Quantitative Biology Lecture 1 (Introduction + Probability)

Transcription

1 21 st Sep 2015 Quantitative Biology Lecture 1 (Introduction + Probability) Gurinder Singh Mickey Atwal Center for Quantitative Biology

2 Why Quantitative Biology? (1. Models) Galileo Galilei The Book of Nature is written in the language of mathematics. Charles Darwin "I have deeply regretted that I did not proceed far enough at least to understand something of the great leading principles of mathematics; for men thus endowed seem to have an extra sense.

3 Why Quantitative Biology? (2. Data Analysis) Explosion of biological data by recent developments in high throughput biotechnology. Moore s Law: doubling of transistors on a chip every 18 months.

4 Quantitative Biology at CSHL Systems Neuroscience Developmental Biology QB Genetics and Genomics

5 Why Statistics? Molecular biology is noisy Biological molecules diffuse Small number fluctuations Genetics is stochastic Genetic variation mainly due to random events (drift) Large data analysis Experimental measurements are noisy Require probabilistic models

6 Key Numbers in Biology See handout Database of useful biological numbers ( Quantified biological properties can Test our understanding of the biology Aid in experimental design

7 Key Numbers Example 1 (Class Exercise) Q: Is there enough time to replicate the E. coli genome? Bionumbers

8 Key Numbers Example 2 (Class Exercise) Q: How far can macromolecules move by diffusion? It takes about 10 seconds on average for a protein to traverse a HeLa cell. An axon 1 mm long is about 100 times longer than a HeLa cell, and as the diffusion time scales as the square of the distance it would take 10 5 seconds or 2 days for a molecule to travel this distance by diffusion. This demonstrates the necessity of mechanisms other than diffusion for moving molecules long distances. A molecular motor moving at a rate of 1 µm/s will take a reasonable time (15 min) to traverse an axon 1 mm long.

9 Classic Examples of Quantitative Reasoning in Biology Two-hit Tumor Suppressor Hypothesis Tumor Subtyping from Expression Profiles Delbruck-Luria Experiment (next lecture)

10 Case study 1 Tumor suppressor gene hypothesis

11 Case Study 1 Incidence of childhood retinoblastoma Retinoblastoma : childhood form of retinal cancer Alfred Knudson (geneticist) studied 48 patients and tabulated age of diagnoses Tumor occurred either in one eye (unilateral) or both eyes (bilateral) Knudson 1971

12 Case Study 1 Incidence of childhood retinoblastoma Probability distribution of retinoblastoma cases by type and laterality Bilateral Unilateral Total Hereditary Nonhereditary Total Knudson observed that bilateral hereditary cases of retinoblastoma occurred at a younger age and showed a curve (previous slide) consistent with a single-mutation process.

13 Case Study 1 Tumour Suppressor Genes Statistical analyses by Knudson (1971) lead to the concept of tumor suppressor genes and the two-hit hypothesis The gene s normal function is to regulate cell division. Both alleles need to be mutated or removed in order to lose the gene activity. The first mutation may be inherited or somatic. The second mutation will often be a gross event leading to loss of heterozygosity in the surrounding area.

14 Case Study 1 Knudsen s two hit hypothesis

15 Case Study 2 Cancer subtyping by genome-wide expression analysis

16 Case Study 2 Cancer Classification Leukemia: AML, ALL Unusual morphology of lymphoblasts in ALL

17 Case Study 2 Microarray Clustering Clustering of genomewide expression date revealed a new subtype of ALL with poor prognosis from conventional therapy Armstrong et al, Nature Genetics ALL and MLL are indistinguishable from a pathologist s perspective

18 Case Study 2 MLL translocations specify a distinct gene expression profile that distinguishes a unique leukemia a, Principal component analysis (PCA) plot of ALL (red), MLL (blue) and AML (yellow) carried out using 8,700 genes that passed filtering. b, PCA plot comparing ALL (red), MLL (blue) and AML (yellow) using the 500 genes that best distinguished ALL from AML.

19 Now let s dive into some mathematics: Probability theory

20 Probabilities and Ensembles An ensemble is set of of probabilities describing a particular random process. An ensemble is defined by x, value of a random variable, taken from the alphabet A x ={a 1,a 2,a 3, a I }, alphabet P X ={p 1,p 2,p 3,..p I }, probabilities All the probabilities have to be positive and add up to one. P(x=a i )=p i p i 0 # a i "A x p(x = a i ) =1 Example: Probability of a single nucleotide in the human genome p A 0.29 p C 0.21 p G 0.21 p T 0.29

21 Joint Probabilities JOINT PROBABILITY: P(x, y) P(x, y) = P(x)P(y) if X and Y are INDEPENDENT MARGINAL PROBABILITY: " y P(x) = P(x, y)

22 Joint Probability: Dinucleotides Single-strand frequencies of human genes A 2 nd position A C G T MARGINAL st position C G T

23 Conditional Probability P(x y) = P(x, y) P(y) Probability of x given y

24 Probability Rules Product Rule (Chain Rule) Sum Rule P(x, y) = P(x y)p(y) P(x) = " y " y P(x,y) = P(x y)p(y)

25 Thomas Bayes ( ), English Presbyterian minister Bayes Theorem P(y x) = P(x y)p(y) P(x) P(y x) = " y P(x y)p(y) P(x y)p(y) Bayes Theorem tells you how to switch the order of variables in a condition probability, i.e., how to convert P(x y) into P(y x).

26 Class Exercise: Am I diseased? There is a rare disease that affects 0.1% of the population. You take a test, which is marketed to be 99% correct, i.e., in 99/100 cases test shows positive when user actually has disease, and in 99/100 cases test shows negative when user actually does not have disease. Test shows positive for you. What is the probability that you have the disease given the test result?

27 Class Exercise continued Write down all the probabilities i) P(disease)=0.001 ii) P(test positive disease)=0.99 iii) P(test negative no disease)=0.99 We want to evaluate P(disease test positive). Use Bayes Theorem (disease test positive) = P (test positive disease)p (disease) P (test positive)

28 Class Exercise continued Use sum rule to obtain P(test positive): (test positive) =P (test positive disease)p (disease)+ P (test positive no disease)p (no disease) =( ) + ( ) = Plugging this into Bayes theorem: P (disease test positive) = =0.09

29 Class Exercise continued Therefore the actual probability of you having the disease given the test result is only 9%, i.e. unlikely you are diseased. Note that the p-value of getting the test result, with the null hypothesis that you are diseasefree, is Thus a frequentist statistician would conclude that the null hypothesis can be safely rejected, and you likely are diseased. This is wrong!

30 Class Exercise continued This picture may help in understanding the counterintuitive Bayesian result Sample of 1000 typical individuals Individual that truly has disease and tests positive (true positive) Individual that does not have disease and tests positive (false positive) Individual that does not have disease and tests negative (true negative)

31 Class Exercise continued In a sample of 1000 typical individuals you may have one individual who likely has the disease [P(disease)=0.01] and likely tests positive [P(positive disease)=0.99]. Of the remaining 999 disease-free individuals, approximately ten will likely test as a false positive [P(positive no disease)=0.01]. Therefore, since there are only 11 individuals who likely test positive P (disease test positive) %