macintosh
Forum Elite
Topics: 38
Posts: 178
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| Tiff
Forum Guru
Topics: 46
Posts: 450
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D?
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| hottie99
Forum Elite
Topics: 21
Posts: 344
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this is the stuff i hate on step 1, i didn't know i needed a degree in molecular and celllular bio for this test 
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| macintosh
Forum Elite
Topics: 38
Posts: 178
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The graph shows beta galactosidase activity against time. The experiment started by culturing E coli in a medium rich in both glucose and lactose. The graph has two parts. In the first part (representing the intial half of the experiment) the glucosidase acitivity is low. This is expplained by the fact that in the presence of sufficient glucose the bacteria makes enough ATP to keep the ATP/cAMP ratio high. The CAP site can not be activated in the presence of low cAMP, therefore a low glucosidase activity.
In the second half of the experiment, glucose is consumed by bacterial metabolism. Consequently not enogh ATP can be produced and the ATP/cAMP ratio is low. When cAMP is high the CAP site becomes activated and therfore high glucsidas activity. Plaease note that the medium at this point is still rich in lactose as it can only be broken down in the presence of high glucosidase activity. The presence of lactose induces formation of inactive repressor, which results in an unbound and active operator.
The correct answer therefore is choice B.
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