Chapter 21 Genomics Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Chapter 21Genomics

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

A

1

Cytogenetic map:

Linkage map:

Physical map:

sc (1A8) w (3B6)

wsc

1.5 mu

wsc~ 2.4 x 106 bp

1 2 2 3 3 4

B C D E FA A AB B BC CC D DE EF F


Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Brooker, Figure 21.1

Obtained from analysis of polytene

chromosomes

The results from each type of mapping technique may be slightly different

Three types of maps associated with the Genome

Figure 21.2Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Sisterchromatids

Treat cellswith agentsthat makethem swelland fixesthem ontoslide.

DenaturedDNA (notin a double-helixform)

Add single-strandedDNA probes that havebiotin incorporatedinto them.

DenaturechromosomalDNA.

Hybridizedprobe

View with afluorescencemicroscope.

Fluorescentmoleculebound toprobe

Add fluorescentlylabeled avidin, whichbinds to biotin.

Fluorescence in situ hybridization

21 - 13


© : From Ried, T., Baldini, A., Rand, T.C., and Ward, D.C. "Simultaneous visualization of seven different DNA probes by in situ hybridization usingcombinatorial fluorescence and digital imaging microscopy. PNAS. 89: 4.1388-92. 1992. Courtesy Thomas Ried

Figure 21.4

EcoRI sites PRESENT on both

chromosomes

EcoRI sites ABSENT from both

chromosomes

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display21 - 17


Region of two homologous chromosomes from individual 1

EcoRI


EcoRI EcoRI EcoRI EcoRIEcoRI

EcoRIEcoRI EcoRI EcoRI EcoRIEcoRI

EcoRIEcoRI EcoRI EcoRIEcoRI


2000 bp 5000 bp 1500 bp 3000 bp 2500 bp

2000 bp 5000 bp 1500 bp 3000 bp 2500 bp

2000 bp 5000 bp 4500 bp 2500 bp

2000 bp 5000 bp 4500 bp 2500 bp

EcoRI site found only on one

chromosome

The three individuals share many DNA fragments that

are identical in size.Indeed, if these segments

are found in 99% of individuals in the population,

they are termed monomorphic


Figure 21.421 - 18

1 2 3

5000 bp

4500 bp

3000 bp

2000 bp

2500 bp

1500 bp

Polymorphicbands areindicated atthe arrows.

Separate the DNAfragments by gelelectrophoresis.

Cut the DNA fromall 3 individualswith EcoRI.


EcoRIEcoRI EcoRI EcoRI EcoRIEcoRI

2000 bp 5000 bp 1500 bp 3000 bp 2500 bp


2000 bp 5000 bp 4500 bp 2500 bp


Southern Blotting of RFLP

Figure not in Brooker

RFLPs (and other markers) can be mapped

Figure not in Brooker

Short Tandem Repeats (STR)

TTTTC = (TTTTC)1

TTTTCTTTTCTTTTC = (TTTTC)3

TTTTCTTTTCTTTTCTTTTCTTTTCTTTTC = (TTTTC)5

Cataloging the world’s SNP variation

www.hapmap.org

http://www.hapmap.org/

Gel electrophoresis

Many cycles of PCRproduce a large amountof the DNA fragmentcontained betweenthe 2 primers.

Add PCRprimers.

The PCRprimers specificallyrecognize sequenceson chromosome 2.

Set ofchromosomes

22

Figure 21.5

The two STS copies in this case are different in length.

Therefore, their microsatellites have different numbers of CA repeats

PCR of microsatellites

Figure 20.12 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

1 2 3 4 5bp

154

150

146

140

(b) Electrophoretic gel of PCR products for a polymorphicmicrosatellite found in the family in (a).

(a) Pedigree

Fra

gm

ent

len

gth

Parents

Offspring

1

3

2

4 5


The likelihood of linkage between two RFLPs is determined by the lod (logarithm of the odds) score method

– Computer programs analyze pooled data from a large number of pedigrees or crosses involving many RFLPs

– They determine probabilities that are used to calculate the lod score

• lod score = log10Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Probability of a certain degree of linkage

Probability of independent assortment

In human genetics, computer algorithms can be used to determine linkage

Figure 21.7


The numbers denote the order of the members of the contig

21 - 31

A B

A B F

F

GH

GH

C D

C

B C

D

D E

E F GH I

I

I

J

J

K

K

K L

L M

M

NO

NO

P

P

PQ R

Q R

Vector

Clone individual piecesinto vectors.

1 3 5

8642

7 9

10

A collection of overlappingclones, known as a contig

M

Physical Mapping

Figure 21.8Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

• Genes A and B had been mapped previously to specific regions of chromosome 11– Gene A was found in the insert of clone #2– Gene B was found in the insert of clone #7

• So Genes A and B can be used as genetic markers (i.e., reference points) to align the members of the contig

21 - 33

1.5 mu

Gene A

Region of chromosome 11

Gene A Gene B

Gene B

1 2 3 4 5 6 7


SelectableMarkergeneTEL

ORI (E.coliorigin ofreplication)

ORI

CEN(yeastcentromere)

CEN

Selectablemarkergene



TEL(yeast telomere)

TEL TEL

Yeastartificialchromosome(YAC)

Cut withEcoRI andBamHI.

Leftarm

Mix and add DNA ligase.

Note: This is not drawn to scale. The chromosomalDNA is much larger than the YAC vector.

Chromosomal DNA

Rightarm

Fragment notneeded in yeast

Cut(occasionally)with a low concentrationof EcoRI to yield verylarge fragments.

EcoRI site

ARS(yeast originof replication)

ARS Large piece of chromosomal DNA

BamHIsite

BamHIsite

+

+


Figure 21.9

EcoRI is used at low

concentrations so only some sites

are digested

Each arm has a different

selectable marker.

Therefore, it is possible to

select for yeast cells with YACs that have both

arms

21 - 36

21 - 37

lacZ

HindIII BamHI SphI

BACvector

parC

parB

parA

oriS

repE

cm R


Figure 21.1121 - 39

Chromosome 1695 million bp

Lo

w r

eso

lutio

n(3

–5

mill

ion

bp

)

Hig

h r

eso

lutio

n(1

–1

00

,00

0 b

p)

p q

13

.2

13

.21

3.1

31

3.1

21

3.1

11

2.3

12

.2

12

.2

12

.1

12

.1

11.2

11.2

11.1

13

.0

21

.0

22

.12

2.2

22

.32

3.1

23

.22

3.3

24

.12

4.2

24

.3

D1

6S

85

D1

6S

60

D1

6S

15

9

D1

6S

48

D1

6S

15

0D

16

S1

49

D1

6S

16

0

D1

6S

40

D1

6S

14

4

500 times expansion

219 bp

Cytogenetic map(resolution of in situhybridization 3–5megabases)

Physical map ofoverlapping cosmidclones (resolution5–10 kilobases)

Sequence-tagged site(resolution 1 base)

YAC N16Y1150,000 bp

STS N16Y1-10

Primer

Primer

3′

5′

5′

*

*

GATCAAGGCGTTACATGA

5′—GATCAAGGCGTTACATGA — 3′

CTAGTTCCGCAATGTACT

Cosmidcontig 211

310C4

N16Y1-29

N16Y1-18

N16Y1-10

N16Y1-19

309G11

312F1

5F3

N16Y1-30

N16Y1-16

N16Y1-12

N16Y1-14

N16Y1-13

= (GT)n

Linkage map(resolution 3–5 cM;not all linkagemarkers are shown)


D1

6A

C6

.5

3′

5′AGTCAAACGTTTCCGGCCTA3′

TCAGTTTGCAAAGGCCGGAT

AGTCAAACGTTTCCGGCCTA

*

*

1 2 3 4 5

Numbers indicateregions that aresubcloned.

(Starting clone)

Subclone.

Cosmidvector

Subclone.

Screen a library.

Screen a library.

(Third clone)

Repeat subcloning andscreening until gene Ais reached.

(Second clone)

....n

Gene B

Gene B

Gene A

Gene A


1 2

2

2 3

n

Figure 20.18

The number of steps required to reach the gene of interest

depends on the distance between the start and end

points

21 - 42

Figure 21.14 21 - 46

ChromosomalDNA

ChromosomalDNA

BACvector

BAC contig

Clones fromone BAC insert:

Vector

Vector

Clone large chromosomalDNA fragments into BACs andcreate a contig for eachchromosome.

Shear DNA into small andlarge pieces. Clonechromosomal DNA piecesinto vectors.

For each BAC, shear intosmaller pieces and clone DNApieces into vectors.

From the clones of each BAC,determine the chromosomalDNA sequence, usually atone end, by shotgunsequencing. The resultsbelow show the sequencesfrom three chromosomal DNAclones.

Based on overlappingregions, create onecontiguous sequence.

ChromosomalDNA

(a) Hierarchical genome shotgun sequencing (b) Whole-genome shotgun sequencing

CCG A C C T T A CCG A C C A

C T T A CC C GA C C GA C CA C CC GA T T A A T C GC GA A T T G

GA CC A CCCGA T T A A TT T A A T CGCGA A T T G

Determine the chromosomalDNA sequence, usually atboth ends, by shotgunsequencing. The resultsbelow show sequences ofthree chromosomal DNAclones.

Based on overlappingregions, create onecontiguous sequence.

T T A CC GGT A GGC A C C T

GGT A GT T A CC GC A C C T G T T A CG GGT C A A A CC T A GG

C A CC T GT T A CGGGT CGGGT CA A A CC T A GG

IsolatechromosomalDNA

IsolatechromosomalDNA


21 - 48

Emulsify the beads so there is onlyone bead per droplet. The dropletsalso contain PCR reagents thatamplify the DNA.

Deposit the beads into a picotiterplate. Only one bead can fit intoeach well.

Add sequencing reagents:DNA polymerase, primers,ATP sulfurylase, luciferase,apyrase, adenosine 5′phosphosulfate, and luciferin.Sequentially flow solutionscontaining A, T, G, or C into thewells. In the example below, Thas been added to the wells.

PPi (pyrophosphate) is releasedwhen T is incorporated into thegrowing strand.

Isolate genomic DNAand break into fragments.

Covalently attach oligonucleotideadaptors to the 5′ and 3′ ends ofthe DNA.

Denature the DNA into singlestrands and attach to beads viathe adaptors. Note: Only one DNAstrand is attached to a bead.

Fragment ofgenomic DNA

Adaptors

PPi + Adenosine 5′phosphosulfate

ATP + luciferin

ATP sulfurylase

Light

Light is detected by a camerain the sequencing machine.

Luciferase


CA

TG

CA

T T

Primer

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Chapter 21 Genomics Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display