Genetic Fine Structure Genetic Fine Structure
Nature of the Gene at Nature of the Gene at the Molecular Levelthe Molecular Level
Bead TheoryBead Theory
The gene is the fundamental unit of The gene is the fundamental unit of
1.1. StructureStructure
The gene is indivisible by crossing over.The gene is indivisible by crossing over.
Crossing over occurs only between Crossing over occurs only between genes. genes.
Bead TheoryBead Theory
The gene is the fundamental unit ofThe gene is the fundamental unit of
2. Change2. Change
The whole gene must change from one The whole gene must change from one allelic form to another, there are no allelic form to another, there are no smaller components within gene can smaller components within gene can change by mutation.change by mutation.
Bead TheoryBead Theory
The gene is the fundamental unit of The gene is the fundamental unit of
3. Function3. Function
The gene functions as a unit, parts of a The gene functions as a unit, parts of a gene cannot function on their own.gene cannot function on their own.
Revised Bead TheoryRevised Bead Theory
The The nucleotide pairnucleotide pair is the fundamental unit is the fundamental unit of of
1.1. StructureStructure
2.2. ChangeChange
The The genegene is the fundamental unit of is the fundamental unit of
3. Function3. Function
How Can the Expression How Can the Expression
of a Gene by Altered By:of a Gene by Altered By:
1. Intragenic recombination?1. Intragenic recombination?
2. Mutation?2. Mutation?
3. Complementation? 3. Complementation?
Intragenic RecombinationIntragenic Recombination
Recombination within a gene is shown Recombination within a gene is shown by recombination between two mutants by recombination between two mutants to give a wild type (non-mutant) form of to give a wild type (non-mutant) form of the gene. the gene.
OO
OO XXMutant 1Mutant 1
Mutant 2Mutant 2OOOO
Wild typeWild type
Double MutantDouble Mutant
Application: Application: Deletion MappingDeletion Mapping
Deletions prevent recombination. Deletions prevent recombination. 1.1. If no wild type recombinants can be If no wild type recombinants can be
produced in a cross between two produced in a cross between two deletion mutants, the deletions are deletion mutants, the deletions are overlapping.overlapping.
2.2. Regions of a gene can be defined by Regions of a gene can be defined by deletion mutations, and point mutations deletion mutations, and point mutations can be located within those regions. can be located within those regions.
Application: Application: Deletion Mapping Deletion Mapping
XXNon-overlapping deletionsNon-overlapping deletions
Overlapping deletionsOverlapping deletions
Unable to achieve Unable to achieve recombination to restorerecombination to restorewild typewild type
Wild TypeWild Type
Double MutantDouble Mutant
Application: Application: Deletion Mapping Deletion Mapping
Deletion and Point MutationDeletion and Point Mutationdo not overlap do not overlap
Unable to achieve Unable to achieve recombination to restorerecombination to restorewild typewild type
XXOO
Wild TypeWild Type
Double MutantDouble Mutant
OO
Deletion and Point MutationDeletion and Point Mutationoverlap overlap
OO
Deletion mapping of the rII regionDeletion mapping of the rII regionof Bacteriophage T4.of Bacteriophage T4.
Application: Deletion MappingApplication: Deletion Mapping
Problem 1, page 3-4Problem 1, page 3-4
In a particular bacteriophage, In a particular bacteriophage, four deletion mutants are four deletion mutants are crossed in pairwise crossed in pairwise combinations to test for their combinations to test for their ability to produce wild-type ability to produce wild-type recombinants. The results are recombinants. The results are given beside where + indicates given beside where + indicates that recombinants were found. that recombinants were found. Draw a deletion map for these Draw a deletion map for these mutations and divide it into mutations and divide it into subdivisions according to subdivisions according to overlapping mutations. overlapping mutations.
11 22 33 44
11 ---- ---- ++ ++
22 ---- ---- ++
33 ---- ----
44 ----
Deletion MutantsDeletion Mutants
Del
etio
n M
utan
tsD
elet
ion
Mut
ants
Application: Application: Deletion MappingDeletion Mapping
SolutionSolutionProblem 1, page 3-4Problem 1, page 3-4
22
11 33
44
Application: Deletion MappingApplication: Deletion MappingProblem 1, page 3-4Problem 1, page 3-4
There are several site-specific point mutations (A, B and There are several site-specific point mutations (A, B and C) that map in the region covered by the deletions. By C) that map in the region covered by the deletions. By coinfection of phage with one of the deletions and phage coinfection of phage with one of the deletions and phage with each of the site-specific mutations, recombinant with each of the site-specific mutations, recombinant phage are observed in the following cases. Assign each phage are observed in the following cases. Assign each site-specific mutation to one of the subdivisions of the site-specific mutation to one of the subdivisions of the deletion map. deletion map.
11 22 33 44
AA ++ ---- ---- ++
BB ---- ---- ++ ++
CC ++ ++ ---- ----
Deletion MutantsDeletion Mutants
Site
-Spe
cific
S
ite-S
peci
fic
Mut
atio
nsM
utat
ions
Application: Application: Deletion MappingDeletion Mapping
SolutionSolutionProblem 1, page 3-4Problem 1, page 3-4
22
11 33
44
1 & 2 overlap1 & 2 overlap 2 & 32 & 3 3 & 43 & 4
AABB CC
How Can the Expression How Can the Expression
of a Gene by Altered By:of a Gene by Altered By:
1.1. Intragenic recombination?Intragenic recombination?
Recombination between two mutant Recombination between two mutant forms gives a wild type version of the forms gives a wild type version of the gene --- changes in both gene --- changes in both genotypegenotype and and phenotypephenotype occur. occur.
MutationsMutations
SilentSilent AGGAGG CGG CGG
Arg ArgArg Arg
SynonymousSynonymous AAAAAA AGA AGA
Lys ArgLys Arg
MissenseMissense AAAAAA GAA GAA
Lys GluLys Glu
NonsenseNonsense CAGCAG UAG UAG
Gln StopGln Stop
Frameshift Frameshift AA(A)GACUUACCAAAA(A)GACUUACCAA
Lys-asp-leu-pro Lys-asp-leu-pro Lys-thr-tyr-glnLys-thr-tyr-gln
Change in a nucleotide can lead to change Change in a nucleotide can lead to change in amino acid found in the protein.in amino acid found in the protein.
How Can the Expression How Can the Expression
of a Gene by Altered By:of a Gene by Altered By:
2. Mutation?2. Mutation?
Change in DNA triplet can alter amino Change in DNA triplet can alter amino acid sequence of protein. acid sequence of protein.
ComplementationComplementation
Production of the wild type phenotype Production of the wild type phenotype when two different mutations are when two different mutations are combined in a diploid or heterokaryon. combined in a diploid or heterokaryon.
PRECURSORPRECURSOR INTERMEDIATEINTERMEDIATE PRODUCTPRODUCT
AA bb
enzyme Aenzyme A
mutant 1 mutant 1
aa BB
enzyme Benzyme B
mutant 2mutant 2
Application: Application: Complementation TestsComplementation Tests
1.1. If a wild type phenotype cannot be If a wild type phenotype cannot be produced in a cross between two produced in a cross between two mutants, the mutations are in the same mutants, the mutations are in the same gene (cistron).gene (cistron).
2.2. If wild type phenotype can be produced, If wild type phenotype can be produced, the mutations are in different genes. the mutations are in different genes.
Application: Complementation TestsApplication: Complementation Tests
Problem 2, page 3-4 Problem 2, page 3-4
Five mutant strains of Five mutant strains of NeurosporaNeurospora give the give the following results in complementation tests following results in complementation tests where a plus signifies complementation and where a plus signifies complementation and a minus shows a minus shows no complementation.no complementation.
Determine how many cistrons are Determine how many cistrons are represented by these mutations and represented by these mutations and indicate which mutants belong to each indicate which mutants belong to each cistron.cistron.
Application: Complementation TestsApplication: Complementation Tests
Problem 2, page 3-4Problem 2, page 3-4
11 22 33 44 55
11 ---- ++ ++ ++ ----
22 ---- ++ ++ ++
33 ---- ---- ++
44 ---- ++
55 ----
Mutant StrainMutant Strain M
utan
t S
trai
nM
utan
t S
trai
n
Application: Application: Complementation TestsComplementation Tests
SolutionSolution
Problem 2, page 3-4Problem 2, page 3-4
(1, 5) (2) (3, 4)(1, 5) (2) (3, 4)
How Can the Expression of a How Can the Expression of a Gene by Altered By: Gene by Altered By:
3. Complementation?3. Complementation?
Production of the wild type Production of the wild type phenotypephenotype when two different mutations are when two different mutations are combined in a diploid or heterokaryoncombined in a diploid or heterokaryon——genotypes are unchangedgenotypes are unchanged. .
Application: Determining the Application: Determining the Order in a Biochemical PathwayOrder in a Biochemical Pathway
Application: Observing Complementation Application: Observing Complementation for Genes within the Same Pathwayfor Genes within the Same Pathway
Suppose we have two different haploid cells, each with Suppose we have two different haploid cells, each with mutations in two of the genes in the pathway, and mutations in two of the genes in the pathway, and these haploid cells fuse to form a diploid cell. Which of these haploid cells fuse to form a diploid cell. Which of the following diploid cells can grow on minimal the following diploid cells can grow on minimal medium? medium?
1.1. ARG-EARG-E__, ARG-G, ARG-G
__ combined with ARG-F combined with ARG-F
__, ARG-H, ARG-H
__
2.2. ARG-EARG-E__, ARG-F, ARG-F
__ combined with ARG-F combined with ARG-F
__, ARG-H, ARG-H
__
1.1. ARG-EARG-E__, ARG-G, ARG-G
__ combined with ARG-F combined with ARG-F
__, ARG-H, ARG-H
__
X X
X X
2. ARG-E2. ARG-E__, ARG-F, ARG-F
__ combined with ARG-F combined with ARG-F
__, ARG-H, ARG-H
__
X X
X X
X
Suppose we have two different haploid cells, each Suppose we have two different haploid cells, each with mutations in two of the genes in the pathway, with mutations in two of the genes in the pathway, and these haploid cells fuse to form a diploid cell. and these haploid cells fuse to form a diploid cell. Which of the following diploid cells can grow on Which of the following diploid cells can grow on minimal medium? minimal medium?
1.1. ARG-EARG-E__, ARG-G, ARG-G
__ combined with ARG-F combined with ARG-F
__, ARG-H, ARG-H
_ _
YESYES
2.2. ARG-EARG-E__, ARG-F, ARG-F
__ combined with ARG-F combined with ARG-F
__, ARG-H, ARG-H
_ _
NONOOrnithine will build up because the ARG-F product is Ornithine will build up because the ARG-F product is missing.missing.
SOLUTION SUMMARYSOLUTION SUMMARY
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