Notes
Slide Show
Outline
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GENETIC LINKAGE
  • Genes are connected together in a linear order on chromosomes.
  • The connectedness means all possible gametic genotypes are not produced EQUALLY frequently.
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Independent Assortment
  • In a double heterozygote, AaBb, which includes both Ab/aB and AB/ab, independent assortment means that the segregation of A and a is independent of the segregation of B and b, i.e., the AB, Ab, aB & ab arrangements in the gametes are expected to be equally frequent.
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LINKAGE
  • Two (or more) loci have to be heterozygous to ‘see’ linkage (in transmission genetics).
  • Linkage is non-independent assortment of alleles at separate loci during meiosis.
  • Linkage is a reflection of the fact that loci occupy specific places on the DNA of chromosomes.


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LINKAGE & parental types
  • An individual produces more gametes that have its parental configuration than it produces gametes with one allele from the father and one from the mother.


  • An Ab/aB individual will produce more Ab and aB gametes (the arrangement in the gametes that united to form it, and therefore called parental) than AB and ab gametes (which are called recombinants).
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Recombination fraction, r, is a variable with a different value for each pair of loci.
  • The closer together two loci are the less the proportion of recombinant types.
  • r takes values between 0 and ½.


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Estimating r
  • For a pair of loci, the expected proportion of the two parental types together is 1 – r, and the expected proportion of the two recombinant types will sum to r.
  • Each parental type is expected to be (1-r)/2.
  • Each recombinant type has the expected frequency of r/2, as they are expected to be equally frequent.
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An Example of Linkage
  • We will look at an example from the pea.
  • To look at linkage we have to be studying two heterozygous loci, so normally we have variation in two different characters.
  • Variety K TALL PUBESCENT leaves
  • Variety L SHORT GLABROUS leaves


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Linkage of height and hairiness (leaf pubescence)
  • Variety K TALL PUBESCENT
    •  SSGG
  • Variety L SHORT GLABROUS
    • ssgg
  • The F1 genotype SG/sg slash notation
  • Possible gametic genotypes of F1:
    • SG Sg sG sg
  • Expected proportions:
    • ¼ ¼ ¼ ¼
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EVALUATING GAMETIC PROPORTIONS with test cross
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Observed proportions are not compatible with equal amounts of each gametic genotype.
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Expected equalities
  • Two parental gametes should be about equally frequent and two recombinant types should also be about equally frequent.


  • If those equalities are not true, then one should suspect differential viability of the types or some other process.
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Calculating the Value of r,
the distance between two loci
  • rSG = number of recombinant gametes divided by total number of gametes.
  • rSG = (18+22)/(178+18+22+182)
  • rSG = 40/400 = 0.10
  • The r value is multiplied by 100 to convert r to ‘map units’.
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If the F1 had Sg/sG genotype (instead of SG/sg) then the recombinant/parental types would have been reversed in the test cross.
  • Tall Tall Short Short
  • Pubesc.. Glabrous Pubescent Glabrous
  • 19 136 130 15


  • In this case, the parental gametes are Sg (TALL GLABROUS) & sG (SHORT PUBESCENT) and the recombinant gametes are SG & sg. In this example, r is estimated to be 0.113 = 34/300, a small amount larger than the 10 map unit estimate of the original data.
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Three heterozygous loci
  • Consider the genotype Hqr/hQR
  • Eight possible gametic genotypes can be produced
    • HQR, HQr, HqR, Hqr, hQR, hQr, hqR & hqr
  • There are only two parental types no matter how many loci are considered, in this case the parental types are Hqr & hQR
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Genes are arranged linearly
  • When the distances between many pairs of loci were determined, it was discovered that the genes could be ordered along a line.
  • The MAP is order of genes (loci) on chromosomes.
  • The process of generating the map is called mapping.
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A mapping problem
  • The recombination fraction, r, has been estimated for 5 pairs of 4 loci.
  • ED = 6 map units
  • IT = 2 map units
  • DT = 6 map units
  • DI = 4 map units
  • EI = 10 map units


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CONSTRUCTING THE MAP
of these four loci
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Chromosomal basis of sex determination
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Bacterial GENES
  • In bacteria the genes are all linked together into a single large circle.
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Bacterial SEX
  • In bacteria sex does not involve the union of two equal (more or less) genomes. Instead, one cell serves as a donor and the other cell functions as a recipient. Typically only part of the genome is transferred.
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LINKAGE RECAPITULATION
  • Among gametes there is always an excess of the two parental arrangements of genes no matter how many loci are studied.
  • Studies of many pairs of loci lead to the realization that all genes were arranged in a linear order.
  • The order of the loci is the same in all individuals in a species.
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Take Home Problem
  • Consider two loci on a circular genome. If many genomes are broken up haphazardly into smaller pieces, how do you expect the probability that two loci are on the same piece to be related to the distance between the loci in the genome?
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Vocabulary
  • Recombination fraction
  • Parental gametic type
  • Recombinant gametic type
  • Double heterozygote
  • chromosome
  • Slash notation
  • Meiosis
  • Map unit
  • Sex linkage
  • Sex determination
  • mapping