Bft1033 6 linkage genes_print

(GENE MAPPING)

INTRODUCTION

Most chromosomes consist of very large numbers of genes

Genes that are part of the same chromosome are said linked

These genes demonstrate linkage in genetic crosses

During meiosis, they are not free to undergo independent assortment They are transmitted as a unit

Crossover results in reshuffling or recombination of alleles between homolog.

Introduction

No crossover: two genetically different gametes are formed Each gamete receive the alleles present on one

homolog or the other

Illustrate complete linkage

Produce parental or noncrossover gametes

Crossover: produce four types of gametes Two parental gametes

Two recombinant or crossover gametes

Introduction

The Linkage Ratio

Complete linkage in Drosophila melanogaster: Mutants: brown (bw) eye and heavy (hv) wing vein

Normal alleles: bw+ (red eye) and hv+ (thin wing vein)

Cross between brown eye and thin vein with red eyes and heavy vein

P:

F1:

bwhv bw hv

bwhv bw hv

red, thin

Brown-thin Red-heavy

bwhv

bw hv

The Linkage Ratio

When the F1 generation is interbred, the F2 generation will be produced in a 1:2:1 phenotypic and genotypic ratio.

When the F1 is tescrossed, it will produce a 1:1 ratio of brown thin and red heavy

The Linkage Ratio

The Linkage Ratio

Morgan crossed yellow bodied white eyed female and wild type male

P: yellow, white ♀ X wild-type ♂

F1: ♀: wild type

♂: expressed both mutant traits

F2: 98.7% parental types (gray bodied, red eyed)

1.3% either: yellow bodied with red eyed , or

gray bodied with white eyed

Crossover and Gene Distance

Morgan made crosses involving other X-linkage genes

P: White-eye, miniature wing ♀ X wild type ♂

F1: even more puzzling

F2: phenotypes differed

62.8%: parental types

37.2%: either: white eyed or

miniature wing

Crossover and Gene Distance

Crossover and Gene Distance

Morgan crossed yellow bodied white eyed female and wild type male:

yw/yw Xy+w+Y

Morgan made crosses involving other X-linkage genes:White-eye, miniature wing ♀X wild type ♂

Morgan postulated that exchange occurred between the mutant genes on the two X chromosomes of the F1 females

Lead to 1.3 and 37.2 recombinant gametes

The closer two gene are, the less likely genetic exchange will occur between them

Morgan proposed the term crossing over to describe the physical exchange leading to recombination.

Crossover and Gene Distance

Crossover and Gene Distance

Crossover and Gene Distance

Concept of a genetic map

Two arrangements of alleles exist for an individuals heterozygous at two loci:

Cross-over of cis results in trans and vice versa

Frequency of recombinants (%) is a characteristic of each gene pair, regardless of cis or trans arrangements

‘cis’ or coupling

w+ m+

w m

‘trans’ or repulsion

w+ m

w m+

In cross A Parental types (yellow-white, wild type): 98.7%

Recombinant types (white, yellow): 1.3%

Distance between genes: 1.3 mu

In cross B Parental types (white-miniature, wild type): 62.8%

Recombinant types (white, miniature): 37.2%

Distance between genes: 37.2 mu

Concept of a genetic map

Cross-over is more likely to occur between distant genes than close genes

Calculating Recombination frequency

Sturtevant (1913) recognized that recombination frequencies could be used to create a map

1% cross-over rate = 1 map unit (mu) or centiMorgan (cM)

Map units (mu) and centiMorgans (cM) are relative measures.

# of recombinant progenyRecombination frequency X 100%

total # of progeny

Example

The test cross Ab/aB x ab/ab is performed. The following numbers of progeny of each genotype are obtained: 87 AaBb, 409 Aabb, 390 aaBb, 114 aabb. What is the approximate distance (in map units)

between the two genes in question?

RF = (87 + 114)/(87 + 409 + 390 + 114) x 100%

= 201/1000 x 100%

= 20.1%

So the distance between the two genes is 20.1 cM

# of recombinant progenyRecombination frequency X 100%

total # of progeny

First genetic map was for Drosophila:3 sex-linked genes

w = white-eyes

m =miniature wings

y = yellow body

Recombination frequencies:w x y = 0.5%

w x m = 34.5%

m x y = 35.4%

0.5 34.5

35.5

Single Crossover

Single crossover

Double Crossover

Double crossover

Three-Point Mapping

The genotype of the organism producing the crossover gametes must be heterozygous at all loci

The cross must be constructed so that genotypes of all gametes can be accurately determined by observing the phenotypes of the resulting offspring

A sufficient of number of offspring must be produced in the mapping experiment to recover a representative sample of all crossover.

Three-Point Mapping

Males hemizygous for all three wild type alleles are crossed to female with three mutant traits (yellow body, white eyes, and echinus eye shape)

F1 consists of females heterozygous at all loci and males hemizygous for all three mutant alleles

When the F1 is intercrossed to produce F2, it produces 8 different classes: Two classes of parental types (the biggest

proportion)

Two classes from single crossover in region I

Two classes from single crossover in region II

Two classes from double crossover (the smallest proportion).

Determining Gene Sequence

Phenotypes

white, echinus

yellow

yellow, white

echinus

yellow, echinus

white

A Mapping Problem in Maize

In maize, the recessive mutant genes: bm (brown midrib), v (virescent seedling), and pr (purple aleurone)

are linked on chromosome 5

A female plant is heterozygous for all three traits is crossed with a male homozygous for all three mutant alleles

F1 data: [+ v bm] 230

[pr + +] 237

[+ + bm] 82

[pr v +] 79

What is the correct sequence of genes?

What is the distance between each pairs of gene?

[+ v +] 200

[pr + bm] 195

[pr v bm] 44

[+ + +] 42

The Five Steps to Solve the Problem

1. Determine the parental and dco types

2. Examine the gene in the middle

3. Re-order the genes (if necessary)

4. Examine sco in region I and II

5. Calculate the distance

A Mapping Problem in Maize

The Five Steps to Solve the Problem

1. Determine the parental and dco types

2. Examine the gene in the middle

3. Re-order the genes (if necessary)

4. Examine sco in region I and II

5. Calculate the distance

A Mapping Problem in Maize

Determine the parental and dco types

The parental types are the biggest number, and dco types are the smallest [+ v bm] 230

[pr + +] 237

[+ + bm] 82

[pr v +] 79

[+ v +] 200

[pr + bm] 195

[pr v bm] 44

[+ + +] 42

Determine the parental and dco types

The parental types are the biggest number, and dco types are the smallest [+ v bm] 230 parental type

[pr + +] 237 parental type

[+ + bm] 82

[pr v +] 79

[+ v +] 200

[pr + bm] 195

[pr v bm] 44 dco type

[+ + +] 42 dco type

The Five Steps to Solve the Problem

1. Determine the parental and dco types

2. Examine the gene in the middle

3. Re-order the genes (if necessary)

4. Examine sco in region I and II

5. Calculate the distance

A Mapping Problem in Maize

Examine the gene in the middle

[+ v bm] 230 parental type

[pr + +] 237 parental type

[pr v bm] 44 dco type

[+ + +] 42 dco type

v

v

v

vv

vbm

bm

bmbm

+

bm

+

+

+

+

+

+

+

+

pr

+

pr

pr

pr

pr

pr

bm+ +

+

++

+

++

The Five Steps to Solve the Problem

1. Determine the parental and dco types

2. Examine the gene in the middle

3. Re-order the genes (if necessary)

4. Examine sco in region I and II

5. Calculate the distance

A Mapping Problem in Maize

Re-order the genes (if necessary)

Temporary Order

[+ v bm] 230 [pr + +] 237 [+ + bm] 82 [pr v +] 79 [+ v +] 200 [pr + bm] 195 [pr v bm] 44 [+ + +] 42

Correct Order

v + bm 230 + pr + 237 + + bm 82 v pr + 79 v + + 200 + pr bm 195 v pr bm 44 + + + 42

The Five Steps to Solve the Problem

1. Determine the parental and dco types

2. Examine the gene in the middle

3. Re-order the genes (if necessary)

4. Examine sco in region I and II

5. Calculate the distance

A Mapping Problem in Maize

Examine sco in region I and II

v

+

bm

+

+

pr

v

+

bm

+

+

pr

v

+

bm

+

+

pr

v bm+

++ pr

v +pr

bm+ +

v ++

bm+ pr

I

II

v + bm 230 parental type

+ pr + 237 parental type

v pr + 79 scoI type

+ + bm 82 scoI type

v + + 200 scoII type

+ pr bm 195 scoII type

The Five Steps to Solve the Problem

1. Determine the parental and dco types

2. Examine the gene in the middle

3. Re-order the genes (if necessary)

4. Examine sco in region I and II

5. Calculate the distance

A Mapping Problem in Maize

Calculate the distance

The formula to calculate the distance between two genes:

In region I =

In regio II =

100xTotal

dcoscoI

100xTotal

dcoscoII

v + bm 230 parental type

+ pr + 237 parental type

+ + bm 82 scoI type

v pr + 79 scoI type

v + + 200 scoII type

+ pr bm 195 scoII type

v pr bm 44 dco type

+ + + 42 dco type

Total 1109Calculate the distance

Distance between v-pr =

Distance between pr-bm =

Distance between v-bm =

And the map is

cM 22.27 100x

1109

42 44 79 82

cM 43.37 100x

1109

42 44 195 200

Calculate the distance

cM 65.64 43.37 22.27

v + bm 230 parental type

+ pr + 237 parental type

+ + bm 82 scoI type

v pr + 79 scoI type

v + + 200 scoII type

+ pr bm 195 scoII type

v pr bm 44 dco type

+ + + 42 dco type

v+/v pr+/pr bm+/bm

22.27 cM 43.37 cM

Interference and Coincidence

Interference: a crossover at one spot on a chromosome decreases the likelihood of a crossover in a nearby spot

where c: coefficient of coincidence

obs dco: observed data

exp dco: expected dco = sco I x sco II

obsdco

expdcoc

I = 1 – c

From the data:

obs dco = 86/1109 = 0.0775

exp dco = 0.2227 x 0.4337 = 0.0966

c = 0.0775/0.0966 = 0.80

I = 1 – 0.80 = 0.20

v+/v pr+/pr bm+/bm

22.27 cM 43.37 cM

v + bm 230 parental type

+ pr + 237 parental type

+ + bm 82 scoI type

v pr + 79 scoI type

v + + 200 scoII type

+ pr bm 195 scoII type

v pr bm 44 dco type

+ + + 42 dco type

Interference and Coincidence

Any questions?Thank you