DNA and Genomics

Question 1

state the structure of a nucleotide

Answer 1

consists of pentose (5-carbon) sugar, phosphate group and the nitrogenous base

Question 2

describe the differences in pentose sugar in RNA and DNA

Answer 2

• the pentose sugar is deoxyribose in DNA, whereas it is ribose in RNA
• difference between a deoxyribose and a ribose lie in the functional group attached to the 2’ carbon atom. a hydrogen atom is found at the 2’ carbon atom in deoxyribose, while a hydroxyl group is found at the 2’ carbon in ribose

Question 3

describe the structure and function of phosphate groups in DNA

Answer 3

• attached to the 5’ carbon atom of the pentose sugar
• gives nucleic acids an acidic characteristic, making DNA negatively-charged

Question 4

describe nitrogenous bases in DNA and RNA

Answer 4

• attached to 1’ carbon atom of pentose sugar
• purines (two fused rings) and pyrimidines (one-ringed structures)
• DNA contains adenine and guanine (purines), cytosine and thymine (pyrimidines)
• RNA contains adenine and guanine (purines), cytosine and uracil (pyrimidines)
• Twice Pure As Gold: 2 ring, purines, adenine, guanine
• CUT the Pye: 1 ring, pyrimidines, cytosine, uracil, thymine

Question 5

describe the formation of nucleotides

Answer 5

• pentose sugar, phosphate group and nitrogenous base are linked together to form a nucleotide due to condensation reactions
• nitrogenous base is joined to 1’ carbon atom of pentose sugar molecule to form a nucleoside (occurs with loss of water molecule during condensation reaction, glycosidic bond formed)
• phosphate group attached to 5’ carbon atom of sugar molecule forms nucleotide (occurs w elimination of a water molecule during condensation reaction, phosphoester bond formed)

Question 6

describe the formation of polynucleotides

Answer 6

• condensation reaction between 5’ phosphate group of one nucleotide and 3’ hydroxyl group of another to form a phosphodiester bond, two nucleotides join to form a dinucleotide
• this repeats to form a polynucleotide
• phosphodiester bonds between 5’ and 3’ carbon atoms of sugars form a linear, unbranched sugar-phosphate backbone
• sugar-phosphate backbone is made up of alternating sugar and phosphate groups, with nitrogenous bases projecting from the backbone
• 5’ end of DNA polynucleotide ends with phosphate group attached to 5’ carbon atom of sugar of first deoxyribonucleotide
• 3’ end of DNA polynucleotide ends with -OH group of 3’ carbon atom of sugar of last deoxyribonucleotide

Question 7

describe the structure of DNA

Answer 7

• two polynucleotide chains twisted around each other to form a double helix
• two strands are anti-parallel (run in opposite direction): one strand runs in the 3’ to 5’ direction while other strands runs in the 5’ to 3’ direction
• two strands held tgt by hydrogen bond between nitrogenous bases of opposite strands. a purine will always pair with a pyrimidine (complementary base pairing)
• two hydrogen bonds formed between adenine and thymine, three hydrogen bonds formed between cytosine and guanine
• ratio of A to T is 1:1 and ratio of C to G is 1:1
• constant width between two sugar-phosphate backbones of DNA = width of purine and pyrimidine (2nm)
• one complete turn of DNA comprises 10 base pairs and spans a length of 3.4 nm
• sugar-phosphate backbones of both strands face outwards of DNA, while nitrogenous bases occupy centre of molecule (hydrophobic)
• hydrophobic interactions between stacked nitrogenous bases contribute to stability of double helical structure

Question 8

describe the packing of DNA

Answer 8

• in eukaryotes, DNA is associated with histone proteins to form chromatin
• most DNA is wound around histone octamers to form nucleosomes. histone proteins have long tails which are rich in lysine (positively-charged)
• DNA and histones are held together by ionic bonds. remainder of DNA (linker DNA) joins adjacent nucleosomes, nucleosomes and linker DNA form ‘beads-on-a-string’ structure
• nucleosomes coiled into a solenoid structure, forming the 30nm chromatin fibre (six nucleosomes in one turn of solenoid)
• 30nm chromatin fibre forms loops anchored to a protein scaffold
• further coiling of 30nm chromatin fibre gives rise to metaphase chromosome

Question 9

explain the importance of complementary base pairing in DNA replication

Answer 9

• essential for DNA to replicate itself accurately prior to nuclear division so daughter nuclei have genetically idental copies of DNA as original parental nucleus
• double helical structure of DNA enables semi-conservative replication to occur
• during replication, two strands separate
• both strands act as templates to which a complementary set of nucleotides will attach by complementary base pairing via hydrogen bonding
• original parental DNA gives rise to 2 daughter DNA molecules with identical base sequences

Question 10

explain the importance of complementary base pairing in DNA repair

Answer 10

• DNA constantly subjected to environmental factors that cause mutations
• structure of DNA allows for a repair mechanism to operate in the event of mutation
• in event of mutation, the intact strand is used as a template to guide the repair via complementary base pairings
• ensure base sequence of DNA remains unchanged

Question 11

explain the importance of complementary base pairing in transcription of DNA

Answer 11

• RNA polymerase can break the hydrogen bonds between nitrogenous bases to unzip the DNA molecule, allowing for transcription
• one polynucleotide DNA strand can act as a template strand for the synthesis of a complementary strand of RNA

Question 12

state the semi-conservative mechanism of DNA replication

Answer 12

• by semi-conservative mechanism, both strands of a DNA molecule are separated and act as templates for the synthesis of two new daughter strands
• hydrogen bonds form between the bases of one old strand and one newly-synthesised daughter strand to form a daughter DNA molecule

Question 13

state the process of semi-conservative DNA replication

Answer 13

1. before start of DNA replication, free nucleotides are synthesised in the cytoplasm and transported into the nucleoplasm via nuclear pores in nuclear envelope
2. intiation of DNA replication
3. synthesis of new DNA strands

Question 14

describe initiation of DNA replication

Answer 14

• dna replication begins at origins of replication. here, enzyme helicase breaks hydrogen bonds between nitrogenous bases, causing DNA molecule to unwind and unzip. a replication bubble is formed. at each end of a replication bubble is a replication fork. the separated DNA strands are parental strands and serve as templates for synthesis of new DNA strands
• as the DNA molecule unwinds and unzips, supercoils are created ahead of the replication due to the tighter twisting of the double helix. the enzyme, topoisomerase, makes a transient double-stranded break in the DNA molecule ahead of the replication fork and then reseals it by catalysing the formation of a phosphodiester bond, hence preventing the formation of supercoils
• single-stranded DNA binding proteins bind with the separated parental DNA strands to stabilise the separated DNA strands, and prevent them from re-forming the double-stranded DNA molecule (reannealing). it also helps prevent the single-stranded DNA from being degraded
• separated parental DNA strands are now available o act as templates for synthesis of new complementary daughter DNA strands. however, the enzyme that synthesises new DNA, DNA polymerase III, cannot initiate the synthesis of a polynucleotide on its own and can only add deoxyribonucleotides to the free 3’ OH end of a growing DNA strand
• enzyme primase must first catalyse the synthesis of a short RNA chain that is complementary to the DNA template strand. this short RNA chain is known as the RNA primer

Question 15

describe the synthesis of new DNA strands

Answer 15

• DNA polymrease III binds to the 3’ end of the RNA primer, and moves along the parental DNA template strand, reading in a 3’ to 5’ direction, as it adds new deoxyribonucleotides to the RNA primer in a 5’ to 3’ direction. the alignment of incoming nucleotides to the growing daughter DNA strand is not random. it is determined by the sequence of DNA template since complementary base pairing must take place. with an RNA primer anchoring the start of the daughter DNA strand, DNA polymerase III catalyses the polymerisation of the strand. DNA polymerase III catalyses the formation of a phosphodiester bond between a growing daughter DNA strand and an incoming nucleotide
• the RNA primer section is then subsequently hydrolysed by DNA polymerase I. removal of RNA primer is done nucleotide-by-nucleotide, and the gap is then filled with complementary deoxyribonucleotides
• DNA polymerase III only adds nucleotides to the free 3’ end of a growing DNA strand, never to the 5’ end. hence both daughter strands are synthesised in the 5’ to 3’ direction. however, the parental strands of DNA have an antiparallel arrangement, hence, the leading strand is synthesised continuously, while the lagging strand is synthesised discontinuously in the form of short fragments (Okazaki fragments)
• DNA ligase catalyses the formation of a phosphodiester bond between two Okazaki fragments, so that the “nick” between the two fragments are sealed. the enzyme requires a free hydroxyl group at the 3’ end of one DNA strand and a phosphate group at the 5’ end of another. ATP is required for the reaction
• at the end of replication, both the parental and daughter strands rewind into a double helix. each resultant daughter DNA double helix molecule consists of one old parental strand and one new daughter strand
• in prokaryotes and viruses, due to the small genome size, DNA molecules are usually small and only one origin of replication is required. in eukaryotes, DNA molecules of a chromosome is very large and many origins of replication are present. during replication, many short daughter DNA strands are formed in various replication bubbles, and these eventually link together to form a continuous daughter DNA strand. this increases the rate of replicating the entire DNA molecule

Question 16

describe the end-replication problem

Answer 16

• the usual dna replication machinery is unable to complete the 5’ ends of daughter DNA strands
• when the lagging DNA strand is synthesised, many Okazaki fragments are formed
• each fragment has an RNA primer made by primase
• DNA polymerase I removes the primers and fills the gap with complementary deoxyribonucleotides
• however, the gap right at the 5’ end of the lagging strand cannot be filled because of the absence of an existing 3’ end which DNA polymerase I can add nucleotides to
• repeated rounds of replication produce shorter and shorter DNA molecules
• to protect genes at the ends of chromosomes from being eroded through multiple rounds of DNA replication, non-coding DNA known as telomeres are found at the ends of eukaryotic chromosomes

Question 17

define a gene

Answer 17

a specific nucleotide sequence along a DNA molecule that codes for a specific sequence of amino acids ina polypeptide chain, or for a functional RNA molecule (rRNA and tRNA)

Question 18

what is the genetic code?

Answer 18

a set of rules by which the nucleotide sequences of a gene, through the intermediate molecule, mRNA, is translated into the amino acid sequence of a polypeptide

Question 19

what are the 5 features of the genetic code?

Answer 19

• triplet code
• degenerate
• punctuated
• non-overlapping
• universal

Question 20

what is a triplet code

Answer 20

• every amino acid in a protein is coded for by a sequence of three bases along a DNA molecule
• triplets of bases on the corresponding mRNA are called codons, which are complementary to the triplets of bases found on the DNA template

Question 21

what is a degenerate code

Answer 21

• more than one codon can code for the same amino acid
• 64 codons but only 20 different amino acids to be coded for

Question 22

what is a punctuated code

Answer 22

• genetic code is punctuated with start and stop codons, for the initiation and termination of translation respectively
• codon AUG is the start codon, it signals the initiation of translation of an mRNA into a polypeptide, AUG codes for the amino acid methionine
• UAA, UGA, UAG are stop codons, they signal the termination of translation to the ribosome, they do not code for any amino acids

Question 23

what is a non-overlapping code?

Answer 23

nucleotides in the genetic code are read in triplets, and each nucleotide in a codon is only read once

Question 24

what is a universal code?

Answer 24

• same set of triplet base codes for the same amino acid, across all organisms
• universal genetic code allows human genes to be expressed in bacteria. e.g. scientists genetically-engineer bacteria to synthesise human insulin by inserting the human insulin gene into prokaryotes

Question 25

differentiate between prokaryotes and eukaryotes in transcription and translation

Answer 25

• in prokaryotes, both transcription and translation occur in the same location (cytoplasm), within the cell. in eukaryotes, transcription occurs in the nucleus whereas translation occurs in the cytoplasm
• in prokaryotes, the sequence of bases along a gene is transcribed directly onto an mRNA. in the genes of eukaryotes, both coding sequences (exons) and non-coding sequences (introns) are transcried onto a primary RNA transcript (pre-mRNA). pre-mRNA is extensively processed into mature mRNA

Question 26

describe the role of mRNA

Answer 26

• carries genetic information out of the nucleus to the cytoplasm in the form of a specific sequence of codons so that protein synthesis can take place
• synthesised in the nucleus via transcription and is complementary to DNA template strand of gene
• transported from nucleus to cytoplasm through nuclear pores
• serves as a template which is read by ribosome for translation, for the synthesis of a polypeptide
• each codon codes for an amino acid
• codons on mRNA base-pairs with complementary anticodons on tRNA by hydrogen bonds
• specific sequence of codons on mRNA is translated to specific amino acid sequence of polypeptide
• mRNA also contains stop and start codons which ensures translation begins and ends at the correct point

Question 27

describe the role of tRNA

Answer 27

• has a specific 3D conformation, its clover-leaf structure allows for interaction with aminoacyl-tRNA synthetase/large ribosomal subunits
• at the anticodon loop of tRNA, three unpaired bases form the anticodon which undergoes complementary base pairing with the codon on the mRNA
• tRNA carries the amino acid to the ribosome for translation
• 3’ end of tRNA contains the three bases CCA, which is the site of attachment for an amino acid that is specific to the anticodon of the tRNA molecule

Question 28

describe the roles of rRNA

Answer 28

• rRNA folds into a unique 3D conformation, made up of many hairpin loops, as a result of complementary base pairing within the rRNA molecule
• rRNA binds with ribosomal proteins to form the large and small ribosomal subunits. during translation, ribosomal subunits assemble to form the ribosome
• part of rRNA forms peptidyl transferase, which catalyses the formation of peptide bonds between amino acids during translation
• rRNA helps align mRNA in the ribosome, so that anticodons of incoming aminoacyl-tRNA molecules can complementary base pair/form hydrogen bonds with mRNA codons
• rRNA orientates the aminoacyl-tRNA in the correct position within the ribosomal binding sites (A, P, E sites)

Question 29

describe the role of ribosomes

Answer 29

• made up of a small and large subunit, small subunit contains an information processing region while large subunit contains an “A” site, “P” site, and “E” site
• mRNA molecule is usually translated simultaneously by several ribosomes (polyribosomes) forming several identical polypeptides, this increases rate of translation and hence rate of polypeptide synthesis

Question 30

state the sequence of events in transcription

Answer 30

1. initiation
2. elongation
3. termination

Question 31

describe initiation in transcription

Answer 31

• in eukaryotes, proteins called general transcription factors bind to the promotor to facilitate the binding of RNA polymerase. the association of transcription factors and RNA polymerase forms the transcription initiation complex
• RNA polymerase binds to the promotor sequence (region of DNA at start of gene, short sequence that is not entirely transcribed by RNA polymerase – non coding DNA)
• binding of RNA polymerase to the promoter causes the DNA double helix to unwind and unzip
• one of the two exposed strands of DNA, called the template strand, is used as a template for transcription
• the other strand which is not transcribed, non-template strand, have the same base sequence as the mRNA synthesised, thymine is replaced by uracil on mRNA

Question 32

describe elongation in transcription

Answer 32

• RNA polymerase reads the template strand from 3’ end to 5’ end
• free ribonucleotides, which are complementary to DNA bases, bind to the DNA template strand by complementary base pairing
• the new RNA strand is synthesised in the 5’ to 3’ direction
• adjacent ribonucleotides are joined together by phosphodiester bonds
• transcription rate can be increased, as several molecules of RNA polymerase follow one after another on the same gene

Question 33

describe termination

Answer 33

• RNA polymerase transcribes a terminator sequence
• newly-formed mRNA (prokaryotes)/pre-mRNA (eukaryotes) is released, RNA polymerase detaches from DNA, and DNA double helix is reformed

Question 34

describe post-transcriptional modification

Answer 34

enzymes in the eukaryotic nucleus modify pre-mRNA before the final mature mRNA leaves the nucleus

addition of modified guanosine cap to 5’ end
* 5’ end is protected with a 5’ modified guanosine cap
* this modified guanosine cap is a methylated guanosine nucleotide, and cannot be degraded by exonuceases (enzymes which hydrolyse nucleic acids starting from its ends)

RNA splicing
* introns are cut out at splice sites which are located at ends of introns, exons are spliced together by spliceosomes, to form mature mRNA
* necessary because eukaryotes have genes that are made up of both exons and introns

polyadenylation of 3’ end
* 3’ end of pre-mRNA is then polyadenylated with a poly-A tail
* the longer the poly-A tail, the greater the mRNA stability, as it will take a longer time for exonucleases to degrade the poly-A tail

• mNA sequence before the start codon, and after the stop codon are not translatewd into the amino acid sequence of the polypeptide (untranslated regions: UTR)
• 5’ UTR is located before the start codon and the 3’ UTR is located after the stop codon

Question 35

state the sequence of events in translation

Answer 35

amino acid activation, binding of ribosome to mRNA, chain initiation, chain elongation, chain termination

Question 36

describe amino acid activation in translation

Answer 36

• tRNA molecules are attached to specific amino acids by enzymes (aminoacyl-tRNA synthetases), ATP is consumed in the process
• each aminoacyl-tRNA synthetase has an active site, which binds one specific type of amino acid to its tRNA molecule
• an amino acid is attached to a specific tRNA molecule to form an aminoacyl-tRNA. the 3’ end of tRNA ends with the base sequence of CCA. the specific mino acid is attached to the 3’ -OH group of the last ribonucleotide in the tRNA

Question 37

explain how the information on mRNA is converted into polypeptide

Answer 37

• ribosome reads the mRNA sequence from 5’ to 3’ direction and translates it into a sequence of amino acids in a polypeptide, where 1 codon on mRNA specifies for each amino acid on the polypeptide
• aminoacyl-tRNA carrying methionine with anticodon UAC binds to the start codon AUG on mRNA, and is positioned in the P site of the ribosome
• a second aminoacyl-tRNA with anticodon complementary to the next codon on mRNA is positioned in A site
• peptidyl transferase catalyses formation of peptide bond between adjacent amino acids (first amino acid is transferred to tRNA at A site)
• ribosome translocates downstream by 1 codon in a 5’ to 3’ direction
• tRNA at P site is transferred to E site and released and A site is now free to receive the next aminoacyl-tRNA
• steps are repeated till stop codon (UAA, UGA, UAG) is reached, release factor protein will bind and the ribosome disassembles, releasing the polypeptide chain

Question 38

describe binding of ribosome to mRNA

Answer 38

• small ribosomal subunit binds to the eukaryotic mRNA at the 5’ modified guanosine cap. the small subunit then moves downstream, in the 5’ to 3’ direction, in search for the AUG start codon

Question 39

define gene mutation

Answer 39

a permanent change in the nucleotide sequence of a gene

Question 40

describe how gene mutation changes phenotype of an organism

Answer 40

• an alteration in the sequence of nucleotides in a gene will lead to a change in codon sequence in mRNA
• this may change the sequence of amino acids in a polypeptide
• if the physical and chemical properties of the amino acid’s R group differ, this can change the 3D conformation of the protein and its function, and hence, subsequently affects the phenotype of an organism

Question 41

define point mutation

Answer 41

a gene mutation that involves a change in only one base

Question 42

describe base substitution, base insertion and base deletion

Answer 42

• base substitution: replacement of a single nucleotide base pair with another in the DNA sequence
• base insertion: adding of one or more nucleotide base pairs in the DNA sequence
• base deletion: removal of one or more nucleotide base pairs in the DNA sequence

Question 43

describe silent mutation

Answer 43

• due to the degeneracy of genetic code, change in the nucleotide sequence (base substitution) may transform one codon into another codon that codes for the same amino acid. this is especially so, if it involves changing the 3rd base in a codon
• this would have no effect on the amino acid sequence of the encoded protein
• there is no observable change in the organism’s phenotype

Question 44

describe missense mutation

Answer 44

• a change in the nucleotide sequence results in a codon that codes for a different amino acid
• organisms’ phenotype is usually altered

Question 45

describe nonsense mutation

Answer 45

• a change in the nucleotide sequence changes a codon that codes for an amino acid into a stop codon
• translation will be terminated prematurely, and the resulting polypeptide will be truncated, leading to a non-functional protein
• nonsense mutations may be a result of frame-shift

Question 46

describe frame-shift mutation

Answer 46

• deletion or insertion of nucleotides not in multiples of three results in the shifting of reading frame. hence, the ribosomes begin to read incorrect triplets from the point of deletion or insertion
• this often results in the production of a non-functional protein
• frame-shift mutations can have drastic consequences, causing extense missense mutation ro nonsense mutation

Question 47

how does sickle cell anaemia come about?

Answer 47

• normal haemoglobin A consists of two alpha-globin chains and two beta-globin chains, which are coded for by two different genes found on two different chromosomes
• in sickle-cell anaemia, haemoglobin S (HbS) is produced instead
• due to a single base substitution, in the mRNA produced, one of the codons is changed from coding for glutamic acid to coding for amino acid valine
• glutamic acid is hydrophilic whereas valine is hydrophobic

Question 48

consequences of sickle cell anaemia

Answer 48

• at low oxygen concentrations, the hydrophobic portion of HbS molecules interact with each other, causing the HbS molecules to polymerise into fibres. this in turn causes the RBCs to change from a circular biconcave shape to a sickle-shape
• sickle-shaped RBCs interfere with circulation, as blood flow will be slower and the cells may obstruct blood capillaries. this deprives multiple organs of oxygen, resulting in their damage
• these cells are also more rigid and fragile than normal RBCs. thus, they have a shorter life span, as they haemolyse readily resulting in anaemia. they also accumulate in spleen for destruction leading to an enlargement of the spleen

Question 49

explain how aminoacyl-tRNA synthetase is adapted to its function

Answer 49

• has an active site with a 3D conformation that is complementary to both a tRNA with a specific anticodon and its respective amino acid
• holds the tRNA and amino acid in correct orientation for formation of tRNA-amino acid high-energy bond (ester bond) at the 3’ end of tRNA