Lecture 1

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Unformatted text preview: tosomal Recessive T raits • Most human genetic disorders are recessive • Frequently show up in consanguineous matings (partners are blood related) What patterns of inheritance would one see in a pedigree for an autosomal recessive trait? - Affected individuals appear in progeny of unaffected parents - All progeny from two affected individuals are affected Autosomal Recessive Disorders • Predict the probability of affected offspring likely to result from the marriage between 2 carriers. Normal Dd Parents: Normal Dd D Eggs DD Normal d Offspring: D Dd Normal (carrier) Dd Normal (carrier) dd Deaf Sperm d Tips for recessive trait pedigree analysis •  For an individual to be affected (homozygous), both parents need to carry at least one copy of the recessive allele. -Therefore, to calculate the chance that an unknown individual will be affected, calculate the chances that the parents carry the recessive allele, then multiply these by the chance that the offspring will get the recessive alleles from both parents. • Note: Now for rare traits, make the simplifying assumption that unrelated, unaffected individuals are not carriers. Cystic Fibrosis - Afflicted individuals (Northern Europeans) often die by the age 20. - Gene encodes chloride transport channel protein - Excess mucus in several organs disrupts their function by causing ducts or tubes to be clogged. - Increased susceptibility to infections Autosomal Recessive Disorders: Cystic Fibrosis What is the chance the baby will be affected by cystic fibrosis? AA aa what is the chance the diamond will be affected? aa AA AA Aa AA Aa AA AA Aa Aa Aa Aa A- 1/2 Aa A- A- 1/4 Aa aa aa ? P [V_1: aa] = P [IV_1 Aa] ~ P [IV_2 Aa] x 1/4 1/4 2/3 Autosomal Recessive Disorders PKU is a rare recessive disorder. John whose brother has the disease marries Jane whose aunt (mom’ side) has the disease. If they have one child, what is the probability that the child will have PKU? The important word is rare, and since it is recessive, it must be homozygous recessive. because there is no information on dad side, you assume he is homozygous dominant AA Aa 2/3 Aa Aa aa P [ Diamond = aa] : what event must occur for diamond must occur? Both John and Jane needs to be a carrier. P [ John = Aa] x P [ Jane = Aa] x 1/4. and the chance that both of them are carriers multiplies with the chance the child is affected (1/4) aa John 2/3 Aa Jane P [Diamond = aa] 2/3 x 1/3 x 1/4 = 1/18 1/2 Aa If their first child has PKU, what is the probability that their second child will have PKU? Since the first child is affected, it is known that both John and Jane are carriers P [ II_2 = aa] =1/4...
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This note was uploaded on 01/16/2014 for the course LIFESCI 4 taught by Professor Ribaya during the Spring '08 term at UCLA.

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