E. coli cells with a mutated z gene of the lac operon cannot grow in a medium containing only lactose as the source of energy because-

1. They cannot synthesize functional beta-galactosidase
2. They cannot transport lactose from the medium into the cell
3. The lac operon is constitutively active in these cells
4. In the presence of glucose, E. coli cells do not utilize lactose

Subtopic:  Gene Regulation: Lac Operon |
 77%
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During DNA replication, Okazaki fragments are used to elongate:

1. The lagging strand towards the replication fork.
2. The leading strand away from the replication fork.
3. The lagging strand away from the replication fork.
4. The leading strand towards the replication fork.
Subtopic:  DNA Replication:III | DNA Replication |
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Which of the following RNA should be most abundant in an animal cell?

1. t-RNA 2. m-RNA
3. mi-RNA 4. r-RNA
Subtopic:  Transcription: I | Transcription |
 72%

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Gene regulation governing lactose operon of E. coli that involves the lac I gene products is:

1. positive and inducible because it can be induced by lactose.
2. negative and inducible because repress or protein prevents transcription.
3. negative and repressible because repress or protein prevents transcription.
4. feedback inhibition because excess of β galactosidase can switch off transcription.

Subtopic:  Gene Regulation: Lac Operon |
 72%

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Telomere repetitive DNA sequences control the function of eukaryotic chromosomes because they:

1. act as replicons

2. are RNA transcription initiators

3. help chromosome pairing

4. prevent chromosome loss

Subtopic:  DNA Replication: II | DNA Replication:III | DNA Replication |

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During transcription, RNA polymerase holoenzyme binds to a gene promoter and assumes a saddle-like structure. What is its DNA-binding sequence?

1. TIAA 2. AATI
3. CACC 4. TATA
Subtopic:  Transcription:II | Transcription |
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What would happen if in a gene encoding a polypeptide of 50 amino acids, 25th codon (UAU) is mutated to UAA?

1. A polypeptide of 25 amino acids will be formed
2. Two polypeptides of 24 and 25 amino acids will be formed
3. A polypeptide of 49 amino acids will be formed
4. A polypeptide of 25 amino acids will be formed
Subtopic:  Genetic Code: III |
 57%

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Statement I: The codon AUG' codes for methionine and phenylalanine.
Statement II: AAA' and 'AAG are both codons that code for the amino acid lysine.


In light of the above statements, choose the correct answer from the options given below:

1. Statement I is correct but Statement II is false.
2. Statement I is incorrect but Statement II is true.
3. Both Statement I and Statement II are true.
4. Both Statement I and Statement II are false.
Subtopic:  Genetic Code |
 56%
From NCERT
NEET - 2021

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Identify the correct statement:

1. The coding strand in a transcription unit is copied to an mRNA.
2. Split gene arrangement is characteristic of prokaryotes.
3. In capping, methylguanosine triphosphate is added to the 3' end of hnRNA.
4. RNA polymerase binds with the Rho factor to terminate the process of transcription in bacteria.

Subtopic:  Transcription |
 66%
From NCERT
NEET - 2021

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During translation initiation in prokaryotes, a GTP molecule is needed in: 

1. Formation of formyl-met-tRNA
2. Binding of 30S subunit of the ribosome with mRNA
3. Association of 30 S-mRNA with formyl met-tRNA
4. Association of 50 S subunit of the ribosome with the initiation complex
Subtopic:  Translation Mechanism |
AIPMT - 2003

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