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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During transcription in eukaryotes, the 5' cap added to pre-mRNA consists of:
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 (m7G).
Step 3: Recall the functions of the cap
The m7G 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.
Which vitamin is essential for the carboxylation of glutamate residues in clotting factors II, VII, IX, and X?
Understanding:
We need to identify which vitamin is required for the post-translational modification (carboxylation) of specific glutamate residues in coagulation factors.
Step 1: Identifying the biochemical reaction
Clotting factors II (prothrombin), VII, IX, and X require gamma-carboxylation of glutamate (Glu) residues to form gamma-carboxyglutamate (Gla) residues. This reaction is catalysed by gamma-glutamyl carboxylase, which requires reduced Vitamin K (hydroquinone form) as a cofactor.
Step 2: Mechanism
Vitamin K acts as an essential cofactor in the carboxylation reaction. During this process, Vitamin K is oxidised to its epoxide form and must be recycled by Vitamin K epoxide reductase. Warfarin inhibits this recycling step, thereby acting as an anticoagulant.
Step 3: Ruling out other options
Vitamin E is an antioxidant with no direct role in coagulation factor carboxylation. Vitamin D is involved in calcium homeostasis and gene regulation. Vitamin A is involved in vision and epithelial differentiation.
Answer:
Vitamin K is the essential cofactor for gamma-carboxylation of glutamate residues in coagulation factors II, VII, IX, and X.
Quick Tip:
Vitamin K-dependent clotting factors can be remembered as "1972" — factors I (fibrinogen, though not truly K-dependent), II, VII, IX, X — along with Protein C and Protein S.
Which of the following hormones uses cyclic AMP (cAMP) as its second messenger?
Understanding:
We need to identify which listed hormone signals through the adenylyl cyclase–cAMP pathway.
Step 1: Classifying hormones by signalling mechanism
Hormones can be broadly classified by the receptors and second messengers they use:
Step 2: Identifying glucagon's pathway
Glucagon binds to a Gs-protein-coupled receptor on hepatocytes and adipocytes. Gs protein activates adenylyl cyclase, which converts ATP to cAMP. cAMP then activates Protein Kinase A (PKA), leading to glycogenolysis and gluconeogenesis.
Step 3: Ruling out other options
Aldosterone, cortisol, and testosterone are all steroid hormones derived from cholesterol. They use nuclear receptor-mediated gene regulation, not cAMP.
Answer:
Glucagon signals through the cAMP second messenger pathway via Gs-protein-coupled receptors.
Quick Tip:
Other hormones using cAMP include PTH, ADH (V2 receptor), TSH, LH, FSH, ACTH, and adrenaline (beta receptors). A common exam trap is confusing ADH — its V1 receptor uses IP3/DAG, but its V2 receptor uses cAMP.
Biotin serves as a coenzyme in which of the following reactions?
Understanding:
We need to identify which metabolic reaction requires biotin (Vitamin B7) as a cofactor.
Step 1: Role of biotin
Biotin is covalently attached to carboxylase enzymes and functions as a carrier of activated carbon dioxide (CO2) in carboxylation reactions. It is covalently bound to the epsilon-amino group of a lysine residue in the enzyme (forming biocytin).
Step 2: Key biotin-dependent enzymes
The major biotin-dependent carboxylases in humans are:
Step 3: Ruling out other options
Oxidative decarboxylation of pyruvate uses the pyruvate dehydrogenase complex (requiring TPP, lipoic acid, FAD, NAD+, and CoA — not biotin). Transamination requires pyridoxal phosphate (Vitamin B6). Hydroxylation of proline requires Vitamin C (ascorbic acid) and requires Fe2+ as a cofactor.
Answer:
Biotin is the coenzyme required for carboxylation of acetyl-CoA to malonyl-CoA by acetyl-CoA carboxylase.
Quick Tip:
A helpful mnemonic for biotin-dependent enzymes is "ACC PP" — Acetyl-CoA Carboxylase, Pyruvate Carboxylase, Propionyl-CoA Carboxylase. All are carboxylases; biotin never participates in decarboxylation or transamination.
A deficiency of which vitamin leads to pellagra, characterized by the classic triad of dermatitis, diarrhea, and dementia?
Understanding:
We need to identify the vitamin whose deficiency causes pellagra with the classic triad of dermatitis, diarrhea, and dementia (the "3 Ds").
Step 1: Identifying Pellagra
Pellagra is caused by deficiency of Vitamin B3 (Niacin/Nicotinic acid) or its precursor, the amino acid tryptophan. Niacin is a precursor for NAD+ and NADP+, which are essential coenzymes in numerous oxidation-reduction reactions.
Step 2: Clinical features
The classic triad of pellagra is:
Some sources add a 4th D — Death — if untreated.
Step 3: Ruling out other options
Thiamine (B1) deficiency causes beriberi and Wernicke-Korsakoff syndrome. Riboflavin (B2) deficiency causes angular stomatitis, glossitis, and corneal vascularisation. Pyridoxine (B6) deficiency causes peripheral neuropathy, sideroblastic anemia, and glossitis.
Step 4: Additional association
Carcinoid syndrome and Hartnup disease can also cause pellagra due to impaired tryptophan availability for niacin biosynthesis.
Answer:
Vitamin B3 (Niacin) deficiency causes pellagra, presenting with dermatitis, diarrhea, and dementia.
Quick Tip:
Remember: in Hartnup disease, a defect in neutral amino acid transport impairs tryptophan absorption, leading to secondary niacin deficiency and pellagra-like symptoms despite an adequate diet.
Calcitriol (1,25-dihydroxycholecalciferol), the active form of Vitamin D, exerts its primary genomic effects by binding to which receptor type?
Understanding:
We need to identify the receptor through which calcitriol (active Vitamin D) mediates its primary genomic actions.
Step 1: Nature of Vitamin D
Vitamin D3 (cholecalciferol) is a fat-soluble, steroid-derived hormone. It is hydroxylated in the liver to 25-hydroxycholecalciferol and then in the kidney (by 1-alpha-hydroxylase) to 1,25-dihydroxycholecalciferol (calcitriol), its biologically active form.
Step 2: Receptor mechanism
Because calcitriol is lipid-soluble, it freely crosses the plasma membrane and binds to the Vitamin D Receptor (VDR), a member of the nuclear receptor superfamily. The calcitriol-VDR complex heterodimerises with the Retinoid X Receptor (RXR) and binds to Vitamin D Response Elements (VDREs) in the promoter regions of target genes, regulating their transcription.
Step 3: Genomic effects
Target genes include those encoding calcium-binding proteins (e.g., calbindin), TRPV6 calcium channels, and RANKL, which collectively mediate increased intestinal calcium absorption, renal calcium reabsorption, and bone mineralisation.
Step 4: Ruling out other options
G-protein-coupled receptors, receptor tyrosine kinases, and ligand-gated ion channels are used by water-soluble messengers that cannot cross the lipid bilayer. Calcitriol is lipid-soluble and uses a nuclear receptor.
Answer:
Calcitriol binds to the nuclear Vitamin D Receptor (VDR), which acts as a transcription factor to regulate gene expression.
Quick Tip:
All steroid hormones and thyroid hormone use nuclear receptors — this is a consistent exam theme. Remember: lipid-soluble = nuclear receptor; water-soluble = cell-surface receptor.
Thyroid hormone synthesis requires adequate dietary iodine. The step in which iodide is oxidised and incorporated into thyroglobulin tyrosine residues is catalysed by which enzyme?
Understanding:
We need to identify the enzyme that catalyses iodide oxidation and its organification onto thyroglobulin tyrosine residues.
Step 1: Thyroid hormone biosynthesis steps
The key steps in thyroid hormone synthesis are:
1. Iodide trapping: I− is actively transported into follicular cells via the Na+/I− symporter (NIS).
2. Oxidation and organification: Iodide is oxidised by thyroid peroxidase (TPO) using H2O2, and the reactive iodine is incorporated onto tyrosine residues of thyroglobulin to form monoiodotyrosine (MIT) and diiodotyrosine (DIT).
3. Coupling: TPO also couples MIT and DIT to form T3 (MIT + DIT) and T4 (DIT + DIT).
4. Secretion: Thyroglobulin is retrieved by endocytosis, proteolysed in lysosomes, releasing T3 and T4.
Step 2: Role of thyroid peroxidase
Thyroid peroxidase (TPO) is the key enzyme responsible for both organification and coupling reactions. Antithyroid drugs such as propylthiouracil (PTU) and methimazole act by inhibiting TPO.
Step 3: Ruling out other options
Deiodinase enzymes convert T4 to the active T3 peripherally. Adenylyl cyclase is activated downstream of TSH receptor signalling (Gs pathway) but does not directly catalyse iodination. "Thyroglobulin synthase" is not a recognised enzyme.
Answer:
Thyroid peroxidase catalyses the oxidation and organification of iodide onto thyroglobulin tyrosine residues.
Quick Tip:
Propylthiouracil (PTU) has a dual advantage in thyrotoxicosis — it inhibits TPO AND blocks peripheral conversion of T4 to T3 by inhibiting Type 1 deiodinase, making it the preferred agent in thyroid storm.
Vitamin B12 (cobalamin) deficiency leads to megaloblastic anaemia partly because it impairs the conversion of which metabolite, thereby trapping folate in an unusable form?
Understanding:
We need to identify the metabolic reaction impaired in B12 deficiency and explain how it leads to folate trapping.
Step 1: The methylfolate trap
Vitamin B12 is required as a coenzyme for methionine synthase, which catalyses the transfer of a methyl group from N5-methyltetrahydrofolate (N5-methyl THF) to homocysteine, generating methionine and regenerating tetrahydrofolate (THF).
Step 2: Consequence of B12 deficiency
When B12 is deficient, methionine synthase cannot function. N5-methyl THF accumulates and cannot be converted back to THF. Since N5-methyl THF is the principal circulating form of folate, this traps the folate pool in the form of N5-methyl THF — a form that cannot participate in nucleotide synthesis. The consequence is functional folate deficiency and impaired DNA synthesis, leading to megaloblastic anaemia.
Step 3: Ruling out other options
The conversion of methylmalonyl-CoA to succinyl-CoA is also B12-dependent (adenosylcobalamin form), but this causes neurological disease (subacute combined degeneration), not folate trapping. Dihydrofolate reductase converts DHF to THF and is inhibited by methotrexate, not B12 deficiency. The serine hydroxymethyltransferase reaction uses N5,N10-methylene THF but is not impaired by B12 deficiency.
Answer:
B12 deficiency impairs the conversion of homocysteine to methionine, trapping folate as N5-methyl THF and causing functional folate deficiency.
Quick Tip:
This is why giving folate alone to a B12-deficient patient corrects the anaemia but DOES NOT prevent neurological damage — the methylfolate trap is bypassed, but the adenosylcobalamin-dependent myelin synthesis pathway remains impaired.
Insulin promotes glucose uptake in muscle and adipose tissue primarily by stimulating the translocation of which glucose transporter to the plasma membrane?
Understanding:
We need to identify the specific glucose transporter isoform whose plasma membrane expression is acutely regulated by insulin in muscle and adipose tissue.
Step 1: Glucose transporter isoforms and their tissue distribution
Step 2: Mechanism of insulin action on GLUT4
In the basal (fasting) state, GLUT4 is sequestered in intracellular vesicles. When insulin binds its receptor tyrosine kinase, autophosphorylation occurs, activating the PI3K-Akt signalling cascade. Akt phosphorylates AS160 (TBC1D4), which releases the inhibitory brake on GLUT4 vesicle fusion, causing GLUT4 translocation to the plasma membrane and increased glucose uptake.
Step 3: Clinical relevance
In Type 2 diabetes, insulin resistance impairs this GLUT4 translocation, leading to hyperglycaemia despite normal or elevated insulin levels.
Answer:
Insulin stimulates translocation of GLUT4 to the plasma membrane in muscle and adipose tissue to promote glucose uptake.
Quick Tip:
Exercise also stimulates GLUT4 translocation via an AMP-activated protein kinase (AMPK) pathway, independently of insulin. This is why exercise improves glycaemic control even in insulin-resistant states.
Which of the following correctly pairs a hormone with its site of synthesis and its primary chemical nature?
Understanding:
We need to identify which pairing correctly matches a hormone, its site of synthesis, and its chemical class.
Step 1: Evaluating each option
Step 2: Confirming Option C
Aldosterone is the primary mineralocorticoid. It is produced in the adrenal cortex (zona glomerulosa), regulated by the renin-angiotensin-aldosterone system (RAAS), and acts on the distal nephron to promote Na+ reabsorption and K+ excretion.
Answer:
Aldosterone is correctly paired with the adrenal cortex as its site of synthesis and is a steroid hormone.
Quick Tip:
Remember the adrenal cortex layers from outside in: Glomerulosa (mineralocorticoids — aldosterone), Fasciculata (glucocorticoids — cortisol), Reticularis (androgens). Mnemonic: "GFR" — same as Glomerular Filtration Rate.
Vitamin C (ascorbic acid) is essential for the activity of prolyl hydroxylase in collagen synthesis. Which of the following correctly describes its biochemical role in this reaction?
Understanding:
We need to identify the precise biochemical role of Vitamin C in the prolyl hydroxylase reaction during collagen synthesis.
Step 1: The prolyl hydroxylase reaction
Prolyl hydroxylase converts proline residues in collagen to 4-hydroxyproline, which is essential for the stability of the collagen triple helix through hydrogen bonding. The enzyme belongs to the family of iron- and 2-oxoglutarate-dependent dioxygenases, requiring:
Step 2: Role of Vitamin C
During the catalytic cycle, the Fe2+ cofactor at the active site becomes oxidised to Fe3+ (ferric state). Vitamin C (ascorbic acid) is required to reduce Fe3+ back to Fe2+, thereby regenerating the active form of the enzyme. Without adequate Vitamin C, prolyl hydroxylase becomes inactive (Fe3+ cannot be recycled), hydroxyproline cannot be formed, and the collagen triple helix is destabilised.
Step 3: Clinical consequence
Scurvy results from Vitamin C deficiency. Defective collagen causes perifollicular haemorrhages, gum disease, poor wound healing, and corkscrew hairs. The symptoms reflect the widespread requirement for stable collagen.
Step 4: Ruling out other options
Vitamin C does not act as a classical coenzyme, does not donate the hydroxyl group itself (O2 is the source via the dioxygenase mechanism), and does not activate the enzyme by phosphorylation.
Answer:
Vitamin C maintains the iron cofactor of prolyl hydroxylase in the active Fe2+ state by reducing Fe3+ back to Fe2+.
Quick Tip:
Lysyl hydroxylase, which hydroxylates lysine residues in collagen (essential for cross-linking), also requires Fe2+ and Vitamin C by the same mechanism. So Vitamin C deficiency impairs both proline and lysine hydroxylation.
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.