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Steps and process of DNA

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Steps and process of DNA
1. Outline the Steps of DNA:
a. DNA Replication begins at the Origin of Replication
b. Helicase cuts hydrogen bonds and separates DNA in half
c. Semiconservative replication produces two copies that each contained one of the original strands and one entirely new strand.
d. Topoisomerases catalyze and guide the unknotting of DNA
e. Single Strand Binding Proteins attach to the halves and keep the DNA molecules separated (they are needed because the sides are attracted to each other and with out the Single Strand Binding Proteins they would move back together)
f. The Replication Fork is formed with the Leading and Lagging strands.
g. In the leading strand, RNA Primase moves along nucleotides and coats with a RNA Primer that will be used as a homing beacon for the DNA Polymerase.
h. DNA polymerase synthesizes the new DNA by adding complementary nucleotides to the template strand.
i. DNA polymerase attaches post primase to DNA nucleotides and move along towards the Replication Fork in 5 prime --> 3 Prime direction with short DNA segments called Okazaki fragments. It adds nucleotides to the half DNA molecules. (Continuous and discontinuous synthesis)
j. Another type of Polymerase replaces the primer with DNA nucleotides so all DNA is there.
k. Ligase stitches up the gaps.
2. List the functions of DNA Polymerase:
a. The primary role is to add the A's, C's, T's and G's to the elongating new strands of DNA.
b. It also proofreads the new copies
c. Changes the RNA primers used to initiate replication into DNA. (short single stranded RNA primers find the start points of DNA replication, but they need to be converted into DNA before the pieces can be ligated together.
3.
a. Amino acid is converted to aminoacyladenylic acid after being activated by ATP.
b. A covalent bond forms between the 5' phosphate group of ATP and the carboxyl end of the amino acid.
c. This reaction is mediated by the synthetaste enzyme
d. Amino acid covalently binds to the 3' end of the tRNA.
e. The enzyme complex separates leaving the charged tRNA with the amino acid.
4. Three major phases of Translation:
a. Initiation:
i. Translation initiation is a process in which mRNA, initiator tRNA, and small and large ribosomal subunits associate with each other to form a complex. ii. Initiation of translation in bacteria begins when the small ribosomal subunit bings to the mRNA near its 5’ end. iii. Initiation factor 3 facilitates the binding of the mRNA to the small ribosomal subunit. iv. Once the 30S subunit complex has formed, initiation factor 2 binds to the complex, and promotes the binding of the initiator transfer RNA to the complex.
v. Translation initiation is completed when the large ribosomal subunit binds, and IF2 and IF3 are released.
b. Elongation:
i. Chain elongation begins with the binding of tRNA, which recognizes the codon in the mRNA, to the A site of the ribosome. ii. This is catalyzed by the EF-Tu transcription factor, and requires the hydrolysis of a GTP. iii. Once the tRNA binds in the A site of the ribosome, the polypeptide chain is moved from the tRNA in the P site to the amino acid attached to the tRNA in the A site. iv. Peptidyl transferase, a protein/RNA complex present in the 50S ribosomal subunit, catalyzes the formation of this new peptide bond between the amino acids.
v. The ribosome then translocates to the next codon. This process is promoted by the elongation factor G and requires another GTP. vi. This places the empty tRNA molecule in the E site of the ribosome, and moves the tRNA containing the growing polypeptide chain in the P site. The next codon in the mRNA chain is positioned in the A site. vii. The uncharged or empty tRNA in the E site then leaves the ribosome and a cycle of chain elongation is completed. viii. Through subsequent cycles of chain elongation the polypeptide chain continues to elongate one amino acid at a time.
c. Termination:
i. Termination begins when a stop codon appears in the A-site. Since there is no tRNA corresponding to the stop codon, a release factor binds in the A-site. ii. The binding of the release factor causes the polypeptide chain to be cleaved from the tRNA, completing the synthesis of the polypeptide. iii. The polypeptide is released, and then the tRNA is released. iv. In the last step, the two ribosomal subunits and the mRNA dissociate from each other. This completes the termination process.
5. Chart

DNA TTC mRNA codon
Anticodon CAG UAG
Amino acid Methionine

6. In a prokaryotic cell, transcription and translation are coupled; that is, translation begins while the mRNA is still being synthesized. In a eukaryotic cell, transcription occurs in the nucleus, and translation occurs in the cytoplasm.
7. A point mutation is when there is a single nucleotide base change in the DNA. (A single base pair is altered by the substitution of one base for another). In a frame shift mutation, one or more bases are inserted or deleted. A frame shift mutation changes the amino acid sequence from the site of the mutation, which can make the DNA meaningless and often results in a shortened protein. The frame shift mutation would leave the mutated DNA vastly more different from its parent DNA (that it was replicated from), then a point mutation would, though in both cases the DNA would still be different/mutated. This is because the codons would be more considerably changed with the frame shift (where now the sequence in every codon in different), rather than in a point mutation (where only one of the codons would be different). 

8. An operon is a unit of genetic material that functions in a coordinated manner by means of an operator, a promoter, and structural genes that are transcribed together. Inducible operons are used to produce proteins only under specific conditions. Most bacteria use glucose as their main energy source. Therefore, the unusual arrival of the milk sugar lactose in the cell turns the operator switch to the "on" position, which allows transcription to occur. Repressible operons have operators that are always on (transcription occurring) unless a signal (for example, too much product) turns them off. An inducer is a molecule that starts gene expression. In negative inducible operons, a regulatory repressor protein is normally bound to the operator, which prevents the transcription of the genes on the operon. If an inducer molecule is present, it binds to the repressor and changes its conformation so that it is unable to bind to the operator. This allows for expression of the operon.
a. An example of an inducible operon is the lac operon. The lac operon is required for the transport and metabolism of lactose in Escherichia coli and some other enteric bacteria. The lac operon is regulated by the availability of glucose and of lactose. Gene regulation of the lac operon was the first complex genetic regulatory mechanism to be elucidated and is one of the foremost examples of prokaryotic gene regulation.
b. Trp is an example of negative regulation of gene expression. Within the operon's regulatory sequence, the operator is blocked by the repressor protein in the presence of tryptophan (thereby preventing transcription) and is

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