IB Biology Quiz: Understand Nucleic Acids
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Understand Nucleic AcidsQuestion 1 of 20

The genome of a newly identified virus is analysed. The base composition is found to be 20% A, 35% U, 15% G, and 30% C. What can be deduced about the genetic material of this virus?

It is double-stranded DNA.
It is single-stranded DNA.
It is double-stranded RNA.
It is single-stranded RNA.
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IB Biology Quiz

IB Biology Quiz: Understand Nucleic Acids

Practice Understand Nucleic Acids in IB Biology with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Understand Nucleic Acids, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Biology.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

The genome of a newly identified virus is analysed. The base composition is found to be 20% A, 35% U, 15% G, and 30% C. What can be deduced about the genetic material of this virus?

  1. It is double-stranded DNA.
  2. It is single-stranded DNA.
  3. It is double-stranded RNA.
  4. It is single-stranded RNA. (correct answer)
Explanation: The presence of uracil (U) instead of thymine (T) indicates that the nucleic acid is RNA. For a nucleic acid to be double-stranded, the base pairing rules must apply, meaning the percentage of adenine should equal its complementary base, and the percentage of guanine should equal its complementary base. In this sample, A (20%) does not equal U (35%), and G (15%) does not equal C (30%). Therefore, the molecule must be single-stranded.

Question 2

Which bond must be broken by a helicase enzyme during DNA replication, and which bond is formed by DNA polymerase when adding a new nucleotide?

  1. Broken: Phosphodiester; Formed: Hydrogen
  2. Broken: Hydrogen; Formed: Phosphodiester (correct answer)
  3. Broken: Glycosidic; Formed: Hydrogen
  4. Broken: Hydrogen; Formed: Glycosidic
Explanation: Helicase functions to 'unzip' the DNA double helix by separating the two strands. This requires breaking the hydrogen bonds that exist between the complementary base pairs (A-T and G-C). DNA polymerase then synthesizes a new strand by adding nucleotides. It does this by catalysing the formation of a phosphodiester bond between the 3' hydroxyl group of the growing strand and the 5' phosphate group of the incoming nucleotide.

Question 3

A nucleic acid sample was analysed and found to contain the sugar ribose. Which other feature would be expected for this molecule if it is messenger RNA (mRNA)?

  1. The presence of thymine base pairing with adenine.
  2. A double-helical structure formed by two antiparallel strands.
  3. A base composition where the amount of guanine equals cytosine.
  4. The presence of uracil instead of thymine. (correct answer)
Explanation: The presence of ribose sugar is characteristic of RNA. RNA molecules contain the nitrogenous base uracil (U) in place of thymine (T), which is found in DNA. Messenger RNA (mRNA) is typically single-stranded, so it does not form a double helix and does not follow Chargaff's rules (G≠C and A≠U).

Question 4

A single DNA nucleotide is composed of three chemical groups. Which option correctly identifies the bond joining the nitrogenous base to the pentose sugar and the bond joining that sugar to a phosphate group?

  1. Base-Sugar: Hydrogen bond; Sugar-Phosphate: Phosphodiester bond
  2. Base-Sugar: Phosphoester bond; Sugar-Phosphate: Glycosidic bond
  3. Base-Sugar: Glycosidic bond; Sugar-Phosphate: Phosphoester bond (correct answer)
  4. Base-Sugar: Peptide bond; Sugar-Phosphate: Glycosidic bond
Explanation: Within a single nucleotide, the nitrogenous base is covalently linked to the 1' carbon of the pentose sugar by a glycosidic bond. The phosphate group is linked to the 5' carbon of the same sugar by a phosphoester bond. A phosphodiester bond is the linkage that connects two adjacent nucleotides in a polynucleotide chain (linking the 3' carbon of one sugar to the 5' carbon of the next).

Question 5

The four nitrogenous bases in DNA can be arranged in any order, leading to vast genetic diversity. For a short DNA segment that is 8 base pairs long, how many unique sequences are theoretically possible?

  1. 32
  2. 4 × 8
  3. 8⁴
  4. 4⁸ (correct answer)
Explanation: At each of the 8 positions in the DNA segment, there are 4 possible choices for the base (A, T, C, or G). Since the choice at each position is independent of the others, the total number of possible unique sequences is calculated by multiplying the number of options for each position. This results in 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4, which is expressed as 4⁸.

Question 6

In the Hershey-Chase experiment, bacteriophages were labelled with either radioactive sulfur (³⁵S) or radioactive phosphorus (³²P). What was the key observation that supported the conclusion that DNA, and not protein, is the hereditary material?

  1. The majority of the ³⁵S was found inside the bacterial cells, while the majority of the ³²P remained outside.
  2. The majority of the ³²P was found inside the bacterial cells, while the majority of the ³⁵S remained outside. (correct answer)
  3. Both ³⁵S and ³²P were found in equal amounts inside the bacterial cells, indicating both components are necessary.
  4. Neither ³⁵S nor ³²P entered the bacterial cells, suggesting an unknown component carries the genetic information.
Explanation: The experiment's logic was to label protein and DNA differently. Protein contains sulfur (in cysteine and methionine) but not phosphorus, so it was labelled with ³⁵S. DNA contains phosphorus (in the phosphate groups) but not sulfur, so it was labelled with ³²P. The experiment showed that the bacteriophage injected its genetic material into the bacterium. The results indicated that the radioactive phosphorus (³²P), corresponding to DNA, entered the cells, while the radioactive sulfur (³⁵S), corresponding to the protein coat, remained outside with the phage ghosts. This strongly supported DNA as the genetic material.

Question 7

What is the primary functional consequence of organizing eukaryotic DNA into nucleosomes?

  1. It ensures that the DNA remains in a double-helical structure, preventing denaturation at physiological temperatures.
  2. It provides direct access for RNA polymerase to initiate transcription at any point along the chromosome.
  3. It separates the coding sequences (exons) from non-coding sequences (introns) prior to protein synthesis.
  4. It allows for the compaction of the long DNA molecule into the nucleus and contributes to gene regulation. (correct answer)
Explanation: A nucleosome consists of DNA wrapped around a core of histone proteins. This fundamental unit of chromatin structure allows for the extensive compaction of the very long eukaryotic DNA molecule to fit inside the small volume of the nucleus. Furthermore, the state of the chromatin (whether it is condensed or relaxed) is a key mechanism for regulating gene expression, by controlling the accessibility of genes to transcription factors and RNA polymerase.

Question 8

A segment of double-stranded DNA from a newly discovered bacterium is found to contain 18% thymine. What percentage of the nucleotides in this segment are purines?

  1. 18%
  2. 32%
  3. 36%
  4. 50% (correct answer)
Explanation: According to Chargaff's rules for double-stranded DNA, the percentage of thymine (T) equals the percentage of adenine (A). Therefore, if T = 18%, then A = 18%. The total percentage of A and T is 18% + 18% = 36%. The remaining percentage must be guanine (G) and cytosine (C), so G + C = 100% - 36% = 64%. Since G = C, the percentage of G is 32% and C is 32%. Purines are adenine and guanine. The total percentage of purines is A + G = 18% + 32% = 50%.

Question 9

The structure of the DNA double helix is stabilized by specific pairing between purines and pyrimidines. Which statement correctly explains the structural reason for this pairing rule?

  1. Pairing a two-ring purine with a one-ring pyrimidine maintains a consistent diameter for the double helix. (correct answer)
  2. Purines can only form two hydrogen bonds, while pyrimidines can form three, which dictates the A-T and G-C pairs.
  3. The glycosidic bonds that attach the bases to the sugar backbone are only stable when a purine is opposite a pyrimidine.
  4. Two purines are too small to span the distance between the backbones, while two pyrimidines are too large to fit.
Explanation: Purines (adenine and guanine) have a two-ring structure, while pyrimidines (cytosine and thymine) have a one-ring structure. If two purines paired, the helix would bulge. If two pyrimidines paired, the helix would constrict. Pairing one purine with one pyrimidine ensures that the distance between the two sugar-phosphate backbones remains constant, maintaining the uniform helical structure. Distractor D incorrectly reverses the sizes of purines and pyrimidines.

Question 10

A short, double-stranded DNA molecule is composed of 80 base pairs and contains 25 guanine bases. How many hydrogen bonds are present in this entire molecule?

  1. 160
  2. 185 (correct answer)
  3. 205
  4. 240
Explanation: The molecule has 80 base pairs in total. If there are 25 guanine (G) bases, there must be 25 cytosine (C) bases, forming 25 G-C pairs. The remaining base pairs must be adenine-thymine (A-T) pairs. The number of A-T pairs is 80 total pairs - 25 G-C pairs = 55 A-T pairs. G-C pairs are connected by three hydrogen bonds, and A-T pairs are connected by two. Total hydrogen bonds = (25 G-C pairs × 3 H-bonds/pair) + (55 A-T pairs × 2 H-bonds/pair) = 75 + 110 = 185 hydrogen bonds.

Question 11

The DNA of a thermophilic bacterium living in hot springs is compared to the DNA of E. coli. What difference in nucleic acid composition would provide greater stability at high temperatures?

  1. The thermophile's DNA would likely have a higher proportion of guanine-cytosine (G-C) base pairs. (correct answer)
  2. The thermophile's DNA would likely have a higher proportion of adenine-thymine (A-T) base pairs.
  3. The thermophile's DNA would have more phosphodiester bonds per unit length to strengthen the backbone.
  4. The thermophile would substitute uracil for thymine to increase the number of hydrogen bonds.
Explanation: The stability of the DNA double helix is partly determined by the number of hydrogen bonds between base pairs. Guanine-cytosine (G-C) pairs are held together by three hydrogen bonds, whereas adenine-thymine (A-T) pairs are held together by only two. A higher proportion of G-C pairs results in a more stable molecule with a higher melting (denaturation) temperature, which is an adaptation for life at high temperatures.

Question 12

Which statement represents a valid conclusion that Erwin Chargaff could make from his base composition data alone, before the structure of DNA was proposed by Watson and Crick?

  1. DNA is a double helix with adenine pairing with thymine and guanine pairing with cytosine.
  2. The base composition of DNA varies between species, but within a species, the ratio of A to T and G to C is always close to 1. (correct answer)
  3. DNA is the universal genetic material because the base pairing rules are consistent across all organisms studied.
  4. The sequence of bases in a DNA molecule carries the genetic code for protein synthesis.
Explanation: Chargaff's data showed two key things: (1) the base composition (the percentage of each of the four bases) varies from one species to another, and (2) in the DNA of any given species, the amount of adenine is equal to the amount of thymine (A=T), and the amount of guanine is equal to the amount of cytosine (G=C). The interpretation of this as evidence for a double helix with specific base pairing was made by Watson and Crick. Chargaff's work itself did not propose a structure or prove DNA was the genetic material.

Question 13

Imagine a hypothetical DNA-like molecule where the hydrogen bonds between base pairs are replaced by strong covalent bonds. What would be the most significant consequence for its biological function?

  1. The two strands would be inseparable under biological conditions, preventing replication and transcription. (correct answer)
  2. The overall negative charge of the molecule would be neutralized, affecting its interaction with proteins.
  3. The genetic information stored in the base sequence would be altered and become unreadable.
  4. The molecule would be unable to form a stable double helix due to the rigidity of the new bonds.
Explanation: Essential biological processes like DNA replication and transcription require the two strands of the double helix to be separated (unzipped) by enzymes like helicase. This separation is possible because the hydrogen bonds holding the strands together are relatively weak. If they were replaced by strong covalent bonds, the strands would be permanently linked, making it impossible to access the base sequence information for templating new strands. Therefore, the molecule could not be replicated or transcribed.

Question 14

What is the key structural feature of a transfer RNA (tRNA) molecule that is absent in a typical messenger RNA (mRNA) molecule?

  1. It contains a sequence of three bases, known as a codon, which specifies a particular amino acid.
  2. It is composed of ribonucleotides linked by phosphodiester bonds and contains the base uracil.
  3. Its single polynucleotide chain folds into a specific three-dimensional structure with an anticodon loop. (correct answer)
  4. It is synthesized using a DNA template in the nucleus before being exported to the cytoplasm.
Explanation: While both mRNA and tRNA are single-stranded RNA molecules, tRNA has a characteristic folded structure. The single chain folds back on itself to form a 'cloverleaf' in 2D and a compact 'L' shape in 3D, stabilized by hydrogen bonds. This structure includes several key functional regions, most notably the anticodon loop that pairs with the mRNA codon. mRNA is typically a linear chain without such extensive and specific folding.

Question 15

What property of the sugar-phosphate backbone of DNA is most responsible for its overall net negative charge at physiological pH?

  1. The hydroxyl groups on the 3' carbon of the deoxyribose sugars, which are weakly acidic.
  2. The nitrogen atoms within the purine and pyrimidine rings, which readily accept protons.
  3. The deprotonation of the phosphoric acid groups, leaving them as negatively charged phosphate ions. (correct answer)
  4. The cumulative effect of polar C-O and P-O bonds throughout the backbone structure.
Explanation: The 'phosphate' in the sugar-phosphate backbone originates from phosphoric acid (H₃PO₄). At the neutral pH of the cell (around 7.4), the phosphate groups are deprotonated (lose H⁺ ions), leaving them with a net negative charge (PO₄³⁻ is linked in the chain, resulting in a negative charge on one of the oxygen atoms). This repeats for every nucleotide, giving the entire DNA molecule a strong overall negative charge.

Question 16

If a mutation caused DNA polymerase to insert a pyrimidine opposite a pyrimidine during replication, what would be the most direct structural consequence for the DNA double helix at that location?

  1. The helix would be wider than normal due to the pairing of two large bases.
  2. The helix would be narrower than normal because two small bases would not span the full distance. (correct answer)
  3. The number of hydrogen bonds in that pair would increase, making the helix more stable.
  4. The sugar-phosphate backbone would break, leading to a single-strand nick.
Explanation: Pyrimidines (cytosine and thymine) are single-ring structures and are smaller than purines (adenine and guanine), which are double-ring structures. The constant diameter of the DNA helix is maintained by pairing a larger purine with a smaller pyrimidine. If two smaller pyrimidines were paired opposite each other, they would not be able to span the normal distance between the sugar-phosphate backbones, causing the helix to constrict or become narrower at that point.

Question 17

Which statement correctly compares the structure of DNA and RNA?

  1. DNA is a polymer of deoxyribonucleotides and is typically single-stranded, while RNA is a polymer of ribonucleotides and is double-stranded.
  2. Both DNA and RNA have a sugar-phosphate backbone, but the sugar in DNA has one fewer hydroxyl group than the sugar in RNA. (correct answer)
  3. RNA contains the pyrimidine base cytosine, whereas DNA contains the pyrimidine base uracil in its place.
  4. The phosphodiester bonds in DNA are stronger than those in RNA, making DNA a more permanent storage molecule.
Explanation: Both nucleic acids have a backbone made of alternating sugar and phosphate groups. The key difference is the sugar: DNA contains deoxyribose, while RNA contains ribose. Deoxyribose lacks a hydroxyl (-OH) group at the 2' carbon, having only a hydrogen atom (-H) instead, hence the name 'deoxy'. Ribose has hydroxyl groups at both the 2' and 3' positions. This difference in the sugar makes DNA more stable than RNA. Distractor A incorrectly states the strandedness, and C incorrectly swaps uracil and cytosine's locations.

Question 18

Which chemical groups are found at the free 5' and 3' ends, respectively, of a linear DNA strand?

  1. A hydroxyl group at the 5' end and a phosphate group at the 3' end.
  2. A carboxyl group at the 5' end and an amino group at the 3' end.
  3. A phosphate group at the 5' end and a hydroxyl group at the 3' end. (correct answer)
  4. A nitrogenous base at the 5' end and a deoxyribose sugar at the 3' end.
Explanation: The directionality of a DNA or RNA strand is defined by the numbering of the carbons in the pentose sugar. The 5' end of the strand has a free phosphate group attached to the 5' carbon of the terminal sugar. The 3' end has a free hydroxyl (-OH) group attached to the 3' carbon of the terminal sugar. This 3' hydroxyl group is where the next nucleotide is added during polymerization.

Question 19

Which statement correctly distinguishes the types of covalent bonds that form the sugar-phosphate backbone of a single DNA strand?

  1. Phosphodiester bonds link the 3' carbon of one deoxyribose sugar to the 5' carbon of the next via a phosphate group. (correct answer)
  2. Glycosidic bonds link the phosphate group of one nucleotide to the deoxyribose sugar of the adjacent nucleotide.
  3. Phosphodiester bonds link the 2' carbon of one deoxyribose sugar to the 5' carbon of the next, forming the backbone.
  4. Hydrogen bonds form between the sugar of one nucleotide and the phosphate of the next, creating a strong backbone.
Explanation: The sugar-phosphate backbone is formed by phosphodiester bonds. Specifically, a phosphate group forms a bridge between the 3' carbon of one sugar molecule and the 5' carbon of the next sugar molecule. Glycosidic bonds link the nitrogenous base to the 1' carbon of the sugar. The 2' carbon of deoxyribose has a hydrogen atom, not a hydroxyl group available for bonding in the backbone. Hydrogen bonds form between complementary bases on opposite strands, not within the backbone.

Question 20

The ratio of (A+T)/(G+C) is determined for the DNA of two different organisms. Organism 1 has a ratio of 1.5, and Organism 2 has a ratio of 0.8. Which deduction is most plausible?

  1. The total percentage of purines in Organism 1 is significantly higher than in Organism 2.
  2. The DNA of Organism 2 would require a higher temperature to denature than the DNA of Organism 1. (correct answer)
  3. Organism 1 must be a eukaryote and Organism 2 must be a prokaryote.
  4. The DNA strands of Organism 1 are longer than the DNA strands of Organism 2.
Explanation: A higher (A+T)/(G+C) ratio means a higher proportion of A-T base pairs. A lower ratio means a higher proportion of G-C base pairs. Organism 2 has a lower ratio (0.8) than Organism 1 (1.5), so Organism 2 has a higher G-C content. Since G-C pairs have three hydrogen bonds compared to two in A-T pairs, DNA with higher G-C content is more stable and requires more energy (a higher temperature) to separate the strands (denature). The percentage of purines (A+G) is always 50% in any double-stranded DNA, making distractor A incorrect.