Q.21 Adenine constitutes 0.16 mole fraction in a given single–stranded DNA. What is
the mole fraction of uracil in the resultant RNA, if this entire DNA fragment is
transcribed?
(A) 0.16
(B) 0.32
(C) 0.34
(D) 0.68
Adenine at 0.16 mole fraction in single-stranded DNA means every adenine (A) base on this DNA template strand directs incorporation of uracil (U) during transcription to RNA. Thus, the mole fraction of U in the resultant RNA directly equals the mole fraction of A in the DNA template, which is 0.16. This follows standard transcription base-pairing rules where RNA polymerase adds U opposite template A.
Transcription Basics
In transcription, a single-stranded DNA serves as the template, read 3′ to 5′, while RNA synthesizes 5′ to 3′. DNA template A pairs with RNA U (not T, as in DNA replication). Other pairings: DNA T→RNA A, DNA C→RNA G, DNA G→RNA C. For single-stranded DNA given as the template, no complementary strand assumptions apply—fractions transfer directly.
Option Analysis
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(A) 0.16: Correct. U fraction in RNA = A fraction in template DNA = 0.16, as each template A specifies one RNA U.
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(B) 0.32: Incorrect. Assumes double-stranded DNA where A = T, so total A+T = 0.32 directs 0.32 U, but problem specifies single-stranded DNA.
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(C) 0.34: Incorrect. Arbitrary value with no basis in base-pairing or Chargaff’s rules for ssDNA transcription.
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(D) 0.68: Incorrect. Might imply 1 – 0.32 (if misapplying dsDNA logic), but irrelevant for ssDNA template.
In the realm of adenine mole fraction in single-stranded DNA transcription, understanding base pairing during RNA synthesis proves crucial for CSIR NET Life Sciences aspirants. When adenine constitutes 0.16 mole fraction in a given single-stranded DNA acting as the template, the mole fraction of uracil in the resultant RNA mirrors this value precisely due to direct templating.
Core Concept: DNA to RNA Base Pairing
Transcription uses the DNA template strand to produce complementary RNA. Key rule: Template DNA adenine (A) pairs with RNA uracil (U). For single-stranded DNA, the given A fraction (0.16) equals the U fraction in RNA—no Chargaff’s equality (A=T) applies, unlike double-stranded DNA. This yields mole fraction of uracil = 0.16.
Step-by-Step Solution
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Identify DNA as single-stranded template: A = 0.16 (mole fraction).
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Recall transcription rules: DNA A → RNA U.
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Conclude: RNA U = 0.16 (same proportion as template A).
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Total RNA bases sum to 1; other bases (G, C, A from DNA T/G/C) fill remainder, but U fixed by A.
Why Other Options Fail
| Option | Value | Why Incorrect |
|---|---|---|
| (A) | 0.16 | Correct: Direct A → U transfer |
| (B) | 0.32 | Assumes dsDNA (A+T=0.32 → U), but ssDNA given |
| (C) | 0.34 | No biochemical basis |
| (D) | 0.68 | Misinterprets as 1 – 0.32; ignores ssDNA |
This single-stranded DNA transcription question tests precise knowledge of molecular biology, avoiding dsDNA pitfalls. Practice similar for CSIR NET success.
