Introduction to nucleic acids Flashcards

1
Q

What are the classes of nucleic acids in the cell?

A

90%- ribonucleic acid RNA:

  • 80% are ribosomal RNA
  • 15* transfer RNA
  • 5% messenger RNA

10%- Deoxyribonucleic acid DNA

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2
Q

What are the functions of nucleic acids?

A

DNA: Repository of genetic information

  • Directs its own replication
  • Directs transcription of complementary molecules of mRNA

mRNA: Carriers of genetic information
- Directs translation of genetic information into proteins

tRNA: translator of genetic information
- delivers amino acids during protein synthesis

rRNA: components of ribosomes
- have structural and functional roles

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3
Q

what are the building blocks of nucleic acids?

A

LOOK AT PICTURE/DOC

sugars (ribose or dexoyribose), phosphate and purine/pyrimidine bases.

Deoxyribose has a hydrogen atom at the 2’ position whereas ribose has a hydroxyl group.

Purine bases: adenine (in RNA + DNA), guanine (in RNA, DNA)

pyrimidine bases: cytosine (in RNA + DNA), Thymine (DNA), uracil (RNA)

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4
Q

what is the difference between a nucleoside and a nucleotide?

A

• Different building blocks of nucleic acids are linked together to form nucleosides and nucleotides
• Nucleosides are formed when a base is connected to a 5-Carbon-Sugar (either ribose or deoxyribose)
• Nucleotides are phosphorylated nucleosides (5’ position of the sugar)
o Depending on how many phosphate groups
o E.g. nucleoside: Adenosine (if base is adenine), nucleotide: Adenosine monophosphate (AMP), Adenosine diphosphate (ADP) or Adenosine triphosphate (ATP)

(LOOK AT PIC)

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5
Q

what is the general structure of nucleic acids?

A

• Sugar molecules are linked together via phosphodiester bond
• Each sugar molecule is connected to a base
At the 1’ position, the sugar is connected to the base
• At the 3’ and 5’ position, each sugar is connected to a phosphate group
• By convention, DNA and RNA are read from 5’ –> 3’ end

(look at pic)

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6
Q

Describe watson-crick base pairing

A

• A–> T via 2 hydrogen bonds
• G–> C via 3 hydrogen bonds
-GC rich regions are more stable than AT rich regions in nucleic acids

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7
Q

Describe the structure of B DNA

look at pic

A
  • The most common structure of DNA is B-DNA
  • B-DNA is a double stranded alpha helix
  • 2 grooves; major and minor grooves
  • The width of the molecule is 2nm
  • Base pairs are planar and are orientated nearly at a right angle to the axis of the helix
  • Distance between 2 base planes is 0.34 nm
  • One complete turn is 3.4nm and contains 10 base pairs
  • Orientation of the strands are antiparallel (the strands run in opposite directions)
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8
Q

Give an overview on DNA replication

A

• Double helix is unwound by helicase enzyme–> splits the parental strand into a leading strand and lagging strand
• The point where the parental strand is split into these 2 strands is known as the replication fork
• The enzyme DNA primase then intervenes
• The DNA primase and RNA primers are used by DNA polymerase as starting points for the replication.
o Free DNA nucleotides needed to form the new strands
o DNA polymerase- an enzyme that adds new nucleotides to a growing strand of DNA
o Primers- a short strand of nucleotides that will binds to the 3’ end of the template DNA strand allowing DNA polymerase to add free DNA nucleotides

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9
Q

the leading strand is replicated in a continuous fashion. what about the lagging strand?

A

• As DNA synthesis proceeds in 5’–>3’ direction, DNA polymerase can’t synthesise the lagging strand in one continuous piece; the lagging strand is synthesised into small fragments known as okazaki fragments.

Okazaki fragments length: 1000-5000 base length

  • RNA primers are removed by the exonuclease activity of DNA polymerase and are replaced with DNA
  • DNA ligase seals the gaps in the DNA to complete the synthesis of the lagging strand
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10
Q

what has the knowledge of DNA replication helped?

A

• The knowledge of how DNA is replicated has helped to develop antiretroviral chemotherapeutic agents

e.g. RETROVIR

  • These nucleoside analogues are phosphorylated at the 5’ position to form triphosphate analogues. Reverse transcriptase incorporates the phosphorylated analogues into the viral genome. when the analogues have been incorporated into the DNA; DNA synthesis is blocked, and chain elongation no longer occurs.
  • This is because these analogues do not have a hydroxyl group at the 3’ position!

(PICTURE)

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11
Q

Describe the structure of RNA molecules?

A

• Single-stranded molecules
• Consist of adenine, guanine, cytosine and uracil nucleotides (joined by phosphodiester bonds)
• Start: 5’ end, termination: 3’ end
• Extensive secondary structures
o Intramolecular double-stranded regions
o E.g. the most common secondary structures are the hairpin loops

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12
Q

how many RNA polymerases do eukaryotes have and describe them

A
  • Eukaryotics have 3 different types of RNA polymerases
  • Distinguished from each other by the class of RNA molecules which directs the synthesis

RNA polymerase I directs the synthesis of rRNAs
RNA polymerase II directs the synthesis of mRNAs
RNA polymerase III directs the synthesis of tRNAs

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13
Q

what is RNA polymerase II sensitive to?

A

• RNA polymerase II is sensitive to the inhibition of alpha-amanitin, which is found in the mushroom Amanita phalloides, The Death Cap.
• Symptoms of Amanita phalloides poisoning:
o Slow, 8-16h after ingestion
> As it takes time for all the old RNA molecules to be degraded

o Stomach pain, vomiting, diarrhoea
o Death after 7-10 days (in 10-15% of patients)

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14
Q

Describe transcription

A

Transcription is the process by which DNA sequences are used by RNA polymerase to produce complementary RNA molecules.
THE STAGES:
1. Initiation: interaction of RNA polymerase with specific sites on the DNA
o Known as promotors; characteristic sequence of the DNA that is upstream/ in front of the gene that has to be transcribed.

  1. Elongation: RNA polymerase selects appropriate ribonucleotides and forms phosphodiester bridges between them.
    o It is a rapid process: 40 nt per seconds
    o Double stranded DNA must be unwounded–> topoisomerases I and II
    - These enzymes are associated with the transcription complex
    - These enzymes are targets for anticancer chemotherapy
  2. Termination:
    o RNA polymerase I–> uses specific proteins
    o RNA polymerase II–> specific termination sequence and protein factors
    o RNA polymerase III–> specific termination sequence
  • The double stranded DNA template is separated into a coding strand and a template strand
  • Coding strand has the same sequence as the RNA transcript except T replaces U
  • Template strand is complementary to the RNA transcript
  • Complentary nucleotides are added by RNA polymerase to the growing RNA molecule

• The template strand is read from 3’-5’ while the RNA transcript is synthesized from 5’-3’

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15
Q

what is the structure of prokaryotic mRNAs like?

A

polycistronic–> encode multiple proteins

no chemical modifications

no splicing

= In prokaryotes, transcription resides immediately in the generation of mature mRNA molecules

• In prokaryotes, transcription and translation occurs simultaneously; transcripts may have been partially transcribed into proteins even before translation has been completed

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16
Q

what is the structure of eukaryotic mRNAs like?

A

Monocistronic; encode a single protein

• In eukaryotes, transcription leads to a primary transcript which has to be further modified before it becomes a mature mRNA.
o It has to be capped and polyadenylated—>
chemical modifications:
5’. methylated guanine nucleotide cap

3’: polyA tail

o Splicing; introns are removed
o Exons are joined together to form the mature mRNA

17
Q

Describe the structure of tRNAs

A
  • They all have a similar folded structure with 4 distinct groups
  • The last 3 bases are unpaired and form the acceptor stem
  • 3’ end is the point at which an amino acid is attached via an ester bond between the 3’ hydroxyl group of the adenosine and the carboxyl group of the amino acid
  • D loop on the right
  • Trinucleotide anticodon directs the specific interaction with the appropriate codon in the mRNA
18
Q

describe the prokaryotic ribosome subunits etc as well as the eukaryotic ribosome

A

PROKARYOTIC RIBOSOME: 70S
subunit–> large–> 50S–> rRNA: 23s, 5s
subunit–> small–> 30S–> rRNA: 16s

EUKARYOTIC RIBOSOME: 80S
Subunit–> large–> 60S–> rRNA: 28S, 5.8S, 5S

subunit–> small–> 40S–> rRNA:18s

19
Q

talk about ribosome-inhibiting antibiotics

A
  • As prokaryotic ribosomes are different to eukaryotic ribosomes, compounds have preferentially target
  • e.g. streptomycin targets the 30S subunit

ANTIBIOTIC AND TARGET:

  • streptomycin–> 30S subunit
  • erythromycin–> 50S subunit
  • tetracycline–> 30S subunit
  • Chloramphenicol–> 50S subunit