bis101kliebensteinfinal2009

bis101kliebensteinfinal2009 - 1 BIS 101 - 003 FINAL...

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1 BIS 101 - 003 FINAL December 8, 2009 READ THE FOLLOWING INSTRUCTIONS: Write your name legibly on each page of this exam. All work must be done in ink . Show all your work. Points will be subtracted for incorrect answers in multiple-choice questions. However, you will not get a negative total score for such a question. This exam is worth 200 points. You have 110 min. to complete the exam. Good Luck!
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2 Name:______________________________ P a g e : 3 ________ 4 ________ 5 ________ 6 ________ 7 ________ 8 ________ 9 ________ 10 ________ 11 ________ 12 ________ 13 ________ 14 ________ Total ________
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3 Name:______________________________ (1) The question involves the lactose operon of E. coli where I = lacI , P = the promoter, O = lacO, Z = lacZ (the galactosidase gene) and Y = lacY (the permease gene). Complete the table below indicating by + that the specific enzyme is going to be produced and by - that it is not going to be produced. (10 points) Genotype Inducer absent Inducer present galactosidase Permease galactosidase Permease I + P + O + Z + Y + - - + + I + P + O + Z - Y + - - - + I + P + O c Z + Y + + + + + I + P - O + Z + Y + - - - - I s P + O + Z + Y - F’I + P + O + Z - Y - - - - - I + P + O c Z - Y + F’I + P + O + Z + Y - - + + + I - P + O + Z + Y + F’I + P + O + Z - Y - - - + + (2) Amino acid serine is encoded by a total of six codons. They are: 5’ UCU 3’ 3’AGG5’ 5’ AGU 3’ 3’UCG5’ 5’ UCA 3’ 3’AGU5’ 5’ UCG 3’ 3’AGU5’ 5’ UCC 3’ 3’AGG5’ 5’ AGC 3’ 3’UCG5’ There are three serine t-RNAs. They have anticodons 3’UCG5’, 3’AGG5’, and 3’AGU5’. Write beside each codon the anticodon that will recognize that codon and label the 5' and 3' ends on the codons and anticodons. (6 points)
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4 Name:______________________________ (1) Which of the following Neurospora crassa asci show a gene conversion and which show postmeiotic segregation? (6 points) Conversion: 6 Postmeiotic segregation: 3, 4, and 7 1 2 3 4 5 6 7 + + + + + + arg + + + + + + + + arg + + arg arg + + arg Arg arg arg arg + arg arg Arg + + arg arg arg arg Arg arg + arg arg arg + Arg arg arg arg arg arg + Arg arg arg arg + ( 2 ) The figure below is part of a scheme illustrating the molecular basis of crossing over in eukaryotes. The illustration shows a double Holiday junction. Enzymatic cutting of DNA strands followed by local DNA resynthesis and ligation results either in no crossing over or crossing over. Show one scenario of cutting DNA strands and resynthesis that would result in crossing over. (6 points)
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5 Name:______________________________ (1) A certain type of deafness in humans is inherited as an X-linked recessive . A man who suffers from this type of deafness marries a normal woman, and they are expecting a child. They find that they are related. Part of the family tree is shown below. How would you advise the parents about the probability of their first child being: (a) a deaf boy: (3 points) (1/2) 4 x 1/2 (boy)
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This note was uploaded on 10/23/2011 for the course BIS 101 taught by Professor Simonchan during the Spring '08 term at UC Davis.

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bis101kliebensteinfinal2009 - 1 BIS 101 - 003 FINAL...

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