Question

This question refers to the lac operon. Select the correct phenotype for the genotype: I--P--OcZ--/I+P--O+Z+

[Terminology: β Galactosidase activity = β Gal activity]

Select one:

a-no β Gal activity, with no lactose (--) , β Gal activity with lactose (+)

b.no β Gal activity, with no lactose (--) , no β Gal activity with lactose (--)

c-β Gal activity, with no lactose (+) , β Gal activity with lactose (+)

d-β Gal activity, with no lactose (+) , no β Gal activity with lactose (+)

Answer 1

Answer is:

b.no β Gal activity, with no lactose (--) , no β Gal activity with lactose (--)

Explanation:

The I+ encodes wild type subunits. Where as I- encodes mutant type sub units. The both sub units combine to form tetramer which will acts as wild-type repressor. When the lactose is absent in the cytoplasm of E-Coli. Then this active repressor binds to the operator (O+). But it is not able to bind to the O^c. But It has Z-. Hence there will be no expression of structural genes from both DNA.

When lactose is present, then it will bind to the repressor and inactivate the repressor. Henec it can not bind to the operator. RNA polymerase binds to the promoter in O^C DNA. But it has Z-. That's why there is not found any expression of lac Z gene. Similarly, the lower strand has P-. It does not allow RNA polymerase to bind to it. Henec it can not proceed further. There will be no expression of structural gene occurs.

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