Protein synthesis

Question 0

Direction of Transcriptiob

Answer 0

5^1 –> 3^1

Nucleotides join at the 3^1 end

Question 1

Transcription

Answer 1

1. transcription factors-gene activating chemicals that bind to promoter
2. trams factors help bind RNA polymerase to the promoter
3. INITIATION: RNA polymerase breaks apart dna and transcription begins at promoter
4. ELONGATION: RNA polymerase aligns incoming RNA nucleotides with complimentary based on the template strand
5. TERMINATION: RNA polymerase teachers terminator(sequence signalling end point) mRNA separates from DNA template.
6. EDITING: introns (non-coding base sequences) are removed; exons (base sequences) that code for amino acids are spliced
7. MODIFICATION: 5^1 end is capped to prevent degradation of copied section; Poly (A) Is added to 3^1 end (aids in translation)

Question 2

Translation - Initiation

Answer 2

INITIATION:
Small ribosomal unit binds to mRNA
Small ribosomal unit scans mRNA for the start codon AUG(start triplet) -> large ribosomal unit combines with smaller unit to form functional ribosome.
mRNA now positioned between ribosomal units

Question 3

Translation: elongation

Answer 3

mRNA codons are matched to tRNA anticodon (which carries the correct amino acid).
mRNA moved through ribosome in one direction adding one amino acids at a time to the growing polypeptide chain

Question 4

Translation termination

Answer 4

mRNA strand is read until a stop codon is reached (UGA, UAA or UAG).
Protein release factor binds to stop codon and water is added instead of amino acid -> breaks bond between polypeptide and tRNA
Polypeptide released from ribosome.
Released protein then folds to form functional unit
Left over mRNA is degraded.