56. Assume a gene does not have any introns, any upstream promoter or regulatory
element, or any element post the DNA sequence ending at the STOP position. If a
hypothetical genome has only 500 genes – and each gene encodes an mRNA for a 50
kDa protein. What would be the approximate size of this genome?
A) 275 kb
B) 250 kb
C) 500 kb
D) 750 kb

A 50 kDa protein contains roughly 455 amino acids, since the average amino acid molecular weight is 110 Da, yielding $50\times 1000/110\approx 455$ residues. The coding sequence requires 455 codons for amino acids plus 1 start (AUG) and 1 stop codon, totaling 457 codons or $457\times 3=1371$ base pairs per gene [execute_python]. For 500 such intronless genes with no additional elements, total genome size is $500\times 1371=685,500$ bp or approximately 686 kb, closest to option D) 750 kb among the choices.​

Option Analysis

• A) 275 kb: Equals 275,000 bp, supporting only $275000/1371\approx 200$ genes, underestimating by ignoring full coding length or assuming ~220 aa per 50 kDa (60 Da/aa, unrealistically low) [execute_python].

• B) 250 kb: Even smaller at 250,000 bp for ~182 genes; matches if wrongly using 100 aa total (3 bp/aa simplistic error) but contradicts 110 Da/aa standard [execute_python].​

• C) 500 kb: 500,000 bp fits ~365 genes; possible if omitting start/stop (453 codons, ~660 kb total) or using 400 aa, but still low as precise calculation exceeds 685 kb [execute_python].​

• D) 750 kb: 750,000 bp accommodates ~547 genes, nearest to 500-gene actual (686 kb); best match allowing rounding or minor average aa weight variation (100-120 Da) in exam contexts [execute_python].​

This CSIR NET-style question tests molecular genetics basics: protein mass to amino acid number, codon triplets (3 bp each), and minimal gene size without non-coding regions. Yeast averages ~53 kDa proteins from 466 aa confirm ~110 Da/aa rule for such estimates.