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Genetics (BIOL 4003) > Exam 2 > Flashcards

Exam 2 Flashcards

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1
Q

4 characteristics of genetic material/DNA

A
  • needs to have the information
  • transmissionable
  • replication
  • variation, must be capable of change and account for change
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2
Q

Exaplin experiment that explained transformation

A

Bacteria that was deathly was killed and put into a mouse with alive bad bacteria but the mouse still died.

took DNA from the dead bacteria somehow

transforming principle

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3
Q

Exaplain how they knew that DNA is genetic material

A

During the mouse experiment with S/R bacteria, only the extract with DNA transformed R bacteria into S bacteria

Also, DNAse made is so the transformation didn’t happen but RNAase didnt kill the DNA

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4
Q

4 levels of nucleic acid structures

A
  1. Nucleotides
  2. Single stranded DNA
  3. Double helix
  4. 3d structure
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5
Q

the builiding blocks of DNA is

A

nucleotides

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6
Q

_ are linked _ to make DNA strands

A

nucleotides are linked covalently

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7
Q

Nucleotide components

A
  • phosphate group
  • pentose sugar (ribose or deoxyribose)
  • nitrogenous base
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8
Q

nucleoside components

A

base + sugar

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9
Q

DNA strands consist of _ bonds

A

ester bonds that link nucleotides together

ester bonds are the P=O bonds in the phos group

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10
Q

phsophediester linkage

A

nucleotides 5’ to 3’ linkages

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11
Q

nucleotides 5’ to 3’ linkages are called

A

phosphodiester linkages

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12
Q

in a DNA strans all sugars are orientated….

A

in the same direction

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13
Q

what forms the backbone of the DNA strand

A

phosphate and sugar molecules

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14
Q

the _ project from the backbone of DNA

A

bases

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15
Q

DNA stores info in…

A

the base sequence

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16
Q

the _ are the _ ring bases

A
  • purines - double ring
  • pyrimidines - single ring
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17
Q

purine and pyramidine bases

list them

A

purine: A,G
pyrimidines: T, U, C

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18
Q

difference between DNA and RNA

A

DNA sugar has a hydrogen not an OH on the 2nd carbon

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19
Q

Linus Pauling used _ to discover the _ helix

A

ball and stick models to discover the alpha helix

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20
Q

Franklin used _ to see the _ of DNA

A

X-ray diffraction to see the molecular patterns of DNA

showed that DNA is helical, more than one strand, and has around 10 base pairs per turn

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21
Q

Chargaff analyzed _ to discover _

A

the base concentrations of DNA from several specices to discover Chargaff’s rule

A = T
C = G

base pairing

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22
Q

Watson and crick used _ to discover _

A

ball and stick models to discover the double helix structure of DNA

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23
Q

Key features of DNA double helix

A
  • clockwise, right handed helix
  • bases in opposite strands hydrogen bond AT/CG
  • 2 strands are antiparallel
  • 10 base pairs in each turn of the helix
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24
Q

base stacking

A

in DNA, flat parts of bases face each other and stabalize by the hydrophobic effect

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25
Q

A is bonded to T with

A

2 H bonds

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26
Q

C is bonded to G with

A

3 H bonds

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27
Q

important for protein binding to DNA are

A

the major and minor groove, which you need to recognize the bases to see

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28
Q

Z dna characteristics

A
  • left handed
  • bases tilted alot
  • backbone is zigzag
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29
Q

RNA double helix features

A
  • right handed
  • 11 to 12 base pairs per turn
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30
Q

Types of RNA secondary structures

A
  • Bulge loop
  • internal loop
  • multibrached loop
  • stem loop

draw these

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31
Q

factors contributing to tertiary structure of RNA

A
  • base pairing and base stacking within RNA
  • interactions with ions, molecules, etc
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32
Q

features of bacterial chromosome

A
  • one origin of replication
  • genes
  • intergenic locations
  • repetitive sequences
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33
Q

nucleoid

A

place where bacterial chromosome is found
* not bound by a membrane

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34
Q

microdomains

A

loop domains to compact bacterial DNA

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35
Q

bacteria use _ to form micro and macrodomains

A

nucleoid associated proteins (NAPs)

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36
Q

NAPs function

nucleoid associated proteins

A
  • fascilitate chromosome compaction
  • bend DNA for it to bind
  • chromosome segregation fascilitation
  • gene regulations
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37
Q

ways DNA in bacteria becomes compact

A
  • NAPs
  • supercoiling
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38
Q

_ supercoils are good for

A

negative supercoils food for transcription and compacting DNA

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39
Q

how does negative supercoiling help dna transcription

A

it creates tension that is released by strand seperation

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40
Q

the control of supercoiling is done by

A
  • DNA gyrase
  • DNA topoisomerase 1
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41
Q

DNA gyrase function

A
  • uses ATP to make negative supercoils
  • relaxes supercoils
  • untangles DNA

increase transcription, very active

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42
Q

DNA topoisomerase 1 function

A

relaxing negative supercoils

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43
Q

complex eukaryotes have _ genes with many _

A

longer genes with many introns

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44
Q

3 types of DNA sequences required for eukaryotic replication

A
  • origins of replication
  • centromeres
  • telomeres
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45
Q

eukaryotic chromosome key features

A
  • linear
  • occurs in sets
  • very long with many genes
  • multiple origins of replication
  • centromere
  • telomeres
  • repeatitive sequences
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46
Q

repeatitive sequences are usually found near

A

centromeric and telomeric regions

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47
Q

sequence complexity

A

the number of times a particular base sequence appears in the genome

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48
Q

three main types of repeative sequences

listed

A
  • unique or non repeatitive
  • modereatly repeatitive
  • highly repeatitive
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49
Q

unique sequences

A

includes protein encoding genes and intergenic regions
* most of the genome

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50
Q

moderately repeatitive sequences

A
  • genes for rRNA and histones
  • regulate gene expression and translation
  • transposable elements
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51
Q

highly repeatitive sequences

A
  • fairly short
  • some interspersed throughout genome
  • some are clustered together in arrays
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52
Q

most of eukaryotic genomes is

A

repeatitive DNA

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53
Q

transposition

A

the integration of small segments of DNA to a new location

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54
Q

two general types of transposition pathways

brief description

A
  • simple transposition - TE moves to a new site using a transposase
  • retrotransposition - TE moves to a new site using an RNA intermediate

TE = transposable elements

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55
Q

proliferation of retrotransposons happens with…

A

reverse transcriptase and integrase
OR
target site primed reverse transcription

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56
Q

in simple transposition and retrotransposition, how many copies of the TE are there

A

simple: 1 copy (just moves)
retro: 2 (make copy then move)

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57
Q

direct repeats and why they exists

A
  • repeats that flank the TE and same direction
  • exist so that transposase can do a staggered cut when inserting gene
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58
Q

inverted repeats

function and defin

A
  • repeats that are opposite directions (palindroms)
  • exist so the transposes can cut TE out
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59
Q

insertation element has

A
  • direct repeat
  • inverted repeat
  • transposase gene
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60
Q

simple transposon has

A
  • direct repeat
  • inverted repeat
  • transposase gene
  • gene for another thing like antibiotic resistance
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61
Q

LTR retrotranspoons contain

A
  • direct repeat
  • long terminal repeat
  • reverse transcriptase gene
  • integrase gene
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62
Q

non LTR retrotransposon contains

A
  • direct repeat
  • reverse transcriptase or endonuclease gene

no LTR and no integrase

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63
Q

transposable elements are considered to be _ when _

A

autonomous elements when they contain all info for transposition

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64
Q

steps of simple transposition

A
  • transposon binds to inverted repeats
  • transposon dimerizes and creates a loop
  • transposase cleaves outside inverted repeats
  • transposase cleaves target DNA at staggered sites
  • the transposable element is inserted
  • direct repeats are made

draw it out!

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65
Q

simple transposition can increase _ by _

A

increase copy number by inserting ahead of the replication fork

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66
Q

LTR retrotransposition process

A
  • transcription of TE with reverse transcriptase multiple times
  • integrase places into target DNA sites

can proliferate in multiple sites

67
Q

non LTR retrotransposition

A
  • retrotranspoon is transcribed
  • one strand of target DNA is cleaved by endonuclease
  • the retrotranspoon RNA with a poly-A tail attaches to the cut DNA like a primer using
  • reverse transcriptase copies RNA into DNA to finish the DNA cut part by adding the TE with transcription
  • endonuclease cuts other strand
  • RNA gets yeeted off, new TE DNA added into the target sequence
  • DNA polymerase fills in gaps
68
Q

abundance of TEs is most in

A

complex eukaryotic multicellular organisms

69
Q

selfish DNA theory

A

TEs exist bc they can

70
Q

hypothesis for why TEs exist

A
  • selfish DNA theory
  • TE offer some sort of advantage
  • TE increase geneic variablility
  • exon shuffling (may lead to genes with more diverse functions)
71
Q

nucleosome structure

A

double stranded DNA segmenet wrapped around octomer of histones
* 2 copies of each histone type: H1, H2A, H2B, H3

72
Q

one histone is composed of

A
  • globular domain
  • an amino terminal tail
  • many positively charged amino acids (Lys, Arg)
73
Q

histones and dna binding

A

histones have many posistive AAs which bind to the negative phosphate groups in the DNA backbone

74
Q

H1 histone

A
  • linker histone
  • less tightly bound
  • helps to organize adjacent nucleosomes
75
Q

Strings on a bead model was tested by

A

using DNase 1 to cut DNA into fragments, hypothetically linker region should cut which they were

76
Q

30 nm fiber

A

at low salt concentrations, H1 remains bound and the beads of nucleosome come closer together, shortens DNA by alot

77
Q

3 ways loop domains are formed

to further compact DNA

A
  • 2 CCCTC binding factor (CTCF) binds to DNA then bind to each other
  • SMC protein forms a dimer than wraps itself around DNA and makes a loop
  • or both of the above happens

drawing chap 10 slide 88

78
Q

each chromosome can be found in

A

chromosome territory

79
Q

the predominant structure in the chromosome territories is

A

loop domains

80
Q

2 levels of compation in interphase

list

A
  • heterochromatin
  • euchromatin
81
Q

hetrochromatin

A
  • tightly compact region
  • no transcription
  • loop domains v compact
82
Q

euchromatin

A
  • less compact regions
  • transcriptionally active
  • 30 nm fiber loop make loops
83
Q

two types of hetrochromatin

list and properties

A
  • consitutive: regions that are permenantly hetrochromatic, usually have lots of repeats, around centromere and telomere
  • facultative: can be both euchromatin or hetrochromatin
84
Q

process of entering metaphase

A
  1. DNA double helix
  2. wrapping DNA to make nucleosomes
  3. formation of zigzag into 30 nm fibers
  4. 30nm fibers anchor to proteins to make loop domains
  5. loop domains come closer to one another
  6. chromosome
85
Q

condensin

role

A
  • condensin 2 enters nucleus during interphase and helps with condensing
  • 1 stays in cytoplasm
  • at end of prophase both help to compact loops
86
Q

cohesin

role

A
  • found along length of chromatid and holds chromatids together
  • degreaded when prophase begins
87
Q

Due to the _ DNA can be replicated to produce two double helices with the identical base sequences?

A

Due to the AT/GC rule

88
Q

What are the expected results for the Meselson and Stahl experiment after 4 generations (i.e, 4 rounds of DNA replication in the presence of light nitrogen)? Note: during generation zero, the DNA is all heavy, and subsequent generations only make light DNA.

A

1/8 half heavy
7/8 light

89
Q

How many replication forks are formed at an origin of replication?

A

2

90
Q

Primase is needed during DNA replication because DNA polymerase is not able to

A

begin synthesis on a bare template strand.

91
Q

In the leading strand, DNA is made in the direction _ the replication fork and is made as _

A

toward, one continuous strand

92
Q

During proofreading, DNA polymerase

A

cuts out a mismatch by digesting DNA in the 3’ to 5’ direction.

93
Q

Which ends of DNA are extended by telomerases?

A

3’ end

94
Q

DNA replication relies on

A

AT/GC rule

95
Q

3 different models proposed for the replication of DNA

list and explain

A
  • conservative model: both parental strands stay together
  • semiconservative model: parental strands split
  • dispersive model: random parental and daughter DNA segments
96
Q

if conservative model was true, the Mendelson and Stahl experiments 1st and second generation (start with heavy DNA)

A
  • generation 1: 50% heavy, 50% light
  • generation 2: 25% heavy, 75% light
97
Q

semi conservative model, Mendelson and Stahl experiments 1st and second generation (start with heavy DNA)

A
  • generation 1: all half heavy
  • gen 2: 50% half heavy, 50% light
98
Q

if dispersive model was true, Mendelson and Stahl experiments 1st and second generation (start with heavy DNA)

A

gen 1: all half heavy
gen 2: all 1/4 heavy

99
Q

DNA synthesis begins at

A

origin of replication

100
Q

oriC 3 types of important DNA sequences

list

A
  • DnaA boxes
  • AT rich regions
  • GATC methylation sites
101
Q

DnaA boxes

A

sites for the binding of DnaA protein

part of oriC

102
Q

AT-rich regions

A

sites where DNA strands seperate

part of oriC

103
Q

GATC methylation sites

A

sites that make sure DNA replication doesnt start again too soon

part of oriC

104
Q

events that occur at oriC

A
  • DnaA proteins bind to DnaA boxes and to each other
  • DNA bends
  • strands seperate at AT rich regions
  • DnaB/helicase binds to the origin to further seperate DNA strands
  • repliation occurs in both direction
105
Q

DNA helicase is combosed of

A

6 subunits

106
Q

DNA helicase travels along DNA in the _ direction

A

5’ to 3’

107
Q

how do GATC sites regulate replication

A

Dam methylates the A on both strands, DNA replication only starts of fully methylated DNA and the daughter cells are not, so replication doesn’t restart too soon

108
Q

DNA helicase seperates strands by _ and in turn generates a _

A

by breaking H bonds between them and generating a positive supercoiling ahead of each replication fork

109
Q

DNA gyrase travels ahead of the helicase and

A

alliviates positive supercoils

110
Q

_ bind to DNA strands to keep them apart

A

SS binding proteins

111
Q

after DNA strands are seperated,

A

RNA primers are synthesized by primase

112
Q

the leading strand has _ primers

A

1

113
Q

the lagging strand has _ primers

A

multiple

114
Q

_ are responsible for synthesizing DNA

A

DNA polymerase

115
Q

the lagging strand leads _ from fork

A

away

116
Q

DNA pol 1-5 in e coli functions

A
  • normal replication: 1 & 3
  • DNA repair: 2, 4, 5
117
Q

DNA pol 3

A
  • responsible for most DNA replication
  • alpha unit catalyzes bond formation
  • DNA pol haloenzyme is 10 units
118
Q

DNA pol 1

struc and func

A
  • composed of a single polypeptide
  • removes the RNA primers and replaces them with DNA
119
Q

features/problems of DNA polymerase

2

A
  • cannot initiate replication on bare strand (needs primer)
  • can only attach 5’ to 3’ but strands are anti parallel (fragments)
120
Q

DNA pol 1 uses _ activity to _

A
  • endonuclease activity to digest RNA primers
  • polymerase activity to replace it with DNA
121
Q

after DNA pol fills DNA in primer spots…

A

DNA ligase comes in to form covalent bonds

122
Q

DNA ligase

function

A

forms covalent bonds between primer and DNA and fragments

123
Q

DNA helicase and primase form a complex _ in order to _ , this complex also forms _ with two DNA holoenzymes

A

DNA helicase and primase form a primosome to coordinate, the primosome comes with a holenzyme to form a replisome

124
Q

how does the replisome avoid hoping

A

creates a loop of where the next okazaki fragment will be synthesized

125
Q

what stops movement of replication forks in e coli

A

tus bond to ter, termination sequences

126
Q

T1 stops

A

counterclockwise forks

127
Q

T2 stops

A

clockwise forks

128
Q

catenanes

A
  • intertwined DNA molecules after replication
  • seperated by topoisomerase 2
129
Q

DNA pol catalyzes the formation of _ between

A

a ester bond between P of new base and 3’OH of old base

2 Ps are released

130
Q

in formation of an ester bond, _ is release

A

PPi

2 P s

131
Q

processive feature of DNA pol 3 is due to

A

DNA pol 3 stays attached to template because of the beta subunit forming a clamp around the template DNA

132
Q

in absense of the beta subunit…

A

DNA pol 3 would fall off the DNA template, very inefficient

133
Q

DNA replication has a high degree of

A

fidelity (accuracy)

134
Q

why is fidelity high in DNA replication

A
  • stability of base pairing
  • structure of DNA polymerase at active site (incorrect pair causes weird shape)
  • proofreading of DNA polymerase
135
Q

proofreading is done with

A

3’ to 5’ endonuclease activity of the DNA pol
* digests new strand until the mismatch is found

136
Q

why is eukaryotic DNA replication more complex

A
  • larger linear chromosomes
  • tightly packed
  • cell cycle highly regulated
137
Q

eukaryotes have long linear chromosomes, so

A

they require more origins of replication so that replication is done quickly

138
Q

characteristics of origin sites in simple eukar

A
  • lots of A and T
  • consensus sequence
139
Q

ARS elements

A

origins of replication in simple eukaryotes

140
Q

in eurkar, replication begins with

A
  • the assembly of preplication complex (preRC)
141
Q

the preRC includes

list

A
  • origin recognition complex
  • MCM helicase
142
Q

origin replication complex

struc and func

A
  • six subunit complex
  • first initiator of eukar DNA replication
143
Q

binding of MCM completes

A

DNA replication licensing

144
Q

eukar strand seperation done by

A

MCM helicase

145
Q

explain how primers are removed in eukary

A
  • Polymerase δ starts the removal of primers by making an Okazaki fragment longer and creating a flap in the adjacent fragment.
  • Flap endonucleases then remove the flap.
  • This process repeats until the DNA primer is completely removed
146
Q

DNA polymerases in mammals and which DNA they act on

A
  • nuclear DNA: alpha, delta and epsilon
  • mitochondrial DNA: gamma
147
Q

DNA pol alpha is the only to

A

associate with primase

148
Q

what makes primer in eukary

A

DNA pol alpha and primase complex synthesizes a short RNA-DNA hybrid primer

10 RNA nucleotides followed by 20-30 DNA

148
Q

what makes primer in eukary

A

DNA pol alpha and primase complex synthesizes a short RNA-DNA hybrid primer

10 RNA nucleotides followed by 20-30 DNA

149
Q

polymerase switch

A
  • exchange of DNA pol alpha for DNA pol epsilon or delta
  • required for elongation
150
Q

_ is needed for elongation in eukar

A

exchange of DNA pol alpha for DNA pol epsilon or delta

151
Q

DNA pol epsilon is used for

A

elongation of leading strand

152
Q

DNA pol delta is used for

A

elongation of lagging strand

153
Q

_ are involved in replication of damaged DNA in eukar and they can _

A

translesion-replicating polymerases, they can synthesize a complmentary strand over an abnormal region

154
Q

if the flap is too long in RNA primer removal in eukar,

A

DNA2 helicase/nuclease cuts the flap shorter

155
Q

why do eukar chromosomes have telomers

A

because 3’ end of DNA cannot be replicated due to the lack of a primer, a primer cannot be added to nothing

156
Q

telomeres consist of

A
  • 3’ repeatitive tandem overhands
  • several G
  • many T
157
Q

if there were no telomeres…

A

linear chromosomes would get shorter every round of DNA replication

158
Q

telomerase creates _ that is

A

adds DNA sequences to the 3’ end that is RNA that is complmentary to the DNA in the telomeric repeat

159
Q

telomerase steps

A
  1. binding to 3’ overhang region
  2. polymerization: synthesis of a repeat
  3. translocation: telomerase moves the repeat to the right and makes another repeat
  4. the completion of this primer can be used now

chap 11 page 72

160
Q

telomerase tend to

A

shorten in actively dividing cells

161
Q

when telomeres are short

A

cells lose their ability to divide

162
Q

cancer cells carry

A

mutations that increase activity of telomerase in order to prevent telomere shortening and continue dividing

Genetics (BIOL 4003) (7 decks)

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