Transcript o F rr

Mendelian genetics in Humans: Autosomal and Sexlinked patterns of inheritance
Obviously examining inheritance patterns of specific traits in humans is
much more difficult than in Drosophila because defined crosses cannot be
constructed. In addition humans produce at most a few offspring rather than
the hundreds produced in experimental genetic organisms such as Drosophila
It is important to study mendellian inheritance in humans because of the
practical relevance and availability of sophisticated phenotypic analyses.
Therefore the basic methods of human genetics are observational rather
than experimental and require the analysis of matings that have already
taken place rather than the design and execution of crosses to directly test
a hypothesis
To understand inheritance patterns of a disease in human genetics you often
follow a trait for several generations to infer its mode of inheritance --dominant or recessive?
Sex-linked or autosomal?
For this purpose the geneticist constructs family trees or pedigrees (genetic
analyses and interviews with family members)
Pedigrees trace the inheritance pattern of a particular trait through many
generations. Pedigrees enable geneticists to determine whether a trait is
genetically determined and its mode of inheritance (dominant/recessive,
autosomal/sex-linked)
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Why do Pedigrees?
Punnett squares and 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.
Goals of Pedigree Analysis
1. Determine the mode of inheritance: dominant,
recessive, partial dominance, sex-linked, autosomal,
mitochondrial, maternal effect.
2. Determine the probability of an affected
offspring for a given cross.
Pedigree symbols:
Male
Female
Sex Unknown
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Affected individual
Spontaneous
abortion
Number of individuals
Deceased
Termination
of pregnancy
3
Pedigree symbols:
relationship line
Sibship line
line of descent
individual’s lines
consanguinity
Monozygotic
Dizygotic
4
Characteristics of an autosomal dominant trait:
1. Every affected individual should have at least one affected
parent.
2. An affected individual has at least a 50% chance of transmitting
the trait
3. Males and females should be affected with equal frequency
4. Two affected individuals may have unaffected children
5. All unaffected individuals are homozygous for the normal
recessive allele.
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Autosomal Dominant
dd
dd
Dd
Dd
dd
Dd
dd
Probability of genotypes
Son = ½ dd and ½ Dd
Daughter = ½ dd and ½ Dd
Probability of GrandDaughter = ¾ dd and ¼ Dd
Two possible matings
½ mating is dd x dd and outcome is 0% D- ,
½ mating is Dd x dd and outcome is ½ dd and ½ Dd,
adding these two you get (1/2*0) + (1/2* 1/2)
= ¾ dd and ¼ Dd
Probability of GrandSon = 100% D-
Dd
dd
Dd
Two possible matings
½ mating is dd x Dd and outcome
is 1/2 Dd and 1/2 dd ,
½ mating is Dd x Dd and outcome
is 1/4 dd and 3/4 D-,
adding these two you get
(1/2*1/2) + (1/2* 3/4)
=3/8dd and 5/8 Dd
DD
Characteristics of an autosomal recessive trait:
There are several features in a pedigree that suggest a recessive
pattern of inheritance:
1.
Rare traits, the pedigree usually involves mating between two
unaffected heterozygotes with the production of one or more
homozygous offspring.
2. The probability of an affected child from a mating of two
heterozygotes is ~25%
3. Two affected individuals usually produce offspring all of whom
are affected
4. Males and females are at equal risk, since the trait is autosomal
5. In pedigrees involving rare traits, consanguinity is often involved.
In the pedigree shown below, an autosomal recessive inheritance
pattern is observed:
I
II:1
II:2
III:9
7
Autosomal Recessive
rr
RR
RR
Rr
Rr
rr
1/2
All
1/4
Normal Affected Affected
X-Linked Dominant
Mothers pass their X’s to both sons and daughters
Fathers pass their X to daughters only.
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
XDY
XdY
XDXd
¼ XDXd (affected)
¼ XdXd (Normal)
¼ XDY (affected)
¼ XdY (Normal)
X dX d
XdY
XDXd
The following pedigree outlines an inheritance pattern
Does this fit an autosomal recessive or autosomal dominant
pattern of inheritance?
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Pedigree of Queen Victoria and the transmission of hemophilia.
Albert
Victoria
Alice
carrier
Irene
carrier
Beatrice
carrier
Alix
carrier
Alice
carrier
Victoria
carrier
carrier
carrier
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Characteristics of a X (sex)-linked recessive trait:
Hemizygous males and homozygous females are affected
Phenotypic expression is much more common in males than in
females, and in the case of rare alleles, males are almost
exclusively affected
Affected males transmit the gene to all daughters but not to any
sons
Daughters of affected males will usually be heterozygous and
therefore unaffected.
Sons of heterozygous females have a 50% chance of receiving the
recessive gene.
GG
gY
GY
gG
GY
GY
GY
gG
gG
GY
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X-Linked Recessive
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
Xr Y
XrY
XRXr
½ XrY ½ XrXr
X RX R
XRY
XrXr
1 XrY 1 XRXr
Y chromosome
Y
Y
Y
Y
All males in this pedigree will have the SAME Y-chromosome!!!
X1/Y1; A1/A2
(grandpa)
x
X2/X3; A3/A4
(grandma)
X2/Y1; A2/A4 (dad)
x
X4/X5; A5/A6 (mom)
X5/X2
A2/A6
Daughter
X4/Y1
Son
A4/A6
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 Ylinked gene (hemizygous).
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Large Pedigrees
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.
Dominant vs. Recessive
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.
4. If two unaffected mate and have an affected
child, both parents must be Rr heterozygotes.
5. Recessive outsider rule: outsiders are those whose parents are unknown. In a recessive autosomal
pedigree, unaffected outsiders are assumed to be RR, homozygous normal.
6. 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.
7. Unaffected children of Rr x Rr have a 2/3
chance of being Rr and a 1/3 chance of being RR.
Assigning Genotypes for Autosomal Recessive Pedigrees
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.
(A-)
(AA)
(Aa)
(A-)
(Aa)
(aa)
(aa)
(Aa)
(Aa)
(Aa)
(A-)
(aa)
(aa)
(Aa)
(A-)
(aa)
(aa)
Steps in assigning genotype
Generations labelled roman numerals I, II, ...
Individuals labelled arabic numerals 1, 2, ...
Shaded = affected
Mated individuals connected by "marriage line".
()
()
()
()
()
()
()
()
()
()
()
()
()
()
()
()
Steps in assigning genotype
Try simplest hypothesis first: 1 gene, 2 alleles, complete
dominance, affected are homozygous recessive.
Fill in genotypes in steps:
(1) All affected are homozygous recessive.
(aa)
()
()
()
()
()
()
()
()
(aa)
()
(aa) (aa)
(aa)
()
()
Steps in assigning genotype
Try simplest hypothesis first: 1 gene, 2 alleles, complete
dominance, affected are homozygous recessive. Fill in
genotypes in steps:
(1) All affected are homozygous recessive.
(2) All unaffected have at least one dominant allele.
(aa)
(A)
(A)
(A)
(A)
(A)
(A)
(A)
(A)
(aa)
(A)
(aa) (aa)
(aa)
(A)
(A)
Steps in assigning genotype
Try simplest hypothesis first: 1 gene, 2 alleles, complete
dominance, affected are homozygous recessive. Fill in
genotypes in steps:
(1) All affected are homozygous recessive.
(2) All unaffected have at least one dominant allele.
(3) All homozygous recessive must get one recessive allele from
each parent.
(aa)
(A)
(A)
(A)
(Aa)
(A)
(A)
(Aa)
(Aa)
(aa)
(Aa)
(aa) (aa)
(aa)
(A)
(A)
Try simplest hypothesis first: 1 gene, 2 alleles, complete
dominance, affected are homozygous recessive. Fill in
genotypes in steps:
(1) All affected are homozygous recessive.
(2) All unaffected have at least one dominant allele.
(3) All homozygous recessive must get one recessive allele from
each parent.
(4) All offspring of homozygous recessive must have at least
one recessive allele.
(aa)
(Aa)
(A)
(Aa)
(Aa)
(Aa)
(A)
(Aa)
(aa) (aa)
(Aa)
(aa)
(Aa)
(aa)
Steps in assigning genotype
(A)
(A)
Steps in assigning genotype
Try simplest hypothesis first: 1 gene, 2 alleles, complete
dominance, affected are homozygous recessive. Fill in
genotypes in steps:
(1) All affected are homozygous recessive.
(2) All unaffected have at least one dominant allele.
(3) All homozygous recessive must get one recessive allele from
each parent.
(4) All offspring of homozygous recessive must have at least
one recessive allele.
(5) For rest of genes, use – = allele unknown.
(aa)
(Aa)
(A-)
(Aa)
(Aa)
(Aa)
(A-)
(Aa)
(Aa)
(aa)
(Aa)
(aa) (aa)
(aa)
(A-)
We have filled in the pedigree without finding any internal
contradictions, i.e. without contradicting our hypothesis.
(A-)
Assigning Genotypes for Dominant Pedigrees
1. All unaffected are aa.
2. Affected children of an affected parent and
an unaffected parent must be heterozygous Aa,
because they inherited an a allele from the
unaffected parent.
3. The affected parents of an unaffected child
must be heterozygotes Aa, since they both
passed an a 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 (Aa).
5. If both parents are heterozygous Aa x Aa,
their affected offspring have a 2/3 chance of
being Aa and a 1/3 chance of being AA.
I
(Aa)
(aa)
II
aa
aa
aa
Aa
Aa
Aa
III
aa
aa aa
Aa
aa aa
Aa
A- aa A-
Conditional Probability- Autosomal dominant
Determining the probability of an affected offspring for most
crosses is simple:
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.
dd 1
Father is dd (probability = 1)
Mother has a 1/2 chance of being DD and a 1/2 of being Dd.
This is a dominant autosomal pedigree,
Determine the probability of an affected child:
1. determine the probability of an affected offspring for
each possible set of parental genotypes.
2. Combine them
Mom has a 1/2 chance of being Dd and a 1/2 chance of being
DD, and dad is dd.
There are thus 2 possibilities for the cross:
DD x dd, or
Dd x dd.
If the cross is DD x dd, all of the offspring will be Dd, and
since the trait is dominant, all will be affected.
On the other hand, if the cross is Dd x dd, ½ the offspring
are Dd (affected) and ½ are dd (normal).
So, probability of mating DD x dd = (½ x 1), with all offspring
affected = (1), and probability of the mating being Dd x dd is
(½ x 1), with ½ the offspring affected.
((1/2*1) * 1) + ((1/2*1) * 1/2)
=3/4 overall probability
DD 1/2
Dd 1/2
Conditional Probability- Autosomal recessive
More complicated recessive pedigree: Grandparent genotype
known, parent genotype is unknown
Father has a ½ chance of being RR and a ½ chance of being Rr,
Mother has a 1/3 chance of being RR and a 2/3 chance of being
Rr.
In this case there are 4 possible matings:
1. There is a 1/2 * 1/3 = 1/6 chance that the mating is RR x
RR. In this case, 0 offspring will be affected (rr).
2. There is a 1/2 * 2/3 = 2/6 = 1/3 chance that the mating is
RR x Rr. In this case, 0 offspring are affected.
3. There is a 1/2 * 1/3 = 1/6 chance that the mating is Rr x
RR. In this case, 0 offspring will be affected (rr).
4. There is a 1/2 * 2/3 = 1/3 chance that the mating is Rr x
Rr. In this case, 1/4 offspring will be affected (rr).
Combining all possibilities:
(1/6 * 0 ) + (1/3 * 0) + (1/6 * 0) + (1/3 *1/4) =
0 + 0 + 0 + 1/12 = 1/12
RR
Rr
Rr
½ RR
½ Rr
Rr
1/3 RR
2/3 Rr
?