Human Genetics

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Human Genetics

• Geneticists are primarily interested in humans to
  establish the pattern of transmission of
  inherited traits specifically those associated
  with disease
• Mendelian ratios do not apply in individual human
  families because of the small size
• Controlled matings cannot be made as is possible
  in experimental genetics

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Determining how diseases are inherited in Humans
Pedigrees

• consists in collecting information on affected
  and nonaffected persons in a family, preparing a
  pedigree chart, and looking for exceptions to
  standard transmission patterns
• Pedigree A family tree drawn with standard
  genetic symbols, showing inheritance patterns for
  specific phenotype characters.
• Used to test various hypothesis and reveal
  allelic determination
• Determine if it is a rare inherited disorder

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Generations in a pedigree diagram are numbered,
by convention, using Roman numerals, starting
with the parental generation, at the top of the
diagram as generation I. For convenience, the
members of each generation are numbered across
the line, from left to right, using normal
numerals
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• Autosomal recessive disorders
• Phenylketonuria (PKU), due to mutations (loss of
  function) in an enzyme called phenylalanine
  hydroxylase - which converts phenylalanine into
  another aminio acid called tyrosine. In a child
  with PKU there is no chemical reaction to convert
  phenylalanine to tyrosine leading to a build up
  of phenylalanine in the blood and other body
  tissues. PKU is treated by a low protein diet. If
  left untreated it can result in mental
  retardation.
• Albinism Albino individuals do not produce
  pigment melanin, which protects skin from UV
  radiation, making their skin sensitive to
  sunlight.

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• Autosomal dominant disorders
• Brachydactyly Malformed hands with short
  fingers. Indian hedgehog gene (Ihh) is expressed
  in the prehypertrophic chondrocytes of cartilage
  elements, where it regulates the rate of
  hypertrophic differentiation. Misexpression of
  Ihh prevents proliferating chondrocytes from
  initiating the hypertrophic differentiation
  process.
• BRACHYDACTYLY TYPE A1 IHH, GLU95LYS In all
  affected individuals of a 4-generation Chinese
  pedigree affected with brachydactyly type A1
  (112500).
• Nat Genet 2001 Aug28(4)386-8. Mutations in IHH,
  encoding Indian hedgehog, cause brachydactyly
  type A-1. Gao B, Guo J, She C, Shu A, Yang M, Tan
  Z, Yang X, Guo S, Feng G, He L.

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Dominant relationships

• A dominant trait is the easiest to recognize.
• A dominant trait will not occur in an individual
  unless it also appears in at least one of the
  parents.
• Exceptions
• A new mutation
• Incomplete penetrance
• A fully dominant trait will not skip generations.
  It will often appear relatively common in a
  pedigree.
• Unaffected sibs will have only unaffected
  offspring.

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Recessive relationships

• In a marriage of two affected individuals, all of
  the offspring will be affected.
• A recessive trait commonly skips one or more
  generations because it is masked in heterozygotes.

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Autosomal inheritance

• An autosomal trait can be passed from a father to
  his son.
• Especially for a recessive autosomal trait,
  approximately the same number of males and
  females will be affected.

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X-linked inheritance

• Can never be passed from a father to his son
  since fathers X is passed to daughters.
• If the trait is recessive, all sons of a female
  who express the trait will also be affected.
• If recessive, the trait will occur most
  frequently in males.
• If dominant, it may occur more often in females.

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Could this trait be autosomal recessive?
YES
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Could this trait be autosomal dominant?
NO
If the trait were an autosomal dominant, the
affected child would have to have an affected
parent who could pass the trait down to the
child.
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Could this trait be X-linked recessive?
YES
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Could this trait be X-linked dominant?
NO
If the trait is a dominant (x-linked or
autosomal), an affected child must have an
affected parent. In this case, the affected male
child would have to have an affected mother if
the trait is inherited as an X-linked dominant
trait.
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Could this trait be autosomal recessive?
NO
The parents would have to be homozygous (aa) and
could only produce homozygous, affected children.
This pedigree contains two, unaffected children.
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Could this trait be autosomal dominant?
YES
If the this trait is inherited as an autosomal
dominant both parents could be heterozygous Aa
and could produce affected children AA or Aa or
they could produce unaffected children aa.
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http//www.people.virginia.edu/rjh9u/pedhint.html
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RECESSIVE TRAIT (A-) unaffected and (aa)
affected What is the genotype of the mother? What
is the genotype of the father? What are the
genotypes of the children?
1
2
I
II
1 2 3 4 5
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RECESSIVE TRAIT (A-) unaffected and (aa)
affected Mothers genotype Aa Fathers genotype
aa Aa, aa, Aa, Aa, aa
Aa
aa
1
2
I
II
3 4 5 6 7
aa
Aa
Aa
Aa
aa
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DOMINANT TRAIT (A-) affected and (aa)
unaffected What is the genotype of the
mother? What is the genotype of the father? What
are the genotypes of the children?
1
2
I
II
3 4 5 6 7
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DOMINANT TRAIT (A-) affected and (aa)
unaffected Mothers genotype aa Fathers
genotype Aa aa, Aa, aa, aa, Aa
aa
Aa
1
2
I
II
3 4 5 6 7
aa
Aa
aa
aa
Aa
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A couple has a female child with Tay Sachs
disease, and three unaffected children. Neither
parent nor any of the four biological
grandparents of the affected child has had this
disease. The most likely genetic explanation is
that Tay Sachs disease is inherited as a(n)
______________ disease. WHY?
autosomal recessive The disease is recessive
because both parents are unaffected, and
autosomal because a female child is affected but
her father is not.
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Individual Independent Events

• Example a gamete carrying a dominant allele
  being formed in a heterozygote.
• ½
• Example a gamete carrying a dominant allele
  forming in a homozygote.
• 1

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Sequence of independent events where the order is
set or irrelevant

• Example a family of three children with
  boy-boy-boy
• Multiply individual probabilities
• ½ x ½ x ½ 1/8

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Sequence of events in which different orders must
be pooled

• A six child family composed of 4 girls and 2 boys
  in any order.
• Probability formula n!/s!t! (p)s(q)t
• n number of individual in the family
• P probability of the first event
• Q probability of the second event
• S number of cases in the first event
• T number of cases in the second event
• 6!/4!2! (0.5)4(0.5)2

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Problem

• A husband and wife, both heterozygous for PKU
  gene (autosomal recessive), plan to have 6
  children. What is the probability of that four
  of the offspring will be normal and two will have
  PKU.
• 6!/4!2!(0.75)4(0.25)2 0.297

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