Brainscape's Knowledge GenomeTM

Browse over 1 million classes created by top students, professors, publishers, and experts.


See full index

Biochemistry 501 > Unit 2 > Flashcards

Unit 2 Flashcards

1
Q

monomers of nucleic acids

A

nucleotides

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

3 components of a nucleotide

A

1) base
2) sugar
3) phosphate

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

2 types of nucleotide bases

A

1) purine
2) pyrimidine

longer word, smaller structure

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q
A

Adenine

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q
A

Guanine

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q
A

Cytosine

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q
A

Thymine (DNA)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q
A

Uracil (RNA)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

nucleoside

A

base connected to a pentose sugar
-for RNA, the sugar is ribose
-for DNA, the sugar is deoxyribose because it lacks the 2’OH

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

What is the structural (sugar) difference between RNA and DNA?

A

DNA lacks the 2’OH of ribose (making it deoxyribose), while RNA does not

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

connection between base and sugar

A

glycosidic bond (covalent) forms between the 1’ position on the sugar and a nitrogen on the base

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

nucleotides

A

the building block of nucleic acids
-a nucleoside with phosphoryl group(s) attached via ester linkage

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

nucleoside monophosphates

A

what the final nucleic acid polymer is composed of

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

nucleoside triphosphates

A

serve as high energy building blocks used by the cell to synthesize the nucleic acid polymer

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

phosphodiester linkage

A

covalent bond that connects two nucleotides to form polymer (nucleic acid)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

mechanism of phosphodiester bond formation

A

1) Base activates 3’OH
2) 3’OH acts as a nucleophile, attacking the alpha phosphate (phosphate closest to the sugar) of a nucleoside triphosphate
3) Pyrophosphate acts as a leaving group to drive the reaction forward

Note: The nucleic acid polymer always grows in the 5’ to 3’ direction (i.e., nucleotides always added at 3’ end)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
17
Q

How do two strands of nucleotides interact?

A

complementary base pairing (Watson-Crick-Franklin)
-bases of nucleotides have hydrogen bond donors/acceptors
-C/G and A/T (DNA) or A/U (RNA)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
18
Q

How many hydrogen bonds are formed in A/T complementary base pairing?

A

2 hydrogen bonds (less energy required to separate; lower melting point)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
19
Q

How many hydrogen bonds are formed in C/G complementary base pairing?

A

3 hydrogen bonds (more energy required to separate; higher melting point)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
20
Q

B-form double-stranded DNA double-helix

A

-right-handed turn
-strands are anti-parallel
-sugar-phosphate backbone on the outside
-nucleobases on the inside
-asymmetrical

Also:
-orthogonal base-pairing
-base-stacking interactions

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
21
Q

dsDNA base-stacking interactions

A

-van der Waals interactions between hydrophobic nucleobase faces (steric effects)
-pi-stacking (electronic effects)

These interactions have the effect of re-enforcing the individual A/T and C/G hydrogen-bonding interactions to drive massive stabilization in the larger context of the dsDNA helix

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
22
Q

major groove

A

wide and deep groove in B-form dsDNA that provides access to the nucleobases from the outside
-proteins can bind through the major groove in a sequence specific manner

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
23
Q

minor groove

A

shallow and narrow groove in B-form dsDNA that provides access to the nucleobases from the outside

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
24
Q

puckered nucleotide sugar

A

in three dimensions, the sugar is puckered and can exist in a C3’-endo or C2’-endo conformation
-RNA nucleotides prefer the C3’-endo conformer because the C2’-endo conformation creates steric problems due to the 2’OH
-DNA nucleotides are predominated by the C2’-endo conformation, but the 2’ position is a hydrogen so either conformation is possible

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
25
Q

RNA nucleotide conformation (sugar puckering)

A

the C3’-endo conformation is preferred because the C2’-endo conformation creates steric problems due to the 2’OH

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
26
Q

DNA nucleotide conformation (sugar puckering)

A

the C2’-endo conformation predominates, but the 2’ position contains a hydrogen (as opposed to OH in RNA) so either conformation is possible
-B-form is C2’-endo (phosphates farther apart)
-A-form is C3’-endo (phosphates closer together)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
27
Q

C2’-endo vs C3’-endo in dsDNA

A

C2’endo places phosphates farther apart than C3’-endo
-B-form is C2’-endo (phosphates farther apart)
-A-form is C3’-endo (phosphates closer together); RNA prefers A-form DNA for this reason

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
28
Q

syn- vs anti- base conformations

A

syn- has the base “over” the sugar, while anti- has the base swung out like a flag; the anti-conformation is energetically more favorable
-only purines can adopt the syn conformation (although it’s very rare and results in Z-form DNA)
-traditional Watson-Crick-Franklin base pairs are anti/anti, but Hoogsteen base pairs can form

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
29
Q

Z-form DNA

A

results from the syn-conformation of purines and is very rare
-left-handed and has a zig-zag shape

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
30
Q

Hoogsteen base pairs

A

forming of base pairs from syn-nucleotides (only one base needs to be syn)
-allows 3-4 strands to be present in a helix as part of some nucleic acid structures
-found in damaged DNA and DNA bound by drugs, but may also be present normally to regulate gene expression

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
31
Q

tertiary structure in DNA/RNA

A

regions of secondary structure in nucleic acids can fold into a complex 3D structure (tertiary); possible in DNA but mostly seen in RNA molecules
-stabilized by metal ions and unusual base pair geometries

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
32
Q

polynucleotide backbone

A

sugar-phosphate via phosphodiester linkage forms polynucleotide backbone

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
33
Q

Is RNA typically single- or double-stranded?

A

typically single-stranded

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
34
Q

What do the 5’ and 3’ ends refer to?

A

5’ = the 5’ carbon of a nucleotide
3’ = the 3’ hydroxyl of a nucleotide

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
35
Q

Why is the genome DNA and not RNA?

A

RNA is more susceptible to degradation because the 2’OH can lead to breakdown of the chain

Other reason: B-form DNA allows easier access to info via the major groove

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
36
Q

Why do we need to separate annealed DNA strands in the genome?

A

the annealed form is great for long-term storage but we also need to separate the strands to make use of base-pairing for replication, transcription, etc.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
37
Q

Tmelting(m)

A

the temperature at which the helix is half double-stranded, half single-stranded (50% denatured)
-stable helix = high Tm
-unstable helix = low Tm

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
38
Q

linking number (Lk)

A

describes the topology of a DNA fragment; the sum you get from counting twist and writhe in a strcuture (Lk=Tw+Wr)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
39
Q

twist (Tw)

A

number of times each of the curves rotates around the central axis C of the double helix

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
40
Q

writhe (Wr)

A

the number of times the intact B-form helix twists about itself

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
41
Q

Can linking number be changed by simply deforming the structure?

A

No, if you try to increase or decrease twist, you will introduce writhe to ensure that the linking number remains unchanged; conversely, if you introduce writhe, the number of twists will adapt to endure that the linking number remains unchanged

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
42
Q

When can we say DNA molecules have the same topology?

A

any time the linking number is the same; if linking number is different, the configuration of DNA is a different topology

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
43
Q

natural twist (Lk0)

A

the helical structure of B-form DNA imparts a natural twist to the DNA polymer

Lk0 = (# of base-pairs)/10.4
This is essentially just the number of turns

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
44
Q

Lk, Lk0, and DNA stability

A

B-form DNA is extremely energetically favorable, so for a given Lk, the Tw component will be ~Lk0, and the Wr will account for the difference between Lk and Lk0

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
45
Q

difference between linking number and natural twist

A

deltaLk = Lk - Lk0
-if deltaLk < 0, the DNA is underwound and negative-supercoiling will result
-if deltaLk > 0, the DNA is overwound and positive-supercoiling will result
-if deltaLk = 0, the DNA is relaxed

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
46
Q

How do cells change DNA topology (linking number)?

A

topoisomerase enzymes cut a strand, allowing the unbroken DNA strand to pass through break in first strand, then the cleaved strand is religated

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
47
Q

What is the coiled state of most DNA in a cell?

A

as a general rule, most DNA inside the cell is negatively supercoiled

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
48
Q

supercoiling

A

results from changing twist of dsDNA

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
49
Q

nucleosomes

A

DNA wound in histones; this is how DNA is compacted in eukaryotes

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
50
Q

How is DNA compacted in eukaryotes?

A

via the wrapping of DNA in histones and additional compaction; this drives chromosome assembly

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
51
Q

histone acetylation

A

if histones are acetylated, chromatin opens and info becomes accessible; i.e., euchromatin
-histones can be acetylated on lysine residues by HATs
-acetylation can be removed from histones by HDACs
-acetylation reduces affinity for DNA
-acetylation marks can recruit transcriptional activators

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
52
Q

histone methylation

A

if histones are methylated, heterochromatin proteins such as HP1 bind across them to promote chromatin compaction; i.e., heterochromatin
-histones can be methylated on lysines by histone methyltransferases
-histone methylation can be removed by histone demethylases
-methylated histones interact with heterochromatin proteins that oligomerize to coat and compact methylated regions

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
53
Q

euchromatin

A

accessible, open regions of chromosomes; transcription can occur here (active)
-marked by histone acetylation

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
54
Q

heterochromatin

A

inaccessible, compact regions of chromosomes; transcription cannot occur here (silent)
-marked by histone methylation and binding of heterochromatin proteins

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
55
Q

parent strand

A

serves as a template to generate the daughter strand in replication; daughter helix composed of one strand from parent and one entirely new strand

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
56
Q

semi-conservative genome replication

A

each daughter helix gets one strand from the parent and one completely new strand

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
57
Q

What catalyzes nucleotide addition?

A

DNA polymerase

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
58
Q

How does DNA polymerase catalyze nucleotide addition?

A

-adds dNTPs (nucleotide triphosphates) to the 3’ end of that last nucleotide in the polynucleotide strand
-synthesis can only proceed off an already existing double-stranded fragment

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
59
Q

DNA polymerase (general)

A

catalyzes nucleotide addition by adding dNTPs to the 3’ end of a polynucleotide strand
-requires a primer and Mg2+ (metal)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
60
Q

Mg2+ and DNA pol/transcription

A

-activates 3’OH attack on dNTP phosphate, makes alpha phosphate more electrophilic
-stabilizes phosphate’s negative charge

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
61
Q

Why does DNA polymerase require a primer?

A

base-pairing interactions are more specific in the context of an existing double-stranded structure than on their own
-dsDNA/primer facilitates addition of complimentary base-pairs to the growing strand by taking advantage of the energy of base-stacking

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
62
Q

tautomerization of nucleobase

A

rare situation in which the position of H-bond donors/acceptors are different
-allows for C/A and T/G pairing

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
63
Q

nontautomeric mispairing

A

wobble mispairing resulting from bases with an extra proton that can still bind but often with a mismatched nucleotide

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
64
Q

“normal” mispairing

A

wobble mispairing between normal bases that nonetheless bond inappropriately because of a slight shift in position of the nucleotides in space

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
65
Q

wobble mispairing

A

mispairings that occur because DNA double helix is flexible enough to accomodate slightly misshaped pairings
-nontautomeric (bases with extra proton) mispairings
-“normal” mispairings

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
66
Q

DNA polymerase proofreading

A

DNA polymerase has 3’ to 5’ proofreading activity to correct mistakes
-the exonuclease domain removes errant nucleotides from the end of a DNA strand

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
67
Q

two domains of DNA polymerase

A

1) polymerase - adds nucleotides
2) exonuclease - removes nucleotides

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
68
Q

How does the DNA pol exonuclease correct errors?

A

an errant base destabilizes the dsDNA structure, causing the 3’ end to “flop” into the exonuclease site where the terminal nucleotide can be removed

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
69
Q

DNA polymerase III

A

the “workhorse” polymerase (main replicative polymerase) that is built for speed and processivity (once it gets going it’s going to run for as long as it can)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
70
Q

DNA polymerase I

A

the “handyman” polymerase (odd-jobs polymerase) that polymerizes small stretches of DNA as part of cleanup or repair jobs
-built for accuracy instead of speed

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
71
Q

ori

A

origin of replication
-prokaryotes have a single ori
-euaryotes have multiple oris

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
72
Q

helicases

A

unwind DNA using energy of ATP hydrolysis

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
73
Q

primase

A

synthesizes a short complementary RNA sequence to serve as a primer for DNA polymerase

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
74
Q

Okazaki fragments

A

the fragments that make up the discontinuous strand

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
75
Q

How are primers removed in replication?

A

DNA pol I recognizes the RNA primers and removes them, replacing them with DNA sequence

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
76
Q

DNA ligase

A

joins together Okazaki fragments of the lagging strand and nicks created by DNA pol I

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
77
Q

How does DNA ligase catalyze the reaction joining Okazaki fragments?

A

1) Ligase uses an ATP to adenylate itself, forming Ligase-AMP
2) AMP is transferred from ligase to the 5’ phosphate of the nicked DNA strand
3) Base catalyzed nucleophilic attack by 3’OH can seal the strand

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
78
Q

single-stranded binding proteins (SSBs)

A

help keep ssDNA from reannealing after being melted by helicase

79
Q

telomerase

A

prevents ends from shrinking by extending leading strand (brings its own RNA primer)

80
Q

telomeres

A

regions of repetitive sequences at the end of eukaryotic chromosomes

81
Q

What are the two main paths to mutations in the genome?

A

1) polymerase makes a mistake
2) chemistry takes its toll

82
Q

mismatch repair pathway

A

1) Cells specifically mark the parental strands of DNA with methylation marks
2) Protein complex detects mismatch, indentifies which strand is methylates, and cleaves the OTHER (daughter) strand error
3) Resulting gap is filled in by DNA pol III and the nick is sealed by DNA ligase

83
Q

alkylation chemical modification

A

adds a methyl or ethyl group to the nucleobase, disrupting a position of H-bond donor/acceptor
-can be mutagenic or not
-corrected by methyltransferase

84
Q

mutagenic

A

chemical modification that disrupts base-pairing

85
Q

deamination chemical modification

A

replaces a nucelobase amine with a carbonyl group, changing the nucleotide as a whole
-thymine cannot undergo deamination
-corrected by base-excision repair
-makes alternative base-pairing between deaminated bases possible

86
Q

depurination chemical modification

A

loss of the entire purine base from the nucleotide
-can’t base-pair (literally doesn’t have a base)
-corrected by base-excision repair

87
Q

UV light chemical modification

A

UV light can cause adjacent, stacked pyrimidines to dimerize
-these dimers produce a distorted DNA structure that is often misread by DNA pol and thus results in replication errors
-corrected by nucleotide-excision repair
-pyrimidines only

88
Q

direct repair by methyltransferase (alkylation repair)

A

the alkyl groups can be transferred from the nucleobase to an alkyltransferase, often becoming permanently stuck; therefore, an entire protein takes on the damage from an alkylated base in a sacrificial manner

89
Q

base-excision repair (depurination repair)

A

1) AP nuclease recognizes abasic sites and nicks the offending backbone at the site of damage, exposing a stretch of dsDNA with a free 3’OH and single nucleotide of ssDNA (this provides a template for DNA pol to extend)
2) DNA pol hops on, fills in the gap, and displaces some DNA
3) Ends are ligated together

Note: exact same as deamination repair but without glycosylase enzyme

90
Q

base-excision repair (deamination repair)

A

1) Simply convert the offending nucleobases into an abasic site using a glycosylase enzyme
2) AP nuclease recognizes abasic sites and nicks the offending backbone at the site of damage, exposing a stretch of dsDNA with a free 3’OH and single nucleotide of ssDNA (this provides a template for DNA pol to extend)
3) DNA pol hops on, fills in the gap, and displaces some DNA
4) Ends are ligated together

Note: exact same as depurination repair after glycosylase enzyme is applied

91
Q

glycosylase enzyme

A

removes base from a nucleotide
-used to correct deamination in the base-excision repair pathway

92
Q

nucleotide excision repair (UV dimer repair)

A

1) Surveillance complex scans DNA structure and locates error
2) Excinuclease cuts both sides of the lesion, strand is unwound
3) DNA pol I fills in the gap and ligase connects the ends

93
Q

endonuclease vs exonuclease

A

endonuclease cleaves the phosphodiester backbone within a nucleotide chain, while exonuclease cleaves the phosphodiester backbone at the 3’ end of a nucleotide chain

94
Q

homologous recombination

A

1) 5’->3’ exonuclease generates free 3’ overhangs (allows for DNA polymerase activity)
2) 3’ overhang performs base pairing interactions with target (homologous segment), displacing one strand of DNA (“strand invasion”)
3) Polymerase extends off free 3’ end using homology target as template
4) Four-way DNA structure called a “Holliday junction” results
5) Holliday junction resolved by cutting either vertically or horizontally (either way, “origins” of DNA in each chromosome is altered)

95
Q

Holliday junction

A

formed by two crossovers in homologous recombination
-B-form is preserved
-helicases can translocate the position of the cross-over point in a process called branch migration
-can be resolved by cutting either vertically or horizontally

96
Q

branch migration

A

process by which helicases can translocate the position of the cross-over point in a Holliday junction

97
Q

two options following cleavage of a Holliday junction

A

1) DNA only “exchanged” between chromosomes along the site of repair
2) An entire arm of chromosome is exhanged outside the repair site

98
Q

homologous recombination and variety

A

homologous recombination also functions to generate variety in the distribution of paternal and maternal genes on your chromosomes:

1) Provides a way to ensure that homologous chromosomes correctly pair during gamete production (must have same sequences to initiate recombination)
2) Resolution of Holliday junctions leads to exchange between chromosomes

99
Q

sequence specific recombination

A

induces DNA rearrangements at defined sites; can be carried out by either tyrosine recombinase or serine recombinase
-orientation of recombinase leads to different recombination products

100
Q

tyrosine recombinase mechanism

A

1) Tyrosine -OH attacks phosphate in DNA backbone, nicking the backbone and forming phosphodiester bond between tyrosine and DNA
2) DNA -OH attacks phosphodiester bond between tyrosine and DNA, kicking off tyrosine and forming a Holliday junction between fragments
3) Tyrosine -OH in recombinase attacks phosphate in DNA backbone again
4) DNA -OH agains attacks phosphodiester bond between tyrosine and DNA, kicking off tyrosine
5) Final product

101
Q

tyrosine recombinase takeaways

A

1) Two sequential single strand breaks
2) Holliday junction intermediate

102
Q

serine recombinase mechanism

A

1) Two double strand breaks produced simultaneously - one across each dsDNA fragment
2) Fragments on one side of the break are rotated so that they switch places
3) Double strand break is then repaired by religation of the broken pieces

i.e., cleave, rotate, connect back together

103
Q

serine recombinase takeaways

A

1) Two double strand breaks introduced simultaneously
2) Repair proceeds without a Holliday junction

104
Q

tissue-specific promoter

A

a recombinase protein can be placed under a tissue-specific promoter so that the recombination only occurs within certain locations in the body

105
Q

two sequence specific recombination outcomes

A

1) insertion/excision
2) inversion

106
Q

transposons

A

transposable elements that copy/paste and cut/paste themselves into the genome; selfish - only purpose is to use your genome to propogate themselves

107
Q

Does transposon movement always cause changes in gene expression?

A

No, only if inserted into a gene coding region or a sequence that regulates gene transcription; this in turn creates genetic diversity (corn example from slides)

108
Q

two major transposon types

A

1) Class I: retrotransposons (copy/paste) - copied to an RNA intermediate that is used to make a second DNA copy for insertion
2) Class II: DNA transposon (cut/paste) - DNA sequence is excised and pasted elsewhere in the genome

109
Q

genetic mosaics

A

the movement of transposons within a genome means that different cells in the ame tissue can have different genomes from each other; they are genetic mosaics

110
Q

What strand do we copy in transcription?

A

an mRNA copy of the coding strand is made using base-pairing interactions with the template strand

111
Q

nucleotide addition by RNA polymerase

A

RNA polymerase is used to add nucleotides to mRNA in transcription (proceeds like DNA polymerase)

-Base activation of 3’OH allows it to act as a nucleophile, attacking the 5’ alpha phosphate of an NTP
-MG2+ ons are critical to activate -OH and stabilize the negative charge on the phosphates during the reaction

112
Q

DNA polymerase vs RNA polymerase

A

-RNA polymerase has an intrinsic helicase (it unwinds/melts DNA on its own) and does not require a primer
-RNA polymerase does not have a separate exonuclease site for proofreading, but can “backtrack,” allowing the same active site used for polymerization to act as a nuclease and give a second chance to correctly transcribe

113
Q

Can both strands of DNA be transcribed?

A

Yes, but not at the same time; only one strand at a time is transcribed
-this is because RNA polymerase holds onto both the template strand and coding strand in the complex

114
Q

Why might RNA polymerase not need a primer but DNA polymerase does?

A

1) DNA replication needs to be more accurate to prevent deleterious mutations from locking in the genome
2) RNA polymerases initiate transcription at promoters

115
Q

gene promoter

A

“landing strip” of DNA that specifies where RNA polymerase will initiate transcription
-positions the RNA polymerase/DNA complex to initiate at a specific location and to use the correct strand as a template

116
Q

sigma subunit

A

facet of RNA polymerase in bacteria that recognizes the promoter in transcription

117
Q

How does RNA polymerase initiate transcription?

A

-in bacteria, sigma subunit recognizes promoter sequence, RNA polymerase binds here, and the sequence positions the complex to initiate at a specific location and use the correct strand as a template
-RNA polymerase melts the DNA strand and begins to add complementary nucleotides, causing sigma to dissociate and the polymerase to “escape” the promoter

118
Q

elongation state

A

state of RNA polymerase in transcription caused by mRNA extension proceeding rapidly

119
Q

RNA polymerase in prokaryotes vs eukaryotes

A

-eukaryotes have 3 forms of RNA polymerase, and each type has its own initiation specificity (recognizes different DNA sequences as promoters)
-transcription in eukaryotes is also regulated by enhancer elements and other sequences that might be far from the transcription site

120
Q

Rho-dependent termination (prokaryotes)

A

Rho factor is a helicase that translocates from the Rho binding site to the 3’ end of the transcript (“climbs” to RNA polymerase), where it unwinds/”peels” RNA polymerase off the template strand

121
Q

Rho factor

A

helicase that acts as a terminator for transcription by translocating (“climbing”) towards the 3’ end of the transcript, where it unwinds/”peels” RNA polymerase off the template strand

122
Q

Rho-independent termination (prokaryotes)

A

if RNA polymerase transcribes through a stretch of DNA that will produce a G/C rich region and a U-rich segment downstream, a hairpin could form, causing RNA polymerase to stall; the U-rich region downstream can only form weak base-pairing interactions with the template strand, and the combination of the stall and weak interaction causes the polymerase to fall off

123
Q

hairpin structure

A

forms when the template strand in transcription produces a G/C rich region in mRNA, stalling the RNA polymerase and thereby acting as a terminator when paired with a U-rich segment downstream

124
Q

When does a bacterial (prokaryotic) cell use the lac operon to produce lactose?

A

in the absence of glucose, as glucose is the preferred energy source

125
Q

repressor

A

binds to operator sequence and blocks RNA polymerase binding, preventing transcription
-operon is only repressed in the absence of a ligand metabolite, as the ligand binds repressor and prevents it from binding to DNA

126
Q

What is the only situation in which the lac genes are not expressed (transcription does not occur)?

A

when a repressor is present but not lactose metabolite
-any time the repressor is absent or a lactose metabolite is present, gene expression occurs

127
Q

lac operon activator synthesis (switch from glucose production to lactose production)

A

1) Glucose inhibits cAMP production, so as glucose levels decrease, cAMP levels increase
2) cAMP binds to a protein called CAP
3) CAP binds to a promoter proximal site and stimulates RNA polymerase recruitment, making it an activator of transcription

128
Q

CAP

A

activated by elevated levels of cAMP in response to low levels of glucose, binds to a promoter proximal site and stimulates RNA polymerase recruitment to the promoter, making it an activator of transcription

129
Q

two mechanisms for turning a gene on in the presence of a ligand

A

1) Repressor (no ligand) = gene off —> repressor + ligand = gene on
2) Activator (no ligand) = gene off —> activator + ligand = gene on

130
Q

two mechanisms for turning off a gene in the presence of a ligand

A

1) Activator - ligand = gene on —> activator + ligand = gene off
2) Repressor - ligand = gene on —> repressor + ligand = gene off

131
Q

Why is eukaryotic gene regulation/transcription so much more complex?

A

the fact that DNA is wrapped in nucleosomes means it will always take extra effort to transcribe any gene

132
Q

eukaryotic transcription activators

A

activators bind to enhancer elements to recruit coactivators and the basal machinery to stimulate transcription
-basal transcription machinery is needed to be able to assemble eukaryotic RNA polymerase II at promoter sites

133
Q

relationship between eukaryotic activators and repressors

A

eukaryotic activators and repressors are modular proteins that can infuence transcription and influence each other; allows for combinatorial control of gene expression
-Activators:
1) recruit coactivators
-Repressors:
1) block coactivator recruitment
2) block activator binding to DNA
3) recruit corepressors

134
Q

sequence-specific interaction of repressor and DNA

A

hydrogen bonds form between specific amino acids and nucleotide bases based on patterns of H-bond donors and acceptors in each molecule; interactions occur in the major groove of B-form dsDNA
-in this way, proteins can read the bases without needing to “melt” the DNA

135
Q

What determines sequence-specificity of DNA binding proteins (e.g., repressor)

A

motifs in the protein structure that make direct contact with the nucleobases through the major groove of B-form dsDNA

136
Q

How do researchers test if two proteins interact?

A

exploiting the modularity of transcriptional activators (DNA binding activity is often separate from regulatory activity); allows researchers to screen protein-protein interactions quickly

137
Q

epigenetics

A

changes in gene expression not caused by changes in DNA sequence
-3 focal types of epigenetic regulation:
1) Regulation of chromatin structure
2) Chemical modification of nucleosides
3) microRNA-mediated regulation

138
Q

two regions of chromatin (accessibility)

A

1) heterochromatin - transcriptionally silent
2) euchromatin - transcriptionally active

139
Q

5-methyl cytosine

A

epigenetically inherited DNA modification that causes methylation with each replication, thereby leading to transcriptional silencing (useful for silencing transposons by turning off gene expression)

140
Q

histone methyltransferases

A

methylate histones on lysines

141
Q

histone acetyltransferase (HAT)

A

acetylate histones on lysines

142
Q

transcription factors

A

activators and repressors
-bind to enhancers
-stabilize or destabilize RNA polymerase on the promoter

143
Q

final mRNA transcript length

A

in eukaryotes, the final mRNA transcript is shorter than the DNA template from which it is transcribed due to removal of introns

144
Q

introns

A

sequences removed from the RNA transcript

145
Q

exons

A

sequences that remain in the RNA transcript and are expressed

146
Q

spliceosome

A

complex enzymatic machine (RNP) composed of both proteins and non-coding RNAs called snRNAs that facilitates splicing, the removal of introns
-only eukaryotes have spliceosomes

147
Q

splicing

A

the process of removing introns from transcribed RNA

148
Q

5’ modification to transcribed mRNA

A

7-methyl guanosine (m7G) cap is added to the 5’ end of eukaryotic mRNA
-aids in formation of translation initiation complex
-protects 5’ end from degradation
-signal for nuclear export

149
Q

3’ modification to transcribed mRNA

A

poly-A tail added to 3’ end of eukaryotic mRNA by polyadenylate polymerase (PAP)
-signals nuclear export
-stabilizes mRNA
-promotes translation

150
Q

alternative isoforms from mRNA processing

A

alternative splicing contributes to the idea that many proteins can be coded from a single gene

151
Q

mRNA processing and surveillance

A

mRNA processing ensures bona fide mRNA has features that provide proof to the cell that the RNA was made by it (rather than it being a virus)

152
Q

Why are pre-mRNAs spliced and processed?

A

1) Alternative isoforms
2) Surveillance against invaders

153
Q

ribonucleoproteins (RNPs)

A

any machine with protein/RNAs working together
-spliceosome is an RNP

154
Q

spliceosome pre-mRNA splicing mechanism

A

snRNAs use base-pairing interactions to locate the intron/exon boundaries within the pre-mRNA; once located, the spliceosome activates and proceeds in 2 steps, both involving nucleophilic attack of a hydroxyl on the phosphodiester backbone:
1) 2’OH from the ribose of an adenosine within the intron cleaves the backbone at the 5’ splice site
2) Free 3’OH of 5’ first exon attacks the phosphate connecting the intron to the second exon, releasing an RNA lariat

155
Q

group I intron mechanism

A

1) Free guanosine nucleotide attacks phosphate at 5’ splice site
2) 3’OH of 5’ exon initiates nucleophilic attack on the phosphodiester backbone at the 3’ exon, releasing a linear fragment

156
Q

group II intron mechanism

A

1) 2’OH of an adenosine in the intron attacks the phosphate at 5’ splice spite
2) 3’OH of the 5’ exon attacks the phosphate connecting the intron to the second exon, releasing an RNA lariat

157
Q

microRNAs

A

small ~22nt genome-encoded RNAs that regulate gene expression
-base-pairing is used to locate miRNA targets; degree of base-pairing sets type of regulation:
1) if 100% miRNA form WCF base-pairs, RNA is destroyed
2) imperfect match - repression but not destruction of RNA

158
Q

three main types of RNA in the cell

A

1) mRNA - encodes protein sequences
2) tRNA
3) ribosomal RNA (rRNA) - non-coding

159
Q

translation

A

going from mRNA to protein

160
Q

codons

A

3 nucleotides that code for a single amino acid

161
Q

properties of the genetic code

A

1) triplet - codon corresponds to one amino acid
2) redundant - except for methionine and tryptophan, each amino acid is encoded by more than one codon
3) universal - same code is used in all organisms

162
Q

tRNA

A

adapter molecules that link codons to associated amino acids by presenting an anti-codon that uses WCF base-pairing to read triplet codons in mRNA

163
Q

charging of tRNA

A

amino acid is activated by adenylation of the carboxyl group, then tRNA synthetase selects the matching tRNA and “charges” it, transferring the amino acid to the tRNA 3’OH
-i.e., a charged tRNA is simply a tRNA with an amino acid attached

164
Q

Why are their fewer than 64 species of tRNA but 64 possible codons?

A

some tRNA can pair with multiple, synonymous codons, all of which encode the same animo acid
-wobble hypothesis

165
Q

wobble hypothesis

A

first two bases create coding specificity, while third base does not make strong interaction and does not need to be a canonical base-pairing interaction
-can occur only at the 3’ end of RNA (5’ end of anti-codon)
-this allows the cell to get around needing 64 tRNAs

166
Q

inosine

A

placed in the anti-codon stem by some tRNAs, inosine can base-pair with multiple different nucleobases, making it well-suited to place in the wobble position (5’ end of anti-codon)

167
Q

reading frame

A

consecutive, non-overlapping codon sequences translated into a polypeptide; reading fram start is set by AUG (START codon)

168
Q

UTR

A

untranslated region that does not code for protein
-outside the START and STOP codons

169
Q

silent mutation

A

changes the mRNA sequence to a synonymous codon and therefore has no effect on resulting polypeptide

170
Q

missense mutation

A

changes the mRNA sequence from a codon for one amino acid to a codon for a different amino acid, changing the polypeptide sequence

171
Q

nonsense mutation

A

changes the mRNA sequence from a codon for an amino acid to a STOP codon, terminating translation and resulting in a truncated polypeptide

172
Q

frameshift mutation

A

disrupts the normal reading frame by inserting or deleting nucleotides that are not multiples of 3 (in this case frame isn’t shifted)
-insertion or deletion

173
Q

ribosome in translation

A

1) Initiates synthesis at START codon
2) Synthesis: forms peptide bonds between amino acids to make a polypeptide
3) Terminating synthesis at STOP codon

174
Q

ribosomal initiation of translation (prokaryotes)

A

ribosome RNA base-pairs with mRNA at the Shine-Dalgarno sequence, aligning ribosome at intended AUG START codon

175
Q

Shine-Dalgarno sequence

A

sequence of mRNA in prokaryotes that aligns ribosome RNA at intended AUG START codon

176
Q

ribosomal initiation of translation (eukaryotes)

A

1) Ribosome assembes initiation complex mediated by 5’m7G cap and poly-A tail interactions
2) Ribosome scans from 5’ cap toward 3’ end of mRNA searching for AUG
Note: very poorly understood beyond this

177
Q

three sites for tRNA/mRNA interactions on ribosome

A

1) A site - holds an aminoacyl tRNA
2) P site - holds the tRNA with growing polypeptide attached
3) E site - holds a tRNA that will exit

178
Q

ribosomal condensation reaction

A

A-site N-terminal amine attacks P-site’s ester linkage to polypeptide’s C-terminal end and the polypeptide is transferred from the P-site tRNA to the A-site tRNA (ribosomal RNA catalyzes the formation of a peptide bond between the amino sites bound to tRNAs at P and A sites; one amino acid added to the end of tRNA in the process)

179
Q

release factor

A

protein that recognizes STOP codon and promotes polypeptide release and translation termination
-not a tRNA, but protein component that mimics tRNA
-reads STOP codon sequence using amino acids, not bases

180
Q

tRNA translocation in ribosome

A

tRNA moves from A-site (amino acid) to P-site (protein) to E-site (exit) in a ribsome

181
Q

N-terminal end of protein

A

corresponds to the 5’ end of mRNA; polypeptide grows from N-terminus to C-terminus

182
Q

physical coupling of transcription and translation (prokaryotes)

A

transcription and translation are physically coupled in prokaryotes, meaning the mRNA begins being translated before its finished being transcribed); but this is not so in Eukaryotes where mRNAs are made in the nucleus and translated in the cytoplasm

183
Q

oligonucleotide synthesis (solid state synthesis)

A

chemical (laboratory) synthesis of short (10-80nt) DNA fragments that are used as primers for PCR
-unique in that fragments are synthesized from 3’->5’

184
Q

polymerase chain reaction (PCR)

A

uses oligos as primers to amplify defined DNA fragments
-repeated cycles of denaturation, primer annealing, and polymerase extension lead to exponential amplification of a target region of DNA

185
Q

Gibson assembly

A

mimics a recombination reaction by assembling smaller overlapping fragments into larger genes
-5’->3’ exonuclease
-high-fidelity DNA polymerase
-taq DNA ligase

186
Q

plasmid

A

a circular DNA molecule that replicates separately from the host chromosome; designed genes can be cloned into plasmids for propagation and downstream applications

Key components:
1) Ori - recruits cellular enzymes to initiate replication
2) Selectable genetic marker - enables selection of cells with the recombinant plasmid
3) Polylinker - allows vector to be cut with restriction enzymes

187
Q

Sanger method

A

involves the incorporation of ddNTPs to terminate synthesis, allowing multiple products to be generated that allow us to identify the locations of a base
-highly accurate, but can only sequence one clonal DNA at a time

188
Q

Illumina method

A

single DNA molecules are spotted on their “own location” on a slide, locally amplified and sequenced using microscopy to read out each base that is added

189
Q

PacBio method

A

single molecules of DNA are trapped in tiny wells, and the sequence of DNA is read out by repeatedly resequencing the same fragment over and over again
-allows us to read longer sequences

190
Q

Nanopore method

A

DNA is fed through a protein-based pore in a membrane and the changes in conductance as the DNA goes through provides the identity of different bases
-many more bases sequenced, but less accuracy

191
Q

Cas9

A

RNA-programmable DNA endonuclease that uses WCF base pairing between “guide RNA” and dsDNA to locate the target site

192
Q

Cas9 gene modification

A

1) Use Cas9 to induce a break at a desired location
2) Use synthetic DNA fragment as repair template

193
Q

dCas9

A

“dead” Cas9 that binds target DNA but doesn’t cut

Biochemistry 501 (4 decks)

Brainscape helps you reach your goals faster, through stronger study habits.
© 2024 Bold Learning Solutions. Terms and Conditions