Transcription and translation: initation, elongation, and termination review

Sorry, I just found this on this site:

Summary

1. Initiation of transcription commences when the RNA polymerase binds to the
promoter region of the gene to be transcribed. At this point, the DNA is unwound
and the double helix is disrupted.
2. RNA polymerase moves past the promoter until it reaches the start sequence of
the gene to be transcribed.
3. A complementary RNA strand is synthesized in the direction of 5 to 3, using
one strand of DNA as a template. This step is known as elongation. The complement
of adenine in RNA is uracil.
4. Once the terminator sequence is reached by the RNA polymerase, transcription
ceases. The mRNA is separated from the DNA and the RNA polymerase falls off
the DNA molecule, constituting termination of transcription. The DNA molecule
reforms its double-helical shape.
5. The mRNA at this point is not in final form; posttranscriptional modifications
need to be made. 7-methyl guanosine, known as the 5 cap, is added to the 5
end. A string of 200 to 300 adenine ribonucleotides are added to the 3 end by
the enzyme poly-A polymerase. This is known as the poly-A tail.
6. Introns (noncoding regions) are cut out of the primary transcript by particles
known as spliceosomes. The spliceosomes then rejoin the remaining exons
(coding regions).
7. The mRNA transcript is ready to exit the nucleus.

I still have a question tho: Where does Elongation and termination begin?

For this too: Summary of Translation:

1. Ribosome subunits (large and small) bind to the 5  cap of the mRNA transcript,
sandwiching the mRNA between them, and translation commences. The large
ribosome subunit contains two sites, the A (acceptor) site and the P (peptide) site.
2. The ribosome moves along the mRNA, reading the code in triplets known as
codons. Translation does not occur until it reads the triplet AUG, which is the
start codon. The start codon corresponds to the amino acid methionine.
3. When the start codon is in the P site, a tRNA delivers the amino acid methionine.
tRNA recognizes the codon because of the complementary anticodon.
4. The second codon is now in the A site. The appropriate tRNA delivers the next
amino acid in the protein sequence. A peptide bond is formed between methionine
and the second amino acid. The ribosome shifts over one codon. The tRNA
that delivered methionine is released to obtain another methionine amino acid.
The second amino acid tRNA shifts over to the P site. The third amino acid, coded
by the third codon, is brought in to the A site by the next tRNA. A peptide bond is
formed between the second and third amino acid.
5. The process of elongation continues until a stop codon is read in the A site. The
stop codons are UAG, UGA, and UAA. The ribosome stalls.
6. A protein known as the release factor recognizes that the ribosome has stalled
and causes the ribosome subunits to disassemble, releasing the mRNA and newly
formed protein.
7. The protein is folded and modified and then targeted to the area of the cell
where it is required.

Thank you very much!

O_o

I also need to know what Wobble hypothesis and redundancy of DNA code is. Please, brief but good explanations Thanks alot to everyone!

What grade of bio are you in?

Im in grade 12.

Oh and sorry for the late reply:(

The proposal that a tRNA can recognize more than one codon by unusual pairing between the first base of the anticodon and the third base of a codon is known as the wobble hypothesis.

More than one codon can code for a single amino acid, indicating a redundancy in the genetic code. For example, UUU, UUC,UCU, and UCC all code for the amino acid phenylalanine. This redundancy minimizes errors that may lead to serious mutations.