Question

Compare the following wild type and mutated mRNA sequences. Write out the amino acid polypeptide for both the wild type and mutated genes, note directionality of each fragment, and describe the type of mutation that is occurring and the consequences to the protein structure.

Wild type mRNA:  3’- AUG CUU CAA GUU GGU CCU CCU UCG– 5’

Mutant mRNA:   3’- AUG CUU CAA GUU GGC CCU CCU UCG– 5’

Answer 1

The given wild type mRNA is : 3’- AUG CUU CAA GUU GGU CCU CCU UCG 5’

The translation is the process in which the polypeptide chain is synthesized with mRNA as a template. The tRNA and ribosomes assist this process. The mRNA consists of the nitrogenous base sequence which acts as the template. A codon is a three-base group in the mRNA which is used in the translation process. These codons are used as a recognition site for the anticodons to bind to them and the tRNA (where the anticodons are already located on the other side) release the one corresponding amino acid (located on that tRNA molecule)  to the growing polypeptide chain. This process goes on and a polypeptide chain is produced.
The translation is the process in which the polypeptide chain is synthesized with mRNA as a template. The tRNA and ribosomes assist this process. The mRNA consists of the nitrogenous base sequence which acts as the template. A codon is a three-base group in the mRNA which is used in the translation process. These codons are used as a recognition site for the anticodons to bind to them and the tRNA (where the anticodons are already located on the other side) release the one corresponding amino acid (located on that tRNA molecule) to the growing polypeptide chain. This process goes on and a polypeptide chain is produced.

The three-letter codons are the base sequence and the code for a particular amino acid. It may possible that two or more codons codes for the same amino acid because we have 64 possible codons and only 20 amino acids.
Also, the mRNA direction is from 5' to 3' end, because protein synthesis occurs in this direction.

So, wild type mRNA sequence: 5' GCU UCC UCC UGG UUG AAC UUC GUA 3'

Amino acid encodes by this mRNA sequence: ALA SER SER TRP LEU ASN PHE VAL

Now, mutant mRNA sequence:  3’- AUG CUU CAA GUU GGC CCU CCU UCG 5’

mRNA sequence: 5' GCU UCC UCC CGG UUG AAC UUC GUA '3

AMINO ACID sequence encoded by mutant mRNA: ALA SER SER ARG LEU ASN PHE VAL

Here, we can see that due to the on-base change from U to C in the mRNA sequence, there is a change in the final one amino acid in the polypeptide chain, this a point mutation( due to one base change) and type of mutation is a missense mutation.

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