Isolation of DNA

DNA isolation extracts DNA from cells. In exams treat it as an early step of recombinant DNA technology.

The governing concept is the initial sample-selection step in DNA isolation. Before cells are lysed and DNA is purified, an appropriate biological sample must be collected and identified. The sample should contain the target cells or tissue and should be handled in a way that limits contamination and degradation. Its quality directly affects the amount, purity, and integrity of the DNA obtained later. Therefore option B is correct. Protein staining is not the first essential extraction step, burning DNA destroys the target molecule, and making RNA produce light is unrelated to isolation. Correct sample collection provides the material on which all subsequent extraction steps depend.

The cell wall is a rigid outer layer, especially in plant, fungal, and bacterial cells. DNA is present inside the cell, so the cell wall and then the cell membrane must be disrupted to release the cellular contents containing DNA. Staining DNA may be used later for analysis, but it is not the purpose of breaking the cell wall. Exam tip: In DNA isolation, cell-wall disruption is done for cell lysis and release of DNA.

In lysis the cell membrane breaks and internal material comes out. It is a key step in DNA isolation.

Detergent dissolves the lipids of the cell membrane and nuclear membrane, causing membrane lysis and releasing DNA into the solution. It does not cut DNA into fragments; restriction enzymes perform that function. Exam tip: associate detergent in DNA isolation with membrane lysis.

SDS (sodium dodecyl sulfate) is a detergent that disrupts the lipids of cell and nuclear membranes, causing lysis and releasing DNA. It can also help denature proteins. Ethanol is used to precipitate DNA, whereas the primary role of SDS is membrane disruption. Exam tip: associate detergents in DNA isolation with membrane lysis.

Proteases are enzymes that break proteins into smaller peptides and amino acids. During DNA isolation, proteins from cells can remain mixed with the DNA, including structural proteins and DNA-associated proteins. Treating the sample with a protease digests these unwanted proteins, making it easier to separate and purify the DNA. Removing protein contamination is important because leftover proteins can interfere with later steps such as restriction digestion, amplification, or other analysis.

Thus option C is correct: protease-like enzymes help remove protein impurities. They do not color DNA, transform a cell into a plant, or freeze water. In a typical isolation procedure, protease treatment is combined with cell disruption and other purification steps. The enzyme acts on proteins, while the DNA is retained for collection and subsequent experiments. This is why proteases improve the quality of the isolated DNA.

RNase is an enzyme that degrades RNA. During DNA isolation, it removes RNA contaminants from the sample and helps obtain purer DNA. Proteins are usually removed using proteases, not RNase. Exam tip: Remember that RNase removes RNA, whereas DNase degrades DNA.

The governing concept is inhibition of metal-dependent nucleases during DNA extraction. EDTA is a chelating agent that binds divalent metal ions, especially magnesium ions. Many DNases require such ions as cofactors for catalytic activity. When EDTA sequesters them, nuclease activity is reduced and the isolated DNA is less likely to be degraded. Therefore option A is correct. EDTA does not cut DNA directly; restriction enzymes perform sequence-specific cutting. It does not illuminate the sample or convert protein into sugar. Its protective effect depends on binding the ions needed by enzymes that would otherwise damage DNA.

DNase is a nuclease enzyme that can cleave phosphodiester bonds in DNA, degrading it into smaller fragments. Therefore, its presence during DNA isolation can reduce both the quantity and quality of the DNA sample. DNase does not convert DNA into protein. Exam tip: Chelating agents such as EDTA are used during DNA isolation to inhibit DNase activity.

During DNA isolation, salt helps neutralize the negative charges on DNA’s phosphate backbone. This reduces the DNA’s solubility, especially when cold alcohol is added, so DNA molecules clump together and precipitate visibly. Therefore, option C is correct. Salt does not translate DNA into protein, create DNA from water, or merely remove cell colour.

In alcohol DNA solubility decreases and it can precipitate. Cold alcohol helps better precipitation.

Cold ethanol lowers the solubility of DNA, causing it to come out of solution as a precipitate. In the presence of salt, the charge effect of DNA phosphate groups is reduced, so DNA can aggregate into visible thread-like strands. Option B is incorrect because ethanol is used to isolate DNA, not to destroy it. Exam tip: Remember cold ethanol primarily for DNA precipitation during DNA isolation.

When alcohol is added, DNA becomes less soluble and may appear as a white, thread-like precipitate. Therefore, the material is precipitated DNA. Water vapour is gaseous and does not form such thread-like precipitates. Exam tip: Cold alcohol is commonly used to precipitate DNA during isolation.

Centrifugation spins a sample at high speed so that components with different densities separate. During DNA isolation, cellular debris or a precipitate may form a pellet at the bottom, while the liquid supernatant can be removed, helping in DNA purification. Staining a membrane is associated with microscopy methods, not centrifugation. Exam tip: the pellet is the material settled at the bottom, whereas the supernatant is the liquid above it.

After centrifugation DNA can collect at the bottom as a pellet. It is washed and dissolved again.

The direct answer is A: the DNA pellet is washed to remove impurities such as salts and residual alcohol. During DNA precipitation, DNA collects as a pellet, but some salts, buffer components, and alcohol can remain around it. A suitable wash, commonly involving alcohol under the protocol, helps remove these contaminants. The reasoning is: precipitation concentrates DNA; unwanted soluble chemicals may remain with the pellet; washing dissolves or carries away many of them; careful drying then helps remove remaining wash liquid; the DNA is finally redissolved for later work. Option A is correct because clean DNA performs better in downstream reactions. Option B is wrong because washing does not cut DNA; cutting requires restriction enzymes or another fragmentation method. Option C is wrong because DNA is not converted into protein by washing; protein synthesis is a cellular process involving RNA and ribosomes. Option D is wrong because a DNA pellet is not a living cell, so washing cannot make a cell alive. Excessive washing or incomplete drying can cause problems, so the protocol must be followed carefully. Memory cue: “wash the pellet to wash away contaminants, not to change the DNA.”

The direct answer is B: DNA can finally be dissolved in TE buffer or nuclease-free water. After isolation, DNA must be placed in a clean, aqueous solution that does not contain harmful nucleases or contaminants. TE buffer contains Tris, which helps maintain a suitable pH, and EDTA, which binds certain metal ions needed by many nucleases; this can help protect DNA during storage. Nuclease-free water is also commonly used, especially when downstream reactions may be sensitive to EDTA. Step by step: purify the DNA; remove unwanted chemicals; transfer it to TE buffer or nuclease-free water; mix gently and allow it to dissolve. Option B is correct. Option A is wrong because hot oil is not a compatible aqueous medium and heat may damage DNA. Option C is wrong because soil suspension contains contaminants and nucleases. Option D is wrong because chlorophyll solution is not a standard DNA solvent and may interfere with later tests. The best choice between TE and water depends on storage and downstream use. Memory cue: purified DNA is usually rehydrated in clean buffer or nuclease-free water.

In TE buffer, Tris acts as a buffer and maintains a stable pH. A stable pH helps preserve the structure and stability of DNA. EDTA binds metal ions and reduces DNase activity, so it is not the component mainly responsible for pH maintenance. Exam tip: Remember T = Tris (pH buffer) and E = EDTA (nuclease protection).

The governing concept is long-term protection of stored DNA. TE buffer contains Tris to help maintain a suitable pH and EDTA to chelate divalent metal ions such as magnesium. Nucleases often need these ions for activity; by binding them, EDTA reduces nuclease-mediated degradation and helps preserve DNA during storage. Thus option D is correct. EDTA does not directly amplify DNA, produce ethanol, or construct a membrane. Amplification requires a separate reaction containing template, primers, nucleotides, polymerase, and suitable conditions. The protective function of EDTA explains why it is included in TE buffer even though it is not itself a DNA-amplifying reagent.

Cellulose is a major component of the plant cell wall. Cellulase hydrolyses cellulose and helps weaken or break the cell wall, making DNA isolation easier. Pectinase acts on pectin in the middle lamella, but it does not directly digest cellulose. Exam tip: remember cellulase for cellulose and pectinase for pectin.

Lysozyme is an enzyme used during bacterial DNA extraction to help open the bacterial cell. Bacteria have a strong cell wall made mainly of peptidoglycan. Lysozyme breaks or weakens this supporting material, so the cell membrane and internal contents become easier to release. This is important because DNA must be freed from inside the cell before it can be purified. Therefore, option B is correct. It does not stain DNA, remove ethanol, or produce RNA; those are unrelated to the specific function of lysozyme in this procedure.

During lysis, weakening the wall makes the later disruption steps more effective and reduces the chance that the DNA remains trapped inside intact cells. Lysozyme does not directly act as a dye or a solvent. Its role is structural: it attacks peptidoglycan and assists cell lysis. Thus the phrase “to weaken the bacterial cell wall” describes its purpose accurately. The supplied answer B and explanation are scientifically consistent and sufficiently clear.

The fungal cell wall mainly contains chitin. Chitinase breaks down chitin and helps disrupt the cell wall, making DNA isolation easier. Protease may help remove proteins, but it is not the primary enzyme for breaking a chitin-containing cell wall. Exam tip: Choose the enzyme according to the major cell-wall component—for fungi, choose chitinase.

Plants may contain polysaccharides and polyphenols that interfere with DNA purification. CTAB method helps reduce them.

During phenol-chloroform extraction, proteins and several other cellular impurities partition into the organic phase, while DNA usually remains in the aqueous phase. Thus, it is used to obtain relatively pure DNA. Restriction enzymes cut DNA; phenol-chloroform does not. Exam tip: Associate phenol-chloroform in DNA isolation with protein removal and purification.