Biochemistry - MCQ Practice Questions
Biochemistry sits at the point where chemistry stops being abstract and starts describing living systems. Practice covers carbohydrates, proteins and amino acids, lipids, nucleic acids, enzymes and enzyme kinetics, metabolic pathways, and vitamins and coenzymes. Pathway questions include the regulation step in the explanation, because that is usually what the question is really testing rather than the sequence itself.
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Which enzyme is used to synthesize a complementary DNA (cDNA) strand from an mRNA template in recombinant DNA technology?
Understanding:
This question asks which enzyme converts mRNA into cDNA, a key step in constructing cDNA libraries used in genetic engineering.
Step 1: Role of reverse transcriptase
Reverse transcriptase is an RNA-dependent DNA polymerase originally found in retroviruses (e.g., HIV, Moloney Murine Leukemia Virus). It reads an mRNA template in the 3' to 5' direction and synthesizes a complementary DNA strand (first strand cDNA) in the 5' to 3' direction using deoxyribonucleotides.
Step 2: Why the other options are incorrect
DNA polymerase I is a prokaryotic enzyme involved in nick translation and DNA repair; it requires a DNA template. RNA polymerase II transcribes protein-coding genes from a DNA template to produce pre-mRNA; it does not make DNA. Terminal transferase adds homopolymeric tails to the 3' ends of DNA molecules and is used in linker addition, not in mRNA-to-cDNA conversion.
Step 3: Significance in genetic engineering
After reverse transcriptase produces the first-strand cDNA, RNase H degrades the mRNA, and DNA polymerase I synthesises the second strand, yielding double-stranded cDNA that can be cloned into a vector.
Answer:
The enzyme that synthesises cDNA from an mRNA template is reverse transcriptase.
Quick Tip:
cDNA libraries are derived from mRNA and therefore represent only the expressed genes of a particular tissue; they lack introns — a key advantage when expressing eukaryotic genes in prokaryotic hosts.
Restriction endonucleases recognise specific palindromic sequences and cleave DNA. EcoRI recognises the sequence 5'-GAATTC-3'. If EcoRI cuts both strands, which type of ends are generated?
Understanding:
This question asks about the type of DNA ends produced when EcoRI cleaves its recognition sequence.
Step 1: EcoRI cleavage pattern
EcoRI recognises the palindromic sequence:
It cuts between G and A on both strands, but at staggered positions — on the top strand after the G and on the bottom strand after the complementary G.
Step 2: Resulting ends
After cleavage, each fragment carries a 4-nucleotide single-stranded overhang:
The overhang projects from the 5' end (5'-AATT-3' single-stranded tail), making these 5' protruding or 5' sticky ends.
Step 3: Distinction from blunt and 3' ends
Blunt ends arise from enzymes like SmaI that cut at exactly the same position on both strands. 3' protruding ends are generated by enzymes like KpnI. EcoRI's staggered cut leaves a 5' overhang, not a 3' overhang.
Answer:
EcoRI generates 5' protruding (sticky) ends with a 4-nucleotide 5' overhang (5'-AATT).
Quick Tip:
Any restriction enzyme that cuts to the left of the axis of symmetry (closer to the 5' end) generates 5' overhangs; cutting to the right generates 3' overhangs; cutting exactly at the centre gives blunt ends.
In the polymerase chain reaction (PCR), what is the primary purpose of the denaturation step carried out at approximately 94–96°C?
Understanding:
This question concerns the role of the high-temperature denaturation step in the PCR thermal cycling protocol.
Step 1: Basis of denaturation
Double-stranded DNA is held together by hydrogen bonds between complementary base pairs and by base-stacking interactions. Heating to 94–96°C breaks these non-covalent interactions, unwinding the double helix and producing two single-stranded DNA templates.
Step 2: Why single strands are needed
Primers can only anneal to single-stranded DNA. Without strand separation, the primers and Taq polymerase have no accessible template, so amplification cannot occur.
Step 3: Why the other options are incorrect
Primer annealing occurs at 50–65°C — the annealing step, not denaturation. Taq DNA polymerase is already active; it does not require a 94°C activation (unlike hot-start polymerases that have an extended 95°C activation at the very start). New strand synthesis (extension) occurs at 72°C — the extension step.
Answer:
The denaturation step at 94–96°C separates the double-stranded DNA into two single-stranded templates.
Quick Tip:
The three PCR steps — denaturation (~95°C), annealing (~55°C), extension (~72°C) — correspond to the optimal temperatures for strand separation, primer binding, and Taq polymerase activity, respectively.
A plasmid vector used in recombinant DNA technology typically contains which of the following essential elements?
Understanding:
This question asks about the minimum essential features that a plasmid cloning vector must carry to function in recombinant DNA work.
Step 1: Origin of replication (ori)
The ori allows the plasmid to replicate autonomously within the host cell, independent of chromosomal replication. Without it, the plasmid would be lost after cell division.
Step 2: Selectable marker
A selectable marker (commonly an antibiotic resistance gene such as ampicillin or kanamycin resistance) allows researchers to identify and maintain only those cells that have taken up the plasmid. Cells without the plasmid die on selective medium.
Step 3: Multiple cloning site (MCS)
The MCS (polylinker) contains clusters of unique restriction enzyme recognition sequences, providing multiple options for inserting foreign DNA into the plasmid.
Step 4: Why other options are incorrect
Centromeres and telomeres are features of yeast artificial chromosomes (YACs), not simple plasmid vectors. A poly-A signal alone is insufficient and is a eukaryotic mRNA processing element. Cos sites and lambda integrase are features of cosmid or lambda phage vectors, not standard plasmids.
Answer:
A standard plasmid cloning vector must carry an origin of replication, a selectable marker, and a multiple cloning site.
Quick Tip:
pUC19 and pBR322 are classic examples: pBR322 carries ampicillin and tetracycline resistance genes, while pUC19 uses the lacZ-alpha complementation system for blue-white screening in addition to ampicillin resistance.
The Ti plasmid of Agrobacterium tumefaciens is widely used to introduce foreign genes into plant cells. Which region of the Ti plasmid is actually integrated into the plant nuclear genome?
Understanding:
This question asks which specific segment of the Ti plasmid becomes stably incorporated into the host plant chromosome during Agrobacterium-mediated transformation.
Step 1: Structure of the Ti plasmid
The Ti (Tumour-inducing) plasmid of Agrobacterium tumefaciens contains four functionally distinct regions: T-DNA, vir genes, ori (origin of replication), and genes for opine catabolism.
Step 2: Role of the T-DNA
The T-DNA (transferred DNA) is a defined segment flanked by 25-bp direct repeat border sequences. These borders are recognised by the Vir proteins, which nick the T-DNA, coat it as a single-stranded nucleoprotein complex, and transfer it into the plant cell nucleus where it integrates stably into the plant genome.
Step 3: Role of the vir region
The vir genes encode the enzymatic machinery (VirA, VirG, VirD1/D2, VirE2, etc.) that processes and exports the T-DNA. They act in trans but are not themselves transferred or integrated into the plant genome.
Step 4: Practical implication
In disarmed binary vector systems, the oncogenes within T-DNA are deleted and replaced with the gene of interest; only the border sequences are needed for transfer and integration.
Answer:
Only the T-DNA region of the Ti plasmid is integrated into the plant nuclear genome.
Quick Tip:
The border sequences (left and right borders) are the only cis-acting elements absolutely required for T-DNA transfer — the vir proteins act in trans, which is why binary vector systems work.
Southern blotting is a technique used to detect specific DNA sequences. Which of the following correctly describes the order of steps in Southern blotting?
Understanding:
This question asks about the correct sequential steps in Southern blotting, a foundational technique in molecular biology for detecting specific DNA sequences.
Step 1: Restriction digestion and electrophoresis
Genomic DNA is first digested with a restriction endonuclease to produce fragments of varying sizes. These fragments are separated by size using agarose gel electrophoresis, with smaller fragments migrating farther from the wells.
Step 2: Denaturation and transfer to membrane
The gel is treated with NaOH to denature the double-stranded DNA into single strands. The denatured DNA fragments are then transferred (blotted) from the gel onto a nitrocellulose or nylon membrane by capillary action (or vacuum/electric transfer), preserving the size-based pattern.
Step 3: Hybridisation with a labelled probe
The membrane is incubated with a labelled (radioactive or fluorescent) single-stranded DNA or RNA probe complementary to the target sequence. The probe hybridises specifically to the complementary band on the membrane.
Step 4: Detection
Unbound probe is washed away, and the hybridised band is visualised by autoradiography (for radiolabelled probes) or chemiluminescence/fluorescence.
Answer:
The correct order is electrophoresis, followed by transfer to membrane, hybridisation with a labelled probe, and finally detection.
Quick Tip:
A useful mnemonic — Southern = DNA (developed by Edwin Southern); Northern = RNA; Western = Protein. In all three, electrophoresis always precedes transfer.
In genetic engineering, the enzyme DNA ligase is used to join two DNA fragments. Which chemical bond does DNA ligase form to seal nicks in the DNA backbone?
Understanding:
This question asks about the specific type of covalent bond that DNA ligase catalyses to join DNA strands.
Step 1: Structure of the DNA backbone
The backbone of a DNA strand consists of alternating deoxyribose sugars and phosphate groups connected by phosphodiester bonds. Each phosphodiester bond links the 3'-OH of one nucleotide to the 5'-phosphate of the next.
Step 2: Action of DNA ligase
When a nick (a break in one strand of a double-stranded DNA) is present, there is a free 3'-OH group on one side and a 5'-phosphate on the other. DNA ligase catalyses the formation of a new phosphodiester bond between these two groups, using either NAD+ (in prokaryotes) or ATP (in eukaryotes and bacteriophages) as a cofactor. This seals the nick and restores the continuous phosphodiester backbone.
Step 3: Why other bonds are incorrect
Hydrogen bonds hold the two complementary strands together but are non-covalent; ligase does not form them. Glycosidic bonds link nitrogenous bases to deoxyribose sugars — these are formed during nucleotide biosynthesis. Peptide bonds are formed by ribosomes during protein synthesis and have no role in DNA joining.
Answer:
DNA ligase seals nicks in the DNA backbone by forming a phosphodiester bond between the 3'-OH and 5'-phosphate of adjacent nucleotides.
Quick Tip:
DNA ligase can only seal a nick — a break where both ends are held in close proximity by base-pairing with the opposite strand. It cannot join two completely separate DNA molecules unless they share complementary (sticky or blunt) ends.
The CRISPR-Cas9 system achieves site-specific genome editing. Which component of the system is directly responsible for determining the specific genomic location to be cut?
Understanding:
This question asks which component of the CRISPR-Cas9 system confers sequence specificity to direct the complex to a particular genomic locus.
Step 1: Components of the CRISPR-Cas9 system
The system consists of two main components: the Cas9 endonuclease and the single guide RNA (sgRNA). The sgRNA is an engineered fusion of the CRISPR RNA (crRNA) and the trans-activating crRNA (tracrRNA).
Step 2: Role of the sgRNA in targeting
The sgRNA contains a ~20-nucleotide spacer sequence at its 5' end that is designed to be complementary to the target DNA strand. Watson-Crick base pairing between this spacer and the target DNA strand directs the Cas9-sgRNA complex to the specific genomic location. Changing the spacer sequence changes the target location.
Step 3: Role of PAM and Cas9
The PAM sequence (5'-NGG-3' for Streptococcus pyogenes Cas9) is required for Cas9 to unwind and interrogate the DNA, but it is a fixed requirement on the target DNA — it does not change with each experiment. The Cas9 protein provides the nuclease activity (via its HNH and RuvC domains) to cut both strands but cannot choose the location on its own. The RuvC domain specifically cleaves the non-complementary (non-template) strand.
Answer:
The single guide RNA (sgRNA) determines the specific genomic location by complementary base pairing with the target DNA strand.
Quick Tip:
To target a new gene, only the 20-nucleotide spacer sequence of the sgRNA needs to be redesigned — the Cas9 protein and the scaffold of the sgRNA remain unchanged, making CRISPR-Cas9 far easier to programme than earlier tools like zinc-finger nucleases.
A researcher digests a circular plasmid (4,000 bp) with two restriction enzymes, EcoRI and HindIII, separately and together. EcoRI alone gives fragments of 1,500 bp and 2,500 bp. HindIII alone gives fragments of 1,000 bp and 3,000 bp. The double digest gives fragments of 500 bp, 1,000 bp, and 2,500 bp. How many EcoRI and HindIII sites are present in the plasmid, respectively?
Understanding:
This question involves restriction mapping — determining the number and positions of restriction sites from digest fragment data.
Step 1: Count the restriction sites from single digests
A circular plasmid linearised by a restriction enzyme yields as many fragments as there are cut sites. EcoRI gives 2 fragments, so there are 2 EcoRI sites. HindIII gives 2 fragments, so there are 2 HindIII sites.
Step 2: Re-examine using the double digest
The double digest yields 3 fragments (500 + 1,000 + 2,500 = 4,000 bp ✓). With 2 EcoRI + 2 HindIII sites on a circular map, we expect 4 fragments from a double digest (total cuts = 4 on a circle = 4 fragments). But only 3 fragments are obtained, which is inconsistent with 2 sites each.
Step 3: Re-evaluate — 1 EcoRI site and 2 HindIII sites
With 1 EcoRI + 2 HindIII sites, total cuts on the circle = 3, giving 3 fragments. This matches. Let the EcoRI site divide the circle into the 1,500 bp and 2,500 bp arcs. One HindIII site lies in the 2,500 bp arc (splitting it into 2,500 − 1,000 = 1,500 and 1,000 bp portions), and one lies in the 1,500 bp arc (splitting it into 1,500 − 1,000 = 500 and 1,000 bp portions). Double digest fragments: 500 + 1,000 + 2,500 — but we need to map this carefully: the 2,500 arc cut by one HindIII gives 2,500 and... this needs trial. Placing HindIII sites 500 bp and 1,500 bp from the EcoRI site (within respective arcs) gives double-digest fragments of 500, 1,000 (from the 1,500 arc), and 2,500 (intact arc) = 500 + 1,000 + 2,500 = 4,000 ✓. This is fully consistent.
Answer:
There is 1 EcoRI site and 2 HindIII sites in the plasmid.
Quick Tip:
For a circular molecule, the number of fragments from a single enzyme equals the number of cut sites. For a double digest, the number of fragments equals the total number of sites from both enzymes combined (assuming no sites coincide). Use fragment totals as an arithmetic check: all fragments must sum to the plasmid size.
Which of the following best describes the purpose of a 'selectable marker' in the context of recombinant DNA cloning?
Understanding:
This question asks about the functional role of selectable markers in recombinant DNA cloning experiments.
Step 1: The challenge of transformation
When recombinant plasmids are introduced into competent host cells (transformation), the uptake efficiency is low — only a small fraction of cells actually take up the plasmid. Selectable markers provide a way to distinguish these successful transformants from the vast majority of untransformed cells.
Step 2: How selectable markers work
A commonly used selectable marker is an antibiotic resistance gene (e.g., ampicillin resistance — bla gene encoding beta-lactamase). After transformation, all cells are plated on growth medium containing the antibiotic. Only cells harbouring the plasmid — and thus expressing the resistance gene — survive and form colonies. Cells without the plasmid are killed.
Step 3: Insertional inactivation as a refinement
In vectors like pUC19, the lacZ-alpha gene serves as a second marker. If the foreign insert disrupts the lacZ gene, colonies appear white (no functional beta-galactosidase) rather than blue (functional enzyme), allowing blue-white screening to distinguish recombinant from non-recombinant plasmids.
Step 4: Why other options are incorrect
The origin of replication is a separate functional element. Restriction enzymes are supplied externally in the experiment, not encoded by the vector marker. Promoter sequences for gene expression are separate elements from selectable markers.
Answer:
A selectable marker allows identification and selection of host cells that have successfully taken up the recombinant plasmid.
Quick Tip:
Insertional inactivation into the ampicillin resistance gene can also be used — cells with intact plasmid (no insert) are AmpR, while those with insert-disrupted AmpR gene are AmpS. However, blue-white screening via lacZ is now more common in modern vectors.