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Pedigree Analysis
 A pedigree is a diagram of family relationships
  that uses symbols to represent people and lines to
  represent genetic relationships.
 These diagrams make it easier to visualize
  relationships within families, particularly large
  extended families.
 Pedigrees are often used to determine the mode
  of inheritance (dominant, recessive, etc.) of
  genetic diseases.
Why do Pedigrees?

   Punnett squares and chi-square tests work well for
    organisms that have large numbers of offspring and
    controlled matings, but humans are quite different:
    1. small families. Even large human families have 20
    or fewer children.
    2. Uncontrolled matings, often with heterozygotes.
    3. Failure to truthfully identify parentage.


                                                           3
   Individuals may wish to be tested if:

   A.)There is a family history of one specific disease.
   B.) They show symptoms of a genetic disorder,
   C.) They are concerned about passing on a genetic
    problem to their children.




                                                            4
5
6
   We will now look at how
    various kinds of traits are
    inherited from a pedigree
    point of view.
   Traits on the Y
    chromosome are only
    found in males, never in
    females.
   The father’s traits are
    passed to all sons.
   Dominance is irrelevant:
    there is only 1 copy of each
    Y-linked gene
    (hemizygous).

                                   7
   Mitochondria are only
    inherited from the
    mother.
   If a female has a
    mitochondrial trait, all
    of her offspring inherit
    it.
   If a male has a
    mitochondrial trait,
    none of his offspring
    inherit it.
                               8
 In any pedigree there are people whose
  parents are unknown. These people are
  called “outsiders”, and we need to make
  some assumptions about their genotypes.
 Sometimes the assumptions are proved
  wrong when the outsiders have children.
  Also, a given problem might specify the
  genotype of an outsider.



                                            9
 Outsider rule for dominant pedigrees:
  affected outsiders are assumed to be
  heterozygotes.
 Outsider rule for recessive pedigrees:
  unaffected (normal) outsiders are assumed
  to be homozygotes.
   Both of these rules are derived from the
    observation that mutant alleles are rare.


                                                10
 Assume that the trait is
  recessive, in a complete
  dominance situation.
 Also assume all
  “outsiders” (people with
  unknown parents) are
  homozygous for the allele
  they are expressing : the
  dominant allele if they are
  unaffected, and the
  recessive allele if they are
  affected.

                                 11
   Assume that the trait is
    dominant in males but
    recessive in females.
   Assume all outsiders are
    homozygotes.
   Thus:
     DD is always affected
     dd is always normal
     Dd is affected in males, but
      normal in females




                                     12
 There are several
  possibilities for
  dominance, but for this
  problem assume the trait
  is dominant but only
  expressed in males.
 Affected outsider males
  are heterozygous;
  unaffected males are
  homozygous normal
 Assume that outsider
  females are homozygous
  normal.
                             13
   Mothers pass their X’s to
    both sons and daughters
   Fathers pass their X to
    daughters only.
   Normal outsider rule for
    dominant pedigrees for
    females, but for sex-linked
    traits remember that males
    are hemizygous and express
    whichever gene is on their
    X.
   XD = dominant mutant allele
   Xd = recessive normal allele

                                   14
   males get their X from their
    mother
   fathers pass their X to
    daughters only
   females express it only if
    they get a copy from both
    parents.
   expressed in males if
    present
   recessive in females
   Outsider rule for recessives
    (only affects females in sex-
    linked situations): normal
    outsiders are assumed to be
    homozygous.
                                    15
   Assume affected
    outsiders are assumed
    to be heterozygotes.
   All unaffected
    individuals are
    homozygous for the
    normal recessive allele.




                               16
   All affected are
    homozygotes.
   Unaffected outsiders
    are assumed to be
    homozygous normal
   Consanguineous
    matings are often (but
    not always) involved.



                             17
 We are now going to look at detailed analysis
  of dominant and recessive autosomal
  pedigrees.
 To simplify things, we are going to only use
  these two types.
 The main problems:
    1. determining inheritance type
    2. determining genotypes for various
  individuals
    3. determining the probability of an affected
  offspring between two members of the chart.       18
 Is it a dominant pedigree or a recessive pedigree?
 1. If two affected people have an unaffected child, it must
  be a dominant pedigree: D is the dominant mutant allele
  and d is the recessive wild type allele. Both parents are Dd
  and the normal child is dd.
 2. If two unaffected people have an affected child, it is a
  recessive pedigree: R is the dominant wild type allele and r
  is the recessive mutant allele. Both parents are Rr and the
  affected child is rr.
 3. If every affected person has an affected parent it is a
  dominant pedigree.


                                                                 19
I

                          1           2




 II

          1           2           3           4       5           6




III

      1       2   3           4       5   6       7       8   9       10


                                                                           20
 1. All unaffected are dd.
 2. Affected children of an affected parent and an
  unaffected parent must be heterozygous Dd, because
  they inherited a d allele from the unaffected parent.
 3. The affected parents of an unaffected child must be
  heterozygotes Dd, since they both passed a d allele to
  their child.
 4. Outsider rule for dominant autosomal pedigrees: An
  affected outsider (a person with no known parents) is
  assumed to be heterozygous (Dd).
 5. If both parents are heterozygous Dd x Dd, their affected
  offspring have a 2/3 chance of being Dd and a 1/3 chance
  of being DD.


                                                                21
22
   1. all affected are rr.
   2. If an affected person (rr) mates with an unaffected
    person, any unaffected offspring must be Rr
    heterozygotes, because they got a r allele from their
    affected parent.
   3. If two unaffected mate and have an affected child, both
    parents must be Rr heterozygotes.
   4. Recessive outsider rule: outsiders are those whose
    parents are unknown. In a recessive autosomal pedigree,
    unaffected outsiders are assumed to be RR, homozygous
    normal.
   5. Children of RR x Rr have a 1/2 chance of being RR and a
    1/2 chance of being Rr. Note that any siblings who have an
    rr child must be Rr.
   6. Unaffected children of Rr x Rr have a 2/3 chance of
    being Rr and a 1/3 chance of being RR.                       23
 Determining the probability of an affected offspring for
  most crosses is quite simple: just determine the parents’
  genotypes and follow Mendelian rules to determine the
  frequency of the mutant phenotype.
 In some cases, one or both parents has a genotype that is
  not completely determined. For instance, one parent has
  a 1/2 chance of being DD and a 1/2 of being Dd.
 If the other parent is dd and this is a dominant autosomal
  pedigree, here is how to determine the overall probability
  of an affected phenotype:
  1. determine the probability of an affected offspring for
  each possible set of parental genotypes.
  2. Combine them using the AND and OR rules of
  probability

                                                               24
   In our example, one parent has a 1/2 chance of being Dd
    and a 1/2 chance of being DD, and the other parent is dd.
   There are thus 2 possibilities for the cross: it could be DD x
    dd, or it could be Dd x dd. We have no way of knowing for
    sure.
   If the cross is DD x dd, all the offspring as Dd, and since the
    trait is dominant, all are affected.
   On the other hand, if the cross is Dd x dd, ½ the offspring
    are Dd (affected) and ½ are dd (normal).
   So, there is a ½ chance that the mating is DD x dd, with all
    offspring affected, and a ½ chance that the mating is Dd x
    dd, with ½ the offspring affected.
   Or: (1/2 * 1) + (1/2 * 1/2) = overall probability
   = 1/2 + 1/4 =3/4

                                                                      25

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Pedigree Analysis

  • 1.
  • 2. Pedigree Analysis  A pedigree is a diagram of family relationships that uses symbols to represent people and lines to represent genetic relationships.  These diagrams make it easier to visualize relationships within families, particularly large extended families.  Pedigrees are often used to determine the mode of inheritance (dominant, recessive, etc.) of genetic diseases.
  • 3. Why do Pedigrees?  Punnett squares and chi-square tests work well for organisms that have large numbers of offspring and controlled matings, but humans are quite different: 1. small families. Even large human families have 20 or fewer children. 2. Uncontrolled matings, often with heterozygotes. 3. Failure to truthfully identify parentage. 3
  • 4. Individuals may wish to be tested if:  A.)There is a family history of one specific disease.  B.) They show symptoms of a genetic disorder,  C.) They are concerned about passing on a genetic problem to their children. 4
  • 5. 5
  • 6. 6
  • 7. We will now look at how various kinds of traits are inherited from a pedigree point of view.  Traits on the Y chromosome are only found in males, never in females.  The father’s traits are passed to all sons.  Dominance is irrelevant: there is only 1 copy of each Y-linked gene (hemizygous). 7
  • 8. Mitochondria are only inherited from the mother.  If a female has a mitochondrial trait, all of her offspring inherit it.  If a male has a mitochondrial trait, none of his offspring inherit it. 8
  • 9.  In any pedigree there are people whose parents are unknown. These people are called “outsiders”, and we need to make some assumptions about their genotypes.  Sometimes the assumptions are proved wrong when the outsiders have children. Also, a given problem might specify the genotype of an outsider. 9
  • 10.  Outsider rule for dominant pedigrees: affected outsiders are assumed to be heterozygotes.  Outsider rule for recessive pedigrees: unaffected (normal) outsiders are assumed to be homozygotes.  Both of these rules are derived from the observation that mutant alleles are rare. 10
  • 11.  Assume that the trait is recessive, in a complete dominance situation.  Also assume all “outsiders” (people with unknown parents) are homozygous for the allele they are expressing : the dominant allele if they are unaffected, and the recessive allele if they are affected. 11
  • 12. Assume that the trait is dominant in males but recessive in females.  Assume all outsiders are homozygotes.  Thus:  DD is always affected  dd is always normal  Dd is affected in males, but normal in females 12
  • 13.  There are several possibilities for dominance, but for this problem assume the trait is dominant but only expressed in males.  Affected outsider males are heterozygous; unaffected males are homozygous normal  Assume that outsider females are homozygous normal. 13
  • 14. Mothers pass their X’s to both sons and daughters  Fathers pass their X to daughters only.  Normal outsider rule for dominant pedigrees for females, but for sex-linked traits remember that males are hemizygous and express whichever gene is on their X.  XD = dominant mutant allele  Xd = recessive normal allele 14
  • 15. males get their X from their mother  fathers pass their X to daughters only  females express it only if they get a copy from both parents.  expressed in males if present  recessive in females  Outsider rule for recessives (only affects females in sex- linked situations): normal outsiders are assumed to be homozygous. 15
  • 16. Assume affected outsiders are assumed to be heterozygotes.  All unaffected individuals are homozygous for the normal recessive allele. 16
  • 17. All affected are homozygotes.  Unaffected outsiders are assumed to be homozygous normal  Consanguineous matings are often (but not always) involved. 17
  • 18.  We are now going to look at detailed analysis of dominant and recessive autosomal pedigrees.  To simplify things, we are going to only use these two types.  The main problems: 1. determining inheritance type 2. determining genotypes for various individuals 3. determining the probability of an affected offspring between two members of the chart. 18
  • 19.  Is it a dominant pedigree or a recessive pedigree?  1. If two affected people have an unaffected child, it must be a dominant pedigree: D is the dominant mutant allele and d is the recessive wild type allele. Both parents are Dd and the normal child is dd.  2. If two unaffected people have an affected child, it is a recessive pedigree: R is the dominant wild type allele and r is the recessive mutant allele. Both parents are Rr and the affected child is rr.  3. If every affected person has an affected parent it is a dominant pedigree. 19
  • 20. I 1 2 II 1 2 3 4 5 6 III 1 2 3 4 5 6 7 8 9 10 20
  • 21.  1. All unaffected are dd.  2. Affected children of an affected parent and an unaffected parent must be heterozygous Dd, because they inherited a d allele from the unaffected parent.  3. The affected parents of an unaffected child must be heterozygotes Dd, since they both passed a d allele to their child.  4. Outsider rule for dominant autosomal pedigrees: An affected outsider (a person with no known parents) is assumed to be heterozygous (Dd).  5. If both parents are heterozygous Dd x Dd, their affected offspring have a 2/3 chance of being Dd and a 1/3 chance of being DD. 21
  • 22. 22
  • 23. 1. all affected are rr.  2. If an affected person (rr) mates with an unaffected person, any unaffected offspring must be Rr heterozygotes, because they got a r allele from their affected parent.  3. If two unaffected mate and have an affected child, both parents must be Rr heterozygotes.  4. Recessive outsider rule: outsiders are those whose parents are unknown. In a recessive autosomal pedigree, unaffected outsiders are assumed to be RR, homozygous normal.  5. Children of RR x Rr have a 1/2 chance of being RR and a 1/2 chance of being Rr. Note that any siblings who have an rr child must be Rr.  6. Unaffected children of Rr x Rr have a 2/3 chance of being Rr and a 1/3 chance of being RR. 23
  • 24.  Determining the probability of an affected offspring for most crosses is quite simple: just determine the parents’ genotypes and follow Mendelian rules to determine the frequency of the mutant phenotype.  In some cases, one or both parents has a genotype that is not completely determined. For instance, one parent has a 1/2 chance of being DD and a 1/2 of being Dd.  If the other parent is dd and this is a dominant autosomal pedigree, here is how to determine the overall probability of an affected phenotype: 1. determine the probability of an affected offspring for each possible set of parental genotypes. 2. Combine them using the AND and OR rules of probability 24
  • 25. In our example, one parent has a 1/2 chance of being Dd and a 1/2 chance of being DD, and the other parent is dd.  There are thus 2 possibilities for the cross: it could be DD x dd, or it could be Dd x dd. We have no way of knowing for sure.  If the cross is DD x dd, all the offspring as Dd, and since the trait is dominant, all are affected.  On the other hand, if the cross is Dd x dd, ½ the offspring are Dd (affected) and ½ are dd (normal).  So, there is a ½ chance that the mating is DD x dd, with all offspring affected, and a ½ chance that the mating is Dd x dd, with ½ the offspring affected.  Or: (1/2 * 1) + (1/2 * 1/2) = overall probability  = 1/2 + 1/4 =3/4 25