L4Biol261Recombination2014

8/9/2019 L4Biol261Recombination2014

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Lecture 4: When 2 or more genes are on the

same chromosomeor lets relax the assumption ofindependence, central to prediction in Classical

Mendelian Models.


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What is the probability of having two girls and a

boy ?

(GGB or GBG or BGG)

a) 1/2b) 1/4c) 1/8

d) 1/16e) 3/8

In genetics you often use both multiplicative and additive rules in oneproblem


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The chromosomal theory of inheritance: first proposed by W. Sutton, T. Boveri

1902, based on correlation between chromosome division into gametes and the

inheritance of sex. For example, W. Sutton wrote that the association of paternal

and maternal chromosomes (of the grasshopper Brachystoma magna) in pairs and

their subsequent separation during the reducing division (of meiosis) may explain

Mendelian segregation and that maternal and paternal pairs are indifferent to

their position relative to others when they assort and segregate in meiosis.

T.H. Morgan and his student C. Bridges experiments on Drosophila showed

phenotype ratios are informative even when they deviate from Mendels classic

ratio because they correlate with chromosome segregation.This work strongly suggested, that units of inheritance of different

phenotypic characters -genes- are in chromosomes

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Ifn is the basic number of chromosomes.

In humans 2n = 46

n = 23There are 22 autosomepairs

and a (1) sex chromosomepair


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Chromosomes differ in size and the position of their

centromere


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Prophase -Metaphase chromosome

Interphasechromosomes

Each diploid humancell contains about 2m

of DNA in nuclei with

an average diameter of

6 micrometers (10-6m)

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To represent the cellular behavior of chromosomes at the cytological

scalein meiosis we use the following model of chromosome structure, a

double stranded DNA molecule with associated histone proteins.

The centromere is part of a DNA molecule, it is hundreds of kilobases

long, with no functional genes. Count chromosomes by counting

centromeres.


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Figure 2-15

Key stages of meiosis and mitosis

Homologous

chromosomes

Sister chromatids

Sister Chromatids

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Synaptonemal complexes at meiosis

(3)Pachytene(synaptonemal complex,crossover, recombination).

Recombinationoccurswhere maternal &

paternalchromosomes

arms break, andrecombine with theirhomologue.

There aregeneticrepair mechanisms

associated with thepairing of chromosomesin Prophase1.

(3)Pachytene(synaptonemal complex,

crossover, recombination).

Recombination (1)

occurs where maternal &paternalchromosomes

arms break, and recombine

with their homologue.

There aregenetic repairmechanisms associatedwith the pairing ofchromosomes in Prophase1.

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Recombinants (1) are produced by crossovers thathappen in prophase 1 of meiosis

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Synaptonemal complex


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Crossover, recombination and

repair has finished

in (4)Diplotene, thechromosomes are beginning to

repel and the chiasmata (points

of recombination)are

beginning to migrate to thetelomeres.

In (5) Diakinesis, chiasmata

have migrated to the

chromosome ends (telomeres)

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Recombination 2: metaphase: paired homologues

orientrandomlyon the metaphase plate Mendel 2

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Understand This !


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Genes for

flower, leaf and

stem shape and

color

There is more than

one per

chromosome, or

they are linked andpotentially non-

independent


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Linked genes:genes are linked when they are on the same chromosome.

Nomenclature: AB/ab , you might see

AB//ab or, AB or ++/ ab, or

ab

Separate chromosomes?A/a ; C/c (known), A/a . C/c (unknown),

Key terms:Meiotic recombination, dyads, tetrads, crossover,

chiasma(ta), parentals, recombinants, recombinant frequency, map unit,

centimorgan.

Key Breeding designs-

Dihybrid Parental cross, eg. Ab/Ab X aB/aB

and F1testcross AB/ab X ab/ab

AB ab

ab X

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Gene linkage can be: very close, close or distant

A B

a b

(2) But, if they are far enough apart on a chromosome, they may

act as if they assort independently

A B

a b

Genes on the same chromosome are linked.

(1) Closelylinked alleles assort into the same gametes and are more

frequently inherited togetherthan you would expect than if they

were on different chromosomes (not linked).

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(1a) Very closely linked allelesusually

assort into the same gametesor, they are

almostalways co-inherited.

A B

ab

A B

ab

A B

a b

A B

a b

A Prophase 1 tetrad-2 synapsed replicated chromosomes

tetrad

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AB ab

AB / ab ab / abab

(1) Dihybrid Parentals (AA BB & aa bb) cross

(2) F1testcross, showing parental, homozygous test

gametesand the expected F1 offspring genotypesfor

(a)very closely linked genes in

(b)

a small - sized sample.

ratios ?

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tester


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Linked alleles are inherited together unless,

(2 )there is a cross-over between the locations (loci) of 2genes on homologous chromatids in prophase 1 of meiosis.

A B

a b

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Gene linkage Gene recombination in Pachytene, a sub-stage in

prophase 1 of meiosis.

Linked alleles are inherited together, except when there is across-over between linked loci on homologous chromatids

A B

a b

There may be a cross-over, but it cannot be detectedunless there are different alleles on each dyad(a pair

of sister chromatids joined at the centromere).

chiasma

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Gene linkage and recombination in Pachytene

(1) Linked alleles are inherited together except when there is a

cross-over between linked loci on homologous chromatids.A B

a b

Crossovers involve the breaking of at least one

chromatid on each dyad and the rejoining of maternalwith paternal fragments.

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Gene linkage and recombination:

Linked alleles are inherited together exceptwhen there

is a cross-over between homologous chromatids or

closely - linked genes sometimes recombine.

A B

a b

a B

A b

A B

a B

A b

a b

Parentals Recombinant

gametes

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In GENERAL

(1)There is always at least one chiasma(ta)-(a crossover point) per

chromosome, per meiosis, regardless of chromosome length.

(2)Thegreater the distancebetween two gene locations (loci)on a

chromosome the greater the chance of a cross over between them.

(3) The position of the crossover between two loci, is at first

approximationrandom.

A B C

a b c

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With CLOSE LINKAGE Parental types will be more frequent-

(in excess of 50 %)than recombinant types (less than 50%).

>1/4

>1/4


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Mapping closely linked genes

using their linkage relationship

Complete linkage between loci (absence of a chiasma in

meiosis) decreases with increasing distance between

markers (loci) on a chromosome, or, as the distance

increases between 2 loci on a chromosome, the probability

of a crossover and recombinant gametes increases.

AlfredSturtevantfirst suggested using % recombinants as a

linear measure of chromosomal map distance

One map unit = 1% recombinants

1 map unit= 1 centimorgan (cM)

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Crossover between 2 markers keeping it organized

Testing for linkage between genes for fruit shape (beaked bk)

and internal division number (1 locule Lc or many lc)

in tomatoes, and if linked, measuring the map distance between

them.

Bk bk lc Lc

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Measuring linkage between genes for beaked fruit and many

locules in tomatoes.

Bk bk lc Lc

P: Bk /Bk . Lc /Lc x bk /bk . lc/ lc

Test cross: Bk/ bk . Lc/ lc x bk /bk . lc /lc

Bk bk Lclc 170

Bk bk lc lc 26

bk bk Lc lc 30bk bk lc lc 168

400

The ratio of the 4 classes of

offspring is clearly not

1:1:1:1 suggesting

The two genes are closely

linked on the same

chromosome.29


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Measuring linkage between genes for beaked fruit and many

locules in tomatoes.

Bk bk lc Lc

P: Bk Lc / Bk Lc x bk lc / bk lc

F1

Bk bk Lc lc 170

bk bk lc lc 168

Bkbk lc lc 26

bkbkLclc 30

400

The % recombinant offspring is:

(26 + 30) / 400 = 56 / 400

= 14%

The distance between Bk and Lc is 14 map units

Bk Lc x bk lcbk lc bk lc

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Test cross:


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AB Ab aB ab

AB/ab Ab/ab aB/ab ab/abab

Dihybrid testcross, showing parental,

recombinant, homozygous testgametes and theexpected genotypesin the F2.

If a & b loci are so far apart that there is always a

crossover between them,then the ratios will be the

same as independent assortment of A & B loci.

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If they are close, then recombinants < parentals

If they are very close then all you see is parentals

F1:AB/ab

SUMMARY


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Double crossover between two markers (gene loci), in this

case there will not appear to be a crossover between v and cv.

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So all you will see are the genotypes of the F1

and the testers- no recombinants.


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Double crossover between v and cv will appear the same

as no crossover, but a third, intermediate markerhas

recombined. Use the third intermediate marker toestimate double frequency crossover.

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The outside markers are still the same as theparental, but the inside marker changes

chromosome association (switches).


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Three markers or loci on the same arm of a

chromosome allows you to:

(1)

detecting double crossovers between theoutside markers (loci)

(2)estimate the distance between loci

(3) determine the order of loci

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The three point cross is used, if they are linked, to map the

distance between three genes and determine their order.

F faciated, ff - two fasciaA purple stem, aa green stem

Hl hairless stems hl hl hairy stems

Parents F A Hl / F A Hl x f a hl / f a hl

F1 Test Cross: F A Hl// f a hl x f a hl // f a hl

Three genes controlling locule form, stem color and stem

hairiness in tomato are linked .

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The three point crossF fasciated ff - two fascia

A purple stem aa green stem


P: F A Hl//F A Hl x f a hl // f a hl

Test X: F A Hl// f a hl x f a hl // f a hl

F A Hl 302

f a hl 308

F a hl 101

f A Hl 109

F A hl 82f a Hl 77

F a Hl 10

f A hl 11

1000

Parental types

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Double crossover gametes

Single recombinants


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The three point crossF fasciated (many locules) ff - two locules

A purple stem X aa green stem


P: F A Hl//F A Hl x f a hl // f a hl

Test X: F A Hl // f a hl x f a hl // f a hl

Parental types

F A recombinants are unlike

parental configuration.

% recombination =

101 + 109 + 10 + 11 / 1000 =

231/ 1000 = 23.1 %

F-A-find recombinants

Work 2 loci at a time

F A Hl 302

f a hl 308

F a hl 101

f AHl 109

F A hl 82f a Hl 77

F a Hl 10

f Ahl 11

100037


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The three point crossF many locules ff - two locules



Parental types

F A recombinants 23.1 %

A Hl recombinants

82 + 77 + 10 + 11 / 1000 =180 / 1000 = 18%

Work 2 loci at a time A, Hl

F A Hl 302f a hl 308

F a hl 101

f AHl 109

FA hl 82

f a Hl 77

F a Hl 10

f A hl 11

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Parental types

F Hl , % recombinants =

101 + 109 + 82 + 77 / 1000 =369 / 1000

= 36.9%

P: F A Hl / F A HlX f a hl /f a hl

Test X: F A Hl / f a hl X f a hl /f a hl

Work 2 loci at a time F-Hl

F A Hl 302

f a hl 308

Fa hl 101

f AHl 109

FAhl 82

f a Hl 77

F a Hl 10

f A hl 11

1000

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A Hl recombinants 18%

F Hl recombinants 36.9%

? Outside loci (markers)

F A Hl 302

f a hl 308

F a hl 101

f AHl 109F A hl 82

f a Hl 77

F a Hl 10

f A hl 11

1000

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F a hl 101

f A Hl 109

F A hl 82

f a Hl 77

F a Hl 10

f A hl 11

1000

Distance between the outside loci =36.9 or 23.1+18= 41.1 ?

F HlA




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F a hl 101

f AHl 109

F A hl 82

f a Hl 77

F a Hl 10

f A hl 11

1000




F A

23.1 18

36.9 + 4.2 = 41.1 map units

Hl

2.1% These classes are double cross-overs

for F and Hl and must be counted twice.

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F A Hl

f a hl

F A Hl302

f a hl 308F a hl 101

f AHl 109

F Ahl 82

f aHl 77

Fa Hl 10

f Ahl 11

(1)

Why calculate all 3 distances to get the order ?

(2) For the order, identify the double recombinant class,

look at which locus has flipped, relative to the parentalorder - this one is in the middle.

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F A Hl

f a hl

There are two ways to calculate the F to Hl distance. The answer

should beidentical.

1. Add the distances between the outside and middle markers or

loci:F to A distance, and A to Hl distance.

2. Identify the middle locus or marker(A), estimate the distance

between the 2 outside markers(F - Hl recombinants), and add 2times the number of double recombinants- because double

recombinants have single crossovers in each outside region.44


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F A Hl

f a hl

Mapping and recombination - the 4 step method when you do not have a data table

(1)Pair gametic types by complement & frequency

(2) ID parental chromosomes: F A Hl & f a hl (a) # highest ?

(b) complementary pairs ?(c)# similar ?

(3) Organize a table with the 2 or 3 mutant symbol headingthe column tops,

corresponding parentals at the top, below each heading.Remember, you are

always organizing & comparing recombinants with parentals.

(4)If the problem involves 3 markers (loci), identify the double recombinant

chromosomes F aHl (a) # lowest ?

(c) complementary pairs? f AHl (c) # similar ?

Then you can ID the middle marker(pair that has flipped to parentals)

Then you have the gene order. Put the double -X -over pair at the bottom

(5) Table the single recombinant classes (middle): (a) # similar ?

Estimate the required distances (b) complementary pairs ?

Parental F A Hl

Parental f a hl

recombinant

recombinant

2

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Interference between chiasmamay decrease the

observed crossover frequency and the apparentgenetic distance.

1. The coefficient of coincidenceis the observed frequency of

double recombinants divided by (standardized by) the expected

number of double recombinants.

2. The expected number of recombinants is the productof the

expectation of a crossover in each region (outside to middle).

3. Interference = 1- coefficient of coincidence

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P cn c px 296R1 cn c + 63

R2 cn + + 119

D cn + px 10

R2 + c px 86

D + c + 15P + + + 329

R1 + + px 82

1000

1. Calculate the individual distances make the map.

ORCompare double cross-overs with parental lines, the one gene

that is different is in the middle (order) then estimate the other distances.

C- curved wings, px - extra wing veins, cn - cinnabar eyes

In the lab, the numbers will not be pre - paired - Look at

magnitudes, check by locus complements (+ + + / cn c px)

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Recombination Mapping in General

As map distance increases over 10 map units, multiple crossovers

cause recombination frequency to underestimate crossover

frequency or the actual recombination map distance. For example,

when the recombination frequency is 30% the actual map distance is

50 mu in Drosophila.

Mapping functions help(p=1/2 (1-e-2d) d=crossover frequency

Or, map closely linked genes and add the distances.

Very Closely linked = ? Closely Linked = ? Distantly linked = ?

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L4Biol261Recombination2014