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.


For this problem, there are two possible genotypes of the unaffected mother:


2. An autosomal dominant trait with an affected mother and an unaffected father.

DD

ddDdDd Dd

Ddd dd ddD

ddDd Dd



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:


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



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:


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


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


Given that this trait is X-linked, which individuals must be heterozygous?a. the motherb. individual 1c. individual 2d. the mother and individual 1



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



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.


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.


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.

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.



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




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.



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.



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.


d dd ddd

dddd dd

Only unaffected offspring could be

produced.

An affected son is observed, so hypothesis 3 is rejected.



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.


R RR RRR

RRRR RR


produced.


Family parented by 3 and 4, where 3 is homozygous

r Rr RrR

RRRR RR


produced.


Family parented by 3 and 4, where 3 is heterozygous


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


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%


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%


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.


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



Continue by solving the Topoisomerase defect.Complex problem 1 - solution pt 5a.


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.


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


Family parented by 3 (heterozygous) and 4

Continue by solving the Topoisomerase defect.Complex problem 1 - solution pt 5c


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.


Continue by solving the Topoisomerase defect.Complex problem 1 - solution pt 5d


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




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.



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.


d Dd Ddd

DDDd Dd Only unaffected

offspirng could be produced.

An affected daughter is observed, so hypothesis 3 is rejected.


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.


r Rr Rrr

RRRr Rr


produced.

An affected daughter and son are observed, so hypothesis 4 is

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?a.hypothesis 1: X-linked dominantb. hypothesis 2: X-linked recessivec. hypothesis 3: autosomal dominantd. hypothesis 4: autosomal recessive


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%

pedigree analysis through genetic hypothesis testing chapter 16 mendelian inheritance chapter 17...

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recessive mutations

following key

linked genes

observed phenotypes

following parents

phenotypes of offspring

genetic testing

pedigree analysis