Biochemical Techniques — MCQs

Biochemical Techniques Indian Medical PG Practice Questions and MCQs

Question 51: Which test requires an intact peptide bond?

A. UV diffraction (Correct Answer)
B. Ninhydrin
C. Diazo reaction
D. All of the above

Explanation: ### Explanation The correct answer is **A. UV diffraction**. **1. Why UV Diffraction is Correct:** UV diffraction (specifically UV absorption spectroscopy) in proteins relies on the presence of the **peptide bond** itself, as well as aromatic side chains. The peptide bond (—CO—NH—) exhibits a characteristic absorption peak in the "far-UV" range (approximately **190–220 nm**). This absorption is due to the electronic transitions within the amide group. Therefore, if the peptide bonds are hydrolyzed or the protein is completely broken down into individual amino acids, this specific spectral signature is lost. **2. Why Other Options are Incorrect:** * **B. Ninhydrin:** This reagent reacts with the **free alpha-amino group** ($NH_2$) of amino acids. While it can react with the N-terminus of a protein, it is primarily used to detect free amino acids. It does not require an intact peptide bond; in fact, it is the standard reagent used to visualize amino acids after a protein has been fully hydrolyzed. * **C. Diazo Reaction (Pauly’s Test):** This test is used to detect specific amino acids like **Histidine and Tyrosine**. It involves the coupling of a diazotized sulfanilic acid with the imidazole or phenolic rings of these amino acids. It depends on the side chain structure, not the peptide backbone. **3. High-Yield Clinical Pearls for NEET-PG:** * **Biuret Test:** This is the most common colorimetric test that **requires at least two peptide bonds** (tripeptide or larger) to form a purple coordination complex with $Cu^{2+}$ ions in an alkaline medium. * **Aromatic Absorption:** While peptide bonds absorb at 190-220 nm, proteins also show a distinct absorption peak at **280 nm** due to aromatic amino acids (**Tryptophan** > Tyrosine > Phenylalanine). * **Ninhydrin Color:** Most amino acids give a **Ruhemann's Purple** color; however, **Proline and Hydroxyproline** (imino acids) yield a characteristic **yellow** color.

Question 52: Which of the following is the best technique to screen neonates for inherited metabolic disorders?

A. Tandem mass spectrometry (Correct Answer)
B. Northern Blotting
C. PCR
D. ChIP technique

Explanation: **Explanation:** **Tandem Mass Spectrometry (TMS/MS-MS)** is the gold standard for newborn screening (NBS) because of its high sensitivity, specificity, and "multiplexing" capability. Unlike traditional methods, TMS can detect over 30–50 different metabolic disorders (including amino acid disorders, organic acidemias, and fatty acid oxidation defects) from a single dried blood spot in just a few minutes. It works by identifying and quantifying diagnostic metabolites (acylcarnitines and amino acids) based on their mass-to-charge ratio. **Analysis of Incorrect Options:** * **Northern Blotting:** Used specifically for the detection and quantification of **RNA** sequences. It is a research tool and not suitable for rapid, high-throughput clinical screening. * **PCR (Polymerase Chain Reaction):** Used to amplify specific **DNA** sequences. While useful for confirming a genetic diagnosis, it is not the primary screening tool for metabolic metabolites in neonates. * **ChIP (Chromatin Immunoprecipitation) technique:** A method used to investigate the interaction between **proteins and DNA** (e.g., transcription factor binding). It has no role in routine metabolic screening. **High-Yield Clinical Pearls for NEET-PG:** * **Guthrie Test:** The historical "heel prick" test used for Phenylketonuria (PKU) screening; it has largely been replaced by TMS. * **Timing:** Newborn screening is ideally performed **24–48 hours after birth** to allow for the accumulation of metabolites after protein ingestion. * **Commonly Screened Disorders:** PKU, Maple Syrup Urine Disease (MSUD), Galactosemia, Congenital Hypothyroidism, and Congenital Adrenal Hyperplasia (CAH).

Question 53: The presence of two lone pairs of electrons on the oxygen atom in a water molecule results in what?

A. Water acting as a non-polar solvent
B. Formation of covalent bonds in ice
C. An overall negative charge on the water molecule (Correct Answer)
D. An overall positive charge on the water molecule

Explanation: **Explanation** **1. Why Option C is Correct:** The water molecule ($H_2O$) has a **bent/V-shaped geometry** due to $sp^3$ hybridization. Oxygen is highly electronegative compared to hydrogen. The oxygen atom possesses two lone pairs of electrons that are not involved in bonding. These lone pairs occupy more space and exert a strong repulsive force, pulling electron density away from the hydrogen atoms. This results in an **asymmetric distribution of charge**, creating a partial negative charge ($\delta^-$) near the oxygen atom and a partial positive charge ($\delta^+$) near the hydrogens. This permanent dipole allows water to act as a universal solvent for polar and ionic substances in biological systems. **2. Why Incorrect Options are Wrong:** * **Option A:** Water is a **polar solvent**, not non-polar. Its polarity is the reason it can dissolve electrolytes and polar biomolecules (like glucose). * **Option B:** Ice is held together by **Hydrogen bonds**, not covalent bonds. While covalent bonds hold the O and H together *within* a molecule, hydrogen bonds are the intermolecular forces responsible for the lattice structure of ice. * **Option D:** The oxygen atom’s high electronegativity and lone pairs ensure the "negative end" of the dipole is at the oxygen, making a positive charge on the oxygen atom physically impossible. **3. Clinical Pearls & High-Yield Facts for NEET-PG:** * **Dielectric Constant:** Water has a high dielectric constant (~78.5), which reduces the attractive forces between ions, facilitating the dissociation of salts (e.g., NaCl) in the cytosol. * **Hydrogen Bonding:** Each water molecule can form a maximum of **four hydrogen bonds**. This property is responsible for the high specific heat and surface tension of water. * **Amphoteric Nature:** Water can act as both an acid and a base, which is fundamental to the **Bicarbonate Buffer System** maintaining blood pH at 7.4.

Question 54: ELISA is:

A. A name of a scientist
B. A radiologic procedure
C. A pioneer in dentistry
D. An enzymatic immune reaction (Correct Answer)

Explanation: **Explanation:** **ELISA (Enzyme-Linked Immunosorbent Assay)** is a fundamental biochemical technique used to detect and quantify substances such as peptides, proteins, antibodies, and hormones. **Why the correct answer is right:** The core principle of ELISA is an **enzymatic immune reaction**. It utilizes the specificity of **antigen-antibody binding** (the "immuno" part) coupled with an **enzyme-mediated color change** (the "enzyme-linked" part). In a typical assay, an enzyme (like Horseradish Peroxidase or Alkaline Phosphatase) is conjugated to an antibody. When a specific substrate is added, the enzyme reacts to produce a visible color change; the intensity of this color is proportional to the amount of the target substance present. **Why the incorrect options are wrong:** * **Option A & C:** ELISA is an acronym, not the name of a scientist or a dental pioneer. * **Option B:** It is a laboratory-based biochemical assay, not a radiologic procedure (which would involve X-rays, CT, or MRI). **High-Yield Clinical Pearls for NEET-PG:** * **Types of ELISA:** * **Direct:** Detects antigens using a primary labeled antibody. * **Indirect:** Detects antibodies (e.g., HIV screening). * **Sandwich:** Detects antigens between two layers of antibodies (highly sensitive). * **Competitive:** Used when the antigen is small. * **Clinical Use:** It is the standard **screening test** for HIV (confirmed by Western Blot). * **Key Enzymes used:** Horseradish Peroxidase (HRP), Alkaline Phosphatase, and β-galactosidase.

Question 55: Hopkins-Cole test is performed on proteins to detect the presence of which group?

A. Imidazole group
B. Indole group (Correct Answer)
C. Guanidine group
D. Phenolic group

Explanation: **Explanation:** The **Hopkins-Cole test** (also known as the Glyoxylic acid test) is a specific biochemical color reaction used to detect the presence of the amino acid **Tryptophan** in a protein solution. 1. **Why the Indole group is correct:** Tryptophan is the only proteogenic amino acid that contains an **indole ring** in its side chain. In this test, glyoxylic acid reacts with the indole ring in the presence of concentrated sulfuric acid to form a characteristic **violet or purple-colored ring** at the junction of the two liquids. 2. **Why the other options are incorrect:** * **Imidazole group:** This is found in **Histidine**. It is detected by the **Pauly’s test**. * **Guanidine group:** This is found in **Arginine**. It is detected by the **Sakaguchi test**. * **Phenolic group:** This is found in **Tyrosine**. It is detected by the **Millon’s test**. **High-Yield Clinical Pearls for NEET-PG:** * **Gelatin and Casein:** Gelatin gives a negative Hopkins-Cole test because it lacks Tryptophan, whereas Casein (milk protein) gives a strongly positive result. * **Tryptophan Derivatives:** Tryptophan is the precursor for clinically significant molecules like **Serotonin (5-HT)**, **Melatonin**, and **Niacin (Vitamin B3)**. * **Hartnup Disease:** A defect in the transport of neutral amino acids (primarily Tryptophan) leads to pellagra-like symptoms due to Niacin deficiency. * **Xanthoproteic Test:** While Hopkins-Cole is specific for the indole ring, the Xanthoproteic test detects all aromatic amino acids (Phenylalanine, Tyrosine, Tryptophan) by nitration, yielding a yellow color.

Question 56: Which of the following tests can distinguish between monosaccharides and disaccharides?

A. Barfoed's test (Correct Answer)
B. Bial's Test
C. Seliwanoff's test
D. Hydrolysis test

Explanation: **Explanation:** **1. Why Barfoed’s Test is Correct:** Barfoed’s test is specifically designed to distinguish **monosaccharides from reducing disaccharides**. The reagent consists of copper acetate in dilute acetic acid (acidic medium). While both monosaccharides and reducing disaccharides can reduce cupric ions to cuprous oxide, **monosaccharides are stronger reducing agents** and react much faster. In an acidic medium, monosaccharides produce a red precipitate within 1–3 minutes, whereas reducing disaccharides require prolonged boiling (7–12 minutes) because they must first undergo slight hydrolysis. **2. Analysis of Incorrect Options:** * **Bial’s Test:** Used to detect **Pentoses** (e.g., ribose). It uses orcinol and HCl to produce a blue-green condensation product. * **Seliwanoff’s test:** Used to distinguish **Ketohexoses** (e.g., fructose) from aldohexoses. It uses resorcinol and HCl to produce a cherry-red complex. * **Hydrolysis test:** This is a general procedure used to break down glycosidic bonds in disaccharides or polysaccharides into their constituent monosaccharides; it is not a specific diagnostic color test for differentiation. **3. High-Yield Clinical Pearls for NEET-PG:** * **Benedict’s vs. Barfoed’s:** Benedict’s test is performed in an **alkaline** medium (detects all reducing sugars), while Barfoed’s is in an **acidic** medium (distinguishes based on the rate of reaction). * **Sucrose:** It is a non-reducing sugar and will give a **negative** result for both Benedict’s and Barfoed’s tests unless hydrolyzed first. * **Osazone Test:** Another method to differentiate sugars based on crystal morphology (e.g., Needle-shaped for Glucosazone, Sunflower-shaped for Maltosazone).

Question 57: What is true about Polymerase Chain Reaction (PCR)?

A. Amplification of a target sequence of DNA (Correct Answer)
B. Uses a thermolabile enzyme
C. A less sensitive technique
D. Uses one primer

Explanation: **Explanation:** **Polymerase Chain Reaction (PCR)** is a fundamental molecular biology technique used to make millions of copies of a specific DNA segment. It is essentially "in vitro cloning." 1. **Why Option A is Correct:** The primary objective of PCR is the **exponential amplification** of a specific target DNA sequence. This is achieved through repeated cycles of denaturation, annealing, and extension, allowing clinicians to detect even minute quantities of genetic material. 2. **Why Other Options are Incorrect:** * **Option B:** PCR requires a **thermostable** enzyme (not thermolabile), most commonly **Taq Polymerase** (derived from *Thermus aquaticus*). This enzyme must survive the high temperatures (94–96°C) used during the denaturation step. * **Option C:** PCR is an **extremely sensitive** technique. It can amplify DNA from a single cell or a tiny drop of blood, making it superior to many traditional culture or staining methods. * **Option D:** PCR requires **two primers** (forward and reverse). These are short synthetic oligonucleotides that are complementary to the sequences flanking the target DNA region on both strands. **High-Yield Clinical Pearls for NEET-PG:** * **Steps of PCR:** 1. Denaturation (~95°C), 2. Annealing (~55°C), 3. Extension (~72°C). * **RT-PCR:** Uses Reverse Transcriptase to convert RNA into cDNA before amplification (Gold standard for **COVID-19** and HIV viral load). * **Real-Time PCR (qPCR):** Allows for the quantification of DNA in real-time using fluorescent dyes. * **Applications:** Diagnosis of genetic mutations (e.g., Sickle cell anemia), forensic medicine (DNA fingerprinting), and detecting infectious agents (TB, Hepatitis).

Question 58: The Biuret reaction is specific for which functional group?

A. Peptide bonds (-CONH-) (Correct Answer)
B. -CSNH2 group
C. -NHNH2 group
D. All of the above

Explanation: **Explanation:** The **Biuret reaction** is a fundamental colorimetric test used for the detection and quantification of proteins. **1. Why Option A is correct:** The test relies on the presence of **peptide bonds (-CONH-)**. In an alkaline medium, cupric ions ($Cu^{2+}$) react with the nitrogen atoms of at least two peptide bonds to form a characteristic **violet or purple-colored coordination complex**. For a positive result, a molecule must contain at least two peptide bonds (i.e., it must be a tripeptide or larger). This is why free amino acids (except histidine) and dipeptides give a negative Biuret test. **2. Why Options B and C are incorrect:** While the Biuret reagent is named after the compound "biuret" (formed by heating urea), the reaction is specifically used in biochemistry to identify the repeating peptide backbone of proteins. * **-CSNH2 (Thiocarbonamide) and -NHNH2 (Hydrazine):** While certain non-protein compounds containing these groups (like oxamide or malonamide) can technically react with copper salts, they are not the physiological targets of this test in a medical biochemistry context. In the scope of the NEET-PG syllabus, the Biuret test is defined by its specificity for the **peptide linkage**. **Clinical Pearls & High-Yield Facts:** * **Minimum Requirement:** A minimum of **two peptide bonds** is required for a positive test. * **Color Intensity:** The intensity of the purple color is directly proportional to the number of peptide bonds present, making it useful for quantitative protein estimation. * **Interference:** High concentrations of ammonium salts can interfere with the reaction. * **Clinical Use:** It is commonly used in laboratories to measure total serum protein levels. * **Note:** Histidine is the only amino acid that may give a positive result due to its imidazole ring.

Question 59: Substitution of which amino acid in the place of glycine would result in an increase in UV absorption of a protein solution?

A. Lysine
B. Arginine
C. Tryptophan (Correct Answer)
D. Histidine

Explanation: ### Explanation **Core Concept: UV Absorption of Proteins** The UV absorption of proteins at **280 nm** is primarily due to the presence of **aromatic amino acids**. These amino acids contain conjugated double bonds in their ring structures that absorb light energy. The three aromatic amino acids are Phenylalanine, Tyrosine, and Tryptophan. **Why Tryptophan is Correct** Among the aromatic amino acids, **Tryptophan (Trp)** has the highest molar absorptivity (extinction coefficient) because of its bulky indole ring. While Tyrosine also absorbs at 280 nm, Tryptophan’s contribution is significantly greater (about 4 times that of Tyrosine). Glycine is a simple amino acid with no side chain and zero UV absorbance at 280 nm. Therefore, substituting Glycine with Tryptophan will drastically increase the solution's absorbance. **Analysis of Incorrect Options** * **A. Lysine & B. Arginine:** These are basic amino acids. They lack aromatic rings and do not contribute to light absorption at 280 nm. * **D. Histidine:** Although Histidine contains an imidazole ring, it does not absorb significantly at 280 nm. Its absorption peak is much lower (around 211 nm). **High-Yield NEET-PG Pearls** * **Order of UV Absorbance (at 280 nm):** Tryptophan > Tyrosine > Phenylalanine. * **Beer-Lambert Law:** This principle is used to quantify protein concentration based on UV absorbance ($A = \varepsilon cl$). * **Peptide Bonds:** These absorb UV light in the "far-UV" range (**190–220 nm**), whereas aromatic side chains absorb in the "near-UV" range (**280 nm**). * **Fluorescence:** Tryptophan is also the primary contributor to the intrinsic fluorescence of proteins.

Question 60: Which statement is true regarding Restriction Fragment Length Polymorphism (RFLP)?

A. Restriction enzymes cut DNA at the nucleotide level.
B. Restriction enzymes act at specific recognition sites. (Correct Answer)
C. Only cohesive ends are produced by restriction enzymes.
D. Only blunt ends are produced by restriction enzymes.

Explanation: ### Explanation **1. Why the Correct Answer is Right:** Restriction Fragment Length Polymorphism (RFLP) is a technique used to detect variations in homologous DNA sequences. It relies on **Restriction Endonucleases (RE)**, which are bacterial enzymes that function as "molecular scissors." These enzymes do not cut DNA randomly; they recognize and bind to specific, usually palindromic, sequences of nucleotides known as **recognition sites** (typically 4–8 base pairs long). A mutation within this site can abolish or create a new cutting point, resulting in DNA fragments of varying lengths, which are then visualized via gel electrophoresis. **2. Analysis of Incorrect Options:** * **Option A:** Restriction enzymes do not cut at the "nucleotide level" (which implies breaking down DNA into individual monomers). Instead, they cleave the **phosphodiester backbone** between specific nucleotides within a double-stranded DNA molecule. * **Options C & D:** Restriction enzymes can produce two types of cuts. Some (like *EcoRI*) produce **cohesive (sticky) ends** with overhanging single-stranded tails, while others (like *HaeIII*) produce **blunt ends** with no overhangs. Both types are useful in recombinant DNA technology, but neither is produced exclusively. **3. High-Yield Clinical Pearls for NEET-PG:** * **Palindromic Sequences:** Recognition sites read the same 5' $\rightarrow$ 3' on both strands (e.g., 5'-GAATTC-3'). * **Applications of RFLP:** Historically used for DNA fingerprinting, paternity testing, and detecting carrier status for genetic diseases like **Sickle Cell Anemia** (where a mutation destroys the *MstII* enzyme recognition site). * **Nomenclature:** The first letter is the Genus, the next two are the species, and the Roman numeral denotes the order of discovery (e.g., *EcoRI* from *E. coli*).

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Spectrophotometry and Colorimetry

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Chromatography Techniques

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Electrophoresis and Applications

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Centrifugation and Ultracentrifugation

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Radioisotope Techniques

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Enzyme-Linked Immunosorbent Assay (ELISA)

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Polymerase Chain Reaction (PCR)

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Blotting Techniques: Southern, Northern, Western

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Mass Spectrometry in Biochemistry

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Recombinant DNA Technology

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DNA Sequencing

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Proteomics and Metabolomics

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