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Expression in poor amount and in inactive form of cDNA of a eukaryotic protein in Escherichia coli using its expression vector is due to<br>P. The absence of capping mechanism of mRNA.<br>Q. Codon bias.<br>R. Absence of polyadenylation.<br>S. Absence of proper glycosylation.

Correct Answer: Q, S

If the coli form bacteria is present in a sample of water, then the coli-form test to be conducted is
(i) Presumptive coli-form test
(ii) Confirmed coli-form test
(iii) Completed coli-form test

cDNA made from the mRNA of an organism was used to make a cDNA library in a vector that allows the expression as a fusion with a reporter tag. What percentage of the cDNA clones is likely to give rise to correct gene products?

cDNA libraries have been prepared that have the same sequences as the mRNA molecules produced by organisms, or else cDNA copies are sequenced directly by RT-PCR (copying of mRNA by reverse transcriptase followed by sequencing of the cDNA copy by the polymerase chain reaction).

If an expression vector contains all three, viz. a controllable promoter, a ribosome binding site and an initiation codon, then a hybrid protein is produced. The _______ region of hybrid protein is encoded by vector and the rest is expressed by the sequence inserted.

In order to isolate cDNA for a specified protein from a mice cDNA library prepared in λgt11 vector which of the following methods you will choose for screening?

A recombinant protein is to be expressed under the control of the lac promoter and operator in a strain of E. coli having the genotype lacI+ crp+. Even in the absence of inducer IPTG, low levels of expression of the recombinant protein are seen (leaky expression). Which one of the following should be done to minimize such leaky expression?

For the aldotetroses I-IV, the combination of true statements, among P-T, is
Basics of Organic Reaction Mechanism mcq question image
P = I and II are diastereomers and II and III are enantiomers
Q = I and IV are mesomers and are optically inactive
R = I and III can be interconverted by a base catalysed isomerisation
S = only I and IV are HlO4 cleavable
T = I and III are D-sugars and II and IV are L-sugars

Phytase, an enzyme produced by Aspergillus niger can be adsorbed on microcrystalline cellulose powder (MCCP). The adsorption follows a Langmuir isotherm and the maximum concentration of the protein that can be obtained on the adsorbent is 70 mg cm-1. At a concentration of 50 mg L-1 of protein in the solution, the concentration of protein on the adsorbent reaches 35 mg cm-1. It is desired to recover 90% of the protein from 1.5 liter of the cell free culture filtrate containing 220 mg L-1 protein by addition of MCCP to the solution. The concentration of the protein adsorbed on the solid at equilibrium will be:

Escherichia coli (E.coli) এর জন্য কোনটি সঠিক নয়?

E. coli (Escherichia coli) bacteria