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BIOL2030 > MT2 > Flashcards

MT2 Flashcards

1
Q

For a molecule to serve as genetic material it must be able to (3)

A
  1. replicate accurately
  2. store large amounts of information
  3. allow for phenotypic variation
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2
Q

Johann Meischer

A

Discovery of DNA - 1869

Isolated weakly acidic substance, ‘nuclein’ from nuclei in human WBCs. Later renamed nucleic acid.

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3
Q

Fredrick Griffith

A

The Transforming Principle - 1928

- Showed that cells can be transformed - uptake genetic material from an external source resulting in new traits

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4
Q

Transforming Principle - Discovery + experiment

A

Fredrick Griffith 1928
Injected heat killed virulent bacteria + non-virulent into mice and mice died. Concluded that a substance in heat-killed virulent bacteria genetically transformed the non-virulent bacteria.
- Non-virulent = R cells
- Virulent = S cells
- The substance responsible causes a permanent, heritable genetic change referred to as TRANSFORMING PRINCIPLE

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5
Q

Avery, McLeod, McCarty

A

DNA Carry Genetic Information - 1944
EXPERIMENT: head killed S strain and extracted cell contents treated with (1) RNAse, (2) DNAse, (3) Protease. These extracts then mixed w non-virulent R strain. Found that transforming activity destroyed only by DNAse, so DNA must be the transforming principle

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6
Q

Alfred Hershey and Martha Chase

A

DNA Carry Genetic Information - 1952

  • DNA, no protein, transmitted to progeny
  • Bacteria infected with T2 phages with 35S labeled coat protein or 32P labelled DNA
  • Labelled protein not in progeny, but labelled DNA was
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7
Q

Singer and Fraenkel-Conrat

A

RNA as Genetic Material - 1956

  • Some viruses contain RNA, not DNA, and still reproduce
  • EXPERIMENT: Separated ssRNA from protein in 2 samples tobacco mosaic virus. Applied protein coat from A to DNA in B. Hybrid progeny were identical to original B.
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8
Q

Aaron Levene

A

DNA is made of repeating units called NUCLEOTIDES

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9
Q

Albrecht Kossel

A

Nucleic acid contains for nitrogenous bases: Adenine (A), Cytosine (C), Guanine (G), Thymine (T)

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10
Q

Edwin Chargaff

A

Analyzed the nucleotide composition of DNA: A=T ; G=C

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11
Q

Watson and Crick

A

Using data collected from others, they proposed the 3D structure of DNA

  • Rosalind Franklin and Maurice Wolkins produced x-ray diffraction data with B form DNA that showed DNA had constant diameter
  • Used (Linus Pauling) model building technique
  • Initially has bases facing out, but Franklin saw that PO4s on inside would be unstable
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12
Q

3 forms of DNA

A

A, B, Z
A and B are the major forms.
A is what comes out when you try to purify DNA

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13
Q

Key characteristics of double helix

A
  • Phosphates outside, bases inside
  • Double helix
  • Anti-parallel strands held together by H bonding
  • Specific base pairing
  • Constant diameter
  • Bases flat, perp to axis ; stacked 0.34 nm apart with 10 bases per turn
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14
Q

Nucleotides linked together by

A

3’ - 5’ phosphodiester bonds

polarity of 5’ phosphate end and 3’ hydroxyl end

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15
Q

A form DNA

A 
Right hand turns
11 bases per turn
Narrower major groove
Forms under low humidity
Found in DNA-protein complexes
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16
Q

B form DNA

A

Right hand turns
10 residues/turn
Form usually found in cells

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17
Q

z form DNA

A

Left handed turns
12/residues/turn
No major grooves
Biological significance unknown

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18
Q

Hairpin and stem formation in DNA

A

Occurs in ssDNA where it has inverted complementary sequence.

  • hairpin has loop at top where the bases don’t quite match up
  • stem has all bases matching up to create a perfect fold
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19
Q

cruciform

A

forms in dsDNA with inverted repeats.

- Normal dna forms two mirrored hairpins/stems where the bases are complementary in the ssDNA

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20
Q

Ways to denature DNA (4)

A
  1. Increase temp
  2. Reduce salt []
  3. Increase pH disrupts H bonding
  4. Sovlents
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21
Q

Factors that determine Tm of DNA (4)

A
  1. G/C content
  2. Ionic strength of buffer
  3. Length of the DNA molecule
  4. Higher [salt] => higher Tm bc it stabilizes the -ve charge of the phosphate groups
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22
Q

Ways you can use Tm (3)

A
  1. CLASSIFY ORGANISMS - GC content in DNA is species specific
  2. DETECT SEQUENCE DIFFS IN 2 NUCLEIC ACIDS OF DIFF ORIGIN - hybrid molecule formed and differences seen in disruptions in base pairing
  3. DETECT RARE GENETIC MUTATIONS - mutated DNA melts at diff temp than normal ranges
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23
Q

Ways to disrupt H bonding in DNA to decrease Tm

A

Solvents: formamide and DMSO

Strong alkaline conditions (high pH)

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24
Q

Conservative replication

A
  1. II
  2. II 11
  3. II 11 11 11
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25
Q

Dispersive replication

A
  1. II
  2. 2 double helices all mixed up
  3. 4 double helices all mixed up
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26
Q

Semiconservative replication replication pattern

A
  1. II
  2. I1 I1
  3. I1 11 11 I1
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27
Q

Meselson and Stahl Experiment

A

Grew E. coli on 15N media for many generations. Switched some cells to 14N medium. Used equilibrium density gradient centrifugation to determine isotope composition of DNA
Found 50% DNA at 14N level and 50% DNA at mixed 14N and 15N level
Proved conservative replication

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28
Q

Semiconservative replication basics

A
  • Separation of the two DNA strands of the parental molecule
  • Each parental strand serves as template for a newly synthesized strand
  • Each daughter DNA molecule consists of one parental strand and one newly synthesized strand
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29
Q

DNA Polymerase

A

Uses deoxyribonucleoside 5’ TRIphosphates
Catalyzes phosphodiester bonds
Requires a preexisting 3’ OH
Cannot make DNA de nova - can only extend chain
Elongates a chain in 3’ to 5’ direction always

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30
Q

Holliday Model

A

Model for recombination in meiosis.

  • single stranded break in each of the DNA double helices
  • single strands cross over and create a holiday junction
  • Junction migrates and you get the Holiday intermediate.
  • Vertical cleavage leads to crossover recombinants and and horizontal cleavage leads to noncrossover recombinants
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31
Q

Models of recombination

A

Holliday Model

Double Strand Break Model

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32
Q

Double Strand Break Model for recombination

A

ds break in one double helix.

  • enzymatic degradation of broken ends and then invasion into the intact double helix creating a loop of ssDNA that becomes a template for the broken DNA helix.
  • 4 places to cleave. Matching cuts = non-crossover recombinants and non-matching cuts = crossover recombinants
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33
Q

RecBCD

A

Enzymes required in the DSB model of recombination
-RecBCD binds a ds break. Unwinds the helix and degrades. RecD is on top and RecB is on bottom
-RecB falls behind allowing a ss loop to form
Complex encounters chi, which increases degradation of 5’ end, leaving 3’ overhang.
-RecBCD loads RecA onto 3’ overhand and then dissociates.

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34
Q

RecA

A

Is loaded onto 3’ overhang by RecBCD in double strand break model recombination
- Promotes strand invasion to make D loop and pairing with homologous DNA

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35
Q

RuvA
RuvB
RuvC

A

Promotes branch migration and heteroduplex formation.

  • RuvA recognizes the Holliday Junction
  • RuvB binds to RuvA/DNA and drives DNA unwinding and rewinding in branch migration
  • RuvC nicks strands for either horizontal or vertical resolution. Works with A and B to cleave the Holliday Junction
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36
Q

Gene Conversion

A
  • Occurs in association with homologous recombination as a result of heteroduplexes
  • Leads to abnormal segregation ratios
  • Heteroduplexes with mismatched bases are repaired, using one strand or the other as a template for correction.
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37
Q

Transcriptional Unit

A 
Region of DNA that codes for an RNA molecule
3 critical regions:
1. Promoter
2. RNA coding region
3. Termination site
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38
Q

Promoter

A

DNA sequence that is recognized by the transcriptional apparatus (RNA poly. and other proteins). Indicates the directions of transcription. When it binds RNA polymerase, it directs the enzyme toward the start site

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39
Q

RNA Polymerase

A
  • Binds to the promoter on DNA
  • Synthesizes RNA 3’ - 5’ using ribonucleosideTRIphosphates
  • Unwinds the DNA helix and forms phosphodiester bonds
  • Only 1 variety in prokaryotes, but multiple in eukaryotes
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40
Q

Bacterial consensus sequences

A
  • 10 5’ TATAAT 3’

- 35 5’ TTGACA 3’

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41
Q

Holoenzyme

A

Prokaryotic RNA polymerase + sigma factors.
Sigma factors help the enzyme recognize the promoter and specific genes to be transcribed (multiple types of sigma factors)

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42
Q

Rho-dependent prokaryotic transcription

A

Rho binds to the RUT on RNA. It moves toward the 3’ end.
When RNA poly runs into the terminator it pauses and Rho catches up.
Rho uses helicase activity to unwind RNA/DNA and ends transcription
About 1/2 the time a hairpin structure forms.

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43
Q

Rho-independent prokaryotic transcription

A

Terminator contains an inverted repeat followed by a bunch of Us that are transcribed.
Us cause RNA poly to pause, causing the inverted repeat to fold into a hairpin
RNA transcript separates

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44
Q

Regulatory promoter

A

Part of the RNA Polymerase II promoter
Located upstream of the core promoter
Transcriptional activator proteins bind to consensus sequences and affect the rate of transcription (these vary)

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45
Q

Core Promoter

A

Part of the RNA Polymerase II promoter
Extends upstream/downstream of the transcription start site
Minimal sequence required for accurate transcription initiation
Includes a number of consensus sequences (TFIIB, TATA, Initiator and DCE) for transcription factor binding

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46
Q

Basal Transcription Apparatus (3)

A
  1. RNA polymerase II
  2. General transcription factors
  3. Mediator Protein - RNA poly can’t bind DNA alone
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47
Q

Rat1

A

Termination of eukaryotic transcription.
RNA Polymerase II transcribes well past the gene coding region.
RNA is cleaved at a specific site and RAT1, a 5’-3’ endonuclease attaches to the 5’ end of the remaining RNA continuing to be transcribed and degrades it. It catches up with RNA polymerase II and transcription is terminated

48
Q

Exons

A

Segments of eukaryotic DNA that are translated

49
Q

Introns

A

Segments of eukaryotic DNA that are transcribed to pre-mRNA and then spliced out to produce mRNA

50
Q

OPeron

A

continuous array of genes in prokaryotes. A single transcription start site for multiple genes.

51
Q

Modifications to eukaryotic pre-mRNA

A
  1. 5’ cap
  2. Polyadenylation of the 3’ end
  3. Splicing of introns
52
Q

5’ capping of eukaryotic pre-mRNA

A

A methylated guanine is added to the 5’ tail:

  1. Phosphate group removed from 5’ end of pre-mRNA
  2. GMP added with 5’-5’ likage with 3 phosphate groups
  3. Methyl groups added to the G and 2’ position of the first 2 nucleotides and possibly the base of the first RNA nucleotide

Helps efficient translation, transport of mRNA from nucleus and protection from degradation.

53
Q

Polyadenylation of eukaryotic pre-mRNA

A

~50-250 A nucleotides added to the 3’ end of pre-mRNA

Important for efficient translation and protects mRNA from degradation

54
Q

How mRNA transcription and processing are coupled.

A

C-Terminal repeat domain (CTD) os RNA Pol II largest subunit mediates coupling.

  • mRNA processing enzymes recruited to the CTD during transcription.
  • 5’ CAP added as soon as 5’ end of pre-mRNA emerges from polymerase
  • RNA that is finished transcribing is cleaved at Poly(A) 3’ cleavage site so Rat1 can terminate transcription`
55
Q

mRNA processing leading to oncogene

A

If extra exons are left in that were supposed to be spliced out, the mRNA is too long and an oncogene can form.

56
Q

beta-thallessemia

A

Mutaions in the beta-globin gene disrupt normal splicing so the spliceosome can’t recognize the splice site and an intron is left in part.

  • causes disorder in hemoglobin
  • children develop anemia
57
Q

Alternative splicing possibilities (4)

A
  1. Exon skipped (spliced out like an intron
  2. Intron retention
  3. Alternative 5’ or 3’ splice site. (some of the intron left in or some of an exon spliced out)
  4. Mutually exclusive exons (either A or B is spliced, never in together)
58
Q

Isoforms of proteins

A

Different forms of a protein produced due to alternative splicing.

59
Q

Alternative Poly(A) 3’ cleavage site

A

Pre-mRNA contains multiple 3/ cleavage sites
This determines the length of the trancript
Each mRNA when translated produces similar proteins of different sizes

60
Q

Alternative processing of mRNA important for

A
  1. switch between the production of functional and non-functional proteins (in drosophila this allows male and female phenotypes)
  2. Different forms of a proteins produced by different cell types, allowing them to specialize.
61
Q

Value of RNA nucleotide editing

A

Makes it possible to have multiple proteins made from the same gene. Provide slightly different functions under different circumstances without increasing the genome size.

62
Q

Substitution editing

A

mRNA

Chemical alteration of an individual nucleotide by specific enzymes.

63
Q

Insertion editing

A

mRNA
Catalyzed by enzymes under direction of GUIDE RNA
gRNA base pirs with the mRNA to be edited and serves a template for addition of nucleotides. Folds back on itself and wherever it folds, a cut is made and a nuceotide can be edited.

64
Q

Guide RNA

A

leads the enzyme catalysis of insertion editing of pre-mRNA

65
Q

RNA interference (RNAi)

A

Short regulatory RNAs silence the expression of homologous genes.
Uses siRNA and miRNA

66
Q

3 mechanisms that siRNA and miRNA use to inhibit the expression of homologous genes

A
  1. mRNA degradation
  2. Inhibition of transcription
  3. Inhibition of translation
67
Q

siRNA comes from

A

produced by cleavage of dsRNA by DICER to produce small sequences.

68
Q

miRNA comes from

A

Inverted repeat in DNA leads to SMALL HAIRPIN RNA that is cleaved into miRNA by Dicer to be small sequences.

69
Q

siRNA silencing by RNA cleavage and degradation

A

siRNAs combine with proteins to make RISC complex, which base pairs perfectly with mRNA, cleaving it and leading to degradation.

70
Q

siRNA silencing by inhibition of transcription.

A

ss siRNA combines with proteins to make RITS complex
Base-pairing bt RITS and target DNA.
Recruits methylating enzymes that add methyl groups to histones so they can’t be transcribed easily because enzymes for transcription can’t access.
Thought to protect against viral interaction

71
Q

miRNA silencing by inhibition of translation

A

ss miRNA combines with proteins to make RISC complex
Partial base-pairing with miRNA and target mRNA
RISC can’t cut so it just sits there and blocks initiation causing premature termination

72
Q

Importance of RNAi

A

regulate many human genes especially during development
Important for gene knock out
helpful for cancer and virus treatments.

73
Q

Dicer

A

Cleaves siRNA and miRNA into their actual forms

74
Q

RISC

A

RNA-induced silencing complex

miRNA silencing by inhibition of translation and siRNA silencing by RNA cleavage and degradation

75
Q

RITS

A

RNA-induced transcriptional silencing

siRNA silencing by inhibition of transcription by methylating at histones on DNA.

76
Q

Nirenberg and Leder

A

developed a technique using ribosome-bound tRNA to decipher the code.

77
Q

nonsense codons

A

stop codons

78
Q

degeneracy of the genetic code

A

Most amino acids are specified by multiple codons.

79
Q

synonymous codons

A

codons that code for the same amino acids

80
Q

How is the genetic code not ambiguous, even though it is degenerate

A

amino acids are coded by multiple codons, but single codons don’t code for more than one amino acid

81
Q

partial degeneracy

A

changing the third base in a codon from a purine to purine, but pyrimidine to pyrimidine

82
Q

complete degeneracy

A

changing the third base in a codon to any of the four bases

83
Q

ways to accommodate code degeneracy

A
  1. Isoaccepting tRNA

2. Wobble effect

84
Q

Isoaccepting tRNAs

A

More tRNAs than there are amino acids. DIfferent tRNAs accept the same amino acid

85
Q

Wobble Effect

A

More codons than there are tRNAs. So the same tRNA is able to accept multiple codons.
Occurs between the 1st base of the anticodon and the 3rd base of the codon.

86
Q

Open reading frame

A

a portion of DNA that, when translated into amino acids contains no stop codons.

87
Q

Prokaryotic vs eukaryotic ribosomes

A

Prokaryotic has 3 RNA molecules

88
Q

Amino acid attachment site

A

The site on a tRNA where the amino acid attached during charging.
5’ CAA 3’

89
Q

aminoacyl-tRNA

how is it formed?

A

tRNA with an amino acid attached to it after charging.

amino acid attached to the A of the amino accid acceptor site on the tRNA using aminoacyl-tRNA synthetase with ATP energy

90
Q

aminoacyl-tRNA synthetase

A

attaches an amino acid to tRNA at the acceptor site using ATP energy
If it attaches the wrong amino acid to the tRNA it has edit abilities to go back and fix its mistake.

91
Q

ribozyme

A

enzyme activity in the large subunit of the ribosome by peptidyl transferase referred to as ribozyme

92
Q

Differences between prokaryotic and eukaryotic translation

A
  • No consensus sequence for ribosome binding in eukaryotes just first AUG. Prokaryotes use the Shine Dalgarno sequence
  • 5’ cap and 3’ poly(A) tail facilitate ribosome binding in eukaryotes.
  • Eukaryotic mRNA is monocystronic
93
Q

Aneuploidy

A

Increase or decrease in number of individual chromosomes

94
Q

Polyploidy

A

Increase or decrease in number of sets of chromosomes

95
Q

4 most common kinds of aneuploidy in humans

A
  1. nullisomy
  2. monosomy
  3. trisomy
  4. tetrasomy
96
Q

Nullisomy

A

Loss of both members of a pair of homologous chromosomes

97
Q

Monosomy

A

loss of a single chromosome

usually not viable, except for in sex chromosomes

98
Q

Trisomy

A

addition of a single chromosome

sometimes viable

99
Q

Tetrasomy

A

gain of two homologous chromosomes

100
Q

Origins of aneuploidy

A
  1. nondisjunction in meiosis or mitosis

2. deletion of a centromere leads to loss of a chromosome

101
Q

CTD

A

C terminal repeat domain is on a tail on RNA polymerase II.

Recruits enzymes necessary for post-transcription modification of the mRNA such at 5’ cap and poly A tail.

102
Q

consequences of aneuploidy

A

In plants it can lead to phenotypic alteration. In humans, more gametes are not viable.
Account for ~30% of miscarriages
Those that survive usually have trisomies of smaller chromosomes or sex chromosomes

103
Q

Primary Down Syndrome

A

Account for most cases of downsyndrome
Random nondisjunction in meiotic division
Usually from mother because primary oocytes are suspended in prophase 1 at birth, continued at ovulation and only finished at fertilization.

104
Q

Familial down syndrome

A

Extra copy of chromosome 21 is attached to another chromosome after ROBERTSONIAN TRANSLOCATION

105
Q

Autopolyploid is…
cause
effect

A

multiples of the same genome
can result from nondisjunction of all chromosomes during meiosis
Effect: usually sterile, genetically unbalanced

106
Q

Allopolyploid is…
Reproduction
Usefulness

A

Multiples of closely related genomes.
Must undergo mitotic nondisjunction before they can self-fertilize.
Show traits of both species and can be useful for agriculture.

107
Q

Replication of circular genomes (2)

A
  1. Theta Replication (bacteria)

2. Rolling circle replication (viruses)

108
Q

Theta replication

A

Single replicon for entire chromosome with bidirectional replication at both forks.

109
Q

Rolling circle replication

A

viruses
No replication bubble
Uncoupling of the two strands of DNA molecule

110
Q

Linear replication (as opposed to theta and rolling circle replication)

A

Multiple replicons, origins of replications and replication bubbles.

111
Q

Problem with replication at the 3’ ends in eukaryotes

A

Telomeres are made up of a G-rich, short, repeated sequence that stabilizes chromosomes.
Each round or replication leaves out ~200 bp at the 3’ end
Telomerase is a specialized reverse transcriptase that extends the parental DNA by RNA-templated DNA synthesis
Loss leads to aging

112
Q

When does crossing over occur

A

Prophase 1 of meiosis

113
Q

Benefits of crossing over (2)

A

(1) Generates diversity

(2) Maintains somatic stability by promoting repair of double stand breaks caused by radiation and such

114
Q

Interphase 1
Prophase 1
Metaphase 1
Anaphase 1

A

Chromosomes duplication
Homo chromosomes pair + crossing over
Tetrads line up
Tetrads split up

115
Q

Hyperchromatic shift

A

as the 2 DNA strands separate there is an increase in absorbance of UV light

116
Q

Importance of 5’ cap and 3’ poly A tail (3)

A
  1. Ribosome binding help
  2. Nuclear transport
  3. Stability from degredation
117
Q

allohexaploid would be…

A

from 2 diff species
6n
polyploidy

BIOL2030 (1 decks)

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