CINQA Workshop Probability Math 105 Silvia Heubach Department of Mathematics, CSULA Thursday, September 6, 2012

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1 CINQA Workshop Probability Math 105 Silvia Heubach Department of Mathematics, CSULA Thursday, September 6, 2012 Silvia Heubach/CINQA 2012

2 Workshop Objectives To familiarize biology faculty with one of the new topics taught in Math 105 To lay a foundation for connections between the MATH 105 course and BIOL 100B, BIOL 300 (Biometrics) and various genetics courses (BIOL 340, 412, 415, 416) To facilitate discussion on examples that connect the courses To start a discussion on how to incorporate more quantitative aspects into biology courses

3 Introduction to Probabilistic models A stochastic model describes a (biological) process that includes unpredictable or chance events. Such a model allows us to give a more realistic description of the underlying process. Generally, the answers we get from a stochastic model are not as clear-cut as those from a deterministic model. Mathematical tools from probability theory will be used to analyze the stochastic model and to determine its behavior.

4 Example 1: Stochastic Production The deterministic model of population growth with fixed per capita production r is b t+1 = r b t or b t = r t b 0 The solution is an exponential function. Example: b 0 =10, r =

5 Example 1: Stochastic Production The stochastic model of population growth has a per capita production that varies with time and r t the model is given by b t+1 = r t b t We cannot derive an explicit solution that gives an exact answer. Example: b 0 =10, r t varies randomly from 1.0 to 1.2 (has average 1.1). Simulations have roughly an exponential shape

6 Example 2: Diffusion Random movement of enzymes or toxins in and out of a cell. We cannot determine whether a molecule is inside or outside at any given time, only the likelihood or probability that it is. We can describe such a diffusion by a Markov chain model. Example: In each one minute time interval, a certain molecule leaves a cell with probability 0.2, and remains inside with probability 0.8. If the molecule is outside the cell, it enters the cell with probability 0.1, and remains outside with probability 0.9.

7 Example 2: Diffusion Visualization: States (inside/outside) are drawn as circles Allowed transitions from one state to another are drawn as arrows Respective probabilities listed on arrows Typical question: What is the probability that a molecule that was inside the cell at time t=0 is still inside the cell at say time t= 5?

8 Worksheet Give at least one example of a biological process where randomness plays an important role. List as many as you can together with the biology course in which the relevant topic is taught.

9 Probability Theory Experiment = chance process with well-defined outcomes Sample space S = set of all possible outcomes Simple event = outcome Set = list of elements written with curly braces Event = a set of outcomes = subset of the sample space. An event occurs if one of its outcomes occurs. Venn diagram = visualization of sample space and events.

10 Sample Space and Events Experiment = Roll of a single die Sample space = Event A = rolling an even number Venn Diagram

11 Operations on Sets: Intersection The intersection of two sets A and B consists of the elements that are common to both. Notation: A B English translation: and. If the intersection is empty (no elements in common), then A and B are mutually exclusive or disjoint.

12 Operations on Sets: Union The union of two sets A and B consists of the elements that are in either one of them or in both Notation: A B English translation: or.

13 Operations on Sets: Complement The complement of a set A consists of the elements that are not in A and is denoted by A c English translation: not. A c A

14 Set operations S = {1, 2,., 19, 20} A = {1, 2,, 9, 10}; B = {1, 2, 3, 5, 7, 11, 13, 17, 19} AÇ B = AÈ B = A c = B c =

15 Assigning Probabilities P(A) = probability that event A occurs Rules P(S) = 1; for any event A, 0 P(A) 1 If A and B are disjoint, then P(A B) = P(A)+P(B) If A and B are not disjoint, then P(A B) = P(A)+P(B)-P(A B) P(A c ) = 1-P(A) If the sample space S has finitely many elements, and all outcomes are equally likely, then P(A) = A S

16 Example 3: Two Dice Experiment: Roll two dice

17 Example 3: Two Dice A= sum is seven; B = at least one 6 P(A) = P(B) = P(A B) = P(A B) =

18 Example 3: Two Dice P(no 6) = Note: If event is described by at least, at most, more than, less than, then consider the probability of the complement.

19 Example 4 At a particular school with 200 male students, 58 play football, 40 play basketball, and 8 pay both. What is the probability that a randomly selected male student plays neither sport? (Hint: Draw a Venn diagram).

20 Worksheet For the experiment of rolling two dice, compute the probabilities of the following events: A = an even sum

21 Worksheet For the experiment of rolling two dice, compute the probabilities of the following events: B = a product less than 10

22 Worksheet For the experiment of rolling two dice, compute the probabilities of the following events: C = largest number rolled is a 4

23 Worksheet For the experiment of tossing three coins, Write down the sample space Compute the probability of getting at least one head Compute the probability of having more heads than tails

24 Worksheet The probability that a tourist goes to an amusement park is 0.47, and the probability she goes to the water park is If the probability that she goes to either the water park or the amusement park is 0.95, what is the probability that she visits both of the parks on her vacation?

25 Stochastic Models of Genetics Inheritance has a random component, so is best described by a stochastic model. Diploid plant has two copies of each gene (one from the ovule and one from the pollen for self-pollinating plants, or in general, one copy of the gene from each parent). Each gene has different variants or alleles, which may result in different observable phenotypic traits or phenotypes such as height or eye color. Plants with two different alleles of a gene are called heterozygous, and those with two copies of the same allele are called homozygous.

26 The Genetics of Inbreeding Simplest form is selfing (or self-pollination). Selfing implies pollen and ovule have the same genotype. Use Punnett Square to determine the genotypes of the offspring of a heterozygous parent. Random mating, so all offspring are equally likely Genotypes: P(AA I ) = ¼ P(Aa I ) = ½ P(aa I ) = ¼

27 The Genetics of Inbreeding Genotypes of the offspring of a homozygous parent? Parent is AA genotypes of offspring? Parent is aa genotypes of offspring?

28 Dominant Genes Dominant allele heterozygous and homozygous of the dominant type have the same phenotype. Dominant allele denoted with capital letter. Suppose plant height is governed by a single dominant allele B, and that there are two phenotypes, short and tall. Genotypes BB and Bb produce tall plants, and genotype bb produces short plants.

29 Worksheet 3 Determine the genotypes and their probabilities of the offspring of a heterozygous father and a homozygous dominant (BB) mother. Genotypes:

30 Worksheet 3 Determine the genotypes and their probabilities of the offspring of two homozygous parents with different genotypes. Genotypes:

31 Worksheet 3 Determine the genotypes and their probabilities of the offspring of a heterozygous mother and a homozygous recessive (bb) father. Genotypes:

32 Conditional Probability Conditional probability allows us to adjust the probability of an event A based on knowledge that an event B has occurred. The probability of event A conditional on event B (with P(B) 0) is defined as The vertical bar is read as given. If all outcomes are equally likely, we can compute the conditional probability as P(A B ) = A B / B.

33 Example 5 Suppose plant height is governed by a single dominant allele B, and that there are two phenotypes, short and tall. Two heterozygous plants (genotype Bb) are crossed. What is the probability that a tall offspring has genotype BB?

34 Example 5 P(BB) = P(bb) = ¼ and P(Bb) = ½ (from selfing example) Wanted: P(BB T), where T = tall plant. Tall plant has to have genotype BB or Bb P(BB T) =

35 P(BB T) = 1 3 Example 5

36 Conditional Probability Two special cases: If A and B are disjoint, then P(A B ) = 0. If A contains B, then P(A B ) = 1. B A Definition of conditional probability yields a formula for the probability of intersections: P(AÇ B) = P(B)P(A B) = P(A)P(B A) A B means that both A and B occurred. This requires that B has occurred and that A occurred conditional on B. Likewise for the second version.

37 Independence If knowledge that event B has occurred does not change the probability for the occurrence of event A, then we say that A is independent of B. Formally: P(A B )= P(A). Do not confuse being independent and being mutually exclusive! Two events A and B that are mutually exclusive are always dependent. Multiplication Rule for independent events: P(A B)=P(A)P(B). Application of independence in genetics probabilities for alleles from each parent are multiplied.

38 Comparison For unions, events being mutually exclusive makes life simple P(A B) = P(A) + P(B) For intersections, events being independent makes life simple P(A B) = P(A)P(B).

39 The Genetics of Inbreeding - revisited P(Aa I ) = ½

40 The Genetics of Inbreeding - revisited P(AA I ) = ¼ Using independence to compute probabilities or proportions of offspring genotypes is particularly useful when dealing with multiple genes at the same time.

41 Worksheet 4 An ecologist is looking for the effects of eagle predation on the behavior of jack rabbits. She sees an eagle with probability 0.2 during an hour of observation, a jack rabbit with probability 0.5, and both with probability Draw a Venn diagram to illustrate the situation.

42 Worksheet 4 She sees an eagle with probability 0.2 during an hour of observation, a jack rabbit with probability 0.5, and both with probability P(R E) = P(E R) =

Law of Total Probability Sets E 1, E 2,, E n form a partition of the sample space S if they are mutually exclusive and collectively exhaustive For any event A we have A = (A E 1 ) (A E 2 ) (A E n )

43 Law of Total Probability Sets E 1, E 2,, E n form a partition of the sample space S if they are mutually exclusive and collectively exhaustive For any event A we have A = (A E 1 ) (A E 2 ) (A E n ) Law of Total Probability P(A) = P(A E 1 ) + P(A E 2 ) + + P(A E n ) = P(E 1 ) P(A E 1 ) + P(E 2 ) P(A E 2 ) + + P(E n )P(A E n ) The sets E i cover different cases. Often, there are just two cases.

Example 6 Selfing 2 nd generation Starting with a selfing heterozygous plant Aa (generation 0), we have offspring with genotype AA I, aa I, and Aa I, with P(AA I ) = ¼, P(aa I ) = ¼, and

44 Example 6 Selfing 2 nd generation Starting with a selfing heterozygous plant Aa (generation 0), we have offspring with genotype AA I, aa I, and Aa I, with P(AA I ) = ¼, P(aa I ) = ¼, and P(Aa I ) = ½. Probabilities give proportions we expect to see in a large number of plants. Question: What genotypes do we have in the second generation and what are the respective proportions?

45 Example 6 Aa I produces all three genotypes with proportions P(AA II Aa I ) = ¼ = P(aa II Aa I ), P(Aa II Aa I ) = ½. AA I produces only AA II P(AA II AA I ) = 1, P(aa II AA I )= P(Aa II AA I ) = 0. Similar for aa II P(Aa II ) =?

46 Example 6 P(Aa II ) =? Three cases for parental genotype - use Law of Total Probability. P(Aa II ) = P(Aa I ) P(Aa II Aa I ) + P(AA I ) P(Aa II AA I ) + P(aa I ) P(Aa II aa I ) = ½ ½ + ¼ 0 + ¼ 0 = ¼ = 0.25

47 Example 6 P(aa II ) =? Three cases for parental genotype - use Law of Total Probability. P(aa II ) = P(Aa I ) P(aa II Aa I ) + P(AA I ) P(aa II AA I ) + P(aa I ) P(aa II aa I ) = ¼ ½ + ¼ 0 + ¼ 1 = = 0.375

48 Selfing n th generation P(Aa n ) = 0.5 n P(Aa n ) + P(AA n ) + P(aa n ) =1 P(AA n ) = P(aa n ) Put together: P(AA n ) = P(aa n ) =(1-0.5 n )/2 In the long run, heterozygous plants disappear, and the homozygous plants occur in equal proportions

49 Example 7- Additive Genes Additive effect of genes the heterozygous plant has a phenotype that is a mixture of the homozygous phenotypes. Same probabilities for genotypes in generation two, but we have three phenotypes. P(tall) = 0.25, P(medium) = 0.5 P(short) = In the additive phenotype example, there is a most common type, and it is in the middle. A similar phenomenon occurs when a trait (like height) is determined by many genes.

50 Example 7 Example: Height depends on 10 genes, Bi = tall allele of gene i, bi = short allele of gene i. A short plant with only short alleles has height 40 cm, and each tall allele adds 1 cm Crossing plants with heights 40 cm and 60 cm results in offspring of height 50 cm (10 short and 10 tall alleles). Offspring in the second generation can range in height from 40 cm to 60 cm Most plants will have heights close to 50 cm

51 Example 8 A lab is attempting to stain many cells. Young cells stain properly 90% of the time and old cells stain properly 70% of the time. If 30% of the cells are young, what is the probability that a cell that stains properly is a young cell?

52 Example 8 What is the probability that a cell that stains properly is a young cell? P(Y)= 0.3 P(O) = 0.7 P(S Y) = 0.9 P(S O)=0.7 P(Y S) = P(S Y)P(Y)/P(S) P(S) =

53 Bayes Theorem Bayes Theorem is often used when we want to compute a conditional probability where the given information consists of a conditional probability with opposite order. Bayes Theorem: For any events A and B where P(A) 0, P(B A) = P(B)P(A B) P(A) where P(A) is often computed using the total law of probability and B is one of the cases in the law of total probability.

54 Example 9 Rare Disease Suppose a rare disease affects only 1% of the population. A diagnostic test correctly diagnoses the disease in 100% of the cases, but produces false positives in 5% of the cases. You have received a positive test result. How likely are you to have the disease? D = has disease; N = does not have disease; + = positive test result Want: P(D +)

55 Example 9 Rare Disease Suppose a rare disease affects only 1% of the population. A diagnostic test correctly diagnoses the disease in 100% of the cases, but produces false positives in 5% of the cases. Want: P(D +) - compute using a population of 10,000

56 Example 9 Rare Disease Suppose a rare disease affects only 1% of the population. A diagnostic test correctly diagnoses the disease in 100% of the cases, but produces false positives in 5% of the cases. D = has disease; N = does not have disease; + = positive test result P(D +) =?

57 Worksheet New cells stain properly with probability 0.95, 1-day old cells stain properly with probability 0.9, 2-day old cells stain properly with probability 0.8, and 3-day old cells stain properly with probability 0.5. Suppose 40% of the cells are new, 30% are 1 day old, and 20% are 2 days old.

58 Worksheet Let D denote the event of an individual having the disease, N the event of not having the disease, and + the event of a positive test result. Compute P(D +). P(D)=0.2, P(+ D) = 1, and P(+ N)=0.05.

59 Worksheet b) Investigate the effects that the prevalence of the disease has on the conditional probability of P(D +), as well as the effect of false positives. Given potential negative side effects of the test, should everybody be tested when a disease is rare? P(D) P(+ D) P(+ N) P(D +)

60 Worksheet b) Investigate the effects that the prevalence of the disease has on the conditional probability of P(D +), as well as the effect of false positives. What is the effect of false positives? P(D) P(+ D) P(+ N) P(D +)

61 Worksheet Consider a dominant gene where plants with genotype BB and Bb are tall, and those with genotype bb are short. Find the probability that a plant has genotype Bb when it results from the following crosses. Two offspring from the cross between a BB and a Bb plant are crossed with each other.

62 Worksheet Consider a dominant gene where plants with genotype BB and Bb are tall, and those with genotype bb are short. Find the probability that a plant has genotype Bb when it results from the following crosses. Two tall offspring from a cross between a Bb plant and a Bb plant are crossed with each other.

63 Markov Chains A (discrete time) Markov chain is a stochastic process where the probability to be in a certain state at a given time only depends on the state at the previous time.

64 Example 10: Diffusion p t = P(molecule inside cell at time t) p t+1 =

65 Example 10: Diffusion DDS: p t+1 = 0.7 p t If molecule starts inside the cell (p 0 =1), then p t = ⅓ + ⅔ (0.7) t Equilibrium: p* = 0.7 p* + 0.1, so p* = ⅓ Solution approaches the equilibrium value in the long run

66 Example 10: Diffusion Interpretation of equilibrium: The probability of seeing the molecule inside after long period of time is 1/3 If many molecules are observed, then about 1/3 of them would be inside the cell after a long period of time

Probability- describes the pattern of chance outcomes

Chapter 6 Probability the study of randomness Probability- describes the pattern of chance outcomes Chance behavior is unpredictable in the short run, but has a regular and predictable pattern in the long