University:
California State University, Northridge
Course:
BIOL 360 | Genetics
Academic year:

2007
Views:

193

Pages:

22
Author:

Davon Sims

Mendelian Analysis !Mendel followed single genes through multiple generations – Law of Segregation – Law of Independent Assortment !Observed ratios of !3:1 and 1:2:1 in monohybrid crosses and !9:3:3:1 in dihybrid crosses !These ratios explain some inheritance patterns Mendelian Analysis in Humans " Most traits are affected by more than one gene " cannot be analyzed using simple Mendelian genetics. " Some are relatively simple genetically " follow the pattern of Mendelian inheritance. " Many single gene traits that can be followed " cause rare diseases " rather than common phenotypes. Mendelian Analysis in Humans " Genetics is not straightforward in humans: " no true-breeding humans " generation time is long (impractical) " controlled matings??? " siblings rarely mate to generate F2 (EEEW!) " small family sizes How can one study human genes, and find genes that cause human diseases? " Geneticists make use of the variation that already exists in the population " Pedigree analysis: " rare traits can be studied by carefully analyzing families that are affected. " the trait may be rare in the population but common in a particular family " Important assumption: " when a trait is rare, and there is a random mating, one can always assume that a person who marries into the affected family is NOT a carrier. Pedigree Analysis • Track several generations of a family for specific characteristic(s) • Standardized symbols • You must know and use these on Exams – do not make up your own please! In humans we must use pedigrees to study inheritance • Pedigrees are an orderly diagram of a families relevant genetic features extending through multiple generations • Pedigrees help us infer if a trait is from a single gene and if the trait is dominant or recessive Dominant (rare) Traits How one can look at a pedigree and tell that the disease is caused by a dominant gene? " Affected individual always has at least one affected parent " Affected individual can have one normal parent " Affected individuals appear in every generation • Two unaffected parents NEVER have affected offspring " Two affected parents may have unaffected progeny " This NEVER happens in recessive traits Polydactyly Dominant human genetic disorder in which affected individuals have extra digits Dominant Single Gene Inheritance -- Heterozygotes show the phenotype -- Affected individuals have at least one affected parent Dominant Traits dd dd Dd dd dd Dd dd Dd Dd Dd Dd dd dd dd dd dd dd dd Dd dd dd Dd dd Dd 50% Dd Dd Dd Dd dd dd dd dd What is the probability that diamond will be affected? Recessive (rare) Traits " Unaffected parents can have Affected progeny " This can NEVER happen in dominant traits " Frequently show up in consanguineous mating (where partners are blood related) " Eeew! " All progeny from two affected individuals are always affected (not all have to be in Dominant) " Can skip generations Question #1 = Two individuals affected with a rare recessive trait have children. #2 = Two other individuals affected with a rare dominant trait have children. Which pedigree belongs to which pair? A. B. Tips for recessive trait pedigree analysis " For an individual to be affected (homozygous), both parents need to be carriers (heterozygous). " Therefore, to calculate the chance that an unknown individual will be affected " calculate the chances that the parents will be heterozygous " then multiply these by the chance that the offspring will get both of the mutant alleles. " Use the product rule! (Chance of events occurring together) " For rare traits, make the simplifying assumption that unrelated, unaffected individuals are not carriers. Rare recessive disease: What is the chance the child will be affected? aa AA *If a is rare: unrelated person is AA A Aa AA A!Aa Aa Aa AA !Aa A2/3Aa ? "aa a A AA Aa a Aa aa aa (1/2)(1/2)(2/3)(1/4) = 1/24 Jack has a brother with cystic fibrosis (rare recessive trait). Jack marries Jill whose aunt has the disease. # If they have one child, what is the probability that the child will have cystic fibrosis? Aa Aa AA Aa Aa A- aa p(Aa)= 2/3 aa Jack Ap(Aa)= 2/3 p(aa)= 1/4 Jill Ap(Aa)= 1/2 p(? aa) = p(Jack Aa) X p(Jill and fam. Aa) X p(?aa) = (2/3) X (2/3 X !) X 1/4 = 1/18 If their first child has cystic fibrosis: • what is the probability that their second child will have cystic fibrosis? Aa aa AA Aa Jack Ap(Aa)= 1 aa Aa Aa A- aa Jill Ap(Aa)= 1 ? p(aa)= 1/4 1 X 1 X 1/4 = 1/4 Sex Chromosomes " discovered between 1891-1909 " males: mismatched pair of chromosomes (XY) " females: matched pair (XX) X Linked Recessive Disorders Pedigree Features • More males than females show the phenotype • NONE of the progeny of affected male show the phenotype – (unless mother is carrier or affected) • ALL of daughters of affected male are carriers (heterozygotes) • Daughter will show phenotype ONLY IF both her mother and her father have the allele X Linked Recessive Disorders Pedigree Features • Affected males pass the allele to • ALL daughters but • NONE of their sons •Affected females pass the allele to •ALL their sons •ALL their daughters • Heterozygous female carriers pass the allele to • HALF their sons and • HALF their daughters Y-Linked Inheritance • Males inherit genes on the differential region of the Y chromosome • Father’s transmit to all their sons - none of their daughters • SRY gene - primary role in Maleness • Maleness is Y linked • No Y linked genes are convincingly associated with nonsexual phenotypic variants Possible Nonsex Y linked gene • Hairy Ear Rim gene: • Proposed to be caused by a Y linked gene • Extremely rare in most countries • Less rare in India • In some, but not all, families hairy ear rims are transmitted from father to son

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