1.

एक पोलीपेप्टाइड श्रृंखला में 50 अमीनो अम्ल कोड करने के लिए किसी cistron में न्युनतम कितने न्युक्किलयोटाइड्स1) 50 (2) 153 (3) 306 (4) 300

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Answer:- (2) 153 nucleotideExplanation:- Each amino acid is coded by a 'codon' which is a triplet, i.e. 3 adjacent nucleotides. e.g. UGG codes for amino acid Tryptophan.The mRNA which gets translated into a chain of amino acids(polypeptide) has a 'start codon'(AUG-which (mostly) codes for an amino acid Methionine), a 'coding sequence' (code for the corresponding amino acid) and a 'stop codon' (UAG/UAA/UGA- terminates the translation process but do not code for any amino acid).Let's do some maths...1 amino acid = 1 codonand, 1 codon = 3 nucleotidetherefore, 1 amino acid = 3 nucleotide50 amino acid = 3 × 50 nucleotide = 150 nucleotideTherefore for 50 amino acids 150 nucleotides are required + a stop codon, i.e. 150 nucleotide + 3 nucleotide = 153 nucleotide

but answer given options no 3

I am sorry, I completely missed the term 'cistron'.

Cistron- the smallest sequence ofDNA needed to direct thesynthesisof afunctionalpolypeptide; agene.

So, in my previous answer i was talking in terms of mRNA, but the question has been asked in terms of the DNA. As we know, the DNA is double-stranded molecule and RNA is single-stranded. The mRNA is formed by the 'transcription' of the DNA and so has only half set of nucleotides.The 'gene or cistron' is name given to the coding sequence of a DNA molecule, the DNA has complimentary base pairing(double- stranded). So, the cistron will have double number of nucleotides than mRNA.Therefor...153 × 2 = 306 nucleotides

I hope this will help you.Please revert back in case of any doubt.

thank you



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