Q.65 A hypothetical plant gene ADSH22 is encoded by the nuclear genome. The length of the mature mRNA for ADSH22 is 2150 nucleotides (nts). This mRNA has a 270 nts long 5′ UTR and 200 nts long 3′ UTR. Taking average molecular weight of an amino acid as 115 Dalton (Da), the calculated molecular weight of ADSH22 protein is ________ kDa (round off to 1 decimal place).

Q.65 A hypothetical plant gene ADSH22 is encoded by the nuclear genome. The length
of the mature mRNA for ADSH22 is 2150 nucleotides (nts). This mRNA has a 270
nts long 5′ UTR and 200 nts long 3′ UTR. Taking average molecular weight of an
amino acid as 115 Dalton (Da), the calculated molecular weight of ADSH22 protein
is ________ kDa (round off to 1 decimal place).

The mature mRNA of the hypothetical plant gene ADSH22 is 2150 nucleotides long, with defined UTR regions excluding them from the coding sequence yields a protein molecular weight of 64.4 kDa. This calculation follows standard molecular biology principles used in CSIR NET Life Sciences exams. Each step relies on the genetic code and average amino acid mass provided.

mRNA Structure Breakdown

Mature mRNA comprises 5′ UTR, coding sequence (CDS), and 3′ UTR. The 5′ UTR (270 nts) precedes the start codon, while the 3′ UTR (200 nts) follows the stop codon; neither translates into protein. CDS length = total mRNA – 5′ UTR – 3′ UTR = 2150 – 270 – 200 = 1680 nts.

CDS to Amino Acids

The CDS encodes amino acids via triplets (codons), with each of the 1680 nts forming 1680 ÷ 3 = 560 codons. In eukaryotic nuclear genes like ADSH22, the start codon (AUG) codes for methionine, and the stop codon does not add an amino acid, so protein length equals codon number minus one—but exam problems typically approximate as nts ÷ 3 for direct calculation. Thus, ADSH22 protein has 560 amino acids.

Molecular Weight Calculation

Each amino acid residue in proteins averages 115 Da as given (slightly higher than the common 110 Da benchmark due to residue mass after water loss in peptide bonds). Total mass = 560 × 115 = 64,400 Da = 64.4 kDa (rounded to one decimal). This ignores post-translational modifications, which are unspecified.

CSIR NET Exam Insights

No options are provided (numerical answer type), but common pitfalls include forgetting UTR subtraction or using free amino acid averages (~138 Da) instead of residues. Practice confirms 3 nts per amino acid universally.

Introduction to ADSH22 Protein Molecular Weight Calculation

In CSIR NET Life Sciences, questions like Q.65 test precise ADSH22 protein molecular weight calculation from hypothetical plant gene mRNA data: 2150 nucleotides total, 270 nts 5′ UTR, 200 nts 3′ UTR, and 115 Da average amino acid mass. This mirrors real exam numericals requiring UTR exclusion, codon division, and mass summation—key for molecular biology sections. Understanding mRNA coding sequence length from total mRNA ensures 64.4 kDa accuracy.

Step-by-Step ADSH22 mRNA to Protein Breakdown

  • Identify CDS: Subtract untranslated regions—2150 – 270 (5′ UTR) – 200 (3′ UTR) = 1680 nts CDS.

  • Codons to Amino Acids: 1680 ÷ 3 = 560 amino acids (standard genetic code).

  • Apply Mass: 560 × 115 Da = 64,400 Da = 64.4 kDa (round to 1 decimal).

This method uses residue weights post-peptide bond (H2O loss), aligning with CSIR standards.

Why UTRs Don’t Count in Protein Molecular Weight

5′ UTR regulates translation initiation; 3′ UTR affects stability—both untranslated. Including them inflates CDS (error: ~157 kDa), a frequent trap. Plant nuclear genes like ADSH22 follow eukaryotic norms.

CSIR NET Tips for Similar Questions

  • Verify units: nts to codons (÷3), Da to kDa (÷1000).

  • Average varies: 110-115 Da common; use given value.

  • No options? Direct calc as here.

Master ADSH22 protein molecular weight calculation for exams—practice yields precision.

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