Biochemistry - MCQ Practice Questions

Understanding:

We are asked to identify the correct statement of Chargaff's base equivalence rule for double-stranded DNA.

Step 1: Recall Chargaff's observations

Chargaff analysed the base composition of DNA from multiple species and made two key observations:

Step 2: Evaluate each option

The statement "total purines equal total pyrimidines" is the direct consequence of Watson-Crick base pairing and is universally true for all double-stranded DNA, regardless of species.

The ratio A/G is NOT fixed; it varies between species — this is actually what Chargaff used to distinguish species. A = C is incorrect because adenine pairs with thymine, not cytosine. The A+T to G+C ratio (also called the AT/GC ratio) varies widely between species and is NOT constant.

Answer:

Chargaff's rule states that in double-stranded DNA, total purines always equal total pyrimidines, i.e., A + G = T + C.

Quick Tip:

Chargaff's rules: A = T and G = C (first parity rule) for dsDNA. The A+T/G+C ratio differs between species — that is how Chargaff showed DNA composition is species-specific.

Understanding:

We need to identify which sequence has the highest melting temperature (Tm) based on its GC content.

Step 1: Recall the relationship between Tm and GC content

G-C base pairs are held together by three hydrogen bonds, whereas A-T base pairs are held by only two hydrogen bonds. A higher proportion of G-C pairs means more energy is needed to separate the strands, resulting in a higher Tm.

Step 2: Calculate GC content of each option

Step 3: Determine the highest Tm

The sequence with 100% GC content (5'-GCGCGCGC-3') will have the highest melting temperature because every base pair contributes three hydrogen bonds.

Answer:

The sequence 5'-GCGCGCGC-3' has 100% GC content and therefore the highest melting temperature.

Quick Tip:

A rough formula for short oligonucleotides is $T_m=2(A+T)+4(G+C)$ °C. For the GCGCGCGC 8-mer: $T_m=4\times 8=32°C$ (approximate). For the ATATATAT 8-mer: $T_m=2\times 8=16°C$. The GC-rich sequence always wins.

Understanding:

We are asked to identify the correct chemical differences between RNA and DNA.

Step 1: Identify the sugar component

DNA contains 2'-deoxyribose (lacks a hydroxyl group at the 2' carbon), while RNA contains ribose (has a 2'-OH group). This is a fundamental structural difference.

Step 2: Identify the pyrimidine base difference

DNA contains the pyrimidine thymine (5-methyluracil), while RNA contains uracil (lacks the methyl group at position 5). Both pair with adenine, but their structures differ by a methyl group.

Step 3: Combine both distinguishing features

RNA = ribose sugar + uracil (replacing thymine).

DNA = deoxyribose sugar + thymine (replacing uracil).

Answer:

RNA differs from DNA in that it contains ribose sugar and uracil in place of thymine.

Quick Tip:

Remember: RNA has an extra -OH at the 2' position of ribose — this makes RNA more susceptible to alkaline hydrolysis than DNA, which is why DNA is a more stable genetic storage molecule.

Understanding:

We need to recall the structural parameters of B-form DNA, specifically the rise per base pair (the axial distance between consecutive base pairs).

Formula:

Key structural parameters of B-DNA:

Step 1: Distinguish rise from pitch

The pitch (3.4 nm) is the distance for one complete 360° turn of the helix, which contains 10 base pairs. The rise per base pair is the distance between two successive base pairs:

Step 2: Verify the diameter

The diameter of B-DNA is approximately 2.0 nm — this is a common distractor but refers to the width, not the rise per base pair.

Answer:

The distance between two consecutive base pairs in B-DNA is 0.34 nm.

Quick Tip:

The classic mnemonic: B-DNA pitch = 3.4 nm, 10 bp per turn, rise = 0.34 nm per bp, diameter = 2.0 nm. Do not confuse pitch with rise per base pair — a common exam trap.

Understanding:

We are asked to identify the RNA species with the shortest half-life that also carries genetic information from the nucleus to ribosomes.

Step 1: Review the functions of RNA types

Step 2: Assess stability

Among all RNA classes, mRNA is the most short-lived. In prokaryotes, the average mRNA half-life is only 2–3 minutes. In eukaryotes it ranges from minutes to hours, but it is still far shorter than the very stable rRNA and tRNA molecules. This instability allows cells to rapidly change their gene expression in response to stimuli.

Answer:

mRNA has the shortest half-life among the major RNA classes and serves as the direct carrier of genetic information from the nucleus to ribosomes.

Quick Tip:

Remember the relative stability order: rRNA > tRNA > mRNA. rRNA makes up ~80% of total cellular RNA precisely because it is so stable and abundant.

Understanding:

We are asked about the specific biological role of the 3' to 5' exonuclease activity of DNA polymerase I.

Step 1: Understand the two exonuclease activities of DNA Pol I

DNA Polymerase I (Pol I) in E. coli possesses three enzymatic activities:

1. 5' to 3' polymerase activity — synthesises DNA in the 5'→3' direction.

2. 3' to 5' exonuclease activity — removes nucleotides from the 3' end of the growing chain.

3. 5' to 3' exonuclease activity — removes RNA primers ahead of the enzyme.

Step 2: Assign the 3' to 5' activity

The 3' to 5' exonuclease activity acts in the direction opposite to synthesis. If an incorrect nucleotide is incorporated, the polymerase can pause, excise the mismatched nucleotide from the 3' end using this activity, and then re-synthesise with the correct nucleotide. This is the proofreading (error-correction) function.

Step 3: Assign the 5' to 3' activity

It is the 5' to 3' exonuclease activity (nick translation ability) of Pol I that removes RNA primers and replaces them with DNA — this is a separate and distinct function.

Answer:

The 3' to 5' exonuclease activity of DNA polymerase I is responsible for proofreading — excising incorrectly incorporated nucleotides during DNA replication.

Quick Tip:

A simple rule: the direction of exonuclease activity is always OPPOSITE to the direction of synthesis. 3'→5' exonuclease proofreads (corrects errors at the 3' end). 5'→3' exonuclease removes primers (acts ahead of synthesis).

Understanding:

We need to determine the anticodon sequence of the tRNA that base-pairs with the start codon 5'-AUG-3'.

Step 1: Write the mRNA codon in 5' to 3' direction

The mRNA codon is: $\text{5’-AUG-3’}$

Step 2: Write the complementary antiparallel sequence

The anticodon on tRNA is antiparallel and complementary to the mRNA codon. Base pairing rules (A pairs with U; G pairs with C) applied antiparallel:

Step 3: Express the anticodon in the conventional 5' to 3' direction

Reversing the anticodon to read 5'→3':

So the tRNA anticodon is 5'-CAU-3'.

Answer:

The tRNA anticodon that recognises 5'-AUG-3' is 5'-CAU-3', written in the conventional 5' to 3' direction.

Quick Tip:

Always remember that the anticodon is written 3'→5' when aligned with the 5'→3' mRNA codon. When asked for the anticodon conventionally, reverse it to 5'→3'. The initiator tRNA (Met-tRNA) always carries the anticodon 5'-CAU-3'.

Understanding:

We are asked about the histone composition of the core nucleosome particle.

Step 1: Recall nucleosome structure

Each nucleosome consists of two distinct components:

1. The core particle: ~147 bp of DNA wrapped ~1.65 turns around a histone octamer.

2. The linker: DNA connecting adjacent nucleosomes, associated with histone H1.

Step 2: Identify the histone octamer composition

The histone octamer contains exactly 8 histone proteins:

These four core histones are highly conserved across eukaryotes. They first form an H3-H4 tetramer, and two H2A-H2B dimers then associate on either side.

Step 3: Role of H1

Histone H1 is the linker histone; it binds to the DNA entering and exiting the nucleosome and helps compact chromatin into higher-order structures. It is NOT part of the octamer core.

Answer:

The histone octamer consists of two copies each of H2A, H2B, H3, and H4.

Quick Tip:

A useful mnemonic for core histones: "H2A, H2B, H3, H4 — the four core histones, always in pairs." H1 is the odd one out — it stays outside the octamer as the linker histone.

Understanding:

We are asked to identify the correct chemical description of the 5' cap structure added to eukaryotic pre-mRNA.

Step 1: Understand the capping reaction

Shortly after transcription initiation, the 5' end of the nascent pre-mRNA is modified. The enzyme guanylyltransferase adds a GMP residue to the 5' triphosphate end of the transcript. This creates an unusual 5'-5' triphosphate linkage (not the normal 3'-5' phosphodiester bond found in the RNA chain).

Step 2: Identify the methylation step

The added guanosine is then methylated at its N-7 position by a methyltransferase using S-adenosylmethionine (SAM) as the methyl donor, producing 7-methylguanosine (${\text{m}}^7\text{G}$).

Step 3: Recall the functions of the cap

The ${\text{m}}^7\text{G}$ 5' cap:

Note: The poly-A tail (described in option D) is a 3' modification, not the 5' cap — a common confusion in exams.

Answer:

The 5' cap is a 7-methylguanosine residue attached via an unusual 5' to 5' triphosphate bridge.

Quick Tip:

The 5'-5' linkage is unique — it is the only such bond in the entire mRNA molecule. All other inter-nucleotide bonds in RNA are standard 3'-5' phosphodiester bonds. This unusual linkage also makes the cap resistant to most cellular nucleases.