©2000 Timothy G. Standish John 15:4 4Abide in me, and I in you. As the branch cannot bear fruit of itself, except it abide in the vine; no more can ye,

©2000 Timothy G. Standish

John 15:4

4 Abide in me, and I in you. As the branch cannot bear fruit of itself, except it abide in the vine; no more can ye, except ye abide in me


Linkage, Crossing Linkage, Crossing Over And Mapping Over And Mapping

In EukaryotesIn EukaryotesTimothy G. Standish, Ph. D.


Chromosomal Theory Chromosomal Theory of Inheritanceof InheritanceFath

er

Mother

e

N

E

nn

Ee

N

e

n

E

N

Telophase II

E

n

e

NN

eE

n

Replicatione

N

E

n

Prophase ICrossing Over

N

E

n

e e

n

E

N

Telophase I


En

eN

en

EN

e ne NE nE N

Independent AssortmentIndependent Assortment

Sperm

n

E

e

N

e

n

N

E

EeNnEeNNEENnEENN

EennEeNnEEnnEENn

eeNneeNNEeNnEeNN

eenneeNnEennEeNn

As long as genes are on different chromosomes, they will assort independently

Eggs

n

Ee

n

e

NN

E


Two Genes On One Two Genes On One ChromosomeChromosome

Father

Mother

e

n

E

N

E

N n

e

Replication

e

n

E

N

Genes linked together on the same chromosome should segregate together during meiosis.

Telophase I

n

e

n

e

NN

E E

Telophase II

E

N N

E

n

e

n

e


En

eN

en

EN

e ne NE nE N

Linked GenesLinked Genes

Sperm

en

N

E

Eggs e

nN

E

EeNnEeNNEENnEENN

EennEeNnEEnnEENn

eeNneeNNEeNnEeNN

eenneeNnEennEeNn

Because genes co-segregate, a 3:1 ratio results instead of the expected 9:3:3:1


Two Genes On One Chromosome Two Genes On One Chromosome With Crossing OverWith Crossing Over

Father

Mother

e

n

E

N

Telophase II

E

N N

e E

n n

e

E

N n

e

Replication

e

n

E

N

Prophase I

E

NN n

e

n

As long as genes on the same chromosome are located a long distance apart, they will assort independently due to crossing over during Prophase I of meiosis

Telophase I

N

eE

N

E

n

e

n


En

eN

en

EN

e ne NE nE N

With Crossing OverWith Crossing Over

EeNnEeNNEENnEENN

EennEeNnEEnnEENn

eeNneeNNEeNnEeNN

eenneeNnEennEeNn

As long as crossing over occurs between genes they will assort independently

Sperm

en

N

E

eN

n

E

Eggs e

nN

Ee

N

E

n


Still Not 9:3:3:1Still Not 9:3:3:1 Linked genes, unless they are far apart on the

chromosome, still do not exhibit the phenotypic ratio expected from independent assortment

Deviation from the expected ratio allows mapping of genes

The further apart two genes are, the more probable a crossover event is between them

CACloser to independent assortment

BATightly linked


Linkage In A Test CrossLinkage In A Test Cross

All F1 progeny should be phenotypically wild type

If a test cross was done crossing the F1 flies with homozygous ebony mahogany flies, the expected outcome would be:

e+e mah+mah

Imagine a situation where true breeding ebony body (e) mahogany eyes (mah) D. menanogaster are crossed with wild type flies:

e e mah mah X e+e mah+mah

Expected F1 would be:e e mah mah X e+e+mah+mah+

e mah e e mah mahe e mah+mahe+e mah mahe+e mah+mah

e mahe mah+e+mahe+mah+

1:1:1:1


Linkage In A Test CrossLinkage In A Test Cross If the actual numbers were: ebony mahogany 41

Wild type 42ebony 8mahogany 9

Linkage would naturally be suspected The ebony and mahogany flies must have resulted from crossing over As there are a total of 100 flies in the sample and 17 represent crossover events, these genes are said to be 17 map units (or centimorgans) apart

mah

17 mu

eChromosome


Mapping A Three Point CrossMapping A Three Point Cross Two point crosses do not tell us much about gene

order on chromosomes Three point crosses allow determination of both

gene sequence and distances between genes Imagine the following cross between an ebony

bodied, curly winged, cardinal eyed male and a female heterozygous for the same traits:

e+e cu+cu cd+cd X ee cucu cdcd Expected F1 would be:

e cu ro

e cu roe cu+ro+e+cu+roe+cu+ro+ e+cu roe+cu ro+ e cu+ roe cu ro+

ee cucucd cd

ee cu+cucd+cd

e+ecu+cucd cd

e+ecu+cucd+cd

e+ecucucd cd

e+ecucu cd+cd

ee cu+cucd cd

ee cucucd+cd


Mapping A Three Point CrossMapping A Three Point Cross

ebony curled cardinal

ebony cardinalebony curled

curled cardinalcurled

cardinalWild type

Phenotype

ebony378

102175

10119

375Observed

3125

125125125125125125

Expected

125




ebony cardinalebony curled

curled cardinalcurled

cardinalWild type

Phenotype

ebony378

103184

10119

375Observed

2

Rearrange in descending order of observance




ebony cardinal

ebony curledcurled cardinal

curledcardinal

Wild type

Phenotype

ebony

378

103

184

10119

375

Observed

2

Because crossing over is an infrequent event (and because we know the parents genotypes) the most commonly appearing class represent no crossing over

Double crossovers would be expected much less frequently than single crossovers. Thus the least frequently observed class must represent crossover between the gene in the middle and the two flanking genes

Intermediate classes represent single crossover events




ebony cardinal


curledcardinal

Wild type

Phenotype

ebony

378

103

184

10119

375

Observed

2

ecd cuWhich is exactly the same as:

e cdcuKnowing that ebony is in the middle allows construction of a tentative map




ebony cardinal


curledcardinal

Wild type

Phenotype

ebony

378

103

184

10119

375

Observed

2e cdcu

e cdcu

+ ++




ebony cardinal


curledcardinal

Wild type

Phenotype

ebony

378

103

184

10119

375

Observed

2e cdcu

e

cdcu

+

++




ebony cardinal


curledcardinal

Wild type

Phenotype

ebony

378

103

184

10119

375

Observed

26

e cdcu

e

cdcu

+

++




ebony cardinal


curledcardinal

Wild type

Phenotype

ebony

378

103

184

10119

375

Observed

2e cdcu

e cd

cu

+ +

+

6




ebony cardinal


curledcardinal

Wild type

Phenotype

ebony

378

103

184

10119

375

Observed

2e cdcu

e cd

cu

+ +

+

204

6

= 21

Double cross

Total

+Single cross = MU

1,000(204 + 6) x 100

21




ebony cardinal


curledcardinal

Wild type

Phenotype

ebony

378

103

184

10119

375

Observed

2e cdcu

e

cd

cu

+

+

6

+

Double cross

Total

+Single cross = MU

21




ebony cardinal


curledcardinal

Wild type

Phenotype

ebony

378

103

184

10119

375

Observed

2e cdcu

e

cd

cu

+

+

37

+

6

Double cross

Total

+Single cross = MU

1,000(37 + 6) x 100

4.3

= 4.3

21


Mapping In Haploid OrganismsMapping In Haploid Organisms Haploid organisms are, in some ways, easier to

work with because all genes impact phenotype On the other hand, they tend to be ugly smelly

things that provide many other challenges Fungi in the phylum Ascomycota are easiest to

work with as they show the order of division in their asci


Zygote

Aa

Meiosis In An AscusMeiosis In An Ascus

a

A

a

A


A

a


a

A

a

A

a

A

A

a


A

a

a

A


a

A

a

A

a

A

A

a


AA

AA

aa

aa


a

A

a

A

a

A

A

a


Meiosis In An Ascus:Meiosis In An Ascus:If Crossing Over OccursIf Crossing Over Occurs

A

Aa

a

A

aA

a


Aa

Aa


A

Aa

a

A

aA

a

a

A

A

a


Aa

Aa


A

Aa

a

A

aA

a

a

A

A

a


AA

aa

AA

aa


A

Aa

a

A

aA

a

a

A

A

a


AA

aa

AA

aa


A

Aa

a

A

a

a

AA

a

A

a

A 4:4 pattern in asci indicates no crossing over has occurred (First-Division Segregation)A 2:4:2 or 2:2:2:2 pattern indicates crossing over (Second-Division Segregation)



Problem 1Problem 1 In Drosophila, vermilion (v) is recessive to red (V) eyes and miniature (m)

wings are recessive to normal (M) wings. The following cross was made:

Male VVMM x vvmm Female

A What was the phenotype of the F1 generation?

B What F2 phenotypic ratio would you expect?

C If the actual F2 phenotypic numbers were:

– 147 red-eyed normal winged – 49 vermilion-eyed miniature winged, – 2 red-eyed miniature winged, – 2 vermilion-eyed normal winged,

How would you explain this?


Solution 1Solution 1AWhat was the phenotype of the F1 generation?

VVMM makes VM gametes

vvmm makes vm gametes

Thus the F1 must be VvMm

BWhat F2 phenotypic ratio would you expect?9 red-eyed normal winged (V_M_)

3 red-eyed miniature winged (V_mm)

3 vermilion-eyed normal winged (vvM_)

1 vermilion-eyed miniature winged (vvmm)


Solution 1 ContinuedSolution 1 ContinuedC If the actual F2 phenotypic numbers were:

– 147 red-eyed normal winged – 49 vermilion-eyed miniature winged, – 2 red-eyed miniature winged, – 2 vermilion-eyed normal winged,

How would you explain this?

mv

m+v+

0.49

0.01

0.49

0.01mv+

m+vmv

m+v+

mv+

m+v

F1 Gametes


Solution 1 ContinuedSolution 1 Continued

0.0001vvm+m+

0.0049vvm+m

0.0049vvm+m

0.2401vvmm

0.01vm+

0.49vm

0.49v+m+

0.01v+m

0.24 vvmm(0.24*200=48)

0.01vm+

0.49vm

0.49v+m+

0.01v+m

0.0049v+vmm

0.0049v+vmm

0.0001v+v+mm

0.0049v+vm+m+

0.2401v+vm+m

0.0001v+vm+m

0.0049v+vm+m+

0.0001v+vm+m

0.2401v+v+m+m+

0.0049v+v+m+m

0.0049v+v+m+m

0.2401v+vm+m

0.01 vvm+_(0.01*200=2)

0.01 v+_mm(0.01*200=2)

0.74 v+_m+_(0.74*200=148)

mv

m+v+

0.49

0.01

0.49

0.01mv+

m+vmv

m+v+

mv+

m+v


Solution 1 ContinuedSolution 1 Continued

Vermillion and miniature winged are closely linked genes on the same chromosome

The distance between vermilion and miniature is 1 centimorgan The reason numbers in the cross do not fit the prediction of 1

centimorgan exactly is that the numbers are the result of chance and thus would not be expected to fit the predicted ratio perfectly

mv

1cM

Documents

©2000 Timothy G. Standish John 15:4 4Abide in me, and I in you. As the branch cannot bear fruit of itself, except it abide in the vine; no more can ye,