2.1 DNA Structure And Replication Flashcards
Components of deoxyribonucleotide vs ribonucleotide (3)
1) deoxyribose vs. ribose (pentose) sugar
2) nitrogenous base (ATCG vs. AUCG)
3) phosphate group
Pentose sugars occur as ____ forms
Ring
In nucleic acids, the 5’ carbon of the pentose sugar is linked to the ________ ____ in an ____ bond
Phosphate group; ester
In nucleic acids, the 1’ carbon of the pentose sugar is linked to the ___________ ____ in an _________ bond.
Nitrogenous base; glycosidic
Difference in structure of deoxyribose VS ribose
At the 2’ carbon, deoxyribose has a hydrogen atom, instead of a hydroxyl group like in ribose
Properties pentose sugars confer on the nucleotide (2):
1) Provides OH groups that contribute to its hydrophilicity
2) Allows for the addition of nucleotides due to a free 3’ OH group
Explain why there is a difference in structure between RNA and DNA (hint: helix)
RNA is a less tightly coiled helix as DNA, as the partial neg. charge of the -OH grp in ribose repels the neg. charge of the phosphate
S->Property of RNA
As RNA exists as a less tightly coiled helix, it is more susceptible to chemical and enzyme degradation
2 types of nitrogenous bases
Purines (AG)
Pyrimidines (CT/U)
Structure of purines (hint: rings)
Have a 6 membered (nitrogen-containing) ring fused to a 5 membered ring
Structure of pyrimidines (hint: ring)
Have a 6-membered ring
Pure As Gold
Purines — Adenine, Guanine
Pyrimidines are CUT from purines
Pyrimidines — cytosine, uracil, thymine
— one ring
Thymine: the meTHYl is MINE
Add. Methyl substituent at C5
Nucleoside =
pentose sugar + nitrogenous base
Glycosidic bond: the __’ carbon of the pentose is linked to the nitrogenous base via a __________ ______.
1; condensation reaction
Nucleotide =
Nucleoside + phosphate grp
A ________ bond is formed between the _’ carbon of the pentose and the ________ _____, in a __________ _______.
Phosphoester; 5; phosphate group; condensation reaction
dATP
deoxyadenosine triphosphate
GMP
guanosine monophosphate
The _’ phosphate group of one nucleotide and the _’ hydroxyl group of the other is joined in a ___________ bond via _________.
5, 3, phosphodiester, condensation
Phosphodiester bonds between 5’ phosphate and 3’ hydroxyl group of nucleotides form a ________, ________, sugar-phosphate backbone.
Linear, unbranched
Role of phosphodiester bonds (2)
1) They confer strength and stability to the polynucleotide chain, as they are strong covalent bonds
2) prevents breakage of chain during DNA replication
5’ end of DNA/RNA chain:
Free 5’ carbon carrying a phosphate group
3’ end of DNA/RNA chain:
Free 3’ carbon carrying a hydroxyl group
DNA consists of two a_________ ___________ _____, which ____ around each other to form a _______ _____.
Antiparallel polynucleotide strands; coil; double helix
Antiparallel
One strand is oriented in the 5’ to 3’ direction while the other in the 3’ to 5’ direction.
Significance of antiparallel SP backbone
1) DNA polymerase moves in opposite directions along the two strands, as it only adds nucleotides in the 5’ to 3’ direction.
2) allows for complementary base pairs to fit together.
Each strand forms a _____-_______ helix
Right-handed
Diameter of helix is
Uniformly 2 nm
The centre of the double-helix has enough space for
1 purine (AG) and 1 pyrimidine (CT)
Orientation of phosphate grps
Project outside the double helix, as they are hydrophilic
Orientation of nitrogenous bases
They orientate inwards toward the central axis at almost right angles, as they are hydrophobic
The bases of opposite strands are bonded tgt by _________ _____, according to rules of _____________ ____ ______.
Hydrogen bonding; complementary base pairing
s______ complementary base pairing occur between &, and between &
A and T (2 H bonds)
C and G (3 H bonds)
Significance of complementary base pairing
The base sequence in one strand determines the base sequence in the complementary strand
S->P: weak hydrogen bonds
They make it relatively easy to separate the two strands of the DNA e.g. by heating
0.34nm
Distance between Base pairs stacked along The Central axis.
The double helix makes a complete turn every
10 base pairs
Each turn is ____nm
3.4nm
Definition of major and minor grooves, and their S—>Property
Grooves of unequal sizes between sugar-phosphate backbones, which are large enough to allow proteins to gain access and make contact with the bases
Combination of bases assuming x no. Of nucleotides
4 to the power of x
The linear sequence of the four bases can be ______ in countless ways.
Varied
Due to the wide variety of linear bp sequence, each gene
Has a unique base sequence, which contains information to make a unique protein
Function of DNA
Stores and transmits genetic information
P->F of DNA: store of genetic information
The base sequence is stable and invariant, making DNA a stable store of genetic information (relatively resistant to spontaneous changes)
S->P: stability of DNA double helix (both prok. and euk.) (4)
1) extensive hydrogen bonds between bp
2) hydrophobic interactions between stacked bp
3) Only the sugar-phosphate backbone is exposed to outside influences
4) nitrogenous bases are safely tucked inside double helix
S->P: stability of double helix (euk. only)
DNA double helix is tightly wound around histones to form repeating array of nucleosomes, which are eventually folded into chromosomes. This protects DNA from thermal and physical damage (prevents degradation)
S->P: invariant base sequence (3)
1) Structure: there is complementary base pairing between DNA strands
2) Hence, genetic info is redundant (helps DNA maintain integrity)
3) A cell can discard a damaged strand with an altered base sequence, and synthesise a new strand using the remaining strand as a template, using rules of complementary base pairing.
Summary of how add. of nucleotides works (enzyme, bonds, reaction) (3)
1) DNA pol. reads the template DNA strand
2) It catalyses the addition of [nucleotide], based on rules of complementary base pairing
2) Phosphodiester bonds are formed via condensation reaction between the free 3’ OH group of the last nucleotide and 5’ phosphate group of the incoming nucleotide
Semi-conservative model of DNA replication (2)
1) Parental DNA strands separate and each acts as a template of the synthesis for a new DNA strand by CBP.
2) Each of the two daughter DNA molecules consist of one parental DNA strand and one newly-synthesised daughter DNA strand
Conservative model of replication
Parental DNA molecule emerges from the replication process intact, and generate DNA copies consisting of entirely new molecules
Dispersive model of replication
All four strands of DNA in daughter DNA molecules have a mixture of parental and new DNA
Semiconservative model: Percentage of DNA after 1st and 2nd replication
1st: all hybrid DNA
2nd: half light DNA and half hybrid DNA
Conservative model: Percentage of DNA after 1st and 2nd replication
1st: half light DNA and half heavy DNA
2nd: 3/4 light DNA and 1/4 heavy DNA
Dispersive model: Percentage of DNA after 1st and 2nd replication
1st and 2nd: all hybrid DNA
Characteristics of DNA replication process (5)
1) Complex and precise: double helix must unwind while the DNA pol. assembles the new antiparallel strands simultaneously
2) Fast: few hours to copy ~3x10^9 base pairs
3) Accurate: mutation rate is ~1 nucleotide change per 10^9 nucleotide each replication
4) Requires cooperation of large team of enzymes and other proteins
5) requires expenditure of ATP
Where does DNA replication begin
Origin of replication
OriR structure and composition
Each oriR is a specific sequence of nucleotides, which is generally A-T rich
Why is oriR A-T rich?
Only 2 H bonds between A-T base pair, hence it is easier for helicase to disrupt the bonds as less energy is needed to overcome them
_________ ________ recognise the oriR sequence and bind to it. The DNA double helix _________ into two strands, forming a __________ _______.
Initiator proteins; separates; replication bubble
Replication fork (structure, location, process there)
Y-shaped structure at each end of the replication bubble, where new strands of DNA are synthesised
Two replication forks _____ ____ from the oriR as replication proceeds ____________, until the entire DNA molecule is ________.
Move away; bidirectionally; separated
DNA replication in prokaryotes (DNA, oriR, directional)
1) small circular DNA molecule
2) single oriR
3) proceeds bidirectionally to a termination site ~halfway around the circular chromosome
Diff between prok. and euk. (shape of DNA, oriR)
1) small circular DNA molecule VS linear DNA molecule
2) one oriR VS multiple oriR
Replication bubbles expand ________, as DNA replication proceeds __________, and they eventually _____, thus ending DNA replication.
Laterally; bidirectionally; fuse
Advantage of multiple oriR
Speed (it speeds up copying of very long DNA molecule)
Would have taken 100 times longer with one oriR
After initiation, helicases bind to __ ______ of the DNA molecule
1 strand
Function of helicase
To unwind the DNA double helix and separate the parental DNA strands
Mechanism of helicase
They break the hydrogen bonds (between complementary bases) holding the two DNA strands together, using ATP as an energy source
Function of single-strand DNA binding proteins (SSB proteins) (2)
1) They bind to and stabilise the unwound ss portion of the DNA double helix, which prevents the ss DNA from reannealing to reform the double helix
2) straighten the DNA chain, facilitating DNA replication
Significance of stability of ss portion of DNA double helix (2)
1) Keeps the two parental strands in the appropriate ss condition to act as template for synthesis of new DNA strand
2) protects the ssDNA, which is very unstable, from being degraded
Mechanism of Topoisomerase
Cleaves a strand of the helix to create a transient ss nick
Why is topoisomerase needed? (Problem, solution)
1) Unwinding causes tighter twisting ahead of the RF, resulting in strain
2) Cleaving relieves strain on the DNA molecule by allowing free rotation around the intact strand, before the broken strand is resealed
Limitations of DNA polymerase (2)
1) DNA polymerase cannot initiate DNA synthesis independently, as they require a free 3’ OH end
2) They only add dNTPs to the free 3’ end of a growing DNA strand => a DNA strand can only elongate in the 5’ to 3’ direction
To initiate synthesis of a DNA strand, an ____ ______ is needed.
RNA primer
A portion of the parental DNA strand serves as a _______ for making the RNA ______ with the _____________ ____ ________.
template; primer; complementary base sequence
Primase function
Joins ribonucleotides (about 10 nucleotides long) to synthesise the primer
Significance of RNA primer
It provides a free 3’ OH end that DNA polymerase can extend
Function of DNA pol. I
It has 5’ to 3’ exonuclease activity and it later replaces the RNA nucleotides of the primer with dNTPs
Direction in which DNA polymerase reads the template DNA
3’ to 5’
DNA pol. assembles the dNTPs for the new daughter DNA strand based on
Complementary base-pairing
Describe how DNA pol. 3 conducts proofreading
When incorrect base pair is recognised, DNA pol. reverses its direction by one base pair (3’ to 5’) and excises the b.p. using 3’->5’ exonuclease activity
DNA pol. catalyses [general]
phosphodiester bond formation between growing DNA daughter strand and incoming nucleotide
DNA pol. catalyses [specific]
phosphoester bond formation between free 3’OH group of the last nucleotide in the growing daughter strand and the free 5’ phosphate group of an incoming dNTP
Why can synthesis of daughter DNA strands only occur in the 5’ to 3’ direction? (hint: structure of DNA pol.)
Active site of DNA pol. 3 has specific 3D conformation that is complementary only to 3’OH group of the growing polynucleotide strand
incoming dNTP loses a __________ group when they form a phosphoester bond with the growing DNA daughter strand
pyrophosphate
Why is continuous synthesis of both DNA strands (i.e, okazaki fragments needed) not possible? (4)
1) DNA strands are antiparallel
2) DNA pol. can only add nucleotides to the free 3’OH end of a growing DNA strand
3) Thus, elongation of daughter DNA strand can only occur in 5’ to 3’ direction
4) As replication fork opens, it exposes the template for the lagging strand at the 5’ end of the new DNA strand, hence a new primer is needed.
Leading strand synthesis
complementary daughter DNA strand is continuously synthesised in a 5’ to 3’ manner, towards the replication fork (does not require ligase)
Lagging strand synthesis
discontinously synthesised as a series of short Okazaki fragments in a 5’ to 3’ manner, against overall direction of the RF (requires ligase)
No. of nucleotides per Okazaki fragment
100-200
How are Okazaki Fragments linked to form a continuous DNA strand? (DNA pol. and DNA ligase)
1) DNA pol. replaces RNA primer with dNTPs
2) DNA ligase catalyses formation of phosphoester bond between 3’ OH end of Okazaki fragment and 5’ end of growing daughter DNA strand
Describe the end replication problem.
occurs when the DNA pol. is incapable of completely replicating the ends of linear chromosomes, resulting in the shorter telomeres / 3’ overhang
Reason for end replication problem
DNA polymerase can only extend in 5’ –> 3’ direction. When FINAL RNA primer of the terminal Okazaki fragment is removed by DNA pol., there is no upstream strand with a free 3’ OH grp available to which it can add dNTPs to
After first round of replication…
the newly synthesisesd daughter DNA strands are shorter than its template strand
Chargaff’s Rules
amount of A=T and amount of G= C, as adenine and thymine are complementary, and same goes for cytosine and thymine
Base composition of DNA is constant (__________ _______ ____) for that organism and is characteristic for a (______ _______).
throughout somatic cells; certain species
How does the structure of DNA enable semi-conservative replication?
Both strands can act as templates, for the assembly of nucleotides based on CBP.
Thus, each DNA molecule consists of one parental strand and one new strand
