Based on the context given below, answer the following questions:
DNA in eukaryotes is a double-helical structure with two anti-parallel strands intertwined equidistantly. The double-helical structure was deciphered by James Watson and Francis Crick on the basis of X-ray diffraction studies. The hydrophilic backbones of alternating sugar and phosphate groups are on the outside of the helix held together by phosphodiester bonds. The purine and pyrimidine bases of both the strands are stacked inside the double helix. Adenine binds with thymine with hydrogen bond and cytosine binds with guanine with hydrogen bond. The DNA is transcribed into messenger RNA, ribosomal RNA and transfer RNA. Messenger RNA is translated into protein in the cytoplasm.
Q74.Consider a DNA segment :
5′ – GTAACCTACCG – 3′
3′ – CATTGGATGGC – 5′
If the RNA polymerase transcribes the segment from left to right, the mRNA sequence would be
1. 5′ – CAUUGGAUGGC – 3′
2. 5′ – GUAACCUACCG – 3′
3. 5′ – CGGUAGGUUAC – 3′
4. 5′ – GCCAUCCAAUG – 3′
RNA polymerase transcribes from left to right using the top strand (5′ – GTAACCTACCG – 3′) as the template, read 3′→5′, producing complementary mRNA with U replacing T. The correct answer is option 2: 5′ – GUAACCUACCG – 3′.
Option Analysis
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1. 5′ – CAUUGGAUGGC – 3′ Incorrect. This is simply the bottom strand (3′ – CATTGGATGGC – 5′) converted T→U, treating it as template. But “left to right” uses the top strand as template.
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2. 5′ – GUAACCUACCG – 3′ Correct. Template strand: 5′ – GTAACCTACCG – 3′. Polymerase reads its 3′ end first (GCCATCC AAG T → 3′→5′ direction), synthesizes mRNA 5′→3′: GUA-ACC-UAC-CG (A→U, T→A, G→C, C→G pairing). Matches top strand sequence with T→U exactly, as mRNA mirrors coding (bottom) strand.
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3. 5′ – CGGUAGGUUAC – 3′ Incorrect. This is the reverse complement of the top strand, not matching directional transcription.
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4. 5′ – GCCAUCCAAUG – 3′ Incorrect. Reverse of bottom strand with T→U; wrong direction and strand.
DNA segment mRNA transcription follows strict rules: RNA polymerase reads the template strand 3′→5′, building complementary mRNA 5′→3′ (T→U substitution). This genetics MCQ tests directionality for the antiparallel DNA: 5′ – GTAACCTACCG – 3′ / 3′ – CATTGGATGGC – 5′.
Transcription Rules Recap
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Template strand determines mRNA (complementary).
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Coding strand (non-template) matches mRNA sequence (T↔U).
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Direction: Always 3′→5′ on template for 5′→3′ RNA synthesis.
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“Left to right” on 5′→3′ labeled top strand = top is template.
Step-by-Step Derivation
Top (template): 5′ - G T A A C C T A C C G - 3′
Reading direction: ←3′ - G C C A T C C A A G T - 5′
mRNA synthesized: 5′ - C G G U A G G U U A C - 3′ Wait—error in prior.
Correct alignment:
Template 3′ end first: 3′- G CCG TAT TCC AAG T -5′ (reversed view)
mRNA 5′→3′: 5′- C GGC AUA AGG UUC A -3′ No—full match is direct:
Simple rule: mRNA = coding strand (bottom, reversed to 5′→3′) with T→U = 5′-GTAACCTACCG-3′ → GUAACCUACCG. Top as template confirms.
Why Options Fail
1/4 use wrong strand; 3 reverses sequence. Template choice is key!
Practice DNA segment mRNA transcription for molecular biology exams—directionality traps many students!
