Question

2) Given the following DNA sequence, identify the template strand, transcribe the template strand, and translate the mRNA. 5' GCGATGAAACGCCCGACGTAGGGC 3' 3' CGCTACTTTGCGGGCTGCATCCCG 5'

Answer 1

IN DNA 4 TYPE OF NITROGENOUS BASES PRESENT A, T, G, AND C. A PAIRS WITH T AND G PAIRS WITH C AND VICE VERSA. TRANSCRIPTION OCCURS IN 5' TO 3' DIRECTION. IN RNA T IS ABSENT, INSTEAD OF T, U IS PRESENT. TRANSLATION STARTS FROM START CODON AND ENDS AT STOP CODON. START CODON CODE TO AMINO ACID BUT STOP CODON DO NOT CODE TO ANY AMINO ACID.

Page) 2. DNA sequence - S'OCGATGAAACGCOCGACC TAGGGC -3' 3' CGCTACTTTGCG GCC TGCATCCCG-S' → Transcription direction · S0 should be in occur in s' to 3' DNA template strand 3' to si direction. so Template strand is - 3 CGCTACTTT GCG GGC TGC ATCCCG-S' Canswer) → in pairs In Transcription, c pairs with and vice versa. In RNA, T is absent and DNA, U is absent : So A with U and vice-versa. Hence, U of RNA pairs with of DNA and of RNA pairs with 7 of DNA. A

Page 2 so Transcribed RNA strand → s' GCG AUGA AA CACCCGACGUAGGGC 3' Canswer) Translation of start codon codon UAA, mRNA occurs AUG or GuG VAG, UGA. from to stop mRNA Strand - ACG DAGGGC 3' SGCGAUGA AA CGC CCG ACGUAGGGC Stop Start codon codon Stop codon do not code to any amino acid. Start codon Aug code to Methionine Met) AAA cac cca code to code to code to Lysine (lys) Arginine (Arg) Proline (pro)

ACG Page 3 code to Threonine Thr) Hence, Translation of mRNA or . or amino quid polypeptide sequence sequence is - Methionine – Lysine - Arginine – Proline, Threonine (answer)