Isolation of DNA

The governing concept is sequential DNA purification. Cell lysis releases DNA together with proteins, membrane lipids, RNA, salts, and other cellular components. The next main objective is therefore to separate these contaminants while keeping the DNA in solution or precipitating it in a controlled way. Option A is correct because removal of impurities improves DNA purity and makes the sample suitable for PCR, restriction digestion, electrophoresis, or other analyses. Options B, C, and D describe transformations that do not occur during extraction and have no laboratory purpose in this step. Depending on the protocol, impurity removal may involve detergents, proteases, organic extraction, binding to silica, washing, or selective precipitation.

RNase breaks down RNA, so it is used during DNA isolation to remove RNA contamination from the sample. Lipase acts on fats, cellulase on cellulose, and amylase on starch. Exam tip: remember RNase for RNA removal and protease for protein removal.

The direct answer is C: proteinase K is used to degrade proteins. It is a proteolytic enzyme that digests many proteins during DNA extraction, including proteins associated with cells and enzymes that might damage DNA, such as nucleases. The reasoning is: cells are broken open; proteins and nucleases are released; proteinase K breaks those proteins into smaller pieces; removal of protein improves DNA purity and helps protect DNA from enzymatic degradation. Option C is correct because this is the established role of proteinase K in many extraction protocols. Option A is wrong because methylating DNA requires DNA methyltransferase, not proteinase K. Option B is wrong because copying RNA into DNA or another nucleic acid requires polymerase-based processes, not a protease. Option D is wrong because cellulose production is a biological synthesis process and is unrelated to proteinase K. Proteinase K does not selectively remove every contaminant, so other steps such as extraction, washing, and purification may still be needed. Memory cue: “proteinase K attacks protein,” including unwanted nucleases.

Nucleases are enzymes that can cut phosphodiester bonds in DNA and break it into smaller fragments. Inactivating them during DNA isolation protects the extracted DNA and keeps it intact, so option D is correct. Removing salts is part of purification, not the purpose of nuclease inactivation. Exam tip: DNases are nucleases that specifically degrade DNA.

The governing concept is chelation, in which EDTA binds metal ions through several coordinating groups and makes them unavailable in solution. In DNA extraction, EDTA commonly binds divalent cations such as Mg2+ and Ca2+. Magnesium is especially important because many nucleases require Mg2+ as a cofactor; removing it can reduce nuclease activity and help protect DNA from degradation. Therefore option A is correct. Helium is a gas rather than the relevant divalent metal ion, while chlorophyll is a pigment and starch is a carbohydrate granule, so options B, C, and D cannot be chelated in this context. EDTA must nevertheless be compatible with later enzyme reactions.

Detergent helps DNA release and EDTA helps inhibit nucleases. Both are useful in isolation buffer.

The phosphate groups of DNA are negatively charged. Cations from salt partly neutralise this charge, while alcohol lowers DNA solubility. Consequently, DNA forms a precipitate that can be separated. Alcohol alone may precipitate DNA less effectively because of the remaining charge repulsion. Exam tip: In DNA extraction, salt plus cold alcohol is mainly used for DNA precipitation.

Both isopropanol and ethanol can be used to precipitate and isolate DNA from a solution. In the presence of salt, alcohol decreases DNA solubility, allowing DNA to be collected as strands or a pellet. DNA replication, in contrast, is carried out by enzymes in cells or in an in vitro system, not by alcohol. Exam tip: In DNA isolation, remember that alcohol is mainly used for precipitation.

A 70 percent ethanol wash helps remove salts and small impurities. Then the pellet is dried properly.

Residual ethanol can inhibit enzyme reactions. Therefore the pellet is dried without over-drying.

Excessive drying can make a DNA pellet hard and compact, so it becomes difficult to resuspend in buffer or water. It does not immediately degrade or turn into protein. Exam tip: after ethanol washing, dry the pellet only long enough to remove residual ethanol.

Protein contamination can affect enzyme reactions and purity readings. Protein removal is necessary.

A low A260 by A280 ratio can indicate protein or phenol contamination. Remember the ratio in quality check.

The A260 by A230 ratio can indicate impurities like salts phenol and carbohydrates. It is useful in purity assessment.

DNA absorbs at 260 nm and the reading can estimate concentration. It helps in sample normalization.

Intact genomic DNA is very large, so it migrates slowly through agarose gel and usually appears as a high-molecular-weight band near the well. In contrast, broken or degraded DNA contains fragments of many sizes and commonly produces a smear in the lane. Exam tip: a distinct heavy band near the well suggests intact DNA, whereas a smear suggests degradation.

DNA is negatively charged due to its phosphate backbone. Therefore it moves toward the positive electrode in an electric field.

The negative charge of DNA is important in steps like precipitation and gel movement. Salt helps in charge shielding.

The direct answer is D: protecting DNA from excessive mechanical force helps prevent DNA shearing. Long DNA molecules can be broken by vigorous pipetting, vortexing, shaking, or repeated rough handling. The reasoning is: genomic DNA has very long molecular chains; mechanical force places stress on those chains; stressed chains may snap into smaller fragments; therefore gentle pipetting and mixing help retain high-molecular-weight DNA. Option D is correct because preventing fragmentation is the main reason for gentle handling during isolation. Option A is wrong because gentle handling does not increase RNA; RNA quantity is controlled by different biological and purification factors. Option B is wrong because DNA handling force does not make protein; protein production requires gene expression, translation, and appropriate cellular machinery. Option C is wrong because removing salt requires washing or purification, not simply avoiding mechanical force. The question says “excessive” force: ordinary careful handling is not automatically harmful, but unnecessary vigorous treatment is risky. Memory cue: “handle long DNA like a delicate thread—pulling or shaking can break it.”

Gentle pipetting protects genomic DNA from shearing. It is important for high molecular weight DNA.

In a fresh sample, DNA is generally less likely to undergo degradation or oxidation, so DNA of better quality may be obtained. RNA does not automatically disappear; it may need to be removed using treatment such as RNase. Exam tip: Fresh or properly preserved material is important for obtaining good-quality DNA during isolation.

The governing concept is nucleic-acid stability during storage. An old or poorly stored biological sample may experience repeated warming, moisture exposure, unsuitable pH, microbial activity, or action by nucleases. These factors can fragment DNA or chemically damage it, lowering its integrity and sometimes reducing the recoverable yield. Hence option A is correct. Degraded DNA may appear as a smear rather than a sharp high-molecular-weight band and may perform poorly in long-amplicon PCR or other applications. Option B is false because storage does not guarantee purity. Option C is unrelated, since proteins may remain or degrade independently. Option D is also false because lysis is still required to release DNA from cells.

The governing concept is physical disruption before chemical lysis. Homogenization breaks a tissue or solid sample into smaller, more uniform particles and increases the surface area exposed to the lysis buffer. As a result, detergents, salts, and protective reagents can contact more cells efficiently, improving the release and recovery of DNA. Therefore option A is correct. Homogenization does not stain DNA, manufacture ethanol, or add smell to proteins, so options B, C, and D are unrelated. It should be performed carefully: excessive grinding can produce heat or shear DNA, while insufficient disruption may leave intact tissue and reduce yield. The exact tool depends on the sample type.

The direct answer is A: liquid nitrogen quickly freezes plant tissue and makes it brittle for grinding. Plant cells have tough walls, so efficient disruption is needed before DNA can be released. Step by step: fresh tissue is exposed to very cold liquid nitrogen; water in the tissue freezes rapidly; the tissue becomes hard and brittle; grinding breaks the cells into a fine powder; DNA can then be extracted more efficiently. Rapid freezing can also slow enzyme activity, including nucleases that might otherwise degrade DNA, although careful handling and a suitable extraction method remain necessary. Option A is correct. Option B is wrong because liquid nitrogen is used to cool and preserve material, not burn DNA. Option C is wrong because it does not convert protein into sugar; that would require different biochemical processes. Option D is wrong because ethanol removal is achieved by washing, evaporation or other protocol steps, not by the main purpose of liquid nitrogen. Liquid nitrogen is extremely cold and requires trained handling and proper protective equipment. Memory cue: liquid nitrogen makes plant tissue “freeze, become brittle, and grind.”

The governing concept is contamination by plant secondary metabolites. When plant cells are disrupted, polyphenols can be released and oxidized into reactive compounds. Some of these compounds associate with nucleic acids or co-purify with the DNA, producing colored or impure preparations and interfering with polymerases and other enzymes. Thus option A is correct. The problem is not that polyphenols improve DNA, convert DNA into RNA, or remove the cell wall. Those claims make options B, C, and D incorrect. Plant extraction protocols may use antioxidants, suitable detergents, high-salt conditions, polyvinylpyrrolidone, or additional purification steps to reduce polyphenol contamination. The exact treatment depends on the plant tissue.