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BI10.1-7 | Molecular Biology — Gate Quiz

Graded 10 questions · 20 min · 3 attempts

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Q1 BI10.1 1 pt

In the Watson-Crick model of DNA, which base pairing rules apply?

A Adenine pairs with Cytosine; Guanine pairs with Thymine
B Adenine pairs with Guanine; Cytosine pairs with Thymine
C Adenine pairs with Thymine; Guanine pairs with Cytosine
D Adenine pairs with Uracil; Guanine pairs with Cytosine

Correct! In DNA: A pairs with T (2 hydrogen bonds) and G pairs with C (3 hydrogen bonds). This is Chargaff's rule: A=T and G=C in double-stranded DNA.

Chargaff's rules: [A]=[T] and [G]=[C] in dsDNA. G-C pairs have 3 hydrogen bonds (stronger), A-T have 2. Higher G-C content = higher melting temperature. In RNA: Uracil replaces Thymine, so A pairs with U (in RNA-DNA hybrids or RNA structure).

Incorrect. In DNA: A-T (2 H-bonds) and G-C (3 H-bonds). Chargaff's rule: %A = %T, %G = %C.

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Q2 BI10.2 1 pt

DNA replication is described as semiconservative. This means:

A Each new DNA molecule has two newly synthesized strands
B Each new DNA molecule has one original (parental) strand and one newly synthesized strand
C The original DNA is completely degraded and two new molecules are synthesized
D DNA replication proceeds only at one replication fork per chromosome

Correct! Semiconservative replication means each daughter DNA molecule retains one parental strand and has one newly synthesized strand. Proven by Meselson-Stahl experiment using 15N/14N density labelling.

Three models of replication: Conservative (both new strands together, original intact) — disproven. Semiconservative (one old + one new per daughter) — proven correct. Dispersive (fragments of old and new interspersed) — disproven. The Meselson-Stahl (1958) experiment using density gradient centrifugation proved the semiconservative model.

Incorrect. Semiconservative = one old strand + one new strand per daughter molecule.

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Q3 BI10.2 1 pt

DNA polymerase can only synthesize DNA in which direction?

A 3' to 5'
B 5' to 3'
C Either direction with equal efficiency
D Randomly, determined by template strand orientation

Correct! DNA polymerase can only add nucleotides to the 3'-OH end of the growing chain, so synthesis proceeds 5' to 3'. This requires one strand (lagging strand) to be synthesized discontinuously as Okazaki fragments.

Because polymerase works 5' to 3', the two strands are synthesized differently: Leading strand = continuous 5' to 3'. Lagging strand = discontinuous Okazaki fragments (each requiring an RNA primer). DNA ligase joins Okazaki fragments. Telomerase adds telomeric repeats using built-in RNA template.

Incorrect. DNA polymerase synthesizes ONLY in the 5' to 3' direction.

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Q4 BI10.3 1 pt

In eukaryotes, which RNA polymerase is responsible for synthesizing mRNA (messenger RNA)?

A RNA polymerase I
B RNA polymerase II
C RNA polymerase III
D All three polymerases equally

Correct! RNA polymerase II synthesizes pre-mRNA (heterogeneous nuclear RNA) which is processed to mature mRNA. It is the target of many transcription factors and the site of action of alpha-amanitin (mushroom toxin).

Eukaryotic RNA polymerases: Pol I (nucleolus): rRNA (28S, 18S, 5.8S). Pol II (nucleoplasm): pre-mRNA, snRNA. Pol III: tRNA, 5S rRNA, small RNAs. Bacterial RNA polymerase is a single enzyme (target of rifampicin). Alpha-amanitin (Amanita phalloides mushroom) specifically inhibits Pol II.

Incorrect. RNA polymerase II synthesizes pre-mRNA/mRNA in eukaryotes.

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Q5 BI10.4 1 pt

A eukaryotic pre-mRNA undergoes 5'-capping, 3'-polyadenylation, and splicing to form mature mRNA. The primary function of the 5' cap (7-methylguanosine) is:

A Providing an anchor for the ribosome and protecting against 5' exonucleases
B Signalling the stop codon for translation termination
C Providing a poly-A binding site for mRNA stability
D Directing the mRNA to the nucleus for processing

Correct! The 5' cap (m7G) protects the mRNA from 5' exonucleases and is recognized by eIF4E (translation initiation factor) for ribosome binding, facilitating translation initiation.

mRNA processing in eukaryotes: (1) 5' capping with 7-methylguanosine — protects and initiates translation. (2) 3' polyadenylation (poly-A tail, ~200 A residues) — protects from 3' exonucleases, aids nuclear export, enhances translation. (3) Splicing (removal of introns by spliceosomes). Prokaryotes lack these modifications.

Incorrect. The 5' cap protects against exonucleases and aids ribosome binding for translation initiation.

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Q6 BI10.5 1 pt

The genetic code is described as 'degenerate.' What does this mean?

A Some codons do not code for any amino acid
B One amino acid can be coded by multiple codons
C Multiple amino acids can be coded by a single codon
D The code differs between species

Correct! Degeneracy (redundancy) means multiple codons can specify the same amino acid. There are 64 codons (4^3) for only 20 amino acids + 3 stop codons. Most amino acids have 2-6 codons.

Properties of the genetic code: Universal (same in all organisms with few exceptions), Non-overlapping (each nucleotide belongs to only one codon), Commaless (no spacers between codons), Degenerate/redundant (multiple codons per amino acid), Unambiguous (each codon specifies only one amino acid). Start codon: AUG (methionine). Stop codons: UAA, UAG, UGA.

Incorrect. Degeneracy means one amino acid can be encoded by multiple codons.

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Q7 BI10.6 1 pt

During translation elongation, which site on the ribosome holds the growing polypeptide chain attached to the tRNA?

A A site (aminoacyl site)
B P site (peptidyl site)
C E site (exit site)
D mRNA binding groove

Correct! The P (peptidyl) site holds the peptidyl-tRNA (tRNA with the growing polypeptide chain). The A site accepts the new aminoacyl-tRNA. Peptidyl transferase transfers the polypeptide to the amino acid at the A site.

Ribosome sites: A site (Aminoacyl) — incoming aminoacyl-tRNA binds. P site (Peptidyl) — tRNA bearing the growing polypeptide. E site (Exit) — deacylated tRNA exits. Peptide bond formation transfers polypeptide from P-site tRNA to A-site amino acid. Translocation moves tRNA from A to P to E.

Incorrect. The P (peptidyl) site holds the tRNA carrying the growing polypeptide.

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Q8 BI10.7 1 pt

The polymerase chain reaction (PCR) amplifies a specific DNA sequence in vitro. Which enzyme is used in PCR and why must it be thermostable?

A E. coli DNA polymerase I; because it is highly accurate
B Taq polymerase; because PCR uses high temperatures (95°C) for denaturation that would destroy non-thermostable enzymes
C RNA polymerase; because PCR uses RNA primers
D Reverse transcriptase; because PCR requires RNA templates

Correct! Taq polymerase (from Thermus aquaticus, a thermophilic bacterium) is thermostable, surviving the 94-95°C denaturation step in each PCR cycle that would inactivate ordinary DNA polymerases.

PCR steps: Denaturation (94-95°C, separates strands), Annealing (50-65°C, primers bind), Extension (72°C, Taq elongates). Each cycle doubles the target DNA. After 30 cycles, approximately 10^9-fold amplification. RT-PCR: reverse transcriptase converts RNA to cDNA first, then PCR amplifies. Used in COVID-19 diagnosis.

Incorrect. Taq polymerase from Thermus aquaticus is used because PCR denaturation steps at 95°C would inactivate mesophilic DNA polymerases.

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Q9 BI10.7 1 pt

Restriction endonucleases are essential tools in recombinant DNA technology. They recognize specific DNA sequences called:

A Promoter sequences
B Palindromic sequences
C Enhancer sequences
D Operator sequences

Correct! Restriction endonucleases recognize specific palindromic DNA sequences (sequences that read the same on both strands in 5' to 3' direction) and cut at or near these sites.

Palindromic sequences: e.g., EcoRI recognizes GAATTC (complement read 5' to 3' is also GAATTC). Type II restriction enzymes cut within or near the recognition site, producing blunt or sticky ends. Sticky ends (cohesive ends) facilitate directional cloning into vectors. Used in Southern blotting, gene cloning, and RFLP analysis.

Incorrect. Restriction enzymes recognize palindromic sequences in double-stranded DNA.

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Q10 BI10.3 1 pt

A single base substitution in a codon changes CGA (Arg) to TGA (stop codon). This type of mutation is called:

A Missense mutation
B Silent (synonymous) mutation
C Nonsense mutation
D Frameshift mutation

Correct! A nonsense mutation converts an amino acid codon to a stop codon (UAA, UAG, UGA), causing premature termination of translation and usually a non-functional truncated protein.

Point mutation types: Silent/synonymous — codon change does not alter amino acid (degeneracy). Missense — codon change alters amino acid (e.g., sickle cell: GAG to GTG = Glu to Val). Nonsense — codon change creates a stop codon (premature termination). Frameshift — insertion/deletion shifts reading frame (usually devastating). Nonsense mutations in collagen, Duchenne muscular dystrophy, etc.

Incorrect. A substitution that creates a premature stop codon is a nonsense mutation.

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