Biotechnology: Principles And Processes

In PCR, the denaturation step separates the double-stranded DNA into single strands. The Taq polymerase then extends the DNA in the to direction. PCR does not involve the formation of Okazaki fragments or discontinuous synthesis because it operates on single-stranded templates, and the extension is continuous on each strand.

To confirm the presence of the cloned DNA fragment in the plasmid, we consider the digestion by different restriction enzymes:Case (a): Digestion with Not I and EcoRI.The recombinant plasmid contains three Not I sites due to the insert and one EcoRI site.This digestion results in four DNA fragments:1900 base pairs1700 base pairs500 base pairs400 base pairsCase (b): Digestion with Not I and Hind III.In this scenario, the plasmid is cut at the three Not I sites and one Hind III site.This digestion also yields four fragments, but of different sizes:1900 base pairs1100 base pairs900 base pairs600 base pairsBy analyzing the gel electrophoresis pattern for each case, these fragment sizes help confirm the presence and length of the inserted DNA in the plasmid.

The restriction sites for BamHI and EcoRI should not be too close to the ori site or gene for ampicillin resistance to ensure proper and safe cutting of the plasmid.

After 30 cycles you have$ 2^{10} $So at the end of the 20th cycle you’d have about1 million copies(Option C).

Solution:Given data from the gel:Plasmid size =10 kbLaneX→one band at 5 kbLaneY→one band at 5 kbLaneX + Y→three bands at 5 kb, 3 kb, and 2 kbInterpretationSingle digestion with Xgivesone band (5 kb)⟹ EnzymeX cuts the plasmid at two sites, dividing it intotwo equal 5 kb fragments.Single digestion with Yalso givesone band (5 kb)⟹ EnzymeY also cuts at two sites, again producingtwo 5 kb fragments.Double digestion (X + Y)gives3 fragments: 5 kb, 3 kb, 2 kb⟹ The four total cuts (2 by X, 2 by Y) arenot alternating evenly, and one fragment remains5 kb.Constructing the restriction mapSinceX alone → 5 kb + 5 kb, the two X sites must beopposite each other.SinceY alone → 5 kb + 5 kb, the two Y sites must also beopposite each other.In thedouble digest, cuts must partition one 5 kb half into3 kb + 2 kb, while the other 5 kb half remains intact.This arrangement is satisfiedonlyif:OneX sitelies within the segment between twoY sites, andThe otherX sitelies exactly opposite.Checking optionsA→ Produces fragments5 kb, 3 kb, 2 kb✔️B→ Fragment sizes mismatch ❌C→ Wrong enzyme pairing ❌D→ Does not yield observed pattern ❌✅Correct Answer: Option A