Pedigree analysis through genetic hypothesis testing Chapter 16 Mendelian Inheritance Chapter 17...
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Transcript of Pedigree analysis through genetic hypothesis testing Chapter 16 Mendelian Inheritance Chapter 17...
Pedigree analysis through genetic hypothesis testing
Chapter 16 Mendelian InheritanceChapter 17 Inheritance of Sex Chromosomes, Linked
Genes, and Organelles
Outline of ActivityI. Introduction with Outline (slide 2)II. Handout, 3 pages (slides 3–5)III. Worksheet I with Solutions (slides 6–11)IV. Simple Pedigree Practice Problems (slides 12–19)V. Complex Pedigree Program Part A: Separase Defect (slides 20–33)VI. Complex Pedigree Problem Part B: Topoisomerase Defect (slides 34–48)
HandoutPage 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.
HandoutPage 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.3.Reject the hypothesis if the observed phenotypes of the offspring do not match the phenotypes predicted by the hypothesis.4.Remember that observed phenotypes that are consistent with predictions do not ‘prove’ that hypothesis to be correct, but rather just fails to reject the hypothesis. Observations from other families in the pedigree can reinforce the support for a hypothesis and provide very strong support if all other hypotheses have been rejected.
The first step in genetic hypothesis testing is to understand the relationships 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.
genotypeRRRrrr
phenotypeunaffectedunaffected
affected
Autosomal recessive traits have the following KEY relating genotype and phenotype.
X-linked recessive traits have the following KEY relating genotype and phenotype.
phenotypeunaffectedunaffected
affected
genotypeRRRrrr
Femalesphenotypeunaffected
affected
genotypeRYrY
Males
Dominant mutations use the letter “D or d”. D represents the mutant allele. d represents the nonmutant allele.
genotypeDDDddd
phenotypeaffectedaffected
unaffected
Autosomal dominant traits have the following KEY relating genotype and phenotype.
X-linked dominant traits have the following KEY relating genotype and phenotype.
phenotypeaffectedaffected
unaffected
genotypeDDDddd
Femalesphenotype
affectedunaffected
genotypeDYdY
Males
HandoutPage 3
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.
Worksheet 1
1. An autosomal recessive trait with an unaffected mother and an affected father.
R Rr
Rr
r rrRrRr Rr
RrRr Rrrr rr
If the mother is heterozygous, both
affected and unaffected
offspring could be produced.
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.
For this problem, there are two possible genotypes of the unaffected mother:
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.
DD
ddDdDd Dd
Ddd dd ddD
ddDd Dd
If the mother is heterozygous, both
affected and unaffected
offspring could be produced.
If the mother is homozygous, only affected offspring
could be produced.
For this problem, there are two possible genotypes of the affected mother:
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 R
Y
Rr
r Yr
Rr
Rr RY
RYRr RY
rr rY
If the mother is heterozygous, both
affected and unaffected
daughters and sons could be produced.
If the mother is homozygous, only
unaffected daughters and sons could be produced.
For this problem, there are two possible genotypes of the unaffected mother:
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.
dY
d
DDd dYDd 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.
Is it possible that the trait shown in this pedigree is dominant?
a. yesb. no
Simple pedigree practice problem 1
F M
1 2 3 4 5 6
F = fatherM = mother
Simple pedigree practice problem 2
Is it possible that the trait shown in this pedigree is X-linked recessive?
a. yesb. no
F M
1 2 3 4 5 6
F = fatherM = mother
Given that this trait is X-linked, which individuals must be heterozygous?a. the motherb. individual 1c. individual 2d. the mother and individual 1
Simple pedigree practice problem 3
F = fatherM = mother
F M
1 2 3 4 5 6
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
Simple pedigree practice problem 4
F = fatherM = mother
F M
1 2 3 4 5 6
Determine the chance that the third child of individuals 11 and 12 will be affected by both traits.
Complex problem 1.
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 dominantb. hypothesis 2: X-linked recessivec. hypothesis 3: autosomal dominantd. 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 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 dominantConsider 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.
Start by solving just the separase defect.Complex problem 1 - solution pt 1.
d Dd dYd
YDDd dY
Only affected daughters and
unaffected sons could be produced.
Affected and unaffected daughters and sons are observed,
so hypothesis 1 is rejected.
Family parented by 1 and 2
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 sons and daughters.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.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. Finally, if we look at the family with 11 and 12 as parents, if 11 is heterozygous, they could have an unaffected son and affected daughter. This hypothesis is not rejected.
Start by solving just the Separase defect.Complex problem 1 - solution pt 2.
r rr rYR
YrRr RY
With a heterozygous mother and affected father, affected and
unaffected daughters and sons could be
produced.
Affected and unaffected offspring are observed, so hypothesis 2 can
not be rejected.
r rr rYR
YRRR 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.
Family parented by 1 and 2 Family parented by 3 and 4
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 sons and daughters.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.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. Finally, if we look at the family with 11 and 12 as parents, if 11 is heterozygous, they could have an unaffected son and affected daughter. This hypothesis is not rejected.
Start by solving just the Separase defect.Complex problem 1 - solution pt 2.
r rr rYr
Yrrr 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.
Family parented by 8 and 9
r rr rYR
YrRr RY
With a heterozygous mother and affected father, affected
daughters and unaffected sons could be produced.
Family parented by 11 and 12
An affected daughter and unaffected son are observed, so
hypothesis 2 can not be 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 basic hypotheses to test: X-linked dominant and recessive, and autosomal dominant and recessive.
Hypothesis 2: X-linked recessiveLook 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.
Start by solving just the Separase defect.Complex problem 1 - solution pt 2a.
r rr rYR
YrRr RY
With a heterozygous mother and affected father, affected and
unaffected daughters and sons could be
produced.
Affected and unaffected offspring are observed, so hypothesis 2 can
not be rejected.
Family parented by 1 and 2
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 recessiveIf 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.
Start by solving just the Separase defect.Complex problem 1 - solution pt 2b.
r rr rYR
YRRR 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.
Family parented by 3 and 4
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 recessiveIf 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.
Start by solving just the Separase defect.Complex problem 1 - solution pt 2c.
r rr rYr
Yrrr 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.
Family parented by 8 and 9
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 recessiveFinally, 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.
Start by solving just the Separase defect.Complex problem 1 - solution pt 2d.
r rr rYR
YrRr 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.
Family parented by 11 and 12
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 dominantLook 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.
Start by solving just the Separase defect.Complex problem 1 - solution pt 3.
d dd ddd
dddd dd
Only unaffected offspring could be
produced.
An affected son is observed, so hypothesis 3 is rejected.
Family parented by 3 and 4
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 recessiveLook 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.
Start by solving just the Separase defect.Complex problem 1 - solution pt 4.
R RR RRR
RRRR 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 homozygous
r Rr RrR
RRRR 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
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 dominantb. hypothesis 2: X-linked recessivec. hypothesis 3: autosomal dominantd. 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.
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
The chance of having an affected child is the sum of the chances of having an affected son and having an affected daughter.
Chance of daughter = 50% Chance daughter is affected = 50% Chance of affected daughter = 25%
Chance of son = 50% Chance son is affected = 50% Chance of affected son = 25%
Chance of having a child affected by the separase defect: 25% + 25% = 50%
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 dominantb. hypothesis 2: X-linked recessivec. hypothesis 3: autosomal dominantd. hypothesis 4: autosomal recessive
Continue by solving the topoisomerase defect.Complex problem 1 - solution pt 5.
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 dominantLook 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 daughters and sons.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.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. 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.
D Dd DYD
YdDd DY
With a homozygous mother, only affected
sons and daughters could be produced.
Affected and unaffected offspring are observed, so hypothesis 1 with a
homozygous mother can be rejected.
d dd dYD
YdDd DY With a heterozygous mother,
affected and unaffected daugthers and sons could be produced. d dd dY
DYd
Dd DY With a heterozygous mother, affected sons and daughters could
be produced.
An affected son and daughter are observed, so hypothesis 1 still can not
be rejected.
Family parented by 1 and 2 (homozygous) Family parented by 3 (heterozygous) and 4Family parented by 1 and 2 (heterozygous)
Affected and unaffected daughters and sons are observed, so hypothesis 1 with a
heterozygous mother can not be rejected. Individual 2 must be heterozygous.
Continue by solving the topoisomerase defect.Complex problem 1 - solution pt 5.
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 dominantLook 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 daughters and sons.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.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. 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.
d Dd dYD
YDDDDY With a heterozygous
mother, affected and unaffected sons could
be produced.
Affected and unaffected sons are observed, so hypothesis 1 can
not be rejected.
Family parented by 8 and 9
d Dd dYd
YDDd 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.
Family parented by 11 and 12
Continue by solving the Topoisomerase defect.Complex problem 1 - solution pt 5a.
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 dominantLook 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.
D Dd DYD
YdDd 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.
d dd dYD
YdDd DY
With a heterozygous mother, affected and unaffected
daughters and sons could be produced.
Family parented by 1 and 2 (homozygous) Family parented by 1 and 2 (heterozygous)
Affected and unaffected daughters and sons are observed, so hypothesis
1 with a heterozygous mother can not be rejected. Individual 2 must be
heterozygous.
Continue by solving the Topoisomerase defect.Complex problem 1 - solution pt 5b.
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 dominantIf 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.
d dd dYD
YdDd 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.
Family parented by 3 (heterozygous) and 4
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 dominantIf 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.
d Dd dYD
YDDDDY With a heterozygous
mother, affected and unaffected sons could
be produced.
Affected and unaffected sons are observed, so hypothesis 1 can
not be rejected.
Family parented by 8 and 9
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 dominantFinally, 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.
d Dd dYd
YDDd 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.
Family parented by 11 and 12
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 recessiveLook 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.
Continue by solving the Topoisomerase defect.Complex problem 1 - solution pt 6.
r Rr rYr
YRRr rY
Only unaffected daughters and
affected sons could be produced.
Affected and unaffected daughters and sons are observed,
so hypothesis 2 is rejected.
Family parented by 1 and 2
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 dominantLook 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.
Continue by solving the Topoisomerase defect.Complex problem 1 - solution pt 7.
d Dd Ddd
DDDd Dd Only unaffected
offspirng could be produced.
An affected daughter is observed, so hypothesis 3 is rejected.
Family parented by 11 and 12
Hypothesis 4: Autosomal recessiveLook 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.
Continue by solving the Topoisomerase defect.Complex problem 1 - solution pt 8.
r Rr Rrr
RRRr Rr
Only unaffected offspring could be
produced.
An affected daughter and son are observed, so hypothesis 4 is
rejected.
Family parented by 3 and 4
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.
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 dominantb. hypothesis 2: X-linked recessivec. hypothesis 3: autosomal dominantd. 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.100%b.50%c.25%d.0%