Need answer for question (F), No. 2,3 & 4 Successful probe the following hypothe

Question

Need answer for question (F), No. 2,3 & 4

Successful probe the following hypothetical gene. Indicate the mRNA codons that it forms and the 0 &Jids; produced in the polypeptide chain. ici sequence of amino mRNA codons Amino acid sequence YLETPKT r the following mutations, show the effect by indicating the mRNA codons and the amino acid sequence in th polypeptide 1. Base substitution -T-A-CHT-T-A+G-A-GtA-T-AHC-C-GHA-A-GtA-C-T mRNA codons Amino acid sequence 2. Triplet addition -T-A-CtT-T-A+G-G-T+G-A-AtA-T-A C-C-GtA-A-GtA-C-T- mRNA codons Amino acid scquence 3. Base addition -T-A-CtT-T-A+G-A-A+A-T-C-A C-C-GHA-A-GHA-C.T mRNA codons Amino acid sequence 4. Base deletion (site indicated by arrow) mRNA codons

Explanation / Answer

Ans) The information about the amino acid sequence of a protein is located in the DNA. All the proteins which expressed in a cell are coded by the genes which are present in DNA. The RNA polymerase bind to the promoter sequences of DNA and copy the genetic information present in DNA into mRNA by Transcription processes and then ribosome’s translate this information located in mRNA (in the form of codons) as a protein by a process called translation. The template DNA strand or negative strand or nonsense strand of DNA is complementary to mRNA. The RNA polymerase binds to the promoter region which is located at an upstream portion of a gene and initiates the synthesis of mRNA by denovo fashion. From the transcription bubble (DNA-RNA hybrid), 5 end of mRNA is released into cytoplasm during transcription processes. The ribosomes bind to mRNA and initiate translation. The first codon on mRNA to be translated is AUG (initiation codon) and it codes for formyl methionine in prokaryotes but in eukaryotes it is methionine. Once the synthesis of protein initiated from the AUG codon it will translate mRNA until it reaches the stop codon (UAA, UAG, UGA). The stop codons will not code for any amino acids and translation processes terminated.

3’-TACTTAGAAATACCGAAGACT-5’                             (template DNA strand or negative strand)

5’-UACUUAGAAAUACCGAAGACU-3’                         (mRNA codons)

YLEIPKT (Amino acid sequence)

1. Base substitution

3’-TACTTAGAGATACCGAAGACT-5’                             (template DNA strand or negative strand)

5’- UACUUAGAGAUACCGAAGACU -3’ (mRNA codons)

YLEIPKT                                                                  (Amino acid sequence)

2. Triple addition

3’-TACTTAGGTGAAATACCGAAGACT-5’                    (template DNA strand or negative strand)                

5’-UACUUAGGUGAAAUACCGAAGACU-3’                (mRNA codons)

YLGEIPKT (Amino acid sequence)

3. Base addition

3’-TACTTAGAAATCACCGAAGACT-5’                          (template DNA strand or negative strand)

5’- TACTTAGAAATCACCGAAGACT -3’ (mRNA codons)

YLEITED (Amino acid sequence)

4. Base deletion

3’-TACTTAGAAAT CCGAAGACT-5’                              (template DNA strand or negative strand)

5’- UACUUAGAAAUCCGAAGACU -3’                          (mRNA codons)

YLEIRR Amino acid sequence)

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