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Sexdependent inheritance or trait expression
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(horns in some sheep breeds) Sex linkage. Genes on sex chromosomes. usually on X but not Y ... Tortoiseshell cat. Males. XOY orange. XoY black. Females. XOXO ... – PowerPoint PPT presentation
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Title: Sexdependent inheritance or trait expression
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Sex-dependent inheritance or trait expression
- Sex-limited traits
- Sex-influenced traits
- Sex-linked inheritance
- x-linked
- y-linked (holandric inheritance)
- Of these, only sex-linkage necessarily involves
the X, Y chromosomes.
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Sex-limited trait
- Expressed in only one sex (hormonal influence).
- Genes affecting usually on autosomes.
- E.g., milk traits, egg production
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Sex-influenced trait
- Genes on autosomes.
- Trait is expressed by both sexes.
- Degree of dominance or penetrance varies by sex.
- Examples
- horns in some breeds of sheep.
- scurs in some breeds of cattle.
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Sex-influenced trait example(horns in some sheep
breeds)
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Sex linkage
- Genes on sex chromosomes
- usually on X but not Y
- Most mammals
- XY male
- XX female
- males only carry a single gene.
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Red-green color blindnessgenetic key
- Males
- XBY ? normal
- XbY ? afflicted
- Females
- XBXB ? normal
- XBXb ? normal
- XbXb ? afflicted
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Example mating XBY XBXb
- Genotypic ¼ XBXB, ¼ XBXb,
- ¼ XBY, ¼ XbY
- Phenotypic ½ normal ?,
- ¼ normal ?, ¼ afflicted ?,
- or ? all normal
- ? 1/2 normal, 1/2 afflicted
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Tortoiseshell cat
- Males
- XOY ? orange
- XoY ? black
- Females
- XOXO ? orange
- XOXo ? tortoiseshell
- XoXo ? black
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Holandric inheritance
- Y-linked.
- Apparently, some genes reside on Y but not X.
- Trait expressed only in males.
- Much less common than X-linked.
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Imprinting
- Non-Mendelian
- Epigenetic modifications (methilation of the DNA)
- Parent of Origin effects
- Maternally or Paternally expressed
- IGF2 Paternally Expressed
- IGF2R Maternally Expressed
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Polar overdominance
- Non-Mendelian.
- Expression depends on which parent an allele came
from. - Expression depends on Genotype (Heterozygous)
- Example muscle hypertrophy in sheep callipyge
- Gene mapped to telomeric region of 18 chromosome
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Inheritance of callipyge
- 1) Founder callipyge ? normal ? ?
- 1/2 normal1/2 callipyge
- Initial assumption was dominant mutation
- Cc ? cc ? ? 1/2 Cc (callipyge)
- 1/2 cc (normal)
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2nd test mating
- Normal ram mated to daughters of founder.
- cc ? Cc ? ? 0 callipyge
- 35 normal
- Disagreed from expected 5050 phen. ratio.
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3rd test mating
- het. callipyge ? het. callipyge ?
- (son of founder daughters of founder)
- (Cpatcmat Cpatcmat)
- Result 29 callipyge, 71 normal
- Expected was 75 callipyge, 25 normal
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New hypothesis
- Cmat inactivates Cpat (polarity).
- Assumed genetic key for 3rd mating
- Cpat Cmat normal (parents dont matter)
- Cpat cmat callipyge
- Cmat cpat normal
- cpat cmat normal
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4th test mating
- Normal CC ? normal cc ? ? all Cpatcmat
- Expect all callipyge, but 3/30 were normal,
possibly because of grandparental imprinting.
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Single-gene abnormalities
- Brain hernia
- hydrocephalus
- tail-less calf
- horns in some breeds
- protoporphyria
- Mule foot
- extra dewclaw in pigs
- unwanted coat color
- (black versus red)
- dwarfism in cattle
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Single-gene abnormalities
- Why are most recessive?
- Unwanted dominant traits easily eliminated by
selection. - Recessive alleles difficult to eliminate.
- Hidden in heterozygous carriers.
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Embryonic Death
- WW embryonic death.
- Ww true white (pink skin and dark eyes),
regardless of which alleles are present at other
loci. - ww non-white (depends on other loci, at least
11 loci, and can be black, bay, cremelo,
appaloosa) - Changes Expected Mendelian Proportions
- 121 (.25/.5/.25) become 12(.33/.66) for Ww x
Ww - White horses occasionally result from dark
parents, possibly due to mutation.
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Snorter dwarf in cattle
- DD normal
- Dd normal?
- dd dwarf
- Problem in mid-1950s, especially Hereford.
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Detecting heterozygous carriers
- Especially important to test elite A.I. Sires.
- DNA tests are available for a few traits.
- Otherwise, can set up test matings.
- Only takes 1 afflicted offspring to prove a
parent is a carrier.
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1. Mate to known carriers.
- R? Rr
- If sire is a carrier (Rr), then we expect 3/4
normal and 1/4 afflicted offspring. - One afflicted offspring means sire is Rr.
- All normal offspring means??
- Prob. Of all normal is (3/4)n.
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Probability of failure to detect a carrier (Rr)
of unwanted recessive
- Number normal Probability (.75)n
- offspring (n)
- 1 .75
- 2 .56
- 5 .24
- 10 .056
- 15 .013
- 20 .003
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2. Mate sire to normal daughtersof known
carriers.
- Daughters expected to be 1/2 RR, 1/2 Rr.
- Thus, matings with sire in question are
- 1/2 R? RR
- 1/2 R? Rr
- Expect 1/8 afflicted offspring if sire is Rr.
- (2x as many matings needed compared to 1.)
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3. Mate sire to afflicted females.
- R? rr
- If sire is Rr, then expect 1/2 Rr, 1/2 rr
- P(normal) .5 for a given offspring.
- P(n/n normal) (.5)n
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Probability of failure to detect a carrier (Rr)
of unwanted recessive
- Number normal Probability (.5)n
- offspring (n)
- 1 .50
- 2 .25
- 3 .125
- 5 .031
- 10 .001
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4. Mate to own daughters
- Cant compute exact probabilities, but
- If sire is a carrier, then 1/2 of daughters
should be also. Then, - 1/2 Rr RR ? all normal
- 1/2 Rr Rr ? 3/4 normal, 1/4 rr
- Overall 7/8 normal, same at mating 2.
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Note
- These matings are generally not useful to test
for defects resulting in embryonic mortality. DNA
tests would be useful.
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What to do if you suspect genetic defect in your
herd?
- Report to breed association.
- Eliminate carriers (parents of affected animals)
- Set up controlled test matings (for important
sires). - Use semen from sires known to be normal
homozygous - To completely eliminate disease is necessary to
use only normal animals. - The progeny of known normal animals will be
normal (no need to test them anymore)
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Linkage and RecombinationException to Law of
Independent Assortment
