Announcements

-First midterm exam will be in this room on Friday (4-25) from 10:30AM-12:20PM

-Exam will cover material presented in lecture and quiz section through the end of last week, however…

-this weeks material will reinforce some of the previous concepts you have learned

-this weeks material WILL be covered on the next midterm exam

Leo Pallanck’s office hours: Friday afternoons by appointment (pallanck@u.washington.edu)

Today

Inheritance

Mutant analysis

Genomicswhat can we learn by studying whole genomes?

how are biological processes studied by analyzing mutants?

Throughout the quarter . . .how and why are model organisms used in genetics?

how does that information apply to humans?

What this course is about

how are unique physical traits determined by genes?

how are traits transmitted to progeny?

how is genetic information read within cells?

From Lecture 1

More to come!

Mutant analysis (AKA Genetic Analysis)

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

Very powerful - can be used to study metabolic pathways, animal development, neurobiology, cell division, etc.

A simple analogy…

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

Analysis of pizza synthesis

Analagous to genes

Analysis of pizza synthesis

No red sauce?!

Analagous to a mutation

The mutant phenotype

Analagous to genes

What is a model organism?A species that one can experiment with to ask a biological question

Why bother with model organisms?- All organisms are related at the molecular

level

- Not always possible to do experiments on the organism you want

- If the basic biology is similar, it may make sense to study a simple organism rather than a complex one

Mutant analysis involves model organisms

Which of these cameras do you think would be easier to understand?

Box camera

IMAX 3-D camera

Features of a good model organism• Short generation time

• Small, easy to maintain

Telomeres

96 million telomeres per cell!

• Large numbers of progeny

• Well-studied life cycle, biology

• Appropriate for the question at hand• Mendel used a model organism—the garden pea- relatively short generation time—one

per year- lots of progeny per cross- self-pollination and out-crossing possible

- true-breeding varieties readily available from local merchant

Some commonly used model organisms

- Bacteria — Escherichia coli

- Budding yeast — Saccharomyces cerevisiae

- Fruit fly — Drosophila melanogaster

- Nematode — Caenorhabditis elegans

- Mouse — Mus musculus

QuickTime™ and aTIFF (Uncompressed) decompressorare needed to see this picture.

Quiz Section this week:

Complementation analysis of yeast mutants

An introduction to yeast . . .

Mutagenesis is easier in single-cell organisms with haploid lifestyles

Budding yeast—a single-celled fungus that divides by buddingYeast cells can exist as haploids…

Haploid life cycle:

Yeast as a model “genetic” organism

mitosis

cytokinesis

The haploid life cycle (1n)

mutation

mating

diploid zygote

haploid

a haploid

1n

The haploid life cycle (1n)

The life cycle of “budding” yeast

Yeast cells can also exist as diploids…

meiosis

A tetrad with 4 haploid spores(“gametes”)

Mendelian segregation occurs here

2 cells

2 a cells

1n

The life cycle of “budding” yeast (cont)

a/ diploid life cycle (2n)

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

(Why might we care about adenine?)

Conducting a mutant analysis with yeast

-adenine plate

“complete” plate

sterile piece of velvet

Adenine-requiring colonies

(ade mutants)

m2

m1m3

“Replica-plating”

plate cells

Treat wt haploid cells with a mutagen:

Identifying yeast mutants that require adenine

m1 wild-type

“complete” plate

-adenine plate

replica-plate using velvet

That is, are they LOF mutations? Why do we care?

What do you conclude?

Genotypes:

ADEade

diploids

ADE is dominant over ade

“” mating type“a” mating type

Are the adenine-requiring mutants recessive?

What would you predict if…

• only one enzyme is needed for synthesis of adenine?

• many enzymes are needed for synthesis of adenine?

How to find out how many different genes we have mutated?

Are m1 and m2 alleles of the same gene?

Do complementation test to ask: are the mutations alleles of the same gene or of different genes?

All mutants would be alleles of the same gene.

Different genes might be mutant.

Are all of the mutations in one gene?

m1 m2

“complete”

diploids

-adenine

replica-plate

“” mating type“a” mating type

Do m1 and m2 complement, or fail to complement?

Are m1 and m2 alleles of the same gene, or alleles of different genes?

Performing a complementation test

What do you conclude from the pair-wise crosses shown below?

m1

m2

m3

m4

m5

m6

m7m

1m2m3m4m5m6m7

o

o

o

o

o

o

o

x

o+ + + + o

Complementation tests with ade mutants

m1, m5, m7 are mutations in one gene

Conclusion?

o = no growth on -ade+ = growth on -ade

m1

m2

m3

m4

m5

m6

m7m

1m2m3m4m5m6m7

x

+ + + +o+

o

+ + +

+ + + +

+ + +

+

+

m1, m5, m7 are mutations in one gene

Conclusion?

m2, m4 are in one gene

Four complementation groups

o

o

o

o

o

o

o

o o

What do you conclude from the pair-wise crosses shown below?

o = no growth on -ade+ = growth on -ade

m3

m6

Usually means four genes

Complementation tests with ade mutants

Yeast cells can normally grow on a sugar called galactose as the sole carbon source. Seven mutant “” haploid yeast strains have been isolated that are unable to grow on galactose (“gal”) plates.

Six of these mutant strains were each cross-stamped on a gal plate with a wild type “a” strain. The resulting pattern of growth on the gal plates is depicted below (shading = growth). In all plates, the wild type strain is in the horizontal streak.

On the leftmost plate, mark the location of the a/ diploid with a circle.What is the mode of inheritance of mutant phenotype in mutants 1-6? How can you tell?

Diploids grow on gal plate… so, wild type is dominant

Practice Question

Each of the seven “” mutant strains was cross-stamped on gal plates against “a” versions of the seven mutants. The results are depicted below:

Looking just at mutants 1–6 for now… group these six mutants by complementation group.

Mutant 1 Mutant 2 Mutant 3 Mutant 4 Mutant 5 Mutant 6 Mutant 7

Mutant1

Mutant2

Mutant3

Mutant4

Mutant5

Mutant6

Mutant7

m1, m2, m5

m3, m6

m4

Practice question (continued)

Now consider mutant 7. What is surprising about the result in the complementation table?

Mutant 7 was cross-stamped on gal plate with wild type as you saw with the other six mutants earlier:

What do you conclude about the mode of inheritance of mutant 7? How does that help you explain the complementation test result for mutant 7?

What can you conclude about how many genes are represented in this collection of seven mutants?

Fails to complement any of the others… how could it be an allele of 3 different genes?

Complementation test fails with a dominant mutation… heterozygote will always show the mutant phenotype

At least 3 genes (can’t tell about m7)

Practice question (continued)

Complementation is relevant to humans

Family A Family B= deaf

Within each family, does deafness look like it’s dominant or recessive?

Assign genotypes (A, B, etc.) to the deaf individuals in these pedigrees.

recessive

aaBB AAbb

AaBb

Complementation is relevant to humans

Niemann Pick Type C disease (NPC): a recessive human neurodegenerative disease resulting in premature death

Cellular cholesterol accumulation accompanies the disease (can be detected using a chemical called ‘filipin’ which fluoresces upon binding cholesterol)

QuickTime™ and aTIFF (Uncompressed) decompressor

are needed to see this picture.

QuickTime™ and aTIFF (Uncompressed) decompressor

are needed to see this picture.normal NPC

QuickTime™ and aTIFF (Uncompressed) decompressor

are needed to see this picture.

NPC1:NPC2 NPC1:NPC3

Do NPC1 & NPC2 complement?Do NPC1 & NPC3 complement?

Practice Question

Hearing mice… independently assorting genes A and B, both needed for hearing:

AaBb AaBbx

WITHOUT drawing a Punnett square, predict the progeny phenotypes and proportions with respect to hearing ability.

a, b: complete LOF, recessive

N

R

CH

NHC

C

H2N

N

R

CH

NC

C

H2N

-OOCN

R

CH

NC

C

H2N

CHCHN

OCH2

COO-

COO-

C

O

NHC

HN

NH

CH

NC

C

using mutants to order the steps in a pathway

Adenine

(AIR)

(CAIR)

(SAICAR)

...

For example, this molecule accumulates in an ade13 mutant.

ADE13

encodes the enzyme that carries out the next step

...

ADE4 ADE5*ADE8 ADE6 ADE7 ADE2 ADE1

ADE13

ADE17ADE16

ADE17ADE16

ADE12

ADE13

AMP

ADE3*

X

redpigment

The Yeast Adenine Biosynthetic Pathway

A B

CD E F G

H

I

J

K

Y

A second phenotype of some ade mutants…

plate cellsMutagenize:

-adenine plate

Replica-plate

complete plate

Some of the adenine-requiring mutants are red!

Are the red ade mutations recessive?

m8 wild-type

complete plate

-adenine plate

replica-plate using velvet

Genotypes:

ADEade

White color is

dominant

Ability to make adenine is dominant

How can one LOF mutation generate two very different phenotypes?

some intermediate

“X”adenine

ADE1

ade1

LOF mutation

Xred

pigment

UNK1*

*not a real gene name! This gene has not yet been identified

X

Hypothesis: Two phenotypes/one LOF mutation

another intermediat

e “Y”

adenine

Y

Suppose we isolate LOTS of independent red mutants:

Are all red mutants defective in the SAME GENE?

How to tell?

m9 m10

mutations fail to complement = same gene

mutations complement = different genes

Must modify the hypothesis.

All pairwise combinations reveal two complementation groups.

diploids

m9 m11

white diploids

Complementation tests of red mutants

XADE2

ade2

X

YADE1

ADE1

red pigmen

t

“UNK1”

Modified hypothesis for red phenotype

adenine

adenine

X Y

But how do mutations in ADE1 result in a build-up of X?

YXADE2 ade1

YX

Mutations in either ADE1 or ADE2 lead to a defect in adenine biosynthesis and lead to the build-up of intermediates in the pathway. Excess “X” is converted to a red pigment.

red pigmen

t

“UNK1”

adenine

Modified hypothesis for red phenotype

YXade2

ade1

YXred

pigment

Same as ade2 single mutation! Red and adenine-requiring.

“UNK1”

1. Phenotype of ade1 ade2 double mutation?

Test your understanding

adenine

ADE7

ADE13

2. Phenotype of ade2 ade7 double mutation?

3. Phenotype of ade2 ade13 double mutation?Same as ade7 single mutation! White and adenine-requiring.

Same as ade2 single mutation! Red and adenine-requiring.4. Phenotype of unk1?White and able to grow on -ade plates.

Practice Questions

YXADE2 ADE1

YX

red pigment

“UNK1”

adenineADE3

MATa ade2 ADE3MAT 2 3ADE adeDiploid

:strain color on growth on complete - ?adenine plate ?plate ( )yes or no1A. A MATa ade2 ADE3 mutant was mated to a MAT ADE2 ade3 mutant to create a diploid. What are the phenotypes of the three strains? Assume all other genes are wild type.

MATa ade2 ADE3MAT 2 3ADE adeDiploid

:strain color on growth on complete - ?adenine plate ?plate ( )yes or nored no

white no

white yes

Practice Questions

YXADE2 ADE1

YX

red pigment

“UNK1”

adenineADE3

1A. A MATa ade2 ADE3 mutant was mated to a MAT ADE2 ade3 mutant to create a diploid. What are the phenotypes of the three strains? Assume all other genes are wild type.

1B. ADE2 and ADE3 assort independently. Draw the chromosomes at metaphase of meiosis I such that the two WILD TYPE alleles face the same pole. Place a crossover on the other chromosome arm relative to the ADE2 and ADE3 genes.

A

CD

B

Tetrad on complete plates

1C. Recall that each tetrad contains the products of a single meiosis. Predict the genotypes and growth properties of each spore resulting from this meiosis.1D. Analysis of many tetrads demonstrates that three types are found, depending on the behavior of the chromosomes in meiosis. Which tetrad best fits the meiosis you just drew? Letter the spores below to match the genotypes in your table.

#1 #2 #3

red

Spore grow without adenine? complete genotype?

A

B

C

D

noade2 ade3

yesADE2 ADE3 yesADE2 ADE3

noade2 ade3

ADE2

ADE2ade2

ade2

ADE3

ADE3ade3

ade3

A

BC

D

1E. Now draw the chromosomes at metaphase of meiosis I such that one wild type and one mutant allele face each pole. Place a crossover on the other chromosome arm relative to the Adenine genes.

Spore grow without adenine?complete genotype?

A

B

C

D

Tetrad on complete plates

1F. Predict the genotypes and growth properties of each spore resulting from this meiosis.

1G. Which tetrad best fits the meiosis you just drew? Letter the spores below to match the genotypes in your table.

#1 #2 #3

red

ADE2

ADE2

ade3

ade3

A

B

1E. Now draw the chromosomes at metaphase of meiosis I such that one wild type and one mutant allele face each pole. Place a crossover on the other chromosome arm relative to the Adenine genes.

Spore grow without adenine?complete genotype?

A

B

C

D

Tetrad on complete plates

1F. Predict the genotypes and growth properties of each spore resulting from this meiosis.

1G. Which tetrad best fits the meiosis you just drew? Letter the spores below to match the genotypes in your table.

#1 #2 #3

red

ADE2

ADE2ade2

ade2

ade3

ade3

ADE3

ADE3

nonono

noade2 ADE3ADE2 ade3ade2 ADE3

ADE2 ade3

A

CB

D

A

CD

B

1H. Now draw the chromosomes at metaphase of meiosis I such that one wild type and one mutant allele face each pole. On one chromosome, place a crossover on the other chromosome arm relative to the Adenine gene. On the other chromosome, place a crossover BETWEEN the centromere and the Adenine gene.

Spore grow without adenine? complete genotype?

A

B

C

D

Tetrad on complete plates

1I. Predict the genotypes and growth properties of each spore resulting from this meiosis.

#1 #2 #3

red

1J. Which tetrad best fits the meiosis you just drew? Letter the spores below to match the genotypes in your table.