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Summary Lecture Notes Genetics Maps & Linkage | Week 4 | Wageningen

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Lecture notes from Week 4 of Fundamentals of Genetics at Wageningen University, covering Chapter 4 on genetic mapping and linkage analysis. Topics include diagnostics of linkage, recombination frequencies, mapping by recombinant frequency, gene order determination through three-point testcrosses, molecular markers, and gene interactions including complementation tests. Essential for understanding how genetic maps are constructed and used in strain building and evolutionary analysis - ideal preparation for exams on classical and molecular genetic mapping techniques.

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Week 4
Ch 4
 Genetics maps are useful for strain building, for interpreting evolutionary
mechanisms, and for discovering a gene’s unknown function.
 Discovering a gene’s function is facilitated by integrating information on
recombination-based and physical maps
4.1 diagnostics of linkage
 When two genes are close together on the same chromosome pair (that is, when they
are linked), they do not assort independently but produce a recombinant frequency of
less than 50 percent. Hence, a recombinant frequency of less than 50 percent is a
diagnostic for linkage
 Two equally frequent nonrecombinant classes totaling in excess of 50 percent
 Two equally frequent recombinant classes totaling less than 50 percent
 For linked genes, recombinants are produced by crossovers between nonsister
chromatids during meiosis. (the two new combinations are called crossover products)
 Chiasmata are the visible manifestations of crossovers (a cross-shaped structure
often formed between two nonsiters chromatids)
 Crossovers produce recombinants for linked genes
o Homologous chromosomes pair at meiosis
o Chromosomes get ds DNA breaks
o Ds breaks are repaired by recombination between non-sister chromatids
o  crossing over visible as a chiasma (chiasmata)
o Each chromosome pair has at least 1, often a few chiasmata
o Very precise process: recombination does not introduce mutations
 Cis conformation: in a heterozygote having two mutant sites within a gene or within a
gene cluster, the arrangement AB/ab
 Trans conformation: in a heterozygote with two mutant sites within a gene or gene
cluster, the arrangement Ab/aB
 A crossover is the breakage of two DNA molecules at the same position and their
regioning in two reciprocal recombinant combinations
 Non crossover: only parental type gametes
 Single crossover: 2 parental and 2 recombinant type gametes
 Reciprocal types are equally frequent: the proportion of recombinants correlates to
the distance separating two gene loci on a chromosome map
 If 10% of meiosis there is crossing over between A-B, there will be 5% recombinant
meiotic products
 Multiple (two or more) crossovers can produce both recombinant and parental
chromatids (so you can have crossing over between just two chromomes or three or
four making a difference in what new genotypes can be created)
 If you cross ABC x abc (for example)
o Two: ABC, AbC, aBc and abc
o Three: ABc, abC, aBC, abc
o Four: ABc, Abc, aBC, abC

4.2 mapping by recombinant frequency
 The farther apart the genes are, the more likely that a crossover will take place and
the higher the proportion of recombinant products will be

,  Thus the proportion of recombinants is a clue to the distance separating two gene loci
on a chromosome map
 Recombinant frequencies for different linked genes range from 0 to 50%, depending
on their closeness (farther apart closer to 50%)
 Distance in m.u. based on recombinant fraction of markers ABC
o A-B: 2/16 = 0,125  12,5% rec = 12.5 cM
o B-C 6/16 = 0,375  37,5% rec = 37,5 cM
o A-C ong 50 cM (12,5 + 37,5)
 Distance based on average chiasma frequency (better)
o A-B: ¼ chiasma/meiosis  1/4x50 cM = 12,5 cM
o B-C: 1 chiasma/meiosis  1x50cM = 50 cM
o A-C ong. 62,5 cM
 A region with 1 crossover on average has a genetic distance of 50 cM, so 66 cM
results from 66/50 = 1,3 crossovers on average per meiosis  1 or 2 per meiosis
expected for this chromosome
 Linkage map: a chromosome map; an abstract map of chromosomal loci that is
based on recombinant frequencies
 The greater the chance of crossovers in the region between the genes the greater the
proportion of recombinants that would be produced
 Thus by determining the frequency of recombinants, we can obtain a measure of the
map distance between the genes
 Recombination between linked genes can be used to map their distance apart on a
chromosome. The unit of mapping (1 m.u.) is defined as a recombinant frequency of
1 percent. Map distances are roughly additive
 Three-point (and higher) trestcrosses enable geneticist to
evaluate linkage between three (or more) genes and to
determine gene order, all in one cross
 Single gene inheritance and two-gene inheritance (linked
and unlinked) can be inferred form diagnostic phenotypic
ratios in both selfing and testcrossing
4.3 mapping with molecular markers
 Molecular markers: a DNA sequence variant that can be
used to map an interesting phenotype to a specific region
of DNA
 Loci of any DNA heterozygosity can be mapped and used as molecular chromosome
markers or milestones
4.4 using the chi-square test to infer linkage
 The chi-square test is useful in testing the significance of deviations form a 1:1:1:1
ratio in deducing linkage between two genes
 Interference: are crossovers in adjacent regions independent evens, or does c.o. in
one region affect the likelihood of c.o. in the other region?
 Generally, there is interference inhibition of nearby crossovers interference (I)
 I = 1 – observes number of DCO/expected number of DCO
o Observed: 8
o Distance between v-ct = 13,2 and ct-cv = 6.4
o Total amount of gametes: 1448
o So I = 1- (8/(0,132x0,064x1448) = 1/3

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