Topic D: DNA and GENOMICS

Question 1

What is the structure of nucleotides?

Answer 1

A nucleotide consists of 3 components: A pentose sugar, a nitrogenous base, and one or more phosphate groups.
They are joined by condensation reactions, the N base is bonded to 1’ C of the pentose sugar, while the P group is bonded to 5’ C of the pentose.
A nucleoSide is a moleq consisting of a pentose sugar and N base.
A nucleoTide is a moleq containing a nucleoside and a phosphate.

Question 2

What is the structure of the various components of a nucleotide?

Answer 2

Pentose sugar:
5 C sugar, 2 different types of pentose sugars: ribose and deoxyribose. In RNA, the pentose sugar is ribose.
In DNA, the pentose is deoxyribose.
Nitrogenous bases:
2 categories: Purines (2 rings) and Pyrimidines. Purines are ADENINE AND GUANINE. Pyrimidines (1 ring) are CYTOSINE, THYMINE and URACIL.

Question 3

What is the structure of polynucleotides?

Answer 3

Many nucleotides are joined together to form polynucleotides, joined by strong covalent bonds called PHOSPHODIESTER bonds.
These are formed by condensation rxn, with -OH group on 3’ C of pentose of one nucleotide and P group on 5’ C of pentose of next moleq.
The two ends of the polynucleotides are different from each other: 5’ end of a polynucleotide has a phosphate group attached to 5’ C of sugar. 3’ end of a polynucleotide has a -OH group on 3’ of sugar

Question 4

What is the structure of DNA?

Answer 4

DNA molecule consists of 2 polynucleotide chains spiralled around an imaginary axis to form a double helix.
The 2 polynucleotide chains are antiparallel. (5’-3’/3’-5’)
Uniform width of 2nm.
The N bases are stacked 0.34nm apart and the helix makes one full turn every 3.4nm.
The hydrophilic sugar-P backbones are on the outside of the helix.
The hydrophobic nitrogenous bases are paired in helix interior.
The stacking of base pairs results in hydrophobic interxns.
The 2 polynucleotides chains or strands are held together by hydrogen bonds between paired N bases.
Adenine always pairs with Thymine with 2 H bonds via CBP
Guanine always pairs with Cytosine with 3 H bonds via CBP.

Question 5

What is the function of DNA?

Answer 5

DNA is the genetic material that organisms inherit from their parents.
Most DNA molecules are very long and each DNA molecule contains numerous genes. Each gene occupies a specific region along the DNA moleq.
The gene is a unit of inheritance, which store coded instructions for the synthesis of proteins and RNA.
During cell div, replication of DNA occurs, the structure of DNA make it possible to accurately copy the entire DNA.

Question 6

What is the structure of RNA?

Answer 6

The RNA molecule is a single polynucleotide chain.
RNA has much fewer nucleotides compared to DNA.
RNA contains ribose as its pentose sugar.
RNA is less stable than DNA because it is more prone to hydrolysis by intracellular enzymes. rNA molecule can contain 4 different nitrogenous bases similar to those in DNA except that in RNA thymine is replaced by uracil.
RNA can develop secondary structures which are formed by complementary base pairing within the RNA molecule. The secondary structures function to promote stability of the molecules
adenine (A) always pairs with uracil (U) with two hydrogen bonds.
guanine (G) always pairs cytosine (C) with three hydrogen bonds.

Question 7

What is mRNA?

Answer 7

A mRNA molecule is a single-stranded RNA. mRNA is synthesised by the transcription of DNA in the nucleus.mRNA is transported to cytoplasm for translation by ribosomes.
mRNA is used as an template to synthesise proteins.

Question 8

What is rRNA?

Answer 8

rRNA molecule is single-stranded RNA. rRNA is synthesized in the nucleolus.
rRNA and ribosomal proteins are then assembled into the large and the small subunits within the nucleolus.
The two subunits are then transported out of the nucleus into the cytoplasm where they associate to form ribosomes.
Ribosomes are the sites of protein synthesis.

Question 9

What is tRNA?

Answer 9

trNA molecule is a single-stranded RNA molecule.
triplet base sequence known as the anticodon is present on the tRNA.
Anticodon on tRNA forms hydrogen bonds with a codon (triplet sequence) on mRNA via complementary base pairing.
The 3’ end is known as the acceptor stem, the site for AA attachment.
The clover leaf structure of the tRNA further folds into a 3-dimensional structure, tRNA is able to fit into the ribosome during translation, tRNA transfers the correct AA to the ribosome during translation.

Question 10

What is the semi-conservative model and what is the evidence?

Answer 10

proposes that that the double helix parental molecule replicates and each of the daughter molecules has a parental strand and a newly synthesised strand.
The two strands of the parental molecule separate and each strand acts as template for synthesis of newly synthesised strand.
Each daughter molecule has one strand conserved from the parental molecule and the other newly synthesised strand.
semi-conservative replication, DNA of first generation would be of intermediate density as all DNA molecules comprise one 15N strand and one 14N strand.
Half of DNA molecules from second generation would be of intermediate density and half of DNA molecules would be of light density as 50% of DNA molecules Comprise one 15N strand and one 14N strand and 50% of DNA molecules comprise two 14N strands.

Question 11

What is the unwinding stage of DNA replication?

Answer 11

A portion of double helix is unwound and unzipped at the origin of replication by DNA helicase. As helicase moves along the double helix just in front of the DNA polymerase, the two parental strands are separated by breaking hydrogen bonds between nitrogenous bases. Helicase uses energy from ATP to unwind and unzip the DNA helix
Replication begins at a specific sequence of nucleotides along the DNA molecule called origins of replication.
As the two DNA strands separate, a replication bubble is formed with two replication forks, a Y-shaped region at the two ends of the bubble where the parental molecule is being unzipped.
DNA replication proceeds in both directions from each origin of replication. For a prokaryotic chromosome, only a single origin of replication is present. However, a eukaryotic chromosome may have multiple origins of replication. Multiple replication bubbles form and eventually fuse, thus speeding up the copying of the very long DNA molecules.
Each strand is bound and stabilised by single-stranded binding proteins, preventing them from rewinding behind the replication fork.
The unwinding of double helix causes tighter twisting and strain in front of the replication fork resulting in a positive supercoil. DNA topoisomerase introduces a break in a single strand, thus allowing the strand to rotate around the break, and reseals the strand, eliminating the positive supercoil in front of the replication fork.
Each parental strand acts as a template for the synthesis of daughter strand. However, DNA polymerase cannot initiate the synthesis of a polynucleotide; they can only add nucleotides to the end of an already existing chain.

Question 12

What is the priming stage of the DNA rep?

Answer 12

RNA primers are short segments of RNA of about 5 to 10 nucleotides required for DNA polymerase to initiate elongation as the RNA primers provide the 3’OH end.
Prìmase (a specialized RNA polymerase) binds to the single-stranded DNA template and synthesises RNA primers in the 5 to 3 direction.
Ribonucleotides are added one at a time via complementary base pairing, using the parental DNA strand as a template on both sides of a replication fork.
The single-stranded binding proteins are displaced where the RNA primers are.

Question 13

What is the elongation stage of DNA replication?

Answer 13

Before the start of DNA replication, free deoxyribonucleotides are synthesised in the cytoplasm and transported into the nucleoplasm via nuclear pores.
DNA polymerase adds deoxyribonucleoTides (more specifically deoxyribonucleoSide triphosphates) to the free 3’OH end of the RNA primer as its active site is specific for the -OH group on the nucleotide.
DNA polymerase catalyses the synthesis of a new strand of DNA in the 5 to 3 direction via conmplementary base pairing. A=T and G=C with the parental strand.
DNA polymerase catalyses the formation of a phosphodiester bond between the 3OH end of the primer and 5’ phosphate group of the dNTP added. The energy for this process comes from the two phosphate groups which are removed as the nucleotide joins the growing end.
As the two parental DNA strands are antiparallel and polymerization proceeds in the 5’ to 3’ direction on both sides of the replication fork, One daughter strand is synthesised towards the replication fork (leading strand).
Another daughter strand is synthesised away from the replication fork (lagging strand).

Question 14

What is the leading strand?

Answer 14

The leading strand is synthesised continuously (in the 5’ to 3 direction) towards the replication fork.
Only one primer is required for DNA polymerase to synthesise the leading strand per replication fork.

Question 15

What is the lagging strand?

Answer 15

The lagging strand is synthesised discontinuously vía a series of Okazaki fragments (in the 5 to 3 direction) away from the replication fork.
Each Okazaki fragment needs to be primed separately.
The RNA primers are excised and replaced with deoxyribonucleotides by another DNA polymerase.
The adjacent Okazaki fragments must be linked together. The 3’OH end of one fragment is adjacent to the 5 phosphate end of the previous fragment.(DNA ligase catalyses the formation of phosphodiester bond between the two Okazaki fragments)

Question 16

What is the termination stage of DNA replication?

Answer 16

The product of replication is thus two daughter DNA molecules formed from one original parental DNA molecule. Each of the daughter DNA molecules is identical to the original parental molecule.
Each daughter molecule contains one strand conserved from the parental molecule and one newly synthesised strand.

Question 17

What is the end-replication problem in DNA replication?

Answer 17

When a eukaryotic linear DNA molecule is replicated during the S phase, the DNA polymerase complex is unable to completely replicate to the end of the chromosome.
DNA polymerase requires a free 3OH group to add deoxyribonucleotides to.
The RNA primers at the 5’ end of the newly synthesized strands are excised but cannot be replaced due to the absence of 3’OH for the polymerisation reaction.
The newly synthesized strand has a shorter 5’ end due to the removed primers, while the parental template strand at the 3’ end is longer.
Hence it is often described as a single-stranded 3’ overhang.
As a result, repeated rounds of DNA replication produce shorter and shorter DNA molecules.

Question 18

What is the importance of base-pairing and H bonding in DNA?

Answer 18

In the DNA double helix, the nitrogenous bases are held together by hydrogen bonds two hydrogen bonds between adenine and thymine, and three hydrogen bonds between cytosine and guanine. Hydrogen bonds together with hydrophobic interactions between the stacked bases, stabilise the structure of the double helix. The adjacent nucleotides within each polynucleotide strand are held together by strong covalent phosphodiester bonds, which are not easily broken. In this way. the integrity of the DNA base sequence is maintained.
DNA replication
DNA is the hereditary material and thus the structure of DNA has to allow for its own replication. It is essential that DNA replicates itself accurately prior to mitosis so that the daughter nuclei have identical copies of DNA as the parent nucleus. The double helical structure of DNA enables semi-conservative replication to occur. During replication, the two parental strands separate and each strand acts as a template for synthesis of new strand via complementary base pairing. Each original DNA molecule will give rise to two daughter DNA molecules with identical base sequence. DNA polymerase might improve the specificity of complementary base pairing at two stages: scrutinise the incoming nucleotide for the proper complementarity with the template (pre-synthetic error control). scrutinise the nucleotide against the template as soon as it is added to the growing strand and DNA polymerase removes the incorrectly paired nucleotide and resumes synthesis (proofreading).
DNA is subjected to environmental factors that can cause changes in base sequence, i.e. mutations. The structure of DNA allows for a repair mechanism to operate in the event of such a mutation. In the event of a mutation, the intact complementary strand can be used as a template to guide the repair by DNA repair enzymes. Such repair mechanisms ensure that the integrity of the base sequence of the DNA molecule is maintained.

Question 19

What is the initiation stage of transcription?
(prokaryotes and eukaryotes)

Answer 19

Prokaryotes: Sigma factor of RNA polymerase recognises and binds to the double stranded DNA at the both the -35 and -10 sequences of promoter. The prokaryotic promoter has two consensus sequences located at 35 and 10 base pairs upstream of the transcription start site. The -10 sequence is a consensus sequence of 5-TATAAT-3 known as a Pribnow box. Sigma factor is then released from the core enzyme. RNA polymerase unwinds and separates the two strands of DNA by breaking hydrogen bonds between bases. The template strand is available for complementary base pairing with ribonucleotides.
Eukaryotes: TATA binding protein (TBP) recognises and binds to the TATA box of promoter
General transcription factors and RNA polymerase are recruited to form the transcription initiation complex. RNA polymerase unwinds and separates the two strands of DNA by breaking hydrogen bonds between bases. The template strand is available for complementary base pairing with ribonucleotides.

Question 20

What is the elongation stage of transcription?

Answer 20

In prokaryotes and eukaryotes,
RNA polymerase adds ribonucleotides (more specifically ribonucleoside triphosphates, rNTP), to the free 3’OH end of the growing RNA chain
RNA polymerase catalyses the synthesis of a new strand of RNA in the 5 to 3 direction via complementary base pairing with the template strand.
RNA polymerase catalyses the formation of a phosphodiester bond between the 3OH end of the RNA and 5 phosphate group of the NTP added.
The energy for this process comes from the two phosphate groups which are removed as the nucleotide joins the growing end.
As the enzyme moves, it separates the DNA helix to expose a new segment of the template strand.
A short RNA-DNA hybrid is formed in the unwound region. Behind the unwound region, the DNA template strand pairs with its non-template strand to refom the double helix. The RNA emerges as a free single strand.

Question 21

What is the termination stage of transcription?

Answer 21

For prokaryotes,
Termination occurs after a terminator sequence found on the DNA template strand is transcribed. The short RNA-DNA hybrid is separated, releasing the newly synthesised RNA transcript and RNA polymerase. No further modification is required before translation
For eukaryotes,
Termination occurs after the terminator sequence, polyadenylation signal sequence found on the DNA template strand. is transcribed It codes for a polyadenylation signal sequence (AAUAAA) in the pre-mRNA At a point about 10-35 nucleotides downstream from the AAUAAA sequence, pre mRNA transcript is cleaved, releasing the pre-mRNA and RNA polymerase After the pre-mRNA is released, post-transcriptonal modification occurs before the mature mRNA is transported to the ribosomes in cytoplasm via the nuclear pore.

Question 22

What is the 5’ Cap of RNA modification?

Answer 22

Structure:
Methyl guanosine nucleoside triphosphate is added to the first nucleotide by a 5 -5 triphosphate linkage. This structure is called a 5 methylguanosine cap Addition of the 5 methylguanosine cap is catalysed by a nuclear enzyme, Quanylyl transferase. 5’ cap is added after transcription of the first 20-40 nucleotides.
Function:
Facilitate the export of the mature mRNA from the nucleus into the cytoplasm. Protects the mRNA from 5 exonucleases (enzymes that hydrolyses phosphodiester bonds between nucleotides from the end of polynucleotide chain, hence conferring stability to mRNA. Facilitates in binding of ribosomes to mRNA.

Question 23

What is the 3’ Poly-A-Tail?

Answer 23

Structure:
The polyadenylation signal sequence, AAUAAA is a signal for transcription termination and polyadenylation of the 3’end of mRNA About 200 adenine residues are added to the 3’ end of the pre-mRNA called the 3 poly-A tail. Addition of poly-A tail is catalysed by the enzyme poly-A polymerase. The poly-A sequence is not coded in the DNA but is added to the RNA in the nucleus after transcription.
Function:
Facilitate the export of the mature RNA from the nucleus into the cytoplasm. Pre-mRNA 5 Cap.
Slow down degradation by 3 exonucleases. The longer the poly-A tail, the longer is the half-life of mRNA.

Question 24

What is RNA splicing and small nuclear riboproteins?

Answer 24

RNA Splicing is the removal of introns and joining of the exons together.
Non-coding sequences on mRNA are called introns. They are interspersed between coding sequences called exons.
Exons are eventually translated into amino acid sequences (except UTRS of the exons at the ends of the RNA.)
The introns are removed from the pre-mRNA and the exons are ligated together, forming an mRNA molecule with a Continuous coding sequence.
Small nuclear ribonucleoproteins (snRNPs) are located in the nucleus and comprise RNA and proteins. Specific snRNPs recognises and bind to 5 splice site and the 3 splice site. Additional proteins interact with snRNPs to form spliceosome. As the exons are brought closer together, intron will loop. Spliceosome excise the introns and join the exons that flanked the intron. releasing introns in a lariat structure (which is subsequently degraded).

Question 25

What is alternative splicing?

Answer 25

When pre-mRNAs are processed by joining exons in different combinations, different mature mRNAs are produced from the same pre-mRNA.
Thus, a single gene can code for more than 1 kind of polypp.
It increases number and variety of proteins in the cell without increasing genome size.

Question 26

What is the characteristics of the Genetic Code?

Answer 26

1. Triplet: Three consecutive nucleotide bases specify for one amino acid
2. Degenerate / redundant : More than one codon can code for the same amino acid The codons that code for the same amino acid usually vary in the 3rd base
3. Unambiguous : No codon specifies for more than one amino acid
4. Unpunctuated:There are no gaps between adjacent codons
5. Non-Overlapping : Successive triplets are read in order with no overlapping of nucleotide bases
6. Universal: The genetic code is common to almost all organisms, although there are a number of exceptions to this rule, particularly in unicellular eukaryotes and organeles genes (e.g. mitochondrion) of some species.

Question 27

What is amino acid activation?

Answer 27

Each amino acid is joined to the correct tRNA by a specific enzyme called an aminoacyl-tRNA synthetase.
The active site of aminoacyl-tRNA Synthetase recognises and binds to a specific pair of amino acid and tRNA.
There are at least 20 different synthetases, one for each amino acid.
The synthetase catalyses the covalent attachment of an amino acid to the 3’ acceptor stem of the corresponding tRNA.
Each tRNA has a specific anticodon which binds to the codon on mRNA by complementary base-pairing. This process is driven by energy from the hydrolysis of ATP.
The resulting aminoacyl tRNA, also called an activated amino acid, is released from the enzyme and delivers its amino acid to a growing polypeptide chain on a ribosome. This process is genetically controlled and takes place continuously in the cytoplasm.

Question 28

What is the pairing of codon and anticodon?

Answer 28

The second recognition step involves correct complementary base pairing between the tRNA anticodon and an mRNA codon. If one tRNA existed for each mRNA codon specifying an amino acid, there would be 61 tRNAs. However, there are only about 45, meaning some tRNAs must be able to bind to more than one codon. The rules for base pairing between the third nucleotide base of a codon and the corresponding base of a tRNA anticodon are not as strict. This relaxation of the base-pairing rules is called wobble effect. The wobble effect explains why the synonymous codons for a given amino acid can differ in their third base, but not usually in their other bases. For example, the base G at the 5 end of a tRNA anticodon can pair with either C or U in the third position of an mRNA codon.

Question 29

What is the initiation stage of translation?

Answer 29

small ribosomal subunit binds to both 5’ end of mRNA and a specific initiator tRNA, which caries the amino acid methionine, with the aid of proteins known as initiation factors.
Initiator tRNA with anticodon, 3-UAC-5 carries N-formyl methionine in prokaryotes. Initiator tRNA with anticodon, 3-UAC-5’ carries methionine in eukaryotes.
The small subunit scans, downstream along the mRNA until it reaches the start codon, AUG, which signals the start of translation; this establishes the start of a reading frame for the mRNA. Hydrogen bonds are formed between the complementary bases of the methionine-tRNA anticodon and the start codon on A large ribosomal subunit attaches and binds to mRNA, forming the translation initiation complex. At the completion of the initiation process, the initiator tRNA sits in the P (peptidyl) site of the ribosome, and the vacant A (aminoacyl) site is ready for the next aminiacyl tRNA. A polypeptide is always synthesized in one direction, from the initial methionine at the amino end (i.e. N-terminus), toward the final amino acid at the carboxyl end (i.e. C-terminus).

Question 30

What is the elongation stage of translation?

Answer 30

Amino acids are added one by one to the preceding amino acid via formation of peptide bonds. Codon recognition:
The anticodon of an incoming aminoacyl tRNA complementary base pairs with mRNA codon in A site
Energy released by hydrolysis of GTP is required for codon recogition.
Peptide bond formation:
Peptidyl transferase, found in a section of the rRNA molecule of the large subunit catalyses the formation of a peptide bond
Peptide bond is formed between the new amino acid in the A site and the carboxyl end of the growing polypeptide in the P site.
This step transfers the polypeptide to the tRNA in the A site
Translocation:
The ribosome translocates by advancing 3 nucleotides
The tRNA carrying the polypeptide in the A site is now found in the P site The empty tRNA in the P site is now in the E (exit) site, where it is released
The ribosome then has an empty A site ready for entry of the amino acyl tRNA Corresponding to the next codon on the RNA. Energy released by hydrolysis of GTP Is required for translocation

Question 30

What is the termination stage of translation?

Answer 30

Elongation continues until a stop codon in the mRNA reaches the A site of the ribosome.
The base triplets UAG, UAA and UGA do not code for amino acids but instead signal to stop translation. A protein called a release factor binds directly to the stop codon in the A site.
The release factor causes the addition of a water molecule instead of an amino acid to the polypeptide chain. This reaction hydrolyses the completed polypeptide from the tRNA in the P site. releasing the polypeptide through the exit tunnel of the ribosome’s large subunit.
The translation assembly then comes apart. Energy released by hydrolysis of GTP is required.

Question 31

What is the polyribosome?

Answer 31

A single mRNA can be used to make many copies of a polypeptide simultaneously because several ribosomes can translate the message from one mRNA at the same time.
Once a ribosome moves past the start codon, a second ribosome can attach to the mRNA; thus, a number of ribosomes can translate one mRNA at the same time Such strings of ribosomes are called polyribosomes /polysomes
They are found in both prokaryotic and eukaryotic cells. enabling a cell to make many copies of a polypeptide rapidly