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BIOL 112 - Practice Qs from 2008

BIOL 112 - Practice Qs from 2008 - Additional Practice...

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Additional Practice Final Exam questions for Biol 112 April 2008 Question #59 (1 point) A mutant of E. coli is isolated in which the genes of the lactose operon are always expressed (constitutively) at high levels, whether lactose is present or not. This could be due to a mutation in (1) lactose (2) lacI (3) the lac terminator (4) the lac operator Mark (a) if only ONE of the above could explain the phenotype of the mutant. (b) if TWO of the above could explain the phenotype of the mutant. (c) if THREE of the above could explain the phenotype of the mutant. (d) if FOUR of the above could explain the phenotype of the mutant. (e) if NONE the above could explain the phenotype of the mutant. Question #60 (1 point) An E. coli mutant that transcribes lac Z constitutively is shown to have a defective operator. Which of the following explains what the defect may be: (a) The gene for the lac operator may have a base substitution mutation that causes it to code for a much shorter protein. (b) Lactose can no longer bind the operator sequence. (c) RNA polymerase blocks the binding of the defective operator to the promoter. (d) The defective operator can no longer bind the repressor Lac I in the absence of lactose. (e) The repressor Lac I binds irreversibly to the defective operator in the absence of lactose. Question #61 – 64 (4 x 0.25 points) A bacterium has a functional lac operon. For each statement below specify whether it is true or false by marking (a) for true and (b) for false. 61. The lac Z, lac Y and lac A sequence of genes contains only 1 promoter preceding the lac Z gene. The sequence is therefore an operon.
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