Transcript Document

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
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
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
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.
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 Aug;28(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.
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.
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.
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.
X-linked inheritance
• Can never be passed from a father to his son
since father’s 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.
Could this trait be autosomal recessive?
YES
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.
Could this trait be X-linked recessive?
YES
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.
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.
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.
http://www.people.virginia.edu/~rjh9u/pedhint.html
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?
2
1
I
II
1
2
3
4
5
RECESSIVE TRAIT:
(A-) unaffected and (aa) affected
Mother’s genotype: Aa
Father’s genotype: aa
Aa, aa, Aa, Aa, aa
I
Aa
aa
1
2
II
3
4
5
6
7
Aa
aa
Aa
Aa
aa
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?
2
1
I
II
3
4
5
6
7
DOMINANT TRAIT:
(A-) affected and (aa) unaffected
Mother’s genotype: aa
Father’s genotype: Aa
aa, Aa, aa, aa, Aa
I
aa
Aa
1
2
II
3
4
5
6
7
aa
Aa
aa
aa
Aa
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.
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
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
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
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