Introduction to Interference By: Nickolay Dovgolevsky Itai Sharon 29/05/03.

Introduction to Introduction to InterferenceInterference

By:Nickolay DovgolevskyItai Sharon

29/05/03

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AgendaAgenda

Biological contextThree point analysisChi-square testResult analysisExercise execution

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Biological Context: MeiosisBiological Context: Meiosis

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Biological Context: Biological Context: RecombinationRecombination

Recombination (crossing over) occurs between homologous chromosomes

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Recombination (crossing over) occurs between homologous chromosomes

Each two loci S and T have recombination rate ST

Biological Context: Biological Context: RecombinationRecombination

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Biological Context: Biological Context: RecombinationRecombination

Recombination (crossing over) occurs between homologous chromosomes

Each two loci S and T have recombination rate ST

If recombination rate between loci A and B is independent of rate between B and C, then AC = AB + BC – 2*AB*BC

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InterferenceInterference

Interference is the phenomenon whereby crossovers do not occur independently along a chromosome

This means:AC AB+BC–2AB*BC

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Measuring InterferenceMeasuring InterferenceCoefficient of Coincidence (c) –

c = AB+BC – AC

2AB*BC

Interference (I) – I = 1 – c

c > 1: Negative interferencec = 1: No interference0 < c < 1: Partial interferencec = 0: Complete interference

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Three Point AnalysisThree Point AnalysisAssuming three co-linear loci A, B and C, estimate AB, BC and AC by calculating percentage of recombinants for each (Ott, 1991). No restriction on absence of interference ILINK is capable of analyzing three-point data

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Three Point AnalysisThree Point Analysis

Where: = 1 - - - Also: AC = +

A-B

B-C

Recombinants

Non-Recombinan

ts

Total

Recombinants

BC

Non-Recombinan

ts

1-BC

Total AB 1-AB

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Results Estimation for Results Estimation for Three Point AnalysisThree Point Analysis

We can estimate the significance of the results by calculating

L(AB, BC, c) 2 ln L(AB, BC, c=1) ~ 2

(1)

where L(AB, BC, c) – likelihood of AB, BC and c L(AB, BC, c=1) – likelihood of AB, BC and

c=1 2

(1) is the chi-square value with df=1

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The The 22 (Chi-Square) Test (Chi-Square) Test

Given An hypothesis H Expected numbers E according to H Observed numbers O

The 2 test gives the probability of seeing numbers greater than, or equal to O instead of E, assuming H

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The The 22 Test: an Example Test: an Example

We cross two pure lines of plants, one with yellow petals and one with red.

The following numbers are observed on F2:orange: 182yellow: 61red : 77Total: 320

P

F1

F2

yellow X red

orange

yellow, red, orange

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The The 22 Test: an Example Test: an ExampleHypothesis: Incomplete dominance2 alleles, G1 (yellow) and G2 (red) with similar dominance; G1/G2 gives orange.

Observed numbers Expected numbers on F2: on F2:orange: 182 orange: 160 yellow:

61 yellow: 80 red : 77 red : 80

Total: 320 Total: 320

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The The 22 Test: an Example Test: an ExampleCalculating the 2 value:

O E (O – E)2 / Eorange: 182 160 3.0 yellow: 61 80 4.5red : 77 80 0.1

2 = 7.6Degrees of freedom: setting the sizes of two of the three phenotypic classes sets the third.This means: df = 2

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The The 22 Test TestThat’s how it works:Calculate (O – E)2/E for each classCalculate the 2 value ( (O – E)2/E)Conclude amount of degrees of

freedomFind the probability according to the

2 distribution tableAccept the hypothesis if p < 0.05

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The Chi-Square TestThe Chi-Square Test

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Estimating InterferenceEstimating InterferenceAssume we have 3 loci, with the following ST calculated: 13 = 0.270 12 = 0.197 23 = 0.314

then 0.270+0.197–0.314 2*0.270*0.197 c = 1.44

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We know that:

Estimating InterferenceEstimating Interference

In this example we’ll get: 2 = 878.94 – 878.75 = 0.19 p = 0.66

How significant is this result?

ILINK computes –2ln(L(AB, BC, c))We can execute ILINK with “no

interference” for –2ln(L(AB, BC,c=1))

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Estimating Interference Estimating Interference with sex differencewith sex difference

In general – if there are differences between recombination rates among males and females, cT tends to be greater than both

It is possible to calculate cm and cf separately

It is possible to assume constant sex difference or varying sex difference

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RemarksRemarksOtt (1991) determined that for

Kosambi level interference and three equally spaced markers ( = 0.15), 847 meiosis are required to reject the “no human crossover interference” hypothesis

Broman et. Al. (2000) found strong evidence for positive interference in the levels implied by Kosambi and Carter-Falconer map functions

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ExerciseExercise

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Loci SelectedLoci Selected

5 – (0.075) – 2 – (0.075) – 8 – (0.225) – 3 – (0.075) – 6 – (0.075) – 4 – (0.075) – 9 – (0.075) – 7

5 – (0.150) – 8 – (0.225) – 3 8 – (0.225) – 3 – (0.150) – 4 8 – (0.225) – 3 – (0.225) – 9 3 – (0.150) – 4 – (0.150) – 7