During your study of the gene for an enzyme you have isolated an amber mutation in the middle of the gene which produces a truncated form of the enzyme. By placing a +1 frameshift mutation a short distance before the amber mutation and a –1 frameshift mutation a short distance after the amber mutation you create a triple mutant that restores expression of a functional, full-length enzyme. 

a) +1 frameshift puts the amber mutation out of frame so that it is no longer read as a stop.

b) -1 frameshift is inserted after the amber so that no longer read as a stop.

c)  -1 frameshift is inserted after the amber to restore the frame of the end of the protein.

d) +1 frameshift puts the amber mutation out of frame so that it restore the frame of the end of the protein.

Which of the following is the correct prediction about the frame shift mutation?

TLS Online TPP Program

#Question id: 26448

#Unit 3. Fundamental Processes

Choose the correct statement regarding DNA photolyase 

a.MTHF polyglu absorb blue light photon of 300-500 nm

b.And the MTHF polyglu radicalize FADH cofactor

c.And this FADH cofactor donate single electron to CBD 

d.And this FADH cofactor donate many electron to CBD

TLS Online TPP Program

#Question id: 26449

#Unit 3. Fundamental Processes

Ability of core and holoenzyme to transcribe DNAs differs, so choose which of the following is correct

a.With  subunit, the holoenzyme worked equally well on both intact as well as nicked types of DNA

b.With  subunit, the holoenzyme worked unequally on both intact as well as nicked types of DNA

c.Core subunit α2ββσ can not perform polymerization on to the intact DNA while perform some of the activity on to the nicked DNA

d.Core subunit α2ββσ can not perform polymerization on to the intact DNA AS WELL AS  to the nicked DNA

TLS Online TPP Program

#Question id: 26450

#Unit 3. Fundamental Processes

Which of the following statements is true regarding bacterial RNA polymerase   

a. Core enzyme can efficiently transcribe both intact and nicked DNA 

b. Sigma destabilizes nonspecifically bound core 

c. Holoenzyme bind tightly to DNA 

d. Higher disassociation rate at lower temperature 

e. Most of the cellular sigma are house keeping gene

TLS Online TPP Program

#Question id: 26451

#Unit 3. Fundamental Processes

Which of the following characteristics of sigma affect non specific binding of the core

TLS Online TPP Program

#Question id: 26452

#Unit 3. Fundamental Processes

Holoenzyme shows highest affinity at

TLS Online TPP Program

#Question id: 26453

#Unit 3. Fundamental Processes

Some of the sigma subunit of E. coli and function is given, Choose which of the following is incorrectly matched