This is about the central dogma Part 6. Translation Translation is the process o

Question

This is about the central dogma Part 6. Translation Translation is the process of protein production. Remember that the primary structure of proteins is the sequence of amino acids. Genes (in DNA) encode the recipe to make this amino acid sequence. Transcription of protein-coding genes produces Messenger RNAs (mRNAs) hich are temporary, disposable copies of these protein recipes. mRNAs are then translated to make proteins. Here are some of the important components of the translational mac mRNA (Messenger RNA): disposable copy of a protein recipe that is written in nucleotide "words" called codons. tRNA (Transfer RNA): read one or a few specific mRNA codons. Ribosomes: cellular parts that are the locations where mRNAs and tRNAs come together army of RNAs that each carry a specific amino acid and each e . See Chapter 09 RNAProteins Powerpoint, Slides 7-21, for a presentation of the interactions among these parts of the translational machinery. Briefly, mRNAs are read in the 5' to 3' direction beginning with the Start Codon (AUG) and then read three nucleotides (codons) at a time until a Stop Codon is reached. To decode the codons, we use the dictionary of genetics called The Genetic Code. 6. Refer to the following DNA sequence (called the Template): Template 3'- CCGTATACCCTTGACAGTTAATCGC CATCGATCCCTCGCGA-5 UCA mRNA Protein Leu Pro CAA CCA AAU AGU a. Transcribe the Template DNA and write the mRNA sequence in the space above. Codon in the mRNA sequence above. acid sequence encoded by the mRNA above. undergoes mutation and is changed to a T. How will that change the mRNA? How will ACA ewe ING}-Aa)Arg AUA AAG AGG b. What is the Start Codon? Locate the Start rneci c. Use the Genetic Code to produce the amino d. Suppose the 10h nucleotide from the 3'-end of the template DNA (the underlined C) GUC UA GUG GCG that change the protein?

Explanation / Answer

3’ – CCGTATACCCTTGACAGTTAATCGCCATCGATCCCTCGCGA - 5’

A) mRNA: 5’- GGCAUAUGGGAACUGUCAAUUAGCGGUAGCUAGGGAGCGCU – 3’

B) mRNA: 5’- GGCAUAUGGGAACUGUCAAUUAGCGGUAGCUAGGGAGCGCU – 3’

AUG in the sequence is the start codon.

C) mRNA: 5’- GGCAUAUGGGAACUGUCAAUUAGCGGUAGCUAGGGAGCGCU – 3’

If we consider AUG as the start codon and UAG as stop codon then amino acid sequence will be:

Met – Val – Thr – Val – Asn

D) If 10th nucleotide from the 3’ end of the template DNA changed to a T then:

mRNA: 5’- GGCAUAUGGAAACUGUCAAUUAGCGGUAGCUAGGGAGCGCU – 3’

Amino acid sequence: Met – Glu – Thr – Val – Asn

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