Population Genetics LabLab Instructor: Ran Zhou PhD student

Department of Biology

Office: Life Sciences Building, Room 5206

Office Hours: T – 2:30 PM – 3:30 PMW – 3:30 PM – 4:30 PMor by appointment

Email ID: razhou@mix.wvu.edu

Introductions

Please indicate the following on the provided sheet:

NameMajorCareer goal/goal for class.

Be honest!

Probability and Population Genetics

Population genetics is a study of probabilitySampling alleles from population each

generation

A

AA

A

A

aa a

a

A

Probability

Frequentist Approach• Determine how often you

expect event A to occur given a LONG series of trials

Bayesian Approach• Determine the plausibility

of event A given what you already know (prior).

ProbabilityMeasure of chance.

P(E) = # of favorable outcome / Total # of possible outcome

It lies between 0 (impossible event) and 1 (certain event). Ex. What is the probability of getting a head in one toss of a balanced coin.

Total possible outcomes = 2 (H, T) # of Heads = 1 (H) P(H) = 1 / 2 = 0.5 = 50 %

Sample- point method :1.Define sample space (S): Collection of all possible outcomes of a

random expt.

Ex. S (Coin tossed twice)

2. Assign probabilities to all sample points

Ex. P(HH) = ¼ ; P(HT) = ¼ ; P(TH) = ¼ ; P(TT) = ¼

Outcome 1 2 3 4

First Toss H H T T

Second Toss H T H T

Shorthand HH HT TH TT

Sample- point method :3.Determine event of interest and add their probabilities.

Ex. Find the probability of getting exactly one head in two tosses of a balanced coin.

i. S (Coin tossed twice) { HH, HT, TH, TT}. ii. P(HT) = ¼ ; P(TH) = ¼

iii. P(HT) + P (TH) = ¼ + ¼ = 2/4 = ½ .

If all sample points have equal probabilities then –

P(A) = na / N

where, na = # of points constituting event A and N= Total # of sample points.

Sample- point method :

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 = 0.375.

In above example, find the probability of getting at least two heads.Solution: 1/8 + 1/8 + 1/8+ 1/8 = 1/2

Outcome Toss 1 Toss 2 Toss 3 Shorthand Probabilities1 Head Head Head HHH 1/82 Head Head Tail HHT 1/83 Head Tail Head HTH 1/84 Tail Head Head THH 1/85 Tail Tail Head TTH 1/86 Tail Head Tail THT 1/87 Head Tail Tail HTT 1/88 Tail Tail Tail TTT 1/8

Problem 1: The game of “craps” consists of rolling a pair of balanced dice (i.e., for each die getting 1, 2, 3, 4, 5, and 6 all have equal probabilities) and adding up the resulting numbers. A roll of “2” is commonly called “snake eyes” and causes an instant loss when rolled in the opening round. Using the Sample-Point Method, find the exact probabilities of a roll of snake eyes.(Time : 10 minutes)

Probability

For large sample space: Use fundamental counting methods.

1. mn rule : If there are “m” elements from one group and “n” elements from another group, then we can have “mn” possible pairs, with one element from each group.

mn= 6*6= 36 .

Second die1 2 3 4 5 6

First die

1

2

3

4

5

6

For large sample space : Use fundamental counting methods.

2. Permutation: Ordered set of “r” elements, chosen without replacement, from “n” available elements.

Remember: n! = n*(n-1)*(n-2)*…………*2*10! = 1 (By definition)

Example: How many trinucleotide sequences can be formed without repeating a nucleotide, where ATC is different from CAT?

Solution: n = 4 (A, T, C and G) r = 3 (trinucleotide sequence)

= 24.

)!(

!

rn

nPnr

)!34(

!4

n

rP

For large sample space : Use fundamental counting principle.

3. Combination: Unordered set of “r” elements, chosen without replacement, from “n” available elements.

Example: How many trinucleotide sequences can be formed without repeating a nucleotide , where ATC is the same as CAT.

Solution: n = 4 ( A, T, C and G) r = 3

= 4

)!(!

!

rnr

nC nr

)!34(!3

!4

n

rC

For large sample space : Use fundamental counting principle.

Problem 2: There are 36 computer workstations in this lab. If there are 18 students in the class, how many distinct ways could students be arranged, with one student per workstation? (10 minutes)

Problem 3: A local fraternity is organizing a raffle in which 30 tickets are to be sold – one per customer. (10 minutes)

(a) What is the total number of distinct ways in which winners can be chosen if prizes are awarded as follows?

(b) If holders of the first four tickets drawn each receive a $30 prize?

Order of Drawing Prize

First $100

Second $50

Third $25

Fourth $10

Laws of Probability1. Additive law of probability:

B)P(A - P(B) P(A) B)(A P

P(B) P(A) B)P(A then 0, B)P(A events, exclusivemutually For

uslysimultaneo B andA event of occurrence ofy ProbabilitB)P(A

Bor A event of occurrence ofy ProbabilitB)P(AWhere,

B)(A Pfor diagramVenn B)(A Pfor diagramVenn

A B

A and B are Mutually Exclusive

Laws of Probability1. Additive law of probability:

Example: From a pack of 52 cards, one card is drawn at random. Find the probability that the card is “Heart” or “Ace”.

Four suits are : Spades, Diamonds, Clubs and Hearts. Each suit has 13 cards: Ace,2,3,4,5,6,7,8,9,10,Jack, Queen and King.There are four of each type, like 4 Aces,4 Jacks, 4 Queens, 4 Kings etc.

Solution:

52

16

52

1

52

4

52

13 A)P(H

52

1 A)P(H ;

52

4 P(A) ;

52

13 P(H)

A)P(H - P(A) P(H) A)(H P

Laws of Probability2. Multiplicative law of probability:

P(B) P(A) B)(A P (If A and B are independent events)

B)|P(A P(B) A)|P(B P(A) B)(A P (If A and B are dependent events)

Example: A pond consists of 50 salmon and 25 trout. Two fish are drawn one by one. Find the probability that both fish are Salmon. a.) with replacement andb.) without replacement

111

49

222

98

74

49

75

50

A)|P(B P(A) B)P(A

:treplacemen Without :(b) Case9

4

75

50

75

50

P(B) P(A) B)P(A

:treplacemen With :(a) Case

Salmon. isdrawn fish second that Prob. : P(B)

Salmon. isdrawn fish first that Prob. :P(A) Let,

:Solution

Problem 4. An inexperienced spelunker is preparing for the exploration of a big cave in a rural area of Mexico. He is planning to use two independent light sources and from reading their technical specifications, he has concluded that each source is expected to malfunction with probability of 0.01. What is the probability that:

a) At least one of his light sources malfunctions?b) Neither of his light sources malfunctions?(Time : 15 minutes)

Problem 5. GRADUATE STUDENTS ONLY: Search the literature for an example of an application of basic probability theory to a problem in genetics or genomics. Describe the hypothesis being tested, the results of the test, and the interpretation. Was this a correct implementation of the method? Two points of extra credit will be awarded if you uncover an error in calculation and/or interpretation that was published in a peer-reviewed journal. Be sure to send the original manuscript to Rose when you submit your report.

(Time: as long as it takes)