gene linkage and mutation
TRANSCRIPT
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G A Y A T R I D A V E
Gene linkageAnd their role in genetic analysis
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What is Linkage?
Linkage is defined genetically: the failure of twogenes to assort independently.
Linkage occurs when two genes are close to eachother on the samechromosome.
However, two genes on the same chromosome arecalled syntenic.
Linked genes are syntenic, but syntenic genes arenot always linked. Genes far apart on the same
chromosome assort independently: they are notlinked.
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Discovery of Linkage
William Bateson and R.C. PunnettWorked on two traits of sweet peas,And explain the concept of linked gene.
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If two genes are on different
chromosomes
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Now suppose both gene A and B
were next to each other on thesame chromosome.
What happens to the ratios inthis diagram?
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And half look like they got a mix
of both parents chromosomes
Half look like they got a set ofthe parents chromosomes
If two genes are on different
chromosomes
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Now suppose both gene A and B
were next to each other on thesame chromosome.
What happens to the ratios inthis diagram?
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Gene Mapping
Gene mapping determines the order of genes and the relativedistances between them in map units
1 map unit= 1 cM (centimorgan)
Gene mapping methods use recombination
frequenciesbetween alleles in order to determine the relativedistances between them
Recombination frequencies between genes are inverselyproportional to their distance apart
Distance measurement: 1 map unit = 1 percent recombination (truefor short distances)
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Centimorgan
honor his mentor, T. H. Morgan. -1 centimorgan isdefined as 1% recombinant progeny in a testcross.
Thus, one can calculate the map distance as:
R = (#recombinants/total) x 100 = distance incM
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Fig. 4.6
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Gene Mapping
Genes with recombination frequencies less than 50 percent are onthe same chromosome = linked)
Linkage group = all known genes on a chromosome
Two genes that undergo independent assortment haverecombination frequency of 50 percent and are located on
nonhomologous chromosomes or far apart on the samechromosome = unlinked
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Now cross (AB ab) F1 progeny with (ab ab) testerto look for recombination on these chromosomes.Suppose you Get
ABab 583
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Cross (ABD abd) F1 progeny with (abd abd) tester
Suppose you Get
ABDabd 580
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Cross (ABD abd) F1 progeny with (abd abd) tester
Suppose you Get
ABDabd 580
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Kind s of crossing over
1. Single crossing over: When only one chiasmaoccurs only at one point of the chromosome pair. it iscalled single crossing over. It produces two non-crossover chromatids and two cross over chromatids.2. Double crossing over: When the crossing overoccurs at two points between any two given points inthe same chromosome pair, it is called double
crossing over. It produces four crossovers. In doublecrossing over following two types of chiasma may beformed.
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Continued.
(A) Reciprocal chiasma: In reciprocal chiasmasame two chromatids are involved in the secondchiasma as in the first. It produces only two non-crossover chromatids, because, it restores the order
which was changed by the first chiasma. In thischiasma out of four chromatids only two areinvolved.
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(B) Complimentary chiasma: When both thechromatids taking part in the second chiasma aredifferent from those chromoatids involved in the firstchiasma, the chiasma is called complimentarychiasma. It produces four single crossovers but nonon-cross over. In complimentary chiasma all thefour chromatids of a tetrad are involved.
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3. Multiple crossing over: When crossing overoccurs at three, four, or more points between any twogiven points in the same chromosome pair these arecalled triple, quadruple or multiple crossing over.
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Double cross over
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Two-stranddouble crossover (d.c.o.)
Three-stranddouble crossover (d.c.o.)
Four -stranddouble crossover (d.c.o.)
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Chromosome interference: crossovers in one region decrease theprobability of a second crossover close by
Coefficient of coincidence= observed number of double recombinantsdivided by the expected number
\
Interference = 1-Coefficient of coincidence
If the two crossovers were independent,
we would expect that the probability of seeing two recombination events occur would be
0.132 between A-D AND 0.064 between D-B
0.132 X 0.064 = 0.008For every 1448 progeny, this would be (1448x0.008)=12.23 double recombinants
We actually observed only (5+3)= 8 double recombinants
So the Coefficient of coincidence = observed / expected = 8/12.23 =0.65
Interference = 1-Coefficient of coincidence
= 1- 0.65
= 0.35
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Mutation
Mistakes in the transmission of the geneticinformation
1 in 1,000,000 genes are mutated
types: Chromosomal
Gene
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Chromosomal
Involve segments, whole, and
entire sets of chromosomes, it is a
change in the number and structure of chromosomes
Basic types:
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NonDisjunction: whole chromosome mutations, it isa failure to separate during meiosis
Too many chromosomes: called polyploidy
Trisomy: (2n+1)Downs Syndrome Too Few Chromosomes: from the left over gametes
Turners Syndrome: (2n-1)
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Gene Mutations
Mutations that result from a change involving manynucleotides within a gene, some may involve onlyone nucleotide
Point mutations: smallest change,effect singlenucleotide.
Frameshift mutation: when a point mutation singlebase is inserted or deleted, shifting the entire codon,
this changes every codon following the mutation.
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Mutation and their role in genetic analysis
How do we determine the relationship among severalmutations that cause the same phenotypic change?
What are the smallest units of DNA capable of mutationand recombination?
Are the gene and its protein product colinear?
Relationship betweenphenotype and gene can be studied through mutants
identified by phenotype distinct from wild type.
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Complementation test
Complementation test (cis-trans test) determineswhether independently isolated mutations for thesame phenotype are in the same or different genesby crossing two mutants.a. If mutations are in different genes, phenotype will be wild type
(complementation).
b. If mutations are in the same gene, phenotype will be mutant(no complementation
C l t ti t t t d t i h th t ti
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Complementation test to determine whether 2 mutationresulting in same phenotype are on same or different gene
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Drosophila provides an example. Wild-type body color isgrey-yellow. If two true-breeding mutant black-bodiedstrains are crossed, all F1 are wild type (Figure 13.2).
a. Genes are e (ebony) and b (black). Black parents are
homozygous mutant but in different genes (e/eb+/b+) and(e+/e+b/b).
b. F1 are heterozygous at both loci (e+/eb+/b) and therefore wildtype, showing complementation.
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