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