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Outline the similarities and differences between RNA and DNA.
Differences:
Type of Nitrogenous Bases:
RNA: Adenine and Uracil, Thymine and Guanine
DNA: Adenine and Cytosine, Thymine and Guanine
Type of Sugar:
RNA: Ribose sugar
DNA: Deoxyribose sugar
Length:
RNA: Shorter
DNA: Longer
Association with Histone Proteins
RNA: Does not associate with histone proteins
DNA: Associates with histone proteins
Number of strands:
RNA: Single stranded
DNA: Double, antiparallel strands
Types:
RNA: Many types of RNA, such as rRNA (ribosomal RNA), tRNA (transfer RNA), mRNA (messenger RNA)
DNA: Only one type of DNA.
Ratio of A:U and G:C
RNA: Ratio of A:U and G:C is not always 1:1, varies in different cells
DNA: Ratio of A:U and G:C is always maintained at 1:1 ratio
Similarities
Directionality: Both have a 5’ and 3’ end.
Function: Both play key roles in cellular processes. For example, DNA serves as a template for the synthesis of a daughter DNA molecule via complementary base pairing to the parental strand during the S phase of Interphase (semi-conservative replication). mRNA, on the other hand, is also used as a template for the synthesis of proteins by ribosomes. tRNA ensures that the correct amino acid is transferred accurately to the growing polypeptide chain during translation, while rRNA is involved in protein synthesis during translation.
Both are polynucleotides.
Both have phosphodiester bond present between monomers.
Complementary nitrogenous bases will undergo complementary base pairing and form hydrogen bonds with one another.
Outline the structure of DNA.
Double stranded.
Strands are anti parallel, and forms a double helix.
Made up of adenine, thymine, cytosine and guanine nitrogenous bases.
Hydrogen bonds present between complementary base pairs.
Outline the structure of tRNA.
Single stranded polynucleotide chain.
Has an anticodon that complementary base pairs with codon on mRNA via hydrogen bonds.
Has a 3’ acceptor stem whereby the specific amino acid is covalently attached to, before being transferred to the polypeptide chain during translation process.
Made up of adenine, uracil, cytosine and guanine nitrogenous bases.
Outline the structure of mRNA.
Single stranded polynucleotide chain.
Made up of adenine, uracil, cytosine and guanine nitrogenous bases.
Outline the structure of rRNA.
Single stranded polynucleotide chain.
Made up of adenine, uracil, cytosine and guanine nitrogenous bases.
Outline the functions of DNA.
Heritable genetic material that is passed on from parental generation to the offspring.
Serves as a template during semi-conservative replication, for the synthesis of a complementary daughter strand via complementary base pairing with the original parental template.
During transcription, one DNA strand serves as a template for the synthesis of a complementary mRNA strand via complementary base pairing.
Outline the initiation process of semi-conservative replication of DNA.
At the origin of replication, DNA Helicase enzymes breaks hydrogen bonds between complementary nitrogenous bases → unwinding and unzipping the dsDNA.
The single stranded binding proteins will bind to the single stranded DNA to prevent them from rewinding behind the replication fork.
The DNA Topoisomerase will cause a break in one of the strands, thus allowing the strand to rotate around the break, before resealing the break to prevent positive supercoiling in front of the replication fork.
Outline the elongation process of semi-conservative replication of DNA.
Primase enzyme will add an RNA primer. The DNA polymerase enzyme has an active site that is specific for the free 3’OH end of the RNA primer, and will recognize it, before adding deoxyribonucleotides to the free 3’OH end of the RNA primer.
Using the parental strand as a template, DNA polymerase will add deoxyribonucleotides via complementary base pairing (adenine & thymine, guanine & cytosine).
DNA polymerase thus catalyzes the synthesis of a new strand of DNA from the 5’ to 3’ direction, and catalyzes the formation of a phosphodiester bond between adjacent deoxyribonucleotides of the newly synthesized strand.
As DNA polymerase moves along the previous region, it will proofread the previous region to ensure that the added deoxyribonucleotides are accurate. If not, it will remove the incorrect deoxyribonucleotides and replace them with accurate ones.
Outline the differences between the lagging strand and the leading strand.
Leading Strand:
Synthesized continuously from the 5’ to the 3’ direction towards the replication fork.
Only requires one RNA primer.
Lagging Strand:
Synthesized discontinuously via a series of Okazaki fragments, from the 5’ to 3’ direction, away from the replication fork.
Requires many RNA primers. Each RNA primer has to be separately excised by DNA polymerase and replaced by deoxyribonucleotides. DNA ligase will then join the Okazaki fragments together by catalysing the formation of phosphodiester bond between two adjacent Okazaki fragments.
Outline the termination process of semi-conservative replication of DNA.
In the end, the complementary parental strand and daughter strand will rewind into a double helix.
The resultant strand will consist of one newly synthesized daughter strand, and one parental strand.
Outline the process of initiation of transcription in eukaryotes.
The TATA binding protein will recognise the TATA box on the DNA sequence and bind to it.
General transcription factors will assemble and recruits the RNA polymerase to form the transcription initiation complex.
The RNA polymerase will then break hydrogen bonds between complementary base pairs in order to unwind the dsDNA.
Outline the process of initiation of transcription in prokaryotes.
The sigma factor of RNA polymerase will bind to the -10 and -35 sequence of the promoter.
RNA polymerase will then break hydrogen bonds between complementary base pairs in order to unwind the dsDNA.
Outline the process of elongation of transcription in both eukaryotes and prokaryotes.
After unwinding the dsDNA to obtain ssDNA, one of the single stranded DNA will serve as a template for RNA polymerase.
Using a ssDNA as template strand, the RNA polymerase will catalyse the synthesis of a RNA strand in the 5’ to 3’ direction by adding deoxyribonucleotides via complementary base pairing.
The RNA polymerase will catalyse the formation of phosphodiester bonds between adjacent ribonucleotides.
Outline the process of termination of transcription in eukaryotes.
The elongation process will happen until the termination signal sequence on the template DNA is transcribed onto the newly synthesized strand.
The termination signal sequence will code for the sequence of AAUAAA in the RNA strand.
Downstream of the termination signal sequence, there is disassembling of the transcription complex and release of the RNA polymerase and newly synthesized RNA strand.
The pre-mRNA must undergo post transcriptional modifications before it can be exported into the cytoplasm for translation to occur.
Outline the process of termination of transcription in prokaryotes.
Once the termination signal sequence from DNA is transcribed onto pre-mRNA, the RNA strand is released.
No further modifications is required before it is exported to cytoplasm for translation to occur.
Outline the activation stage that occurs prior to the translation process.
The enzyme, amino-acyl tRNA synthetase will catalyse the covalent attachment of a specific amino acid to the 3’ acceptor stem of the tRNA.
Thereafter, the activated amino acid (in the form of aminoacyl tRNA) will be released from the enzyme.
Outline the initiation stage of the translation process.
The translation initiation factors and the tRNA carrying methionine amino acid (initiator tRNA) will bind to the small ribosomal subunit.
The small ribosomal subunit will then bind to the mRNA, and scan from the 5’ to 3’ direction in search of the start codon, AUG.
Once it has located the start codon (AUG), the anti-codon of the initiator tRNA will complementary base pair with the start codon via hydrogen bonds.
The large ribosomal subunit will then bind to the mRNA to form the translation initiation complex.
The tRNA will now be at the P site of the ribosome, with the A site vacant for the next incoming aminoacyl tRNA with anticodon complementary to the codon of mRNA at the A site to occupy.
Outline the elongation stage of the translation process.
The incoming aminoacyl tRNA with an anticodon complementary for the codon of mRNA at the A site will complementary base pair with one another via hydrogen bonds.
Peptidyl transferase enzyme of the large ribosomal subunit will catalyse the formation of a phosphodiester bond between the new amino acid and the growing polypeptide chain.
Thereafter, the ribosome will translocate by one codon downstream of the mRNA.
The tRNA carrying the polypeptide chain will now be at the P site, while the empty tRNA will now be at the E site, and exit.
Outline the termination process of translation.
The elongation process will occur until a stop codon (UAG, UGA, UAA) reaches the A site of the ribosomes.
This will cause the release factors to bind to the mRNA at the A site, which causes the bond between the newly synthesized polypeptide chain and the tRNA to hydrolyse.
The polypeptide chain is released and the translation complex disassembles.
Outline the post transcriptional modifications that a pre-mRNA has to undergo.
Addition of the 3’Poly-A-Tail
Poly-A-polymerase will catalyse the addition of around 200 adenine residues to the 3’ end of the pre-mRNA with the aims of protecting it from degradation by the 3’ exonucleases, and facilitating its export to cytoplasm to be translated into a polypeptide chain.
Addition of the 5’ cap
Guanylyl transferase will catalyse the addition of the methyl guanosine nucleoside triphosphate to the 5’ end of the pre-mRNA with the aims of protecting it from degradation by the 3’ exonucleases, and facilitating its export to cytoplasm to be translated into a polypeptide chain.
Splicing
Excise the introns and join the exons.
Alternative splicing allows for many mRNA to be produced from a single pre-mRNA.