Linkage and crossing over

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Chapter 7

• Linkage and crossing over

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Where were going

• Some concepts- that of linkage, which isnt hard
• Calculating recombination, and what it means
  crosses involved
• The dreaded 3 point cross.
• Thinking required!!

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Linkage Fig 7-1-

• if two genes are close on the same chromosome,
  they will be linked. Often, the linkage is
  incomplete- the closer two genes are to each
  other, the more complete the linkage

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Individual is AaBb, but the A and B genes are on
the same chromosome!
These will show crossover readily- not very close!
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We find linkage by doing testcrosses

• 7-2 If the gene is on the X chromosome, we use
  a male with the recessive traits for the cross-
  Fig. 7-3.

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NOTE only 2, NOT 4 gamete types!
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NOTE A 121 ratio, NOT 9331 ratio. bw bw/
hvhv Bw bw/hv hv bw bw/hv hv NO bw bw/
hv hv
Only 2, not 4, progeny types!
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Here we have linkage, but not complete- with
crossover!
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• If you have enough genes, you can locate them
  consecutively, through their recombination
  frequencies. Problem 13

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• 8 8 13
  6 17
• adp--------c--------vg-------------pr------b------
  -----------d
• Here, we use the information that adp is at one
  end, then arrange the results consistent with the
  data. e.g. vg has to be between adp and pr,
  since the adp-vg frequency is 16, and the pr-vg
  is 13 if vg was to the right of pr, then the
  adp-vg frequency would be gt 29
• Note these are idealized frequencies tend
  towards 50, the further away two genes are from
  each other!

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• Hi everybody!
• We have a quiz Friday ?
• You will be allowed to work with up to two others
  in doing your quiz- sort of a hybrid between quiz
  and graded group work.

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Mapping genes the (dreaded) three-point cross.

• Here, youre trying to locate the arrangement of
  three genes on the chromosome, as well as the
  distances between them.

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Fig 7-8

• Two classes are obvious the non-crossovers, and
  the double crossovers. If all the wild-type
  alleles are on one parental chromosome, and all
  the mutant alleles on the other, then the DCOs
  will have 2 and one mutant allele, or two
  mutant and one allele (y, w ec, y w ec).
  This determines the gene in the middle. To get
  the recombination frequency between two genes,
  divide the total recombinants by the total of
  offspring. The DCOs get counted twice.
• The best way to do this, is to work problems!

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Here, we know the order- on a problem, we dont
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DCOs get counted twice!
Sum 10,000
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More problemssample 12, 1517
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Fig 7-10
Here, the one that is different from the parental
chromosomes in the DCO is pr
.224 .145 .078!
.434 .356 .078!
This tells you that pr is in the
middle. V--.224----pr--.434--------bm
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coefficient of coincidence Interference

• Coefficient of coincidence C actual frequency
  of DCO/expected frequency of DCOs
• Interference is 1-C Fig. 7-10
• The expected frequency of DCOs is 9.7-
  (.224X.434)
• The actual frequency of DCOs is 7.8- This
  produces a coefficient of coincidence (C) of
  7.8/9.7 0.804. Interference is 1-C. If you
  have few DCOs interference is high, and the
  value is close to 1. If you have as many as
  expected, then C is close to1 and interference is
  close to 0. There are cases where you have
  recombination hot spots, and the interference
  can be negative!

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Things to know

• How to determine the map distance, given a cross
• Three point cross
• Interference (with formulas)