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Question
How many of the following statement(s) regarding insertional inactivation of lacZ gene are incorrect?
1. lacZ gene remains active in non-recombinant transformed bacteria.
2. X-gal gets hydrolysed and produces a blue product in non-recombinant transformed bacteria.
3. Bacterial colonies which are transformed with recombinant DNA appear white.
4. 𝛃-Galactosidase is produced in recombinant bacteria.
5. Bacterial colonies which are transformed with recombinant DNA appear blue.
How many of the following statement(s) regarding insertional inactivation of lacZ gene are incorrect?
1. lacZ gene remains active in non-recombinant transformed bacteria.
2. X-gal gets hydrolysed and produces a blue product in non-recombinant transformed bacteria.
3. Bacterial colonies which are transformed with recombinant DNA appear white.
4. 𝛃-Galactosidase is produced in recombinant bacteria.
5. Bacterial colonies which are transformed with recombinant DNA appear blue.
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Solution
The correct option is B Two
The lacZ gene codes for the protein 𝛃-galactosidase. It is an enzyme that hydrolyses the chromogenic substrate X-gal and produces a blue colored product. The lacZ gene is present in several vectors as a selectable marker. Some of the vectors have a recognition site for a restriction enzyme within the DNA sequence of the lacZ gene.
When the gene of interest is inserted within the lacZ gene of the vector and introduced into the bacteria, the enzyme 𝛃-galactosidase will not be synthesised in the host cell because of insertional inactivation. Thus, X-gal will not be hydrolysed to produce a blue product. The colonies as a result will appear white in colour and are referred to as recombinants as they would have taken up the recombinant vector with the gene of interest during transformation.
Those bacteria that have taken up the vector (transformed) which does not have the gene of interest inserted within its lacZ gene (non-recombinant) will have the lacZ gene active. As a result, it will have the ability to produce the enzyme 𝛃-galactosidase. This enzyme will hydrolyse X-gal to produce a blue coloured product which will result in these bacterial colonies appearing blue.
Thus, the lacZ gene as a selectable marker helps to screen recombinants from non-recombinants.
Therefore statements 4 and 5 are incorrect. Bacterial colonies which are transformed with recombinant DNA appear white, while bacterial colonies which are transformed with non-recombinant DNA (vector without the insert) appear blue.
The lacZ gene codes for the protein 𝛃-galactosidase. It is an enzyme that hydrolyses the chromogenic substrate X-gal and produces a blue colored product. The lacZ gene is present in several vectors as a selectable marker. Some of the vectors have a recognition site for a restriction enzyme within the DNA sequence of the lacZ gene.
When the gene of interest is inserted within the lacZ gene of the vector and introduced into the bacteria, the enzyme 𝛃-galactosidase will not be synthesised in the host cell because of insertional inactivation. Thus, X-gal will not be hydrolysed to produce a blue product. The colonies as a result will appear white in colour and are referred to as recombinants as they would have taken up the recombinant vector with the gene of interest during transformation.
Those bacteria that have taken up the vector (transformed) which does not have the gene of interest inserted within its lacZ gene (non-recombinant) will have the lacZ gene active. As a result, it will have the ability to produce the enzyme 𝛃-galactosidase. This enzyme will hydrolyse X-gal to produce a blue coloured product which will result in these bacterial colonies appearing blue.
Thus, the lacZ gene as a selectable marker helps to screen recombinants from non-recombinants.
Therefore statements 4 and 5 are incorrect. Bacterial colonies which are transformed with recombinant DNA appear white, while bacterial colonies which are transformed with non-recombinant DNA (vector without the insert) appear blue.
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