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BIOLOGICAL MOLECULES

126 questions found

Practice Questions

The function of the 5′ cap and 3′ poly-A tail added to eukaryotic mRNA is to

A. Serve as the start and stop signals for translation of the coding sequence
B. Facilitate the splicing of exons and the removal of introns
C. Protect the mRNA molecule from exonucleolytic degradation in the cytoplasm
D. Provide the template for the synthesis of the protein's primary sequence

The 5' cap and the 3' poly-A tail are not translated. Their primary roles are to increase the stability of the mRNA by protecting its ends from ribonucleases, and to facilitate the initiation of translation by interacting with translation initiation factors.

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Jun 27, 2026

The instability of RNA compared to DNA is primarily due to the

A. Presence of uracil instead of thymine
B. Presence of a 2'-hydroxyl group on its ribose sugar
C. Single-stranded nature of all RNA molecules
D. Inability of RNA to form hydrogen bonds

The 2'-OH group in the ribose sugar of RNA is reactive. It can act as a nucleophile and attack the adjacent phosphodiester bond under alkaline conditions, leading to the self-hydrolysis (cleavage) of the RNA strand. DNA, lacking this 2'-OH, is far more chemically stable.

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Jun 27, 2026

The function of a chelating agent, such as EDTA, in an enzyme reaction is to

A. Bind to the active site as a competitive inhibitor
B. Bind to and remove free metal ion activators or cofactors from the solution
C. Denature the protein by breaking disulfide bridges
D. Increase the substrate concentration to saturating levels

Many enzymes require metal ions (like Mg²⁺, Ca²⁺) as activators or cofactors. EDTA chelates (binds tightly to) these divalent cations, making them unavailable to the enzyme. This effectively inhibits the enzyme's activity. Adding back an excess of the metal ion reverses the inhibition.

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Jun 27, 2026

Bile salts are amphipathic cholesterol derivatives secreted from the liver. Their hydrophobic side associates with lipid droplets, and their hydrophilic side faces the aqueous intestinal fluid. This coating breaks large globules into smaller ones (micelles), vastly increasing the surface area for lipase action.

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Jun 27, 2026

The fundamental reason that proteins are the most diverse class of biological molecules in terms of structure and function is that

A. They are the only molecules built from nitrogen-containing monomers
B. They are synthesized directly from the information encoded in DNA
C. Their monomers (20 amino acids) can be arranged in a vast number of sequences with different R-group chemistries
D. They are synthesized only in the presence of nucleic acids

With 20 different amino acids as monomers, the number of possible sequences and lengths for a polypeptide is astronomically large. Furthermore, the diverse chemical properties of the 20 R-groups (charged, polar, non-polar, etc.) enable a protein to fold into an immense variety of complex 3D shapes.

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Jun 27, 2026

In a solution of DNA, the absorption of ultraviolet light at 260 nm is significantly increased if the DNA undergoes

A. Annealing to a complementary strand
B. Denaturation (melting) into single strands
C. Supercoiling by gyrase enzymes
D. Packaging around histone proteins

The nitrogenous bases in double-stranded DNA are stacked and have a lower absorbance. When the double helix is denatured into two random, single-stranded coils, the bases become unstacked. This unstacking increases their absorbance of UV light at 260 nm, a phenomenon known as the hyperchromic effect.

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Jun 27, 2026

Specificity is the ability of an enzyme to choose exactly one substrate from a pool of similar molecules. This is due to the exact complementary fit and specific chemical interactions (ionic, H-bonding, hydrophobic) between the substrate and the R-groups lining the active site.

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Jun 27, 2026

In covalent catalysis, a powerful nucleophilic R-group in the active site (e.g., the -SH of cysteine or -OH of serine) forms a transient covalent bond with the substrate. This acyl-enzyme intermediate is then resolved by another step, releasing the product and regenerating the free enzyme.

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Jun 27, 2026

The allosteric regulation of an enzyme differs from competitive and non-competitive inhibition in that allosteric modulators

A. Always bind to the active site of the enzyme
B. Bind to a site distinct from the active site, leading to a conformational change
C. Are always irreversible inhibitors of the enzyme
D. Compete with the substrate for binding at the catalytic site

Allosteric regulation is mediated by modulator molecules that bind to a site (allosteric site) physically distinct from the active site. This binding causes a conformational change that can either increase (allosteric activator) or decrease (allosteric inhibitor) the activity of the enzyme at its active site.

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Jun 27, 2026

The reason that an increase in the concentration of a competitive inhibitor does not change the maximum velocity (Vmax) of an enzymatic reaction is that

A. The inhibitor reduces the turnover number of the enzyme
B. The inhibitor permanently denatures a fraction of the enzyme population
C. The inhibitor's binding can be overcome by sufficiently increasing the substrate concentration
D. The inhibitor binds only to the enzyme-substrate complex, not the free enzyme

The definition of competitive inhibition is a "competition" for the active site. At a high enough concentration, the substrate out-competes the inhibitor for the active site, so all enzyme molecules can still bind substrate and reach Vmax. The apparent Km is increased, but Vmax is ultimately unchanged.

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Jun 27, 2026

In the structure of an antibody molecule, the region responsible for the vast diversity that allows binding to a specific antigen is the

A. Constant region of the heavy chain
B. Variable region at the amino-terminal end of both the light and heavy chains
C. Transmembrane anchoring domain
D. The carbohydrate moiety attached to the Fc region

The amino-terminal ends of both the light (VL) and heavy (VH) chains form the antigen-binding site. These variable domains have highly diverse amino acid sequences from one antibody clone to another, creating a unique 3D surface that is specific for a single epitope.

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Jun 27, 2026

Saponification is the base-catalyzed hydrolysis of the ester bonds in a fat or oil. This reaction cleaves the triglyceride, producing glycerol and the salts of the fatty acids (soaps). Lipases perform an analogous enzymatic hydrolysis.

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Jun 27, 2026

Kinases are a class of transferase enzymes that catalyze the transfer of a γ-phosphate group from a high-energy donor molecule like ATP to a specific substrate. Protein kinases phosphorylate specific serine, threonine, or tyrosine residues on target enzymes, regulating their activity. Phosphatases reverse this.

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Jun 27, 2026

The primary difference between α-D-glucose and β-D-glucose is the orientation of the hydroxyl group attached to the

A. 1st carbon atom
B. 4th carbon atom
C. 6th carbon atom
D. 5th carbon atom

When glucose forms a ring, carbon 1 becomes an asymmetric carbon (the anomeric carbon). In the α-anomer, the -OH on C1 is below the plane of the ring (trans to the CH2OH at C5). In the β-anomer, the -OH is above the plane of the ring (cis to the CH2OH).

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Jun 27, 2026

The main structural difference between amylose and amylopectin, the two components of starch, is that amylopectin has a

A. Linear, unbranched structure of glucose linked by α-1,4 bonds
B. Highly branched structure due to the presence of α-1,6-glycosidic bonds
C. Structure composed of β-1,4-linked glucose units only
D. Lower molecular weight and solubility compared to amylose

Amylose is a linear polymer of glucose with α-1,4 linkages. Amylopectin is a much larger, branched polymer that has both α-1,4 linkages in the straight chain and α-1,6 glycosidic bonds at the branch points approximately every 24-30 glucose units.

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Jun 27, 2026

A non-competitive inhibitor’s effect on a Lineweaver-Burk plot of enzyme kinetics is observed as a

A. Decrease in the slope, with Vmax unchanged
B. Change only in the intercept on the substrate axis, with Vmax unchanged
C. Decrease in Vmax, with the Km value remaining unchanged
D. Increase in Vmax, with a decrease in Km

On a double-reciprocal plot, a non-competitive inhibitor produces a line that intersects the control line at the x-axis (Km is unchanged), but has a steeper slope and a higher y-intercept (Vmax is decreased). It reduces the number of functional enzyme molecules.

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Jun 27, 2026

In competitive inhibition, the apparent Km (Michaelis constant) of the enzyme for its substrate is

A. Unchanged
B. Decreased
C. Increased
D. Equal to Vmax

A competitive inhibitor competes for the active site, effectively making it harder for the enzyme to bind its substrate. More substrate is required to reach half the maximum velocity. Therefore, the apparent Km (substrate concentration at 1/2 Vmax) is increased in the presence of a competitive inhibitor.

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Jun 27, 2026

In the process of translation, the specific role of transfer RNA (tRNA) is to

A. Form the structural backbone of the ribosome
B. Provide the template for the sequence of amino acids
C. Act as an adaptor molecule that matches a specific amino acid to its corresponding mRNA codon
D. Catalyze the peptide bond formation between adjacent amino acids

tRNA acts as an adaptor, carrying a specific amino acid at its 3' end and recognizing a specific three-nucleotide codon on the mRNA through its complementary anticodon loop. This bridges the genetic code and the amino acid sequence of a protein.

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Jun 27, 2026

The use of Benedict’s test on a solution of sucrose yields a negative result (no color change) because sucrose

A. Is a monosaccharide, not a disaccharide
B. Has a free aldehyde group that is locked in the furanose ring
C. Lacks a free anomeric carbon capable of reducing Cu²⁺, as both are involved in the glycosidic bond
D. Is a non-reducing sugar that can only be detected by the iodine test

The glycosidic bond in sucrose is formed between the anomeric carbon (C1) of glucose and the anomeric carbon (C2) of fructose. Since neither carbonyl group is free to open into an aldehyde or ketone form, sucrose cannot reduce Cu²⁺ and is thus a non-reducing sugar.

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Jun 27, 2026
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