PracticeFinal-11KEY2

PracticeFinal-11KEY2 - [1(20 points Diagnosis of genetic...

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pg 1 [1] (20 points) Diagnosis of genetic diseases is often confounded by observing phenotypes that range from severe to mild making it unclear whether one gene with multiple alleles is involved or two or more genes are involved (one example is Duchenne & Becker muscular dystrophy). Confusion can arise when some mutations completely eliminate function while others lead to reduced function (e.g. the ß + and ß o globin mutations). Understanding the molecular nature of the mutations can help in understanding the phenotypic symptoms. a] Draw the structure of a generalized eukaryotic gene that has three introns with the protein coding portion of the gene beginning in exon 3 and ending in exon 4. b] Draw the resulting mRNA. c] Identify the location of ALL key regulatory and structural elements e.g. start & end of transcription, translation, splicing etc. Write out and circle all strictly conserved sequences that demarcate functional sites.
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pg 2 2) 25pts Jeffrey (I-1) and his oldest child (II-1) are sufferers of adult polycystic kidney disease (PKD), a rare autosomal dominant disorder. Jeffrey’s other children are at-risk but do not at present exhibit overt symptoms and have not been diagnosed with the disorder. Both are anticipating the birth of their first child, and are worried that they possess and may have transmitted the defective allele. They have elected prenatal diagnosis, although they do not wish their own status to be evaluated. The Southern blot has been typed for the closest known RFLP which unfortunately maps 4 cM from the PKD gene. When probed, allele A is represented by a single band at 4.9 kb, whereas allele B is represented by bands at 3.7 and 1.2 kb. 1 2 1 2 3 4 5 1 2 4.9 3.7 1.2 Use the following notations when solving this problem: Pkd = polycystic kidney disease + = normal A +/B Pkd = genotype illustrating A linked to wildtype and B linked to the Pkd allele 1pt each. Total =9 Answer : I-1 = Pkd B/+ B; I-II = + A/+A; II-1 = Pkd B/+ A; II-3 = + A/+A; II-4 = + A/+A; II-2 must have a plusA from mom & can have either Pkd B or + B from Dad thus, __ B/+ A and his child III-1 definitely inherited the B allele from the Dad and thus would be __ B/+ A. For II-5, we need to back reason. The fetus (III_2) has A & B and must have gotten the B allele from the Dad II-4 because II-4 is + B/ + B. Thus, the A allele must come from II-5 who is + A/ __ B. The fetus got the + A allele from grandmother (I-2) but the A allele could acquire the PkD allele by crossover in II-5 and hence must also be listed in the fetus as _ A/+ B. A ] Next to the symbols in the pedigree write the genotypes of each individual with respect to their linkage between A vs. B and Pkd vs. +. 9pts I II III
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pg 3 B] What is the probability that fetus III-1 will inherit PKD? Explain how you arrived at that number. 3pts
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This note was uploaded on 12/13/2011 for the course BIOSCI 137 taught by Professor Staff during the Fall '11 term at UC Irvine.

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PracticeFinal-11KEY2 - [1(20 points Diagnosis of genetic...

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