This chapter is crucial because it connects core biological principles with real-world applications such as PCR, DNA cloning, genetic engineering, and gene silencing/CRISPR editing. Board and competitive exams repeatedly ask how biotech tools work, how to interpret experiments (restriction digestion, inheritance patterns, expression/contamination control), and how design choices affect outcomes—especially in vector selection, gene regulation, and genome editing strategies.
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15
Questions
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Marking
Q1. Starting with double‑stranded DNA molecules, PCR ideally doubles the number of molecules each cycle. Using , how many double‑stranded DNA molecules will be present after cycles (assume 100% efficiency)?
Q2. A circular plasmid of total size gives the following restriction patterns: EcoRI single digest → fragments of and ; HindIII single digest → fragments of and ; EcoRI + HindIII double digest → fragments of , and . What is the shortest distance (in kb) between an EcoRI site and the nearest HindIII site on the plasmid?
Q3. Assertion (A): Replacing dTTP with dUTP in PCR amplicons and pre‑treating subsequent PCR reaction mixes with Uracil–DNA glycosylase (UDG) effectively prevents carry‑over contamination from previous PCR products without degrading native genomic DNA templates.
Reason (R): UDG recognises and removes uracil from DNA, creating abasic sites in uracil‑containing contaminating amplicons; UDG is heat‑labile and is inactivated during the high‑temperature initial denaturation of PCR, so native genomic templates (which lack uracil) remain intact and newly synthesised amplicons are protected during amplification.
Both A and R are true and R is the correct explanation of A
Both A and R are true but R is not the correct explanation of A
A is true but R is false
Both A and R are false
Q4. A hereditary retinal disorder requires gene delivery to photoreceptor cells (terminally differentiated, essentially non‑dividing) for durable expression with minimal risk of insertional mutagenesis. The patient has no pre‑existing neutralising antibodies to adeno‑associated virus (AAV). Which viral vector choice is most appropriate and why?
Lentiviral vector — integrates into host genome to give permanent expression, so it is preferred for non‑dividing cells
Adenoviral vector — gives very high expression from episomes but is highly immunogenic and typically provides only transient expression
Adeno‑associated virus (AAV) vector — tends to persist episomally in non‑dividing cells, gives long‑term expression with low immunogenicity and a low frequency of random integration, minimising insertional oncogenesis risk
Gamma‑retroviral (oncoretrovirus) vector — integrates efficiently but requires host cell division and carries significant insertional mutagenesis risk
Q5. A gene has Exon 1 = and Exon 2 = separated by an intron of . Two transcript isoforms exist: Isoform A (Exon1+Exon2) and Isoform B (Exon1 only). To detect Isoform A specifically from cDNA (reverse‑transcribed RNA) while avoiding amplification from genomic DNA or from Isoform B, which primer pair is best?
Both primers located entirely within Exon 1
Forward primer in Exon 1 and reverse primer whose 3′ end spans the Exon1–Exon2 junction (part complementary to end of Exon1 and part to start of Exon2)
Forward primer in Exon 1 and reverse primer located inside Exon 2 (internal)
Both primers located entirely within Exon 2
Q6. A PCR reaction is initiated with target DNA molecules. Assuming 100% efficiency so that each cycle doubles the number of molecules, use to calculate the expected number of target molecules after cycles.
3200
800
1600
100
Q7. A researcher expresses a eukaryotic protein in E. coli using two constructs: Vector X has a very strong T7 promoter and expresses the protein in the cytosol; Vector Y has a moderate promoter and includes an N-terminal signal peptide directing the protein to the periplasm. After induction, Vector X gives high total protein but most is insoluble (inclusion bodies), while Vector Y gives lower total yield but most protein is soluble and enzymatically active. Which explanation best accounts for these observations?
The strong promoter in Vector X causes rapid overexpression that overwhelms cytosolic chaperones, causing misfolding and inclusion bodies; targeting to the periplasm plus slower expression from Vector Y allows disulfide bond formation and proper folding, yielding soluble active protein.
Cytosolic expression in Vector X exposes the protein to high protease activity so it aggregates; the periplasm of Vector Y lacks proteases, so protein remains soluble.
Vector X proteins are insoluble because absence of a signal peptide prevents glycosylation in the cytosol; Vector Y proteins are soluble because periplasmic glycosylation stabilizes them.
Vector X triggers an RNA interference-like mechanism in bacteria causing truncated translation products that aggregate, whereas Vector Y avoids this due to the signal peptide.
Q8. At three STR loci the observed allele sizes (in bp) are: Mother — L1: 120, 128; L2: 200, 208; L3: 150, 156. Alleged father — L1: 124, 132; L2: 196, 208; L3: 152, 154. Child — L1: 120, 124; L2: 208, 196; L3: 156, 154. Based on Mendelian inheritance for these loci, which conclusion is best supported?
The alleged father is excluded because the child does not carry allele 132 at Locus 1.
The alleged father can be excluded because at Locus 3 the child has allele 156, which is absent in his genotype.
The data are inconclusive because only three loci are insufficient for statistical paternity testing.
The alleged father is consistent with being the biological father: at each locus the child can be explained as inheriting one allele from the mother and the other allele from the alleged father.
Q9. A transgenic crop containing a Bt gene under the CaMV 35S promoter was integrated as multiple tandem copies. Initially plants show high Bt mRNA and protein and strong pest resistance. After several generations, the transgene DNA is still present, but Bt mRNA levels are greatly reduced; bisulfite sequencing reveals heavy cytosine methylation across the promoter region. Which mechanism most plausibly explains the generational loss of expression?
Increased post-translational degradation of Bt protein by cellular proteases caused loss of protein despite unchanged transcription.
Transcriptional gene silencing (TGS) due to promoter methylation—likely triggered by repeated transgene copies—leading to reduced transcription and lower mRNA.
Post-transcriptional gene silencing (PTGS/RNAi) degrading Bt mRNA without affecting promoter methylation.
Homologous recombination has excised critical coding regions while leaving promoter sequences intact.
Q10. In a plant genome-editing experiment using CRISPR-Cas9 with two gRNAs targeting exon 1 and exon 4, researchers observed frequent large deletions between the two cut sites, numerous off-target edits at sites with 1–2 mismatches, and mosaicism in regenerated plants. To minimize large deletions and off-target modifications while still achieving an efficient knockout, which strategy is most appropriate?
Design a single, highly specific gRNA targeting an early coding exon and deliver a high-fidelity Cas9 as a transient ribonucleoprotein (RNP) complex so that short-lived, precise single-site DSBs produce frameshift indels while minimizing off-targets and avoiding two-cut deletions.
Use a constitutive strong promoter to express Cas9 continuously and include multiple gRNAs targeting several exons to ensure knockout in all cells.
Deliver Cas9 via a replicating viral vector to maintain long-term expression and increase chance of biallelic edits, reducing mosaicism.
Switch to dual gRNAs with standard Cas9 but add antibiotic selection to enrich for edited cells and thereby eliminate mosaicism and off-target concerns.
Q11. A PCR reaction starts with copies of a target DNA. Assuming ideal amplification with exact doubling each cycle (use ), what is the minimum number of cycles required to obtain at least copies?
Q12. In a ligation–transformation experiment of ligated DNA was used to transform competent E. coli. After recovery, l (which is 10% of the total recovery) was plated and gave colonies on LB+amp. A control transformation with vector alone (no insert) produced colonies under identical plating. Estimate (i) the transformation efficiency in transformants per DNA and (ii) the approximate percentage of colonies that are true recombinants after correcting for background. (Use .)
Q13. DNA fingerprinting (mini‑satellite/STR analysis) was done on a mother, an alleged father and a child. Allele sizes (bp) at three independent loci are:
Locus 1 — Mother: 100, 110; Alleged father: 130, 140; Child: 100, 140.
Locus 2 — Mother: 150, 160; Alleged father: 160, 170; Child: 150, 170.
Locus 3 — Mother: 200, 210; Alleged father: 220, 230; Child: 210, 220.
Which interpretation is correct based on Mendelian inheritance of one allele from each parent?
The alleged father is excluded because the child carries allele 100 at Locus 1, which is not present in his genotype.
The test is inconclusive — more loci must be tested before any inference.
The alleged father is unlikely because the child does not carry allele 160 at Locus 2, which should be inherited from him.
The alleged father is consistent with being the biological father: at each locus the child carries one allele that can be derived from the mother and the other allele that can be derived from the alleged father.
Q14. Assertion (A): Adeno‑associated virus (AAV) vectors are frequently chosen for in vivo gene delivery because they tend to elicit relatively low immune responses in the host.
Reason (R): AAV vectors have a limited packaging capacity of approximately .
Both A and R are true and R is the correct explanation of A.
Both A and R are true but R does not correctly explain A.
A is true but R is false.
A is false but R is true.
Q15. CRISPR–Cas9 reagents were microinjected into one‑cell zygotes to correct a recessive disease allele. Several founder animals show the corrected allele in many somatic tissues, but none transmit the corrected allele to their offspring. Which explanation best accounts for this observation?
Off‑target CRISPR mutations killed or sterilized germ cells, so corrected allele cannot be transmitted.
Homology‑directed repair (HDR) works only in somatic cells and never in germline cells, so germline remained unedited.
CRISPR editing occurred post‑zygotically (after several cell divisions), producing mosaic animals in which germline precursor cells remained unedited; hence somatic tissues show correction but gametes do not.
Cas9 and guide RNA were actively degraded or epigenetically silenced specifically in germline cells, preventing editing in gametes.