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suter [353]
3 years ago
7

You hypothesize that a transcriptional regulator may control expression of specific genes in LB medium, and want to find out wha

t these genes are. You have made a set of strains (i.e., library) in which a promoterless lacZ was inserted at varied chromosomal locations. Design an experiment to identify these genes using agar plates with LB and X-gal. Describe the steps in this procedure.
Biology
1 answer:
s344n2d4d5 [400]3 years ago
7 0

Answer:

if the promoterless lacZ is inserted close to any gene(s) or operon, then the laz gene will also be transcribed when the operon gets expressed. The polycystronic RNA will eventually produce a beta-galactosidase protein(product of lacZ gene) which will cleave X-Gal( structural analog to lactose), which will produce blue chromogenic substance.

prepare LB agar. Allow it to cool, add X-gal before it solidifies, pour to plates.

create grids and patch plate each strain on it. Just pick the strain that looks blue, amplify the sequence upstream of the lacZ and sequence it.

Explanation:

Just pick the strain that looks blue, amplify the sequence upstream of the lacZ, and sequence it.

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A DNA fragment is introduced into the lacZ gene of a plasmid, which also contains an ampicillin resistance gene. What is the app
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Answer: The bacteria transformed with this particular plasmid will form white colonies on the plates containing ampicillin and Xgal.

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In recombinant DNA technology, when a plasmid is to be used to transform a host cell, such markers are used to help screen the transformed cells from the ones that have not taken up the plasmid. Xgal present in the plates is an artificial substrate which is hydrolyzed by

β-galactosidase into 5-bromo-4-chloro-indoxyl which will dimerize and oxidise into 5,5'-dibromo-4,4'dichloro-indigo. This is a blue pigment which will give blue color to the bacterial cells. Introducing a DNA fragment in this lacZ gene will make it non-functional so it will not be able to produce the enzyme.

Therefore, when a bacterial cell is transformed with a plasmid containing ampicillin resistance gene and a DNA fragment introduced in the lacZ gene and then grown on plates containing ampicillin and Xgal, white colored colonies will appear. The white colonies will show the bacterial cells that have successfully taken up the plasmid with the DNA fragment incorporated in the lacZ gene as this will render the gene non-functional and will not produce β-galactosidase which will breakdown Xgal to give blue colonies. Since the plates contain ampicillin, only the bacterial cells that have been successfully transformed with the plasmid ( the ones that have the DNA fragment and the ones without it) will grow as the ampicillin resistance will give them resistance against ampicillin in the plates. The bacterial cells that have not taken up the plasmid will not be resistant to ampicillin and will not form colonies on the plate.

This is called blue-white screening which is used to identify successfully transformed host cells. A picture of this is given in the attachment, taken from the following website:

https://www.mun.ca/biology/scarr/Blue_&_White_Colonies.html

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