Match List I with List II
List I | List II | ||
|---|---|---|---|
| A | RNA polymerase III | I | snRNPs |
| B | Termination of transcription | II | Promotor |
| C | Splicing of Exons | III | Rho factor |
| D | TATA box | IV | SnRNAs, tRNA |
Choose the correct answer from the options given below :
Correct Answer :
A-IV, B-III, C-I, D-II
Solution :
To find the correct match, let us analyze each item in List I with its corresponding match in List II:
A. RNA polymerase III: In eukaryotes, RNA polymerase III is the enzyme responsible for transcribing transfer RNA (tRNA), 5S ribosomal RNA (5S rRNA), and some small nuclear RNAs (snRNAs, such as U6 snRNA). Therefore, A matches with IV (SnRNAs, tRNA).
B. Termination of transcription: In prokaryotic cells, transcription termination can be Rho-dependent, requiring a specific termination protein called the Rho factor (a helicase) to dissociate the RNA transcript from the DNA template. Therefore, B matches with III (Rho factor).
C. Splicing of Exons: Splicing is the post-transcriptional modification process in eukaryotic cells where non-coding introns are removed and coding exons are joined. This reaction is catalyzed by a large molecular complex called the spliceosome, which is made up of small nuclear ribonucleoproteins, commonly referred to as snRNPs ("snurps"). Therefore, C matches with I (snRNPs).
D. TATA box: The TATA box is a conserved, adenine- and thymine-rich DNA sequence found in the promoter region of eukaryotic genes. It acts as a key site for transcription factor binding to initiate the assembly of the transcription machinery. Therefore, D matches with II (Promoter).
Combining all the correct matches, we get:
A - IV
B - III
C - I
D - II
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