1. Which DNA binding domain is seen in region 4.2 of sigma factor?
a) Zinc fingers
b) Helix turn helix
c) Leucine zipper
d) Beta barrel
Explanation: Region 4.2 of sigma subunit has a helix turn helix motif that helps it to bind to the major groove of DNA. The similar motif has been seen in region 3 as well. Zinc fingers and leucine zipper are examples of other DNA binding motif
2. Which sigma factor is bound to Rse A?
a) σD
b) σG
c) σF
d) σE
Explanation: Rse A binds σE and prevents it from activating transcription. Under extreme conditions this binding is broken by cleavage of c terminal cytosolic tail of transmambrane Rse A, thus releasing σE for transcribing necessary genes.
3. You artificially increase the level of σ32 in an E. coli culture. What will you expect to see?
a) rpo H transcription permanently inhibited
b) cdc transcription constitutively turned on
c) rpo H temporarily activated
d) cdc 6 permanently turned off
Explanation: σ32 is necessary for transcribing genes like rpo H that helps in surviving heat shock while it doesn’t act on cdc6 that has a role in replication. As σ32 is unstable it temporarily turns on rho H till the concentration ratio of σ32 and σ70 returns to normal.
4. If a bacteria is facing lack of carbon sources in the media, which sigma factor concentration is expected to increase?
a) Sigma E
b) Sigma S
c) Sigma 70
d) Sigma 32
Explanation: Sigma S is synthesized when the cell is under stressed conditions as in stationary phase. It transcribes rpo S gene which helps to overcome carbon starvation in stationary phase. Sigma E works in extreme conditions of shock, while sigma 70 and 43 work in normal conditions.
5. You want your bacterial culture to grow well so you made an enriched media with all forms of carbohydrates. Which of this carbohydrate should you restore first if you want the culture to keep growing at the same rate?
a) Glucose
b) Lactose
c) Galactose
d) Fructose
Explanation: Prokaryotes first use all the glucose in the media to derive energy. Only when they sense a low glucose level they search alternative energy sources like lactose, fructose or galactose.
6. In a mero-diploid you have one I- and one I+ Repressor gene in the two lac operons. I+ is wild type while I- is a mutant which produces a repressor that doesn’t bind. What would be the functionality of the diploid?
a) Function normally
b) Will be non-functional
c) May or may not function
d) Will have impaired functioning
Explanation: The set of gene product produced from the wild type I+ loci will be sufficient to induce repression in both the operons. Thus, the system will not be impaired or non-functional.
7. Which of these Lac operon mero- diploid will be constitutively turned on? Considering the terms has their usual meaning.
a) I+ + P + O+ + Z+ + Y+ + A+ / I+ + P+ O+ + Z+ + Y+ + A+
b) I+ + P + OC + Z+ + Y+ + A+ / I+ + P+ O+ + Z+ + Y+ + A+
c) Is + P + O+ + Z+ + Y+ + A+ / I+ + P+ O+ + Z+ + Y+ + A+
d) I+ + P + O+ + Z+ + Y+ + A+ / I– + P+ O+ + Z+ + Y+ + A+
Explanation: OC is an operon that can’t bind to the repressor. Consecutively it is always turned on. IS produces a repressor which binds tightly to operon and never lets it turn on- This is the case of constitutively turning off.
8. Which of these combinations will result is am operon that can never produce beta galactosidse?
a) I– + P + O+ + Z+ + Y+ + A+ / I+ + P+ O+ + Z+ + Y+ + A+
b) I+ + P + OC + Z– + Y+ + A+ / I+ + P+ O+ + Z– + Y+ + A+
c) Is + P + OC + Z+ + Y+ + A+ / I+ + P+ OC + Z+ + Y+ + A+
d) I-d + P + O+ + Z+ + Y+ + A+ / I– + P+ O+ + Z+ + Y+ + A+
Explanation: This is because although this operon is constitutively turned on, yet it has a mutated lac Z gene which produces beta galactosidase. This mutation exists in both of the loci so it can’t produce this enzyme even if the operon is on.
9. Which of the Ars operon regulation site is actually preventing RNA polymerase from binding and producing the Ara genes BAD?
a) O1
b) O2
c) P
d) Z
Explanation: Although O2 is much farther from the ara promoter P yet it can regulate P as it loops out and reaches the site. On the other hand O1 however being closer doesn’t control the promoter of BAD genes
10. Which of these Ara genes is a mode of feedback autoregulation?
a) Ara A
b) Ara B
c) Ara C
d) Ara D
Explanation: As ara C level rises it itself binds beside its promoter preventing the RNA polymerase to bind to the same there by having a negative feedback control. Ara BAD promoter lacks this type of control and is regulate din different way.