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BIOL 2030 > Section 2 - Module 6 > Flashcards

Section 2 - Module 6 Flashcards

1
Q

What is the significance of 1952?

A

Year that Alfred Hershey and Martha Chase demonstrated that DNA and not proteins is transmitted on to progeny. DNA as the genetics material in Bacteriophages

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2
Q

Frederick Griffin’s work

A

Demonstrated cell can be transformed

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3
Q

What is the transforming principle?

A

DNA

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4
Q

What is Transformation?

A

acquiring new genetic material from uptake of external source

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5
Q

What bacteria did Hersey work with?

A

E/coli and their associated bacteriophage

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6
Q

Aaron Levene’s work?

A

Proposed tetranucleotide theory. Stating that DNA is made of repeating units called nucleotides

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7
Q

Albrecht Kossel’s work?

A

Determined nucleic acids contain four nitrogenous bases.

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8
Q

The four nitrogenous bases in DNA

A

Adenine (A), Cytosine (C), Guanine (G), Thymine (T)

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9
Q

The four nitrogenous bases of RNA

A

Adenine (A), Cytosine (C), Guanine (G), Uracil (U)

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10
Q

Significance of 1953?

A

Year Franklin and Wilkins devise the secondary structure of DNA. The double helix structure of DNA

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11
Q

DNA secondary structure

A

Structure of the set interactions between bases. Such as which parts of the strands are bound to each other and phosphate backbone.

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12
Q

Significance of 1948?

A

Year Erwin Chargaff discovered pattern in DNA base pairs. Nucleotide compositions DNA - A=T; G=C.

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13
Q

Describe nucleotide structure

A

Phosphate group, base (A, G, G, t), and deoxyribose sugar. Purines attached to pyridines

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14
Q

Purines

A

Adenine (A) and Guanine (G). They has an additional ring structure

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15
Q

Pyridines

A

Cytosine (C) and Thymine (T). They have an amide functional group

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16
Q

Phosphate group is attached to the __ carbon of deoxyribose sugar

A

5’

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17
Q

Base is attached to the _ carbon in deoxyribose

A

1’

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18
Q

What makes up deoxyribose sugars of DNA?

A

5 carbons with OH at 3’ carbon. Between 4’ and 1’ there is an ether (O).

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19
Q

What is different in RNA structure?

A

it is a ribose sugar and has an additional OH on carbon 2’. uses the U base T

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20
Q

Nucleoside

A

Sugar + Base (exposed nitrogenous bases)

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21
Q

Nucleotide

A

Sugar + Base + Phosphate group

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22
Q

Chargaff’s Rule

A

Purine(A+G)/Pyrimidines (T+C) = 1.0 approximately

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23
Q

Double stranded DNA

A

(C+T) = (A+G)

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24
Q

Rosalind Franklin and Maurice Wilkins

A

x-ray diffraction lead to discovery of DNA being a helix of constant diameter

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25
Q

Erwin Chargraff

A

Base pairing

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26
Q

Direction DNA spiraling around helix axis

A

anti-parallel (opposite directions)

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27
Q

Direction of reading DNA

A

5’ to 3’

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28
Q

What is perpendicular to the helix axis in DNA?

A

Base pairs

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29
Q

DNA two Grooves

A

Major and minor grooves allow for proteins to bind to and recognize DNA

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30
Q

Another way to refer to denaturing of DNA

A

Melting

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31
Q

What is “Melting”

A

Separation of two DNA strands, to single stranded DNA (ssDNA)

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32
Q

What is reversible posses in separation of DNA

A

Renaturation

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33
Q

Ways to denature or “melt” DNA

A

Increase temperature, reduce salt concentration, increase pH, solvents

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34
Q

Melting Temperature (Tm) of DNA

A

Defined as the temperature when DNA duplex is separated into single strands

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35
Q

What can DNA Tm indicate

A

duplex (hybridized DNA molecules) stability, higher Tm the more stable the DNA helix

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36
Q

How to measure DNA denaturation

A

absorbance. AS DNA duplex separated, the absorbance increases (hyperchromic shift).

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37
Q

What does higher Tm indicate?

A

More stable DNA helix

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38
Q

How does GC content (% G+C), effect melting temperature?

A

Higher GC content increases stability, therefore increasing melting temperature

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39
Q

How does salt concentration effect melting temperature?

A

Higher salt concentration, higher the melting temperature. Salt causes the phosphate backbone to the tightly packed, and this shield the negative charged of the phosphate backbone. Increases stability.

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40
Q

How can GC content in DNA classify organisms?

A

It is species specific

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41
Q

Does mutated of normal DNA have higher melting point?

A

Mutated melting rages are just different than “normal” ranges no specification of higher or lower

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42
Q

Molecular Biology Techniques involving DNA melting

A

polymerase chain reaction (PCR) and southern blotting

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43
Q

Equation for Melting Temperature

A

Tm = 81.5 + 16.6 lg [M] + 0.41(%GC) - 675/L

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44
Q

The three proposed models of DNA replication

A

Conservative, dispersive, and semiconservative replication

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45
Q

Correct model of DNA replication

A

semiconservative

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46
Q

What results of 1st round of Meselson experiment?

A

50% light and 50% heavy so NOT conservative replication

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47
Q

What is results of 2ndround of Meselson experiment?

A

There where some DNA that were 100% intermediate. So NOT intermediate replication

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48
Q

Semi-conservative replication

A

Each daughter cell consists of one parental strand and one newly synthesized strands based off a parental strand.

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49
Q

DNA synthesis Requirements

A

1) ssDNA template
2) all four dNTPs
3) DNA polymerase and other supporting enzymes
4) Free 3’-OH group

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50
Q

dNTPS

A

deoxynucleotide triphosphate, with each using a different DNA base: adenine (dATP), cytosine (dCTP), guanine (dGTP), and thymine (dTTP).

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51
Q

Mechanism of DNA Synthesis

A

Catalysis of phosphodiester bond between dNTPS and bases

52
Q

Direction of DNA chain elongation

A

5’ to 3;

53
Q

Template strand reading direction

A

3’ to 5’

54
Q

Origin of Replication

A

the specific nucleotide sequence where replication begins

55
Q

Where dies synthesis take place

A

Within a replication bubble

56
Q

Where are DNA strands synthesized simultaneously

A

replication fork

57
Q

What is a DNA molecule/region of DNA that replicates from a single origin of replication called?

A

Replicon

58
Q

Replication is ______

A

Semicontinuous

59
Q

Leading strand synthesis is _ and the direction of the fork

A

continuous

60
Q

Lagging strand synthesis is _ and occurs in the opposite direction of the fork

A

discontinuous

61
Q

Bacteria Genome

A

Circular

62
Q

Bacteria Replication

A

Theta replication

63
Q

Virus Genome

A

Circular

64
Q

Virus Replication

A

Rolling circle replication

65
Q

Eukaryotes Genome

A

Linear

66
Q

Eukaryotic replication

A

Linear replication

67
Q

Theta replication

A

1) single replicon (entire chromosome)
2) bidirectional replication w two forks within a single bubble
3) semi discontinuous in both replications forks
4) results in two circular DNA molecules

68
Q

Rolling circle

A

1) no replication bubble
2) uncoupling of the replication of the two DNA molecules
3) replication is continuous
4) results in multiple circular DNA molecules

69
Q

Linear

A

1) multiple replicons, origins or replication, and replication bubbles
2) bidirectional
3) Semi discontinuous at both replication forks
4) results in two linear DNA molecules

70
Q

Four stages of replication

A

1) Initiation
2) unwinding
3) elongation
4) termination

71
Q

Initiation

A

Initiator protein bind to the origin of replication (oriC) and a short section of DNA unwinds and proteins bein the ssDNA. ss-binding-protein keeps DNA separated and Helicases binds to the lagging strand template and breaks hydrogen bonds.

72
Q

What bonds to helicases break

A

hydrogen bonds

73
Q

Unwinding

A

Helicases break the hydrogen bonds between DNA strands while DNA gyrase (a topoisomerase) travels ahead of the fork and alleviates supercoiling caused by unwinding

74
Q

What does unwinding cause without DNA gyrase to reverse

A

supercoiling

75
Q

Chain Elongation

A

RNA primer (RNA nucleotides stretch) is synthesized by Primase. RNA primer provides a free 3’ OH for the DNA polymerases to use. RNA primer is replaced with nucleotides

76
Q

Why do we require an RNA primer?

A

Because the production of RNA does NOT require a 3’ end

77
Q

E. Coli principle replication enzyme

A

Pol III

77
Q

DNA ligase

A

seals the nick in the sugar phosphate backbone

77
Q

What DNA polymerase activity fills in the gap of DNA nucleotides (E. Coli)

A

Pol I

77
Q

Exonucleases

A

Removes primers starting at the 5’ ends. Removes Newley incorporated nucletides that do not match the template strand

77
Q

E. Coli polymerase that removes and replication RNA primers with DNA

A

pol I

78
Q

How many E. Coli DNA polymerases are there

A

Five (Pol I to Pol V)

79
Q

Eukaryotic DNA polymerases

A

Alpha, delta, and epsilon (all 5’ to 3’ activity) (delta and epsilon do 3’ to 3’ exonuclease activity)

80
Q

Delta DNA polymerase

A

Lagging-strand synthesis of nuclear DNA, DNA repair, and translesion DNA synthesis

80
Q

Alpha DNA polymerase

A

Initiation of nuclear DNA synthesis and DNA repair; has primase activity

80
Q

Epsilon DNA polymerase

A

Leading-strand synthesis

80
Q

Benefit of more origins in human DNA synthesis?

A

multiple origins ensure efficient genome replication in limited time

80
Q

Eukaryotic DNA is packaged into ____

A

Chromatin

80
Q

Telomeres

A

1) are the end of linear chromosomes
2) made up of G-rich short repeated sequences
3) stabilize chromosomes
specialized reverse transcriptase
4) extends the end of the parental DNA by RNA-templated DNA synthesis
5) responsible for the replication of the chromosomes ends
6) extends the DNA, filling int he gap due to the removal of the RNA primer

81
Q

Transcription + translation =

A

gene expression

82
Q

DNA replication

A

information transferred from one DNA molecule to another

83
Q

Transcription

A

Information transferred from DNA to an RNA molecule

84
Q

Translation

A

Information is transferred from RNA to a protein through a code that specify the amino acid sequence

85
Q

Prokaryotes gene expression

A

transcription and translation both occur in the cytoplasm

86
Q

Eukaryotic gene expression

A

Transcription in nuclear then pre-mRNA is processed to mRNA and leaves into the cytoplasm for translation

87
Q

Protein coding RNA

A

mRNA

88
Q

RNA unique to prokaryotes

A

CRISPR RNA (crRNA)

89
Q

Folded complex of RNA called _

A

Hairpin-loops and stem-loops

90
Q

Synthesis in respect to DNA template strand

A

complementary and antiparallel

91
Q

Transcription

A

1) initiation does not require a primer
2) ribonucleotides are added to the 3’OH group of the growing RNA chain
3) DNA unwinds at the front of the transcription bubble
4)rewinds

92
Q

Three requirements of transcription

A

1) DNA template
2) RNA nucleotides (rNTP’s)
3) RNA polymerase and other proteins

93
Q

RNA transcription

A

Template is read in 3’ to 5’ direction, while RNA is synthesized in the 5’ to 3’ direction
ONLY the RNA coding region is transcribed

94
Q

Promoter

A

upstream of the start site, adjacent to gene. Indicates the direction of transcription. Orients the enzyme towards the start site

95
Q

RNA coding region

A

downstream of start site

96
Q

Termination site

A

downstream of start site

97
Q

Initiation (prokaryotic)

A

assembly of transcription apparatus on the promoter and begins synthesis of RNA

98
Q

Elongation (prokaryotic)

A

DNA is threaded through RNA polymerase, unwinds the DNA, ass new nucleotides to the 3’ end of the growing RNA strand.

99
Q

Termination (prokaryotic)

A

the recognition of the end of transcription

100
Q

What do bacteria use to recognize promoters?

A

Sigma factors

101
Q

what would happen without sigma?

A

core enzyme would initiate transcription randomly

102
Q

Holoenzyme

A

in prokaryotes it is the complete enzyme complex composed of the core RNA polymerase and the sigma factor

103
Q

Consensus sequences

A

short stretch of DNA that is conserved among promoters of different genes (prokaryotic thing)

104
Q

Common sequences (or elements)

A

-10 (pribnow box) and -35 nucleotides which are upstream of the start site (+1). They are NOT identical in all promoters

105
Q

Promoter sequences strength =

A

frequency of trasnciription

106
Q

Strong promoter

A

recA

107
Q

Down mutations

A

base substitutions that make the sequence less similar to the consensus sequences reduce the rate of transcription

108
Q

up mutation

A

sequence become more similar to the consensus sequences

109
Q

RNA transcription in prokaryotes

A

is initiated when core RNA polymerase binds to the promoter with the help of sigma

110
Q

Terminators in bacteria

A

Rho-dependent (requires Rho protein) and Rho-independent (also called intrinsic terminator)

111
Q

Rho-dependent termination

A

1) rho bind to RNA upstream of terminator
2) RNA polymerase pauses when it reaches terminator and Rho catches up
3) Rho unwinds DNA-RNA hybrid using helicase activity

112
Q

Rho-independent termination

A

1) inverted repeats
2) polymerase pauses at Us
3) hairpin formation destabilize DNA-RNA hybrid
4) RNA transcript dissociated from RNA polymerase, DNA reanneals

113
Q

Consensus sequences in order for eukaryotic transciption

A

TFIIB (-35), TATA(-25), initiator(+1), and DCE (+30)

114
Q

Core promoter

A

extend upstream/downstream of transcription start site. Minimal sequence required for accurate transcription initiation. Includes a number of consensus sequences

115
Q

Regulatory Promoter

A

located upstream of the core promoter, exact location can be variable. Transcriptional activator proteins binds to consensus sequences and affect the rate of transcription.

116
Q

Order of Basal transcription apparatus assembly

A

TATA binding protein, general transcription factors, basal transcription apparatus

117
Q

What is required for termination

A

cleavage of the mRNA at a specific site

118
Q

What degrades the remaining mRNA terminating transcription?

A

5’ to 3’ exonuclease

BIOL 2030 (7 decks)

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