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Exam 3

front 1

When a DNA molecule is described as replicating bidirectionally, that means that it has two:

back 1

replication forks.

front 2

An Okazaki fragment is a:

back 2

segment of DNA that is an intermediate in the synthesis of the lagging strand.

front 3

Which one of the following statements about enzymes that interact with DNA is true?

back 3

Exonucleases degrade DNA at a free end.

front 4

E. coli DNA polymerase III:

back 4

is the principal DNA polymerase in chromosomal DNA replication.

front 5

The proofreading function of DNA polymerase involves all of the following except:

back 5

reversal of the polymerization reaction.

front 6

The 5' → 3' exonuclease activity of E. coli DNA polymerase I is involved in:

back 6

removal of RNA primers by nick translation

front 7

Prokaryotic DNA polymerase III:

back 7

has a β subunit that acts as a circular clamp to improve the processivity of DNA synthesis.

front 8

At replication forks in E. coli:

back 8

RNA primers are synthesized by primase

front 9

Which of these enzymes is not directly involved in methyl-directed mismatch repair in E. coli?

back 9

DNA glycosylase

front 10

The role of the Dam methylase is to:

back 10

modify the template strand for recognition by repair systems.

front 11

The functional unit of genetic information is called:

back 11

a gene

front 12

The process whereby RNA specifies a DNA sequence is known as:

back 12

reverse transcription

front 13

Proteins interact predominantly within which portion of a double-stranded DNA helix?

back 13

major groove

front 14

AT-rich DNA will denature/melt

back 14

at a lower temperature than GC-rich DNA.

front 15

In Bacteria and most Archaea, the enzyme that introduces negative supercoils into DNA is known as

back 15

DNA gyrase

front 16

Ciprofloxacin, nalidixic acid, and novobiocin are effective antiobiotics because they inhibit the activity of

back 16

DNA gyrase

front 17

Most plasmids are

back 17

double-stranded DNA, though a few are not.

front 18

Housekeeping genes are present in

back 18

chromosome

front 19

The precursor of each new nucleotide in a strand of DNA is a

back 19

deoxynucleoside 5'-triphosphate

front 20

DNA replication always proceeds from the ____ of the incoming nucelotide to the _____ of the previously added molecule

back 20

5' phosphate/ 3' hydroxyl

front 21

During DNA synthesis, the RNA primer is removed by a

back 21

Pol I exonuclease

front 22

The template for RNA polymerase is _____, and the RNA chain growth is _____ the chain growth of DNA.

back 22

DNA/ Identical to

front 23

Promoters are specific sequences of _____ that are recognized by ______.

back 23

DNA/ sigma factor

front 24

An example of nucleotide pairing is

back 24

A and T

front 25

Stop codons are also called ____ codons.

back 25

nonsense and termination

front 26

There are about ____ different tRNAs in bacterial cells and about _____ in mammalian cells

back 26

60/100-110

front 27

tRNA is released from the ribosome at the _____ site

back 27

E

front 28

Which statement is generally TRUE regarding protein synthesis?

back 28

the 23S rRNA plays a role in TRANSLOCATION; the 16S rRNA plays a role in INITIATION

front 29

Streptomycin inhibits _____ of protein synthesis, whereas tetracycline inhibits _____ of protein synthesis

back 29

initiation/elongation

front 30

In all cells, genes are composed of

back 30

nucleic acids

front 31

Which of the following is an example of one codon?

back 31

CAG

front 32

Which of the following is NOT correct regarding DNA and RNA synthesis?

back 32

Both processes require an RNA primer to begin.

front 33

Termination of RNA synthesis is ultimately determined by

back 33

specific nucelotide sequences on the template strand.

front 34

GTP provides energy for

back 34

translation 

front 35

Transcription of chaperonins is greatly accelerated when a cell is stressed by

back 35

excessive heat

front 36

The flow of biological information begins with

back 36

DNA

front 37

Most of the genes that encode proteins are found in

back 37

unique-sequence DNA

front 38

Why was the discovery of the structure of DNA so important for understanding genetics?

back 38

Without knowledge of the structure of DNA, it was impossible to understand how genetic information was encoded or expressed.

front 39

If Avery, MacLeod, and McCarty had found that samples of heat-killed bacteria treated with rNase and DNase transformed bacteria, but that samples treated with protease did not, what conclusion would they have made?

back 39

protein is the genetic material

front 40

Could Hershey and Chase have used a radioactive isotope of carbon instead of 32p? Why or why not?

back 40

No: carbon is found in both protein and nucleic acid.

front 41

How do the sugars of rNA and DNA differ?

back 41

the sugar of rNA has a hydroxyl group that is not found in the
sugar of DNA

front 42

the antiparallel nature of DNA refers to

back 42

the opposite direction of the two strands of nucleotides.

front 43

How does Z-DNA differ from B-DNA?

back 43

Z-DNA has a left-handed helix; B-DNA has a right-handed helix. The sugar–phosphate backboneof Z-DNA zigzags back and forth, whereas the sugar– phosphate backbone of B-DNA forms a smooth continuous ribbon

front 44

A DNA molecule 300 bp long has 20 complete rotations. this DNA molecule is

back 44

negatively supercoiled

front 45

How does bacterial DNA differ from eukaryotic DNA?

back 45

Bacterial DNA is not complexed with histone proteins and is circular

front 46

Neutralizing their positive charges would have which effect on the histone proteins?

back 46

they would bind less tightly to the DNA

front 47

Which of the following is a characteristic of DNA sequences at the telomeres? a. One strand consists of guanine and adenine or thymine
nucleotides. b. they consist of repeated sequences. c. One strand protrudes beyond the other, creating some
single-stranded DNA at the end. d. All of the above

back 47

d. All of the above

front 48

how many bands of DNa would be expected in Meselson and Stahl’s experiment after two rounds of conservative replication?

back 48

Two bands

front 49

Discontinuous replication is a result of which property of DNa

back 49

antiparallel nucleotide strand

front 50

Place the following components in the order in which they are first used in the course of replication: helicase, single-strand-binding protein, DNa gyrase, initiator proteins.

back 50

Initiator proteins, helicase, single-strand-binding protein, DNA gyrase

front 51

Primers are synthesized where on the lagging strand?

back 51

At the beginning of every Okazaki fragment

front 52

Which bacterial enzyme removes the primers

back 52

DNA polymerase I

front 53

Initiator proteins

back 53

Bind to origin and separate strands of DNa to initiate replication

front 54

DNa helicase

back 54

Unwinds DNa at replication fork

front 55

Single-strand-binding proteins

back 55

attach to single-stranded DNa and prevent secondary structures from forming

front 56

DNa gyrase

back 56

Moves ahead of the replication fork, making and resealing breaks in the double-stranded helical DNa to release the torque that builds up as a result of unwinding at the replication fork

front 57

DNa primase

back 57

Synthesizes a short rNa primer to provide a 3'-Oh group for the attachment of DNa nucleotides

front 58

DNa polymerase III

back 58

elongates a new nucleotide strand from the 3'-Oh group provided by the primer

front 59

DNa polymerase I

back 59

removes rNa primers and replaces them with DNa

front 60

DNa ligase

back 60

joins Okazaki fragments by sealing breaks in the sugar– phosphate backbone of newly synthesized DNa

front 61

What are some differences from prokaryotes in the genome structure of eukaryotic cells that affect how replication takes place?

back 61

The size of eukaryotic genomes, the linear structure of eukaryotic chromosomes, and the association of DNA with histone proteins

front 62

What would be the result if an organism’s telomerase were mutated and nonfunctional?

back 62

Chromosomes would shorten each generation.

front 63

Why is recombination important?

back 63

Recombination is important for generating genetic variation.

front 64

Stem Loop structures are a critical part of

back 64

Rho- dependent termination and rho-independent termination

front 65

The main purpose of recombination is

back 65

To align truly homologous chromosomes

front 66

The function of gyrase is

back 66

To relieve torsional stress from unwinding DNA

front 67

The molar ratio of Histone H1 to H2 in the core is

back 67

1:2

front 68

If a splice site were mutated so that splicing did not take place, what would the effect be on the mrNA?

back 68

It would be longer than normal.

front 69

What evidence indicated that eukaryotic genes are not colinear with their proteins?

back 69

When DNA was hybridized to the mRNA transcribed from it, regions of DNA that did not correspond to RNA looped out

front 70

What is the function of the sigma factor?

back 70

The sigma factor controls the binding of RNA polymerase to the promoter.

front 71

What is the difference between the template strand and the nontemplate strand?

back 71

The template strand is the DNA strand that is transcribed into an RNA molecule, whereas the nontemplate strand is not transcribed.

front 72

Which class of rNA is correctly paired with its function?

back 72

transfer rNA (trNA): attaches to an amino acid

front 73

In a polyribosome, the polypeptides associated with which ribosomes will be the longest?

back 73

those at the 3' end of mrNA

front 74

In elongation, the creation of peptide bonds between amino acids is catalyzed by

back 74

rRNA

front 75

In elongation, the creation of peptide bonds between amino acids is catalyzed by

back 75

The Shine–Dalgarno sequence

front 76

Amino acids bind to which part of the trNA

back 76

3' end

front 77

through wobble, a single __________ can pair with more than one _____________.

back 77

anticodon, codon

front 78

A codon is

back 78

three nucleotides that encode an amino acid

front 79

What determines the secondary and tertiary structures of a protein?

back 79

The amino acid sequence(primarystructure) of the protein

front 80

Which one of the following statements is not true for all E. coli DNA polymerases?

back 80

They possess 5'to 3' exonuclease activity.

front 81

Which statement is true regarding negative supercoiled DNA ?

back 81

Negative supercoiled DNA is under-rotated and allows for easier strand separation during replication and transcription

front 82

What would Avery, Macleod, and McCarty have concluded if their results had been that only RNAse treatment of the heat-killed bacteria prevented transformation of genetic virulence?

back 82

RNA is the genetic material

front 83

Okazaki fragments are found associated with

back 83

Lagging strand

front 84

After the first round of replication, Meselson and Stahl saw only one DNA band of density intermediate to DNA containing only 15N or 14N. After this observation, which hypothesis for DNA replication could be eliminated ?

back 84

conservative

front 85

DNA synthesis in eukaryotes is?

back 85

semi-conservative

front 86

According to Chargaff’s rules, if a genome is 30% adenine, then what percentage of the genome should be guanine?

back 86

20%

front 87

How many base pairs per turn on DNA

back 87

10

front 88

What is the function of DNA Polymerase I in E.coli

back 88

Removes and replaces primers

front 89

What is the function of DNA Polymerase III in E.coli

back 89

Elongates DNA

front 90

What is the function of DNA Polymerase II, IV, and V in E.coli

back 90

DNA repair for all, ttranslesion DNA synthesis for V, II Halts synthesis, IV just repairs

front 91

Binds to origin and separates strand of DNA to initiate replication

back 91

Initiator protein

front 92

Unwinds DNA at replication fork

back 92

DNA Helicase

front 93

Attach to single-stranded DNA and prevent reannealing

back 93

Single-strand-binding proteins

front 94

Moves ahead of the replication fork, making and resealing breaks in the double-helical DNA to release torque that builds up as a result of unwinding at the replication fork

back 94

DNA gyrase

front 95

Synthesizes short RNA primers to provide a 3-OH group for attachment of DNA nucleotides.

back 95

DNA primase

front 96

Elongates a new nucleotide strand from the 3-0H group provided by the primer

back 96

DNA polymerase III

front 97

Removes RNA primers and replaces them with DNA

back 97

DNA Polymerase I

front 98

Joins Okazaki fragments by sealing nicks in the sugarophoshate backbone of newly synthesized DNA

back 98

DNA ligase

front 99

After crossing over results in recombination and non-recombination products take place after what?

back 99

DNA sytnthesis

front 100

How does Holiday junction predict crossover of recombinant DNA?

back 100

Ir depends if it was cut horizontal or vertical

front 101

Correct order of the Central Dogma

back 101

information from DNA--- transcription to RNA translated to protein

front 102

Transcribed in both eukaryotic and prokaryotic cells.

back 102

Messenger RNA (mRNA), Ribosomal RNA (rRNA), Transfer RNA (trna)

front 103

Transcribed in only eukaryotic celss

back 103

Pre-messenger RNA (pre-mRNA), Small nuclear RNA (snRNA), Small nucleolar RNA (snoRNA), MicoRNA(miRNA), small interfering RNA (siRNA)

front 104

Structural and functional components of the ribosome

back 104

rRNA

front 105

Carries genetic code for proteins

back 105

mRNA

front 106

helps incorporate amino acids into polypeptide chain

back 106

tRNA

front 107

Processing of pre-mRNA

back 107

snRNA

front 108

processing and assembly of rRNA

back 108

snoRNA

front 109

Inhibits the translation of mRNA

back 109

MiRNA

front 110

Triggers the degradation of other RNA Molecules

back 110

siRNA

front 111

In translated RNA what sequence is only found in prokaryotes

back 111

Shine-Delgarno sequence

front 112

In translated RNA what sequence is only found in eurakryotes

back 112

Kozak Sequence

front 113

in excision of transcripts 5 prime is

back 113

the branch point

front 114

in excision of transcripts 3 prime is

back 114

The consensus sequence

front 115

u1 in a spliceosome

back 115

is snrna

front 116

Purines are what base nucleotides

back 116

A and G

front 117

pyridines are what base nucleotides

back 117

T and C

front 118

In mRNA 5' to 3' goes from what chemical group to the other.

back 118

Amino to Carboxyl

front 119

Basic order of charged TRNA

back 119

First Aminoacyl second Peptidyl and E exit.