Linkage and Recombination

Linkage and Recombination

Genes are on chromosomes

Alleles segregate

Alleles for different genes assort independently

Pea plants have 7 pairs of chromosomes. Flies have 4 pairs. Humans have 23 pairs.

Mendel

Morgan and Bridges

Bateson et al.

What is the inheritance pattern for two traits determined by genes on the same chromosome?

“Linked” genes are those that do NOT assort independently.(They reside on the same chromosome.)

We will see that...

DMD

SCIDXIST

XLP

Fragile XHaemophilia

R-G Colorblindness

Human X chromosome

RP3

HPRT

PGKThe corollary isn’t always true!Just because genes are on the same chromosome doesn’t necessarily mean they are linked.

As a student in Morgan’s lab, Sturtevant was the first person to correctly interpret linkage; he created the first genetic maps.

Sturtevant in his lab at Caltech

Alfred Sturtevant

Mapping Genes on Chromosomes

Segregation of genes on the same chromosome

wild typepr+ pr+ vg+

vg+

purple eyesvestigial wings

pr pr vg vg

gamete? gamete?wild type

phenotypepr+ pr vg+ vg

x

testcross

[ ]

testcross parent genotype?

pr+ vg+

pr+ vg

pr vg+ pr vgphenotypes:

For I.A., expect: 1 : 1 : 1 : 1 ratio

observed: 1339 : 151 : 154 : 1195

pr pr vg vg

Sturtevant’s interpretation:

parental type(same as gametes that made the parent)mostl

y

recombinant (non-parental) type

occasional

crossover

wild typepr+ pr+ vg+

vg+

purple eyesvestigial wings

pr pr vg vg

gamete? gamete?wild type

phenotypepr+ pr vg+ vg

x

The genes are on the same chromosome & the parental alleles (mostly) co-segregate

How do we know the genotypes of the gametes?

Testcross parent’s gametes… only recessive alleles present

So testcross progeny phenotype allows us to deduce the heterozygous parent’s gamete genotype

sister chromatids(each is a double helix)

homologues

cohesins

synaptonemal complex

Recombination… a brief review

[see lecture 2 for details]

Single-strand nicks

strand exchange

Holliday junction -No deletions are

caused by recombination.-No mutations are caused by recombination.

Recombination… a brief review (cont)

ALWAYS pull from the centromeres

Segregation of recombinant chromosomes during Meiosis I

3

4

pr +

prvg

vg

1

2

prpr

+

vg +

vg +

1

4

32

Label the chromatids

Telophase I Gametes

Parental

Parental

Recombinant

Recombinant

Telophase II

One recombination event: 2 recombinant and 2 non-recombinant products

Recombination… a brief review (cont)

Which chromatid goes where?

An example with a double crossover:

Test your Understanding

Draw these chromosomes in anaphase I.

Label all alleles appropriately.

ALWAYS pull from the centromeres

3

4

pr +

pr

vg

vg+

1

2

prpr

+

vg +

vg

Other types of crossovers

# xovers resulting gametes

0 parental

parental!2(2 strands)

2 parental2 non-par.

2(3 strands)

4 non-par.

2(4 strands)

Crossing over must occur for faithful segregation at meiosis I

Sturtevant’s findings—summary

Genes on the same chromosome can show linkage instead of independent assortment

Gametes (mostly) have the same allele combinations as the homologs in the parent

Recombination can give rise to gametes with non-parental (=recombinant) allele combinations

Two parental types are more abundant and occur at roughly equal frequency to one another

Non-parental types are less abundant and occur at roughly equal frequency to each other

Identifying the parental type

Option 1. Know the gametes that made the heterozygous parent

pr+ pr+ vg+ vg+

pr pr vg vg

pr+ vg+ pr vggametes:

pr+ pr vg+ vg

pr+ vg pr

vg+gametes:

dominant alleles together on the same chromosome = “cis” configuration

dominant alleles on different chromosomes = “trans” configuration

define the parental type

X

pr+ pr+ vg vg pr pr vg+ vg+X

pr+ vg+

pr vgpr+ vgpr

vg+

The cross from our previous example:

A different cross from our example:

Identifying the parental type

Option 2. The two most abundant progeny types

(assuming the genes are linked)

1287 1204

170 154

Cross: pr+ pr vg+ vg x pr pr vg vg

Progeny:

What were the gametes that made the heterozygous parent? pr+ vg pr

vg+

=

=

pr+ prvg+ vg implie

s

Doesn’t make se

nse!???

A notation system that defines allele configuration

linkage

Configuration(i.e., cis or trans)

separate chromosomes

pr+

pr;

w+

w

Sturtevant’s interpretation of linkage

-Recombination involves the physical exchange of chromosomal segments between homologs.

-The frequency of recombinant types indicates the distance between linked genes.

What is the evidence in support of these

hypotheses?

Test of Sturtevant’s hypothesis

Harriet Creighton and Barbara McClintock, maize Curt Stern, Drosophila

The problem: homologous chromosomes look alike… how to tell if they really exchanged segments?

found genetically marked chromosome with structurally (visually) distinct homologues

Prediction: Chromosome from genetically recombinant plants should show structural rearrangement

Creighton/McClintock test Sturtevant’s hypothesis

C = colored Wx = starchyc = colorless wx = waxy“knob

”translocation

x

Progeny phenotypescolored, waxy

colorless, starchy

colored, starchy

colorless, waxy

Conclusion: genetic recombination exchange of chromosome segments

Structural markers: Genetic markers:

recombinant genotypes and chromosomes!

Mapping genes using recombination

Alfred Sturtevant’s major insight:If crossovers occur at random

Probability of crossover between two genes is proportional to the distance between them

Crossover between A and B much more likely than between B and D

Map distance… example from last time

1287 1204

170 154

Cross: pr+ pr vg+ vg x pr pr vg vg

Progeny:

What is the map distance between these two genes?

non-parental types (least abundant)

parental types (most abundant)

1287 1204

170 154

Step 1. Identify the parental and non-parental types.

parental =

non-parental =

pr+ vg and pr vg+

pr+ vg+ and pr vg

Map distance… example (cont’d)

non-parental types (least abundant)

parental types (most abundant)

1287 1204

170 154

Step 2. Calculate % recombinant products.

% recombinant =

(170 + 154)

(170+154+1287+1204)

x100= 11.5%

Map distance =

11.5 map units = 11.5 centiMorgan’s (cM)

Map distance… example (cont’d)

Number Recombinant Types

Total Number of Progeny

Another Example (recessive b mutation)

Which are the parental types? and the recombinant types?

Recombination Frequency =

4615

red blac

k

4707purple

tan

307red tan

295purpl

e black

pr b Example:

Recombination Frequency =

Total Number of Progeny

pr b Example:307 + 295

4615 + 4707 + 307 + 295= 0.06

Parental types Recombinant types

pr + b pr b + pr + b + pr b

= 6.0 %x 100602

9924

4615

red blac

k

4707purple

tan

307red tan

295purpl

e black

Number Recombinant Types

Another Example (recessive b mutation)

Genetic loci: % Recombinants

Alfred Sturtevant’s mental leap:

11.5 %

% recombinants is directly proportional to distance

purple - vestigial (pr - vg)

6.0 %purple - black (pr - b)16.5 %vestigial - black (vg - b)

He drew a genetic map of the second chromosome:

vgpr

11.5 cM

b

6 cM

Creating a Genetic Map

Revealed from other

crosses

Why 17.5cM and not 16.5cM?

b+

b vg

vg+

pr

pr+

Point to ponder:If you examined a population of meioses, what is the maximum recombination frequency you could see between two genes?

Genes very close together…

Low probability of crossover between them

…very few recombinants

“tight linkage”

Genes further apart… more recombinants

* Crossing-over creates new combinations of traits.

* The frequency of recombinant types indicates the distance between linked

genes.

* Two Parental types in ≈ frequencies.Two Recombinant types ≈ frequencies.

* If genes are linked,Parental types > recombinant

types.

Summary

Practice questionThe pedigree shows segregation of two disorders

one is autosomal dominant (A= disease, a = not) one is autosomal recessive (b = disease, B = not).

I

II

III

1 2

1 2

1

= autosomal dominant= autosomal recessive= both traits

Is the gamete that III-1 received from II-2 parental or non-parental?

BUT FIRST… break down the question:

Talk to your neighbors and devise a systematic, step-by-step strategy to solve the problem

Step 1.

Step 2.

Step 3.

etc.

The pedigree shows segregation of two disordersone is autosomal dominant (A= disease, a = not)

one is autosomal recessive (b = disease, B = not).

I

II

III

1 2

1 2

1

= autosomal dominant= autosomal recessive= both traits


BUT FIRST… break down the question:

Talk to your neighbors and devise a systematic, step-by-step strategy to solve the problem

Step 1. Figure out all the genotypes!

Step 2. What are the gametes that made II-2?

Step 3. What is the gamete that II-2 made?

Step 4. Does the gamete that II-2 made have a different genotype than the gamete(s) that made him?

Practice question

I

II

III

1 2

1 2

1


a ba b

A B

? ?

AB

A B

a b

A b

a b

a B

a b

ab Ab

ab gamete

Gametes that made II-2=

Gamete that II-2 gave to III-1=

AB and ab

Ab= autosomal dominant= autosomal recessive= both traits

The pedigree shows segregation of two disordersone is autosomal dominant (A= disease, a = not)

one is autosomal recessive (b = disease, B = not).

Quiz Section this week:

yeast tetrad analysis (genetic mapping)

An introduction to the yeast tetrad analysis terminology. . .

In yeast…

Tetrad of spores… can examine products of individual meioses!

Tetrad with only the 2 parental types = “parental ditype” (PD).

Tetrad with only the 2 recomb. types = “non-parental ditype” (NPD).Tetrad with all four combinations = “tetratype” (T).

Looking at whole tetrads (PD/NPD/T) is informative.

non-parental

4 haploid spores

1n

meiosis2n

diploid

Example

Suppose URA1 and URA2 are on the same chromosome…

Diploid SporesGrowth on -ura plates?

no

no

no

yes

Parental ditype?Non-parental ditype?Tetratype?

Exercise

Suppose URA1 and URA2 are on separate chromosomes, but each tightly linked to their respective centromeres…

what kinds of tetrads (growth on -ura plates) would this diploid produce?

1+ = URA1

1- =ura1

2+ =URA2

2- =ura2

1+ 2-

1+ 2-

1- 2+

1- 2+

1- 2-

1- 2-

1+ 2+

1+ 2+

and

PD NPD

Using 2 analysis to explore linkage

P1 cross:

F1 :All long, no speck

Xlong, no speck

vestigial, speck

Test cross

Long, no speck: 2929Vestigial, speck: 2921

Vestigial, no speck: 2070Long, speck: 2080

Can these results be explained by chance deviation from independent assortment?

Is this really a 1:1:1:1 ratio as we would expect for independent assortment? Is the deviation from independent assortment due to chance?

2 analysis: Test the “null” hypothesis—that the observed deviation from 1:1:1:1 segregation is due to chance variation.

Why test the null hypothesis?It gives a precise expectation (1:1:1:1)

We cannot test directly for linkage, because (assuming they are linked) we do not know the map distance separating these genes.

Using 2 analysis to explore linkage

Out of 10,000 testcross progeny . . .expectedobserved

long no speck 2929

(o-e)2 (o-e)2

e

(Obs-Exp)2

Exp2 = •

2 analysis

vestigial speck 2921vestigial no speck 2070long speck 2080

P

df123456

0.995

0.0000.0100.0720.2070.4120.676

0.975

0.0000.0510.2160.4840.8311.237

0.900

0.0160.2110.5841.0641.6102.204

0.500

0.4551.3862.3663.3574.3515.348

0.100

2.7064.6056.2517.7799.236

10.645

0.050

3.8415.9917.8159.488

11.07012.592

0.025

5.0247.3789.348

11.14312.83214.449

0.010

6.6359.210

11.34513.27715.08616.912

0.005

7.87910.59712.83814.86016.75018.548

2 table

What does this P value mean?REJECT

NO WAY could these numbers be due to

independent assortment!

What is the P value?

Degrees of freedom?

P1 cross:

F1 :

All long, no speck

Xlong, no speck

vestigial, speck

Test whether the data truly show linkage by doing a Chi-square analysis. What is the null hypothesis?

speck

Draw the chromosomes of the F1 hybrid in prophase of Meiosis I such that the gametes will produce all four progeny types.

Homework

Practice question

Out of 100 testcross progeny . . .

expectedobserved

long no speck 29

(o-e)2 (o-e)2

e

vestigial speck 29

vestigial no speck 21

long speck 21

What does this P value mean?

What is the P value?

What if Sturtevant had analyzed only 100 testcross progeny?

Practice question

Out of 100 testcross progeny . . .expectedobserved

long no speck 29

(o-e)2 (o-e)2

e

= 2.56

(Obs-Exp)2

Exp2 = •

vestigial speck 29vestigial no speck 21long speck 21

25

25

25

25

42

42

42

42

= 0.64

= 0.64

= 0.64

= 0.64

2 analysis The testcross data do

not significantly differ from a 1:1:1:1 ratio.

The “null” hypothesis:

P

df123456

0.995

0.0000.0100.0720.2070.4120.676

0.975

0.0000.0510.2160.4840.8311.237

0.900

0.0160.2110.5841.0641.6102.204

0.500

0.4551.3862.3663.3574.3515.348

0.100

2.7064.6056.2517.7799.23610.645

0.050

3.8415.9917.8159.48811.07012.592

0.025

5.0247.3789.34811.14312.83214.449

0.010

6.6359.21011.34513.27715.08616.912

0.005

7.87910.59712.83814.86016.75018.548

2 table

What does this P value mean?

Can’t Reject

Yes, these numbers could be due to independent assortment!

Degrees of freedom?

Since Sturtevant DID analyze more flies, however, we are certain that vg and sp are linked.

41.5 %

Genetic loci: % Recombinants11.5 %purple - vestigial (pr - vg)

vestigial - speck (vg - sp)


vgpr

11.5 cM sp?41.5 cM

b

6 cMsp?41.5 cM

If to the left, fewer recombinants with pr (and b). If to the right,more recombinants with pr (and b).

Where does speck map?

How do we distinguish these possibilities?

P1 cross:

X

2504

F1 :F1 dihybrid

2498 2501 2497

Parental types Recombinant types

testcross parent

X

Testcross progeny:

redno speck

purplespeck

purpleno speck

redspeck

pr+ sp +

pr+ sp+

pr sppr sp

Genetic loci: % Recombinants11.5 %

41.5 %

purple - vestigial (pr - vg)

vestigial - speck (vg - sp)


vgpr

10.7 cM sp41.5 cM

b

6 cM

Where shall we place speck?

What is the map distance between b and sp?~58.2 cM

Documents

Linkage and Recombination