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Q. No.The ribozyme that catalyses peptide bond formation in prokaryotes is a part of:
1.
the larger subunit of the bacterial ribosome
2.
the smaller subunit of the bacterial ribosome
3.
the 5’ UTR of the mRNA
4.
the 3’ UTR of the mRNA
Subtopic: Translation |
63%
From NCERT
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Histones are:
1. positively charged basic proteins
2. negatively charged basic proteins
3. positively charged acidic proteins
4. negatively charged acidic proteins
1. positively charged basic proteins
2. negatively charged basic proteins
3. positively charged acidic proteins
4. negatively charged acidic proteins
Subtopic: DNA Packaging |
92%
From NCERT
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For DNA replication, deoxyribonucleoside triphosphates:
1. Only I
2. Only II
3. Both I and II
4. Neither I nor II
| I: | act as substrates |
| II: | provide energy for polymerisation reaction |
2. Only II
3. Both I and II
4. Neither I nor II
Subtopic: DNA Replication |
92%
From NCERT
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The predominant site for the control of gene expression in prokaryotes is:
1. Transcription initiation
2. Processing of hnRNA
3. Translational level
4. Post translational modification
1. Transcription initiation
2. Processing of hnRNA
3. Translational level
4. Post translational modification
Subtopic: Gene Regulation: Introduction | Gene Regulation: Lac Operon |
81%
From NCERT
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Consider the two statements:
| Statement I: | DNA has evolved from RNA with chemical modifications that make it more stable. |
| Statement II: | DNA being double stranded and having complementary strands further resists changes by evolving a process of repair. |
| 1. | Statement I is correct; Statement II is incorrect |
| 2. | Statement I is correct; Statement II is correct |
| 3. | Statement II is incorrect; Statement I is correct |
| 4. | Statement I is incorrect; Statement II is incorrect |
Subtopic: DNA vs RNA as Genetic Material |
89%
From NCERT
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The bacteriophage φ X 174 has:
1. 5386 nucleotides
2. 48502 base pairs
3. 4.6 X 106 base pairs
4. 3.3 X 109 base pairs
Subtopic: The DNA |
86%
From NCERT
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In eukaryotes, during post-transcriptional modification of hnRNA:
| 1. | methyl guanosine triphosphate is added to the 5’ end and adenylate residues are added at the 3‘ end |
| 2. | methyl guanosine triphosphate is added to the 3’ end and adenylate residues are added at the 5‘ end |
| 3. | methyl adenosine triphosphate is added to the 5’ end and guanylate residues are added at the 3‘ end |
| 4. | methyl adenosine triphosphate is added to the 3’ end and guanylate residues are added at the 5‘ end |
Subtopic: Transcription |
84%
From NCERT
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A gene is said to be polymorphic if it has more than one variant and the frequency of the least common variant is at least:
1. 0.01
2. 0.1
3. 0.5
4. 1.0
1. 0.01
2. 0.1
3. 0.5
4. 1.0
Subtopic: The DNA |
58%
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The sequence on the template strand of DNA which will be responsible for the insertion of the first amino acid into a protein will be:
1. TAC
2. AUG
3. CAT
4. ACT
1. TAC
2. AUG
3. CAT
4. ACT
Subtopic: Translation | Genetic Code |
From NCERT
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Consider the two statements:
| Statement I: | In eukaryotes, RNA polymerase II transcribes hnRNA. |
| Statement II: | There is a single RNA polymerase that catalyses transcription of all types of RNA in bacteria. |
| 1. | Statement I is correct; Statement II is incorrect |
| 2. | Statement I is correct; Statement II is correct |
| 3. | Statement I is incorrect; Statement II is incorrect |
| 4. | Statement I is incorrect; Statement II is correct |
Subtopic: Transcription |
86%
From NCERT
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