The time taken for synthesis of the mRNA for a 110 kDa protein is 60.0 minutes. This calculation relies on standard bacterial transcription rates and protein composition assumptions relevant to CSIR NET Life Sciences exam problems.[execute_python]​

Calculation Steps

A protein’s molecular weight of 110 kDa equals 110,000 Da. With an average amino acid residue mass of 110 Da, the protein contains 110,000 / 110 = 1,000 amino acids.​

Each amino acid requires a 3-nucleotide codon in mRNA, so the coding sequence totals 1,000 × 3 = 3,000 nucleotides. The problem specifies no upstream sequences or abortive transcripts, so only the coding region is transcribed.​

At 50 nucleotides per minute, transcription time is 3,000 / 50 = 60 minutes, rounded to one decimal place as 60.0.[execute_python]​

Key Assumptions

Bacterial RNA polymerase elongates at rates around 40-80 nucleotides per second (2,400-4,800 nt/min), but the problem uses 50 nt/min, likely to simplify exam calculations.​

The 110 Da average reflects residue masses in proteins (free amino acids average ~138 Da minus 18 Da water), weighted by common occurrences like glycine and alanine.​

No 5′ UTR, promoter, or Shine-Dalgarno sequences are included per the problem’s explicit instructions.​

Common Pitfalls

Trainees often forget to exclude water mass or include non-coding regions, yielding incorrect nucleotide counts like 3,300 nt. Others misconvert kDa to Da or overlook the codon triplet code.​

Using free amino acid averages (138 Da) gives ~798 amino acids and underestimates time at ~47.8 min, a frequent trap in CSIR NET questions.​

Real bacterial rates vary by conditions, but exam problems prioritize given values over literature ranges (e.g., 25-50 nt/s).​

In bacterial transcription rate problems for CSIR NET Life Sciences, determining time for mRNA synthesis of a 110 kDa protein at 50 nucleotides/min is a classic quantitative genetics question. This SEO-optimized guide breaks down the calculation using average amino acid molecular weight of 110 Da, codon rules, and exam assumptions for precise 60.0 min answer.[execute_python]​

Step-by-Step Solution

Start with protein molecular weight: 110 kDa = 110,000 Da. Divide by average amino acid residue mass (110 Da) to get number of amino acids: 110,000 ÷ 110 = 1,000.​

Multiply by 3 nucleotides per codon: 1,000 × 3 = 3,000 nucleotides for the coding sequence. No upstream sequences means full mRNA length matches this.​

Apply transcription rate: 3,000 nt ÷ 50 nt/min = 60.0 min (rounded to one decimal).​

Why 110 Da Average?

Proteins use residue masses (~110 Da) after peptide bond formation removes water (18 Da from ~138 Da free amino acids). Frequent small residues like Ala/Gly lower the effective average.​

Exam Relevance

CSIR NET tests integration of biochemistry (amino acid weights), molecular biology (codon table), and kinetics (transcription rates). Pitfalls include untranslated regions or wrong Da conversion.​

This approach ensures accuracy for similar problems in genetics, biotechnology, and cell biology sections.​