DATA ANALYSIS

Module Code: CA660

Lecture Block 4

2

Examples using Standard Distributions/sampling distributions

)...(ˆ orgametesfractionionrecombinatn

nr

r

ceInterferensoRFpossiblerrrrr BCABBCABAC ,32generallymorerrCrrr BCABBCABAC

*2eCoincidencCoefftCceInterferenC .,1 **

frequencytrecombinandoubletruerwhererr

rC

BCAB12

12*

2

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 = - 0.5 ln (1-2r)

so converting an estimated R.F. to Haldane’s map distance• Thus, for locus order ABC

mAC = mAB + mBC (since mAB= - 0.5ln(1-2rAB) ) 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. mAC vs mAB or mBC

• In practice - too simple/only applies to specific conditions; may not relate directly to physical distance = common Mapping Fn. issue).

6

ExamplesRECOMBINANTS, 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 P1, P2, P3, P4 are probabilities of observing a member of each of 4 classes respectively

rnr

r

nrXP

)1(}{

dcba PPPPdcba

ndXcXbXaXP 43214321 !!!!

!},,,{

7

Sampling and Sampling Distributions – Extended Examples: refer to primer

Central Limit TheoremIf X1, X2,… Xn 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

,...2,1'

i

n

xx

ii

}{ bUaPbx

aPx

x

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

xxx

rxrP

rxrPrxP

}{}{

n

r

n

r FF

F

9

Attribute and Proportionate Samplingrecall 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=X1+ X2+…. + Xn ~ 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

xp̂

xnY

nasNnpq

npY

n

nnY

x

Y

Y

x

x )1,0(

)1,0(~ˆ

N

npq

ppU

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 H0: P1 – P2 =0 so, can write S.E. as

for pooled

X & Y =# successes

2

22

1

1121

ˆˆˆˆ)ˆˆ(2

nqp

nqpUpp

S.E., n1, n2 large.

Small sample n-1

21

11ˆˆ

nnqp

21

2211

21

ˆˆˆ

nn

pnpn

nn

YXp

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.

parameterPoissonX

U ,5.0

pproportionsampleobservedsosampleinNoxnpq

pnxU ˆ/,.

5.0)(

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.

PpExE }ˆ{}{

13

Examples in contextRates 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?

H0: p1=p2 vs H1: p1< p2 (where p1, p2 proportion boys, girls with antibody respectively).

Soln.

Can not reject H0

Actual p-value = P{U ≤ -1.44) = 0.0749

349.0139113

5434ˆ

p388.0

139

54ˆ

301.0113

34ˆ

2

1

p

p

44.1

1391

1131

651.0349.0

388.0301.0

U

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}

0786.0415.149.14

3005.279

UPUP

1423.08588.0107.149.14

3005.315

UPUP

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 1 2 3 4 5 6 7 8 9 10 11 12 13 14 151st (x) 70 80 72 76 76 76 72 78 82 64 74 92 74 68 84 2nd (y) 68 72 62 70 58 66 68 52 64 72 74 60 74 72 74

d =x-y 2 8 10 6 18 10 4 26 18 -8 0 32 0 -4 10

(a) So for 95% C. limits

15

98.1014

)(80.8

15

025.0

2

std

dds

dd ii

16

Contd.Value for t0.025 based on d.o.f. = 14. From t-table, find t0.025 = 2.145

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 14.88

(b) The claim is that > 0, so we look at H0: = 0 vs H1: > 0 ,

So t-statistic as before, but right-tailed (one sided only) Rejection Region. For d.o.f. = 14, t0.01 = 2.624. So calculated value from our data

clearly in Rejection region, so H0

rejected in favour of H1 at = 0.01 Reduction in B.P. after medication strongly supported by data.

95.015

98.10145.280.8

15

98.10145.280.8

DP

10.3

1598.10

80.8

nsd

t

0

t14

Accept Reject = 1%t0.01 = 2.624.