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Molecular Genetics > Exam I - Lecture (1-5) > Flashcards

Exam I - Lecture (1-5) Flashcards

1
Q

Transcription is reversible?

A

yes

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2
Q

translation is reversible?

A

no

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3
Q

Genetic material must ___ very large amounts of ___

A

store; information

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4
Q

Genetic material must have the capacity to be ___ accurately to be ____ “unchanged” to the next generation

A

Replicated; transmitted

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5
Q

Genetic material must ___ _____

A

Encode; phenotype

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6
Q

Transformation principle = DNA

A

Avery MacLeod and Mccarty

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7
Q

Transformation

A

F. Griffith

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8
Q

Bacteriophage genetic material = DNA

A

Hersey and Chase

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9
Q

Griffiths experiment used what organism?

A

Streptococcus pneumoniae

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10
Q

S Strain - Smooth

A

virulent → mouse dies

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11
Q

R strain

A

non virulent → mouse lives

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12
Q

Transformation

A

some cellular component is taken up by live R bacteria (Streptococcus pneumonia) from dead S bacteria, making them virulent.

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13
Q

Who postulated that genes = DNA

A

Avery

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14
Q

Mechanism of genetic transformation

A

CapS (capsule gene) fragment released when the S cell is being heat killed. The CapS fragment enters the chromosome of the CapR cell → recombination and cell division leads to a virulent S cell.

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15
Q

Bacteriophage components

A

50% protein and 50% DNA

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16
Q

Phage attaches to E. Coli and injects its chromosome ..

A
  • Phage chromosome replicates
  • Expression of phage genes produces phage structural components
  • Progeny phage particles assemble
  • Bacterial wall lyses, releasing progeny phages
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17
Q

1952 Hersey and Chase experiment

A

Concluded that DNA - not protein - is the genetic material in bacteriophages

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18
Q

How a centrifuge works

A

Larger, denser molecules move toward the bottom of the tube more quickly.

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19
Q

How did the Hershey and Chase experiment work?

A

32P DNA → infect bacteria → Break bacterial cell walls away from any viral material remaining outside them → centrifuge → little 32P in supernatant → bacteria lyse → 32P labeled DNA in progeny phage

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20
Q

what does a nucleotide consist of

A

5-carbon sugar + phosphate + nitrogenous base (4 options)

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21
Q

Nucleotides (monomer) are connected by ___ bonds to form a nucleic acid (polymer)

A

phosphodiester bonds

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22
Q

What atom is is in the 2’ carbon of ribose (RNA)

A

OH

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23
Q

What atom is is in the 2’ carbon of deoxyribose (DNA)

A

H

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24
Q

The nitrogenous base is attached to carbon ___ of the sugar

A

carbon 1

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25
Q

The phosphate is attached to carbon ___ of the sugar

A

carbon 5

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26
Q

DNA sugar

A

deoxyribose

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27
Q

RNA sugar

A

Ribose

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28
Q

Which group of a nucleotide is an acid?

A

The phosphate group → it releases H + ions

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29
Q

Why are DNA and RNA negatively charged molecules?

A

Because the phosphate releases H+ ions

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30
Q

Ribose has ___ in its 2’ position

A

OH

RNA is not as stable as DNA because OH is reactive

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31
Q

deoxyribose has ___ in its 2’ position

A

H

Makes DNA have very high stability

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32
Q

ribose and deoxyribose are ___ sugars

A

pentose

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33
Q

Purine

A

Has two rings

Adenine and Guanine

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34
Q

Pyrimidine

A

1 ring

Cytosine, Uracil, Thymine

(uracil and thymine differ in a methyl group thats found in thymine)

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35
Q

Nucleoside

A

Pentose sugar + nitrogenous base

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36
Q

Linking nucleotides by phospohodiester bonds or linkages (covalent)

A

5’ phosphate group of one nucleotide is linked to the 3’ hydroxyl group of the next nucleotide

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37
Q

Ester linkage (phosphodiester bond) involves

A

loss of water

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38
Q

Nucleic acids have polarity

A

3’ end ends with hydroxyl (-OH)

5’ end ends with phosphate group

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39
Q

Chargaff’s rules

A

A = T, C = G

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40
Q

Rosalind Franklin - Diffraction data

A

discovered the helical DNA structure - but Watson and Crick received the prestige

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41
Q

width of helix

A

2.0 nm

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42
Q

distance between bases

A

0.34 nm

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43
Q

length of one complete turn of helix

A

3.4 nm

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44
Q

what bond is between the nitrogenous bases

A

hydrogen (weak) → but strong enough to stabilize the molecule

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45
Q

DNA is

A
  • complementary
  • Constant width
  • antiparallel
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46
Q

Ideal DNA structure

A

B Form

  • Spiral staircase with bases being treads
  • flat base pairs perpendicular to the backbone (sugar+phosphate)
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47
Q

The B Form

A
  • Hydrated in vivo conditions
  • Right handed double helix
  • On the outside → major and minor grooves (transcription factors)
  • 10bp/turn (theoretically)
  • In the cell, it is closer to 10.5 bp/turn
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48
Q

Most of our DNA is ___ because our cells are mostly composed of __

A

B; water

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49
Q

A- Form of Double Helix

A
  • Shorter
  • 11 bp/turn
  • bases are tilted
  • Right handed
  • DNA when dehydrated
  • In vivo, dsRNA and RNA/DNA hybrids resemble A-form
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50
Q

Z form - of DNA double helix

A
  • Longer and thinner
  • 12 bp/turn
  • backbone a zigzag
  • Left handed
  • Formation favored by high conc. of +charged ions and long GC or AT stretches (in vitro)
  • Never reported in vivo
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51
Q

DNA triple helix

H DNA

A

Hoogsteen Base pairs - H-DNA - sideways pairing forms

C=G=C or T=A=T

One purine interacting with 2 pyrimidines

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52
Q

Why is it very unlikely to have a triple helix DNA strand in the cell?

A

because it’s pH is very low. Our body’s pH is 7.2

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53
Q

DNA tetraplex or quadruplex

A

Guanosine tetraplex

only DNA sequences with high portions of G (end of chromosomes → telomeres)

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54
Q

Polypeptides made of AA are attached to each other via ___ ___

A

Peptide bonds

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55
Q

Building blocks of proteins

A

amino acids

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56
Q

Amino acid structure

A

R group → distinguishes one aa from another

Amino group

Carboxyl group

Hydrogren

All attached to an Alpha carbon

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57
Q

non-polar, aliphatic R groups

A

Glycine, Alanine, Proline, Valine, Leucine, Isoleucine, Methionine

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58
Q

Polar, uncharged R groups

A

Serine, Threonine, Cysteine, Asparagine (N) , Glutamine (Q)

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59
Q

Polar, Negatively charged R groups

A

Aspartate (D), Glutamate (E)

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60
Q

Polar, Positive Charged R groups

A

Histidine, Lysine (K), Arginine (R)

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61
Q

Non-polar, aromatic, R grou[s

A

Phenylalanine (F) , Tyrosine (Y) , Tryptophan (W)

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62
Q

Peptides are ___linked together by __

A

amino acids; peptide bonds

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63
Q

O=C-NH

A

Peptide bond

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64
Q

R groups face the __ direction of each other

A

opposite

trans-conformation → most common

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65
Q

secondary structure

A

regular repeats resulting from hydrogen bonds involving the backbone

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66
Q

Types of secondary protein structure

A

Alpha helix → coil (10-15 aa long). Right handed spiral (1.2nm)

Beta pleated sheet → a fold (strands 3-10 aa long)

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67
Q

what sequences of aa can form a-helix

A
  • No consecutive bulky or long R groups
  • No consecutive like-charged R groups → would cause repulsion
  • Few with polar R groups (Ser, Thr, Asn, Asp) → these destabilize the helix
  • Infrequent glycine (very flexible) or proline (inflexible)
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68
Q

What type of beta strand is most stable?

A

Antiparallel. Alternating C-N-C-N terminus

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69
Q

Beta sheets consist of at least 2 beta strands

A

Parallel / Antiparallel

70
Q

Super-secondary structural elements

A

Helix-Helix Turn

Coiled-coil

Four-helix-bundle

71
Q

Helix - turn - Helix (HTH)

A

in proteins, HTH is a major structural motif capable of binding DNA

AA sequence forms a turn

72
Q

How does HTH bind DNA?

A

C-terminal binds to a major groove. N-terminal helps to position the complex

73
Q

Types of super-secondary beta structural elements

A

Beta hairpin

Antiparallel Beta sheet

Beta barrel → can form forks in the cell membrane

74
Q

Mixed Super-secondary structural elements

A
  • Beta-alpha-beta
  • A/B barrel → beta sheets on the inside and alpha helices on outside → higher stability
75
Q

Tertiary structure of protein

A

is the overall folded 3D shape of a single polypeptide chain

30 structure is determined by 10 and 20 structures combined with interactions between R groups and the environment

76
Q

What proteins are on the inside of a protein

A

Non-polar AA

Hydrophobic exclusion

77
Q

Interactions important to keep 3D shape + functions

A

Hydrogen bond

Disulfide Bridge

Ionic Bond

Van der Waals → hydrophobic interactions

78
Q

Hemoglobin

A

40 structure

contains 2 alpha + 2 beta subunits

79
Q

Quartenary protein interactions

A

Hydrogen bonds

ionic bonds

Hydrophobic and Van der Waals interactions

80
Q

oligomers

A

proteins composed of multiple peptide chains

  1. Homooligomers
  2. Heterooligomers
81
Q

Homooligomers

A

identical subunits

82
Q

heterooligomers

A

nonidentical subunits

83
Q

Intrinsically unstructured proteins

A

⅓ of all proteins in higher eukaryotes may lack fixed 30 structure entirely or at least contain unstructured sections or regions → diversity in binding

84
Q

example of homooligomer (homodimer)

A

E.Coli - Beta clamp

85
Q

example of heterooligomer

A

hemoglobin

86
Q

Protein function

A
  • Defense
  • communication
  • enzymes
  • transport
  • storage
  • structure
87
Q

karyotype

A

arranges chromosomes by size

88
Q

homo sapiens has __ chromosomes

A

46

89
Q

Bacteria’s chromosomes are __

A

circular

90
Q

In cells, ___ molecules are much longer than the cell diameter (or nucleus) - require ___

A

DNA ; compacting

91
Q

prokaryotes

A

no nuclear membrane separating genetic material from other cellular compartments

92
Q

DNA supercoiling: First level of compaction

A

in bacteria

“a coiled coil”

Relaxed circular DNA

  • add two turns → over-rotate → positive supercoil
  • remove two turns → under-rotate → negative supercoil
93
Q

DNA topoisomerases

A

catalyzes DNA under-winding and relaxation

94
Q

DNA gyrase

A

Type II Topoisomerase → introduces negative supercoils to bacterial chromosomes

95
Q

typical bacterial chromosome contains

A

50 giant loops of supercoiled DNA (- charge) arranged around a protein scaffold (+ charge)

96
Q

Chromatin in eukaryotes

A

DNA and proteins (50/50)

97
Q

basic unit of chromatin

A

nucleosome

98
Q

Nucleosomes: beads on a string

A

Histone core of nucleosome + linker DNA of nucleosome

99
Q

Histone proteins

A

isolate proteins bound to DNA and run on a gel

100
Q

H1

A

half the amount as the other histone proteins

101
Q

what histones are approximately equimolar

A

H2A, H2B, H3, H4

SDS page gel electrophoresis

101
Q

what histones are approximately equimolar

A

H2A, H2B, H3, H4

SDS page gel electrophoresis

101
Q

what histones are approximately equimolar

A

H2A, H2B, H3, H4

SDS page gel electrophoresis

102
Q

Histone proteins are

A

highly conserved, positively charged proteins

103
Q

Histone proteins are

A

highly conserved, positively charged proteins

104
Q

about 25 % of each histone

A

lysine (K) and Arginine (R)

105
Q

Histone protein structure

A

all have N terminal “tails” that are important for the regulation of chromatin structure

N terminus - unstructured domain

106
Q

histone fold domain

A

3 alpha helices

107
Q

histone fold domain

A

3 alpha helices

108
Q

which histones have longer C termini

A

H2A + H2B

109
Q

the histone fold motif

A

supersecondary structure

Loop1, Loop 2, alpha helix 1, alpha helix 2, alpha helix 3

110
Q

histone octamers ____ ____ into repeating units

A

organize; DNA

No DNA present → no histone octamer

111
Q

H2A and H2B form a

A

dimer

112
Q

H3 and H4 form a

A

tetramer

113
Q

what does one histone octamer consist of?

A

2x each core histone

  • H2A, H2B, H3, H4
  • left handed supercoil of 146 bp winds 1.67 times around the histone octamer
114
Q

what is responsible for binding protein interactions between core histones as well as binging DNA to form nucleosomes?

A

Histone folds

Hydrogen bonds

115
Q

Regulation of chromosome structure

A
  • Chromatin remodeling complexes
  • Histone Modifications by enzymes
116
Q

Chromatin Remodeling Complex

A

Nucleosomes can be arranged by this ATP driven complex.

Consists of 2-18 proteins

These complexes change in some way the association between the histone cores with the DNA wrapped around.

“Promoter regions has to be accessible for transcription factors to begin transcription

117
Q

Chromatin Remodeling Complex in humans

A

H. Sapiens

hBRM - ATPase

10 - Number of subunits

Bromodomain - Domain

Activation - Effect on transcription

118
Q

H3.3

A

maintaining transcriptionally active open state

119
Q

Open chromatin (transcriptionally active)

A

exposed promoter - not too condense (chromatin is available)

120
Q

CENPA

A

maintaining kinetochore attachement

121
Q

Kinetochore attachment sites to the spindle fibers

A

separates the sister chromatids

122
Q

H2AX

A

attracting DNA repair enzymes

123
Q

double strand break in histone/chromatin

A

H2AX → phosphorylation attracts repair proteins (when there is a break - in the DNA)

124
Q

modifications of ___ alter chromatin structure

A

histone tails

125
Q

histone modifying enzymes attach ___

A

chemical groups to aa of nucleosome subunits

126
Q

Modifications: closed → open

A

cis: if the outcome is a direct result of the modification

Trans: if the modification attracts another protein that performs the histone modifying function

127
Q

histone tail modifications

A

will either affect the clamping together nucleosomes closer or pushing them further away, altering DNA accessibility

128
Q

Acetylation of Lysine in residues

A

decreases the positivity of histone→ enhances DNA accessibility (looser DNA → enhances transcription of the DNA)

129
Q

HAT

A

histone acetyl transferase

more transcription

130
Q

HD = HDAC

A

histone deacetylase

Less transcription

131
Q

tails compromise ___ of mass of histones

A

25-30%

132
Q

Histone tails are rich in

A

Lys (K)

Ser (S)

Arg (R)

133
Q

Lysine (K) in tails

A

acetylation

methylation

ubiqutination

134
Q

Serine (S) in tails

A

phosphorylation

135
Q

Arg (R) in tails

A

Methylation

136
Q

Linker histone H1 (large protein)

A

binds the nucleosome and “packs” adjacent nucleosomes

137
Q

chromatin is more compacted when ___ binds the nucleosome

A

H1

138
Q

H1 helps nucleosomes condense into a higher level of packaging

A

30 nm filament

139
Q

How is the 30nm. fiber maintained?

A

By interactions between the tails of the adjacent nucleosomes as well as compaction by H1

140
Q

euchromatin

A

undergoes condensation and de-condensation during cell cycle - may become transcriptionally active

141
Q

heterochromatin

A

stays more compacted

142
Q

mitotic chromosomes is ___ shorter than its naked DNA

A

50,000 X

143
Q

basic unit of heredity

A

gene

144
Q

A gene is a …

A

basic unit of heredity and a sequence of nucleotides in DNA and RNA that encodes the synthesis of a gene product, either RNA or protein

145
Q

a gene segment of a nucleic acid

A

carries the code for a protein or a ncRNA (functional non-coding RNA)

146
Q

prokaryote vs eukaryote

A

Both have cell membrane

Prokaryote: circular DNA/chromosome, no nucleus

Eukaryote: linear DNA, nuclear membrane

147
Q

what separates living cells from gene creatures

A

plasma membrane

148
Q

cloning

A

making identical copies

149
Q

General steps of DNA cloning

Recombinant DNA Technology or Genetic Engineering

A
  1. Isolate a fragment of chromosome (DNA segment) to be cloned by digestion with restriction enzymes (endonuclease)
  2. select a small molecule of DNA, capable of self replication (cloning vectors) and digest with restriction enzymes
  3. Join two DNA fragments covalently (DNA ligase) - recombinant DNA
  4. Move recombinant DNA from the test tube to a host cell (most common E.Coli); provide the machinery for DNA replication
    1. Identify/select host cells containing recombinant DNA (selectable markers; permit the growth of a cell or kill the cell under a defined set of conditions).
150
Q

endonucleases → cut the nucleic acids in the middle of the molecule “molecular scissors”

A

restriction enzymes

151
Q

Type II restriction endonuclease

A

Cleaves DNA at specific base sequences

152
Q

DNA ligase

A

Joins two DNA molecules together

153
Q

Nucleases (Exo and Endo) used for degradation of nucleic acids

A

Exo-nuclease → degrades from either 3’ or 5’ end.

154
Q

recognition sequence (type II restriction endonucleases)

A

usually 4 to 8bp long and palindromic

155
Q

plasmids

A

circular DNA molecule that replicates separately from the host chromosome

156
Q

frequency with a particular recognition sequence occurs in the DNA

A

the longer the recognition sequence the lower the probability!

4bp → 4^4 = 256bp

6bp → 4^6 = 4,096bp

157
Q

plasmids(up to 5kb)

A

bacterial artificial chromosomes

158
Q

Origin of replication

A

(ori)

159
Q

what is used to clone DNA

A

pBR322

160
Q

gel electrophoresis

A
  1. DNA samples containing fragments of different sizes are placed in wells in an agarose gel
  2. An electrical current is passed through the gel
  3. All DNA fragments move froward the positive pole; smaller fragments move faster than large fragments. After electrophoresis, fragments of different sizes have migrated different distances
  4. A dye specific for nucleic acids is added to the gel.
  5. DNA fragments appear orange under UV light
161
Q

PCR components

A
  1. template DNA
  2. oligonucleotide primers
  3. DNA polymerase
  4. dNTPs
  5. Buffer to maintain pH and provide Mg2+
162
Q

PCR components

A
  1. template DNA
  2. oligonucleotide primers
  3. DNA polymerase
  4. dNTPs
  5. Buffer to maintain pH and provide Mg2+
163
Q

primers

A

synthetic oligonucleotides

164
Q

Primers (short nucleotide strand/oligonucleotide) synthesize the ______ direction

A

5’ → 3’

165
Q

Taq DNA polymerase

A

stable at very high temperatures

166
Q

stages of PCR (repeat 25-30 times)

A

denaturation, primer annealing and elongation

167
Q

denaturation

A

first, the denaturation of double stranded DNA into single stranded DNA occurs at high temp. (94-96C)

168
Q

annealing

A

the reaction temperature is lowered and DNA primers bind to single-stranded DNA with complementary sequences. The annealing temperature varies for each primer set and is dictated by length and sequence

169
Q

Extention/elongation PCR

A

following annealing, the temperature is raised to 72 for elongation by DNA polymerase. DNA polymerase uses the primers, template and dNTPs to make new DNA strands

Molecular Genetics (9 decks)

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