1. HARDY-WEINBERG EQUILIBRIUM
Population Genetics Lab 2
BINOMIAL PROBABILITY
&
HARDY-WEINBERG EQUILIBRIUM

2. 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:
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 =
Outcome
Toss 1
Toss 2
Toss 3
Shorthand
Probabilities 1
Head
HHH
1/8 2
Tail
HHT 3
HTH 4
THH 5
TTH 6
THT 7
HTT 8
TTT

3. Total # of sample points = 230 = 1,073,741,824
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

4. 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…
Great example!

5. 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).

6. Assumptions of Binomial Distribution
Number of trials are independent, finite, and conducted under the same conditions.
There are only two types of outcome.(Ex. success and failure).
Outcomes are mutually exclusive and independent.
Probability of getting a success in a single trial remains constant throughout all the trials.
Probability of getting a failure in a single trial remains constant throughout all the trials.
Number of success are finite and a non-negative integer (0,n)

7. Properties of Binomial Distribution
Mean or expected # of successes in n trials:
Variance of y:
Standard deviation of y, σ (y) = (nsf)1/2
E(y) = ns
V(y) = nsf
SD(y) = 𝑽(𝒚)

8. 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

9. Example: Find the expected # of heads in 30 tosses of a balanced coin
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

10. Problem 1 (10 minutes)
Problem 1: A nuclear allozyme locus has three alleles, A1 and A2, and A3, with frequencies 0.847, 0.133, and 0.020, respectively. If we sample 30 diploid individuals, what is the probability of:
a) Not finding any copies of A2?
b) Finding at least one copy of A2?
c) GRADUATE STUDENTS ONLY: Finding fewer than 2 copies of A2?

11. 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:
Emphasize flipping of the inequality when dividing by negative number.
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).

12. Problem 2 (15 minutes)
Problem 2: The frequency of red-green color-blindness is 0.07 for men and for women. You are designing a survey to determine the effect of color blindness on educational success. How many males and females would you have to sample to ensure that the probability including at least one color blind individual of each sex would be 0.90 or greater?

13. 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

14. 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

15. Standard Deviation v. Standard Error
Standard deviation: variability of individuals around the sample mean
Standard error: variability of sample means around the population mean
We expect ~68% of the data to fall within 1 standard deviation of the mean.

16. A1 A11 A12 … A1j A2 A21 A22 … A2j … Ai Ai1 Ai2 …Aij SUM 1 𝑗 𝐴1𝑗
Genotype1
Genotype2
…Genotype j
SUM A1
A11
A12
… A1j
1 𝑗 𝐴1𝑗 A2
A21
A22
… A2j
1 𝑗 𝐴2𝑗 … Ai
Ai1
Ai2
…Aij
1 𝑗 𝐴𝑖𝑗
Aij ≥𝟎, 𝒊,𝒋=𝟏,𝟐,𝟑,…

17. 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
If p = what is q?

18. q = 1 – p = 0.43
What do you notice about the standard error of q?

19. Problem 3 (10 minutes)
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

20. Estimation of allele frequency for dominant locus
For dominant loci, we only know the genotype of homozygous recessive individuals (in absence of sequence data).
(Note: this is an estimate)
Where, q = Frequency of Allele A2
N22 = # of individuals with genotype A2A2
N = total # of diploid individuals = N11+N12+N22

21. Problem 4 (15 minutes)
Go to the “Genetics Home Reference” website ( and use the search feature to find a condition caused by a dominant allele in humans. On the main description page, find the frequency of the condition in human populations.
Assuming HWE and Mendelian inheritance of the disease, what is the frequency of the recessive allele in this population?
What is the standard error of this estimate?
How many affected children would you expect in the next generation? (Global/U.S/etc)
What are the assumptions of these estimates?
Eg.

22. Hypothesis Testing
Hypothesis: Tentative statement for a scientific problem, that can be tested by further investigations.
Null Hypothesis(H0): There is no significant difference in observed and expected values.
Alternate Hypothesis(H1): There is a significant difference in observed and expected values.
Example:
H0 = Fertilized and unfertilized crops have equal yields
H1 = Fertilized and unfertilized crops do not have equal yields

23. Remember: In final conclusion after the experiment ,we either –
"Reject H0 in favor of H1" Or
“Fail to reject H0”,

24. 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.
Ho is TRUE
Ho is FALSE
Accept Ho
Fail to reject Ho β
1-α
Type II Error α
1-β
Reject Ho
Type I Error

25. 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

26. Testing departure from HWE
In a randomly mating population, allele and genotype frequencies remain constant from generation to generation.
H0= 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)
“disturbing factors” is a bit of strange phrasing. I don’t think that part is needed. You just list the assumptions of the model separately.

27. HWE Assumptions Random mating No selection
Equal numbers of offspring per parent
All progeny equally fit
No mutation
Single, very large population
No migration

28. χ2 - test
Where,

29. Is this population in Hardy-Weinberg equilibrium ?
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 ?

30. χ2 Genotype Obs. #(O) Exp. #(E) (O-E) (O-E)^2 (O-E)^2/E A1A1 5000 4225
Expected frequency
under HWE
Expected number
A1A1
p2 = =
 = 4225
A1A2
2pq = 0.455
0.455  = 4550
A2A2
q2 =
 = 1225
Genotype
Obs. #(O)
Exp. #(E)
(O-E)
(O-E)^2
(O-E)^2/E
A1A1
5000
4225
775
600625
A1A2
3000
4550
-1550
A2A2
2000
1225 χ2

31. 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 ( ) is greater than the critical value (3.84), we reject the null hypothesis and accept the alternative (Not in HWE).
Explain how to determine df

32. Problem 5 (10 minute)
Based on the observed genotype counts in Problem 3, test whether the population that had been sampled is in HWE.
Critical Chi square values are given in the table to the right. Think carefully about which one you should use (Hint: How many parameters are estimated from the data when the allele frequencies of 3 alleles are estimated?).
What are some possible explanations for the observed results?
d.f.
Critical value of χ2 1
3.8415 2
5.9915 3
7.8147 4
9.4877 5 6