Genes linked and unliked

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Genes linked and unliked

• Pages 92 and 93 study guide
• Chapter 13-3 of your Raven Johnson Book

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Dihybrid Cross with Linkage

• Genetic linkage occurs when particular alleles
  are inherited together.
• alleles that are on the same chromosome are more
  likely to be inherited together, and are said to
  be linked.
• Linkage Animation

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Lets go back and look at some rules

• Recessive Genes The three criteria
• The first appearance of the recessive trait
  within a family usually is in the children of the
  unaffected parents.
• 25 of the children will be express trait.
• Both males and females can express the trait
  unless it is a recessive sex linked gene.

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Dominant Genes 4 rules

• If the trait is dominant, it will be expressed in
  all generations.
• The trait is passed from the affected parent to
  about 50 of his/her children.
• Any parent that does not express the trait does
  not transmit it to any of his/her children.
• Both males and females can express and transmit
  the trait.

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Remember Mendel

• Monohybrid cross has a 31 ratio in the F2
  generation, when parental type is AA x aa
• Dihybrid cross has a 9331 in the F2 generation
  when AA/BB x aa/bb parental type are crossed.

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Morgan

• The Nobel Prize in Physiology or Medicine 1933
• "for his discoveries concerning the role played
  by the chromosome in heredity"  Thomas Hunt
  Morgan USA California Institute of Technology
  Pasadena, CA, USA b. 1866d. 1945

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Gene Linkage

• Genes on the same chromosome are said to be
  linked (they tend to be inherited together)
• Recombination Animation

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linkage

• Wild type fruit fly
• Grey body and normal wings
• b gray
• b black (mutant)
• vg normal wings
• vg vestigial wings

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The cross

• b b vg v heterozygous gray with
  normal wings
• X
• bb vgvg (mutants black bodies vestigial wings)
• Gametes would be bvg and bvg

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According to Mendel

• This cross should have produced a ratio of
  1111
• 1 gray normal wing
• 1 black vestigial
• 1gray vestigial
• 1 black normal

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What did he really get?

• bb vgvg 965 (wild type)
• bb vgvg 944 (black vestigial)
• bb vgvg 206 (gray vestigial)
• bb vgvg 185 (black normal)
• Morgan reasoned that body color and wing shape
  are usually inherited together in a specific
  combination. The genes are located on the same
  chromosome

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So what could it be?

• All the phenotypes are represented and new
  combinations are due to crossing over.

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Crossing over in Prophase I of Meiosis
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Genetic Recombination

• exchange of alleles between homologous
  chromosomes.
• The alleles of parental linkage groups separate
  and new associations of alleles are formed in the
  gametes.
• Off spring showing this new combination are
  called recombinants

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Example recombination

• In peas. Seed coat and seed shape are unlinked
• Cross YyRr yellow round
• With yyrrr green wrinkled

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What did he see?

• ¼ YyRr yellow round
• ¼ yyrr green wrinkled
• ¼ yyRr green round
• ¼ Yyrr yellow wrinkled

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So.

• ½ have parental phenotypes yellow round and
  green wrinkled
• Because the other offspring have different
  combinations of seed shape and color then their
  parents they are recombinants

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Crossing over

• Linked genes do not sort independently because
  they are located on the same chromosome and tend
  to move together through meiosis and
  fertilization
• But recombination does exist between linked genes

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• http//bcs.whfreeman.com/thelifewire/content/chp10
  /1002s.swf

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Take another look

• bb vgvg 965 (wild type)
• bb vgvg 944 (black vestigial)
• bb vgvg 206 (gray vestigial)
• bb vgvg 185 (black normal)
• Morgan reasoned that body color and wing shape
  are usually inherited together in a specific
  combination. The genes are located on the same
  chromosome

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In this case

• When looking at the percentage of offspring that
  did not have parental phenotype we find that 17
  do not.
• This led Morgan to propose some mechanism which
  allowed for the exchange of segments between
  homologous chromosomes.
• Called Crossing Over.

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Now enters Alfred Sturterant a student of
Morgans
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Sturtevant (1917) hypothesis

• Different recombination frequencies reflect
  different distances between genes on a
  chromosome. Or
• If two genes are far apart on a chromosome there
  is a higher probability that a crossover event
  will separate them than if the two genes are
  close together

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Frequency of crossover exchange...            
of chromatids of a homologous pair at synapsis
forming a chiasmata... 

• Frequency is GREATER the FARTHER apart 2 genes
  are is proportional to relative distance
  between 2 linked genesRelative distance is
  established as...
• 1 crossover frequency 1 map unit of map
  distance
• 1   CrossOver  Freq      
• 1  centiMorgan (cM)

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An example of linkage

• Start with two different strains of corn (maize).
  
• one that is homozygous for two traits
• yellow kernels (C,C) which are filled with
  endosperm causing the kernels to be
• smooth (Sh,Sh).
• a second that is homozygous for
• colorless kernels (c,c) that are wrinkled because
  their endosperm is shrunken (sh,sh)

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then

• When the pollen of the first strain is dusted on
  the silks of the second (or vice versa), the
  kernels produced (F1) are all yellow and smooth.
• So the alleles for yellow color (C) and
  smoothness (Sh) are dominant over those for
  colorlessness (c) and shrunken endosperm (sh).

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• To simplify the analysis, mate the dihybrid with
  a homozygous recessive strain (ccshsh). Such a
  mating is called a test cross because it exposes
  the genotype of all the gametes of the strain
  being evaluated.
• According to Mendel's second rule, the genes
  determining color of the endosperm should be
  inherited independently of the genes determining
  texture. The F1 should thus produce gametes in
  approximately equal numbers

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• If the inheritance of these genes observes
  Mendel's second rule i.e., shows independent
  assortment, union of these gametes should produce
  approximately equal numbers of the four
  phenotypes. But as the chart shows, there is
  instead a strong tendency for the parental
  alleles to stay together. It occurs because the
  two loci are relatively close together on the
  same chromosome. Only 3.6 of the gametes contain
  a recombinant chromosome

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again

• During prophase I of meiosis, pairs of duplicated
  homologous chromosomes unite in synapsis and then
  nonsister chromatids exchange segments during
  crossing over. It is crossing over that produces
  the recombinant gametes. In this case, whenever a
  crossover occurs between the locus for kernel
  color and that for kernel texture, the original
  combination of alleles (CSh and csh) is broken up
  and a chromosome containing Csh and one
  containing cSh will be produced.

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• Chromosome Maps
• The percentage of recombinants formed by F1
  individuals can range from a fraction of 1 up to
  the 50 always seen with gene loci on separate
  chromosomes (independent assortment). The higher
  the percentage of recombinants for a pair of
  traits, the greater the distance separating the
  two loci. In fact, the percent of recombinants is
  arbitrarily chosen as the distance in
  centimorgans (cM), named for the pioneering
  geneticist Thomas Hunt Morgan. In our case, the c
  and sh loci are said to be 2.8 cM apart.

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Lets do this again

• Eye color on drosophila
• Cinnabar. A mutant phenotype brighter than red
  (wild)
• cn cinnabar
• cn wild
• b black
• b gray

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• Recombination frequencies reflect the distances
  between genes on a chromosome. The
• farther apart two genes are, the more likely that
  a crossover will occur between them and
• hence the higher the recombination frequency. A
  genetic map, called a linkage map, can
• be constructed based on recombination
  frequencies. 1 map unit , called a centimorgan,
• equals a 1 recombination frequency.

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• Linkage Maps

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• After tallying the results we find a
  recombination frequency between cn and b is 9.
• Remember from a previous slide that we showed
  that the crosses between b and vg loci are twice
  as frequent (17)

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• If we assume that the probability of crossing
  over between 2 genes is directly proportional to
  the distance between them, then the distance
  along the drosophila chromosome between b and vg
  must be twice as great as the distance between b
  and cn.

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• Thus the b and cn loci are separated by 9 map
  units .
• While the b and vg genes are 17 map units apart

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What is the sequence

• We can eliminate the sequence b - vg- cn
• Since we know that cn is closer to b then is vg
• b ------------------------------- vg 17 map
  u.
• b ------------ cn 9 map units or
  centimorgans

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• In order to continue we must determine the
  frequency of crossing over between the vg and cn
  genes.
• Experiment done we find the recombination
  frequency of 9.5 or 9.5 centimorgans
• Or b-------------cn------------vg
• 9cM 9.5cM
• There is some discrepancies due to multiple
  crossing over episodes