This chapter is central for board and competitive exams because it links core biotechnology tools (enzymes, cloning, PCR, vectors) with how we engineer genetic material in practice. Concepts like restriction digestion/ligation, recombinant DNA strategies, PCR efficiency, restriction mapping, and site-directed mutagenesis repeatedly appear in MCQs, and being able to reason from fragment sizes and experimental logic is a key scoring skill.
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Marking
Q1. A restriction enzyme recognizes a specific 6-base palindromic sequence and cleaves wherever it occurs in a long random DNA molecule. Assuming equal base frequencies and randomness, on average how many base pairs apart will successive cleavage sites be?
Q2. A circular plasmid of is cut separately by EcoRI to yield fragments of and , and by HindIII to yield and . A double digest with EcoRI + HindIII yields fragments of and . Which statement about the relative positions of EcoRI and HindIII sites is correct?
HindIII site is within the EcoRI fragment, dividing it into and .
HindIII is located within the EcoRI fragment, dividing it into and .
There are two HindIII sites, one in each EcoRI fragment.
EcoRI and HindIII cut at the same position on the plasmid producing identical fragments.
Q3. In a ligation reaction, a circular plasmid vector of size and an insert of size are used. If of vector is taken and an insert-to-vector molar ratio of is desired, what mass of insert (in ng) should be used? Use .
Q4. Assertion (A): In site-directed mutagenesis using whole-plasmid PCR (e.g., QuikChange), treating the PCR product with DpnI before transformation increases the proportion of mutant plasmids among transformants. Reason (R): DpnI specifically cleaves methylated sequences, so it digests the parental (methylated) plasmid template isolated from E. coli while leaving the PCR-amplified (unmethylated) mutant plasmid intact.
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.
A is false but R is true.
Q5. A circular plasmid of gives the following digestion pattern: with enzyme A → fragments and ; with enzyme B → fragments and ; with A + B (double digest) → fragments . Which of the following best describes the arrangement of A and B sites on the plasmid?
Both B sites lie within the A fragment and split it into and subfragments.
One B site lies within the A fragment dividing it into and , and the other B site lies within the A fragment dividing it into and .
Both B sites are equidistant from both A sites and are positioned so that the double digest yields four equal fragments.
The two A sites are apart, and both B sites flank one A site within a region.
Q6. In an ideal PCR (100% amplification efficiency, each cycle doubles the number of DNA molecules), starting from a single double‑stranded DNA molecule, the number of molecules after 25 cycles will be:
Q7. A ligation reaction to insert a fragment into a plasmid yields an estimated 40% recombinant plasmids (the rest are empty vectors). After transformation and antibiotic selection 1200 colonies appear (both recombinants and empty‑vector transformants). Approximately how many colonies would you expect to contain the recombinant plasmid?
Q8. A circular DNA molecule of total length is digested with enzyme P yielding fragments of and . Digestion with enzyme Q yields fragments of and . A double digest with P and Q yields fragments of , and . Which statement about the relative positions of P and Q sites is correct?
The Q site lies within the fragment produced by P.
The Q site lies within the fragment produced by P.
P and Q cut at the same nucleotide position in the circular DNA.
The double digest data are inconsistent; it should produce four fragments.
Q9. Assertion (A): Two isoschizomer restriction enzymes that recognize the same DNA sequence can give different digestion patterns on genomic DNA.
Reason (R): Methylation of cytosine residues within the recognition sequence can inhibit cleavage by one enzyme but not by its isoschizomer because the enzymes differ in sensitivity to methylated bases.
Both A and R are true and R explains A.
Both A and R are true but R does not explain A.
A is true but R is false.
A is false but R is true.
Q10. A student amplified a eukaryotic gene from genomic DNA by PCR and cloned the fragment into a bacterial expression vector that supplies a suitable promoter and ribosome‑binding site. Transcription from the plasmid in E. coli is confirmed, but SDS‑PAGE shows only smaller truncated polypeptides instead of the expected full‑length protein. What is the most likely reason and the best corrective step?
Cause: PCR introduced frameshift mutations because Taq polymerase lacks proofreading; corrective step: re‑amplify using a high‑fidelity proofreading polymerase.
Cause: The cloned DNA contains introns that E. coli cannot splice out; corrective step: synthesize cDNA from mature mRNA and clone the intron‑free coding sequence.
Cause: The plasmid lacks a functional origin of replication causing unstable maintenance; corrective step: reclone into a vector with a proper origin.
Cause: A eukaryotic promoter in the cloned fragment prevents translation in bacteria; corrective step: insert a eukaryotic promoter upstream of the gene.