Question

Bring this DNA sequence to protein using the transcription (3pts) and translation (4 pts) processes.

note: Pending the direction your DNA is located

Second position UUU Ae UCU UCC cys DU Sey UAU UAC UAA UAG UGU UGC UGA UGG UUA tyr Stop Stop JC Stop CUU CUC his CUA 5 'ATGCCGACGCCATAA 3' Lleve esta secuencia de ADN hasta proteína mediante los procesos de transcripción (3pts) y traducción (4 pts). First position (5'-end) CUC AUU AUC ile AUA AUG met PUUDU< Third position (3'-end) < Bucu AAA AAG GAU GUC vol GCC GAC Osp GGU GGC GCA GGG GUA GUG Сслаа GAA GAG glu GCG Initiation Termination OJO: Pendientes a la dirección en que se encuentra su ADN (4 pts) Second position UUU UAU UGU UUC Phe UCU UCC UCA UAC tyr UGC cys UCG Stop Stop UUA UUG CUU CUC leu CCU UAA UAG CAU CAC CAA CAG his CCC UGA Stop UGG trp CGU CGC arg CGA CGG AGU CUA gin First position (5'-end) AAU POCO DUO DUCUDUCU Third position (3'-end) asn CCA CCG ACU ACC ACA ACG GCU CUG AUU AUC ile AUA AUG met GUU GUC val GUA GUG AGC ser AGA arg AGC AAC AAA AAG GAU GAC GAA GAG asp GCC GGU GGC GGA GGG gly GCA glu GCG

Answer 1

DNA sequence: 5'-ATGCCGACGCCATAA-3'

mRNA sequence: 5'-AUGCCGACGCCAUAA-3'

Protein sequence: N-met-pro-thr-pro-C

The given DNA sequence acts as coding strand, and mRNA will he synthesized using template DNA strand (complementary to coding strand). The final mRNA sequence will he same as that of coding strand except uracil (U) is present in mRNA at the place of thymine (T) in DNA.

Protein synthesis will use mRNA as template. Ribosome will read mRNA from 5' to 3' direction and form amino acid sequence from N-terminal to C-terminal.

At the end of mRNA sequence, stop codon (UAA) is present that does not code for any amino acid.