Quiz: BIO131 Nucleic Acids and Molecular Biology — 35 questions

1. Which combination accurately describes the BIO131 teaching schedule?

A. Twelve lectures, ten three-hour tutorials, and three four-hour practical sessions

B. Fourteen lectures, nine three-hour tutorials, and two four-hour practical sessions

C. Fourteen lectures, eight two-hour tutorials, and two three-hour practical sessions

D. Sixteen lectures, nine two-hour tutorials, and four four-hour practical sessions

Answer: (B)

Explanation

BIO131 includes 14 lectures, nine three-hour tutorial sessions, and two four-hour practical sessions focused on nucleic-acid and protein electrophoresis. The other schedules alter one or more of these specified components.

2. What is the consequence of an unjustified absence from a BIO131 session?

A. It affects attendance records without changing validation decisions

B. It prevents validation of the module, semester, and academic year

C. It reduces the practical-session mark but permits module validation

D. It requires completion of an additional tutorial before validation

Answer: (B)

Explanation

An unjustified absence prevents validation of the module, semester, and year. A justified absence can instead be documented administratively, so the two types of absence have different consequences.

3. A student earns full credit on the practical sessions, the week-43 test, and the week-51 examination. What proportion of the final BIO131 grade comes from the examination?

A. 20%

B. 30%

C. 50%

D. 70%

Answer: (C)

Explanation

The week-51 two-hour final examination contributes 50% of the final module grade. The practical sessions contribute 30% and the week-43 multiple-choice test contributes 20%, so neither represents the examination weighting.

4. What did J. F. Miescher discover in 1869?

A. DNA as the hereditary material through bacterial transformation

B. RNA as the molecule that directs protein translation

C. The DNA double helix formed by two antiparallel strands

D. Nuclein composed of DNA and associated proteins from cell nuclei

Answer: (D)

Explanation

In 1869, J. F. Miescher extracted nuclein, consisting of DNA and associated proteins, from lymph-node cell nuclei. The double helix and DNA’s genetic role were established in later discoveries.

5. Which researchers established in 1952 that DNA carries genetic information?

A. Watson, Crick, and Wilkins

B. Avery, Hershey, and Chase

C. Watson, Hershey, and Miescher

D. Miescher, Franklin, and Crick

Answer: (B)

Explanation

Avery, Hershey, and Chase established in 1952 that DNA carries genetic information. Watson, Crick, and Wilkins are associated with describing the DNA double helix in 1953.

6. Which achievement is associated with Watson, Crick, Wilkins, and Rosalind Franklin’s work in 1953?

A. Formulation of the RNA-to-protein translation process

B. Discovery of nuclein in lymph-node cell nuclei

C. Demonstration that DNA carries genetic information

D. Description of the DNA double-helix structure

Answer: (D)

Explanation

Their work described the DNA double helix in 1953, with Franklin’s contribution being important to that achievement. The discovery of nuclein and the demonstration of DNA’s genetic role refer to earlier milestones.

7. Which sequence correctly applies the central dogma of molecular biology?

A. RNA is translated into DNA, and protein directs RNA transcription

B. Protein is transcribed into DNA, and DNA is translated into RNA

C. DNA is translated into RNA, and RNA is replicated into protein

D. DNA can replicate and be transcribed into RNA, and RNA can be translated into protein

Answer: (D)

Explanation

The central dogma describes DNA replication, DNA-to-RNA transcription, and RNA-to-protein translation, with no transfer from protein to nucleic acid or from one protein to another. Translation therefore does not describe DNA becoming RNA; that process is transcription.

8. Which set of components defines a nucleotide?

A. A pentose sugar, a nitrogenous base, and one or more phosphate groups

B. A pentose sugar, a phosphate group, and a peptide bond

C. A nitrogenous base, a phosphate group, and an amino acid

D. A pentose sugar and a nitrogenous base joined by an N-glycosidic bond

Answer: (A)

Explanation

A nucleotide contains a pentose sugar, a nitrogenous base, and one or more phosphate groups. A sugar linked to a base without phosphate is a nucleoside, not a nucleotide.

9. What distinguishes a nucleoside from a nucleotide?

A. A nucleoside has a base and amino acid, whereas a nucleotide has a sugar

B. A nucleoside has phosphate and base, whereas a nucleotide lacks phosphate

C. A nucleoside contains two sugars, whereas a nucleotide contains one sugar

D. A nucleoside has a sugar and base, whereas a nucleotide also has phosphate

Answer: (D)

Explanation

A nucleoside consists of a pentose sugar covalently linked to a nitrogenous base through an N-glycosidic bond and lacks phosphate. Adding one or more phosphate groups produces a nucleotide.

10. Which base set correctly distinguishes DNA from RNA?

A. DNA contains thymine, whereas RNA contains uracil

B. DNA contains ribose, whereas RNA contains deoxyribose

C. DNA contains uracil, whereas RNA contains thymine

D. DNA contains adenine and guanine, whereas RNA lacks both

Answer: (A)

Explanation

Both DNA and RNA contain adenine, guanine, and cytosine, but DNA contains thymine while RNA contains uracil. The sugar difference is not a base difference, and RNA retains adenine and guanine.

11. During DNA synthesis, how are strand direction and nucleotide addition related?

A. The strand extends from 3′ to 5′ by adding nucleotides to a 5′ hydroxyl end

B. The strand extends from 5′ to 3′ by adding nucleotides to a 3′ hydroxyl end

C. The strand extends from 5′ to 3′ by adding nucleotides to a 5′ hydroxyl end

D. The strand extends from 3′ to 5′ by adding nucleotides to a 3′ hydroxyl end

Answer: (B)

Explanation

DNA polymerization proceeds in the 5′-to-3′ direction because each incoming nucleotide is added to a 3′ hydroxyl end. The antiparallel arrangement of DNA strands does not reverse the direction of polymerization on an individual growing strand.

12. Which comparison between human and Escherichia coli genomes is accurate?

A. Humans have about 3.3 × 10⁹ base pairs, whereas Escherichia coli has about 4 × 10⁶

B. Humans and Escherichia coli each have one circular chromosome of similar size

C. Humans have about 4 × 10⁶ base pairs, whereas Escherichia coli has about 3.3 × 10⁹

D. Humans and Escherichia coli each have about 3.3 × 10⁹ base pairs

Answer: (A)

Explanation

Human cells contain approximately 3.3 × 10⁹ base pairs, while Escherichia coli has approximately 4 × 10⁶ base pairs and 4,288 genes. The comparison involving reversed genome sizes is incorrect because the bacterial genome is much smaller.

13. Which description correctly distinguishes nuclear and organellar DNA in eukaryotic cells?

A. Nuclear DNA is circular, while mitochondrial DNA is linear

B. Nuclear and mitochondrial DNA are both organized as circular chromosomes

C. Nuclear and mitochondrial DNA are both organized as linear chromosomes

D. Nuclear DNA is linear, while mitochondrial DNA is circular

Answer: (D)

Explanation

Animal eukaryotic nuclear DNA forms linear chromosomes, whereas mitochondrial DNA is circular; plant chloroplast DNA is also circular. Treating nuclear DNA as circular confuses it with organellar DNA.

14. What is the first level of compaction formed when eukaryotic DNA wraps around histone proteins?

A. A chromosome containing several fully condensed chromatin fibers

B. A replication fork containing two newly copied DNA strands

C. A nucleosome containing DNA around a histone octamer

D. An RNA–protein complex containing translated messenger RNA

Answer: (C)

Explanation

A nucleosome forms when 146 base pairs of DNA wrap around a histone octamer, initiating DNA compaction. A chromosome is a higher-order condensed structure rather than the first compaction level.

15. Which RNA category accounts for approximately 80% of cytoplasmic RNA?

A. Mitochondrial DNA

B. Ribosomal RNA

C. Messenger RNA

D. Transfer RNA

Answer: (B)

Explanation

Ribosomal RNA represents approximately 80% of cytoplasmic RNA, while messenger RNA represents about 5%. Transfer RNA is another major category but is not the predominant one by this proportion.

16. What does the DNA melting temperature, Tm, represent?

A. The temperature at which every DNA molecule has become single stranded

B. The temperature at which half of the DNA molecules are single stranded

C. The temperature at which DNA reaches its maximum absorbance at 280 nm

D. The temperature at which half of the DNA molecules are fully replicated

Answer: (B)

Explanation

Tm is defined as the temperature at which 50% of DNA molecules are single stranded, marking the midpoint of denaturation. It does not indicate complete conversion of every DNA molecule to single strands.

17. According to the Beer–Lambert law, how does absorbance change when concentration increases while path length and molar absorptivity remain constant?

A. Absorbance increases in direct proportion to concentration

B. Absorbance remains unchanged as concentration increases

C. Absorbance decreases in direct proportion to concentration

D. Absorbance increases in inverse proportion to concentration

Answer: (A)

Explanation

The Beer–Lambert law is $A_\lambda = \varepsilon_\lambda l c$, so absorbance is directly proportional to concentration when the other terms are fixed. The inverse relationship applies to transmittance rather than absorbance.

18. Which measurement pairing is used for nucleic-acid quantification and protein-contamination assessment?

A. Absorbance at 280 nm and the A260/A230 ratio

B. Absorbance at 600 nm and the A230/A260 ratio

C. Absorbance at 260 nm and the A260/A280 ratio

D. Absorbance at 230 nm and the A280/A260 ratio

Answer: (C)

Explanation

Nucleic acids are quantified using absorbance at 260 nm, and the A260/A280 ratio helps assess protein contamination. Absorbance at 280 nm is associated more strongly with proteins than with the primary nucleic-acid measurement.

19. A DNA sample has an A260/A280 ratio of 1.6; what does this result most strongly suggest?

A. A pure DNA preparation

B. Protein contamination

C. Complete DNA denaturation

D. RNA contamination

Answer: (B)

Explanation

A pure DNA solution typically has an A260/A280 ratio of 1.8–2, whereas a value below 1.7 indicates protein contamination. RNA contamination is instead associated with a ratio above 2.

20. What is produced when DNA replication proceeds bidirectionally from one origin of replication?

A. Two replication forks with the complete parental duplex preserved intact

B. One replication fork with both parental strands in one daughter duplex

C. One replication fork with newly synthesized strands copied without templates

D. Two replication forks with one parental strand in each daughter duplex

Answer: (D)

Explanation

Semiconservative replication proceeds bidirectionally from one origin and produces two replication forks, with each daughter duplex retaining one parental strand. Preserving the entire parental duplex would describe conservative rather than semiconservative replication.

21. Why must a DNA polymerase extend an existing primer during DNA synthesis?

A. It can join Okazaki fragments but cannot add nucleotides to a growing strand

B. It can begin synthesis without a primer but cannot extend an RNA-containing template

C. It can add nucleotides to a primer’s 3′ hydroxyl end but cannot initiate a strand independently

D. It can synthesize DNA from a template’s 5′ end but cannot recognize complementary bases

Answer: (C)

Explanation

DNA polymerase requires a primer-bound template and adds nucleotides to the primer’s 3′ hydroxyl end, producing growth in the 5′ to 3′ direction. Its inability to initiate a strand without a primer is the key requirement.

22. How are Okazaki fragments on the lagging strand processed during DNA replication?

A. They begin with short RNA primers and are joined by DNA ligase

B. They form as continuous DNA segments and are separated by helicase

C. They begin with DNA primers and are joined by RNA polymerase

D. They begin at protein primers and are joined by ribosomal RNA

Answer: (A)

Explanation

Lagging-strand synthesis produces 150–200-nucleotide Okazaki fragments, each initiated by a short RNA primer and later joined by DNA ligase. Helicase separates the strands but does not join the fragments.

23. In an ideal PCR reaction, how many copies of the initial DNA duplex are theoretically present after 20 cycles?

A. Approximately $2 \times 20$, or about $40$ copies

B. Approximately $10^{20}$, or about $10^{20}$ copies

C. Approximately $2^{20}$, or about $10^6$ copies

D. Approximately $20^2$, or about $400$ copies

Answer: (C)

Explanation

PCR theoretically doubles the target DNA during each cycle, so 20 cycles produce $2^{20}$ copies, approximately $10^6$. The reaction uses repeated thermal cycles, two primers, and thermostable Taq polymerase to achieve this amplification.

24. During transcription, which DNA strand is copied to produce an RNA sequence that matches the coding strand except that uracil replaces thymine?

A. The coding strand, producing RNA in the 5′ to 3′ direction

B. The template strand, producing RNA in the 3′ to 5′ direction

C. The coding strand, producing RNA in the 3′ to 5′ direction

D. The template strand, producing RNA in the 5′ to 3′ direction

Answer: (D)

Explanation

RNA polymerase reads the template strand and synthesizes RNA in the 5′ to 3′ direction, so the RNA resembles the coding strand with U replacing T. The coding strand is not the strand directly copied during transcription.

25. Which pairing correctly identifies the RNA polymerase used for each eukaryotic gene type?

A. Polymerase I for rRNA, II for mRNA, and III for tRNA

B. Polymerase I for tRNA, II for rRNA, and III for mRNA

C. Polymerase I for rRNA, II for tRNA, and III for mRNA

D. Polymerase I for mRNA, II for tRNA, and III for rRNA

Answer: (A)

Explanation

Eukaryotic RNA polymerase I transcribes rRNA genes, polymerase II transcribes mRNA genes, and polymerase III transcribes tRNA genes. A common confusion is assigning tRNA transcription to polymerase II rather than polymerase III.

26. Which set of processing events converts a eukaryotic pre-mRNA into a stabilized mature mRNA?

A. Adding a 3′ cap, a poly(U) tail, and removing exons by splicing

B. Adding a 5′ cap, a poly(A) tail, and removing introns by splicing

C. Adding a promoter, a poly(A) tail, and replacing introns with ribosomes

D. Adding a 5′ cap, a poly(A) tail, and removing exons by replication

Answer: (B)

Explanation

Eukaryotic pre-mRNA is stabilized by a 5′ cap, an approximately 200-adenine poly(A) tail, and splicing of introns when present. Exons are retained in the mature transcript, so removing them would not describe normal mRNA processing.

27. Which sequence correctly describes the major stages of translation?

A. Elongation at AUG, initiation by DNA replication, and termination at a promoter

B. Initiation at AUG, elongation by amino-acid addition, and termination at a stop codon

C. Initiation at a stop codon, elongation by intron removal, and termination at AUG

D. Initiation at a promoter, elongation by RNA synthesis, and termination at a poly(A) tail

Answer: (B)

Explanation

Translation begins when the ribosome initiates at an AUG codon, continues through codon reading and amino-acid addition, and ends at a stop codon. Promoters, introns, and poly(A) tails are associated with transcription or RNA processing rather than these translation stages.

28. A researcher wants a collection representing every region of an organism’s genome, including regions that are not transcribed. Which library is appropriate?

A. An mRNA library containing mature ribosomal components

B. A cDNA library containing expressed transcripts

C. A protein library containing translated products

D. A genomic library containing the entire genome

Answer: (D)

Explanation

A genomic library represents the entire genome and therefore includes non-transcribed regions. A cDNA library is made from fully transcribed mRNAs and represents expressed transcripts rather than the complete genome.

29. Which sequence best describes molecular cloning of a gene?

A. Inserting the gene into a host chromosome without using a cloning vector

B. Replicating the gene in a test tube without introducing it into a host organism

C. Isolating the gene, examining it directly, and omitting vector-based replication

D. Isolating the gene, inserting it into a vector, replicating it in a host, and analyzing it

Answer: (D)

Explanation

Molecular cloning involves isolating a gene, placing it into a cloning vector, replicating it in a host organism, and analyzing the resulting material. Isolating or examining a gene alone does not constitute the complete cloning process.

30. What characterizes a restriction enzyme?

A. It removes nucleotides progressively from one DNA strand end

B. It cuts both DNA strands at or near a specific recognition site

C. It joins separate DNA fragments by forming phosphodiester bonds

D. It copies a DNA region beginning at a replication origin

Answer: (B)

Explanation

A restriction enzyme is a restriction endonuclease that cleaves both DNA strands at or near a specific recognition site. Removing nucleotides from a strand end is the characteristic action of an exonuclease, not an endonuclease.

31. Which sequence correctly describes plasmid cloning?

A. Digesting vector and insert, ligating them, transforming bacteria, selecting, and amplifying the clone

B. Transforming bacteria, digesting the host genome, ligating proteins, and selecting unmodified cells

C. Ligating vector and insert, digesting them, transforming bacteria, selecting, and degrading the clone

D. Amplifying the insert, removing its recognition sites, transforming bacteria, and preventing recombinant growth

Answer: (A)

Explanation

Plasmid cloning proceeds through restriction digestion, ligation, bacterial transformation, selection, and amplification of the recombinant clone. Restriction enzymes cut DNA, whereas DNA ligase joins compatible DNA ends; reversing these roles would disrupt the procedure.

32. Which description best identifies a plasmid used as a cloning vector?

A. A circular protein complex that carries inserted DNA between bacterial cells

B. A small circular, double-stranded DNA molecule that replicates autonomously outside the bacterial chromosome

C. A short RNA molecule that directs protein synthesis inside the bacterial cytoplasm

D. A large linear, single-stranded DNA molecule integrated permanently into the bacterial chromosome

Answer: (B)

Explanation

A plasmid is a small circular, double-stranded, autonomous, extrachromosomal bacterial DNA molecule that can carry cloned DNA. It is not an RNA molecule, protein complex, or necessarily an integrated part of the bacterial chromosome.

33. What primarily determines how DNA fragments migrate during agarose-gel electrophoresis?

A. The fluorescent label attached to each fragment

B. The length of each DNA fragment

C. The concentration of probe used for detection

D. The nucleotide sequence of each fragment

Answer: (B)

Explanation

Agarose-gel electrophoresis separates DNA fragments primarily by size, so fragment length determines their migration pattern. Nucleotide sequence becomes relevant in sequence-specific detection methods such as hybridization, not as the main variable in ordinary electrophoretic separation.

34. A researcher wants to identify which separated DNA fragment contains a particular complementary sequence; which method is most appropriate?

A. Southern blotting with a labeled probe

B. Molecular-weight calibration with a DNA ladder

C. Agarose electrophoresis without further detection

D. Fluorescence measurement from chain-terminating nucleotides

Answer: (A)

Explanation

Southern blotting transfers separated DNA to a membrane and uses a labeled complementary probe to detect the target sequence. Electrophoresis alone separates fragments by size but does not provide sequence-specific detection.

35. How does Sanger sequencing verify the identity of a cloned insert?

A. By hybridizing a labeled probe to membrane-bound DNA

B. By detecting fluorescence from incorporated chain-terminating nucleotides

C. By measuring the migration distance of fragments through agarose

D. By comparing the insert with a molecular-weight marker

Answer: (B)

Explanation

Sanger sequencing verifies a cloned insert by detecting fluorescence signals generated when chain-terminating nucleotides are incorporated. Southern blotting instead identifies complementary sequences by probe hybridization, while electrophoresis primarily separates fragments by size.