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Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set 4 (pertaining to last weeks lectures) is posted on the course website. -A reading assignment on DNA hybridization has been posted on the course website. Today: -Mutant analysis (screen vs. selection; reversion; suppression; mutation rate; mutagens). -Repair of mutations

Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

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Page 1: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

Announcements

-Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website.

-Problem set 4 (pertaining to last weeks lectures) is posted on the course website.

-A reading assignment on DNA hybridization has been posted on the course website.

Today:-Mutant analysis (screen vs. selection; reversion; suppression; mutation rate; mutagens).

-Repair of mutations

Page 2: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

Mutant analysis (AKA Genetic Analysis)

The use of mutants to understand how a biological process normally works*

*See the Salvation of Doug article at the following site:http://bio.research.ucsc.edu/people/sullivan/savedoug.html

• Start with “unknown” system (e.g., metabolic pathway, embryonic development, behavior, etc.)

• Generate mutations that affect the “unknown” system (i.e., that “break” the “unknown” system)

• Study the mutant phenotypes to reveal the functions of the genes

• Map the genes

• Identify the genes (more on this later)

Page 3: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

Wild-type yeast can survive on ammonia, a few vitamins, a few mineral salts, some trace elements and sugar…

They synthesize everything else they need, including adenine

What genes does yeast need to synthesize

adenine?

Case study: analyzing the adenine biosynthetic pathway by generating and studying “ade” mutants

Conducting a mutant analysis with yeast

Page 4: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

-adenine plate

“complete” plate

sterile piece of velvet

Adenine-requiring colonies

(ade mutants)

m2

m1m3

Replica-plate

plate cells

Treat wt haploid cells with a mutagen:

Identifying yeast mutants that require adenine

This is an example of a “genetic screen”

Page 5: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

Identifying interesting mutations—screen vs. selectionScreen Each member of the population is examined…

does it fit the phenotype criteria that have been set up?Individuals not meeting the criteria don’t survive (or are otherwise eliminated from the population)

Selection

Example 1: Looking for a translator

Example 2: Looking for wingless fly mutants

Primary selection or screen is often followed by secondary selection or screen

Russian EnglishScreen: read resumésSelection: advertise in Russian

Screen: Look at each fly… wings present?Selection: Open vial, let flies fly away

Page 6: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

Reversion and Suppressors

Another approach - start with a mutant

Look for reversion to wild type (or less mutant)-sometimes called “reverse” genetics.

ww

look for red eyes

What kinds of mutations might you find?

Most genetic screens are “forward” screens - start with wild type organism and look for new phenotypes caused by mutation (e.g., screen for yeast ade mutants).

Page 7: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

1)Mutations that restore function to the white gene (revertant).

w gene

X

XX

or

w gene

X

w gene

X

plus suppressor mutation in some other gene

What kinds of suppressor mutations might you imagine?

2) Mutations that bypass (or suppress) the need for a white gene.

Reversion and Suppressors

Page 8: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

intermediate X

adenineADE2

YADE1

red pigment

Partial biosynthetic pathway for adenine in yeast

adenine

ade2

XX Y

ADE1

build up of

Page 9: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

Revert ade- to Ade+… does RED revert to WHITE?

-Ade plate

Ade+ revertant

it’s white

Pretty good proof that one mutation (ade2) has two phenotypes

ade2 mutant

Treat wt haploid cells with a mutagen:

Page 10: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

5’..AUG....UAC....UGA..3’

A working hypothesis… ade2 has reverted to ADE2

ADE2

ColorGrowth on -ade mRNA

sequence

ade2

revertant

(ade2-R)

+ white

- red 5’..AUG....UAG....UGA..3’

+ white 5’..AUG....UAC....UGA..3’

STOP!

Page 11: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

Has the mutant gene changed back to WT?

A test of the hypothesis……do a cross: ADE

2ade2-Rx :)

If ade2-R is “true revertant”… :)

ADE2

ade2-R

ade2-R = ADE2

the diploid is homozygous WT meiosis

Spores from the diploid should be:

All Ade+, white

Most Ade+ revertants… like this. But some exceptions!

Page 12: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

Some revertants behave differently…

ADE2

revertantx

meiosis

Most spores: Ade+, white

Some spores: ade-, red!

Ade+, white Ade+, white

DiploidsAde+, white

Interpretation?

In these revertants…

• ade2 is still in mutated form

• a new mutation somewhere else suppresses the ade2 phenotype!

The cross:

Page 13: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

Summary of revertant types

ade revertants come in two varieties:

1. “True” revertants

2. Suppressors (aka “extragenic suppressors” or “second site suppressors”)

ade2 mutant allele ADE2 (wt)

A mutation in a different gene eliminates the ade2 mutant phenotype

Definition of suppressor?

A mutation in a second gene that eliminates the mutant phenotype of a mutation in the first gene.

Page 14: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

What is this suppressor?

Linkage analysis… mapped to chr XV

SUP3 codes for tRNATyr

huh?

To recap…

ADE2

ade2

sup3 SUP3

Ade+

white!

5’..AUG....UAG....UGA..3’

Explanation?

TyrSTOP mutation

WT

Page 15: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

How does SUP3 suppress ade2?

WT tRNA (sup3)

AUG

Tyr

AUC

Tyr

Mutant tRNA (SUP3)

TyrSTOP5’..AUG....UAG....UGA..3’

ade2 mRNA ade2 mRNA

5’..AUG....UAG....UGA..3’

The mutant tRNA suppresses the nonsense codon! --full-length ADE2 protein is made--Ade+, white colony!

Page 16: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

Doesn’t the suppressor tRNA cause problems for cells?

• Yeast has 8 tRNA-TYR genes• Only one of them has the suppressor mutation.

What reads the normal TYR codons, UAC?

What about genes that normally end in UAG?

• Not all ORFs end with UAG.• For those that do, there’s still a competition between the suppressor tRNA and termination factor.

Even so, a cell with a SUP mutation can be quite sick.

Page 17: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

Another kind of suppression (unrelated to ade2)

WT protein 1 WT protein 2

Mutant protein 1 WT protein 2

Mutant protein 1 Mutant protein 2

Restoration of function!

Page 18: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

Red/White and Ade+/ade-

One gene or two closely linked ones?

1. Isolate new red mutants: do they also require adenine?

2. If the adenine mutation is reverted to wild type (ADE) does the red color also revert to white?

yes

yes

3. If red is reverted to white, does ade- revert to Ade+?

Page 19: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

AdenineYXade2 ADE1

red pigmen

t

gene R

Can we get redwhite revertants that are still ade-?

Some ideas

WADE3ade3

In an ade2 strain (red)… LOF mutation of either gene R or ADE3 white colonies, but still ade-.

Page 20: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

ade2 mutant revert phenotype to white

Using complete media . . .

Some white colonies could be true revertants.Some mutations could be suppressors.Some could be in other genes of the pathway?

Making redwhite revertants

Treat with mutagen

rev#1

rev#2

Page 21: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

Summary…

revert ade2 ADE2

color?

grow without adenine?

mutate ADE3 to LOFmutate gene R to LOF

white yes

white no

white no

from

Growth without adenine

distinguishes

AdenineYXade2 ADE1

red pigmen

t

gene R

WADE3

In ade2 strain:

Page 22: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

How many mutants are like ade3?

How many are like “gene R”?

adenineADE2 ADE1

gene R

• 10 complementation groups are ade- and white

• lots and lots, but “gene R” has never been identified!

red

• Respiration defective cells can’t make red pigment. Respiration mutations are epistatic to red pigment!

*** use up 1 ATP, non-reversible step

• 2 are ade- and red.

Final tally…

Page 23: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

Quiz Section this week:

Genetic Analysis in Caenorhabditis elegans

An introduction to C. elegans. . .

Page 24: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

A bit of background on Caenorhabditis elegans

• 1 mm long nematode worm.

• 3.5 day generation time.

• Predominantly internally selfing hermaphrodite (make sperm and oocytes).

• Rare males arise spontaneously and can cross with hermaphrodite (male sperm fertilize hermaphrodite oocyte).

• Moves by wriggling (like a snake).

Page 25: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

C. elegans hermaphrodite

head tail

Page 26: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

C. elegans generates bends using dorsal and ventral muscle strips.

worm movie

Page 27: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

Inbreeding is important for model organism genetics

• Outbred (wild) populations are genetically heterogeneous.

•Highly inbred strain has little or no genetic variability.

Page 28: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

With each generation, ½ of the previously heterozygous alleles become homozygous.

Inbreeding makes strains homozygous for everything

QuickTime™ and aNone decompressor

are needed to see this picture.

X X XX

XX X X

Page 29: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

Inbreeding is important for model organism genetics

• Outbred (wild) populations are genetically heterogeneous.

•Highly inbred strain has little or no genetic variability.

• Mutant alleles behave simply - only change present in cross.

• E. coli, yeast, fruit fly, C. elegans, zebrafish, mouse are highly inbred.

Page 30: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

To conduct a mutant analysis begin with inbred WT strain, then treat with a mutagen to generate a large population of mutagenized animals

Why mutagenize?

FREQUENCY!!

Spontaneous mutations are VERY RARE.

Mutagenesis can increase frequency by about 10,000 fold.

Mutant Analysis: generating mutants

Page 31: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

X-Rays (H. J. Muller’s X-linked “ClB” system in Drosophila)

C

l

B

How frequently do new mutations appear on this X-chromosome?

Estimation of mutation rate: X-ray-induced mutations

rossover suppressor = X-chromosome with inversions… no recombination

lethal (l) = recessive lethal (XlY males are dead)

Bar (B) = bar-shaped eyes; bar shape is DOMINANT

X-rays

x Bar-eyed femaleB l

x

Page 32: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

How frequently do new mutations appear on this X? x Bar-eyed

female

x wt

look just at sons

If new lethal mutation…

dead

dead!

no sons!

B l

Pick Bar-eyed female progeny B l

1 female/cross; repeat many times

dead

viable

If no new mutations…

B l B l

Estimation of mutation rate: X-ray-induced mutations

Page 33: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

X-ray dose

% X-linkedrecessiveLethalmutations

Certain external agents (mutagens) can drastically increase mutation rates.

no X-raytreatment

Estimation of mutation rate: X-ray-induced mutations

Spontaneous mutation rate (2/1000 X-chromosomes)

Page 34: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

Spontaneous mutation rates

Measurement of spontaneous mutation rates:

2 mutations per 1000 X-chromosomes

2 mutations per 1,000,000 genes

Mutation rate = 2 x 10-6/gene/generation

Very similar rate calculated for humans!

(from assumption of 1000 genes on X)

Rough calculation:

If 35,000 genes in human genome…

2 x 10-6 x 35000 = ~ 0.07 mutations per generation

or 1 mutation (somewhere in the genome) per 14 gametes…

i.e., you would only get ~2 mutants/1,000,000 animals analyzed from spontaneous mutations - using a mutagen can increase this rate to ~2 mutants/100 animals

Page 35: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

Radiation

- X-rays, -rays: ionizing radiation

cause breaks in DNA chromosomal rearrangements!

- Ultraviolet light: non-ionizing radiationthymine dimers

impede DNA polymerase

Some mutagens (electromagnetic radiation)

Page 36: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

C

Chemical mutagens

- Alkylating agents, e.g., ethylmethane sulfonate (EMS)

G

EMS

O6-ethyl-G

T

- intercalating agents, e.g., acridine orangeQuickTime™ and aTIFF (Uncompressed) decompressorare needed to see this picture.

cause frame shift mutations

base substitutions

Some mutagens (chemical mutagens)

Page 37: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

Transposons: jumping genomic segments of DNA

Transposase gene

Small pieces of DNA (a few hundred to a few kbp in length) that can move from one site in the genome to another.

•ALL organisms have them (~45% of our genome: transposon remnants!)•Jumping genes, Selfish DNA•Mechanism for rapid evolutionary change

Transposons can also cause mutations if they hop into or near genes

Allele R

Allele r

Transposon insertion

The wrinkled pea trait that Mendel studied was caused by a transposon insertion that inactivated a gene

Page 38: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

What are the sources of spontaneous mutations?

How are mutations repaired?

Mutation; repair of mutations

Page 39: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

mutation!

Spontaneous mutations

- Base alteration or loss

- Replication errors

CA C G

CA A C GT T G C A T G

AT T G C A T G

new

old

corrected?

yes T A C

no CA A C GT T G C A T G

A C

CA A C GT T G C G T G

A C

replication

probably exceeds 50,000/cell/day

Page 40: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

Damage control

Experimentally observed mutation rate in E. coli (inside the cell):

Expected error rate of E. coli DNA polymerases (from physical/chemical properties of the bases:

Experimentally observed error rate of E. coli DNA polymerases (in the test tube):

Conclusions:

-DNA polymerases must possess a “proofreading” ability.-There must be yet another backup error detection system in the cell.

1 mutation/1010 bases polymerized

1 mutation/105 bases polymerized

1 mutation/107 bases polymerized

Page 41: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

Proof-reading by DNA polymerase

A CA C GT T G C A T G

A A C GT T G C A T G

new

oldTcorrection

DNA polymerase has 3 activities:

- can add bases to 3’ end

- the end must be base-paired

- template must be available

- can excise (remove) bases from 3’ end

Normally, addition rate >> excision rate

- can remove bases from 5’ end (involved in DNA replication and some forms of repair)

(for optimal activity)

(not covered in this course)

Damage control

Page 42: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

Proof-reading (cont’d)

A A CT T G C A T G

TA A C GT T G C A T G

A A C GT T G C A T G

C

3’ end base-paired extension rate high

3’ end NOT base-paired extension rate low

probability of excision high

A A C GT T G C A T G

C

3’ end base-paired again!

Proof-reading corrects 99% of incorporation errors!

DNA pol

Page 43: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

Experimentally observed mutation rate in E. coli (inside the cell):

Expected error rate of E. coli DNA polymerases (from physical/chemical properties of the bases:

Experimentally observed error rate of E. coli DNA polymerases (in the test tube):

Conclusions:

-DNA polymerases must possess a “proofreading” ability.-There must be yet another backup error detection system in the cell.

1 mutation/1010 bases polymerized

1 mutation/105 bases polymerized

1 mutation/107 bases polymerized

mismatch repair system

Damage control

Page 44: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

“mismatched” base

Mismatch repair

Proofreading catches many errors but some still slip by; how are they detected and repaired?

GACGTACATGCTGCATGTAC

GACGTACATGCTGCATGTAC

GACGTATATGCTGCATGTAC

repair is biased; tends to restore normal sequence

GACGTACATGCTGCATGTAC

GACGTATATGCTGCATATAC

repaired unrepaired

Page 45: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

Best understood in bacteria

1. Identify mismatched bases in DNA

2. Recognize the template strand

3. Correct the OTHER strand

mutS protein in E. coli

use methylation state of DNA to identify template strand

TGATCAACTAGT

TGATCAACTAGT

CH3

CH3

deoxyadenosine

methylase (DAM)

Mismatch repair

Page 46: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

DNA replication

DNA replication

transiently hemimethylated template strand can be distinguished from newly synthesized strandtransiently hemimethylated template strand can be distinguished from newly synthesized strand

Mismatch repair

Page 47: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

mutL

mutHmutS

Mismatch repair—the mutSHL system

5'-CACGTTACAAGGTCATGTTTCCGATCTA-3’3'-GTGCAATGTTCCAGGACAAAGGCTAGAT-5'

CH3

mutS protein recognizes mismatch

mutH protein recognizes parental strand

mutL protein promotes mutH activity (make cut in new strand)

TTACAAGGTCCTGTTT

TTACAAGGTCATGTTTexcise mismatch region

5'-CACG CCGATCTA-3’3'-GTGCAATGTTCCAGGACAAAGGCTAGAT-5'

CH3re-synthesize DNA

mismatch

Page 48: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

250nm

0

10

100

% surviving

cells

0 6 12Minutes of UV irradiation

phruvrAuvrBuvrCuvrD

Repair of UV light induced DNA damageWhat genes are involved?

E. coli

WT cellsMutants defective in UV repair

Page 49: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

2 mechanisms of UV damage repair: light-dependent and light-independent.

250nm

# pyrimidine dimers/kb

DNA

time

Blue light (300-

500nm)

in the dark

UV light pulse

5’-CACGTTACAAGGTCCTGTTTCCGATCT-3’3’-GTGCAATGTTCCAGGACAAAGGCTAGA-5’

5’-CACGTTACAAGGTCCTGTTTCCGATCT-3’3’-GTGCAATGTTCCAGGACAAAGGCTAGA-5’

Phr=photolyase (+ blue light)

Pyrimidine dimers in the genome converted to small ss DNA fragments

pyrimidine dimers ‘disappear’

Repair of UV light induced DNA damage

Page 50: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

Phr=photolyase (+ blue light)

Light-dependent UV repair mechanism

Page 51: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

uvrD

uvrBuvrC uvrA

5’-CACGTTACAAGGTCCTGTTTCCGATCT-3’3’-GTGCAATGTTCCAGGACAAAGGCTAGA-5’

excise damaged region

5’-CACGT TTTCCGATCT-3’3’-GTGCAATGTTCCAGGACAAAGGCTAGA-5’

TACAAGGTCCTG

pyrimidine dimer

Repair of UV light induced DNA damage (cont’d)Light-independent mechanism

Page 52: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

uvrBuvrC uvrA

This repair system also corrects alkylation damage induced by chemical mutagens (e.g. EMS, MMS, etc.)

5’-CACGTTACAAGGTCCTGTTTCCGATCT-3’3’-GTGCAATGTTCCAGGACAAAGGCTAGA-5’

TACAAGGTCCTG

5’-CACGT TTTCCGATCT-3’3’-GTGCAATGTTCCAGGACAAAGGCTAGA-5’

TACAAGGTCCTG3’-GTGCAATGTTCCAGGACAAAGGCTAGA-5’

replace damaged region

5’-CACGT TTTCCGATCT-3’

Light-independent mechanism

Repair of UV light induced DNA damage (cont’d)

Page 53: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

mut genes found in humans also… mutations in mut genes associated with colon cancer.

Mechanisms of DNA damage repair are conserved

Xeroderma pigmentosa—defective UV repair system… mutations affect genes that resemble uvrA-D (not clear if there is a phr counterpart in humans).

Page 54: Announcements -Solutions to problem set 3 and from the questions pertaining to last Fridays lecture have been posted on the course website. -Problem set

Testing for mutagens… the Ames test

Premise:

- start with his- bacteria (Salmonella)

- spot test compound on plate

- if the compound causes mutations… sometimes his- will mutate to his+

Interpretation:

Compound #1 = non-mutagenic

#2 = mildly mutagenic

#3 = strongly mutagenicQuestion: some compounds that are known to be mutagenic in mammals only yield positive results if pre-incubated with a liver extract; why?

1

2 3

test compounds