DNA Structure and replication

Question 1

Characteristic properties of DNA as hereditary material

Answer 1

1. high capacity for information storage + chemically stable to encode information without fail, can’t change easily due to age, nutrition or environment
2. replicate accurately
3. be capable of variation

Question 2

Structure of DNA (monomers and polymers)

Answer 2

• nucleic acids exist as polymers called polynucleotides
• each polynucleotide is composed of monomers called nucleotides

Question 3

structure of nucleotide?

Answer 3

1. 5 carbon sugar, ie pentose
2. nitrogenous base
3. phosphate group

5 carbon sugar + nitrogenous base = nucleoside
all 3 = nucleotide

Question 4

what are the two types of nucleic acids

Answer 4

Deoxyribonucleic acid (DNA):
- pentose sugar is deoxyribose
- deoxyribonucleotides are monomers of DNA

Ribonucleic acid (RNA)
- pentose sugar is ribose
ribonucleotides are monomers of RNA

Question 5

pentose sugar structure

Answer 5

• 5-carbon sugars, 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.
• Two types of pentose (ribose and deoxyribose), they distinguish DNA from RNA.

Main difference between the ribose and deoxyribose sugars:
- deoxyribose: At the 2’ carbon, the hydroxyl group (-OH) is replaced by a hydrogen atom (H).
- leads to significant differences in structure and
functions of the two types of nucleic acids.
- in RNA: 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 more susceptible to chemical and enzyme degradation

Question 6

nitrogenous bases

Answer 6

• has nitrogen-containing ring structure
• two types: purines and pyrimidines
• purines: 6-membered ring fused to a 5-membered ring (Adenine and Guanine)
• pyrimidines: 6-membered ring (Cytosine and Thymine/Uracil)

Question 7

what is a nucleoside

Answer 7

• combination of a pentose sugar with a nitrogenous base
• condensation rxn as it occurs with elimination of water
• 1’ carbon of pentose is linked in glycosidic bond to nitrogenous base
  (ref to page 7 for diagram)
• two types: ribonucleosides and deoxyribonucleosides

Question 8

what is a nucleotide + how does number of phosphate groups vary

Answer 8

nucleotide: formed by further condensation between nucleoside and phosphate group, forming a phosphoester bond between 5’ carbon of pentose and phosphate group

number of phosphate groups varies from 1-3
- 1 phosphate group → nucleoside monophosphate (e.g. AMP, adenosine monophosphate)
- 2 phosphate groups → nucleoside diphosphate (e.g. ADP, adenosine diphosphate)
* 3 phosphate groups → nucleoside triphosphate (ATP, adenosine triphosphate)

Question 9

formation of di and polynucleotides

Answer 9

• dinucleotide: condensation between 5’-phosphate group of one nucleotide and 3’-hydroxyl group of the other to form phosphodiester bond
• phosphodiester bonds between 5’ phosphate and 3’ hydroxyl groups of nucleotides form linear, unbranched sugar-phosphate backbone
• phosphodiester bonds are strong covalent bonds, confering strength and stability on the polynucleotide chain, preventing breakage of the chain during dna replication

Question 10

polarity / directionality in polynucleotide

Answer 10

manner in which deoxyribonucleoside triphosphates are added to the 3’ end of a growing chain has resulted in a polynucleotide molecule that has polarity or directionality

each DNA / RNA strand has two free ends that are chemically different from each other
- 5’ end with free 5’ carbon carrying phosphate group
- 3’ end with free 3’ carbon carrying a hydroxyl group

dna / rna base sequence read in 5’ to 3’ direction

Question 11

how many bases form a complete turn of helix

Answer 11

10

Question 12

features of the double helix

Answer 12

1. 2 polynucleotide strands / chains, each strand forms a right hand helix, two strands coil around each other to form a double helix.
2. helix diameter is 2nm, just enough for 1 purine (A,G) and 1 pyrimidine (C, T, U)
3. strands are antiparallel
4. each strand has:
   - sugar phosphate backbone with phosphate groups that project outside double helix since they are hydrophilic
   - nitrogenous bases that orientate inwards towards central axis at almost right angles
   - puts hydrophobic nitrogenous bases in molecule’s interior and away from surrounding aq medium
5. bases of opposite strands are bonded tgt by relatively weak hydrogen bonds

Question 13

how does complementary base pairing come about

Answer 13

A-T and G-C pairs are the only ones that can fit within the physical dimensions of the double helix + they are in accord with Chargaff’s rules

Question 14

significance of complementary base pairing

Answer 14

• 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.
• weak hydrogen bonds make it relatively easy to separate two strands of DNA (separating A-T pair is easier than separating the C-G pair as A-t has two hydrogen bonds, G-C has three hydrogen bonds)

Question 15

more properties of complementary base pairing

Answer 15

1. base pairs stacked 0.34nm apart along centrla axis due to hydrophobic interaction contributing to overall stability of the molecule
2. double helix makes complete turn (3.4nm) every 10 base pairs
3. major and minor grove in sugar phosphate backbones, but both large enoguh to allow protein molecules to gain access and make contact with bases

Question 16

why are there complementary base pairs (A always with T and C always with G?)

Answer 16

1. steric restrictions
   - The sugar-phosphate backbone of each polynucleotide chain has a regular helical structure.
   - The DNA double-helix has a uniform diameter of 2 nm.
   - T and C are pyrimidines, which have a single ring. A and G are purines, which are about twice as wide as pyrimidines -> solution is always to pair a purine with a pyrimidine.
2. Hydrogen Bond Factors
   - Each nitrogenous base has chemical side groups (such as H, N and O) that can form hydrogen bonds with its appropriate partner.
   - Such chemical side groups in purines and pyrimidines have well defined positions.
   - A is capable of forming 2 hydrogen bonds with T, whilst G is capable of forming 3 hydrogen bonds with C.-

Question 17

variation of linear base sequence

Answer 17

base pairing rules do not restrict the base sequence along each dna strand, linear sequence of the 4 bases can be varied!!!

many combinations of bases due to hunas having many nualeotide pairs -> so each gene has unique base sequence

Question 18

what does chromosome contain

Answer 18

chromosome = single dna molecule bundled up with various proteins

Question 19

what is one complete copy of genome

Answer 19

1 set of 23 chromosomes

Question 20

stable, invariant storage of genetic information

Answer 20

• The genetic information that must be stored / preserved lies in the specific order of the base
  pairs, i.e. base sequence must be stable and invariant.
• DNA is a marvellous device for the stable storage of genetic information – it is relatively
  resistant to spontaneous changes (mutations).

Question 21

What structural features stabilises the DNA double-helix?

Answer 21

• Extensive hydrogen bonds between base pairs. (each hydrogen bond is, by itself, weak)
• Hydrophobic interactions (or ‘stacking forces’) between the stacked base pairs.
  - only sugar phosphate backbone is exposed to outside influences
• 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 -> nucleosomes folded into higher order structures such as the chromosome, in which the DNA is prevented from thermal and physical damage.

Question 22

what structural features results in invariant base sequence

Answer 22

**Specific, complementary base pairing between DNA strands, resulting in **
a) genetic information is redundant (i.e. present more than once) in the DNA molecule.
b) if the base sequence in one of the two strands is accidentally altered, the cell discards
the damaged strand, then makes perfectly good strand by using the remaining intact strand as a template, following Chargaff’s rules of complementary base pairing. The
redundancy of genetic information helps to maintain its integrity.

Question 23

how are dna strands able to be templates for replication

Answer 23

1. The two strands of DNA are complementary – each stores the information necessary to reconstruct the other.
2. When a cell copies a DNA molecule, each strand serves as a template for ordering
   nucleotides into a new, complementary strand.
3. 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.

Question 24

what is Watson and Crick’s dna replciation model

Answer 24

• The two DNA strands unwind and separate from each other, i.e. the hydrogen bonds between complementary base pairs are broken.
  -** Each DNA strand then acts as a template for the assembly of a complementary strand.**
• Nucleotides line up singly along the template DNA strand according to Chargaff’s rule of complementary base pairing, i.e. A-T and G-C.
• DNA polymerases join the nucleotides together at their sugar-phosphate moieties.
• Upon completion of DNA replication, 2 identical daughter DNA molecules are produced from a single parental DNA molecule.
• This model of DNA replication is described as SEMI-CONSERVATIVE as each of the two daughter DNA molecules consists of one parental DNA strand and one newly-synthesised daughter DNA strand.

Question 25

what does semiconservative replication mean

Answer 25

2 parental dna strands separate and both strands act as template for asembly of daughter strand

Question 26

properties of mechanism of dna replication (4)

Answer 26

1. **is extremely complex** - The timing of steps is extremely precise – the double-helix must **unwind** whilst the **replication machinery copies the two antiparallel strands simultaneously.** 2. **is extremely fast.** - cell takes few hours to copy 3 x 10^9 base pairs 3. **is extremely accurate.** - The **mutation rate**, approximately 1 nucleotide change per 10^9 nucleotides each time the DNA is replicated, is roughly the same for organisms as different as bacteria and humans. 4. **requires the cooperation of a large team of enzymes and other proteins, as well as the expenditure of ATP.**

Question 27

LOCATION of origins of replication

Answer 27

- DNA replication begins at one or more sites on the DNA molecule called origins of replication (oriR) - Each oriR is a specific sequence of nucleotides, which is 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. Proteins that initiate DNA replication **(initiator proteins)** recognise this sequence 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, there is a 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

Question 28

origins of replication in prokaryotes vs eukaryotes

Answer 28

prokaryotes - small circular DNA molecule with **single origin of replication** - dna replication proceeds bidirectionally from oriR to termination site located approx halfway around circular chromosome eukaryotes - linar DNA molecule with **multiple oriR** - multiple regions of the chromosome undergo replication at the same time - advantage: **speed**, as multiple replication bubbles **form and eventually fuse**, thus speeding up the copying of very long DNA molecules. This is important given the much larger size of a eukaryotic chromosome.

Question 29

separtion of parental dna strands

Answer 29

why is this necessary? - so both dna strands can act as templates for synthesis of daughter dna strands 3 proteins involved in replication process 1. helicases 2. single-stranded DNA binding proteins (SSB proteins) 3. topoisomerases

Question 30

helicases function

Answer 30

- After initiation, "unwinding" enzymes called helicases bind to one strand of the DNA molecule. - Using ATP as an energy source, helicases break the hydrogen bonds holding the two strands of DNA together -> **unwinds the DNA double-helix and separates the parental DNA strands at the region of the replication fork.** - Each of the two parental DNA strands serve as the template for the synthesis of a new DNA strand

Question 31

Single-strand DNA-Binding proteins (SSB proteins) functions

Answer 31

1. binding of SSB proteins temporarily **stabilise** the unwound single-stranded portion of the DNA double-helix 2. This **prevents the single-stranded (ss) DNA from re-annealing** to reform the duplex -> keeps the two parental strands in the **appropriate single-stranded condition to act as template.** 3. **protects** the ssDNA, which is very unstable, **from being degraded**

Question 32

topoisomerases function

Answer 32

- Unwinding causes tighter twisting / 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 resealing of the broken strand

Question 33

2 limitations of dna polymerases

Answer 33

**Limitation 1** - None of the DNA polymerases can initiate the synthesis of a DNA strand on its own. - Solution: To **initiate synthesis of a DNA strand** in the cellular context, an **RNA primer is used.** **Limitation 2:** - DNA polymerases only add dNTPs to the free 3’ end of a growing DNA strand, never to the 5’ end. **Thus, a growing DNA strand can only elongate in the 5’ to 3’ direction.** - Another Factor to Consider: The two strands of a DNA double-helix are **antiparallel**, i.e. their sugar-phosphate backbones run in opposite directions - Problem: **Continuous synthesis of both DNA strands at a replication fork is not possible.**

Question 34

synthesis of RNA primer

Answer 34

RNA PRIMASE SYNTHESISES RNA PRIMER!!! 1. A portion of the **parental DNA strand serves as template for making the RNA primer** with the **complementary** base sequence 2. An enzyme called **primase** joins the ribonucleotides to make the primer. The primer is about 10 nucleotides long in eukaryotes. **Hydrolysis of ATP is involved.** 3.**The RNA primer provides a free 3’ OH end that DNA polymerase can extend, thereby priming the synthesis of the daughter DNA strand. (OBJECTIVE)** 4. **A DNA polymerase with 5’ to 3’ exonuclease activity** (i.e. DNA polymerase I) later replaces the RNA nucleotides of the primers with DNA versions.

Question 35

synthesis of daughter strands

Answer 35

**1. Complementary base pairing between templates and nucleotides** - The parental DNA strands, separated at the replication fork and each primed with an RNA primer, serve as the templates for semi-conservative DNA replication. - **DNA polymerase reads the template and assembles the nucleotide monomers (deoxyribonucleoside triphosphates, dNTPs) for the newly-synthesised daughter DNA strand based on complementary base pairing.** - The error frequency of the newly-synthesised DNA strand must be low to ensure that replication is carried out accurately. - 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). **2. Phosphodiester bond formation between growing daughter DNA strand and incoming nucleotide** - With an RNA primer anchoring the start of the daughter DNA strand, DNA polymerases catalyse the polymerisation of the strand. - All DNA polymerases **catalyse phosphodiester bond formation between a growing daughter DNA strand and an incoming nucleotide**. - Because of active site specificity of the DNA polymerases, syntheses of both daughter DNA strands can only occur in one direction, i.e. in the 5’ to 3’ direction. This is the other limitation of DNA polymerases. - The addition of nucleotide (i.e. dNTP) to the growing daughter DNA strand requires the **formation of a phosphoester bond between the free 3’ hydroxyl group of the last nucleotide in the growing strand and the free 5’ phosphate group of the incoming dNTP.** - in this process, 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.**

Question 36

solution for limitation 2 of dna polymerase (continuous synthesis of both dna strands at a replication fork is not possible)

Answer 36

1. Leading Strand Synthesis: - The complementary daughter DNA strand that is **continuously synthesised** as a single polymer **along the template strand.** - Polymerised in the **mandatory 5’ to 3’ manner** **towards the replication fork.** 2. Lagging Strand Synthesis: - The complementary DNA strand that is **discontinuously synthesised** as a series of short fragments known as **Okazaki fragments.** - Each Okazaki fragment is polymerised in the **mandatory 5’ to 3’ manner** **against the overall direction of the replication fork**

Question 37

iimplication for lagging strand synthesis (initiation and ligation)

Answer 37

1. Each Okazaki fragment requires an RNA primer for strand initiation 2. The Okazaki fragments are then ligated / linked in two steps to produce a continuous DNA strand. (i) DNA polymerase ( DNA polymerase I) **removes the RNA primer and replaces it with dNTPs.** (ii) DNA ligase - a linking enzyme - 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.**

Question 37

what is the end replication problem

Answer 37

1. The end replication problem occurs in linear chromosomes (in eukaryotes) 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.** 2. More specifically, during 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**. * This is because 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.