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Biochem II Self Study > Exam 1 Review > Flashcards

Exam 1 Review Flashcards

1
Q

Which of the following relies on a glycosylase enzyme for its first step?

A. Base excision repair
B. Nucleotide excision repair
C. Mismatch repair
D. Homologous recombination
E. Nonhomologous end joining

A

A. Base excision repair

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2
Q

If a cytosine in chromosomal DNA is damaged by hydrolysis, which repair mechanism is used to repair it?

A. Nucleotide excision repair
B. Non-homologous end-joining
C. Direct chemical reversal repair
D. Base excision repair
E. Mismatch repai

A

D. Base excision repair

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3
Q

The enzyme reverse transcriptase is NOT required for the mobilization of which genetic

A. SINES
B. Transposons
C. Retroviruses
D. LINES
E. All of them require reverse transcriptase

A

B. Transposons

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4
Q

Approximately ______ % of our DNA contains known genes in open reading frames.

A. 1.5%
B. 15%
C. 25%
D. 55%
E. 100%

A

A. 1.5%

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5
Q

Many DNA polymerases have built-in error-correcting activity, which is called:

A. DNA glycosylase
B. DNA polymerase δ
C. Proof-reading exonuclease
D. Reverse transcriptase
E. Helicase

A

C. Proof-reading exonuclease

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6
Q

Which of the following repair pathways can most accurately repair a double-strand break?

A. Base excision repair
B. Nucleotide excision repair
C. Homologous recombination
D. Direct chemical reversal
E. Nonhomologous end joining

A

C. Homologous recombination

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7
Q

Which of the following enzymes the cells use to remove supercoiling?

A. Rag1/Rag2 complex
B. Sliding clamps
C. Rad 51
D. Helicases
E. None of the above

A

E. None of the above
(topoisomerase)

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8
Q

An essential component of PCR is the use of two oligonucleotides

A. Which are complementary to one another
B. At least one of which must contain repetitive sequence
C. Which hybridize to opposite strands on either side of a target sequence
D. Which hybridize on the same strand to either side of a target sequence
E. None of the above

A

C. Which hybridize to opposite strands on either side of a target sequence

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9
Q

In a Cot reassociation experiment, which reassociates faster: poly U/poly A or E. coli DNA?
Explain your answer.

A

Poly U/poly A will reassociate faster because any poly U strand can base pair with any poly A strand. DNA from E. coli is all unique sequence DNA, so each strand must find its original complementary strand, which takes longer

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10
Q

What are the important features of the E. coli oriC sequence?

A

AT rich region, DnaA box, Several GATC sequences

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11
Q

AT rich region (E. coli oriC)

A

duplex opens at this region allowing access of the replication machinery to the origin region

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12
Q

DnaA box (E. coli oriC)

A

Binding site for replication protein DnaA.

Binding of multiple DnaA proteins bends DNA leading to torsional stress and promoting opening of the DNA duplex at the AT rich region

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13
Q

GATC sequences (E. coli oriC)

A

GATC is methylated in bacterial cells.

In newly replicated DNA (post-initiation) these sequences are hemi-methylated because the Dam methylase did not have time to modify the newly synthesized DNA.

SeqA protein bind to the hemi-methylated sites blocking a new round of binding of DnaA to the origin.

This prevents multiple rounds of replication initiation

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14
Q

What features should a useful cloning vector have?

A
  • An origin of replication
  • a selectable maker
  • a unique restriction site(s)
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15
Q

Initiator proteins

A

DnaA in prokaryotes (dnaA binds to oriC)

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16
Q

Helicase

A

Opens DNA

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17
Q

Why does Helicase need ATP?

A

DNA helicases must unwind the DNA before it can be copied

these enzymes use lots of ATP to separate the strands

DNA helicases hydrolyze ATP
they circle one of the two strands and travel along that strand unwinding the helix by forcing apart the H bonds

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18
Q

Primase

A

In E. coli, dnaB is the helicase. It forms a complex with primase (primeosome)

makes RNA primers for DNA polymerase

Primase is a type of RNA polymerase that creates an RNA primer; then DNA polymerase can bind
are about 10 nucleotides long

is one of the most error prone and slow polymerases

RNA primers can be recognized later and replaced with DNA by a less error prone polymerase

Primase needs to act just once on the leading strand, but many times on the lagging strand

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19
Q

Single Strand DNA binding proteins

A

bind to the DNA to prevent reannealing and intramolecular base pairing

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20
Q

Clamp loader

A

a clamp loader protein and ATP are needed to load the sliding clamp

in E. coli, the clamp is actually a subunit of DNA poly III

The clamp loader works hard on the lagging strand

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21
Q

Sliding clamp

A

a sliding clamp increases the speed and processivity of the replicase

keeps DNA polymerase from falling off the DNA

a sliding clamp loads onto the DNA before polymerase, and greatly increases the processivity of polymerase

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22
Q

DNA polymerase III

A

is the replicative polymerase

it synthesizes both the leading and lagging strand

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23
Q

DNA polymerase I

A

removes the RNA primers from the lagging strands and replaces them with DNA

it has two exonuclease activities, one of which is proofreading activity, as well as its polymerase activity

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24
Q

Ligase

A

ligase must act many times to seal nicks on in the sugar-phosphate backbone (phosphodiester) between each pair of okazaki fragments

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25
Q

Topoisomerase

A

relieve overwinding of DNA by helicase

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26
Q

Restriction enzyme

A

bind loosely to DNA, then scan for their recognition sequence. binding changes the conformation of the DNA

the enzyme forms multiple contacts with the backbone, with major groove, and the minor groove

amino acids of the enzyme form H-bonds to the phosphates of the backbone and to the bases

restriction enzymes can leave different ends after cutting DNA
5’ overhang, 3’ overhang, Blunt

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27
Q

Endonuclease

A

cuts within a DNA molecule

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28
Q

Exonuclease

A

digests DNA from a free end

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29
Q

Telomerase

A

an enzyme that adds repeats to the end of a chromosome

its is a ribonucleoprotein: it contains an RNA molecule that serves as an internal template for repeat addition

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30
Q

Telomeric repeats

A

it adds telomeric repeats (TTGGGGTTGGGGTTGGGG) to maintain telomere length (T2G4 Tetrahymena: T2AG3 mammals)

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31
Q

Telomere capping mechanisms

A

G-quadruplexes and capping proteins
t-loops and capping proteins

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32
Q

Shelterin

A

telomeres are protected by the shelterin complex

a complex of 6 proteins: TPP1, TIN2, TRF1, TRF2, RAP1, POT1

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33
Q

Uracil DNA glycosylase

A

DNA repair enzymes that initiate the base excision repair pathway and remove uracil from DNA.

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34
Q

AP endonuclease

A

is an enzyme that is involved in the DNA base excision repair pathway (BER).

Its main role in the repair of damaged or mismatched nucleotides in DNA is to create a nick in the phosphodiester backbone of the AP site created when DNA glycosylase removes the damaged base.

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35
Q

RecA/Rad51

A

plays a key role in DNA double-stranded break repair and homologous recombination.

RecA/Rad51 binds to ssDNA and forms contiguous filaments that promote the search for homologous DNA sequences and DNA strand exchange.

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36
Q

RAG1/RAG2

A

proteins initiate V(D)J recombination by introducing double-strand breaks at the border between a recombination signal sequence (RSS) and a coding segment.

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37
Q

Which proteins work harder on the lagging strand

A

clamp loader

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38
Q

DNA ligase

A

joins DNA molecules end-to-end by catalyzing phosphodiester bond formation between 3’ OH and 5’ PO4

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39
Q

Polynucleotide kinase

A

adds a phosphate to a 5’ OH

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40
Q

phosphatase

A

removes a 5’ phosphate

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41
Q

Role of DNA ligase in cloning

A

Adaptors (short, dsDNA fragments of known sequence) are added to the genomic DNA fragments using DNA ligase

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42
Q

How electrophoresis separates DNA fragments

A

Separated according to their size.

An electric current is applied to pull them through the gel. DNA fragments are negatively charged, so they move towards the positive electrode.

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43
Q

Steps to create a library of human genomic DNA

A

Step 1:
isolate DNA from lymphocytes
partial digest with HindIII, treat fragments with alkaline phosphatase
digest the vector with HindIII

Step 2
mix vector and “insert” (here, the human DNA fragments) DNA and add DNA ligase
the “sticky ends” created by the restriction enzyme will base pair, and ligase will seal the sugar phosphate backbone

Step 3
transform tthe ligated DNA into E. coli and select for cells that took up a vector

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44
Q

cDNA library

A

a collection of cloned DNA sequences that are complementary to the mRNA that was extracted from an organism or tissue

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45
Q

Genomic library

A

Te genomic DNA libraries comprise large DNA fragments.

not all our DNA is transcribed

not all genes are transcribed in every tissue

the cDNA will represent the Expressed Genome, and have multiple copies of some genes and no copies of others

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46
Q

How dideoxy nucleotides are used in DNA sequencing

A

used to terminate growing DNA chains and create subsets of truncated fragments in a sequencing reaction

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47
Q

C-value paradox

A

c-value: amount of DNA (in pg) in a haploid genome

​​refers to the unexpectedly high variations in c-values of different species. In his words, “different species contain different amounts of DNA in their nuclei

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48
Q

Low Cot

A

“Repeated DNA”

indicates sequences that easily find a pairing partner

if a DNA sequence is repeated in the genome, one strand can reassociate with the opposite strand from any other repeat

this does not take long, and indicates low complexity DNA: “Repeated DNA”

49
Q

High Cot

A

“Unique Sequence DNA”

indicates sequences that do not easily find a partner.

for ex. a DNA sequence that is unique in the genome, like a single copy gene, can only reassociate with its original opposite strand.

this takes a long time and indicates high complexity DNA: “Unique Sequence DNA”

50
Q

E. coli genome

A

one circular chromosome, one origin of DNA replication

features:
- wall to wall genes with little intergenic space
- genes organized into operons
- few repeated genes
- no introns
- some genes arrived by Horizontal GT

51
Q

Yeast Genome

A

16 separate linear chromosomes, each has a centromere, two telomeres, multiple origins

features:
- wall to wall genes with little intergenic space
- no operons
- some repeated genes
- a few genes have introns

52
Q

Prokaryotes primary sequencing read

A
  1. Genes have no introns, so there are long regions of coding sequence between the ATG and the first stop.
  2. There are often multiple stop codons in the same frame to be sure that translation stops at the right place.
  3. There are highly conserved promoter features like the Pribnow box that are often found right in front of genes.
53
Q

Eukaryotes primary sequencing read

A
  1. Most genes have multiple introns, with the exons (coding regions) often being less than 100 codons long.
  2. The conservation of splice sites is weak in higher eukaryotes, so it is not often possible to tell intron from exon.
  3. Eukaryotic promoters are highly variable and hard to recognize from sequence alone.
54
Q

Significance of reading frame

A

To find genes, you need to know the reading frame. It begins with the ATG and continues in groups of exactly three nucleotides until it encounters a stop codon.

55
Q

Types of repeated DNA in our genome

A

simple sequences, SINES, LINES, highly expressed genes

56
Q

Simple repeats

A

repeated sequences between 1-30 nucleotides

mononucleotide repeat, dinucleotide repeat, trinucleotide repeat, tetranucleotide repeats, Variable number tandem repeats

57
Q

mononucleotide repeat

A

one nucleotide repeated over and over

Ex. AAAAAA

58
Q

dinucleotide repeat

A

Ex. GT repeated 10 to 50 times, the length of these is highly polymorphic

making these GOOD MARKERS

59
Q

trinucleotide repeats

A

repeats of CAG or CCG are involved in several human diseases such as Huntington’s

classified as polyglutamate (PolyQ) disorders with abnormal CAG repeats in the coding region, and non-polyglutamate (non-polyQ) disorders

60
Q

Variable number tandem repeats (VNTR)

A

a 28-30 nucleotide sequence is repeated 20 to 100 times at various locations throughout the genome

61
Q

DNA-only Transposons

A

cut and paste gene from one area of genome to another: “fossilized” (i.e. no longer active) in our genome

62
Q

Long Interspaced Sequences (LINES)

A
  1. most abundant class (20%)
  2. range: 1-9 kb length
  3. called “non-retroviral retrotransposons”
  4. can move to new places in the genome
  5. There are many scattered throughout our genome
  6. are transcribed into mRNA, some encode reverse transcriptase
63
Q

Most common LINE in the human genome

A

L1 (6.5 kb)

64
Q

LINES: how they move

A

reverse transcriptase copies the L1 RNA into L1 DNA which is inserted into the target DNA forming a new L1 Element there

65
Q

Short Interspaced Sequences (SINES)

A
  1. 2nd Most abundant (13%)
  2. Alu elements are related to 7SL
66
Q

Most common SINE

A

Alu element

67
Q

Alu element

A

contains site for restriction enzyme Alu1

10^6 Alu elements in genome

are transcribed into RNA but do not encode any protein

68
Q

SINES: how Alu moves

A

Alu moves in genome using reverse transcriptase (from LINES) via an RNA intermediate

after transposition, the original copy is still present so the number of elements in the genome increases

69
Q

Alu elements in human disease

A

cause 0.1% of human disease (inherited)

can be insertional mutagens

70
Q

Reaction mechanism: a 3’ OH is always required

A

DNA polymerase catalyzes the formation of a phosphodiester bond between the 3’ OH of the last nucleotide of the primer and the 5’ triphosphate of the incoming nucleotide

71
Q

why one strand is made continuously but the other is made in fragments

A

Because DNA polymerase can only act in the 5’ to 3’ direction, replication of one strand has to be discontinuous

72
Q

Initiation proteins

A

Helicase, SSB, Primase, Clamp loader complex, sliding clamp, DNA poly III (in prokaryotes)

73
Q

Elongation proteins

A

DNA poly III, DNA poly I, DNA ligase, Topoisomerase

74
Q

replication in eukaryotes is similar but is coupled to the cell cycle

A

The initiation of DNA replication is carefully controlled, and coupled to the cell cycle

75
Q

3’ – 5’ exonucleolytic proofreading

A

a cytosine is accidentally incorporate across from an adenine

the C does not base pair properly, leaving the 3’ end unpaired and blocking further polymerization

the exonuclease activity of DNA polymerase chew off the C

the correct nucleotide is inserted and polymerization continues

76
Q

how strand-directed mismatch repair knows which strand is new

A

In E. coli, the mismatch repair system uses the methylation state to tell old from new strand (old is methylated, new not).

In eukaryotes, the mismatch repair system looks for nicks, which are more common on the new strand

The repair system does not just remove the mismatched base. rather, all the new strand in the region is removed and resynthesized

77
Q

how telomerase RNA serves as a template for adding telomeric repeats

A

Telomerase RNA also provides the template for addition of new telomeric repeats by the reverse-transcriptase protein subunit

78
Q

T-loops

A

These telomeres form little loops at the end of the chromosomes, which are called the t-loops. These are formed by inserting the ends of the chromosome, which is usually 3’ overhang back into the DNA of the chromosome.

play an important role in stabilizing the tertiary RNA structure by facilitating long-range interactions between different regions of the molecule.

79
Q

how telomeres shorten in disease and in a cell’s normal life span

A

As a cell begins to become cancerous, it divides more often, and its telomeres become very short. If its telomeres get too short, the cell may die.

Often times, these cells escape death by making more telomerase enzyme, which prevents the telomeres from getting even shorter.

telomere length shortens with age

80
Q

Hayflick limit

A

the number of times a normal cell can divide

in normal cells, telomeres shorten with each cell division

81
Q

the Hayflick limit, its relationship to telomerase, and how this changes in cancer cells

A

a transformed cell has bypassed the hayflick limit (Senescence) and an immortalized cell has survived Crisis, by reactivating telomerase activity.

By definition, all human cancer cells are transformed and immortal. the malignant ones are also tumorigenic

82
Q

Types of mutations

A

silent, missense, nonsense, sense, frameshift

83
Q

silent mutations

A

mutations affect the DNA but not the proteins, therefore no effect on phenotype

Since the genetic code is degenerate, several codons code for the same amino acid. Especially, third base changes often have no effect on the amino acid sequence of the protein

84
Q

missense mutation

A

sub one amino acid for another

substitute one amino acid for another. Some missense mutations have very large effects, while others have minimal or no effect. it depends on where the mutation occurs in the protein’s structure, and how big a change in the type of amino acid it is

85
Q

nonsense mutation

A

sub a codon with a stop codon

convert an amino acid into a stop codon. The effect is to shorten the resulting protein. These are often devastating mutations that result in completely non-functional proteins. however, sometimes nonsense have only a little effect if they occur at the carboxy terminal end of proteins

86
Q

sense mutation

A

stop codon to another amino acid

are the opposite of nonsense mutations. Here, a stop codon is converted into an amino acid codon. Since DNA outside of protein-coding regions contains an average of 3 stop codons per 64, the translation process usually stops after producing a slightly longer protein

87
Q

Frameshift mutation

A

insertion or deletion of a nucleotide causes a frameshift.

frameshift mutations result in all amino acids downstream from the mutation site being completely different from wild type. these proteins are generally non-functional

88
Q

polymorphisms

A

recessive and dominant mutations

you have two copies of each chromosome, but they are not the same. About 1-2% of the nucleotides are different, most of these are inconsequential and are called polymorphisms

89
Q

recessive mutations

A

a mutation that does not manifest a phenotype in the presence of a normal copy.

Ex. mutations that destroy the activity of a protein are often tolerated because we have two copies of most genes, and we can get by with half the amount of most proteins

90
Q

Dominant mutations

A

A mutation that manifests itself to change the outcome for that protein’s role in the system, even in the presence of a normal copy

91
Q

Depurination

A

is the process whereby purine bases (adenine and guanine) are lost because their N-glycosyl linkage to deoxyribose are spontaneously hydrolyzed

about 5,000 purine bases per cell per day are lost this way

breaks the glycosidic bond between the sugar and the A or G base

affects both purines

92
Q

Deamination

A

an extracyclic amino group is lost from the base

deamination can occur on other nucleotides as well as cytosine

deamination occurs on all bases but the deamination of cytosine to uracil is the most common, with a rate of about 100 bases per cell per day

93
Q

Ames Test

A

determines the mutagenicity of chemicals and is the first test typically done when a new drug or chemical is synthesized that will be used on humans

94
Q

how a non-carcinogenic compound can become a carcinogen

A

When a cell oxidizes benzopyrene to diol epoxide, by cytochrome P450 enzymes, can cause the cell to become carcinogenic

95
Q

UV-damage causes

A

pyrimidine dimers, ionizing radiation causes severe damage

96
Q

Repair mechanisms for single strand (nicks, gap)

A

Direct repair (DR)
Mismatch repair (MMR)
Base excision repair (BER)
Nucleotide excision repair (NER)

97
Q

Base excision repair (BER)

A

removes a single nucleotide. three enzymes remove the base, which is then replaced by DNA poly and ligase

98
Q

BER mechanism

A

A family of endonucleases known as DNA glycosylases recognizes altered bases and removes them.

an AP (“apurinic”) endonuclease recognizes that the deoxyribose lacks the base and removes the sugar in conjunction with a phosphodiesterase. DNA polymerase and DNA ligase fill in the missing base using the opposite strand as the template

99
Q

BER enzymes

A

DNA glycosylases and AP endonuclease

100
Q

Nucleotide Excision repair (NER)

A

a large multienzyme complex scans the DNA for distortions in the helix (rather than for a single base pair)

101
Q

NER mechanism

A

a cut is made on either side of the distortion by an endonuclease and an associated DNA helicase then removes the entire portion of the damaged strand

the large gap (12-13 nucleotides in E. coli, 24-32 in eukaryotes) is repaired by DNA polymerase and DNA ligase

102
Q

NER enzymes

A

in e. coli these are uvrA, B, C, and D proteins

in eukaryotes they are called XPA, XPB, XPC, and XPD

103
Q

How Mismatch repair picks the correct strand to fix

A

in e coli, the parent strand is methylated

Human/eukaryotes
nick directed:
lagging strand contains transient nicks (b/w okazaki fragments and newly replicated replacements for RNA primer regions)
how leading strand is nicked is not known

proteins called
Msh: MutS homolog
Mlh: MutL homolog

104
Q

Repair mechanisms for double strand break repair

A

Homologous recombination (HR) and Non-homologous end joining (NHEJ)

105
Q

Homologous recombination (HR)

A

we can repair a double strand break using the homologous copy of our chromosomes as a source of sequence information

106
Q

Non-homologous End joining (NHEJ)

A

The break is repaired by DNA ligation but some nucleotides are lost at the break site. this is better than letting the chromosome break, especially as so little of human DNA encodes protein

107
Q

why homologous recombination is accurate and non-homologous end joining is not

A

HR being a precise mechanism uses extensive homology, while NHEJ is error prone as it utilizes no or limited homology.

108
Q

the functions of recA/Rad51

A
  • Binds single stranded DNA
  • Then binds double stranded DNA and searches for homologous region
  • Opens the double stranded DNA and forms a D-loop
  • Hydrolyzes ATP as it moves along the DNA
109
Q

the significance of 5’ to 3’ exonuclease after a double strand break

A

This leaves a 3’ overhang that can invade another strand and be extended by polymerase

110
Q

the structure of an antibody molecule

A

two identical light chains and two identical heavy chains in a Y-shape

antigen binding sites are on the ends of the arms of the Y

111
Q

light chains are made when

A

one V region is joined to one J region by nonhomologous end joining

111
Q

antibody genes are found in

A

segments in the genome

112
Q

heavy chains are made by

A

joining V, D, and J regions

113
Q

sequences involved in recombination

A

heptamer, nonamer, 12-, and 23-base spacers

114
Q

the specific arrangement of these spacers assures that

A

V will only join to J

115
Q

where RAG proteins bind

A

The RAG proteins bind to the RSSs, and then to each other to bring the nonamers and heptamers together

116
Q

which gene segments encode the amino acids in each part of the Y protein structure

A
117
Q

how more than a million different antibodies can be made from a relatively small number of gene segments

A

there even more ways to generate antibody diversity including changes in mRNA splicing and somatic hypermutation, where mutations occur in rearranged antibody genes of mature B cells to generate more variation

these generate antibodies of even higher affinity

all together, we can make more than 10^8 different antibodies from the Ig locus

Biochem II Self Study (10 decks)

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