Linkage

• Genes linked on the same chromosome may segregate together.

A

b

Independent Assortment2 chromosomes

a

A A

B

B b a B a b

P: AABB x aabb

Parental Gametes: AB & ab

F1

P PR R

MeiosisOne Chromosome No Cross Over Parent Cell

Daughter Cells Have Parental Chromosomes

A a

B b

A

B

a

b

a

b

A

B

2n = 2

MeiosisOne Chromosome With Cross Over Parent Cell

Daughter Cells Have Recombinant Chromosomes

A a

B b

A

B

A

b

a

b

a

B

2n = 2

Meiosis Prophase I

If genes are linked, crossing over must occur for there to be recombination.

A aA a

B B b b

a A

Linkage?

P: AABB x aabb

F1: AaBb

Test cross: AaBb x aabb

1/4: AaBb

1/4: Aabb

1/4: aaBb

1/4: aabb

P: AABB x aabb

F1: AaBb

Test cross: AaBb x aabb

?: AaBb

?: Aabb

?: aaBb

?: aabb

no linkage linkage

Recombination Frequency

…or Linkage Ratio: the percentage of recombinant types,

– if 50%, then the genes are not linked,

– if less than 50%, then linkage is observed.

Linkage

• Genes located on the same chromosome do not recombine,

– unless crossing over occurs,

• The recombination frequency gives an estimate of the distance between the genes.

Recombination Frequencies

• Genes that are adjacent have a recombination frequency near 0%,

• Genes that are very far apart on a chromosome have a linkage ratio of 50%,

• The relative distance between linked genes influences the amount of recombination observed.

A C

In this example, there is a 4/10 chance of recombination.

A B

In this example, there is a 2/10 chance of recombination.

a b

a c

Linkage Ratio

P GGWW x ggww

Testcross F1: GgWw x ggww

recombinant

total progeny

GW Gw gW gw

? ? ? ?

= Linkage Ratio

Linkage Ratio Units

% = mu (map units)

- or -

% = cm (centimorgan)

A C

In this example, there is a 4/10 chance of recombination.

A B

In this example, there is a 2/10 chance of recombination.

a b

a c

cis“coupling”

trans“repulsion”

Fly Crosses(white eyes, minature, yellow body)

• In a white eyes x miniature cross, 900 of the 2,441 progeny were recombinant, yielding a map distance of 36.9 mu,

• In a separate white eyes x yellow body cross, 11 of 2,205 progeny were recombinant, yielding a map distance of 0.5 mu,

• When a miniature x yellow body cross was performed, 650 of 1706 flies were recombinant, yielding a map distance of 38 mu.

Study Figs 4.2, 4.3, and 4.5

Simple Mapping

• white eyes x miniature = 36.9 mu,

• white eyes x yellow body = 0.5 mu,

• miniature x yellow body = 38 mu,

my

38 mu

36.9 mu

w

0.5 mu

Do We have to Learn More Mapping Techniques?

• Yes, – three point mapping,

• Why,– Certainty of Gene Order,– Double crossovers.

Gene Order

• It is often difficult to assign the order of genes based on two-point crosses due to uncertainty derived from sampling error.

A x B = 37.8 mu,A x C = 0.5 mu,B x C = 37.6 mu,

Double Crossovers

• More than one crossover event can occur in a single tetrad between non-sister chromatids,

– if recombination occurs between genes A and B 30% of the time,

• (p = 0.3),

• then the probability of the event occurring twice is 0.3 x 0.3 = 0.09, or nearly 10 map units.

• If there is a double cross over, does recombination occur?

– how does it affect our estimation of distance between genes?

Why Me? Why Map?

• Over 4000 human diseases have a genetic component,

– knowing the protein produced at specific loci facilitates the treatment and testing,

• Facilitates both classical and molecular analysis of organisms.

Classical Mapping

Cross an organism with a trait of interest to homozygous mutants of known mapped genes.

Then, determine if segregation is random in the F2 generation,

• if not, then your gene is linked (close) to the known mapped gene.

target

What recombination frequency do you expect beteen the target and HY2?

What recombination frequency do you expect beteen the target and TT2?

Three Point Testcross

Triple Heterozygous

(AaBbCc )

x

Triple Homozygous Recessive

(aabbcc)

Three Point Mapping Requirements

• The genotype of the organism producing the gametes must be heterozygous at all three loci,

• You have to be able to deduce the genotype of the gamete by looking at the phenotype of the offspring,

• You must look at enough offspring so that all crossover classes are represented.

Representing linked genes...

W G D w g d

x

w g dw g d

w g d

P

Testcross

= WwGgDd

= wwggdd

Representing linked genes...

+ + + w g d

x

w g dw g d

w g d

P

Testcross

= WwGgDd

= wwggdd

Phenotypic Classes

W-

ww

G-

gg

G-

gg

D-

dd

D-

dd

D-

dd

D-

dd

W-G-D-

W-G-dd

W-gg-D

W-gg-dd

wwG-D-

wwG-dd

wwggD-

wwggdd

W-G-D-

W-G-dd

W-gg-D

W-gg-dd

wwG-D-

wwG-dd

wwggD-

Crossovers

0

W G D

w g d1

1

2

#

179

52

46

4

22

22

2

wwggdd 173 0

1

1

2

W-G-D-

W-G-dd

W-gg-D

W-gg-dd

wwG-D-

wwG-dd

wwggD-

#

179

52

46

4

22

22

2

wwggdd 173Parentals

Recombinants,double crossover

Recombinants 1 crossover, Region I

Recombinants 1 crossover, Region II

W G D

w g d

III

W-G-D-

W-G-dd

W-gg-D

W-gg-dd

wwG-D-

wwG-dd

wwggD-

#

179

52

46

4

22

22

2

wwggdd 173Parentals

Recombinants,double crossover

Recombinants 1 crossover, Region I

Recombinants 1 crossover, Region II

W G D

w g d

I

Total = 500

Region I:

46 + 52 + 2 + 4

500x 100

= 20.8 mu

W-G-D-

W-G-dd

W-gg-D

W-gg-dd

wwG-D-

wwG-dd

wwggD-

#

179

52

46

4

22

22

2

wwggdd 173Parentals

Recombinants,double crossover

Recombinants 1 crossover, Region I

Recombinants 1 crossover, Region II

W G D

w g d

II

Total = 500

Region II:

22 + 22 + 2 + 4

500x 100

= 10.0 mu

20.8 mu

W G D

w g d

W-gg-D

wwG-dd 4

2Recombinants,double crossover

Total = 500

10.0 mu 20.8 mu

0.1 x 0.208 = 0.0208

6/500 = 0.012

NO GOOD!

Interference

…the affect a crossing over event has on a second crossing over event in an adjacent region of the chromatid,

– (positive) interference: decreases the probability of a second crossing over,

• most common in eukaryotes,

– negative interference: increases the probability of a second crossing over.

Gene Order in Three Point Crosses

• Find either double cross-over phenotype, based on the recombination frequencies,

• Two parental alleles, and one cross over allele will be present,

• The cross over allele fits in the middle...

–

#

2001

52

46

589

990

887

600

1786

Which one is the odd one?

A C B

a c b

II I

A-B-C-

A-B-cc

A-bb-C-

A-bb cc

aaB-C-

aaB-cc

aabbC-

aabbcc

A-B-C-

A-B-cc

A-bb-C-

A-bb cc

aaB-C-

aaB-cc

aabbC-

#

2001

52

46

589

990

887

600

aabbcc 1786

Region I

A C B

a c b

I

990 + 887 + 52 + 46

6951x 100

= 28.4 mu

A-B-C-

A-B-cc

A-bb-C-

A-bb cc

aaB-C-

aaB-cc

aabbC-

#

2001

52

46

589

990

887

600

aabbcc 1786

Region II

A C B

a c b

28.4 mu

600 + 589 + 52 + 46

6951x 100

= 18.5 mu

II18.5 mu

Today

• Coefficient of Confidence,

• Gene mapping in humans,

• Problems, problems, problems,

– Be sure to at least try them before Friday.