Bacterial and Phage Genetics Questions and Answers

Q: Frederick Griffith used smooth (S) and rough (R) strains of Streptococcus pneumoniae in his classical experiment that showed DNA might be the genetic element. Which ONE of the following observations gave the clue for this discovery?

The clue was that the R strain became S strain when mixed with heat killed S strain.In 1928 Griffith injected mice with live non-virulent rough (R) pneumococci together with heat-killed virulent smooth (S) cells. Neither alone killed the mice, but the mixture did, and live S bacteria were recovered. Something from the dead S cells had permanently transformed the R cells; Avery, MacLeod and McCarty later showed the 'transforming principle' was DNA.R remaining R would show nothing. S becoming R with killed R cells was not the observation, and mixing R with live S cells proves nothing, because live S bacteria are already present.

Q: Entry of λ phage lysogen to lytic phase is triggered by

Induction is triggered by loss of co-operativity in binding of \(\lambda\) repressor.DNA damage activates RecA, which acts as a co-protease and stimulates the cI repressor to cleave itself between its N-terminal DNA-binding domain and C-terminal dimerisation domain. The DNA-binding fragments can no longer dimerise or bind co-operatively to the operators OR and OL, so they fall off, the lytic promoters fire, and the prophage enters the lytic cycle.An increase in repressor would strengthen lysogeny, not end it. A decrease in RecA function prevents induction. A mutation in the phage genome is not the normal trigger.

Q: Which one of the following conjugations will result in formation of merodiploids?

Merodiploids result from an F′ donor × F− recipient cross.An F′ (F-prime) plasmid is an F factor that has picked up a segment of chromosomal DNA. When it is transferred to an F− cell, the recipient gains a second copy of those chromosomal genes on the plasmid, so it is partially diploid, a merodiploid. This is the basis of complementation tests in bacteria.An F+ × F− cross transfers only the F plasmid, with no chromosomal genes. In Hfr × F− crosses, chromosomal DNA does enter but is usually recombined into the recipient chromosome or lost, rather than kept as a stable second copy. An Hfr cell is not a recipient in the usual sense, so the last option is not a productive mating.

Q: Phages are collected from an infected E.coli donor strain of genotype ser+, gly+, leu+ and used to transduce a recipient of genotype ser-, gly-, leu-. The treated recipient population is plated on a minimal medium supplemented with leucine and serine. Many colonies grew. Which one of the following combination of genotypes is appropriate for the colonies that grew ?

Every colony that grew must be gly+, so the answer is the set in which all three genotypes carry gly+: ser-, gly+, leu+; ser+, gly+, leu+; ser+, gly+, leu-.The recipient is ser- gly- leu-. The plate is minimal medium with leucine and serine added, so the only nutrient a cell still has to make for itself is glycine. Only transductants that received the gly+ allele from the donor can grow; whether they also picked up ser+ or leu+ does not matter, because both amino acids are supplied.So the colonies are a mixture of gly+ cells that may or may not be co-transduced for the other two markers. Each of the other three options includes at least one gly- genotype, which could not form a colony on this medium.

Q: A bacterial strain that is lys+ his+ val+ is used as a donor, and another strain that is lys- his- val- as the recipient for setting up an experiment on transformation. Initial transformants were isolated on a minimal medium with supplemented with lysine and valine. Which of the following maximum genotypes will grow on this medium ?

Every transformant that has become his+ will grow, in any combination with lys and val, which is the first option.The medium is minimal medium supplemented with lysine and valine. Lysine and valine are supplied, so the lys and val alleles do not matter; histidine is not supplied, so a cell can grow only if it has become his+. The selection is for his+ alone.Transformation can bring in his+ alone or together with linked donor genes, so all four his+ classes can appear: lys- his+ val+, lys+ his+ val-, lys+ his+ val+ and lys- his+ val-. The other options include his- genotypes, which cannot grow without histidine, or list too few classes.

Frederick Griffith used smooth (S) and rough (R) strains of Streptococcus pneumoniae in his classical experiment that showed DNA might be the genetic element. Which ONE of the following observations gave the clue for this discovery?

A.R strain became S strain when mixed with heat killed S strain
B.R strain remained R strain when mixed with heat killed S strain
C.S strain became R strain when mixed with heat killed R strain
D.R strain became S strain when mixed with live S strain

Exam Year(s)

GATE Life Sciences - 2012,

Answer: A

The clue was that the R strain became S strain when mixed with heat killed S strain.

In 1928 Griffith injected mice with live non-virulent rough (R) pneumococci together with heat-killed virulent smooth (S) cells. Neither alone killed the mice, but the mixture did, and live S bacteria were recovered. Something from the dead S cells had permanently transformed the R cells; Avery, MacLeod and McCarty later showed the 'transforming principle' was DNA.

R remaining R would show nothing. S becoming R with killed R cells was not the observation, and mixing R with live S cells proves nothing, because live S bacteria are already present.

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Entry of λ phage lysogen to lytic phase is triggered by

A.mutation in the \(\lambda\) genome
B.loss of co-operativity in binding of \(\lambda\) repressor
C.increase in the \(\lambda\) repressor concentration
D.decrease in recA function

Exam Year(s)

GATE Life Sciences - 2012,

Answer: B

Induction is triggered by loss of co-operativity in binding of \(\lambda\) repressor.

DNA damage activates RecA, which acts as a co-protease and stimulates the cI repressor to cleave itself between its N-terminal DNA-binding domain and C-terminal dimerisation domain. The DNA-binding fragments can no longer dimerise or bind co-operatively to the operators O_R and O_L, so they fall off, the lytic promoters fire, and the prophage enters the lytic cycle.

An increase in repressor would strengthen lysogeny, not end it. A decrease in RecA function prevents induction. A mutation in the phage genome is not the normal trigger.

Discuss in Forum

Which one of the following conjugations will result in formation of merodiploids?

A.\(F^{+}\) donor × \(F^{-}\) recipient
B.Hfr donor × \(F^{-}\) recipient
C.\(F'\) donor × \(F^{-}\) recipient
D.\(F^{+}\) donor × Hfr recipient

Exam Year(s)

GATE Life Sciences - 2023,

Answer: C

Merodiploids result from an F′ donor × F⁻ recipient cross.

An F′ (F-prime) plasmid is an F factor that has picked up a segment of chromosomal DNA. When it is transferred to an F⁻ cell, the recipient gains a second copy of those chromosomal genes on the plasmid, so it is partially diploid, a merodiploid. This is the basis of complementation tests in bacteria.

An F⁺ × F⁻ cross transfers only the F plasmid, with no chromosomal genes. In Hfr × F⁻ crosses, chromosomal DNA does enter but is usually recombined into the recipient chromosome or lost, rather than kept as a stable second copy. An Hfr cell is not a recipient in the usual sense, so the last option is not a productive mating.

Discuss in Forum

Phages are collected from an infected E.coli donor strain of genotype ser⁺, gly⁺, leu⁺ and used to transduce a recipient of genotype ser^-, gly^-, leu^-. The treated recipient population is plated on a minimal medium supplemented with leucine and serine. Many colonies grew. Which one of the following combination of genotypes is appropriate for the colonies that grew ?

A.ser\(^{-}\), gly\(^{-}\), leu\(^{+}\); ser\(^{+}\), gly\(^{-}\), leu\(^{+}\); ser\(^{+}\), gly\(^{-}\), leu\(^{-}\)
B.ser\(^{+}\), gly\(^{+}\), leu\(^{+}\); ser\(^{+}\), gly\(^{-}\), leu\(^{+}\); ser\(^{+}\), gly\(^{-}\), leu\(^{-}\)
C.ser\(^{-}\), gly\(^{+}\), leu\(^{+}\); ser\(^{+}\), gly\(^{+}\), leu\(^{+}\); ser\(^{+}\), gly\(^{+}\), leu\(^{-}\)
D.ser\(^{-}\), gly\(^{+}\), leu\(^{+}\); ser\(^{+}\), gly\(^{-}\), leu\(^{+}\); ser\(^{+}\), gly\(^{-}\), leu\(^{-}\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

Every colony that grew must be gly⁺, so the answer is the set in which all three genotypes carry gly⁺: ser^-, gly⁺, leu⁺; ser⁺, gly⁺, leu⁺; ser⁺, gly⁺, leu^-.

The recipient is ser^- gly^- leu^-. The plate is minimal medium with leucine and serine added, so the only nutrient a cell still has to make for itself is glycine. Only transductants that received the gly⁺ allele from the donor can grow; whether they also picked up ser⁺ or leu⁺ does not matter, because both amino acids are supplied.

So the colonies are a mixture of gly⁺ cells that may or may not be co-transduced for the other two markers. Each of the other three options includes at least one gly^- genotype, which could not form a colony on this medium.

Discuss in Forum

A bacterial strain that is lys⁺ his⁺ val⁺ is used as a donor, and another strain that is lys^- his^- val^- as the recipient for setting up an experiment on transformation. Initial transformants were isolated on a minimal medium with supplemented with lysine and valine. Which of the following maximum genotypes will grow on this medium ?

A.\(\text{lys}^{-}\text{his}^{+}\text{val}^{+}\); \(\text{lys}^{+}\text{his}^{+}\text{val}^{-}\); \(\text{lys}^{+}\text{his}^{+}\text{val}^{+}\); \(\text{lys}^{-}\text{his}^{+}\text{val}^{-}\)
B.\(\text{lys}^{+}\text{his}^{+}\text{val}^{+}\); \(\text{lys}^{-}\text{his}^{-}\text{val}^{+}\)
C.\(\text{lys}^{-}\text{his}^{-}\text{val}^{-}\) only
D.\(\text{lys}^{+}\text{his}^{-}\text{val}^{+}\) only

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

Every transformant that has become his⁺ will grow, in any combination with lys and val, which is the first option.

The medium is minimal medium supplemented with lysine and valine. Lysine and valine are supplied, so the lys and val alleles do not matter; histidine is not supplied, so a cell can grow only if it has become his⁺. The selection is for his⁺ alone.

Transformation can bring in his⁺ alone or together with linked donor genes, so all four his⁺ classes can appear: lys^- his⁺ val⁺, lys⁺ his⁺ val^-, lys⁺ his⁺ val⁺ and lys^- his⁺ val^-. The other options include his^- genotypes, which cannot grow without histidine, or list too few classes.

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