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FinalExam2009Key1

# FinalExam2009Key1 - Version Key#1 Final Exam BlS101-001...

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Version #1 Key BlS101-001, Spring 2009, Final Exam R. L. T:|3tTz, lnstructor 1. Sanfilippo A syndrome is an autosomal recessive lethal neurological disease. This disease was first reported in the Cayman Islands particularly in a population in the West Bay district of Grand Cayman. Sanfilippo A syndrome leads to neurological degeneration andearly death. Two individuals (III-1 and III-2) are expecting a child (O). Although the woman's father (II-3) is not from a high-risk group, her maternal uncle died of the disease. Furthermore, her husband's sister (III-3) also died of the disease. The woman is concerned that she may be a carrier and that her child will inherit Sanfilippo A syndrome. Using the symbols S (normal) and s (Sanfilippo A syndrome), indicate the genotypes of the individuals I-1, II-3 and II-4 on the pedigree below (6 pts.). What is the probability that the woman's mother is a carrier of the disease (i.e.,looks normal but carries the s allele)? Circlethe correct answer. (6 pts.) What is the probability that the woman will inheritthe s allele? Circle the correct answer. (6 pts.) A. u2 B. u4 c. U3 D.213 E. U9 F. 1/18 G.ll2 H. u4 I. ll3 J. 213 K. u9 L. l/18 U2 u4 u3 2t3 u9 1/18 The mother can apryar normal in 2 ways: SS and Ss. The Ss genotype makes up 2/3 of these possibilities Assuming thefother is SS, the woman has a % chance of getting the S or tlrc s alleles from her mother. The product of 1/z and 2/3 or I/3 is also acceptable. Just lilrB the woman's mother the husband can appear normal in 2 ways: SS or Ss. Because his sister is fficteQ his parents must be Ss. What is the probability that the husband is a carrier of the disease? Circle the correct answer. (6 pts.) M. N. o. P. a. R.

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Version#l Key I I I I I 2. The bacteria E. coli has the ability to synthesize the essential amino acid, histidine. Geneticists used mutational analysis to determine that multiple genes are needed to make histindine. The complementation matrix below was used to analyze these mutants. Complementation Matrix for 9 Mutants 5 Genes: Group A: mutants 1, 3 Group B: mutants 2, 6 Group C: mutant 4 Group D: mutants 5,9 Group E: mutants 8, 7 Order of groups is not important but the mutants should be distributed in among the 5 genes or complementation groups. From the results of this complementation test, indicate the number of genes (complementation groups) in the histidine operon and the gene location for each of the nine mutants in their respective genes. (14 pts.) vl C (! 5 E Mutants 1 2 3 4 5 6 7 8 9 1 + + + + + + + 2 + + + + + + 3 + + + + + + 4 -l + + + + 5 + + + 6 + + + 7 + 8 + 9
Version #l Key 3. The Wobble Hypothesis states that the third base at the 5' side of the anticodon loop can hydrogen bond with certain noncomplementary bases. The Wobble rules are shown below.

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