Exam 1 Review

Question 1

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

Answer 1

A. Base excision repair

Question 2

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

Answer 2

D. Base excision repair

Question 3

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

Answer 3

B. Transposons

Question 4

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

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

Answer 4

A. 1.5%

Question 5

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

Answer 5

C. Proof-reading exonuclease

Question 6

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

Answer 6

C. Homologous recombination

Question 7

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

Answer 7

E. None of the above
(topoisomerase)

Question 8

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

Answer 8

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

Question 9

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

Answer 9

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

Question 10

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

Answer 10

AT rich region, DnaA box, Several GATC sequences

Question 11

AT rich region (E. coli oriC)

Answer 11

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

Question 12

DnaA box (E. coli oriC)

Answer 12

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

Question 13

GATC sequences (E. coli oriC)

Answer 13

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

Question 14

What features should a useful cloning vector have?

Answer 14

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

Question 15

Initiator proteins

Answer 15

DnaA in prokaryotes (dnaA binds to oriC)

Question 16

Helicase

Answer 16

Opens DNA

Question 17

Why does Helicase need ATP?

Answer 17

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

Question 18

Primase

Answer 18

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

Question 19

Single Strand DNA binding proteins

Answer 19

bind to the DNA to prevent reannealing and intramolecular base pairing

Question 20

Clamp loader

Answer 20

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

Question 21

Sliding clamp

Answer 21

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

Question 22

DNA polymerase III

Answer 22

is the replicative polymerase

it synthesizes both the leading and lagging strand

Question 23

DNA polymerase I

Answer 23

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

Question 24

Ligase

Answer 24

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

Question 25

Topoisomerase

Answer 25

relieve overwinding of DNA by helicase

Question 26

Restriction enzyme

Answer 26

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

Question 27

Endonuclease

Answer 27

cuts within a DNA molecule

Question 28

Exonuclease

Answer 28

digests DNA from a free end

Question 29

Telomerase

Answer 29

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

Question 30

Telomeric repeats

Answer 30

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

Question 31

Telomere capping mechanisms

Answer 31

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

Question 32

Shelterin

Answer 32

telomeres are protected by the shelterin complex a complex of 6 proteins: TPP1, TIN2, TRF1, TRF2, RAP1, POT1

Question 33

Uracil DNA glycosylase

Answer 33

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

Question 34

AP endonuclease

Answer 34

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.

Question 35

RecA/Rad51

Answer 35

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.

Question 36

RAG1/RAG2

Answer 36

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

Question 37

Which proteins work harder on the lagging strand

Answer 37

clamp loader

Question 38

DNA ligase

Answer 38

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

Question 39

Polynucleotide kinase

Answer 39

adds a phosphate to a 5' OH

Question 40

phosphatase

Answer 40

removes a 5' phosphate

Question 41

Role of DNA ligase in cloning

Answer 41

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

Question 42

How electrophoresis separates DNA fragments

Answer 42

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.

Question 43

Steps to create a library of human genomic DNA

Answer 43

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

Question 44

cDNA library

Answer 44

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

Question 45

Genomic library

Answer 45

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

Question 46

How dideoxy nucleotides are used in DNA sequencing

Answer 46

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

Question 47

C-value paradox

Answer 47

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

Question 48

Low Cot

Answer 48

“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”

Question 49

High Cot

Answer 49

“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”

Question 50

E. coli genome

Answer 50

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

Question 51

Yeast Genome

Answer 51

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

Question 52

Prokaryotes primary sequencing read

Answer 52

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.

Question 53

Eukaryotes primary sequencing read

Answer 53

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.

Question 54

Significance of reading frame

Answer 54

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.

Question 55

Types of repeated DNA in our genome

Answer 55

simple sequences, SINES, LINES, highly expressed genes

Question 56

Simple repeats

Answer 56

repeated sequences between 1-30 nucleotides mononucleotide repeat, dinucleotide repeat, trinucleotide repeat, tetranucleotide repeats, Variable number tandem repeats

Question 57

mononucleotide repeat

Answer 57

one nucleotide repeated over and over Ex. AAAAAA

Question 58

dinucleotide repeat

Answer 58

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

Question 59

trinucleotide repeats

Answer 59

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

Question 60

Variable number tandem repeats (VNTR)

Answer 60

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

Question 61

DNA-only Transposons

Answer 61

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

Question 62

Long Interspaced Sequences (LINES)

Answer 62

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

Question 63

Most common LINE in the human genome

Answer 63

L1 (6.5 kb)

Question 64

LINES: how they move

Answer 64

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

Question 65

Short Interspaced Sequences (SINES)

Answer 65

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

Question 66

Most common SINE

Answer 66

Alu element

Question 67

Alu element

Answer 67

contains site for restriction enzyme Alu1 10^6 Alu elements in genome are transcribed into RNA but do not encode any protein

Question 68

SINES: how Alu moves

Answer 68

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

Question 69

Alu elements in human disease

Answer 69

cause 0.1% of human disease (inherited) can be insertional mutagens

Question 70

Reaction mechanism: a 3' OH is always required

Answer 70

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

Question 71

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

Answer 71

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

Question 72

Initiation proteins

Answer 72

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

Question 73

Elongation proteins

Answer 73

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

Question 74

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

Answer 74

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

Question 75

3’ – 5’ exonucleolytic proofreading

Answer 75

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

Question 76

how strand-directed mismatch repair knows which strand is new

Answer 76

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

Question 77

how telomerase RNA serves as a template for adding telomeric repeats

Answer 77

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

Question 78

T-loops

Answer 78

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.

Question 79

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

Answer 79

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

Question 80

Hayflick limit

Answer 80

the number of times a normal cell can divide in normal cells, telomeres shorten with each cell division

Question 81

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

Answer 81

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

Question 82

Types of mutations

Answer 82

silent, missense, nonsense, sense, frameshift

Question 83

silent mutations

Answer 83

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

Question 84

missense mutation

Answer 84

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

Question 85

nonsense mutation

Answer 85

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

Question 86

sense mutation

Answer 86

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

Question 87

Frameshift mutation

Answer 87

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

Question 88

polymorphisms

Answer 88

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

Question 89

recessive mutations

Answer 89

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

Question 90

Dominant mutations

Answer 90

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

Question 91

Depurination

Answer 91

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

Question 92

Deamination

Answer 92

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

Question 93

Ames Test

Answer 93

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

Question 94

how a non-carcinogenic compound can become a carcinogen

Answer 94

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

Question 95

UV-damage causes

Answer 95

pyrimidine dimers, ionizing radiation causes severe damage

Question 96

Repair mechanisms for single strand (nicks, gap)

Answer 96

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

Question 97

Base excision repair (BER)

Answer 97

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

Question 98

BER mechanism

Answer 98

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

Question 99

BER enzymes

Answer 99

DNA glycosylases and AP endonuclease

Question 100

Nucleotide Excision repair (NER)

Answer 100

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

Question 101

NER mechanism

Answer 101

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

Question 102

NER enzymes

Answer 102

in e. coli these are uvrA, B, C, and D proteins in eukaryotes they are called XPA, XPB, XPC, and XPD

Question 103

How Mismatch repair picks the correct strand to fix

Answer 103

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

Question 104

Repair mechanisms for double strand break repair

Answer 104

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

Question 105

Homologous recombination (HR)

Answer 105

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

Question 106

Non-homologous End joining (NHEJ)

Answer 106

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

Question 107

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

Answer 107

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

Question 108

the functions of recA/Rad51

Answer 108

- 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

Question 109

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

Answer 109

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

Question 110

the structure of an antibody molecule

Answer 110

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

Question 111

light chains are made when

Answer 111

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

Question 111

antibody genes are found in

Answer 111

segments in the genome

Question 112

heavy chains are made by

Answer 112

joining V, D, and J regions

Question 113

sequences involved in recombination

Answer 113

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

Question 114

the specific arrangement of these spacers assures that

Answer 114

V will only join to J

Question 115

where RAG proteins bind

Answer 115

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

Question 116

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

Question 117

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

Answer 117

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