TLS Online TPP Program

#Question id: 38797


Q. Why might a plant that survives thermotherapy be at a disadvantage later on? 

#Plant Biotechnology
  1.  It becomes genetically modified and cannot reproduce. 
  2. Its defense mechanisms may have been inactivated by the heat. 
  3. It grows too quickly for the soil to provide adequate nutrients. 
  4. It loses its ability to perform photosynthesis at normal temperatures.
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TLS Online TPP Program

#Question id: 40427

#Applied Microbiology

Match the Vector Type (List I) with its unique Feature/Function (List II).

List I (Vector Type)List II (Feature/Function)
(a) Expression Vector(i) Can accomodate large DNA inserts (100-300 kb)
(b) Shuttle Vector(ii) Designed for protein production (transcription/translation)
(c) BAC (Bacterial Artificial Chromosome)(iii) Replicates in two unrelated host species
(d) Cosmid(iv) Hybrid of plasmid and phage lambda 'cos' site

TLS Online TPP Program

#Question id: 40428

#Applied Microbiology

Which of the following stationary phases is most appropriate for the purification of a recombinant protein tagged with a Hexa-histidine (His-tag) sequence?

TLS Online TPP Program

#Question id: 40429

#Applied Microbiology

Read the following statements regarding chromatography techniques:

    Statement I: Ion exchange chromatography separates molecules based on differences in their net surface charge.

    Statement II: Gel filtration chromatography (Size Exclusion) separates molecules based on their specific binding to a ligand.

    Choose the correct answer from the options given below:

    TLS Online TPP Program

    #Question id: 40430

    #Applied Microbiology

      Assertion (A): Total cellular RNA passed through an oligo(dT) column results in the specific elution of mRNA.

        Reason (R): Prokaryotic rRNA and tRNA constitute the majority of total RNA and they lack a 3' Poly-A tail.

        Choose the correct option:

        TLS Online TPP Program

        #Question id: 40431

        #Applied Microbiology

        Match the Chromatography Type (List I) with its Principle of Separation (List II).

        List I (Chromatography)List II (Basis of Separation)
        (a) Ion Exchange(i) Hydrophobicity
        (b) Gel Filtration(ii) Specific Ligand Interaction
        (c) Affinity(iii) Net Charge
        (d) Reverse Phase(iv) Molecular Size / Stokes Radius

        TLS Online TPP Program

        #Question id: 40432

        #Botany

        Match the enzymes in LIST-I with their specific metallic activators/co-factors in LIST-II.

        LIST-I (Enzyme)LIST-II (Activator/Co-factor)
        A. NitrogenaseI. Magnesium (Mg²⁺)
        B. RuBisCOII. Molybdenum (Mo)
        C. Alcohol DehydrogenaseIII. Iron (Fe)
        D. FerredoxinIV. Zinc (Zn²⁺)
        Choose the correct answer from the options given below: