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Structure of DNA

98 questions found

Practice Questions

21. The primary reason why a purine base cannot pair with another purine base inside a stable DNA double helix is that such an orientation would

A. Break the phosphodiester bonds
B. Cause a local distortion that widens the helix beyond 2 nanometers
C. Prevent the formation of any hydrogen bonds
D. Force the DNA to become a single-stranded RNA

Purines are double-ringed; pairing two purines would exceed the 2 nm diameter, while pairing two pyrimidines would make the helix too narrow.

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23. The average distance separating two adjacent, vertically stacked nucleotide base pairs in the B-DNA model is

A. 3.4 nanometers
B. 0.34 nanometers
C. 2.0 nanometers
D. 0.11 nanometers

With 10 base pairs per complete helical turn of 3.4 nm, the step distance between consecutive base pairs is 3.4/10=0.34 nm.

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24. An increase in the collective ratio of G-C base pairs relative to A-T base pairs within a DNA fragment results in

A. A lower melting temperature (Tm​) due to weak bonds
B. A higher melting temperature (Tm​) due to triple hydrogen bonding
C. Spontaneous conversion of the helix into a linear RNA molecule
D. The immediate exclusion of all histone proteins

G-C pairs are linked by three hydrogen bonds, which require more thermal energy to disrupt than the two hydrogen bonds holding A-T pairs together.

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Introns are intervening, non-translated sequences within a gene that are transcribed into pre-mRNA but removed by splicing.

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Transcription generates an RNA transcript that is complementary and anti-parallel to the DNA template strand, swapping thymine for uracil.

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10. In a double-stranded DNA molecule, the base-pairing pattern always obeys Chargaff’s rules, which state that the total amount of

A. Adenine equals cytosine
B. Purines equals pyrimidines
C. Thymine equals guanine
D. Uracil equals thymine

Chargaff’s rules dictate that [A]=[T] and [G]=[C]; therefore, the sum of purines (A+G) must equal the sum of pyrimidines (T+C).

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27. The presence of a hydroxyl group (-OH) exclusively at the 3′ carbon of the deoxyribose sugar is functionally vital during DNA replication because it

A. Acts as the mandatory nucleophile required to attach the next incoming nucleotide
B. Binds directly to the nitrogenous base of the opposing strand
C. Stabilizes the major groove of the double helix
D. Triggers the destruction of the template strand

DNA polymerases require a free 3'-OH group to form a phosphodiester bond with the 5' phosphate group of an incoming dNTP.

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11. The vertical distance covered by one complete turn of the standard B-DNA double helix is

A. 0.34 nanometers
B. 2.0 nanometers
C. 3.4 nanometers
D. 20 nanometers

One complete helical turn spans a longitudinal length of 3.4 nm (34 A˚) and contains approximately 10 base pairs.

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28. The molecular feature that prevents the human gene for insulin from being translated accurately inside a bacterial cell without prior modification is the

A. Difference in the universal genetic code
B. Presence of introns within the human genomic DNA sequence
C. Inability of bacteria to form peptide bonds
D. Complete absence of ribosomes in prokaryotes

Human genes contain introns, which bacteria cannot slice out because they lack spliceosomes, leading to a flawed, non-functional protein.

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A codon is a triplet of bases that specifies a particular amino acid or a stop signal during protein translation.

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13. The nitrogenous bases that possess a single-ring structure and are classified as pyrimidines in DNA are

A. Adenine and Guanine
B. Cytosine and Thymine
C. Thymine and Uracil
D. Adenine and Cytosine

Cytosine and thymine are single-ring pyrimidines, whereas adenine and guanine are double-ring purines.

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14. The functional product coded by a structural gene during the process of gene expression is directly a

A. Polopolysaccharide chain
B. Polypeptide chain
C. Phospholipid bilayer
D. Steroid hormone molecule

Structural genes carry the specific nucleotide blueprints required to assemble amino acids into a polypeptide chain on a ribosome.

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15. The chemical group located specifically at the 3′ terminus of a functional DNA strand is a

A. Phosphate group
B. Free hydroxyl group (-OH)
C. Methyl group
D. Ketone group

The 3' end of a nucleic acid strand terminates at the third carbon of the sugar ring, which carries a free hydroxyl group.

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16. The dynamic force that stabilizes the double helix by stacking the hydrophobic nitrogenous bases on top of one another belongs to

A. Covalent interactions
B. Hydrophobic interactions and van der Waals forces
C. Electrostatic ionic bridges
D. Disulfide linkages

While hydrogen bonds hold the pairs together horizontally, vertical base-stacking interactions insulate the hydrophobic bases from water.

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1. According to the structural model proposed by Watson and Crick, the two strands of a DNA double helix run

A. Parallel to each other
B. Anti-parallel to each other
C. Perpendicular to each other
D. Radially outward from a core

One strand runs in the 5' to 3' direction while its complementary partner runs in the 3' to 5' direction.

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Guanine pairs with cytosine via three hydrogen bonds, creating a more thermally stable bond than the adenine-thymine pair.

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A gene is the fundamental structural and functional unit of heredity, consisting of a nucleotide sequence that carries the code for a protein.

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4. The chemical group located specifically at the 5′ terminus of a single DNA strand is a

A. Hydroxyl group
B. Phosphate group
C. Carboxyl group
D. Amino group

The 5' end of a DNA or RNA strand terminates in a phosphate group attached to the 5' carbon of the pentose sugar.

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Adenine (a purine) always pairs with thymine (a pyrimidine) using two hydrogen bonds in standard Watson-Crick base pairing.

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