dna structure and replication Flashcards
(23 cards)
What are the important characteristics of DNA as hereditary material?
a. have high capacity for information storage and be chemically stable to be able to
encode information without fail. It must not change easily due to age, nutrition, or environment.
b. replicate accurately.
c. be capable of variation.
Describe pentose sugars.
They are five-carbon sugars and occur as ring forms. In nucleic acids, the 5’ carbon is linked in an ester bond to the phosphate group and the 1’ carbon is linked in a glycosidic bond to the nitrogenous base.
Describe the difference between ribose and deoxyribose sugars.
At the 2’ carbon of deoxyribose, OH is replaced by H. partial negative charge of the hydroxyl group in ribose repels the negative charge of
the phosphate, preventing the RNA chain from coiling in as tight a helix as it does in DNA. RNA is more susceptible to chemical and enzyme degradation.
Describe the importance of phosphodiester bonds in DNA.
Phosphodiester bonds between 5’ phosphate and 3’ hydroxyl groups of nucleotides form a linear, unbranched sugar-phosphate
backbone. Phosphodiester bonds are strong covalent bonds. They confer strength and stability on the polynucleotide chain. This is
the basis in preventing breakage of the chain during DNA replication.
Describe the main features of DNA.
- DNA consists of two polynucleotide strands / chains. Each strand forms a right-handed
helix and the two strands coil around each other to form a double-helix. - The diameter of the helix is uniformly 2 nm — there is just enough space for 1 purine
and 1 pyrimidine. - The strands run in opposite directions, i.e. they are antiparallel.
- Each strand has a sugar-phosphate backbone with:
a. phosphate groups that project outside the double-helix since they are hydrophilic; and
b. nitrogenous bases that orientate inwards toward the central axis at almost right angles — the relatively hydrophobic nitrogenous bases
in the molecules interior, away from the surrounding aqueous medium. - The bases of the opposite strands are bonded together by relatively weak but extensive hydrogen bonds.
What is the significance of complementary base pairing?
The base sequence in one strand determines the base sequence in the complementary strand. The weak hydrogen bonds make it relatively easy to separate the two strands of the DNA, for example, by heating.
Since the three-dimensional structure of DNA is only stable when the base pairs are complementary, this meant that the base sequence of one strand could determine the base sequence of its complementary strand. This is necessary in DNA replication and transmission of the genetic information stored.
What is the significance of grooves along the DNA molecule?
There are grooves of unequal sizes between the sugar-phosphate backbones called the major groove and minor groove. Both these grooves are large enough to allow
protein molecules to gain access and make contact with the bases.
Why is form complementary base pairs?
- Steric restrictions
a. The sugar-phosphate backbone of each polynucleotide chain has a regular helical structure.
b. The DNA double-helix has a uniform diameter of 2 nm.
c. Purines are about twice as wide as pyrimidines, which have a single ring.
d. The solution is always to pair a purine with a pyrimidine. - Hydrogen Bond Factors
a. Each nitrogenous base has chemical side groups that
can form hydrogen bonds with its
appropriate partner.
b. Such chemical side groups in purines and pyrimidines have well defined positions.
Why does each gene have a unique base sequence?
Variation of linear base sequence: Although the base pairing rules dictate the combinations of nitrogenous bases that form the ‘rungs’ of the double-helix, they do not restrict the base sequence along each DNA strand. The linear sequence of the four bases can be varied in countless ways. eg human beings have 3 x 10^9 nucleotide pairs — 4 to the power of 3 x 109 combinations of bases
How is DNA a stable, invariant storage of genetic information?
DNA is relatively resistant to spontaneous changes (mutations).
Extensive hydrogen bonds between base pairs. Hydrophobic interactions (or ‘stacking forces’) between the stacked base pairs. Exposure to outside influences only the sugar-phosphate backbone; Nitrogenous bases being safely tucked inside the double-helix. (Eukaryotes only: DNA double-helix being tightly wound around histones to form a repeating array of nucleosomes. The nucleosomes are eventually folded into higher order structures such as the chromosome, in which the DNA is prevented from thermal and physical damage.)
Specific, complementary base pairing between DNA strands: genetic information is redundant (i.e. present more than once), if the base sequence in one of the two strands is accidentally altered, the cell discards the damaged strand. It then makes a perfectly good strand by using the remaining intact
strand as a template, following complementary base pairing. The
redundancy of genetic information helps to maintain its integrity.
How can DNA be replicated accurately?
- The two strands of DNA are complementary – each stores the information necessary to
reconstruct the other. - When a cell copies a DNA molecule, each strand serves as a template for ordering
nucleotides into a new, complementary strand. - Where there was one double-stranded DNA molecule at the beginning of the process, there
are now two – each an exact replica of the ‘parent’ molecule to ensure faithful transmission of genetic instructions.
Describe the 3 models of DNA replication.
Conservative: Parental
DNA molecule emerges from the
replication process intact, i.e. it is
conserved, and generates DNA
copies consisting of entirely new
molecules.
Dispersive: All four
strands of DNA following replication
have a mixture of old and new DNA.
Semi-conservative: Each parental DNA strand acts as a template for the assembly of a complementary strand. Each of the two daughter DNA molecules consists of one parental DNA strand and one
newly-synthesised daughter DNA strand.
How does the Meselson-Stahl experiment prove that DNA replication is semi-conservative?
1st generation: All hybrid DNA, intermediate in density [14N/15N
DNA] 2nd generation: 50% hybrid DNA, and 50% light DNA [14N/14N DNA] These results can only be achieved if replication is semi-conservative, i.e. the 2 strands of the parental heavy DNA separate and both strands acts as a template for the assembly of a daughter light strand
How does DNA replication start?
DNA replication begins at one or more origins of replication (a specific sequence of nucleotides, generally A-T rich (only 2 hydrogen bonds between each A-T base pair, easier to disrupt the bonds as less
energy is needed to overcome them)).
1.initiator proteins recognise
and bind to the oriR sequence. The DNA double-helix is separated into two strands, forming a replication ‘bubble’.
2. The length of DNA unwound to initiate replication is typically ~ 50 bp. ATP is required.
3. At each end of a replication bubble Y-shaped structure called a replication fork, where the
new strands of DNA are synthesised. The two replication forks move away from the
oriR as replication proceeds bidirectionally, until the entire DNA molecule is separated
Describe DNA replication in prokaryotes versus eukaryotes.
Prokaryotes: The prokaryotic chromosome is a small circular DNA molecule, with a single origin of replication. DNA replication proceeds bidirectionally from oriR to a termination site located approximately halfway around the circular chromosome, resulting in the synthesis of two daughter DNA molecules
Eukaryotes: The eukaryotic chromosome is much larger and consists of a linear DNA molecule, with multiple origins of replication, hence multiple regions of the chromosome undergo
replication at the same time.
How does DNA replicate so quickly in eukaryotes despite the larger size of a eukaryotic chromosome?
The advantage of having multiple origins of replication in eukaryotes is speed. Multiple replication bubbles form and eventually fuse, thus speeding up the copying of very long DNA molecules. Replication takes approximately eight hours in human cells
with multiple origins of replication. If it had only one origin of replication, it would take 100
times longer. This is important given the much larger size of a eukaryotic chromosome.
What proteins are involved in the separation of parental DNA strands?
Helicase: Using ATP, helicases break the hydrogen bonds
holding the two strands of DNA
together. This unwinds the DNA
double-helix and separates the
parental DNA strands at the region of the replication fork.
Single strand DNA binding proteins: The unwound single-stranded portion of the DNA double-helix is temporarily stabilised
by the binding of single-strand DNA-binding proteins. This prevents the single-stranded (ss) DNA from re-annealing to reform the duplex. This keeps the two parental strands in the appropriate single-stranded condition to act as template. This also protects the ssDNA, which is very unstable, from being degraded
Topoisomerase: Unwinding causes supercoiling ahead of the replication fork, resulting in tension / torque. Topoisomerases cleave a strand of the helix to create a transient single-stranded nick. This relieves strain on the DNA molecule by allowing free rotation (swiveling) around the intact strand, and then reseal the broken strand
Explain the function of RNA primers in relation to the limitations of DNA polymerase.
None of the DNA polymerases can initiate the synthesis of a DNA strand on its own; DNA synthesis cannot occur de novo. RNA primer is used.
Describe the synthesis of RNA primers, along with DNA replication.
- A portion of the parental DNA strand serves as template for making the RNA primer with the complementary base sequence.
- Primase joins the ribonucleotides to make the primer. Hydrolysis of
ATP is involved. - The RNA primer provides a free 3’OH end that DNA polymerase can
extend. - A DNA polymerase with 5’ to 3’
exonuclease activity later replaces the RNA nucleotides of the primers with DNA versions.
Describe complementary base pairing in the synthesis of daughter DNA strands.
- Complementary base pairing between templates and nucleotides
a. The parental DNA strands serve as the templates for semiconservative DNA replication.
b. DNA polymerase reads the template and assembles the deoxyribonucleoside triphosphates, for the daughter DNA strand based on complementary base pairing.
c. When an incorrect base pair is recognised, DNA polymerase reverses its direction by one base pair of DNA. The 3’→ 5’ exonuclease
activity of the enzyme allows the incorrect base pair to be excised (proofreading.)
Describe phosphodiester bond formation in daughter DNA strand synthesis.
- Phosphodiester bond formation between growing daughter DNA strand and incoming
nucleotide
a. With an RNA primer anchoring the start of the daughter DNA strand, DNA polymerases catalyse phosphodiester bond formation between a growing
daughter DNA strand and an incoming nucleotide (polymerisation). Because of active site specificity of the DNA polymerases, syntheses of both
daughter DNA strand can only occur in the 5’ to 3’ direction. The addition of nucleotide (i.e. dNTP) to the growing daughter DNA strand requires the formation of a phosphoester bond between the free 3’OH of the last nucleotide in the growing strand and the free 5’ phosphate group of the incoming dNTP. The incoming dNTP loses a pyrophosphate group when they form the phosphoester bond with the growing daughter DNA strand. The energy released from pyrophosphate bond breakage is coupled to phosphoester bond
formation.
Due to the limitations of DNA polymerase, how are the anti parallel DNA strands synthesised?
The complementary daughter DNA strand that is continuously synthesised as a single polymer is the leading strand. Polymerised in the mandatory 5’ to 3’ manner
towards the replication.
Lagging strand: The complementary DNA strand that is discontinuously synthesised as a series of short Okazaki fragments. Each Okazaki fragment is polymerised in the mandatory 5’ to 3’ manner against the overall direction of the replication fork. Each Okazaki fragment requires an RNA primer for strand initiation. They are then ligated: DNA polymerase removes the RNA primer and replaces it with dNTPs and DNA ligaments catalyses the formation of a phosphoester bond between the 3’ end of each new Okazaki fragment and the 5’ end of the growing daughter DNA strand.
Describe the end replication problem.
The end replication problem occurs in linear chromosomes (i.e. in eukaryotes only) as DNA
polymerase is incapable of completely replicating all the way to the ends of linear chromosome, leading to shortening of telomeres
with each round of DNA replication.
The very end of the lagging strand is not replicated.
Each time a cell with linear chromosome divides, a small section at the extreme 3’ end of
the parental strand does not undergo DNA replication. When the final RNA primer at the end of lagging strand is removed,
there is no upstream strand onto which DNA polymerase can build to fill the resulting gap. Hence the daughter DNA strand resulting from lagging strand synthesis would be
shortened with each round of replication.
