Human Genetics Geneticists are primarily interested in humans to establish the pattern of transmission of inherited traits – specifically those associated with disease –Mendelian ratios do not apply in individual human families because of the small size –Controlled matings cannot be made as is possible in experimental genetics

Determining how diseases are inherited in Humans – Pedigrees consists in collecting information on affected and nonaffected persons in a family, preparing a pedigree chart, and looking for exceptions to standard transmission patterns Pedigree = A “family tree” drawn with standard genetic symbols, showing inheritance patterns for specific phenotype characters. –Used to test various hypothesis and reveal allelic determination –Determine if it is a rare inherited disorder

Generations in a pedigree diagram are numbered, by convention, using Roman numerals, starting with the parental generation, at the top of the diagram as generation I. For convenience, the members of each generation are numbered across the line, from left to right, using normal numerals

Autosomal recessive disorders Phenylketonuria (PKU), due to mutations (loss of function) in an enzyme called phenylalanine hydroxylase - which converts phenylalanine into another aminio acid called tyrosine. In a child with PKU there is no chemical reaction to convert phenylalanine to tyrosine leading to a build up of phenylalanine in the blood and other body tissues. PKU is treated by a low protein diet. If left untreated it can result in mental retardation. Albinism: Albino individuals do not produce pigment melanin, which protects skin from UV radiation, making their skin sensitive to sunlight.

Autosomal dominant disorders Brachydactyly: Malformed hands with short fingers. Indian hedgehog gene (Ihh) is expressed in the prehypertrophic chondrocytes of cartilage elements, where it regulates the rate of hypertrophic differentiation. Misexpression of Ihh prevents proliferating chondrocytes from initiating the hypertrophic differentiation process. BRACHYDACTYLY TYPE A1 [IHH, GLU95LYS] In all affected individuals of a 4-generation Chinese pedigree affected with brachydactyly type A1 (112500). Nat Genet 2001 Aug;28(4): Mutations in IHH, encoding Indian hedgehog, cause brachydactyly type A-1. Gao B, Guo J, She C, Shu A, Yang M, Tan Z, Yang X, Guo S, Feng G, He L.

Dominant relationships A dominant trait is the easiest to recognize. A dominant trait will not occur in an individual unless it also appears in at least one of the parents. –Exceptions: A new mutation Incomplete penetrance A fully dominant trait will not skip generations. It will often appear relatively common in a pedigree. Unaffected sibs will have only unaffected offspring.

Recessive relationships In a marriage of two affected individuals, all of the offspring will be affected. A recessive trait commonly skips one or more generations because it is masked in heterozygotes.

Autosomal inheritance An autosomal trait can be passed from a father to his son. Especially for a recessive autosomal trait, approximately the same number of males and females will be affected.

X-linked inheritance Can never be passed from a father to his son since father’s X is passed to daughters. If the trait is recessive, all sons of a female who express the trait will also be affected. If recessive, the trait will occur most frequently in males. If dominant, it may occur more often in females.

Could this trait be autosomal recessive? YES

Could this trait be autosomal dominant? NO If the trait were an autosomal dominant, the affected child would have to have an affected parent who could pass the trait down to the child.

Could this trait be X-linked recessive? YES

Could this trait be X-linked dominant? NO If the trait is a dominant (x-linked or autosomal), an affected child must have an affected parent. In this case, the affected male child would have to have an affected mother if the trait is inherited as an X-linked dominant trait.

Could this trait be autosomal recessive? NO The parents would have to be homozygous (aa) and could only produce homozygous, affected children. This pedigree contains two, unaffected children.

Could this trait be autosomal dominant? YES If the this trait is inherited as an autosomal dominant both parents could be heterozygous Aa and could produce affected children AA or Aa or they could produce unaffected children aa.

I II RECESSIVE TRAIT: (A-) unaffected and (aa) affected What is the genotype of the mother? What is the genotype of the father? What are the genotypes of the children?

I II RECESSIVE TRAIT: (A-) unaffected and (aa) affected Mother’s genotype: Aa Father’s genotype: aa Aa, aa, Aa, Aa, aa Aaaa Aa aa

I II DOMINANT TRAIT: (A-) affected and (aa) unaffected What is the genotype of the mother? What is the genotype of the father? What are the genotypes of the children?

I II DOMINANT TRAIT: (A-) affected and (aa) unaffected Mother’s genotype: aa Father’s genotype: Aa aa, Aa, aa, aa, Aa aa Aa aa Aa

A couple has a female child with Tay Sachs disease, and three unaffected children. Neither parent nor any of the four biological grandparents of the affected child has had this disease. The most likely genetic explanation is that Tay Sachs disease is inherited as a(n) ______________ disease. WHY? autosomal recessive The disease is recessive because both parents are unaffected, and autosomal because a female child is affected but her father is not.

Individual Independent Events Example: a gamete carrying a dominant allele being formed in a heterozygote. –½ Example: a gamete carrying a dominant allele forming in a homozygote. –1

Sequence of independent events where the order is set or irrelevant Example: a family of three children with boy-boy-boy –Multiply individual probabilities: ½ x ½ x ½ = 1/8

Sequence of events in which different orders must be pooled A six child family composed of 4 girls and 2 boys in any order. –Probability formula: n!/s!t! (p) s (q) t n = number of individual in the family P = probability of the first event Q = probability of the second event S = number of cases in the first event T = number of cases in the second event –6!/4!2! (0.5) 4 (0.5) 2

Problem A husband and wife, both heterozygous for PKU gene (autosomal recessive), plan to have 6 children. What is the probability of that four of the offspring will be normal and two will have PKU. –6!/4!2!(0.75)4(0.25)2 = 0.297