11_PPT S2015 (1)

BIOL 102

Lecture 11

Gene Function and Mutational Analysis

Reading pages: 219-226

Problems: Ch6 15-17

Mutational AnalysisMutational Analysis

• Powerful tool for studying gene function– forward genetics: identification of mutants and

description of their heritable phenotypes precedes molecular analysis of products

– reverse genetics: based on genome sequences, gene of potential interest is mutated and the phenotype of mutated gene is studied

• In classical genetics, mutagens were widely used• In a neo-classical approach, insertional mutagens

(e.g., transgenic elements) both disrupt gene and tag it for isolation

• Modern genetics- RNA interference (RNAi)

Forward and Reverse MutationsForward and Reverse Mutations

• Unrelated to forward and reverse genetics• Forward mutation: change away from wild-

type allele

a+ a

D+ D

• Reverse mutation: change back toward wild-type allele (reversion)

a a+

D D+

Genetic Screens (1)Genetic Screens (1)Can be applied to any problem, depending upon

ingenuity and resources• Morphological mutations

– change in shape or form• Biochemical mutations

– screening for auxotrophs from mutagenized prototrophs by supplying various substrates required for growth

Prototroph= an organism that can grow on minimal growth media

Auxotroph= a mutant strain that cannot synthesize molecules required for growth and needs a special substrate(s) to grow

Genetic Screens (2)Genetic Screens (2)

• Lethal mutations– premature death– recessive lethals are more useful than

dominant lethals that are difficult to maintain

• Conditional mutations– display wild-type under permissive

(nonrestrictive) conditions– display mutant phenotype under restrictive

conditionse.g., temperature sensitive mutations

Mutational Analyses

Mutagenize seeds with purple color flower. Self the plants from the mutagenized seeds and screen progeny

Look for any new phenotype or a specific phenotype

21 3

Mutational Analyses

To determine the nature of the mutations, cross each mutant with the wild-type parent

- Similar results were obtained when mutant 2 was crossed to the wild-type, as well as when mutant 3 was crossed to wild-type. - This indicates that all three mutants have recessive mutations in a single gene.

Mutant 1 x wild-type white blue

F1 all blue

F2 3/4 blue and 1/4 white

21

The mutations in mutant 1 and mutant 2 are in the same gene

Cross the homozygous recessive mutants to each other

Complementation Test

No wild-type blue color flower was recovered

No functional copy of w1 gene

w2 gene

21 3

The mutations in mutant 1 and mutant 2 are in the same gene

Cross homozygous recessive mutants to each other


No functional copy of w1 gene

21 3

The mutations in mutant 2 and mutant 3 are in different genes

Cross the homozygous recessive mutants to each other


One functional copy of w1 gene


w1 gene

w2 gene

Wild-type blue color flower was recovered

2 3

The mutations in mutant 2 and mutant 3 are in different genes

Cross homozygous recessive mutants to each other




• Mutation alters gene function by altering structure/function in a product– wild-type: normal allele

• designated by plus (+) sign• example: arg+ or just +

– mutation: change in nucleotide sequence• sometimes designated by minus (–) sign• example: met- or just met

• Nutritional mutants– prototroph: wild-type, synthesizes nutrients– auxotroph: mutant, fails to make essential

nutrient(s). (Example, an amino acid)

Biochemical MutationsBiochemical Mutations

Biochemical PathwaysGeorge Beadle and Edward Tatum performed the first genetic dissection of a biochemical pathway - genes specify enzymes and catalyze specific steps leading to a product. Nobel Prize 1958

Experiment1. Irradiate Neurospora crassa (haploid) to induce mutations.

Cross with opposite mating type and collect progeny spores. 2. Identify auxotrophs that can not grow on minimal media (MM)3. Grow auxotrophs on:

- complete nutrient media- positive control - MM (negative control)- MM + all Vitamins (to identify vitamin auxotrophs)- MM + all amino acids (a.a.) (to identify amino acid

auxotrophs)

Amino acid (a.a.) auxotroph,requires a.a.

i.e. an auxotroph

Amino acid (a.a.) auxotroph,requires a.a.

Tryptopan

Methionine

Methionine auxotroph (met mutant),requires methionine

Biochemical PathwayExperiment4. Identified multiple a.a. auxotrophs5. You are interested in methionine biosynthetic mutants only6. We know that methionine (met) biosynthesis involves certain

intermediates. Grow the mutants on the intermediate substrates.

Cystathionine homocystine methionine

Strain Complete MM MM+aa MM+cystath MM+homocys MM+ met

1 + - + - - -

2 + - + - + +

WT + + + + + +

3 + - + - - + 4 + - + - - +

Growth Response (+ = growth; - = no growth) on Different Media

Biochemical PathwayExperiment4. Identified multiple a.a. auxotrophs5. You are interested in methionine biosynthetic mutants only

- Check which mutant grows when MM is supplemented with methionine



1 + - + - - -

2 + - + - + +

WT + + + + + +

3 + - + - - + 4 + - + - - +


Mutant 1 is not a met mutant; Mutants 2,3 and 4 are met mutants

Biochemical PathwayExperiment6. Identify mutations in methionine pathway and in specific

branches of the pathway.Work backward from methionine.Assume that each mutant is defective in only a single gene.



1 + - + - - -

2 + - + - + +

WT + + + + + +

3 + - + - - + 4 + - + - - +


Supplementing homocystine to MM restores growth of Mutant 2 but not mutant 3 or mutant 4

Biochemical PathwayExperiment6. Identify mutations in methionine pathway and in specific

branches of the pathway-work backward from methionine.Assume that each mutant is defective in only a single gene.



1 + - + - - -

2 + - + - + +

WT + + + + + +

3 + - + - - + 4 + - + - - +


Mutants 3 and 4 require methionine to grow

Biochemical Pathway

Summary of the mutants


• Mutant 1 is not a met mutant • Mutants 2, 3 and 4 are met mutants or met auxotrophs• Mutant 2 is defective in converting cystathionine to homocystine; requires homocystine to growth• Mutants 3 and 4 are defective in converting homocystine to methionine; both require methionine to grow

mutant 3 and mutant 4

Biochemical Pathway

mutant 2

Enzyme 2 Enzyme 3 and enzyme 4 or both have mutation in the same gene?

Biochemical Pathway

• Are mutants 3 and 4 mutated in the same or different genes?

• Complementation Test- In Neurospora complementation test is done by

generating a heterokaryon - fusing cells from 2 distinct mutants to give rise to the heterokaryon.

- Use the heterokaryon to test growth on complete media and media without methionine (or minimal media)

Testing complementation by using a Testing complementation by using a heterokaryonheterokaryon

Biochemical Pathway

• Complementation Test: Generate heterokaryon by fusing the cells from the 2 mutants.

If the heterokaryon does not grow on the media without methionine, then the mutations is in the same gene. i.e. mutant alleles of the same gene.

Will

be

solv

ed in

cla

ss

mutant 3 and mutant 4

Biochemical Pathway

mutant 2

Enzyme 2 Mutations in the same gene which encodes a single enzyme

Biochemical Pathway•Complementation Test: Generate heterokaryon by fusing the cells from the 2 mutants

If the heterokaryon does grow on the media without methionine, then the mutations are in two different genes. Two possible segregations.

1. Two unlinked genes

Will

be

solv

ed in

cla

ss

Biochemical Pathway•Complementation Test:

If the heterokaryon does grow on the media without methionine, then the mutations are in different genes. Two possible segregations.

2. Two linked genes

Will

be

solv

ed in

cla

ss

Biochemical Pathway

mutant 3 and mutant 4mutant 2

Enzyme 2 Enzyme 3 and Enzyme 4

Biochemical Pathway

Conclusions• Mutant 1 differs from mutant 2, 3, and 4

Strain WT Mutant 1 Mutant 2 Mutant 3 Mutant 4

Mutant 1 - + + +

Mutant 2 - + +

WT + + + + +

Mutant 3 - -Mutant 4 -

Complementation test: heterokaryon growth

• Mutant 2 differs from mutant 1, 3, and 4• Mutant 3 differs from mutant 1 and 2 but not from mutant 4• Mutant 3 and 4 do not complement, therefore encode the same

gene product• Mutant 3 and 4 are alleles of the same genes

Biochemical Pathway

mutant 3mutant 2

Enzyme 2 Enzyme 3

•Each step in a biochemical pathway is catalyzed by a specific enzyme•One gene one polypeptide- some enzymes are made of more than one polypeptide•Mutations that result in loss of enzyme activity lead to accumulation of precursors which may be toxic

Enzyme 3

mutant 4

Biochemical PathwayBiochemical Pathway

We considered a linear pathway for our answer. Other interpretations of the pathway is also possible.

Branched pathways

If enzyme 1 is absent, the mutant would require addition of met and threonine for growth

Precursor homoser

threonine

Cystathio homocys metMutant 3 & 4

Mutant 1

Mutant 2


Two or more pathways leading to the same product. These examples are unrelated to the Neurospora mutations we studied

Either enzyme A or enzyme B are required to produce the final product

Precursor 2

Precursor 1Product

Gene A

Enzyme A

Gene BEnzyme B


Two or more products (polypeptides) required to form a single enzyme (multisubunit enzyme)

Mutation in genes encoding subunits A or B results in no product

Precursor Product

Gene A Gene B

Enzyme Subunits A + B

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