6. Note whether in the following strains of E. coli, B-galactosidase will be: (7

Question

6. Note whether in the following strains of E. coli, B-galactosidase will be: (7 Points) (A) Abundant in both the presence and absence of lactose; (B) Abundant only in the absence of lactose; (C) Abundant only in the presence of lactose; (D) Abundant only in the presence of sucrose; or (E) Absent in both the presence and absence of lactose. Strain 1: Wild-type Strain 2: o(c) Strain 3: i(s) p(-) Strain 4: p(-) o(c) Strain 5: i(s) o(c) / ?(+) o(+) Strain 6: i(-) p(+) / i(+) p(-) Strain 7: i(-) p(+) o(c) / ?(+) p(-) 0(+)

Explanation / Answer

Strain 1: Wild-type

(C)Abundant only in the presence of lactose:

Yes it express because lac is present in the medium and it work as induce that enhance the production of beta-galactosidase. Here lactose bind with repressor, product of I-gene, and bring conformational changes in the protein and it unable to bind with operator.

Strain 2: o(c)

(A)Abundant in both the presence and absence of lactose:

As this strain contains Oc mutation, hence, due to constitutive nature due to mutation that will not allow the binding of repressor protein, hence, it will express be galatosidase in all the conditions. Here presence or absence of lactose does not matter.

Strain 3:i(s) p(-)

(E )Absent in both the presence and absence of lactose

In this I is encode for super-repressor and promoter is mutated. In this condition rerpressor bind to operator but promoter will not allow the binding of RNA polymerase hence any condition beta galactosidase will not form.

Strain 4: p(-) o(c)

(E )Absent in both the presence and absence of lactose

In this situation promoter is mutated therefore RNA polymerase does not bind to promoter and move to operator. Although here promoter is constitutive expression but it will not express beta galactosidase.

Strain 5: i(s) o(c) / i(+) o(+)

(A)Abundant in both the presence and absence of lactose:

It is diploid condition where two operons are present.

In this situation here repressor is a super repressor and this will bind only wild type operator and in case of Oc it will not bind to this operator because it get mutated. Here beta galactosidase will synthesize.

Strain 6: i(-) p(+) / i(+) p(-)

(A)Abundant in both the presence and absence of lactose:

In this situation diploid condition is present. As the product of I-gene, repressor, can act on distant operator hence it make i(-) p(+) operon functional. Whereas in second condition, i(+) p(-), promoter will not bind with RNA polymerase. Here it act like wild type. Mentioning the same results of wild type.

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