Download Pedigree analysis through genetics hypothesis testing

Pedigree analysis through genetic hypothesis testing
Outline of Activity
I. Introduction with Outline (slide 1)
II. Handout, 3 pages (slides 2-4)
III. Worksheet I with Solutions (slides 5-10)
IV. Simple Pedigree Practice Problems (slides 11-18)
V. Complex Pedigree Program Part A: Separase Defect (slides 19-32)
VI. Complex Pedigree Problem Part B: Topoisomerase Defect (slides 33– 47)
Handout
Page 1
Pedigrees A and B both represent the same family.
• Genetic testing shows that individual 4 has only nonmutant alleles of both
genes and individual 12 has only mutant alleles of both genes.
• Individuals 6, 8, 9, 12 and 14 have cancer.
Handout
Page 2
Individuals 11 and 12 are concerned because 11 is pregnant with their third child. They just
learned that their daughter also has cancer, has both mutations, and they are worried about their
next child.
How can you determine the chance of that third child inheriting both mutations? To determine
the chance that 11 and 12’s third child will inherit both mutations, it is necessary to determine
the mode of inheritance of each trait.
Are they inherited as dominant or recessive traits? Are the genes autosomal or X-linked?
To determine the answers, you can engage in genetic hypothesis testing.
1. Make a hypothesis that the trait is inherited according to a particular mechanism (for example
autosomal recessive).
2. Determine whether the pattern of inheritance observed in the family is consistent with the
predictions of that hypothesis.
Handout The first step in genetic hypothesis testing is to understand the relationships
Page 3
between genotypes and phenotypes using symbols for alleles.
Recessive mutations use the letter “R or r”.
R represents the nonmutant allele. r represents the mutant allele.
Autosomal recessive traits have
the following KEY relating
genotype and phenotype.
genotype
RR
Rr
rr
phenotype
unaffected
unaffected
affected
X-linked recessive traits have
the following KEY relating
genotype and phenotype.
Females
Males
genotype phenotype
genotype phenotype
RR
unaffected
RY
unaffected
Rr
unaffected
rY
affected
rr
affected
Dominant mutations use the letter “D or d”.
D represents the mutant allele. d represents the nonmutant allele.
Autosomal dominant traits have
X-linked dominant traits have
the following KEY relating
the following KEY relating
genotype and phenotype.
genotype and phenotype.
Females
Males
genotype phenotype
genotype phenotype
genotype phenotype
DD
affected
affected
DD
DY
affected
Dd
affected
affected
Dd
dY
unaffected
dd
unaffected
unaffected
dd
Worksheet 1
Using Punnett squares, determine the phenotypes of offspring that the following
parents could produce.
Note: For some of the situations listed below, there could be more than one genotype for at least one of the parents.
1. An autosomal recessive trait with an unaffected mother and an affected father.
2. An autosomal dominant trait with an affected mother and an unaffected father.
3. An X-linked recessive trait with an affected mother and an unaffected father.
4. An X-linked dominant trait with an unaffected mother and an affected father.
Using Punnett squares, determine the phenotypes of offspring that the following
parents could produce.
Note: For some of the situations listed below, there could be more than one genotype for at least one of the parents.
1. An autosomal recessive trait with an unaffected mother and an affected father.
r r
R Rr Rr
r rr rr
If the mother is
heterozygous, both
affected and
unaffected
offspring could be
produced.
r r
R Rr Rr
R Rr Rr
If the mother is
homozygous, only
unaffected
offspring could be
produced.
Using Punnett squares, determine the phenotypes of offspring that the following
parents could produce.
Note: For some of the situations listed below, there could be more than one genotype for at least one of the parents.
2. An autosomal dominant trait with an affected mother and an unaffected father.
d d
D Dd Dd
d dd dd
If the mother is
heterozygous, both
affected and
unaffected
offspring could be
produced.
d d
D Dd Dd
D Dd Dd
If the mother is
homozygous, only
affected offspring
could be produced.
Using Punnett squares, determine the phenotypes of offspring that the following
parents could produce.
Note: For some of the situations listed below, there could be more than one genotype for at least one of the parents.
3. An X-linked recessive trait with an unaffected mother and an affected father.
r
Y
R
Rr RY
r
rr
rY
If the mother is
heterozygous, both
affected and
unaffected
daughters and sons
could be produced.
r
Y
R
Rr
RY
R
Rr RY
If the mother is
homozygous, only
unaffected
daughters and sons
could be produced.
Using Punnett squares, determine the phenotypes of offspring that the following
parents could produce.
Note: For some of the situations listed below, there could be more than one genotype for at least one of the parents.
4. An X-linked dominant trait with an unaffected mother and an affected father.
D Y
d Dd dY
d Dd dY
The mother must be
homozygous for the
nonmutant allele and the
father must carry the mutant
allele. Only affected
daughters and unaffected
sons could be produced.
Simple pedigree practice problem 1
Is it possible that the trait shown in this pedigree is dominant?
a. yes
b. no
Simple pedigree practice problem 2
Is it possible that the trait shown in this pedigree is X-linked recessive?
a. yes
b. no
Simple pedigree practice problem 3
Given that this trait is X-linked, which individuals must be heterozygous?
a. the mother
b. individual 1
c. individual 2
d. the mother and individual 1
Simple pedigree practice problem 4
Given that this trait is X-linked recessive, what is the chance that these parents will
produce another affected son?
a. 100%
b. 50%
c. 25%
d. 0
Complex problem 1.
Determine the chance that the third child of individuals 11 and 12 will be
affected by both traits.
Reminder:
• Genetic testing shows that individual 4 has only nonmutant alleles of both genes and
individual 12 has only mutant alleles of both genes.
• Individuals 6, 8, 9, 12 and 14 have cancer.
Genetic testing shows that individual 4 has only nonmutant alleles of both genes.
Genetic testing also shows that individual 12 has only mutant alleles of both genes.
Start by solving just the trait caused by the separase mutation.
There are four possible hypotheses to test about the pattern of inheritance of the trait
caused by the separase defect.
Which of these hypotheses can not be rejected?
a. hypothesis 1: X-linked dominant
b. hypothesis 2: X-linked recessive
c. hypothesis 3: autosomal dominant
d. hypothesis 4: autosomal recessive
Complex problem 1 - solution pt 1.
Start by solving just the separase defect.
Genetic testing shows that individual 4 has only nonmutant alleles of both genes.
Genetic testing also shows that individual 12 has pnly mutant alleles of both genes.
There are four basic hypotheses to test: X-linked dominant and recessive, and autosomal dominant and recessive.
Hypothesis 1: X-linked dominant
Consider the family with individuals 1 and 2 as parents. The mother must be dd since she is not
affected. The father must be DY since he is affected. This couple could not produce affected
sons or unaffected daughters. Both are observed. This hypothesis is rejected.
Family parented by 1 and 2
D Y
d Dd dY
d Dd dY
Only affected
daughters and
unaffected sons could
be produced.
Affected and unaffected
daughters and sons are observed,
so hypothesis 1 is rejected.
Complex problem 1 - solution pt 2a.
Start by solving just the Separase defect.
Genetic testing shows that individual 4 has only nonmutant alleles of both genes.
Genetic testing also shows that individual 12 has only mutant alleles of both genes.
There are four basic hypotheses to test: X-linked dominant and recessive, and autosomal dominant and recessive.
Hypothesis 2: X-linked recessive
Look at the family with individuals 1 and 2 as parents. The father must be rY since he is affected. The mother could be RR or Rr
since she is not affected. If the mother is heterozygous, this mating could produce affected daughters and sons.
R
r
Family parented by 1 and 2
With a heterozygous
r Y mother and affected
Rr RY father, affected and
rr rY unaffected daughters
and sons could be
produced.
Affected and unaffected offspring
are observed, so hypothesis 2 can
not be rejected.
Complex problem 1 - solution pt 2b.
Start by solving just the Separase defect.
Genetic testing shows that individual 4 has only nonmutant alleles of both genes.
Genetic testing also shows that individual 12 has only mutant alleles of both genes.
There are four basic hypotheses to test: X-linked dominant and recessive, and autosomal dominant and recessive.
Hypothesis 2 CONTINUED: X-linked recessive
If we look at the family parented by individuals 3 and 4, and we know that 4 has only nonmutant alleles, and if individual 3 is
heterozygous, they could have an affected son.
Family parented by 3 and 4
R Y
R RR RY
r rr rY
With a heterozygous mother
and unaffected father,
unaffected daughters and
affected sons could be
produced.
Unaffected daughters and affected
sons are observed, so hypothesis 2
can not be rejected.
Complex problem 1 - solution pt 2c.
Start by solving just the Separase defect.
Genetic testing shows that individual 4 has only nonmutant alleles of both genes.
Genetic testing also shows that individual 12 has only mutant alleles of both genes.
There are four basic hypotheses to test: X-linked dominant and recessive, and autosomal dominant and recessive.
Hypothesis 2 CONTINUED: X-linked recessive
If we look at the family parented by 8 and 9, who are both affected, they would be rr and rY respectively, and would produce only
affected children, and individual 12 would only have mutant alleles.
Family parented by 8 and 9
r Y
r rr rY
r rr rY
With a homozygous mutant
mother and affected father,
only affected daughters and
sons could be produced.
Only affected offspring are
observed, so hypothesis 2 can
not be rejected.
Complex problem 1 - solution pt 2d.
Start by solving just the Separase defect.
Genetic testing shows that individual 4 has only nonmutant alleles of both genes.
Genetic testing also shows that individual 12 has only mutant alleles of both genes.
There are four basic hypotheses to test: X-linked dominant and recessive, and autosomal dominant and recessive.
Hypothesis 2 CONTINUED: X-linked recessive
Finally, if we look at the family with 11 and 12 as parents, if 11 is heterozygous, they could have an affected daughter and unaffected
son.
Family parented by 11 and 12
r Y
R Rr RY
r rr rY
With a heterozygous mother
and affected father, affected
daughters and unaffected sons
could be produced.
An affected daughter and
unaffected son are observed, so
hypothesis 2 can not be rejected.
Complex problem 1 - solution pt 3.
Start by solving just the Separase defect.
Genetic testing shows that individual 4 has only nonmutant alleles of both genes.
Genetic testing also shows that individual 12 has only mutant alleles of both genes.
There are four basic hypotheses to test: X-linked dominant and recessive, and autosomal dominant and recessive.
Hypothesis 3: Autosomal dominant
Look at the family with individuals 3 and 4 as parents. Both parents must be dd since they are
unaffected. In addition, individual 4 has only nonmutant alleles This couple could not produce
affected sons. An affected son is observed. This hypothesis is rejected.
Family parented by 3 and 4
d d
d dd dd
d dd dd
Only unaffected
offspring could be
produced.
An affected son is observed, so
hypothesis 3 is rejected.
Complex problem 1 - solution pt 4.
Start by solving just the Separase defect.
Genetic testing shows that individual 4 has only nonmutant alleles of both genes.
Genetic testing also shows that individual 12 has only mutant alleles of both genes.
There are four basic hypotheses to test: X-linked dominant and recessive, and autosomal dominant and recessive.
Hypothesis 4: Autosomal recessive
Look at the family with individuals 3 and 4 as parents. Individual 3 could be homozygous for the
nonmutant allele or heterozygous. Individual 4 has only nonmutant alleles. Regardless of whether
the mother is homozygous for the nonmutant allele or heterozygous, they could only produce
unaffected offspring. However, an affected son is observed.
Family parented by 3 and 4, where 3 is homozygous
R R
R RR RR
R RR RR
Only unaffected
offspring could be
produced.
An affected son is observed, so
hypothesis 4 is rejected.
Family parented by 3 and 4, where 3 is heterozygous
R R
R RR RR
r Rr Rr
Only unaffected
offspring could be
produced.
An affected son is observed, so
hypothesis 4 is rejected.
Genetic testing shows that individual 4 has only nonmutant alleles of both genes.
Genetic testing also shows that individual 12 has only mutant alleles of both genes.
Start by solving just the trait caused by the separase mutation.
There are four possible hypotheses to test about the pattern of inheritance of the trait
caused by the separase defect.
Which of these hypotheses cannot be rejected?
a. hypothesis 1: X-linked dominant
b. hypothesis 2: X-linked recessive
c. hypothesis 3: autosomal dominant
d. hypothesis 4: autosomal recessive
Genetic testing shows that individual 4 has only nonmutant alleles of both genes.
Genetic testing also shows that individual 12 has only mutant alleles of both genes.
Having determined which pattern of inheritance can not be rejected, what is
the chance that 11 and 12 would have a child affected by the separase defect?
a. 100%
b 50%
c. 25%
d. 0%
Genetic testing shows that individual 4 has only nonmutant alleles of both genes.
Genetic testing also shows that individual 12 has only mutant alleles of both genes.
There are four possible hypotheses to test about the mode of inheritance of the trait
caused by the topoisomerase mutation.
Which of these hypotheses can not be rejected?
a. hypothesis 1: X-linked dominant
b. hypothesis 2: X-linked recessive
c. hypothesis 3: autosomal dominant
d. hypothesis 4: autosomal recessive
Complex problem 1 - solution pt 5a.
Continue by solving the Topoisomerase defect.
Genetic testing shows that individual 4 has only nonmutant alleles of both genes.
Genetic testing also shows that individual 12 has only mutant alleles of both genes.
There are four basic hypotheses to test: X-linked dominant and recessive, and autosomal dominant and recessive.
Hypothesis 1: X-linked dominant
Look at the family with individuals 1 and 2 as parents. The father must be dY since he is unaffected. The mother could be DD or Dd
since she is affected. Since either could be true, if either one can not be rejected, the hypothesis is not rejected. If the mother is
heterozygous, this mating could produce affected sons and daughters.
Family parented by 1 and 2 (homozygous)
d Y
D Dd DY
D Dd DY
With a homozygous
mother, only affected
daughters and sons
could be produced.
Affected and unaffected offspring
are observed, so hypothesis 1
with a homozygous mother can
be rejected.
Family parented by 1 and 2 (heterozygous)
d Y
D Dd DY
d dd dY
With a heterozygous mother,
affected and unaffected
daughters and sons could be
produced.
Affected and unaffected daughters
and sons are observed, so hypothesis
1 with a heterozygous mother can
not be rejected. Individual 2 must be
heterozygous.
Complex problem 1 - solution pt 5b.
Continue by solving the Topoisomerase defect.
Genetic testing shows that individual 4 has only nonmutant alleles of both genes.
Genetic testing also shows that individual 12 has only mutant alleles of both genes.
There are four basic hypotheses to test: X-linked dominant and recessive, and autosomal dominant and recessive.
Hypothesis 1 CONTINUED: X-linked dominant
If we look at the family parented by individuals 3 and 4 and we know that 4 has only nonmutant alleles, if individual 3 is
heterozygous, they could have an affected son.
Family parented by 3 (heterozygous) and 4
d Y
D Dd DY
d dd dY
With a heterozygous mother,
affected daughters and son
could be produced.
An affected daughter and son are
observed, so hypothesis 1 still
can not be rejected.
Continue by solving the Topoisomerase defect.
Complex problem 1 - solution pt 5c
Genetic testing shows that individual 4 has only nonmutant alleles of both genes.
Genetic testing also shows that individual 12 has only mutant alleles of both genes.
There are four basic hypotheses to test: X-linked dominant and recessive, and autosomal dominant and recessive.
Hypothesis 1 CONTINUED: X-linked dominant
If we look at the family parented by 8 and 9 who are both affected, we know that individual 8 must be heterozygous because her
father has only nonmutant alleles. 8 and 9 could produce affected and unaffected sons.
Family parented by 8 and 9
D Y
D DDDY
d Dd dY
With a heterozygous
mother, affected and
unaffected sons could
be produced.
Affected and unaffected sons are
observed, so hypothesis 1 can
not be rejected.
Continue by solving the Topoisomerase defect.
Complex problem 1 - solution pt 5d
Genetic testing shows that individual 4 has only nonmutant alleles of both genes.
Genetic testing also shows that individual 12 has only mutant alleles of both genes.
There are four basic hypotheses to test: X-linked dominant and recessive, and autosomal dominant and recessive.
Hypothesis 1 CONTINUED: X-linked dominant
Finally, if we look at the family with 11 and 12 as parents, 11 must be dd because she is unaffected and 12 must be DY since he is
affected, they could have an affected son and unaffected daughter. This hypothesis is not rejected.
Family parented by 11 and 12
D Y
d Dd dY
d Dd dY
With an affected father and
unaffected mother, only affected
daughters and unaffected sons
could be produced.
An affected daughter and unaffected
son are observed, so hypothesis 1
can not be rejected.
Complex problem 1 - solution pt 6.
Continue by solving the Topoisomerase defect.
Genetic testing shows that individual 4 has only nonmutant alleles of both genes.
Genetic testing also shows that individual 12 has only mutant alleles of both genes.
There are four basic hypotheses to test: X-linked dominant and recessive, and autosomal dominant and recessive.
Hypothesis 2: X-linked recessive
Look at the family with individuals 1 and 2 as parents. The father must be RY since he is unaffected. The mother must
be rr since she is affected. Those parents could only produce unaffected daughters and affected sons. This couple
could not produce affected daughters or unaffected sons. Both affected and unaffected daughters and sons are
observed.
This hypothesis is rejected.
Family parented by 1 and 2
R Y
r Rr rY
r Rr rY
Only unaffected
daughters and
affected sons could
be produced.
Affected and unaffected
daughters and sons are observed,
so hypothesis 2 is rejected.
Complex problem 1 - solution pt 7.
Continue by solving the Topoisomerase defect.
Genetic testing shows that individual 4 has only nonmutant alleles of both genes.
Genetic testing also shows that individual 12 has only mutant alleles of both genes.
There are four basic hypotheses to test: X-linked dominant and recessive, and autosomal dominant and recessive.
Hypothesis 3: Autosomal dominant
Look at the family with individuals 11 and 12 as parents. Individual 11 is unaffected and must therefore be dd.
Individual 12 has only mutant alleles and must therefore be DD. This couple could not produce unaffected
offspring. An affected daughter is observed.
This hypothesis is rejected.
Family parented by 11 and 12
D D
d Dd Dd
d Dd Dd
Only unaffected
offspirng could be
produced.
An affected daughter is observed,
so hypothesis 3 is rejected.
Complex problem 1 - solution pt 8.
Continue by solving the Topoisomerase defect.
Genetic testing shows that individual 4 has only nonmutant alleles of both genes.
Genetic testing also shows that individual 12 has only mutant alleles of both genes.
There are four basic hypotheses to test: X-linked dominant and recessive, and autosomal dominant and recessive.
Hypothesis 4: Autosomal recessive
Look at the family with individuals 3 and 4 as parents. Individual 3 is affected and must be rr. Individual 4 has only
nonmutant alleles and must be RR. This couple could not produce affected offspring. An affected daughter and son
are observed.
This hypothesis is rejected.
Family parented by 3 and 4
R R
r Rr Rr
r Rr Rr
Only unaffected
offspring could be
produced.
An affected daughter and son are
observed, so hypothesis 4 is
rejected.
Genetic testing shows that individual 4 has only nonmutant alleles of both genes.
Genetic testing also shows that individual 12 has only mutant alleles of both genes.
There are four possible hypotheses to test about the mode of inheritance of the trait
caused by the topoisomerase mutation.
Which of these hypotheses can not be rejected?
a. hypothesis 1: X-linked dominant
b. hypothesis 2: X-linked recessive
c. hypothesis 3: autosomal dominant
d. hypothesis 4: autosomal recessive
Genetic testing shows that individual 4 has only nonmutant alleles of both genes.
Genetic testing also shows that individual 12 has only mutant alleles of both genes.
Having determined which pattern of inheritance can not be rejected, what is the
chance that 11 and 12 would have a child affected by the topoisomerase defect?
a. 100%
b. 50%
c. 25%
d. 0
What is the chance that 11 and 12 will have a child
affected by both separase and topoisomerase
defects?
a.
b.
c.
d.
100%
50%
25%
0%