1. Pedigree analysis through genetic hypothesis testing
Chapter 16 Mendelian Inheritance
Chapter 17 Inheritance of Sex Chromosomes, Linked Genes, and Organelles

2. Outline of Activity
Introduction with Outline (slide 2)
Handout, 3 pages (slides 3–5)
Worksheet I with Solutions (slides 6–11)
Simple Pedigree Practice Problems (slides 12–19)
Complex Pedigree Program Part A: Separase Defect (slides 20–33)
Complex Pedigree Problem Part B: Topoisomerase Defect (slides 34–48)

3. Pedigrees A and B both represent the same family.
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.

4. 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.
Make a hypothesis that the trait is inherited according to a particular mechanism (for example autosomal recessive).
Determine whether the pattern of inheritance observed in the family is consistent with the predictions of that hypothesis.
Reject the hypothesis if the observed phenotypes of the offspring do not match the phenotypes predicted by the hypothesis.
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.
Points 1-4 are meant to clarify the issue and are intended to be guidelines for hypothesis testing, and not questions to consider as activities.

5. The first step in genetic hypothesis testing is to understand the relationships between genotypes and phenotypes using symbols for alleles.
Handout
Page 3
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.
X-linked recessive traits have the following KEY relating genotype and phenotype.
Females
Males
genotype RR Rr rr
phenotype
unaffected
affected
genotype RR Rr rr
phenotype
unaffected
affected
genotype RY rY
phenotype
unaffected
affected
Dominant mutations use the letter “D or d”.
D represents the mutant allele. d represents the nonmutant allele.
Autosomal dominant traits have the following KEY relating genotype and phenotype.
X-linked dominant traits have the following KEY relating genotype and phenotype.
Females
Males
genotype DD Dd dd
phenotype
affected
unaffected
genotype DD Dd dd
phenotype
affected
unaffected
genotype DY dY
phenotype
affected
unaffected

6. 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.
The following are a series of simple practice problems designed to demonstrate the principle and methods of pedigree genetic hypothesis testing. Mastery of this method should allow success solution to the initial complex pedigree, and any other you encounter. The answers to these questions are on the next slide which is hidden for instructor use. These questions with the solutions (slides 7-10) follow the hidden slide.
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.

7. Worksheet 1 - solution sheet
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.
1. An autosomal recessive trait with an unaffected mother and an affected father. R r Rr 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.
2. An autosomal dominant trait with an affected mother and an unaffected father.
This slide is hidden for instructor use. d dd D Dd
If the mother is heterozygous, both affected and unaffected offspring could be produced. d d
If the mother is homozygous, only affected offspring could be produced. D Dd Dd D Dd Dd
3. An X-linked recessive trait with an unaffected mother and an affected father. r Y
If the mother is heterozygous, both affected and unaffected daughters and sons could be produced. r Y
If the mother is homozygous, only unaffected daughters and sons could be produced. R Rr RY R Rr RY r rr rY R Rr RY
4. An X-linked dominant trait with an unaffected mother and an affected father. d Y 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.

8. 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.
An autosomal recessive trait with an unaffected mother and an affected father.
For this problem, there are two possible genotypes of the unaffected mother: R r Rr 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.

9. 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.
For this problem, there are two possible genotypes of the affected mother: d dd D Dd
If the mother is heterozygous, both affected and unaffected offspring could be produced. d d
If the mother is homozygous, only affected offspring could be produced. D Dd Dd D Dd Dd

10. 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.
For this problem, there are two possible genotypes of the unaffected mother:
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. r Y r Y R Rr RY R Rr RY r rr rY R Rr RY

11. 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
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.

12. F M 1 2 3 4 5 6 Simple pedigree practice problem 1
Is it possible that the trait shown in this pedigree is dominant?
yes no
Answer = B (no)
The detailed solution is found on the next slide. It is hidden for instructor use.
F = father
M = mother

13. F M 1 2 3 4 5 6 Simple pedigree practice problem 1 - solution
Is it possible that the trait shown in this pedigree is dominant?
yes
*no
This slide is hidden for instructor use.
Hypothesis 1 The trait is autosomal dominant
Hypothesis 2 The trait is X-linked dominant d d d Y dd dY
Since neither parent is affected, only unaffected offspring could be produced.
Since neither parent is affected, only unaffected offspring could be produced. d dd dd d dd dd
Because both affected and unaffected offspring are observed, hypothesis 1 is rejected.
Because both affected and unaffected offspring are observed, hypothesis 2 is rejected.

14. F M 1 2 3 4 5 6 Simple pedigree practice problem 2
Answer = A (yes)
The detailed solution is found on the next slide. It is hidden for instructor use.
Is it possible that the trait shown in this pedigree is X-linked recessive?
yes no
F = father
M = mother

15. F M 1 2 3 4 5 6 Simple pedigree practice problem 2 - solution
Is it possible that the trait shown in this pedigree is X-linked recessive?
*yes no
This slide is hidden for instructor use.
Hypothesis 1 The trait is X-linked recessive
and the mother is homozygous nonmutant.
Hypothesis 2 The trait is X-linked recessive and the mother is heterozygous r Y r Y
Only unaffected offspring could be produced.
Both affected and unaffected daughters and sons could be produced. R Rr RY R Rr RY R Rr RY r rr rY
Affected and unaffected offspring are observed, so hypothesis 1 is rejected
assuming the mother is homozygous.
Affected and unaffected offspring are observed, so hypothesis 2 cannot be rejected if we assume the mother is heterozygous.
NOTE: When a parent could have more than one genotype, then only one of the possible genotypes needs to make predictions that are consistent with observations to support the hypothesis.

16. F M 1 2 3 4 5 6 Simple pedigree practice problem 3
Answer = D (the mother and individual 1)
The detailed solution is found on the next slide. It is hidden for instructor use.
Given that this trait is X-linked, which individuals must be heterozygous?
the mother
individual 1
individual 2
the mother and individual 1
F = father
M = mother

17. F M 1 2 3 4 5 6 Simple pedigree practice problem 3 - solution
Given that this trait is X-linked, which individuals must be heterozygous?
the mother
individual 1
individual 2
*the mother and individual 1
This slide is hidden for instructor use.
It is not stated whether this trait is dominant or recessive, so both possibilities must be tested.
Hypothesis 1 The trait is X-linked recessive and the mother is homozygous nonmutant.
Hypothesis 2 The trait is X-linked recessive and the mother is heterozygous.
Hypothesis 3 The trait is X-linked dominant. r Y r Y
Affected and unaffected daughters and sons could be produced. D Y
Only unaffected offspring could be produced.
Only affected daughters and unaffected sons could be produced. R Rr RY R Rr RY d Dd dY R Rr RY r rr rY d Dd dY
Affected and unaffected offspring are observed, so hypothesis 1 is rejected.
Affected and unaffected offspring are observed, so hypothesis 2 is not rejected. The mother is heterozygous as is the unaffected daughter 1.
Affected and unaffected daughters and sons are observed, so hypothesis 3 is rejected.

18. F M 1 2 3 4 5 6 Simple pedigree practice problem 4
Answer = C (25%)
The detailed solution is found on the next slide. It is hidden for instructor use.
Given that this trait is X-linked recessive, what is the chance that these parents will produce another affected son?
100%
50%
25%
F = father
M = mother

19. F M 1 2 3 4 5 6 Simple pedigree practice problem 4 - solution
Given that this trait is X-linked recessive, what is the chance that these parents will produce another affected son?
100%
50%
*25%
This slide is hidden for instructor use.
The previous practice problem showed that the only hypothesis not rejected is the one in which the mother is heterozygous.
Calculating the chance of both having a son and that son being affected is done by multiplying the individual chance of each event.
The chance of having a son is 50%.
The chance that the son is affected by inheriting a mutant allele from his mother is 50%.
The total chance is 50% x 50% = 25%.
F = father
M = mother

20. 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.

21. Start by solving just the trait caused by the separase mutation.
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.
It is very difficult to answer this question without working out the Punnet squares for each hypothesis. Each hypothesis is evaluated on the following slides with solutions on the hidden slides.
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?
hypothesis 1: X-linked dominant
b. hypothesis 2: X-linked recessive
c. hypothesis 3: autosomal dominant
d. hypothesis 4: autosomal recessive

22. 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.
This slide may be hidden at instructor discretion. It demonstrates why hypothesis 1 can be rejected.
Family parented by 1 and 2 D Y
Only affected daughters and unaffected sons could be produced. d Dd dY d Dd dY
Affected and unaffected daughters and sons are observed, so hypothesis 1 is rejected.

23. Complex problem 1 - solution pt 2.
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 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.
This slide may be hidden for instructor use and addresses hypothesis 2 completely. Slides 24 through 27 break it down by part.
Family parented by 1 and 2
Family parented by 3 and 4
With a heterozygous mother and affected father, affected and unaffected daughters and sons could be produced.
With a heterozygous mother and unaffected father, unaffected daughters and affected sons could be produced. r Y R Y R Rr RY R RR RY r rr rY r rr rY
Affected and unaffected offspring are observed, so hypothesis 2 can not be rejected.
Unaffected daughters and affected sons are observed, so hypothesis 2 can not be rejected.

24. Complex problem 1 - solution pt 2.
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 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.
This slide may be hidden for instructor use and addresses hypothesis 2 completely. Slides 24 through 27 break it down by part.
Family parented by 11 and 12
Family parented by 8 and 9 r Y r Y
With a homozygous mutant mother and affected father, only affected daughters and sons could be produced.
With a heterozygous mother and affected father, affected daughters and unaffected sons could be produced. R Rr RY r rr rY r rr r rY rr rY
Only affected offspring are observed, so hypothesis 2 can not be rejected.
An affected daughter and unaffected son are observed, so hypothesis 2 can not be rejected.

25. 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.
This slide may be hidden at the instructor’s discretion. It is the first of four slides that are a breakdown of slides
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.
Family parented by 1 and 2
With a heterozygous mother and affected father, affected and unaffected daughters and sons could be produced. r Y R Rr RY r rr rY
Affected and unaffected offspring are observed, so hypothesis 2 can not be rejected.

26. 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.
This slide may be hidden at the instructor’s discretion. It is the second of four slides that are a breakdown of hidden slides
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
With a heterozygous mother and unaffected father, unaffected daughters and affected sons could be produced. R RR RY r rr rY
Unaffected daughters and affected sons are observed, so hypothesis 2 can not be rejected.

27. 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.
This slide may be hidden at the instructor’s discretion. It is the third of four slides that are a breakdown of hidden slides
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
With a homozygous mutant mother and affected father, only affected daughters and sons could be produced. r rr rY r rr rY
Only affected offspring are observed, so hypothesis 2 can not be rejected.

28. 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.
This slide may be hidden at the instructor’s discretion. It is the fourth of four slides that are a breakdown of hidden slides
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
With a heterozygous mother and affected father, affected daughters and unaffected sons could be produced. R Rr RY r rr rY
An affected daughter and unaffected son are observed, so hypothesis 2 can not be rejected.

29. 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.
This slide may be hidden at the instructor’s discretion. It demonstrates why hypothesis 3 can be rejected.
Family parented by 3 and 4 d d
Only unaffected offspring could be produced. d dd dd d dd dd
An affected son is observed, so hypothesis 3 is rejected.

30. 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.
This slide may be hidden at the instructor’s discretion. It demonstrates why hypothesis 4 can be rejected.
Family parented by 3 and 4, where 3 is homozygous
Family parented by 3 and 4, where 3 is heterozygous R R R R
Only unaffected offspring could be produced.
Only unaffected offspring could be produced. R RR RR R RR RR R RR RR r Rr Rr
An affected son is observed, so hypothesis 4 is rejected.
An affected son is observed, so hypothesis 4 is rejected.

31. Start by solving just the trait caused by the separase mutation.
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.
Answer = b
Only the X-linked recessive hypothesis cannot be rejected.
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?
hypothesis 1: X-linked dominant
b. hypothesis 2: X-linked recessive
c. hypothesis 3: autosomal dominant
d. hypothesis 4: autosomal recessive

32. Chance of having a child affected by 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.
Answer = B (50%)
Chance of having a child affected by the separase defect:
25% + 25% = 50%
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%

33. Chance daughter is affected = 50% Chance of affected daughter = 25%
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.
100%
50%
25%
This slide may be hidden at the instructor’s discretion.
Answer = B (50%)
Chance of having a child affected by the separase defect:
25% + 25% = 50%
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%

34. Which of these hypotheses 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.
This slide may be hidden at the instructor’s discretion. It is very difficult to answer this question without working out the Punnet squares for each hypothesis. Each hypothesis is evaluated on the following slides with solutions on the hidden slides.
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?
hypothesis 1: X-linked dominant
b. hypothesis 2: X-linked recessive
c. hypothesis 3: autosomal dominant
d. hypothesis 4: autosomal recessive

35. Complex problem 1 - solution pt 5.
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 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.
This slide may be hidden at the instructor’s discretion.
Family parented by 1 and 2 (homozygous)
Family parented by 1 and 2 (heterozygous)
Family parented by 3 (heterozygous) and 4 d Y d Y d Y
With a homozygous mother, only affected sons and daughters could be produced. D Dd DY D Dd DY
With a heterozygous mother, affected and unaffected daugthers and sons could be produced. D Dd DY
With a heterozygous mother, affected sons and daughters could be produced. D Dd DY d dd dY d dd dY
Affected and unaffected daughters and sons are observed, so hypothesis 1 with a heterozygous mother can not be rejected. Individual 2 must be heterozygous.
Affected and unaffected offspring are observed, so hypothesis 1 with a homozygous mother can be rejected.
An affected son and daughter are observed, so hypothesis 1 still can not be rejected.

36. Complex problem 1 - solution pt 5.
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 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.
This slide may be hidden at the instructor’s discretion.
Family parented by 8 and 9
Family parented by 11 and 12 D Y D Y
With an affected father and unaffected mother, only affected daughters and unaffected sons could be produced.
With a heterozygous mother, affected and unaffected sons could be produced. D DD DY d Dd dY d Dd dY d Dd dY
Affected and unaffected sons are observed, so hypothesis 1 can not be rejected.
An affected daughter and unaffected son are observed, so hypothesis 1 can not be rejected.

37. 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.
This slide may be hidden at the instructor’s discretion. It demonstrates why hypothesis 1 can not be rejected.
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)
Family parented by 1 and 2 (heterozygous) d Y
With a homozygous mother, only affected daughters and sons could be produced. d Y
With a heterozygous mother, affected and unaffected daughters and sons could be produced. D Dd DY D Dd DY D Dd DY d dd dY
Affected and unaffected daughters and sons are observed, so hypothesis 1 with a heterozygous mother can not be rejected. Individual 2 must be heterozygous.
Affected and unaffected offspring are observed, so hypothesis 1 with a homozygous mother can be rejected.

38. 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.
This slide may be hidden at the instructor’s discretion. It demonstrates again why hypothesis 1 can not be rejected.
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
With a heterozygous mother, affected daughters and son could be produced. d dd dY
An affected daughter and son are observed, so hypothesis 1 still can not be rejected.

39. Complex problem 1 - solution pt 5c
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.
This slide may be hidden at the instructor’s discretion. It demonstrates again why hypothesis 1 cannot be rejected.
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
With a heterozygous mother, affected and unaffected sons could be produced. D DD DY d Dd dY
Affected and unaffected sons are observed, so hypothesis 1 can not be rejected.

40. Complex problem 1 - solution pt 5d
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.
This slide may be hidden at the instructor’s discretion. It demonstrates again why hypothesis 1 cannot be rejected.
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
With an affected father and unaffected mother, only affected daughters and unaffected sons could be produced. d Dd dY d Dd dY
An affected daughter and unaffected son are observed, so hypothesis 1 can not be rejected.

41. 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.
This slide demonstrates why hypothesis 2 can be rejected.
Family parented by 1 and 2 R Y
Only unaffected daughters and affected sons could be produced. r Rr rY r Rr rY
Affected and unaffected daughters and sons are observed, so hypothesis 2 is rejected.

42. 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.
This slide demonstrates why hypothesis 3 can be rejected.
Family parented by 11 and 12 D D d
Only unaffected offspirng could be produced. Dd Dd d Dd Dd
An affected daughter is observed, so hypothesis 3 is rejected.

43. 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.
This slide demonstrates why hypothesis 4 can be rejected.
Family parented by 3 and 4 R R
Only unaffected offspring could be produced. r Rr Rr r Rr Rr
An affected daughter and son are observed, so hypothesis 4 is rejected.

44. Which of these hypotheses 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.
Answer = a
Only the X-linked dominant hypothesis can not be rejected.
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?
hypothesis 1: X-linked dominant
b. hypothesis 2: X-linked recessive
c. hypothesis 3: autosomal dominant
d. hypothesis 4: autosomal recessive

45. 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.
Answer = b (50%)
Chance of having a child affected by Topoisomerase
50% + 0 = 50%
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?
100%
50%
25%

46. Chance of having a child affected by Topoisomerase 50% + 0 = 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.
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.
100%
*50%
25% 0%
Answer = b
Chance of having a child affected by Topoisomerase
50% + 0 = 50%
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 = 100%
Chance of affected daughter = 50%
Chance of son = 50%
Chance son is affected = 0%
Chance of affected son = 0%
Chance of having a child affected by Topoisomerase defect =
50% + 0% = 50%

47. What is the chance that 11 and 12 will have a child affected by both separase and topoisomerase defects?
100%
50%
25% 0%
Answer = c (25%)
Total chance of affected child = 50% x 100% x 50% = 25%

48. What is the chance that 11 and 12 will have a child affected by both separase and topoisomerase defects?
100%
50%
*25% 0%
Answer = c (25%)
Total chance of affected child = 50% x 100% x 50% = 25%
All daughters would be affected by topoisomerase defect from 11 and 12, and no sons.
50% of daughters would be affected by separase defect from 11 and 12.
So chance of a child affected by both traits:
Chance of daughter = 50%
Chance she is affected by topoisomerase defect = 100%
Chance she is affected by separase defect = 50%
Total chance of affected child = 50% x 100% x 50% = 25%