Chapter 15

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• Chapter 15 The Chromosomal Basis of Inheritance

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Mendel and Chromosomes
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Genetic Problem

• Cross a white-eyed male fruit fly with a
  wild-type red-eyed female fruit fly. Red is
  dominant to white.
• Show your results for both the F1 and the F2
  generations.
• Fruit fly (Drosophila melanogaster) genetics
• Genetic Symbols
• If mutant is recessive, first letter is lower
  case ( w white eye)
• If mutant is dominant, first letter is capital
  (Cy curly wings)
• Wild-type is designated by a superscript (Cy
  straight, normal wings)
• Wild-type normal or most frequently observed
  phenotype
• Mutant phenotypes alternatives to wild-type due
  to mutations in wild-type gene

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Morgans results

• This exact experiment was conducted by Thomas
  Morgan in the early 1900s.
• Do you think your results match what Morgan found
  for the F1? The F2?
• F1 yes, F2 no!
• What Morgan found in the F2 generation
• Red-eyed females
• Red-eyed males
• White-eyed males
• Where are the white-eyed females? Can you
  explain his results?

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Morgans Results

• Deduced that the gene for eye color is linked to
  sex, and is therefore on the X chromosome
• Possible female genotypes ww, ww, ww
• Possible male genotypes w and w
• Led to the discovery of sex-linked inheritance
• Are there any human traits that could follow a
  similar pattern?

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Sex-linked Inheritance

• A gene located on either sex chromosome is called
  a sex-linked gene
• Females can be carriers, while males are
  hemizygous (one allele)
• Examples in humans include color blindness,
  Duchenne muscular dystrophy, and hemophilia
• If only need one X to survive, what do females do
  with the other X?

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X Chromosome Inactivation

• One of the 2 X chromosomes in every female cell
  will become inactive and condense into a barr
  body
• The genes on this barr body are not expressed
• This process is random and independent of other
  embryonic cells (all subsequent mitotic divisions
  will include that active X chromosome)
• Discovered a gene called XIST (X-inactive
  specific transcript) that is active only in barr
  bodies. The RNA from this gene covers the X.

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Genetic Problem Part II

• Cross a dihybrid wild-type fly (gray body, normal
  wings) with a double mutant (black body,
  vestigial wings).
• Use the following letter combinations
• Gray body b
• Black body b
• Normal wings vg
• Vestigial wings vg
• Show your results for 2300 offspring both
  genotype and phenotype.

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Morgans results

• Expected results
• Wild-type gray normal (bbvgvg) 575
• Black body, vestigial wing (bbvgvg) 575
• Gray body, vestigial wing (bbvgvg) 575
• Black body, normal wing (bbvgvg) 575
• Morgans results
• Wild-type gray normal (bbvgvg) 965
• Black body, vestigial wing (bbvgvg) 944
• Gray body, vestigial wing (bbvgvg) 206
• Black body, normal wing (bbvgvg) 185

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Morgans Results cont.

• How could these numbers be so different?
• Discovered the concept of linked genes
• Genes on the same chromosome tend to be inherited
  together

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Genetic Recombination

• the production of offspring with new combinations
  of traits different from those combinations found
  in the parents results from the events of
  meiosis and random fertilization
• Parental types progeny same as parents
• Recombinants progeny different from parents
• If receive 50 recombinants from a testcross then
  the genes are on different chromosomes.
• Use Morgans data as an example.

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Morgans Example
Phenotype Genotype Expected if unlinked Expected if completely linked Actual results
Black, normal bbvgvg 575 206
Gray, normal bbvgvg 575 1150 965
Black, vestigial bbvgvg 575 1150 944
Gray vestigial bbvgvg 575 185
Recombination Frequency 391 recombinants / 2300
offspring 100 17
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Morgans Results

• Since 17 of the progeny were recombinants, the
  linkage must be incomplete. Meaning they are on
  the same chromosome, but at different locations
  on that chromosome.
• Today, we know that this results from crossing
  over in meiosis.
• Can use this type of information to create a
  genetic map of a chromosome

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Challenge problem

• A wild-type fruit fly (heterozygous for gray body
  and red eyes) is mated with a black fruit fly
  with purple eyes. The offspring are as follows
  wild-type, 721 black-purple, 751 gray-purple,
  49 black-red, 45. What is the recombination
  frequency between these genes for body color and
  eye color?

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Linkage Maps

• Create a gene or linkage map for the following
  Drosophila traits
• Frequency between b and vg loci is 17
• Recombination frequency is only 9 between b and
  cn (cn a third gene locus on the same chromosome
  for cinnabar eyes)
• The recombination frequency is 9.5 between cn and
  vg

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Linkage Map

• The recombination frequency is slightly less for
  b-vg because of double crossovers. A second
  cross over would cancel out the first thus
  reducing the frequency.
• Only an approximation

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Challenge!

• A space probe discovers a planet inhabited by
  creatures who reproduce with the same hereditary
  patterns seen in humans. Three phenotypic
  characters are height (Ttall, tdwarf), head
  appendages (Aantennae, ano antennae), and nose
  morphology (Nupturned snout, ndownturned
  snout). Since the creatures are not
  intelligent, Earth scientists are able to do
  some controlled breeding experiments, using
  various heterozygotes in testcrosses. For tall
  with antennae, the offspring are tall-antennae,
  46 dwarf-antennae, 7 dwarf-no antennae, 42
  tall-no antennae, 5. For antennae and an
  upturned snout, the offspring are
  antennae-upturned snout, 47 antennae-downturned
  snout, 2 no antennae-downturned snout, 48 no
  antennae-upturned snout, 3. A third testcross
  using a heterozygote for height and nose
  morphology was also conducted. The offspring
  are tall-upturned snout, 40 dwarf-upturned
  snout, 9 dwarf-downturned snout, 42
  tall-downturned snout, 9.
• Calculate the recombination frequencies for all
  three experiments.
• Determine the correct sequences of these three
  genes on their chromosome.

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Answer!

• A and T 12
• A and N 5
• T and N 18
• Sequence T-A-N

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Alterations in Chromosome Number

• Usually caused by nondisjunction
• Aneuploidy having an abnormal number of a
  particular chromosome
• Trisomy having one extra chromosome (2n 1)
• Monosomy missing one chromsome (2n 1)
• Polyploidy having an extra set of chromosomes
  (triploidy 3n, tetraploidy 4n, common in
  plants, rare in mammals)

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Alterations to Chromosome Structure

• Deletion removes a chromosomal segment
• Duplication repeats a segment
• Inversion reverses a segment within a
  chromosome
• Translocation segment of one chromosome is
  moved to another chromosome (reciprocal
  translocation exchange, nonreciprocal
  translocation only one chrom.)

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Human Examples

• Down syndrome Trisomy 21
• Klienfelters syndrome XXY
• Turners syndrome X
• Cri du chat specific deletion of chromosome 5
• Chronic myelogenous leukemia translocation of
  chromosomes 9 and 22