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BIOL 3020 > Exam 1 > Flashcards

Exam 1 Flashcards

1
Q

Central Dogma

A

DNA –> Transcription –> RNA –> Translation –> Protein

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2
Q

Exceptions to the Central Dogma

A

1. Reverse Transcription: Certain retroviruses such as HIV are able to convert single-stranded RNA into a double-stranded DNA copy, which is then inserted into the genome of the host cell

2. RNA Replication: some viruses’ RNA genome is copied directly into RNA without the use of DNA as an intermediary

3. RNA Editing: the base sequence of RNA is altered after it is transcribed from the DNA so that it no longer corresponds precisely to the DNA

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3
Q

Prokaryotic

A

Lacks a nucleus (bacteria and archaea)

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4
Q

Eukaryotic

A

contains a nucleus

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5
Q

Recombinant DNA technology (genetic engineering)

A

allows genes to be isolated, sequenced, modified, and transferred from one organism to another

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6
Q

Nucleosides

A

a base covalently bonded to the 1’ position of a pentose sugar molecule

(base + sugar)

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7
Q

Nucleotide

A

nucleoside with one or more phosphate groups covalently bonded to the 3’ or 5’ position

(base + sugar + phosphate)

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8
Q

Phosphodiester bonds

A
  • in nucleic acid polymers, the ribose or deoxyribose sugars are linked by a phosphate between the 5’ position of one sugar and the 3’ position of the next
  • creates sugar-phosphate backbone with a base attached to the 1’ of each sugar
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9
Q

DNA/RNA sequence

A

Consists of bases A,C,G,T/U in the DNA or RNA chain. Conventionally written in the free 5’ to free 3’ end of the molecule

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10
Q

Bonds between Base-Pairs

A
  • three hydrogen bonds between C and G
  • two hydrogen bonds between A and T
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11
Q

Purines

A

Adenine and Guanine

(two rings)

(heterobicyclic molecule)

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12
Q

Pyrimidines

A

Cytosine and Thymine (or Uracil in RNA)

(one ring)

(heterocyclic molecule)

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13
Q

Difference between Uracil and Thymine

A

Thymine has a methyl group at the Carbon-5 position while Uracil just has a hydrogen

Thymine is “5-methyluracil”

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14
Q

Structural Difference between DNA and RNA

A

one hydroxyl group (-OH) on the 2’ position (in RNA, but not in DNA)

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15
Q

Glycosidic bond (or glycosylic bond)

A

bond between bases and sugars

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16
Q

Structure of DNA

A

Nitrogenous bases attatched via a glycosidic bond to pentose sugars that are linked together by phosphodiester bonds between the 3’ and 5’ carbons of those sugars to form a long-chain polymer

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17
Q

Amino Acid Structure

A

a carbon atom is linked to a carboxyl group, a primary amino group, a proton and a side chain (R)

All natural forms exist in the L conformation (chiral), Lewis structures of natural amino (L) acids always have the carboxyl group at the top and the amino group to the left.

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18
Q

Polarity of Amino Acids

A

Amino acids are dipolar ions (zwitterions) in aqueous solution and behave as both acids and bases (they are amphoteric)

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19
Q

Globular Proteins

A

folded compactly and behave in solution more or less as spherical particles (most enzymes are globular in nature)

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20
Q

Fibrous Proteins

A

very high axial ratios (length/width) and are usually important structural proteins, for example in silk fibroin and keratin in hair and wool.

Generally much larger than globular proteins.

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21
Q

Prosthetic groups

A

may act as cofactors in enzyme reactions, or as large associations (lipids in lipoproteins or the carbohydrate in glycoproteins)

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22
Q

Peptide bond

A

bonds amino acids to one another (links carboxyl group of one amino acid covalently to the amino group of the next) to create a polypeptide.

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23
Q

Written directionality of polypeptides

A

N-terminus to C-terminus

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24
Q

Primary Structure of polypeptide

A

Linear amino acid sequence from N-terminus to C-terminus, also includes any covalently bonded prosthetic groups and disulfide links (salt bridges) between cystiene residues (cystines)

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25
Q

Van Der Waals forces

A

noncovalent associations between electrically neutral molecules

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26
Q

Hydrogen Bonds in Proteins

A

formed between a covalently bonded hydrogen atom on a donor group (ex. O-H or N-H) and a pair of non-bonding electrons on an acceptor group (ex O–C or N-)

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27
Q

Protein Functions

A
  1. Enzymes: allmost all enzymes are proteins (a limited number of RNAs act as enzymes)
  2. Signaling: receptor proteins in cell membranes can bind ligands
  3. Transport and Storage: hemoglobin, transferrin, ferritin, lipoproteins
  4. Structure and Movement: Collagen, keratin
  5. Nutrition: Casein and ovalbumin
  6. Immunity: antibodies
  7. Regulation: transcription factors
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28
Q

Which base is not found in DNA?

A

Uracil

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29
Q

Which statements best describe a DNA double helix?

A
  • the sugar-phosphate backbone is on the outside of the helix
  • the backbones of the two strands run in opposite (anitparallel) directions
  • bases are positioned perpendicular to the overall axis of the DNA strand
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30
Q

Which statements about RNA are true?

A
  • RNA secondary structure is more variable than DNA
  • RNA contains uracil instead of thymine
  • RNA differs from DNA by containing ribose, not deoxyribose sugars
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31
Q

Which statements best describe a nucleotide?

a. the phosphate group is attached to the base
b. a nucleotide contains a base, a sugar, and a phosphate group
c. the sugar contains two nitrogen atoms
d. the bases contain extra-cyclic groups such as NH2 or O
e. the base is attached to the sugar by a glycosidic bond

A
  • b. a nucleotide contains a base, a sugar, and a phosphate group
  • d. the bases contain extra-cyclic groups such as NH2 or O
  • e. the base is attached to the sugar by a glycosidic bond
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32
Q

Nucleic Acids: Sugar components

(KNOW THE NUMBERING OF THE SUGARS)

A
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33
Q

Nucleic Acids: nucleosides

A
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34
Q

Structure of 5’ and 3’ ends

A
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35
Q

DNA Characteristics

A
  1. The polynucleotide strands are antiparallel.

(5’–>3’ and 3’–>5’).

  1. Bases in opposing strands are complementary. Base pairs always have one purine and one pyrimidine. (A = T, G = C).
  2. The bases interact by hydrogen bonding.
  3. DNA has 10 (10.5 in vivo) base pairs per turn of helix.
  4. DNA helices have major and minor grooves.
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36
Q

Differences between DNA and RNA

A
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37
Q

Separating the double helix into single strands during replication requires which protein?

a. DNA polymerase I
b. DNA polymerase II
c. helicase
d. topoisomerase
e. primase

A

Helicase

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38
Q

Leading and lagging strand synthesis differs in that:

a. On the leading strand, synthesis is 5’ to 3’ but 3’ to 5’ on the lagging
b. Leading strand synthesis does not require topoisomerase, unlike lagging strand synthesis
c. The leading strand requires only a single primer, while the lagging strand requires multiple primers
d. Helicases open the leading strand faster than the lagging strand

A

c. The leading strand requires only a single primer, while the lagging strand requires multiple primers

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39
Q

The technique used to determine that DNA replication occurs semi-conservatively was…

A

equilibrium density-gradient centrifugation

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40
Q

All of the following are properties of DNA polymerase III except:

a. It completes synthesis of the leading strand before beginning synthesis of the lagging strand
b. It is responsible for incorporating most of the nucleotides in the lagging strand
c. It is responsible for incorporating most of the nucleotides in the leading strand
d. It contains a 3’ to 5’ proofreading activity
e. It has a very low error rate

A

a. It completes synthesis of the leading strand before beginning synthesis of the lagging strand

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41
Q

If the sequence ACATA mutates to become ACTTA, the A to T alteration is called a:

a. depurination
b. transversion
c. transition
d. tautomeric shift

A

Changes from a purine to a pyrimidine, so the answer is:

b. transversion

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42
Q

Which of these DNA repair mechanisms does not use DNA polymerase during repair?

a. homologous recombinational repair of a double-strand break
b. base excision repair of an apurinic site
c. mitotic mismatch repair
d. photolyase repair of a thymine dimer
e. nucleotide excision repair of a thymine dimer

A

d. photolyase repair of a thymine dimer

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43
Q

A tautomeric shift during replication most likely results in a:

a. missense mutation
b. deletion mutation
c. frameshift mutation
d. insertion mutation
e. inversion mutation

A

a. missense mutation

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44
Q

Site-directed mutagenesis:

a. can be used to restore a mutant gene back to wild-type sequence
b. can be used to modify the function of a gene
c. can be used to modify the binding site for a repressor in the 5’ untranslated region of a gene
d. all of the above
e. none of the

A

d. all of the above

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45
Q

Definition of Mutation Rate and facts

A

The term mutation rate is the likelihood that a gene will be altered by a new mutation

  • It is commonly expressed as the number of new mutations in a given gene per generation
  • It is in the range of 10-5 to 10-9 per generation
  • The mutation rate for a given gene is not constant, It can be increased by the presence of mutagens
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46
Q

Mutation Frequency

A

Number of mutant genes divided by the total number of copies of that gene in a population

  • If 1 million bacteria were plated and 10 were mutant

The mutation frequency would be 1 in 100,000 or 10-5

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47
Q

What does mutation frequency depend on?

A
  • Timing of the mutation
  • Likelihood that the mutation will be passed on to future generations
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48
Q

Types of Mutation, by order of severity

A
  1. Genome-level changes: ploidy changes (whole genome duplication/deletion)
  2. Chromosome-level changes: c’some gain/loss
  3. Double-strand break: loss of chromosome telomere-proximal to break
  4. Chromosome rearrangements: large insertions, large deletions, duplications, inversions, translocations, transposition
  5. Locus alterations: point mutations, repeat alterations
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49
Q

Primary ways in which structure of chromosomes can be altered

A
  1. Change in total genetic information (deficiencies/deletions)
  2. Genetic material/information rearranged (inversions/translocations)
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50
Q

Change in total genetic information

A

•Deficiencies/Deletions

–The loss of a chromosomal segment

•Duplications

–The repetition of a chromosomal segment

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51
Q

Genetic material/information rearranged

A
  • Inversions

–A change in the direction of part of the genetic material along a single chromosome

  • Translocations

–Segment of one chromosome becomes attached to a non-homologous chromosome

  • Simple translocations: one way transfer
  • Reciprocal translocations: two way transfer
52
Q
A

Deletion

53
Q
A

Duplication

54
Q
A

Inversion

55
Q
A

Simple Translocation

56
Q
A

Reciprocal Translocation

57
Q

Point mutation

A

change in a single base pair, it involves a base substitution

58
Q

Transition

A

a change of a pyrimidine (C,T) to another pyrimidine or a purine (A,G) to another purine

59
Q

Transversion

A

change of a purine to a pyrimidine or vice versa

60
Q
A

Silent Mutation

61
Q
A

Missense Mutation

62
Q
A

Nonsense mutation

63
Q
A

Frameshift

64
Q

Tautomeric Shifts

A

spontaneous changes in base structure can cause mutations if they occur immediately prior to DNA replication

65
Q

Two most common methods of spontaneous mutations to DNA

A

Depurination and Deamination

66
Q

Depurination

A

purine bases (adenine and guanine) are lost because their N-glycosyl linkages to deoxyribose are spontaneously hydrolyzed

67
Q

Deamination

A
  • occurs on all bases except thymine
  • amino group is removed and replaced with an oxygen atom
68
Q

How deamination causes mutation during DNA replication

A
69
Q

Thymine Dimer

A

Covalent bonds form between adjacent pyrimidines, particularly thymines, as a result of ultraviolet light such as sunlight. This causes problems for DNA replication as the dimers can not be recognized by DNA polymerase.

70
Q

DNA Repair Mechanisms: Genome Level or Chromosome level

A
  • No way to correct - must be prevented
71
Q

DNA Repair Mechanisms: Double Strand Breaks

A

–Homologous recombination

–Non-homologous end-joining

72
Q

DNA Repair Mechanisms: Point alterations

A

–Direct repair

–Base excision repair

–Nucleotide excision repair

–Mismatch repair

73
Q

Three possible models of DNA replication

A
  1. Conservative Model
  2. Semi-conservative Model
  3. Dispersive Model
74
Q

Conservative Model

A

Both parental strands stay together after DNA replication

75
Q

Semiconservative Model

A

The double-stranded DNA contains one parental and one daughter strand following replication

76
Q

Dispersive Model

A

Parental and daughter DNA are interspersed in both strands following replication

77
Q

Meselson-Stahl experiment and how it proved semi-conservative replication

A
  • An ultracentrifuge can separate DNA strands that contain heavy 15N or light14N
  • Grow E. coli in 15N medium for many generations
  • all nitrogen in DNA is 15N
  • Switch E. coli to 14N medium
  • Analyze DNA after each of several generations
  • “old” DNA will have 15N and “new” DNA will contain 14N
  • proved that DNA replication follows the semiconservative model
78
Q

Properties of DNA Polymerases

A
  1. Complex enzymes with high fidelity, high processivity.
  2. Can extend a DNA chain but can not start one de novo. - this means that every DNA polymerase requires a primer.
79
Q

Prokaryotes: DNA Polymerase III

A

synthesizes most of the DNA

80
Q

Prokaryotes: DNA Polymerase I

A

Replaces RNA primer with DNA

81
Q

Prokaryotes: Other DNA Polymerases

A

used for DNA repair

82
Q

Eukaryotes: DNA Polymerase α and δ (delta)

A

make the leading strand

83
Q

Eukaryotes: DNA Polymerase α and ε (epsilon)

A

make the lagging strand

84
Q

DIRECTION OF DNA

A

NEW SUBUNITS ARE ADDED TO THE 3’ END

ONLY

ONLY

ONLY

DNA IS SYNTHESIZED IN THE 5’ TO 3’ DIRECTION

ONLY

ONLY

ONLY

85
Q

Benefits of the Major Groove in DNA

A

allows proteins to search for a certain DNA section

86
Q

Purpose of DNA ligase

A
  • DNA polymerase leaves a gap in the backbone of the DNA, so ligase connects these two segments of the backbone by adding a phosphodiester bond
87
Q

DNA helicase

A

“unzips” the two strands of DNA by breaking the hydrogen bonds between the base pairs

88
Q

DNA primase

A

creates primers as helicase unzips the molecule. Only one primer is needed for the leading strand, but one primer is needed for each Okazaki fragment in the lagging strand

  • fun fact: the primers place by primase are actually RNA, not DNA
89
Q

single strand binding protein (SSBs)

A

blocks a single strand of DNA so that it will not rehybridize with another strand to form a double strand until synthesis is completed

90
Q

Topoisomerase (or DNA gyrase in prokaryotes)

A

untwists the DNA to relieve tension caused by unzipping a coil

91
Q

Telomerase: Why is it necessary?

A

There is no place to put polymerase on the last primer on the 3’ end of the lagging strand, so it will just fall off leaving the lagging strand shorter than it should be (losing DNA)

92
Q

What does telomerase do to fix this issue?

A

it acts as a polymerase and adds base pairs (as RNA) so a primer can attach and synthesize the last remaining chunk of DNA

  • The DNA at telomeres is composed of a simple repeated sequence.
  • A special DNA polymerase called telomerase carries an internal template that directs the synthesis of these repeats.
93
Q

Know the numbering of the sugar and where the base and phosphates are attached

A

Base is attached to the 1’ carbon of the sugar (glycosidic bond) and the phosphate group is attached to the 3’ or 5’ carbon on the sugar

94
Q

Know the details of leading strand and lagging strand replication, and Okazaki fragments

A
95
Q

What are dNTPs and NTPs are required for?

A

Deoxynucleotide triphosphates and nucleotide triphosphates are the substrates for DNA synthesis.

  • basically nucleotides with 2 extra phosphates attached.
  • This is formed when either the 5’ carbon atom of one deoxyribose sugar bonds to an oxygen on the first phosphate. This is a condensation reaction which removes diphosphate from each nucleotide. This is why the nucleotides found in DNA are actually dNMPs, (deoxyribonucleotide monophosphates
96
Q

Where does replication begin?

A

at a special sequence called an origin

  • prokaryotic chromosomes have a single origin
  • eukaryotic chromosomes have multiple origins From the origin, replication proceeds bidirectionally
  • this yields two sites where DNA is synthesized These are called replication forks
97
Q

Replication of Circular DNA

A
98
Q

replicon

A

any piece of DNA that replicates as a single unit

99
Q

terminus

A

the point where replication ends in a piece of circular DNA. usually opposite of the origin.

100
Q

Bacterial DNA Replication Diagram

A
101
Q

lesions

A

alterations to the normal chemical or physical structure of the DNA

102
Q

bulky adducts

A

bulky lesions such as pyrimidine dimers and arylating agent adducts distort the double helix and cause localized denaturation

103
Q

what happens if lesions are not repaired?

A
  • a mutation could arise and become fixed in the DNA by direct or indirect mutagenesis. Mutations in the germ line may be intereted while mutations in somatic cells can lead to altered cell function, including carcinogenesis
  • The chemical change may produce a physical distortion in the DNA that blocks replication and or translation, resulting in cell death
  • DNA lesions may be mutagenic and/or lethal
104
Q

Consequence of Depuration or Deamination

A

the loss of bases results in an apurinic (AP site, only in the case of depuration) which means information encoded in the purine is lost

105
Q

abasic site

A

when either a purine or pyrimidine is lost

106
Q

apurinic (AP) site

A

a purine is lost

107
Q

Oxidative damage

A

reactive oxygen species such as superoxide, hydrogen peroxide, or the hydroxide radical (OH) can attack DNA, producing a range of oxidation products with altered properties

  • can lead to single and double strand breaks
108
Q

alkylation

A

alkylating agents are chemicals that add alkyl (aka methyl) groups to nucleic acids

  • these can be lethal if they interfere with the unwinding of DNA during replication and transcription
109
Q

Bulky adducts: Cyclobutane pyrimidine (mainly thymine) dimers

A

formed by ultraviolet light causing adjacent pyrimidines to form a cyclobutane ring, creating a bulky lesion which may interfere with replication and transcription

110
Q

Physical Mutagens: High-energy ionizing radiation

A

absorption of high-energy ionizing radiation such as X-rays and gamma rays cuases the target molecules to lose electrons, which can cause extensive alterations to DNA, including strand breaks, and base and sugar destruction

111
Q

Physical Mutagens: Nonionizing radiation

A

causes molecular vibrations or promotion of electrions to higher energy levels within the target molecules, which can cause new chemical bonds

  • includes UV light, which can cause pyrimidine dimers
112
Q

Types of Chemical Mutagens

A
  1. Base analogs (derivatives of the normal bases with different base pairing properties) –> direct mutagenesis
  2. Nitrous acid –> deamination of C to U, can cause transitions
  3. Alkylating and arylating agents –> produces lesions which can disrupt transcription and replication–> indirect mutagenesis
  4. most chemical mutagens are carcinogens and cause cancer
113
Q

Direct Mutagenesis

A
  • results from the presence of a stable, modified base with altered base pairing properties in DNA
  • this lesion will become fixed as a mutation if replication occurs
114
Q

Indirect Mutagenesis

A

most lesions in DNA are repaired by error free direct reversal or excision repair mechanisms before passage of a replication fork.

115
Q

Which statement about DNA polymerase III is FALSE?

a. It is responsible for incorporating most of the nucleotides in the leading strand.
b. It contains a 3’ to 5’ proofreading activity.
c. It is responsible for incorporating most of the nucleotides in the lagging strand.
d. It does not require primers to extend the leading strand.
e. It has a very low error rate.

A

d. It does not require primers to extend the leading strand.

116
Q
  1. Which of the following types of mutation is likely to have the greatest consequence to the cell?
    a. a G => A change that creates a stop codon within a gene
    b. trisomy of an autosome
    c. an insertion of a T in a homopolymeric run of Ts
    d. an inversion of the long arm of the X chromosome
    e. a reciprocal translocation between two autosomes
A

b. trisomy of an autosome

117
Q

(True/False) The function of helicases during replication is to:

  1. join the ends of adjacent Okazaki fragments together
  2. separate the DNA strands from each other
  3. relieve tension induced by supercoiling
  4. remove the RNA primers that initiate replication
A
  1. False
  2. True
  3. False
  4. False
118
Q

Which of the following statements about nucleotides are true and which are false? (True / False)

  1. The phosphate group is attached to the nitrogenous base.
  2. A nucleotide contains a base, a sugar, and a phosphate group.
  3. The sugar contains two nitrogen atoms.
  4. The nitrogenous base is attached to the sugar by a glycosidic bond.
A
  1. False
  2. True
  3. False
  4. True
119
Q

In dideoxy DNA sequencing, which carbons on the sugar portion of the nucleotide lack OH groups?

A

2’ and 3’

120
Q

How to fix pyrimidine dimers?

A

Photoreactivation by DNA photolyases.

  • these photoreactivating enzymes absorb blue light and transfer the energy to the cyclobutane ring, which is then cleaved
121
Q

How to fix alkylation?

A

Alkyltransferase.

  • removes the alkyl group
  • can also remove alkylated bases through base excision repair
122
Q

How to fix single strand breaks?

A
  • single strand breaks can simply be resealed by DNA ligase
123
Q

How to fix double strand breaks?

A

if left unrepaired, single strand breaks will convert to double strand breaks, which can be repaired by homologous recombination or nonhomologous end joining.

124
Q

Nucleotide excision Repair (operates mainly on bulky lesions)

A
  1. endonuclease cleaves the DNA a precise number of bases on either side of the lesion
  2. an oligonucleotide containing the lesion is removed, leaving a gap
  3. the gap is filled by DNA Polymerase I
  4. DNA ligase creates the final phosphodiester bond
125
Q

Base excision repair

A
  1. individual modified bases are recognized by one of a group of relatively lesion specific DNA glycosylases that cleave the glycosidic bond between the altered base and the sugar, leaving an abasic site
  2. AP endonuclease cleaves the DNA, then the gap may be furthered by exonuclease activity
  3. Gap is filled by polymerase I
  4. Sealed by ligase

BIOL 3020 (7 decks)

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