1. 1 Mendelian genetics in Humans: Autosomal and Sex- linked patterns of inheritance Obviously examining inheritance patterns of specific traits in humans is much more difficult than in Drosophila because defined crosses cannot be constructed. In addition humans produce at most a few offspring rather than the hundreds produced in experimental genetic organisms such as Drosophila It is important to study mendellian inheritance in humans because of the practical relevance and availability of sophisticated phenotypic analyses. Therefore the basic methods of human genetics are observational rather than experimental and require the analysis of matings that have already taken place rather than the design and execution of crosses to directly test a hypothesis To understand inheritance patterns of a disease in human genetics you often follow a trait for several generations to infer its mode of inheritance --- dominant or recessive? Sex-linked or autosomal? For this purpose the geneticist constructs family trees or pedigrees (genetic analyses and interviews with family members) Pedigrees trace the inheritance pattern of a particular trait through many generations. Pedigrees enable geneticists to determine whether a trait is genetically determined and its mode of inheritance (dominant/recessive, autosomal/sex-linked)

2. Why do Pedigrees? Punnett squares and tests work well for organisms that have large numbers of offspring and controlled matings, but humans are quite different: 1. small families. Even large human families have 20 or fewer children. 2. Uncontrolled matings, often with heterozygotes. Goals of Pedigree Analysis 1. Determine the mode of inheritance: dominant, recessive, partial dominance, sex-linked, autosomal, mitochondrial, maternal effect. 2. Determine the probability of an affected offspring for a given cross.

3. 3 Male FemaleSex Unknown Affected individual 5 Number of individuals Deceased Spontaneous abortion Termination of pregnancy Pedigree symbols:

4. 4 line of descent individual’s lines relationship line Sibship line consanguinity MonozygoticDizygotic

5. 5 Characteristics of an autosomal dominant trait: 1. Every affected individual should have at least one affected parent. 2. An affected individual has at least a 50% chance of transmitting the trait 3. Males and females should be affected with equal frequency 4. Two affected individuals may have unaffected children 5. All unaffected individuals are homozygous for the normal recessive allele.

6. Autosomal Dominant Probability of genotypesSon = ½ dd and ½ Dd Daughter = ½ dd and ½ Dd Probability of GrandDaughter = ¾ dd and ¼ Dd Two possible matings ½ mating is dd x dd and outcome is 0% D-, ½ mating is Dd x dd and outcome is ½ dd and ½ Dd, adding these two you get (1/2*0) + (1/2* 1/2) = ¾ dd and ¼ Dd dd Dd DD dd Probability of GrandSon = 100% D- Dd dd Dd dd Dd dd Two possible matings ½ mating is dd x Dd and outcome is 1/2 Dd and 1/2 dd, ½ mating is Dd x Dd and outcome is 1/4 dd and 3/4 D-, adding these two you get (1/2*1/2) + (1/2* 3/4) =3/8dd and 5/8 Dd

7. 7 Characteristics of an autosomal recessive trait: There are several features in a pedigree that suggest a recessive pattern of inheritance: 1.Rare traits, the pedigree usually involves mating between two unaffected heterozygotes with the production of one or more homozygous offspring. 2. The probability of an affected child from a mating of two heterozygotes is ~25% 3. Two affected individuals usually produce offspring all of whom are affected 4. Males and females are at equal risk, since the trait is autosomal 5. In pedigrees involving rare traits, consanguinity is often involved. In the pedigree shown below, an autosomal recessive inheritance pattern is observed: II:1II:2 III:9 I

8. Autosomal Recessive rr RR Rr rr RR All Normal 1/2 Affected 1/4 Affected

9. X-Linked Dominant Mothers pass their X’s to both sons and daughters Fathers pass their X to daughters only. For sex-linked traits remember that males are hemizygous and express whichever gene is on their X. X D = dominant mutant allele X d = recessive normal allele XDYXDY XdXdXdXd XDXdXDXd XdYXdY XDXdXDXd XdYXdY ¼ X D X d (affected) ¼ X d X d (Normal) ¼ X D Y (affected) ¼ X d Y (Normal)

10. 10 The following pedigree outlines an inheritance pattern Does this fit an autosomal recessive or autosomal dominant pattern of inheritance?

11. 11 Pedigree of Queen Victoria and the transmission of hemophilia. Victoria Albert Alice carrier Beatrice carrier Irene carrier Alix carrier Alice carrier Victoria carrier

12. 12 Characteristics of a X (sex)-linked recessive trait: Hemizygous males and homozygous females are affected Phenotypic expression is much more common in males than in females, and in the case of rare alleles, males are almost exclusively affected Affected males transmit the gene to all daughters but not to any sons Daughters of affected males will usually be heterozygous and therefore unaffected. Sons of heterozygous females have a 50% chance of receiving the recessive gene. gY GG GYgGGY gG GY

13. X-Linked Recessive males get their X from their mother fathers pass their X to daughters only females express it only if they get a copy from both parents. expressed in males if present recessive in females XrYXrY XRXRXRXR XRXrXRXr XRYXRY XrXrXrXr XrYXrY ½ X r Y½ X r X r 1 X R X r 1 X r Y

14. 14 Y chromosome All males in this pedigree will have the SAME Y-chromosome!!! X1/Y1; A1/A2xX2/X3; A3/A4 (grandpa)(grandma) X2/Y1; A2/A4 (dad) x X4/X5; A5/A6 (mom) X5/X2A2/A6X4/Y1A4/A6 DaughterSon Y Y Y Y Traits on the Y chromosome are only found in males, never in females. The father’s traits are passed to all sons. Dominance is irrelevant: there is only 1 copy of each Y- linked gene (hemizygous).

15. Large Pedigrees We are now going to look at detailed analysis of dominant and recessive autosomal pedigrees. To simplify things, we are going to only use these two types. The main problems: 1. determining inheritance type 2. determining genotypes for various individuals 3. determining the probability of an affected offspring between two members of the chart.

16. Dominant vs. Recessive Is it a dominant pedigree or a recessive pedigree? 1. If two affected people have an unaffected child, it must be a dominant pedigree: D is the dominant mutant allele and d is the recessive wild type allele. Both parents are Dd and the normal child is dd. 2. If two unaffected people have an affected child, it is a recessive pedigree: R is the dominant wild type allele and r is the recessive mutant allele. Both parents are Rr and the affected child is rr. 3. If every affected person has an affected parent it is a dominant pedigree. 4. If two unaffected mate and have an affected child, both parents must be Rr heterozygotes. 5. Recessive outsider rule: outsiders are those whose parents are unknown. In a recessive autosomal pedigree, unaffected outsiders are assumed to be RR, homozygous normal. 6. Children of RR x Rr have a 1/2 chance of being RR and a 1/2 chance of being Rr. Note that any siblings who have an rr child must be Rr. 7. Unaffected children of Rr x Rr have a 2/3 chance of being Rr and a 1/3 chance of being RR.

17. Assigning Genotypes for Autosomal Recessive Pedigrees 1. all affected are rr. 2. If an affected person (rr) mates with an unaffected person, any unaffected offspring must be Rr heterozygotes, because they got a r allele from their affected parent. (Aa) (A-)(aa) (Aa) (AA) (aa) (Aa) (A-) (aa) (A-)

18. Generations labelled roman numerals I, II,... Individuals labelled arabic numerals 1, 2,... Shaded = affected Mated individuals connected by "marriage line". Steps in assigning genotype ()

19. Try simplest hypothesis first: 1 gene, 2 alleles, complete dominance, affected are homozygous recessive. Fill in genotypes in steps: (1) All affected are homozygous recessive. Steps in assigning genotype (aa)() (aa) () (aa) ()

20. Try simplest hypothesis first: 1 gene, 2 alleles, complete dominance, affected are homozygous recessive. Fill in genotypes in steps: (1) All affected are homozygous recessive. (2) All unaffected have at least one dominant allele. Steps in assigning genotype (aa)(A) (aa) (A) (aa) (A)

21. Try simplest hypothesis first: 1 gene, 2 alleles, complete dominance, affected are homozygous recessive. Fill in genotypes in steps: (1) All affected are homozygous recessive. (2) All unaffected have at least one dominant allele. (3) All homozygous recessive must get one recessive allele from each parent. Steps in assigning genotype (aa)(A) (Aa) (A)(aa) (Aa) (A) (aa) (A)

22. Try simplest hypothesis first: 1 gene, 2 alleles, complete dominance, affected are homozygous recessive. Fill in genotypes in steps: (1) All affected are homozygous recessive. (2) All unaffected have at least one dominant allele. (3) All homozygous recessive must get one recessive allele from each parent. (4) All offspring of homozygous recessive must have at least one recessive allele. Steps in assigning genotype (aa)(A) (Aa) (A)(aa) (Aa) (A)(aa) (A)

23. Try simplest hypothesis first: 1 gene, 2 alleles, complete dominance, affected are homozygous recessive. Fill in genotypes in steps: (1) All affected are homozygous recessive. (2) All unaffected have at least one dominant allele. (3) All homozygous recessive must get one recessive allele from each parent. (4) All offspring of homozygous recessive must have at least one recessive allele. (5) For rest of genes, use – = allele unknown. We have filled in the pedigree without finding any internal contradictions, i.e. without contradicting our hypothesis. Steps in assigning genotype (aa)(A-) (Aa) (A-)(aa) (Aa) (A-)(aa) (A-)

24. Assigning Genotypes for Dominant Pedigrees 1. All unaffected are aa. 2. Affected children of an affected parent and an unaffected parent must be heterozygous Aa, because they inherited an a allele from the unaffected parent. 3. The affected parents of an unaffected child must be heterozygotes Aa, since they both passed an a allele to their child. 4. Outsider rule for dominant autosomal pedigrees: An affected outsider (a person with no known parents) is assumed to be heterozygous (Aa). 5. If both parents are heterozygous Aa x Aa, their affected offspring have a 2/3 chance of being Aa and a 1/3 chance of being AA. I II III (Aa) (aa) aa Aa aa Aa A-

25. Conditional Probability- Autosomal dominant Father is dd (probability = 1) Mother has a 1/2 chance of being DD and a 1/2 of being Dd. This is a dominant autosomal pedigree, Determine the probability of an affected child: 1. determine the probability of an affected offspring for each possible set of parental genotypes. 2. Combine them Mom has a 1/2 chance of being Dd and a 1/2 chance of being DD, and dad is dd. There are thus 2 possibilities for the cross: DD x dd, or Dd x dd. If the cross is DD x dd, all of the offspring will be Dd, and since the trait is dominant, all will be affected. On the other hand, if the cross is Dd x dd, ½ the offspring are Dd (affected) and ½ are dd (normal). So, probability of mating DD x dd = (½ x 1), with all offspring affected = (1), and probability of the mating being Dd x dd is (½ x 1), with ½ the offspring affected. ((1/2*1) * 1) + ((1/2*1) * 1/2) =3/4 overall probability dd 1 DD 1/2 Dd 1/2 Determining the probability of an affected offspring for most crosses is simple: Determine the parents’ genotypes and follow Mendelian rules to determine the frequency of the mutant phenotype. In some cases, one or both parents has a genotype that is not completely determined.

26. Conditional Probability- Autosomal recessive More complicated recessive pedigree: Grandparent genotype known, parent genotype is unknown Father has a ½ chance of being RR and a ½ chance of being Rr, Mother has a 1/3 chance of being RR and a 2/3 chance of being Rr. In this case there are 4 possible matings: 1. There is a 1/2 * 1/3 = 1/6 chance that the mating is RR x RR. In this case, 0 offspring will be affected (rr). 2. There is a 1/2 * 2/3 = 2/6 = 1/3 chance that the mating is RR x Rr. In this case, 0 offspring are affected. 3. There is a 1/2 * 1/3 = 1/6 chance that the mating is Rr x RR. In this case, 0 offspring will be affected (rr). 4. There is a 1/2 * 2/3 = 1/3 chance that the mating is Rr x Rr. In this case, 1/4 offspring will be affected (rr). Combining all possibilities: (1/6 * 0 ) + (1/3 * 0) + (1/6 * 0) + (1/3 *1/4) = 0 + 0 + 0 + 1/12 = 1/12 RR Rr ? ½ RR ½ Rr 1/3 RR 2/3 Rr