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LS3 > Midterm 2 > Flashcards

Midterm 2 Flashcards

0
Q

Enzyme for DNa replication

A

DNA polymerase

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1
Q

What are the three steps of transcription?

A

Initiation, elongation, and termination

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2
Q

How does the DNA polymerase add a nucleotide? (Chemistry)

A

The 3’ oh attacks the alpha phosphate of an incoming dNTP

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3
Q

Role of the exonuclease

A

It has a high affinity for incorrect pairs and degrades the DNA from the 3’ end

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4
Q

Processive versus non-processive

A

//

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5
Q

When does chromosome replication occur?

A

In the s phase of the cell cycle

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6
Q

End replication problem

A

Since DNA synthesis needs an RNA primer to initiate DNA strands, the ends of DNA will have a hard time replicating.

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7
Q

Telomeres

A

In eukaryotes, to fix the end replication problem, they have telomeres which are TG rich seq that doesn’t code for anything

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8
Q

Telomerase

A

A special DNA polymerase that only creates the telomeres. It has an RNA component and so it is a ribonucleoprotein and doesn’t need an template to add bases. It’s also like a reverse transcriptase because it uses the RNA template to make dna

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9
Q

What end does the telomerase act on?

A

It acts on the 3’ end

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10
Q

General causes of mutation

A

Environmental factors like chemicals and UV light, tautomerization, wrong base pairing, transposons, inaccuracy in replication

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11
Q

Transition mutation

A

When a purine is switched with another purine or pyrimidine with pyrimidine

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12
Q

Transversion mutation

A

Purine to pyrimidine switch

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13
Q

Point mutation

A

Mutations that alter a single nucleotide

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14
Q

Why are DNA micro satellites prone to mutation?

A

DNA microsatellites are long repeating seq like CACA and they are harder to replicate because slippage may occur. Therefore the repeat length may vary depending on the individual

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15
Q

Mutagen

A

A chemical that increases the rate of mutation

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16
Q

Deamination of cytosine

A

This is one of the most common mutations and turns cytosine into uracil

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17
Q

DNA depurination

A

This is when there is spontaneous hydrolysis of the N glycosyl linkage and it produces a deoxyribose without the base

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18
Q

Thymine dimer

A

When two thymine bases fuse to make a cyclobutane ring and this causes DNA polymerase to stop during replication if it reaches this point

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19
Q

How do gamma radiation and x-rays damage DNA?

A

They break the double strand which is really difficult for the cell to repair

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20
Q

Intercalating agents

A

They are able to slip between bases and cause errors in replication. They do this by causing additions, deletions, or even frame shifts

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21
Q

Base excision repair

A

DNA glycosylase removes the base first, and then AP endo and exo nuclease remove the backbone. The gap is then filled with the correct base by DNA polymerase.

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22
Q

Photoreactiviation

A

Reverses the formation of pyrimidine dimers from uv radiation by using the energy from light directly to break the dimer bond

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23
Q

Nucleotide excision repair

A

The DNA is scanned by a tetramer (UvrAB) for distortions and if a distortion is detected, UvrC will cleave a segment with the lesion out.

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24
Q

Methyltransferase

A

It reverses the methylation of guanine by taking the methyl group and putting on its own cysteine

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25
Q

What can mutations in the genes coding for nucleotide excision repair cause?

A

Can lead to UV light sensitivity

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26
Q

Double strand break repair

A

Usually repaired by non homologous end joining but during replication is is fixed by homologous recombination.

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27
Q

Non-homologous end joining

A

Ku70 and Ku80 bind to the broken ends and recruit DNA PKcs which then recruit Artemis, an exo and endonuclease that will process the broken ends. Then lipase is recruited to to seal the ends together.

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28
Q

Homologous recombination

A

The broken strand uses the original parent strand as a template and then the copied area is switched with the template so now the repaired strand has part of the template strand within it.

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29
Q

DNA cloning

A

Selective amplification of a particular gene or DNA segment

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30
Q

The five general steps of DNA cloning

A

1) cut DNA at specific location (restriction endonuclease)
2) select a cloning vector that can self replicate
3) join your DNA fragment with the vector
4) put this recombinant DNA into a host organism
5) select the cells that have the recombinant DNA

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31
Q

What do you need for a successful DNA vector?

A

You need an origin of replication, a selectable marker, and a region in which your DNA can be inserted

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32
Q

Plasmid

A

A circular double stranded DNA molecule with a size from 1 to 200 kb. They replicate separately from the host cells chromosomal DNA

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33
Q

How to insert your DNA into plasmid

A

Use same cleaving enzyme like EcoR1 to make sticky ends. Then allow your DNA fragments to anneal and use ligase to seal them together

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34
Q

How to get vector into host cell

A

Use transformation for bacterial cells and transfection in mammalian cells

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35
Q

BACs and YACs

A

They are used to clone large segments. If you clone these within the lacZ gene, you can tell if the insert is present or not.

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36
Q

How to tell if BAC or YAC has insert

A

If it does have the insert, the lacZ gene will be disrupted and won’t turn color in media with x-gal.

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37
Q

DNA library

A

1) DNA digested with restriction enzyme
2) cleaved DNA mixed with vector ligase
3) this creates a DNA library because each vector has different DNA fragment. You can then screen for your gene of interest

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38
Q

Genomic versus cDNA library

A

A genomic library represents the entire genome while a cDNA library represents only the genes expressed in the cell since the mRNA was copied into DNA

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39
Q

Reverse transcriptase

A

Converts RNA sequences into DNA. Needs a primer and either DNA or RNA template . You can make tissue specific cDNA libraries

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40
Q

RNA polymerase in prokaryotes

A

Has beta core

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41
Q

RNA polymerase in eukaryotes

A

There’s I, II and III but we focus mostly on RNAP II because it is involved in mRNA synthesis. The others make RNA for ribosomes and tRNA

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42
Q

How many metal ions are involved in RNA polymerase?

A

Two Mg2+ ions are involved but one comes with the incoming rNTP

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43
Q

What direction is RNA made in?

A

From 5’ to 3’

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44
Q

Template strand versus the coding strand

A

The template strand is the one the RNA will be made complementary to, while the coding strand is the one the RNA is identical to.

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45
Q

Antisense versus the sense strand

A 
Antisense = template strand 
Sense = coding strand
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46
Q

What does upstream mean?

A

It means toward the 5’ end of a given sequence

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47
Q

RNA polymerase holoenzyme

A

Only in prokaryotes, consists of the RNA polymerase core (5 subunits) and the sigma factor

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48
Q

Job of sigma factor

A

It recognizes the promoter region and helps the RNA polymerase bind to the promoter region.

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49
Q

Promoter region (types)

A

The most common promoter region has a -10 and -35 element. A stronger promoter might have a UP element. The third type has no -35 but instead has an extended -10. The last type has a discriminator next to the -10 region.

50
Q

What recognizes the UP element?

A

Alpha CTD (or the carboxyl terminal domain of the alpha subunit)

51
Q

Steps within transcription initiation

A

Closed complex = binding of sigma factors to promoter region
Open complex = DNA begins to be unwound
Initial transcript = short RNAS are made and released repeatedly until elongation phase begins

52
Q

How does the RNA polymerase transition to the open complex?

A

The sigma 1 factor moves out of the active site and the promoter melts

53
Q

Steps of transcriptional elongation

A

RNA polymerase unwinds the DNA at the front, adds rNTPs, reanneals the DNA in the back, proofreads and dissociates the mRNA from the template

54
Q

Hydrolytic editing*

A

The nuclease within the RNA polymerase cleaves off an entire segment of ribonucelotides

55
Q

Pyrophosphorolytic editing

A

Look in book

56
Q

Terminator

A

A sequence that triggers the RNA polymerase to dissociate from DNA and release the RNA

57
Q

Rho independent terminator

A

A sequence rich in GC that will form a hairpin and cause a release of the polymerase without the need of a protein

58
Q

Rho dependent terminator

A

When the protein factor rho is present, RNA synthesis will stop and every dissociates. These sequences have a rut site that rho binds to.

59
Q

GTF’S

A

They are general transcription factors that are needed in eukaryotes to recognize and bind the promoter. Somewhat similar to the sigma factor in prokaryotes

60
Q

The transcription factors needed for pol II

A

TFIID, TFIIA, TFIIB, TFIIF, TFIIE, TFIIH

61
Q

The main promoter for eukaryotes

A

TATA box

62
Q

What binds to the TATA box first?

A

TBP and TFIID bind first. They then recruit other GTF’s and pol II which forms the ore initiation complex

63
Q

Steps for initiation in eukaryotes

A

Closed complex = melting of the promoter by TFIIH (needs ATP)
Open complex = phosphorylation of the polymerase tail by TFIIH which also needs ATP
Promoter escape = release from promoter and most general transcription factors

64
Q

What is the last step needed before elongation occurs?

A

The mediator complex connects the activator sequence to the initiation complex. Chromatin remodeler and HAT are also needed

65
Q

What happens in eukaryotic elongation?

A

The transcription initiation factors and mediator complex are shed and new proteins are recruited. Different phosphorylation of the tail exchanges initiation factors for the new proteins

66
Q

What phosphorylation signals the capping enzyme in the promoter escape phase?

A

Phosphorylation of serine 5

67
Q

What phosphorylation signals the splicing machinery?

A

Serine 2 phosphorylation

68
Q

5’ cap

A

A modified guanine is added to the 5’ end of the mRNA. The beta phosphate of the chain attacks the alpha phosphate and creates a 5’ to 5’ linkage. This guanine is then methylated (modified). All eukaryotic mRNAs have a cap

69
Q

Polyadenylation

A

Poly A tail is added to the 3’ end. CstF and CPSF bind to the tail, and then to the RNA when signal sequence is transcribed. CstF leaves after it cleaves the RNA and PAP is recruited. PAP adds about 200 Adenines to the RNA without a template and finally polyA binding proteins bind to the tail

70
Q

Torpedo model

A

After the first RNA strand is cleaved off to make the polyA tail, the uncapped RNA is degraded by RNase and this destabilizes the RNA polymerase and causes termination.

71
Q

Allosteric model

A

Once the polyA signal is passed, the RNA polymerase becomes less processive because of a conformational change

72
Q

Function of polyA tail and 5’ cap

A

They stabilizes the mRNA, allow it to be transported out of the nucleus, and help with translation initiation

73
Q

Gene expression is prokaryotes versus eukaryotes

A

In prokaryotes, transcription and translation can occur simultaneously. While in eukaryotes, there is transcription, then RNA processing, then transport, and finally translation.

74
Q

Exon

A

Sequences that contain the genetic information and remain present within the final RNA product

75
Q

Introns

A

Intervening sequences that are removed by RNA splicing

76
Q

hnRNA

A

Heteronuclear RNA. RNA of many different sizes

77
Q

Pulse chase experiment of RNA

A

In the pulse, RNA was labeled with 32P for 30 minutes. The RNA was then separated by rate zonal centrifugation. In the chase, they stopped transcription and allowed the cells to grow for 3 hours. They concluded that mRNA is derived from hnRNA.

78
Q

Signals for RNA splicing

A

There’s the 5’ splice site, 3’ splice site and the branch site. They occur at exon/intron boundaries in most genes

79
Q

The splicing reactions

A

It includes two transesterification reactions.

1) 5’ splice site is cleaved
2) the branching point OH attacks the 5’ end phosphate and forms a loop (lariat) 5’ to 2’ bond
3) the 3’ end is cleaved
4) the 5’ exon attacks the 3’ exon and they join together

80
Q

What is the spliceosome made of?

A

Made of SnRNPs (small nuclear ribonucleoprotein particles) U1,2,4,5,6 which catalyze the reactions. They are ribozymes!

81
Q

Which SnRNP recognizes the 5’ splicing site?

A

Usually U1

82
Q

The steps of the spliceosome mediated splicing reaction

A

E complex, a complex, b complex, and finally the c complex

83
Q

What happens in the E complex of splicing reactions?

A

U1 recognizes the 5’ binding site, BBP binds to the branching point, and U2AF binds to the py tract and the 3’ splice site

84
Q

What happens in the A complex of splicing?

A

U2 comes in and replaces BBP

85
Q

B complex of the splicing reactions

A

U4, U5 and U6 bind to the intron and form a complex with the others. U1 leaves and U6 takes its place at the 5’ splice site. U4 is then released and this allows U6 to interact with U2

86
Q

C complex of RNA splicing

A

U6 and U2 catalyze the first transesterification reaction and then U5 helps with the second one.

87
Q

A rare form of RNA splicing

A

Some rare RNA have the ability to splice themselves. They can do this by folding into a conformation that catalyzes it’s own release from the surrounding exons

88
Q

Alternative RNA splicing

A

Occurs in 95% of human genes with introns. Allows multiple proteins to be made from one gene. It is often cell type specific.

89
Q

Introns early model

A

Originally prokaryotes had introns too but lost them over time in order to replicate more efficiently.

90
Q

Introns late model

A

Introns were inserted into genes that originally did not have introns

91
Q

Exon shuffling

A

Allows the creation if new genes through duplication and recombination. It is one reason why introns have been kept in our genome.

92
Q

What do you need for translation? (Key components)

A

mRNAs, tRNAs, and the ribosome

93
Q

Three stages of translation

A

Initiation, elongation, and termination. Same as transcription!

94
Q

How many codons are there?

A

There are 61 codons coding for 20 amino acids plus 3 stop codons for a total of 64 codons.

95
Q

How many possible reading frames does mRNA have?

A

3 possible reading frames

96
Q

Open reading frame

A

Starts with a start codon and ends with a stop codon

97
Q

Which direction are codons read?

A

5’ to 3’

98
Q

Silent mutation

A

When a nucleotide is switched but it still codes for the same amino acid

99
Q

Missense mutation

A

When a nucleotide is switched and codes for a completely different amino acid

100
Q

Nonsense mutation

A

When a nucleotide is switched and the new codon is now a stop codon

101
Q

Which end of the tRNA holds the amino acid?

A

The 3’ end. It usually ends with CCA

102
Q

Uncharged versus charged tRNA

A

Uncharged doesn’t have an amino acid and charged does

103
Q

Secondary and tertiary structure of a tRNA

A

Secondary is like a cloverleaf with anticodon loop, variable loop, pseudouridine loop, d loop and acceptor arm. The tertiary structure looks like an L with the acceptor arm on one end and the anticodon loop on the other end

104
Q

What enzyme charges the tRNAs and how does it do it?

A

Aminoacyl-tRNA synthetase requires ATP in order to add the amino acid to the tRNA. (20 different synthetases)

105
Q

Adenylation

A

The carboxylic acid of the amino acid attacks the alpha phosphate of the adenine to make an adenylylated amino acid. This process requires ATP

106
Q

How does tRNA charging occur? (Chemical process)

A

The adenylylated amino acid is attacked by the 3’ OH of the tRNA. This process releases the AMP.

107
Q

Size of ribosome subunits in prokaryotes

A

The large subunit is 50S and the small subunit is 30S. This adds to 70S

108
Q

Ribosome subunit size in eukaryotes

A

The large subunit is 60S and the small subunit is 40S to make a total of 80S

109
Q

The ribosome cycle

A

Initiation, elongation, and termination

110
Q

Polysomes

A

When multiple ribosomes can translate a single mRNA molecule simultaneously

111
Q

Function of the small ribosome subunit

A

It mediates the interaction between the codons and tRNA anticodons

112
Q

Function of the large ribosome subunit

A

Catalyzes the peptide bond formation, and has binding sites for G proteins that assist in initiation, elongation, and termination

113
Q

Peptidyl transferase reaction

A

The reaction that adds the next amino acid to the chain (creating a peptide bond)

114
Q

What direction are polypeptides synthesized?

A

From the n terminus to the c terminus

115
Q

The structure of a ribosome

A

Has an a site or aminoacyl site, the p site (peptidyl) and finally the e site (exit)

116
Q

Prokaryotic translation initiation

A

The 16s rRNA be pairs with the ribosome binding site (RBS) and this positions the start codon with the p site of the ribosome

117
Q

Polycistronic

A

When mRNA contains multiple open reading frames

118
Q

Initiation in prokaryotic translation

A

The small subunit binds to the start codon and then the initiator tRNA binds to the AUG and has a special methionine. The large subunit then comes in to form the complete ribosome

119
Q

Function of IF3, IF2, and IF1

A

IF3 binds the e site and doesn’t allow the 50S to bind. IF1 blocks aminoacyl tRNA from binding to the a site. IF2 is a gtpase that interacts with IF1 and the special methionine tRNA. This leaves only the p site open for the initiator tRNA

120
Q

Monocistronic

A

MRNAs with only one open reading frame

121
Q

Why are eukaryotic mRNAs held in circles? And how?

A

This is so the ribosome can easily reassociate to make more protein. It is held in a circle by the binding of elF4A to the poly A tail

122
Q

Release factors for translation in prokaryotes

A

The class 1 RFs release the peptide chain from the last tRNA. Class 2 RFs release the class 1 factors.

123
Q

RRF

A

Ribosome recycling factor helps release the tRNAs and dissociate the ribosome subunits

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