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Molecular Basis of Inheritance - Genetic Code

Grade 12CBSEBiology

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The genetic code is the set of rules by which information encoded within genetic material (DNADNA or mRNAmRNA sequences) is translated into proteins by living cells.

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Triplet Nature: The code is a triplet, meaning three nitrogenous bases constitute one codon. Since there are 44 bases (A,U,G,CA, U, G, C), the total possible permutations are 43=644^3 = 64 codons.

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Codon Specificity: Out of 6464 codons, 6161 codons code for amino acids, while 33 codons do not code for any amino acids and function as stop codons (UAA,UAG,UGAUAA, UAG, UGA).

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Universal Property: The genetic code is nearly universal; for example, the codon UUUUUU would code for Phenylalanine (PhePhe) in both bacteria and humans.

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Degeneracy: Some amino acids are coded by more than one codon (e.g., GGU,GGC,GGA,GGGGGU, GGC, GGA, GGG all code for Glycine).

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Non-overlapping and Commaless: The code is read in mRNAmRNA in a contiguous fashion without punctuations or overlapping of bases.

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Dual function of AUGAUG: It acts as the initiation codon (Start codon) and also codes for the amino acid Methionine (MetMet).

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Stop Codons: UAAUAA (Ochre), UAGUAG (Amber), and UGAUGA (Opal) are terminator codons that signal the end of translation.

📐Formulae

Total Codons=4n=43=64\text{Total Codons} = 4^n = 4^3 = 64

Number of Sense Codons=64−3=61\text{Number of Sense Codons} = 64 - 3 = 61

Length of polypeptide=Number of bases in mRNA3−1 (for stop codon)\text{Length of polypeptide} = \frac{\text{Number of bases in } mRNA}{3} - 1 \text{ (for stop codon)}

💡Examples

Problem 1:

If a sequence of mRNAmRNA has 9090 nitrogenous bases, including the initiation and termination codons, how many amino acids will be present in the resulting polypeptide chain?

Solution:

903−1=30−1=29\frac{90}{3} - 1 = 30 - 1 = 29 amino acids.

Explanation:

Since the genetic code is a triplet, 9090 bases form 903=30\frac{90}{3} = 30 codons. However, the last codon is a stop codon which does not code for any amino acid. Therefore, the polypeptide will contain 2929 amino acids.

Problem 2:

Why did George Gamow argue that the genetic code must be a triplet?

Solution:

If the code were a singlet (414^1), it would only code for 44 amino acids. If it were a doublet (424^2), it would code for 1616 amino acids. Since there are 2020 essential amino acids, a triplet code (43=644^3 = 64) is the minimum required to cover all 2020 amino acids.

Explanation:

Permutations of the 44 bases must exceed the 2020 known amino acids to ensure every amino acid has at least one corresponding codon.

Genetic Code - Revision Notes & Key Diagrams | CBSE Class 12 Biology