Genetic Fine Structure Nature of the Gene at the Molecular Level

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.


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.


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


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


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


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


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



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.



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


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


(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


__, ARG-H, ARG-H

__



__, ARG-H, ARG-H

__

X X

X X

2. ARG-E2. ARG-E__, ARG-F, 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?



__, ARG-H, ARG-H

_ _

YESYES

2.2. ARG-EARG-E__, ARG-F, 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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Genetic Fine Structure Nature of the Gene at the Molecular Level