Q. No.The enzyme aminoacyl-tRNAsynthetase is able to recognize all the following in the tRNA molecule except:
1. The acceptor stem.
2. Bases in the stem loop regions.
3. The anticodon region.
4. 5’ end.
1. The acceptor stem.
2. Bases in the stem loop regions.
3. The anticodon region.
4. 5’ end.
A mutation leading to the formation of a nonfunctional sigma subunit of RNA polymerase holoenzyme in prokaryotes will result in bacteria being unable to:
1. assemble RNA polymerase onto the DNA strand.
2. identify and tightly bind promoter elements.
3. convert a closed complex to an open complex.
4. synthesize RNA once an open complex is formed.
The 5’ cap in the final transcript in eukaryotic mRNA helps in:
1. Protection from inappropriate splicing.
2. Synthesis of the polyA tail.
3. Transport of mRNAs into the nucleus.
4. Recognition of the 5' end of the transcript by a ribosome.
What would be the actual inducer for the lac operon?
| 1. | lactose | 2. | allolactose |
| 3. | ß-galactosidase | 4. | galactose |
3 out of 64 codons are termination codons. These termination codons:
1. recognise a single tRNA
2. are capable of binding to many different tRNAs
3. bind to release factors instead of tRNAs
4. only specify a single amino acid
RNA molecules can have catalytic functions as well. A well known example would be:
1. Polynucleotide kinase.
2. Base pair recognition.
3. Methyl transferase.
4. Peptidyl transferase.
Removal of introns describes which of the following processes in relation to the primary transcript in eukaryotes?
1. Trimming
2. 5' capping
3. Poly-A tailing
4. Splicing
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The primary RNA transcript in eukaryotes contains introns. This can be demonstrated by:
| I: | A comparison of the DNA and mRNA sequences. |
| II: | A comparison of the genomic DNA and cDNA sequences. |
| III: | A hybridization between DNA and mRNA molecules. |
| 1. | I and II only | 2. | I and III only |
| 3. | II and III only | 4. | I, II, and III |
Two amino acids will differ from each other in their properties because they have different:
1. The amino group.
2. The carboxyl group.
3. The R group.
4. Peptide bonds.
Identify the statement that will be incorrect for transcription in both prokaryotes and eukaryotes:
| 1. | RNA polymerase identifies and binds to specific promoter elements in the DNA. |
| 2. | Transcription factors control the binding of RNA polymerase. |
| 3. | RNA polymerase opens a double-stranded DNA to expose the template strand. |
| 4. | RNA polymerase catalyzes RNA synthesis from 3' to 5' of the new strand. |
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