DATA ANALYSIS Module Code: CA660 Lecture Block 4

2 Examples using Standard Distributions/sampling distributions Background Recombinant Interference Greater physical distance between loci  greater chance to recombine - (homologous). Departure from additivity increases with distance -hence mapping. Example: 2 loci A,B, same chromasome, segregated for two alleles at each locus  A,a,B,b  gametes AB, Ab, aB, ab. Parental types AB, ab gives Ab and aB recombinants. Simple ratio. Denote recombinant fraction as R.F. (r) Example: For 3 linked loci, A,B, C, relationship based on simple prob. theory

3 Example cont.- LINKAGE/G.M CONSTRUCTION Genetic Map -Models linear arrangement of group of genes / markers (easily identified genetic features - e.g. change in known gene, piece of DNA with no known function). Map based on homologous recombination during meiosis. If two or more markers located close together on chromosome, alleles usually inherited through meiosis 4 basic steps after marker data obtained. Pairwise linkage - all 2- locus combinations (based on observed and expected frequencies of genotypic classes). Grouping markers into Linkage Groups (based on R.F.’s, significance level etc.). If good genome coverage –many markers, good data and genetic model, No. linkage groups should  haploid no. chromosomes for organism. Ordering within group markers (key step, computationally demanding, precision important). Estimation multipoint R.F. (physical distance - no. of DNA base pairs between two genes vs map distance => transformation of R.F.). Ultimate Physical map = DNA sequence (restriction map also common)

4 STANDARD DISTRIBUTIONS - Examples/Extensions GENETIC LINKAGE and MAPPING Linkage Phase - chromatid associations of alleles of linked loci - same chromosome =coupled, different =repulsion Genetic Recombination - define R.F. (in terms of gametes or phenotypes); homologous case - greater the distance between loci, greater chance of recombining. High interference = problem for multiple locus models. R.F. between loci not additive. Need Mapping Function Haldane’s Mapping Function Assume crossovers occur randomly along chromosome length and average number =, model as Poisson, so P{NO crossover} = e - and P{Crossover} = 1- e -

5 Example - continued P{recombinant} = 0.5  P(Crossover} (each pair of homologs, with one crossover resulting in one-half recombinant gametes) Define Expected No. recombinants in terms of mapping function (m = 0.5 ) R.F. r = 0.5(1-e -2m ) (form of Haldane’s M.F.) with inverse m = ln (1-2r) so converting an estimated R.F. to Haldane’s map distance Thus, for locus order ABC m AC = m AB + m BC (since m AB = - 0.5ln(1-2r AB ) ) etc. Substituting for each of these gives us the usual relationship between R.F.’s (for the no interference situation) Net Effect - transform to straight line i.e. m AC vs m AB or m BC In practice - too simple/only applies to specific conditions; may not relate directly to physical distance = common Mapping Fn. issue).

6 Examples RECOMBINANTS, BINOMIAL and MULTINOMIAL Binomial No. of recombinant gametes, produced by a heterozygous parent for a 2-locus model, with parameters, n and  = P{gamete recombinant} (= R.F.) So for r recombinants in sample of n Multinomial 3-locus model (A,B,C) - 4 possible classes of gametes (non-recombinants, AB recombinants, BC recombinants and double recombinants at loci ABC). Joint probability distribution for r.v.’s requires counting number in each class where a+b+c+d = n and P 1, P 2, P 3, P 4 are probabilities of observing a member of each of 4 classes respectively

7 Sampling and Sampling Distributions – Extended Examples: refer to primer Central Limit Theorem If X 1, X 2,… X n are a random sample of r.v. X, (mean , variance  2 ), then, in the limit, as n , the sampling distribution of means has a Standard Normal distribution, N(0,1) Probabilities for sampling distribution – limits for large n U = standardized Normal deviate

8 Large Sample theory In particular is the C.D.F. or D.F. In general, the closer the random variable X behaviour is to the Normal, the faster the approximation approaches U. Generally, n  30  “Large sample” theory

9 Attribute and Proportionate Sampling recall primer sample proportion and sample mean synonymous Probability Statements If X and Y independent Binomially distributed r.v.’s parameters n, p and m, p respectively, then X+Y ~ B(n+m, p) - (show e.g. by m.g.f.’s) So, Y=X 1 + X 2 +…. + X n ~ B(n, p) for the IID X~B(1, p). Since we know  Y = np,  Y =  (npq) and, clearly then and, further is the sampling distribution of a proportion

10 Difference in Proportions Can use  2 : Contingency table type set-up Can set up as parallel to difference estimate or test of 2 means (independent) so for 100 (1-  C.I. Under H 0 : P 1 – P 2 =0 so, can write S.E. as for pooled X & Y =# successes S.E., n 1, n 2 large. Small sample n-1 2-sided

11 C.L.T. and Approximations summary General form of theorem - an infinite sequence of independent r.v.’s, with means, variances as before, then approximation  U for n large enough. Note: No condition on form of distribution of the X’s (the raw data) Strictly - for approximations of discrete distributions, can improve by considering correction for continuity e.g.

12 Generalising Sampling Distn. Concept - see primer For sampling distribution of any statistic, a sample characteristic is an unbiased estimator of the parent population characteristic, if the mean of the corresponding sampling distribution is equal to the parent characteristic. Also the sample average proportion is an unbiased estimator of the parent average proportion Sampling without replacement from a finite population gives the Hypergeometric distribution. finite population correction (fpc) =  [( N - n) / ( N - 1)], N, n are parent population and sample size respectively. Above applies to variance also.

13 Examples in context Rates of prevalence of CF antibody to P1 virus among given age group children. Of 113 boys tested, 34 have antibody, while of 139 girls tested, 54 have antibody. Is evidence strong for a higher prevalence rate in girls? H 0 : p 1 =p 2 vs H 1: p 1 < p 2 (where p 1, p 2 proportion boys, girls with antibody respectively). Soln. Can not reject H 0 Actual p-value = P{U ≤ -1.44) =

14 Examples – contd. Large scale 1980 survey in country showed 30% of adult population with given genetic trait. If still the current rate, what is probability that, in a random sample of 1000, the number with the trait will be (a) < 250, (b) 316 or more? Soln. Let X = no. successes (with trait) in sample. So, for expected proportion of 0.3 in population, we suppose X ~B(1000,0.3) Since np=300, and √npq = √210 =14.49, distn. of X ~N(300,14.49) (a)P{X<280} or P{X≤279}  (b) P{X≥316} 

15 Examples contd. Blood pressure readings before and after 6 months on medication taken in women students, (aged 25-35); sample of 15. Calculate (a) 95% C.I. for mean change in B.P. (b) test at 1% level of significance, (  = 0.01) that the medication reduces B.P. Data: Subject st (x) nd (y) d =x-y (a) So for 95% C. limits

16 Contd. Value for t based on d.o.f. = 14. From t-table, find t = So, 95% C.I. is: i.e. limits are 8.80  6.08 or (2.72, 14.88), so 95% confident that there is a mean difference (reduction) in B.P. of between 2.72 and (b) The claim is that  > 0, so we look at H 0 :  = 0 vs H 1 :  > 0, So t-statistic as before, but right-tailed (one sided only) Rejection Region. For d.o.f. = 14, t 0.01 = So calculated value from our data clearly in Rejection region, so H 0 rejected in favour of H 1 at  = 0.01 Reduction in B.P. after medication strongly supported by data. 0 t 14 Accept Reject = 1% t 0.01 =