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Biological Chemistry > Nucleic acids > Flashcards

Nucleic acids Flashcards

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1
Q

How does deoxyribose in DNA differ from ribose in RNA

A

Carbon 2 has a hydrogen in DNA but a hydroxyl in RNA

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2
Q

What are purines

A

adenine and guanine - 2 heterocyclic rings

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3
Q

What are pyrimidines

A

Cytosine, thymine and uracil - 1 heterocyclic rings

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4
Q

How is a nucleoside formed

A

when a base is linked to 1’ carbon of a deoxyribose (or ribose) molecule

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5
Q

Nucleoside nomenculture

A

Deoxyribose + adenine –> deoxyadenosine

Deoxyribose + cytosine–> deoxycytidine

Deoxyribose + guanine –> deoxyguanosine

Deoxyribose + thymine –> deoxythymidine

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6
Q

How many H bonds are there between G and C

A

3

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7
Q

How many H bonds are there between A and T

A

2

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8
Q

What is Chargaff’s rule

A

A and T % and C and G % is the same

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9
Q

How is DNA held together

A

H bonds between complementary bases in an antiparallel fashion. 5’ - 3’ phosphodiester bonds

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10
Q

What is a double helix atructure

A

2 antiparallel strands wrapped around each other

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11
Q

Double helix properties

A

Right handed , Strands held together by base-pairing and by hydrophobic interactions between adjacent base-pairs (base-stacking), Base lie flat inside the helix, perpendicular to the sugar phosphate backbone, Diameter 2nm , Height 2.4nm per turn,10 base pairs per turn. Has major and minor grooves

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12
Q

What are major and minor grooves

A

major is more exposed and shows nucleotides allowing DNA binding proteins to bind to it. Minor grooves are closer together

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13
Q

Where is B DNA found

A

in humans

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14
Q

What is A DNA

A

more open structure, more packed bases. Dehydrated DNA. 11bp/turn, right-handed, slant base pairs

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15
Q

What is Z DNA

A

12bp/turn. Left-handed (anticlockwise spiral). Zig Zag backbone, may form if DNA contains long runs of alternating G and C

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16
Q

What is supercoiling

A

If DNA is underwound or overwound, it will become supercoiled – the molecule twists around itself

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17
Q

What generates negative supercoiling

A

underwinding (helps transcription to occur)

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18
Q

What generates positive supercoils

A

overwinding and unwinding a DNA molecule with fixed ends (blocks genetic info)

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19
Q

What occurs during DNA denaturation

A

Breaking non-covalent bonds, breaking hydrogen bonds and base stacking while keeping 2 strands intact

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20
Q

What conditions are required to denature DNA and what occurs

A

When heated to 70-110 degrees (or exposed to alkaline conditions – bases ionised causing H bonds to break) DNA becomes denatured – the strands separate. If allowed to cool slowly the strands will re-anneal.

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21
Q

How can you measure denaturation

A

measuring absorbance of UV light at a wavelength of 260nm. Single stranded DNA absorbs more UV than double stranded DNA. Therefore, UV absorbance rises as DNA denatures (the ‘hyperchromic shift’)

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22
Q

Define melting temperature TM of DNA

A

he temperature needed to denature 50% of the DNA molecules in a sample.

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23
Q

What increases TM

A

DNA with high concentration of GC base-pairs, since there are more H bonds between the strands. Or by the presence of cations (e.g., Na+) - these reduce repulsion between negatively charged phosphate groups on the 2 strands.

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24
Q

How does ribose and deoxyribose differ

A

ribose has extra O on C2 making it more reactive and versatile because the oxygen is an extra binding site

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25
Q

RNA nomenculture

A

cytidine monophosphate, guanosine monophosphate, uridine monophosphate

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26
Q

What can tRNA do

A

RNA molecules may fold back on themselves to form complex secondary structures with intramolecular base-pairing e.g. tRNA

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27
Q

Difference between DNA and RNA

A

RNA has ribose instead of deoxyribose, RNA contains uracil instead of thymine. Like thymine, RNA has different nomenclature. DNA is usually double stranded, RNA is usually single stranded.

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28
Q

Why does RNA have many functions

A

Due to its greater structural flexibility and reactivity

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29
Q

RNA function

A

Carry information (mRNA), Act as a transporter (tRNA), Act structurally or catalytically (rRNA), Act as a regulator of gene expression (micro RNAs)

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30
Q

Define gene

A

The entire stretch of DNA necessary to produce a particular functional product, which may be a protein or an RNA molecule,

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31
Q

Define genome

A

The genetic (I.e. hereditary) material (usually DNA) contained in an organism, cell, virus or organelle

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32
Q

Define chromosome

A

A single long molecule of DNA that includes numerous genes. The DNA of a chromosome is usually associated with proteins. Eukaryotic chromosomes are visible during cell division.

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33
Q

What is chromatin

A

The DNA-protein complex present in the nuclei of eukaryotic cells during interphase

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34
Q

What is proof that genes are made of DNA

A

Alterations in DNA cause genetic diseases. Introduction of foreign DNA into an organism may alter its characteristics

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35
Q

What are prokaryotic chromosomes like

A

have only one chromosome that is usually circular (no free ends)

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36
Q

Why does DNA need to be compressed in bacteria

A

An E. coli cell is only 2um long but contains 1.3nm DNA. By binding to protein to form the nucleoid (but not enclosed by a membrane, I.e. nor a nucleus)

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37
Q

% of DNA in bacteria that is coding

A

85%

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38
Q

Mycoplasma genitalium genome size and coding genes

A

genome size of 0.58 x 10^6bp – 480 protein coding genes.

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39
Q

E coli genome size and coding genes

A

genome size of 4.6 x10^6 bp – 43000 protein coding genes

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40
Q

E coli genome size and coding genes

A

genome size of 4.6 x10^6 bp – 43000 protein coding genes

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41
Q

Streptomyces coelicolor genome size and coding genes

A

genome size of 8.7 x 10^6 bp – 7800 protein coding genes

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42
Q

Properties of plasmids

A

Accessory circular DNA molecules separate from the chromosome, Size range 1kb – 400kb, A cell may contain between 1 and 200 copies of a particular plasmid, Plasmids carry non-essential genes e.g., for antibiotic resistance, Readily passed from cell to cell, Have an origin of replication, Can be exploited in genetic manipulation

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43
Q

What is an intron

A

sequence inserted within the gene

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44
Q

What is intergenic space

A

space before and after the gene – non coding DNA

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45
Q

Features of human mitochondrial genome

A

A circular DNA molecule, about 16500 bp in length, Multiple copies present in each mitochondrion, 37 genes: 22 encoding tRNA, 2 encoding rRNA, 13 encoding mRNA (therefore protein), Short intergenic space, Almost all mitochondrial DNA is organized into genes, Most mitochondrial proteins are encoded by genes in the nucleus

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46
Q

What is repetitive DNA

A

non-coding sequences present in thousands or millions of copies

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47
Q

What % of human DNA encodes protein

A

1.2%

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48
Q

Eukaryotic chromosomes

A

Eukaryotes have a number of linear chromosomes (from 1 pair to over 200) that are only visible at mitosis/meiosis.
Humans – 46 chromosomes – 23 pairs, one of each pair from each parent. The 2 copies of a particular chromosome are homologous.

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49
Q

How does interphase change chromosomes

A

become thin and tall. Darker regions contain heterochromatin (compressed and inactive), lighter regions show euchromatin (active)

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50
Q

How can chromosomes be identified

A

Each chromosomes occupies a distinct territory within the nucleus which can be identified by adding a probe containing a fluorescent protein which binds to a specific chromosomes and can be viewed under a microsocope to observe location and activity.

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51
Q

euk v pro chromsomes

A

euks have multiple chromsomes that are linear. Pro have one chromsome that is circular

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52
Q

euk v pro non-coding DNA

A

euks have extensive non-coding DNA whereas pro has little

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53
Q

euk v pro genes

A

simple euk - 4000 genes. complex euk - 20000 genes. bacterium - 4300

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54
Q

Packing ratio formula

A

length of DNA/ length of structure DNA packed into

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55
Q

Eukaryotic chromosome composition

A

1/3 DNA (one long linear molecule),1/3 histone proteins, 1/3 non-histone proteins

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56
Q

What are miotic chromosomes

A

identical chromosomes (sister chromatids) joined together at the centromere.

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57
Q

What are histone proteins

A

There are 5 types in eukaryotes: Histone H1, H2A, H2B, H3 and H4. Short and have a small amino acid sequence. ALL have a basic (gives positive charge allowing interaction with DNA) amino acid group connected: approx. 20% of amino acids are arginine or lysine). Highly conserved between species

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58
Q

What is the bead on a string structure

A

during interphase the content is lysed and we see ‘bead on a string’ - string is DNA, beads are nucleosomes

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59
Q

What is the structure if nucleosomes

A

Is 146 bp DNA wrapped around “core particle” containing: 2 molecules of histone H2A, 2 molecules of histone H2B, 2 molecules of histone H3, 2 molecules of histone H4 . Packing ratio = 7. Nucleosomes are subunits of chromosome and chromatin

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60
Q

H1 function

A

binds to DNA outside core particle, sealing the nucleosome

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61
Q

Nucleosomes in nucleus

A

packed into the nucleus of interphase cells in disordered chains with a varying density of nucleosomes.

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62
Q

What do chromosomes visible at mitosis/meiosis show

A

scaffold of non-histone proteins (e.g., condensins and topoisomerases) that anchor long loops of nucleosomes

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63
Q

What is condensin

A

A ring-shaped protein that can anchor the ends of a loop of DNA

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64
Q

What is topoisomerase II

A

An enzyme that can remove or add supercoils of DNA

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65
Q

Why do we need compressions

A

because during mitosis the 2m becomes 4m, in order for the DNA to be pushed apart you need compression. Acts as a compression and segregation device (non-hidtone proteins do).

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66
Q

What is chromatin

A

disordered chains of nucleosomes (density of nucleosomes greater in heterochromatin)

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67
Q

How can histone proteins modify eukaryotic genes

A

acetylation of lysine residues

Acetylation (addition of positively charged acetyl group) removes positive charge from side – chain and so weakens interaction between histones and negatively charged DNA

DNA associated with acetylated histones is much more readily transcribed than DNA associated with unmodified histones

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68
Q

How does acetylation make it easier to transcribe a gene

A

Before acetylation the DNA is tightly associated with the nucleosome and not very accessible for transcription – gene switched off. After acetylation DNA is less tightly associated with nucleosome and more accessible for transcription – gene may be switched on

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69
Q

What is semi conservative DNA replication

A

the 2 strands dissociate and each strand acts as a template forming 2 new daughter strands, and 2 parents – the parents have been conserved therefore semi conservative

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70
Q

Main points of how semi conservative DNA replication occurs in vivo

A

Partial unwinding and replication, unwinds progressively, Creates replication fork, Occurs in 5’ to 3’, Further unwinding and replication then occurs

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71
Q

How does DNA replication in E.coli occur

A

1.DNA is circular, so there is an origin of replication (100-150 bp long, rich in A-T base pairs)

  1. Have 2 template strands (top and bottom)
  2. 2 replication forks I.e., bidirectional replication
  3. Once daughter strand starts to synthesis the pairing strand starts to shorten
  4. Causes formation of loops
  5. Creates 2 daughter chromosomes
  6. Eukaryotic chromosomes have multiple points of origin
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72
Q

DNA polymerase function

A

Synthesise DNA

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73
Q

DNA polymerase in E.coli

A

5 types of DNA polymerase (2,4,5 – DNA repair mechanism. 1 and 3 involved in synthesis.

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74
Q

DNA polymerase in humans

A

14 types of DNA polymerase (some in synthesis majority are involved in repair mechanisms)

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75
Q

What does DNA polymerase need in order to synthesize DNA

A
  1. All 4 deoxynucleoside triphosphates
  2. A template
  3. A primer
  4. The splitting of the phosphate molecules
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76
Q

Why does DNA polymerase need all 4 deoxynucleoside triphosphates

A

I.e., dATP, dCTP, dGTP, dTTP. When DNA replication begins 2 phospjate molecules are split off forming the nucleotide.

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77
Q

Why does DNA polymerase need a primer

A

a short piece of nucleic acid base-paired to the template. The primer acts as a start point; it must have a 3’OH group. Without primer there is no reaction. The hydroxyl group is important because it is needed to bind to the inner of the phosphate molecule in order to split the 2 phosphates so its is no longer a triphosphate molecule.

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78
Q

Why does DNA polymerase need the splitting of a phosphate molecule

A

to generate energy

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79
Q

DNA polymerase 1 has a 5’ to 3’ exonuclease.

A

degrades the strand at an open site in a 5’ to 3’ fashion.

The 5’ to 3’ exonuclease activity and DNA polymerase activity can combine to allow nick translation. Can replace a bond by stimulating DNA synthesis while degrading the strand.

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80
Q

What is an exonuclease

A

cuts at a specific site inside the gene – not a function just a difference

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81
Q

DNA polymerase 1 has a 3’ to 5’ exonuclease

A

This allows DNA polymerase 1 (and DNA polymerase III) to remove incorrect nucleotides from newly made DNA (proof-read)

After the incorporation of incorrect nucleotide, DNA synthesis stops because no hydrogen bonds have been formed due to incorrect base so is hanging off.

There is removal of mis-matched nucleotide

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82
Q

DNA polymerase 3 function

A

responsible for making most of the new DNA during replication because of its speed and high processivity

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83
Q

DNA polymerase 1 v 3 polymerasation and processivity rate

A

1 polymerisation - 10 (per second), 3 - 1000 (per second)
1 processivity - 20. 3 - >500,000

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84
Q

What is the leading strand

A

5’ -> 3’

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85
Q

What is the lagging strand

A

3’ -> 5’

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86
Q

Replication fork appearance step 1 - DNA helicase partial unwinding

A

Each base pair needs 2 ATP molecules to unwind.

Unwinding a DNA molecule with fixed ends introduces positive supercoils.

Positive supercoils are removed by DNA gyrase, a type II topoisomerase.

Single stranded DNA binding protein holds the strand apart (1 for every 10 base pairs)

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87
Q

Replication fork appearance step 2 - synthesis of primers using primase

A

One primer on 5’ to 3’, the other on 3’ to 5’ strand. The 3’ to 5’ primer is closer to the fork and moves in the 5’ to 3’ direction

88
Q

Replication fork appearance step 3 - synthesis of DNA using DNA polymerase 3

A

DNA polymerase III and dNTPs form strands in the 5’ to 3’ direction

There is then further unwinding of the strand

5’ to 3’ can continue replication but can’t continue in the lagging strand. So another RNA primer is added, to synthesize the next fragment.

So lagging strand is synthesized as short (approx. 1000 bp) fragments called Okazaki fragments

89
Q

Replication fork appearance step 4 - DNA polymerase 1 removing RNA primers

A

Primer degraded by 5’ -> 3’ exonuclease activity and simultaneously replaced with DNA using upstream Okazaki fragment as primer.

There is a gap (one missing phosphodiester bond)

90
Q

Replication fork appearance step 5 - DNA ligase sealing Okazaki fragment gaps

A

The RNA primer is replaced by DNA polymerase I

The gap is sealed by DNA ligase as a new phosphodiester bond is formed

91
Q

In all species DNA replication is

A

Semiconservative, Bidirectional (2 replication forks per origin), Semi-discontinuous (lagging strand synthesized as Okazaki fragments), Dependent on RNA primers

92
Q

differences in DNA replication between bacteria and eukaryotes

A
  1. DNA polymerase speed: 1000 nucleotides per second (B). 50 nucleotides per second (E)
  2. Size of Okazaki fragments: 1000-2000 nucleotides (B).100 – 200 nucleotides (E)

Minimum time taken to replicate genome: Approx. 40 minutes (B). 3.4 minutes (Drosphilia embryo) 15 minutes (Frog embryo) (E)

Number of origins replication: 1 (B) 10000 – 100000 (humans)

93
Q

mRNA function

A

Carry information from the nucleus to the cytoplasm

94
Q

What is the coding strand (sense strand)

A

the DNA strand which has the same base sequence as the RNA being synthesized (except T for U)

95
Q

What is the non-coding strand (antisense strand)

A

the template strand for RNA synthesis, whose sequence is complementary to the RNA molecule

96
Q

RNA polymerase function

A

the enzyme that carries out transcription

97
Q

What is a promoter

A

A cis-acting (I.e., affects sequences on the same DNA molecule) sequence that signals the start of a gene – a binding site for RNA polymerase

98
Q

What is a terminator

A

DNA sequence that signals the end of a gene – a site where RNA polymerase dissociates from DNA.

99
Q

Organization and transcription if a simple bacterial protein-coding gene

A

Promoter – transcribed gene – terminator

RNA polymerase binds to promoter sequence and initiates transcription from a 5’ to 3’ direction. The complementary template strand is from the noncoding strand.

Either strand of a DNA molecule can act as the coding strand, but there is always one coding and one noncoding strand

100
Q

What does RNA polymerase require for activity:

A
  1. All four nucleoside triphosphates I.e., ATP, CTP, GTP, UTP
    2.A double stranded template DNA molecule that includes a promoter sequence
    3.A primer is not required
    - RNA polymerase uses the same mechanism as DNA polymerases I.e., addition of nucleotides to 3’ end of the RNA molecule so that RNA is synthesized in the direction 5’ -> 3’.
101
Q

RNA polymerase reaction

A

OH group binds to the inner most phosphate group, release pyrophosphate ( 2 phosphate molecules). Initiating energy for transcription to take place.

102
Q

RNA polymerase must

A

Bind to the promoter and then …Synthesize an RNA copy of the coding strand

103
Q

Transcription - initiation

A

RNA polymerase binds to the promoter , DNA strands partially unwind, RNA synthesis begins

104
Q

Transcription - elongation

A

RNA polymerase moves along the DNA molecule synthesizing an RNA copy. Only 15-17bp of DNA are unwound at any time

105
Q

Transcription - termination

A

RNA polymerase dissociates from DNA releasing the new RNA molecule

106
Q

Transcription 1 in E.coli - Finding the promoter

A

We look for consensus sequences - “average” sequences found in association with in many different genes.

  • 35 hexamer (35 nucleotides upstream from the start site of the transcription. Hexamer – 6 nucleotides)

“Pribnow box” - upstream from the transcription start point

Both of those promoter sequence is needed for transcription to start

20 base pairs between promoter sequences which is where RNA polymerase binds

107
Q

What are the 2 forms RNA polymerase exists in E.coli

A

the holoenzyme is a hexamer with the subunit structure of 2 alpha, one beta, one beta prime, one omega and one sigma. The sigma allows specific binding to the promoter sequence (carries out initiation but not elongation). The core enzyme is a pentamer with the subunit structure: 2 alpha, one beta, one beta prime and one omega (carries out elongation but not initiation)

108
Q

How does RNA polymerase work in E.coli

A

the holoenzyme binds to the promoter. The sigma subunit then dissociates behind the leaving core enzyme which carries out elongation. After termination the core enzyme dissociates from DNA and binds to a sigma unit to reform the holoenzyme.

109
Q

Transcription in E.coli

A

1.RNA polymerase (holoenzyme form) binds to promoter sequence
2.Unwinding of the DNA sequence takes place in the middle of your pribnow box
3.Sigma dissociates from holoenzyme to give rise to core enzyme and gives raw materials to start transcription
4.NTPs added in, synthesis continues. By product is pyrophosphate
5.RNA moves down the strand, leaving DNA behind and causes RNA to trail behind
6.RNA molecules are formed. RNA polymerase dissociates.

110
Q

How does transcription in bacteria and eukaryotes differ

A

1.Eukaryotes have separate RNA polymerases for mRNA, rRNA and tRNA.
2.Eukaryotes make a 1 degree transcript (aka pre-mRNA) that is processed in the nucleus to form mRNA
3.Eukaryotic promoters differ from those in bacteria
4.Eukaryotic RNA polymerases cannot directly recognize the promoter

111
Q

RNA polymerase in eukaryotes

A

RNA polymerase I – synthesis of rRNA

RNA polymerase II – synthesis of mRNA

RNA polymerase III synthesis of tRNA

112
Q

What is the 5’ splice site

A

The point where one exon ends and an intron begins

113
Q

What is the 3’ splice site

A

The point where one intron ends and another exon begins

114
Q

What is capping

A

addition of methylated G (the cap) to the 5’ end of the 1 degree transcript. The cap stabilizes the mRNA molecule and is required for translation

115
Q

What is cleavage and polyadenylation

A

the 3’ end of the 1 degree transcript is cleaved and a “poly A tail” of approx. 250 A residues is added. This stabilizes the mRNA molecule

116
Q

What is RNA splicing

A

using an enzyme has the ability to splice specific exons together and splice introns out where introns are eventually degraded.

117
Q

RNA processing in eukaryotes

A

RNA splicing – most eukaryotic genes are split into exons, which are represented in the mRNA and introns which are not

Both introns and exons are present in the 1 degree transcript. The exons are spliced together, and the introns are removed by RNA splicing to generate the mRNA.

Remember, exons are expressed, introns intervene

118
Q

RNA polymerase II - transcription in eukaryotes

A

Promoters recognized by RNA polymerase II usually consists of a core promoter that may include a TATA box, and one or more proximal promoter elements (aka upstream promoter elements – UPEs)

Efficient transcription of eukaryotes genes may also require more distant sequences called enhancers

119
Q

What do promoters recognized by RNA polymerase II require

A

Transcription factors which interact with the core promoter

120
Q

TFID in initiating transcription

A

TFID is a multi-subunit protein that includes the TATA – binding protein (TBP)

121
Q

What forms the pre-initiation complex

A

Binding of TFID and further general transcription factors and RNA polymerase II

122
Q

Role of proximal promoter elements and enhancers

A

Additional transcription factors bind to proximal promoter elements and enhancers stimulating the formation of the pre-initiation complex. Also recruits histone acetylases. Allows the gene to be transcribed

123
Q

What is the genetic code

A

the nucleotide sequence of a gene and its corresponding mRNA molecule determines amino acid sequence of a protein

124
Q

How did Nirenberg (1961) crack the genetic code

A
  1. Synthesized an RNA containing U only, I.e., UUUUUUU.
  2. Incubated RNA with extract of E.coli cells containing ribosomes, amino acids etc…
  3. Found a protein was made consisting entirely of phenylalanine I.e., Phe- Phe – Phe
  4. Therefore, the codon UUU must represent phenylalanine
125
Q

Features of the genetic code

A
  1. Non-overlapping
  2. Degenerate
  3. Translation requires an initiation and termination codon
  4. Universal
126
Q

What is the initiation codon

A

AUG (occasionally GUG) - represents methionine

127
Q

What are the termination codons

A

UGA, UAA, UAG

128
Q

How many possible reading frames does a sequence have

A

3

129
Q

What is an open reading frame

A

The region between an initiation codon and a termination codon. Genes contain long ORFs which encode the protein corresponding to the gene

130
Q

Features of tRNA

A
  1. Small (73-90 nucleotides long)
  2. Contains some unusual bases
  3. Have an anticodon
  4. Have an unpaired sequence CCA at 3’ end
  5. Have extensive internal base pairing
131
Q

What unusual bases does tRNA contain

A

thymine, dihydrouracil, pseudouracil (all U-like) and methylguanine (G-like) - came from post-transcriptional modification

132
Q

Why does tRNA have an unpaired sequence at the end

A

so amino acids can be linked to the 3’ OH the adensone at the 3’ end

133
Q

Describe the secondary and tertiary structure of tRNA

A

have a ‘clove-leaf’ secondary structure with 3 (or 4) loops, and 4 double stranded stems. Tertiary structure resembles an inverted L

134
Q

What is an aminoacyl tRNA

A

a tRNA attached to its amino acid. When it attached the amino acid is ‘activated’

135
Q

What is the base wobble

A

tRNA can recognize more than one codon because some base at 5’ end of anticodon to pair with more than one base at 3’ end if codon

136
Q

What does RNA need for the genetic code to operate

A

The RNA whose anticodon recognizes a codon representing a particular amino acid must only be linked to that amino acid

137
Q

What are the enzymes called that link tRNAs to amino acids

A

aminoacyl-tRNA synthetases. There are 20 different ones (one for each amino acid_

138
Q

Step1 of aminoacyl tRNA synthetase reaction

A

amino acid + ATP –> aminoacyl-AMP + PPi

139
Q

Step 2 of aminoacyl tRNA synthetase reaction

A

aminoacyl-AMP + tRNA –> aminoacyl-tRNA + AMP

140
Q

Overall reaction of aminoacyl tRNA synthetase

A

amino acid + ATP + tRNA —> AA-tRNA + AMP + PPi

141
Q

What are the 3 ribosome binding sites

A

E (exit) site, P (peptidyl) site, A (aminoacyl) site. Provides an interface for mRNA and aminoacyl tRNAs

142
Q

Ribosomes catalystaion role

A

they catalyse the formation of peptide bonds

143
Q

What are ribosomes composed of

A

protein and rRNA (1:2 ratio)

144
Q

What is the size of the ribosome measured in

A

Svedbergs (S) - the sedimentation rate during centrifugation

145
Q

What are the 2 subunits of a ribosome and what is its size in E.coli

A

a small (30S) and a large (50S). Total is 70S

146
Q

Small subunit and large subunit function

A

Small - binds to mRNA. Large - catalyses peptide bond formation

147
Q

The human ribosome

A

80 S (60:40). Mitochondria have their own ribosome (55S) that are more similar to bacterial ribosomes

148
Q

Which direction is the protein synthesised

A

NH3+ –> COO-

149
Q

What is the first amino acid formed in prokaryotes v eukaryotes

A

Pro - N-formyl-methionine. (aldehyde group). Euk - methionine

150
Q

State of the ribosome when not translating mRNA

A

intact ribosomes exists in equilibrium with its subunits

151
Q

What are the stages of translation

A

Initiation (binding of mRNA and first tRNA, assembly of the ribosome). Elongation (addition of amino acids to growing peptide). Termination (dissociation of the completed protein from the ribosome)

152
Q

Translation initiation in E.coli

A

The small (30S) ribosome subunit binds to the ribosome binding site on the mRNA as result of base-pairing between it and the 16S rRNA

F – met – tRNAmet Binds to the intiation codon. AUG codons that encode internal methionines within the protein are not recognised as initiation codons because they lack an associated ribosome binding site

The 50S subunit binds, generating the complete ribosome

153
Q

Translation initiation in eukaryotes

A

The first amino acid is methionine,
The 40S subunit and Met-tRNAmet combine prior to binding to mRNA

There is no ribosome binding site on the mRNA. Instead the 40 S subunit recognises the cap at the 5’ end of the mRNA

The 40S subunit and Met-tRNAmet migrate along the mRNA until the first AUG is reached. The 60S subunit then binds generating an intact ribosome with the first tRNA in the P site

154
Q

Translation elongation - part 1

A

EF-TU (prokaryotes) EF-1 alpha (Eukaryotes) - recruit a tRNA and put it in the A site. Needs hydrolysis of GTP.

Peptidyl transferase reaction forms peptide bond between P and A. (carboxyl end of met binds to amino terminal of A.

One tRNA molecule now no longer has anything bound to it

155
Q

Translation elongation - part 2

A

EF-G (pro) EF-2 (Euk) - pushes ribosome forward (like a spring) so the scarce tRNA is at the E site and A is empty

EF-TU and EF-1a and another moleule of hydrolysed GTP causes another molecule to bind to A site, the tRNA molecule on E site has been released.

Peptidyl-transferase reaction bforms another peptide bond from P to A site

Cycle repeats

156
Q

The elongation cycle

A

1.Binding of aminoacyl-tRNA (needs GTP)
2.The petidyl transferase reaction (I.e., peptide bond synthesis)
3.Translocation (movement of the ribosome one codon along the mRNA – needs GTP)

Each cycle incorporates one amino acid into the protein. Cycles continue until a termination codon is reached

157
Q

Energy expenditure during translation

A

One ATP molecule is used to link the amino acid to its RNA. One GTP molecule is used when the aminoacyl tRNA binds to the ribosome. One GTP molecule is used during translocation

158
Q

NH2 acting as a nucleophile

A

Binds to inner most carbon and splits ester bond so new peptide bond can be formed

159
Q

What component of the ribosome catalyse sthe peptidyl transferase reaction

A

23S rRNA - confirmed by X-ray crystallography

160
Q

What are elongation factors

A

accessory proteins that are not part of the ribosome required for translation

161
Q

EF-TU elongation factor

A

EF-TU (EF-1a in euk) delivers aminoacyl tRNAs to the ribosomes

162
Q

EF-G elongation factor

A

EF-G (EF-2 in euk) uses energy provided by the hydrolysis of GTP to move the ribosome along the mRNA during translocation

163
Q

Differenced in translation between eukaryotes and prokaryotes

A
  1. Eukaryote ribosomes are larger than prokaryotic ribosomes
  2. Initiation is different: recognition of ribosomal binding site (prokaryotes) or cap(eukaryotes)
  3. Elongation is very similar in eukaryotes and prokaryotes
  4. Eukaryotic and prokaryotic ribosomes are sensitive to different antibiotics and inhibitors
164
Q

Erythromycin/ clarithromycin target and action

A

Target - prokaryotic large ribosomal subunit. Inhibits the peptidyl transferase reaction and translocation

165
Q

Chloramphenicol

A

Target - prokaryotic and mitochondrial, large subunit. Inhibits the peptidyl transferase reaction

166
Q

Ricin

A

Target - eukaryotic 60S subunit. Catalytic, removes a base from rRNA. Lethal does in humans approx 10^-4 g

167
Q

What are housekeeping genes

A

encode products that are needed constantly so are expressed at all times and in all tissues

168
Q

Which genes can be switched on or off

A

tissue specific gene in eukaryotes (e.g., insulin) , inducible genes in eukaryotes and prokaryotes

169
Q

What controls gene expression

A

regulating transcription (main mechanism), translation and RNA processing

170
Q

What does Lac Z do in E.coli

A

encodes beta galactosidase (catalyses lactose –> glucose +galactose)

171
Q

What does Lac Y do in E.coli

A

encodes galactoside permase (uptake of lactose)

172
Q

What does Lac A do

A

encodes galactoside transacetylase

173
Q

What does Lac i gene do

A

encodes the Lac repressor protein

174
Q

What is the CAP site

A

binding site for catabolite activator protein (CAP)-cAMP complex. Presence stimulates transcription of Lac Z, Y and A

175
Q

How are the Lac genes organised

A

they are adjacent on chromosomes and are copied into one long run mRNA (operon). This mRNA is translated into beta-galactosidase, galactoside permease and galactoside transacetylase

176
Q

What is an operon

A

cluster of several genes transcribed into a single mRNA

177
Q

Why is only one promoter needed to control all 3 genes

A

because Lac Z, Y and A are all arranged into an operon

178
Q

What occurs when lactose is absent

A

lac repressor protein binds to the operator, RNA polymerase cannot bind to the promoter so transcription does not occur

179
Q

What happens when lactose is present

A

lactose is converted into allolactose which binds to lac repressor and induces a conformational change so lac repressor cannot bind to the operator and RNA polymerase can transcribe lac Z, Y and A

180
Q

Why is transcription inefficient when glucose is present

A

promoter is weak (poor binding site for RNA polymerase). The promoter differs from the ideal promoter in 3 places

181
Q

When does catabolite repression occur

A

when lactose and glucose are present the lac operon genes are only transcribed weakly because lactose breakdown is only present when glucose is not available

182
Q

What does the removal of catabolite repression require

A

the catabolite activator protein (CAP) and cyclic AMP (cAMP)

183
Q

Mechanism of catabolite repression

A
  1. CAMP concentration in the cell depends on glucose levels
  2. When glucose concentration is high the cAMP concentration is low, and vice versa
  3. When the glucose concentration is low, cAMP binds to CAP forming a complex which binds to CAP site and stimulates binding of RNA polymerase, increasing transcription of the lac Z, Y and A genes
  4. When the glucose concentration is high, cAMP is destroyed so cAMP-CAP complex does not form and transcription of lac Z, Y and A remains weak, even if lactose is present
  5. Other operons involved in catabolism (breakdown) are also subject to catabolite repression
184
Q

Structure of CAP site

A

has a helix turn helix DNA binding domain. 2 alpha 0 helicases (one contacting base in the major groove) seperated by a sharp turn

185
Q

How is gene expression controlled in eukaryotes

A

promoters consist of core promoter and several proximal promoter elements (PPEs). Also influenced by distant regulatory sequences (enhancers). Requires PPEs binding to transcription factors

186
Q

Example of gene expression in eukaryotes

A

regulated by availabilty/activity of transcription factors, Muscle specific genes contain PPEs named E-boxes (CANNTG) that are binding sites for MyoD, which is only expressed in muscle

187
Q

Why do Viruses matter

A

They cause many diseases e.g., AIDS, ebola, covid, rabies. They evolve fast and so new disease may appear rapidly and unexpectedly

188
Q

What is a virus

A

one or more nucleic acid molecules within a coat made of protein

189
Q

Features of a virus

A

1)Can replicate and evolve
2)Do not grow
3)Lack genes needed for energy production and (usually) protein synthesis
4)Contain DNA or RNA, not usually both
5)Are insensitive to changes in their environment
6)Can only reproduce inside a living cell (I.e., are obligate parasites)
7)Are found to affect organisms from all domains of life
8)Are very small (usually)

190
Q

What is the general structure of a virus

A

A protein coat (the capsid) surrounding a nucleic acid genome that may be:
1. A double stranded DNA
2.Single stranded DNA
3.Double stranded RNA (e.g., rotaviruses)
4. Single stranded RNA

A virus may be naked, or enveloped – surrounded by a lipid membrane acquired from the cell or nuclear membrane of an infected cell

191
Q

What do enveloped viruses acquire

A

their membrane when leaving the host cell. It encodes for insertion of glycoproteins into the host cell membrane, which it then picks up the glycoprotein spikes along with a potion of the cell membrane bilayer as it leaves

192
Q

Life cycle of a virus - DNA genome

A
  1. Enters the cell and releases its virus DNA
  2. Viral DNA replicates using specific RNA polymerases
  3. DNA also transcribed into mRNA then translated into capsid protein
  4. Capsid protein and viral DNA come together and become incapsid by the lipid membrane to form new virus particles
193
Q

Positive strand ssRNA viruses

A

single stranded RNA genome. Genome can act directly as mRNA

194
Q

Negative strand ssRNA viruses

A

single stranded RNA genome. Genome is complementary to mRNA

195
Q

What is a reterovirus

A

replicates via a DNA intermediate

196
Q

Replication of a + ss RNA virus

A

RNA-dependent RNA polymerase synthesis, RNA complementary to viral genome (-) strand. RNA polymerases use - strand as a template to synthesise new copies of virus genome

197
Q

Replication of HIV

A

Reverse transcriptase makes a DNA copy of their genome which is intergrated into a host cell chromosome

198
Q

What is a viroid

A

infectious agent without protein

199
Q

Potato spindle tuber viroid.

A

Affects growth of potatoes, tomatoes, peppers. The infectious agent is a circular, single stranded RNA molecule about 350 nucleotides in length with substantial internal base pairing. The RNA replicates inside infected cells

200
Q

What is a prion

A

an infectious agent without DNA or RNA

201
Q

What do prions cause and what are the effects

A

transmissible spongiform encephalopathies (TSEs). A group of related diseases leading to fatal and untreatable neurological deterioration

202
Q

What are transmissible spongiform encephalopathies charcterized by

A

very extensive neuronal death and spongiform degeneration (vacuolation) of the brain

203
Q

Spongiform degeneration (vacuolation) of farm animals

A
  1. Scrapie (affects sheep/goats)
  2. Bovine spongiform encephalopathies (cattle)
204
Q

What is Creutzfels-Jakob disease:

A

A transmissible spongiform encephalopathies of Humans. arises in middle and old age; most cases sporadic, a few transmitted (e.g., through treatment with contaminated growth hormone), some cases inherited as a dominant genetic disorder

205
Q

What is variant Creutzfels-Jakob disease:

A

A transmissible spongiform encephalopathies of Humans. Affects younger people, transmitted via BSE infected beef, now virtually eradicated

206
Q

What is fatal familial insomnia

A

A transmissible spongiform encephalopathies of Humans. A dominant genetic disorder that can be transmitted by injection of affected brain tissue into healthy lab animals

207
Q

Features of prions

A
  1. Much smaller than viruses
  2. Very heat resistant
  3. Very radiation resistant
208
Q

Mis folded prions

A

Prions are believed to be mis-folded versions of a normal cellular protein (the prion protein (PrP)). Once introduced into an unaffected brain, prions cause normally folded prion protein to mis-fold in turn. Mutation in the gene encoding the prion protein may make it more likely to mis-fold spontaneously, accounting for inherited transmissible spongiform encephalopathies

209
Q

Normal v misfolded prion protein - protease

A

normal is protease sensitive, mis-fold is resistant

210
Q

Normal v misfolded prion protein - alpha helix content

A

normal - high conent (42%). Mis-fold - low content (30%)

211
Q

Normal v misfolded prion protein - beta sheet content

A

normal - low content (3%). Mis-fold - high content (43%)

212
Q

Normal v misfolded prion protein - solublity

A

Normal - soluble. Mis-fold - insoluble

213
Q

Normal v misfolded prion protein - aggregate formation

A

normal - does not form aggregates. Mis fold - aggregates to form fibres

214
Q

Prion replication in transmissible spongiform encephalopathies

A

a normal PrP undergoes a conformational change. Infecting mis-folded PrP causes correctly folded proteins to mis fold

215
Q

Inherited transmissible spongiform encephalopathies

A

PrP with reduced stability and increase propensity to mis-fold undergoes a sponatneous conformational change and becomes mis-folded

Biological Chemistry (12 decks)

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