How do enzymes spends up. Chemical compounds
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Answer: The bacteria transformed with this particular plasmid will form white colonies on the plates containing ampicillin and Xgal.
Explanation: The lacZ gene produces an enzyme called β-galactosidase which is responsible for the breakdown of lactose into glucose and galactose. The lacZ gene is one of the three genes (the other two being lacA and lacY) of the lac operon which is responsible for the transport and mechanism of lactose in E. coli and many other bacteria.
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
Answer:
The nucleic acid is a macromolecule present in living systems in the form of either DNA or RNA, whose molecule consists of many nucleotides which are linked with each other to form long chains. The presence of DNA or RNA in an individual promotes continuity of life. The variability and complexity of nucleic acid are dependent on its constituent molecules as, sugar, a phosphate group, and a nitrogenous base. However, two monocyclic nitrogenous bases were classified purines (adenine:A, guanine:G) and pyrimidines (thymine:T, cytosine:C, and uracil:U).
In addition, RNA is a single-stranded structure which differs from DNA in two aspects, the sugar content of RNA is ribose, and pyrimidine base uracil replaces the thymine base of DNA. A ratio between two strands of DNA can be calculated by using Chargrff's rule which states that "DNA from any cell of all organisms possess a 1:1 ratio (base Pair Rule) of pyrimidine and purine bases'.
The purine to pyrimidine ratio in RNA with different bases is given below in the following table:
Presence of single strand in RNA does not follow chargaff's rule and alters the purine: pyrimidine ratio. The above ratio of each base in the given information is not equal, due to which the ratio of purines to pyrimidines is different. The ratio of DNA and RNA is not equal because DNA has its complementary DNA strand. However, bases in RNA might vary due to its single-strandedness property, but its ratio is dependent on the species.
Therefore, it can be concluded that the ratio of purines to pyrimidines is dependent on the species due to its genetic makeup. Hence, option (c) is correct.
