biology : class12 Transformation and Recombinant Screening : DNA Isolation, Restriction Enzymes, Vector, Ligation, Host Cell : Recombinant DNA Technology MCQ Questions and Answers
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Transformation and Recombinant Screening
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MCQ Questions
1. A plasmid vector carries ampicillin-resistance and tetracycline-resistance genes. A foreign DNA fragment is inserted within the tetracycline-resistance gene, while the ampicillin-resistance gene remains intact. Which plating strategy most specifically identifies recombinant transformants?
A.
Plate cells directly on tetracycline; sensitive colonies are recombinants
B.
Plate on ampicillin, then replica-plate surviving colonies onto tetracycline; ampicillin-resistant but tetracycline-sensitive colonies are recombinants
C.
Plate on tetracycline, then select colonies that grow after transfer to ampicillin
D.
Plate on both antibiotics simultaneously; colonies resistant to both are recombinants
Correct Answer:
Plate on ampicillin, then replica-plate surviving colonies onto tetracycline; ampicillin-resistant but tetracycline-sensitive colonies are recombinants
Explanation:
Insertion of foreign DNA disrupts the tetracycline-resistance gene but leaves ampicillin resistance functional. Therefore, recombinants survive ampicillin selection but fail to grow on tetracycline during replica plating.
2. In blue-white screening, a foreign DNA fragment is ligated into the multiple cloning site located within the lacZ coding region of a plasmid. After transformation, cells are plated on medium containing antibiotic, X-gal, and IPTG. Which colony phenotype represents a likely recombinant?
A.
Blue colony that grows because lacZ remains functional
B.
White colony that grows because insertion disrupts lacZ while the plasmid marker provides antibiotic resistance
C.
White colony that fails to grow because lacZ disruption prevents plasmid replication
D.
Blue colony that fails to grow because IPTG inhibits transformed cells
Correct Answer:
White colony that grows because insertion disrupts lacZ while the plasmid marker provides antibiotic resistance
Explanation:
Insertional disruption of lacZ prevents functional beta-galactosidase production, so recombinant colonies remain white on X-gal. Their growth on antibiotic-containing medium confirms possession of the plasmid-selection marker.
3. A transformation mixture contains competent bacteria, recombinant plasmids, non-recombinant plasmids, and untransformed cells. After plating on antibiotic-containing medium, which population is primarily eliminated by the selection step?
A.
Cells containing recombinant plasmids
B.
Cells containing non-recombinant plasmids
C.
Untransformed cells lacking the plasmid-borne resistance marker
D.
All cells because antibiotics prevent transformation
Correct Answer:
Untransformed cells lacking the plasmid-borne resistance marker
Explanation:
Antibiotic selection permits growth of cells that acquired a plasmid carrying the corresponding resistance gene. Untransformed cells generally lack this marker and are eliminated, whereas recombinant and non-recombinant plasmid-bearing cells can both survive.
4. A researcher observes many antibiotic-resistant colonies after transformation, but only some are white on X-gal plates. Which conclusion is most justified?
A.
Every antibiotic-resistant colony necessarily contains the desired insert
B.
White colonies are candidates for recombinant plasmids, whereas blue colonies may contain non-recombinant plasmids
C.
Blue colonies must be untransformed cells because X-gal selects against plasmids
D.
White colonies are necessarily dead cells that failed antibiotic selection
Correct Answer:
White colonies are candidates for recombinant plasmids, whereas blue colonies may contain non-recombinant plasmids
Explanation:
Antibiotic resistance indicates plasmid acquisition, but not necessarily insertion of the foreign DNA. White color suggests lacZ disruption by an insert, so white colonies are selected for further confirmation; blue colonies commonly represent vectors without an insert.
5. Why is a recovery period commonly provided after chemically induced transformation but before plating cells on selective medium?
A.
To allow degradation of the plasmid so that only chromosomal transformants survive
B.
To allow expression of the plasmid-encoded antibiotic-resistance gene before antibiotic exposure
C.
To make the foreign DNA undergo restriction digestion inside the host
D.
To prevent plasmid replication until colonies become visible
Correct Answer:
To allow expression of the plasmid-encoded antibiotic-resistance gene before antibiotic exposure
Explanation:
Immediately after DNA uptake, the resistance gene may not yet produce enough resistance protein. Recovery allows transcription and translation of the marker, increasing the survival of genuinely transformed cells during selection.
6. Two plasmids are used in a transformation experiment: plasmid P carries ampicillin resistance and an intact lacZ gene, while plasmid Q carries only ampicillin resistance and a foreign insert but lacks lacZ. On ampicillin-X-gal-IPTG plates, which outcome best distinguishes cells carrying P from cells carrying Q?
A.
P-containing cells are white and Q-containing cells are blue
B.
P-containing cells are blue and Q-containing cells are white
C.
Both types are blue because ampicillin activates lacZ
D.
Both types are white because foreign DNA always inhibits beta-galactosidase
Correct Answer:
P-containing cells are blue and Q-containing cells are white
Explanation:
An intact lacZ gene in plasmid P produces beta-galactosidase, causing blue colonies in the presence of X-gal and IPTG. Plasmid Q lacks functional lacZ, so its transformants appear white while still growing on ampicillin.
7. A plate contains antibiotic-resistant white colonies selected by blue-white screening. Which additional test most directly verifies that a colony contains the intended recombinant plasmid rather than a mutation that merely disrupts lacZ?
A.
Testing whether the colony grows on a medium without any nutrient source
B.
Analyzing plasmid DNA by restriction digestion or insert-specific PCR
C.
Repeating plating on X-gal without antibiotic
D.
Measuring the colony's resistance to a different unrelated antibiotic
Correct Answer:
Analyzing plasmid DNA by restriction digestion or insert-specific PCR
Explanation:
White color indicates disruption of lacZ but does not prove that the correct insert is present. Restriction analysis or insert-specific PCR directly tests the size or sequence of the DNA inserted into the plasmid.
8. A transformation control receives competent bacteria without plasmid DNA and is plated on antibiotic medium. No colonies appear, whereas the experimental transformation plate has colonies. What does this control most strongly demonstrate?
A.
The antibiotic is ineffective against the host cells
B.
The colonies on the experimental plate likely acquired the plasmid-borne resistance marker
C.
The host cells were unable to take up any DNA
D.
The antibiotic selected specifically for recombinant plasmids rather than all plasmids
Correct Answer:
The colonies on the experimental plate likely acquired the plasmid-borne resistance marker
Explanation:
The no-plasmid control confirms that untransformed host cells cannot grow under the antibiotic condition. Thus, growth on the experimental plate is associated with acquisition and expression of the plasmid resistance gene, although further screening is needed to distinguish recombinant from non-recombinant plasmids.
9. A plasmid contains a kanamycin-resistance gene and a lacZ screening region. After transformation, colonies are first selected on kanamycin and then screened on X-gal. A colony grows on kanamycin but remains white. Which interpretation is most appropriate?
A.
The cell is certainly untransformed because white colonies cannot contain plasmids
B.
The plasmid likely entered the cell and lacZ was disrupted, but the intended insert still requires confirmation
C.
The colony must contain an intact lacZ gene expressed at a very high level
D.
Kanamycin resistance proves that the foreign DNA has integrated into the host chromosome
Correct Answer:
The plasmid likely entered the cell and lacZ was disrupted, but the intended insert still requires confirmation
Explanation:
Kanamycin resistance indicates retention and expression of the plasmid marker, while the white phenotype suggests loss of functional lacZ activity. Because mutations or unintended insertions can also disrupt lacZ, molecular confirmation of the desired recombinant is necessary.
10. In a transformation experiment, cells are exposed to plasmid DNA but are not made competent. No colonies grow on selective medium. Which inference is most consistent with this result?
A.
The plasmid necessarily lacked a selectable marker
B.
The cells probably did not efficiently take up the plasmid because competence was not induced
C.
The antibiotic must have caused plasmid replication to stop in every cell
D.
The absence of colonies proves that the plasmid contained a toxic insert
Correct Answer:
The cells probably did not efficiently take up the plasmid because competence was not induced
Explanation:
Competent cells possess conditions that facilitate uptake of extracellular DNA. Without an appropriate competence treatment, plasmid entry is usually inefficient, so cells fail to acquire the resistance marker and cannot form colonies under selection.
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