.40 0.08 1
• p2 0.52
• 2pq 0.40
• q2 0.08

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Calculate Expected Number of Genotypes

• Expected number of genotype
• Genotypic frequency x population number (N)
• Genotype Expected number
• / 0.52 x 38 20
• /- 0.40 x 38 15
• -/- 0.08 x 38 3

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Chi Squared Test

• Chi-square, a statistical test used for comparing
  observed frequencies with expected frequencies.
  The larger the chi-square value, the greater is
  the difference between the observed and the
  expected values.
• When using the Chi-square analysis, we test the
  null hypothesis that there is no difference
  between samples (observed and expected) and we
  assume that if there is any difference, then it
  arose simply by chance and is not real.
• Our null hypothesis is that your class is in
  Hardy-Weinberg equilibrium. Whether or not we can
  accept the null hypothesis is given by a p-value.
  
• If the calculated p-value is less than 0.05, the
  null hypothesis is disproved the population is
  not in Hardy-Weinberg equilibrium.
• If the p-value is greater than 0.05, the
  population may be in Hardy-Weinberg equilibrium
  we can not prove that it is not in Hardy-Weinberg
  equilibrium.

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Chi Squared Test

• As an example, lets say that Chi-square analysis
  of some data gives a p-value of 0.17. This means
  that there is a 17 probability that the
  difference between the observed and the expected
  values is due to chance. It also means that there
  is an 83 (100 - 17 83) probability that the
  difference is not due to chance the difference
  is real.

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Chi Squared Test
Observed Expected
(O-E)2
E / 25 20 1.25
/- 5 15 6.66 -/-
8 3 8.33
Total 17.12 X2
Critical Value (from statistics table)
5.9 17.12 is above 5.9 so the ratio is not
accepted.
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Allele Server

• Cold Spring Harbor Lab
• DNA Learning Center

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http// vector.cshl.org
Click on Resources
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Click on Bioservers
Click on Bioservers
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Enter the Allele Server
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Allele Server
Click on Manage Groups
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Select Type of Data
Select Group
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Your Group
Scroll down to Your Group
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Click on Add Group
Click on Add Group
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Fill Out Form
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Click on Edit Group
Click on Edit Group
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Fill out Completely
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Click on Individuals Tab
Click on Individuals Tab
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Add Each Student With Information

• Add as much info as you can
• Genotype ( /, /-, -/- )
• Gender
• Personal Info

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Click on Done
Done
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Select Your Group
Select and then press OK
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Click to Analyze
Then Click Here
Click Here
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Then Look at the Terse and Verbose Tabs
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Extensions

• Is your class in Hardy-Weinberg Equilibrium?
• Compare your group to other existing groups.
• Form an explanation for the origination of Alu
  and how it spread throughout different
  populations.
• Have students do manual calculations first and
  then compare to the computer generated version.

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Plotting Alu PV92 on a World Map

• The Alu element first appeared tens of millions
  of years ago and since that time, it has been
  increasing within our genome at the rate of about
  one copy every 100 years.
• It is difficult to tell how Alu arose. It shows a
  striking similarity to a gene (called 7SL RNA)
  that performs a vital function in our metabolism.
  But Alu , it seems, has no function. It is
  self-serving and, like a parasite, takes
  advantage of us for its own replication without
  providing us any advantage to our own survival.
• Most Alu elements are fixed they are found at
  the same chromosomal site in every person on the
  planet. Fixed Alu elements must have arose very
  early in our evolution, well before Homo sapiens
  appeared. When modern humans did arise some
  200,000 years ago, the vast majority of our Alu
  insertions came to us already intact in our DNA.
• The Alu PV92 insertion, however, is not fixed.
  This insertion may or may not be present on one
  or both of a persons number 16 chromosomes.
  Since not everyone has the Alu PV92 element, it
  must have arisen after the initial human
  population began growing.
• It is a widely held belief that modern humans
  originated in Africa and then disseminated across
  the planet. Did the Alu PV92 insert arise in
  Africa or on some other continent during our
  spread across the globe?
• In the following exercise, you will plot the
  allele frequencies for various populations on a
  world map and make some determination as to where
  this Alu arose and how it might have spread
  across continents.

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.18
.20
.12
.18
.86
.53
.30
.52
.09
.26
.35
.96
.15
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Classroom How To

• We have worksheets for calculating allelic and
  genotypic frequencies for your class
• BABEC has the materials for the world population
  data