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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
Complementation Test
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
Complementation Test
One functional copy of w1 gene
One functional copy of w2 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
Complementation Test
One functional copy of w1 gene
One functional copy of w2 gene
• 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
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
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.
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
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.
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
Mutants 3 and 4 require methionine to grow
Biochemical Pathway
Summary of the mutants
Cystathionine homocystine methionine
• 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
Biochemical PathwayBiochemical Pathway
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
Biochemical PathwayBiochemical Pathway
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