slide1
Download
Skip this Video
Download Presentation
Dr. Chris Rongo [email protected] GENETICS 502 C. elegans Section, Lecture 2 10:00-11:25, Waksman Auditorium Tu

Loading in 2 Seconds...

play fullscreen
1 / 46

Dr. Chris Rongo [email protected] GENETICS 502 C. elegans Section, Lecture 2 10:00-11:25, Waksman Auditorium Tu - PowerPoint PPT Presentation


  • 142 Views
  • Uploaded on

Dr. Chris Rongo [email protected] GENETICS 502 C. elegans Section, Lecture 2 10:00-11:25, Waksman Auditorium Tuesday, March 3, Spring 2009 Today’s Lecture: Complementation Linkage Analysis Two-factor Mapping Three-factor Mapping. Genetic dissection of biological.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Dr. Chris Rongo [email protected] GENETICS 502 C. elegans Section, Lecture 2 10:00-11:25, Waksman Auditorium Tu' - siusan


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
slide1
Dr. Chris Rongo

[email protected]

GENETICS 502

C. elegans Section, Lecture 2

10:00-11:25, Waksman Auditorium

Tuesday, March 3, Spring 2009

Today’s Lecture: Complementation

Linkage Analysis

Two-factor Mapping

Three-factor Mapping

slide2
Genetic dissection of biological

processes

Wild type animal

WT

WT

WT

mutant

WT

WT

WT

WT

1) Identify a process to study

GOAL:

2) Mutagenize

To obtain mutants that are defective in a specific biological process.

Mutant screens allow researchers to identify the molecular players that underlie a given biological process. By discovering these players, researchers obtain an experimental handle on

the gene or process.

3) Screen for a phenotype

4) Define genes by complementation testing

5) Map gene

6) Clone gene

slide3
Complementation tests

Q: When do you use this?

Note: requires recessive mutations

mut-1

Complementation: the production of a

wild-type phenotype when two different

mutations are combined in trans.

The two mutations are probably in

different genes.

WT phenotype

mut-3

mut-1

Noncomplementation

or failure to complement

between two mutations usually indicates

that the two mutations are in the same gene.

Mut phenotype

mut-2

slide4
Is the new unc* mutation an allele of the gene unc-44?

unc*

+

P0

X

unc*

+

unc-44(e1111)

unc*

F1

X

+

unc-44(e1111)

unc*

+

F2

unc*

+

unc-44(e1111)

unc-44(e1111)

unc-44(e1111)

unc-44(e1111)

Unc or WT?

Unc or WT?

WT

WT

50%

If unc* is an allele of unc-44 _____% of progeny are Unc.

If unc* complements unc-44, it is probably not a mutation in the same gene.

Q: What do you do next if unc* fails to complement unc-44? What molecular experiment could you perform to determine whether it is a new allele of unc-44?

slide5
a

Complex Complementation Tests

Intragenic complementation: the production of a wild-type phenotype

when two mutations in the same gene are combined in trans.

Easy to explain for proteins that have more than one functional domain:

Mutation 1 in Protein AB

Alone, they lack

either

A

or B activity

.

A

B

Mutation 2 in Protein AB

Combined in trans,

they have both

A

and

A

B

B activity

.

Keep in mind that there are other ways that intragenic complementation

can occur (polycistronic messages, alternative splicing, et cetera.)

slide6
Complex Complementation Tests

Non-allelic non-complementation: the production of a mutant phenotype

when two mutations in two different genes are combined in trans.

Two recessive mutations involved in embryogenesis: emb-1 & emb-4.

emb-1 maps to chromosome I

emb-4 maps to chromosome III

GENOTYPE PHENOTYPE

emb-1/emb-1 ; +/+ mutant

emb-1/+ ; +/+ WT

+/+ ; emb-4/emb-4 mutant

+/+ ; emb-4/+ WT

Seems normal for two recessives, but…

emb-1/+ ; emb-4/+ mutant

A wild-type copy of each gene is present,

yet the animal still has a mutant phenotype.

slide7
“Threshold Levels” model for

Non-allelic Non-Complementation

Perhaps EMB-1 and EMB-4 dimerize to form a functional protein,

and the mutations make non-functional subunits.

+/+ ; +/+ 100% Level of Heterodimer

emb-1/+ ; +/+ 50% Level of Heterodimer

+/+ ; emb-4/+ 50% Level of Heterodimer

emb-1/+ ; emb-4/+ 25% Level of Heterodimer

If the threshold limit for the level of the heterodimer in a cell is 50%

before you see a phenotype, then any genotype combination that

makes less than 50% will show a mutant phenotype.

Heterozygosity for either gene alone gives 50%, which is enough.

Heterozygosity for both genes gives 25%, which is

not enough, so mutant phenotype appears.

slide8
“Poison” model for

Non-allelic Non-Complementation

Perhaps EMB-1 and EMB-4 dimerize to form a functional protein, and

the mutant proteins, when together, form a nonfunctional protein that can

interfere with the function of wild-type dimers.

+/+ ; +/+ no Poisoned Heterodimers

emb-1/+ ; +/+ no Poisoned Heterodimers

+/+ ; emb-4/+ no Poisoned Heterodimers

emb-1/+ ; emb-4/+ 25% are Poisoned Heterodimers

For example, perhaps EMB-1/EMB-4 heterodimers

are subunits in a polymer chain.

A poisoned heterodimer might be able to add itself

to the end of the chain, but not allow additional

subunits to be added (i.e., chain terminator).

slide9
Linkage analysis in C. elegans

Q: Why do geneticists map genes?

How to begin mapping:

  • Cross your mutation with
  • known markers on the six
  • chromosomes.
  • Genes on C. elegans
  • chromosomes tend to be in
  • clusters, making the
  • assignment of a new
  • gene to a chromosome
  • easier.
mendel s first law the law of segregation
a+

a+

a-

a-

Mendel’s First Law: The Law of Segregation

Alternative versions (alleles) of genes account for variation during heredity.

For each inherited characteristic (gene), an organism inherits two alleles, one from each parent.

If two alleles differ, then the dominant allele is fully expressed for the characteristic, whereas the recessive is not expressed.

Two alleles for a given gene segregate during gamete production.

1. Alleles are on homologous chromosomes.

2. They line up at metaphase.

3. Each homolog goes into a different gamete.

4. You are drawing Metaphase I every time you write a genotype.

slide11
a

a

a

a

If two genes are closely linked...

unc-5 +

+ mut

P0

X

unc-5 +

+ mut

unc-5 +

F1

+ mut

unc-5 +

unc-5 +

+ mut

F2

+ mut

+ mut

unc-5 +

The mutations repel or exclude each other in the F2 generation:

1/__ are Unc non Mut

1/__ are Mut non Unc

4 4

Question: Do you ever get worms that are both Unc and Mut?

If perfectly linked, then never. If closely linked, then rarely (recombinants)

punnett square for closely linked genes
Punnett Square For Closely Linked Genes

F1: unc-5/mut

Gametes From Sperm

Gametes

From

Oocytes

what if the genes are unlinked
What if the genes are unlinked?

More help from the 19th century:

“During gamete formation the segregation of the alleles

of one gene is independent of segregation of the alleles

of another gene.” - Gregor Mendel, 2nd Law

20th century update:

“This is true IF the genes are on separate chromosomes or

are far apart on the same chromosome.”

- Barbara McClintock, Thomas

Hunt Morgan and others.

slide14
If two genes are unlinked...

+

mut

unc-5

+

X

;

;

P0

+

mut

unc-5

+

unc-5

+

F1

;

Self fertilize

+

mut

1 +/+ ; +/+

F2

2 +/+ ; +/mut

9 nonUnc nonMut

2 unc-5/+ ; +/+

4 unc-5/+ ; +/mut

1 +/+ ; mut/mut

3 Mut nonUnc

2 unc-5/+ ; mut/mut

1 unc-5/unc-5 ; +/+

3 Unc nonMut

2 unc-5/unc-5 ; +/mut

1 Unc Mut

1 unc-5/unc-5 ; mut/mut

a

a

a

punnett square for unlinked genes
Punnett Square For Unlinked Genes

F1: unc-5/ + ; mut/ +

Gametes From Sperm

Gametes

From

Oocytes

slide16
unc-5

+

F1

;

Self fertilize

+

mut

If you pick Unc F2s, what % are also Mut?

25%

If you pick Mut F2s, what % are also Unc?

25%

Why?

Unlinked mutations sort independently.

unc-5

F1

Self fertilize

mut

0%

If you pick Unc F2s, what % are also Mut?

If you pick Mut F2s, what % are also Unc?

0%

Why?

Linked mutations repel each other.

slide18
C. elegans genome is about 20,000 genes, 97Mb

Molecular and genetic data is stored in public databases:

http://www.wormbase.org/

http://elegans.swmed.edu/

The genome project generated large clones that cover the entire genome:

Cosmid ~ 30KB insert (acts like a phage and plasmid)

can contain 5-20 genes.

YAC = Yeast artificial chromosome ~250KB insert

contains hundreds+ of genes

slide20
Connecting the genetic map to the molecular map:

mut

(+)

mnDf10

(+)

mnDf100

(-)

mnDf101

(-)

mnDf102

unc-3

genetic map

molecular map

SNP8

SNP9

YACs

}

Cosmids

Genes

slide21
Deficiency Mapping

1 2 3 4

1 2 3 4

Reannealing

1 2 3 4

1 2 4

Breakpoint

Deficiency: A deletion of a portion of a chromosome

Genotype:mnDf22/mnC1 dpy-10(e128) unc-52(e444)II.

Description: Hets are WT and segregate WT, paralysed DpyUnc

and dead eggs. Maintain by picking WT.

Balancer chromosome: suppresses recombination (above, mnC1).

Breakage

X-rays or gamma rays

1 2 3 4

1 2 4

slide22
+

mDf10

Balancer

unc-3(m22)

lon-2(e43)

The balancer is marked with two recessive mutations:

unc-3 (which is within the deficiency and therefore “uncovered” )

lon-2 (which is outside of the deficiency, but closely linked so that

recombination rarely occurs)

Self fertilization results in the following progeny:

mDf10/mDf10 lethal Q: Why are Dfs lethal?

mDf10/unc-3 lon-2 Unc

unc-3 lon-2/unc-3 lon-2 Unc Lon

slide23
+

mut

P0

mut

X

+

mnDf10

mut

F1

X

+

unc-3 lon-2

mut

+

F2

mut

+

mnDf10

unc-3 lon-2

mnDf10

unc-3 lon-2

phenotypes:

Mut or WT

WT

WT

WT

F3

phenotypes:

____

____

____

____

1/4 Mut

1/4 Dead

1/4 Mut

1/4 Unc Lon

1/4 Dead

1/4 Unc Lon

slide24
MEIOSIS IN C. ELEGANS

Special Feature: Holometabolous chromosomes (multiple centromeres)

pairing

DNA

meiosis

replication

Division I

recombination

meiosis

Division 2

Gametes =

products of meiosis

slide25
recombination distance = map distance

Two Factor Mapping:

  • When do you use it?:
  • - How far is the gene from a known marker?
  • To confirm or establish which chromosome the gene maps to.
  • - When you want to generate more recombinants for mapping.

dpy-9(-)

mut(+) unc-34(-)

;

mut(-) unc-34(+)

dpy-9(+)

25%

Pick Dpys: % Unc: _______

0% if perfectly linked, >0% if partially linked.

Pick Uncs: % Mut: _____

25%

Pick Dpys: % Mut: _____

slide26
Two Factor Mapping: How far is a gene from a marker?

b c

+ +

P0

(BC)

X

(WT)

b c

+ +

b c

F1

+ +

Parental Gametes

Recombinant Gametes

b c

+ +

b +

+ c

a

slide27
MAPPING: where is the gene?

Linkage MAPPING: what is the gene close to or linked to?

dpy-14

unc-31

dpy-9

+6

-27

+12

ced-2

-19

How often does recombination occur between dpy-9 & ced-2?

8%

How often does recombination occur between dpy-9 & dpy-14?

39% (not accurate)

Which mutations will show linkage to dpy-9?

ced-2

slide28
Three Factor Mapping: Where Is A Mutation Compared To An Interval?

unc-31

dpy-14

F1 Heterozygote:

mut

unc-31

dpy-14

Recombinant Gamete:

mut

Unc non-Dpy

non-Unc Dpy

mut

unc-31

dpy-14

F2 Recombinant

/Parental:

unc-31

dpy-14

unc-31

dpy-14

1/2 Unc

1/4 Unc Dpy

1/4 Unc Mut

3/4 Dpy

1/4 Unc Dpy

F3:

Conclusion: All Unc non-Dpy F2 segregate Mut F3, and no non-Unc Dpy F2 segregate Mut F3;

thus, the mut gene is to the right of the unc-31…dpy-14 interval.

slide29
Three Factor Mapping: Where Is A Mutation Compared To An Interval?

unc-31

dpy-14

F1 Heterozygote:

mut

unc-31

dpy-14

Recombinant Gamete:

mut

Unc non-Dpy

non-Unc Dpy

mut

unc-31

dpy-14

F2 Recombinant

/Parental:

unc-31

dpy-14

unc-31

dpy-14

1/4 Unc Dpy

3/4 Unc

1/2 Dpy

1/4 Unc Dpy

1/4 Dpy Mut

F3:

Conclusion: No Unc non-Dpy F2 segregate Mut F3, and all non-Unc Dpy F2 segregate Mut F3;

thus, the mut gene is to the left of the unc-31…dpy-14 interval.

slide30
Three Factor Mapping: Where Is A Mutation Compared To An Interval?

unc-31

dpy-14

F1 Heterozygote:

mut

unc-31

unc-31

dpy-14

dpy-14

Recombinant

Gamete:

mut

mut

Unc non-Dpy

Unc non-Dpy

non-Unc Dpy

non-Unc Dpy

F2 Recombinant

/Parental:

unc-31

mut

unc-31

dpy-14

mut

dpy-14

unc-31

dpy-14

unc-31

dpy-14

unc-31

dpy-14

unc-31

dpy-14

1/4 Unc Mut

1/2 Unc

1/4 Unc Dpy

3/4 Unc

1/4 Unc Dpy

3/4 Dpy

1/4 Unc Dpy

1/4 Dpy Mut

1/2 Dpy

1/4 Unc Dpy

F3:

Conclusion: Some Unc non-Dpy F2 segregate Mut F3, and some non-Unc Dpy F2 segregate Mut F3;

thus, the mut gene is within the unc-31…dpy-14 interval.

slide31
For example, you mate your marker strain, bc/bc, to males

that are homozygous mutant for your gene of interest, a/a.

The F1 cross progeny are wild-type in phenotype, and

genotypically bc/a.

You collect 19 recombinants from the F2 generation by

isolating 10 B nonC and 9 C nonB.

Of the 10 B nonC, 9 picked up a. The ratio 9/10 is large so

a is far away from b.

Of the 9 C nonB, 2 picked up a. The ration 2/9 is small so

a is close to c.

b c

a

9

1

B nonC

7

2

C nonB

16

3

T

otal

a

slide32
How can you use this information to figure out a map position?

b

a

c

?

-15

-20

# recombinants (B nonC):

# recombinants ( C nonB):

# recombinants Total:

Total as % of Interval:

9

7

16

16/(16+3)=84%

1

2

3

3/(16+3)=16%

Size of Interval (in cM) = (-20) - (-15) = 5 cM

Gene a is to the right of gene b, at a distance that is 84% of the b-to-c interval.

This distance is (0.84)(5) = 4.2 cM to the right of gene b.

Thus, position of gene a = (-20) + 4.2 = -15.8

slide33
What if a is so close to c that you don’t get any recombinants?

Q: How would you figure out if a is close or to the right?

slide34
Mapping with SNPs

(Single Nucleotide Polymorphisms)

A

T

Bristol, “WT”

N2

T

C

Hawaiian

HA

Shotgun cloned about 5 Mbp (11,000 clones) of Hawaii

6222 potential polymorphisms

4670 SNPs

1552 small insertions/deletions

1 polymorphism/873 bp aligned sequence-

only 5 Mpb out of 97 Mbp were aligned.

Q: Are there other SNPs in the genome?

slide35
SNIP/SNPs

A

GAATTC

Bristol, “WT”

N2

T

GAACTC

Hawaiian

HA

EcoRI Restriction Enz. sequence

GAATTC

Of the 6222 Hawaiian polymorphisms,

3457 produce RFLPs

(Restriction Fragment Length Polymorphisms)

SNPs that also create RFLPs are called snip-SNPs.

So far, about 500 confirmed snip-SNPs

1 confirmed snip-SNP/220 Kbp

(0.6 cM on genetic map)

slide36
N2 HA

N2 HA

SNP

Genomic DNA

PCR product

PCR with SNP-specific primers

from Bristol (N2) strain

PCR with SNP-specific primers

from Hawaiian strain

RsaI

PCR Products

Digested PCR products

slide37
Clone F59H5 at position 23175

Variation & Probability: VarS=GT Psnp=0.9430

CB4856 read: [email protected],g,62

Verified: Yes

5' primer = TGCTCTTCCTCCTGAACTTC

3' primer = CAGGATGTTTTCGTCTGGAC

enzyme used = ApoI

N2 digest RAATTY ApoI, AcsI, FsiI, XapI

HA digest none -- --

TCCTCCGACACTGGTACTCCAACATCAGCGATCATAGCTTGTCCTTCTCT

CGAACGTCTTTCGACGTGATCGATAACTTTTCAGCTTGTCCTTCTAATCC

CGAAAGTCTTTCGACGTGGTCGATAACTTCTCCTGGTCGCtgctcttcct

cctgaacttcTCCTCCTTGCTGTTGACGAAGATCCAAAATCCAACGGACT

CTGAATGTCCCGAAGCCAACTCCATCGTCCGACGTTGCTGCTTCTCTGAT

GTACGACGTCTTCCGATCTCCGCTGCTGGTGATCCTCAGAGCGCACCGTC

CTGATCCATCCCTTCGTGCGTACAGCACGACACTCCGGGGGAAAAAAGGC

TCGAAATTCAAAGGAATTTTGAGTGGACCAAATTGTACAGAACGTTGCAA

ATGTGCTCGGATCCACGACGACATCCGTATTTCTTGTCATCTTCGACTGT

TCCTCTTCGTCTTCTGGAGTCATAACTTGTGCACGTTGAATCGGAAAATT

[T/G]TTGTTCTTTTTACCGGAATTTGAACTTCTCGCAATGTGTAGACTT

CTGCATTCGATGCCAATTCCAACGAAATTAACTCCATCTTCGGAATGGTA

GATTCTGCTCCACTATTTGTCGGCTTTGTCCTCGCTTTTCGACATCACCA

GTGGCATCCCGATAAAGCCACCCGGATATGAAAAATCCTCTTGATTCTTG

TACAATTGAAGAATTTCTCCTTGTGACTCCGCCAACATCGTCACATTCAT

CATGTAGAGATTCGTCTTGATTTGCTCCTTGATTTGATGCGGAAAATGTC

GATTCCACGAAACGACTTCTATCCATTGAAAGCTGCTGTCTGATTGACAA

ATCGTCCAGCACAATCACATCTTCAAGCCAAGGCGGGTGATAAATCCATC

GATTTTTCGATATCATCCTTTGCCTGCCGAACATCCAACTTCCAACTACA

TCTTTGAATTCCACGAAACGACACGACGAAATACTTTCCATCCAGCAAAA

TTCCA

slide38
Combining Three Factor Mapping With SNPs

X

unc-31 mut dpy-14

(from N2 background)

P0

Hawaiian (HA)

unc-31 mut dpy-14

HA

F1

F2

Unc non-Mut non-Dpy

Unc Mut non-Dpy

non-Unc non-Mut Dpy

non-Unc Mut Dpy

1

2

3

4

5

6

recombinant

unc-31

unc-31 mut dpy-14

unc-31 mut

unc-31 mut dpy-14

dpy-14

unc-31 mut dpy-14

mut dpy-14

unc-31 mut dpy-14

parental

1

2

3

6

5

4

unc-31

mut

dpy-14

unc-31

mut

dpy-14

SNP

SNP

slide39
Combining Three Factor Mapping With SNPs

1

2

3

6

5

4

N2

unc-31

mut

dpy-14

unc-31

mut

dpy-14

SNP

SNP

HA

F2 Rec.

Genotype

recombinant (N2 & HA)

unc-31 HA

unc-31 mut N2 dpy-14

1

parental (N2)

unc-31 mut HA

unc-31 mut N2 dpy-14

2

unc-31 mut N2

unc-31 mut N2 dpy-14

3

PCR/Digestion of F2 Worms

HA dpy-14

unc-31 mut N2 dpy-14

1

2

3

4

5

6

4

N2

HA

N2 dpy-14

unc-31 mut N2 dpy-14

5

mut N2 dpy-14

unc-31 mut N2 dpy-14

6

slide40
Using SNPs To Map A Mutation
  • Steps
  • Make a table that represents the recombinant chromosome for each of your recombinants.
    • Label each marker and SNP according to whether it came from the N2 or the HA chromosome.
    • Label whether the recombinant chromosome contains the mutation.

F2 Rec.

Genotype

recombinant (N2 & HA)

unc-31 HA

unc-31 mut N2 dpy-14

1

parental (N2)

unc-31 mut HA

unc-31 mut N2 dpy-14

2

unc-31 mut N2

unc-31 mut N2 dpy-14

3

HA dpy-14

unc-31 mut N2 dpy-14

4

N2 dpy-14

unc-31 mut N2 dpy-14

5

mut N2 dpy-14

unc-31 mut N2 dpy-14

6

slide41
Using SNPs To Map A Mutation

Steps

2. Mark the crossover on the table.

slide42
Using SNPs To Map A Mutation
  • Steps
  • Figure out on which side of the crossover the mutation lies:
    • If the recombinant chromosome contains the mutation (i.e., the recombinant animal has the mutant phenotype), then the mutation is on the N2 side of the recombination site.
    • If the recombinant chromosome does not contain the mutation (I.e, the recombinant animal does not have the mutant phenotype), then the mutation is on the HA side of the recombination site.

The arrows point to where the mutation is relative to the recombination site

slide43
Using SNPs To Map A Mutation
  • Steps
  • There will be at least two critical recombinants: one that defines the left hand border of where the mutation lies, and one that defines the right hand border of where the mutation lies.
    • For the left hand border, find the recombinant that has a rightward arrow that is the furthest to the right.
    • For the right hand border, find the recombinant that has a leftward arrow that is the furthest to the left.

For the left hand border, recombinants 1 and 6 are the same: between unc-31 and the SNP. Thus, the left hand border is within this interval.

For the right hand border, recombinants 2 and 5 are the same, and more to the left than the arrows in recombinants 3 and 4. The site for recombinant 2 and 5 falls between unc-31 and the SNP. Thus, the right hand border is within this interval.

Conclusion: the mutation falls somewhere between unc-31 and the SNP.

slide44
Which mapping method do you use?

Some considerations…

Deletion mapping:

Also tells you if your mutation is recessive, etc.

Falling out of fashion with the rise of powerful SNP mapping.

Mapping with genetic markers:

Generates marked strains that can be used for

other types of mapping.

No PCR problems.

Mapping with molecular markers (SNPs):

In theory, the single fastest method.

Makes it possible to follow dominant alleles.

slide45
Practice Problem, part a:

You are mapping a new ced (cell death defective) mutation.

slide46
Practice Problem, part b,c:

b. You decide to use SNP mapping to narrow down where ced* is. You generate a dpy-24 ced* unc-75 triple mutant, and cross it to the Hawaiian strain. You collect Dpy nonUnc recombinants that are also Ced in phenotype.

R

e

c

o

m

b

i

n

a

n

t

s

c. You tak

e

5

of

your

Dpy

D

p

y

n

o

n

U

n

c

C

e

d

nonUnc recombin

a

nts

that

N

2

H

A

1

2

3

4

5

have

ced*

and make

DNA

from the Ced animals. Then

you perf

o

rm

PCR

with

S

N

P

-

S

N

P

1

specific primers, digesting the

DNA with the correspond

i

ng

enzyme.

R

e

c

o

m

b

i

n

a

n

t

s

D

p

y

n

o

n

U

n

c

C

e

d

H

A

1

2

3

4

5

N

2

S

N

P

2

W

here is

ced*

relative to the

two SNPs?

Which recombina

n

ts

show

this?

What would be your next

mapping exper

i

ment?

ad