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```									Population Genetics Lab 2

BINOMIAL PROBABILITY
&
HARDY-WEINBERG EQUILIBRIUM
Last Week : Sample Point Methods:
Example: Use the Sample Point Method to find the probability of
getting exactly two heads in three tosses of a balanced coin.
1. The sample space of this experiment is:
Outcome   Toss 1   Toss 2   Toss 3   Shorthand   Probabilities
5      Tail     Tail     Head       TTH           1/8
6      Tail     Head     Tail       THT           1/8
7      Head     Tail     Tail       HTT           1/8
8      Tail     Tail     Tail       TTT           1/8

2. Assuming that the coin is fair, each of these 8 outcomes has a probability of 1/8.

3. The probability of getting two heads is the sum of the probabilities of outcomes
2, 3, and 4 (HHT, HTH, and THH), or 1/8 + 1/8 + 1/8 = 3/8 = 0.375.
Sample- point method :

Example: Find the probability of getting exactly 10 heads in 30 tosses of
a balanced coin.

Total # of sample points = 230 =   1,073,741,824
Need a way of accounting for all the possibilities
Example: In drawing 3 M&Ms from an unlimited M&M bowl that is
always 60% red and 40% green, what is the P(2 green)?

If one green M&M is just as good as another…
Binomial Probability Distribution

Where, n = Total # of trials.
y = Total # of successes.
s = probability of getting success in a single trial.
f = probability of getting failure in a single trial (f = 1-s).
Assumptions of Binomial Distribution
1. # of trials are independent, finite, and conducted under
the same conditions.
2. There are only two types of outcome.(Ex. success and
failure).
3. Outcomes are mutually exclusive and independent.
4. Probability of getting a success in a single trial remains
constant throughout all the trials.
5. Probability of getting a failure in a single trial remains
constant throughout all the trials.
6. # of success are finite and a non-negative integer (0,n)
Properties of Binomial Distribution

Mean or expected # of successes in n trials, E(y) = ns

Variance of y,            V(y) = nsf

Standard deviation of y, σ (y) = (nsf)1/2
Example: Find the probability of getting exactly 10 heads in 30
tosses of a balanced coin.
Solution:

We know,       n = 30
y = 10
s = 0.5
f = 0.5
Example: Find the expected # of heads in 30
tosses of a balanced coin. Also calculate variance.

Solution:

E(Y) = ns = 30*0.5 = 15

V(Y) = nsf = 30*0.5*0.5 = 7.5
Problem 1 (10 minutes)(2 points)
An allozyme locus has three alleles, A1,A2, and A3 with
frequencies 0.947, 0.033, and 0.020, respectively. If we sample
30 diploid individuals, what is the probability of:

•Not finding any copies of A2?

•Finding at least one copy of A2?

•GRADUATE STUDENTS ONLY: Finding fewer than 2 copies
of A2?
Example: How many diploid individuals should be sampled to detect at
least one copy of allele A2 from Problem 1 with probability of at least
0.95?
Solutions:

Thus, to detect at least one copy of allele A2 with probability of 0.95,
one would need to sample at least 90 alleles (i.e., at least 45 diploid
individuals).
Problem 2 (15 minutes)(2 points)
The frequency of red-green color-blindness is 0.07 for men and 0.005
for women. Cities are considering implementing new traffic lights that
can be distinguished by people with this disability. However, the
investment is large, so President Obama has asked you to calculate
the minimum size of a city that would have a probability of 0.9 of
having at least 10 men who are red-green color-blind. Assume equal
population sizes for men and women.

Full credit (2 pts): Solve for at least 1 man
+1 Bonus: Set up the equation for at least 10 men
+3 Bonus: Solve for at least 10 men
Estimation of allele frequency for Co-dominant locus

Where, p = Frequency of allele A1
q = Frequency of Allele A2
N11 = # of individuals with genotype A1A1
N12 = # of individuals with genotype A1A2
N22 = # of individuals with genotype A2A2
N = total # of diploid individuals =N11+N12+N22
Estimation of Standard Error

Where, p = Frequency of allele A1
q = Frequency of Allele A2
SEp = Standard error for frequency of allele A1
SEq = Standard error for frequency of allele A2
N = total # of diploid individuals =N11+N12+N22
Standard Deviation v. Standard Error

We expect ~68% of the data to fall within 1
standard deviation of the mean.

Standard error is the standard deviation of the
mean. We expect sample means to fall within 1
standard error of the true mean ~68% of the time.
Example: What are the allele frequencies of alleles A1 and A2,
if the following genotypes have been observed in a sample of
50 diploid individuals?
Genotype          Count
A1A1              17
A1A2              23
A2A2              10

Solution:   N11 = 17, N12 = 23, and N22 = 10
q = 1 – p = 0.43
Problem 3 (10 minutes) (2 pts)
Estimate the allele frequencies (include their
respective standard errors) for alleles A1, A2, and
A3 if the following genotypes have been observed
in a sample of 200 individuals         Genotype Count
A1A1    19
A2A2    17
A3A3    14
A1A2    52
A1A3    57
A2A3    41
Problem 4 (Time 10 min.)(2 pts)
Tay Sachs disease is an autosomal recessive genetic
disorder causing the death of nerve cells in the brain due
to the steady accumulation of gangliosides. Extensive
genotyping has determined that approximately 1 in 30 of
the 5 million Ashkenazi Jews within the United States is a
carrier.
a) Assuming HWE and Mendelian inheritance of the
disease, what is the frequency of the recessive allele
in this population?
b) What is the SE of this estimate? (Assume 1,000
people were sampled)
c) How many affected children would you expect to be
born in this population?
d) What are the assumptions of these estimates?
Hypothesis Testing
Hypothesis: Tentative statement for a scientific
problem, that can be tested by further investigations.

1.Null Hypothesis(Ho): There is no significant difference in
observed and expected values.

2.Alternate Hypothesis(H1): There is a significant
difference in observed and expected values.

Example:

Ho = Fertilized and unfertilized crops have equal yields

H1 = Fertilized and unfertilized crops do not have equal yields
Remember: In final conclusion after the
experiment ,we either –

"Reject H0 in favor of H1"

Or

“Fail to reject H0”,
Type I error: Error due to rejection of a null hypothesis, when it
is actually true (False positive).

Level of significance(LOS) (α) : Maximum probability
allowed for committing “type I error”.

At 5 % LOS (α=0.05), we accept that if we were to
repeat the experiment many times, we would falsely
reject the null hypothesis 5% of the time.
P- value:

   Probability of committing type I error

   If P-value is smaller than a particular
value of α, then result is significant at
that level of significance
Testing departure from HWE
In a randomly mating population, allele and
genotype frequencies remain constant from
generation to generation.

Ho= There is no significant difference between observed and
expected genotype frequencies (i.e. Population is in HWE)

H1= There is a significant difference between observed and
expected genotype frequencies (i.e. Population is not in
HWE)
HWE Assumptions

1. Random mating
2. No selection
a. Equal numbers of offspring per parent
b. All progeny equally fit
3. No mutation
4. Single, very large population
5. No migration
χ2   - test

Where,
Example: A population of Mountain Laurel at Cooper’s
Rock State Forest has the following observed genotype
counts:
Genotype        Observed number
A1A1            5000
A1A2            3000
A2A2            2000

Is this population in Hardy-Weinberg equilibrium ?
Genotype           Expected frequency     Expected number
under HWE              under HWE

A1A1               p2 = 0.652 = 0.4225    0.4225  10000 = 4225

A1A2               2pq = 0.455            0.455  10000 = 4550

A2A2               q2 = 0.1225            0.1225  10000 = 1225

Genotype Obs. #(O) Exp. #(E)      (O-E)   (O-E)^2 (O-E)^2/E
A1A1    5000       4225          775    600625 142.1598
A1A2    3000       4550         -1550   2402500 528.022
A2A2    2000       1225          775    600625 490.3061
χ2      1160.488
The critical value (Table value) of χ2 at 1 df and
at α=0.05 is approx. 3.84.

Conclusion: Because the calculated value of χ2
(1160.49) is greater than the critical value (3.84),
we reject the null hypothesis and accept the
alternative (Not in HWE).
Problem 5 (Time 10 min) (2 pts)
Based on the observed genotype counts in problem
3, test whether the population that had been sampled
is in HWE. What are some possible explanations for
the observed results?

Genotype Count
A1A1     19
A2A2     17
A3A3     14
A1A2     52
A1A3     57
A2A3     41

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