Genetics Exam 3

0.0(0)
studied byStudied by 0 people
0.0(0)
full-widthCall Kai
learnLearn
examPractice Test
spaced repetitionSpaced Repetition
heart puzzleMatch
flashcardsFlashcards
GameKnowt Play
Card Sorting

1/74

flashcard set

Earn XP

Description and Tags

I wish the professor taught this class better idk.

Study Analytics
Name
Mastery
Learn
Test
Matching
Spaced

No study sessions yet.

75 Terms

1
New cards

A silent mutation is most likely the result of which of the following (hint: look at the genetic code):

 

Transition

 

Transversion

Transition

2
New cards

Which of the following would likely result in a frameshift mutation?

 

Depurination

 

Trinucleotide repeat expansion

 

Base modifier

 

Intercalating agent

Intercalating agent

3
New cards

Which of the following recognizes a mismatch?

 

Uvr Complex

 

MutL

 

MutH

 

MutS

MutS

4
New cards

The Uvr complex can distinguish between the parent strand and the daughter strand when repairing DNA mutations.

 

True

 

False

False

5
New cards

UV radiation is a physical mutagen that most commonly causes:

 

Thymine dimers

 

Base modifications

 

Frameshift

 

Depurinations

Thymine dimers

6
New cards

Of the following mutations, which would lead to a 50% chance of a point mutation when the mutation is identified by DNA repair mechanisms before DNA replication?

 

Depurination

 

Cytosine deamination

 

5-methylcytosine deamination

 

Trinucleotide repeat expansion

5-methylcytosine deamination

7
New cards

Which of the following would repair a depurination?

 

Mismatch repair system

 

Nucleotide excision repair

 

Photolyase

 

Non-homologous end joining

Nucleotide excision repair

8
New cards

The disorder Xeroderma Pigmentosum results in the absence of:

 

Photolyase

 

Uvr proteins

 

MutL

 

BRCA2

Uvr proteins

9
New cards

The sequence below is the TEMPLATE strand of DNA.  Following transcription, which of the following would be the sequence of the RNA?

5’ – A T C G G T A T C G T A – 3’

 

5’ – TAGCCATAGCAT – 3’

 

5’ – UAGCCAUAGCAU – 3’

 

5’ – UACGAUACCGAU – 3’

 

5’ – TACGATACCGAT – 3’

 

5’ – AUCGGTAUCGUA – 3’

5’ – UACGAUACCGAU – 3’

10
New cards

The role of the sigma factor is to:

 

Bind to the TATA box and recruit RNA polymerase

 

Facilitate hairpin formation during termination

 

Recruit RNA polymerase for transcription

 

Recognize the intron/exon boundaries during splicing

Recruit RNA polymerase for transcription

11
New cards

A gene that is transcribed at low amounts, but is incapable of increasing the rate of transcription would most likely be the result of what?

 

A mutation in the promoter

 

A mutation in the regulatory sequence

 

A mutation in the start site

 

An absence of general transcription factors

A mutation in the regulatory sequence

12
New cards

What do both the rho-dependent and rho-independent mechanisms of termination have in common? 

 

Terminate transcription immediately after the stop codon

 

A sequence rich with A-U base pairs

 

Both require a helicase to separate the DNA-RNA complex

 

The formation of a hairpin structure

The formation of a hairpin structure

13
New cards

The polyA tail is responsible for:

 

Alternative splicing

 

Transport of the mRNA out of the nucleus

 

Initiation of translation

 

Preventing degradation

 

All of the above are functions of the polyA tail

Preventing degradation

14
New cards

snRNPs are involved in which of the following?

 

Splicing

 

Transcription

 

Translation

 

Coupling

Splicing

15
New cards

Alternative splicing allows an organism to ___________. 

 

Carry fewer genes

 

Carry more genes

 

Produce fewer gene products

Carry fewer genes

16
New cards

Which of the following is translated into protein?

 

Introns

 

Promoter

 

Exons

 

Terminator

 

All of the above are translated into protein

Exons

17
New cards

Which of the following is NOT used during TRANSLATION?

 

mRNA

 

tRNA

 

Ribosome

 

Terminator

Terminator

18
New cards

The peptidyl transferase is the enzymatic function of the:

 

Splicesome

 

Large ribosome subunit

 

Small ribosome subunit

 

Polymerase

Large ribosome subunit

19
New cards

The role of the small ribosomal subunit is to:

 

Bind the release factor

 

Recognize the start codon

 

Recognize the mRNA

 

Attach ATP to the acceptor stem

Recongize the mRNA

20
New cards

Which of the following will bind to the release factor?

 

UAA

 

AUA

 

GUU

 

AUG

UAA

21
New cards

Which of the following contains a CODON?

 

tRNA

 

mRNA

 

Intron

 

rRNA

 

Both answers A and B contain codons

mRNA

22
New cards

Molecule X interacts with a DNA binding protein, causing the protein to bind to DNA and down-regulate gene expression.  Molecule X is a:

 

Activator

 

Repressor

 

Inducer

 

Co-repressor

 

Inhibitor

Co-repressor

23
New cards

The lac operon contains how many promoters?

 

0

 

1

 

3

 

4

1

24
New cards

Which of the following sites would be a target of a TRANSLATIONAL REPRESSOR?

 

Shine-Delgarno sequence

 

Stop codon

 

Enhancer

 

Silencer

Shine-Delgarno sequence

25
New cards

The inducer molecule for the lac activator protein is:

 

TPP

 

Tryptophan

 

cAMP

 

Glucose

cAMP

26
New cards

What would be the result if the trp codons in region 1 of the trp leader sequence were missing?

 

Transcription would always occur

 

Transcription would always attenuate

Transcription would always attenuate

27
New cards

Which of the following circumstances would favor binding of CAP protein and the lac repressor to DNA?

 

High glucose, low lactose

 

Low glucose, low lactose

 

High glucose, high lactose

 

Low glucose, high lactose

Low glucose, low lactose

28
New cards

A protein that covers and hides the intron/exon boundary site is:

 

Translational regulator

 

Splicing repressor

 

Splicing enhancer

 

RISC

Splicing repressor

29
New cards

Which of the following protects turtles from apoxia (absence of oxygen)?

 

ß-Galactosidase

 

INS

 

APOLD1

 

trpL

APOLD1

30
New cards

Which of the following would NOT occur if an insulator were removed from DNA?

 

Heterochromatin would spread to a region that would otherwise be euchromatin

 

An enhancer or silencer would interact with an inappropriate gene

 

Regulatory transcription factors would bind to the wrong enhancer or silence

 

All of the above are possible if an insulator were removed from DNA

Regulatory transcription factors would bind to the wrong enhancer or silence

31
New cards

Regulatory transcription factors would NOT bind to which of the following?

 

TATA Box

 

Enhancer sequence

 

Glucocorticoid Response Element

 

Silencer

 

Operator

TATA Box

32
New cards

Which transcription factor is the most likely to bind to the following sequence (the underlined nucleotides are binding sites for the alpha-helix of the transcription factor(s))

 

ATCGAATTAGCGTATGATTATACGA

 

Zinc finger

 

Leucine zipper homodimer

 

Helix-loop-helix heterodimer

Leucine zipper homodimer

33
New cards

DNA undergoes a mutation in a region adjacent to a housekeeping gene.  The mutation causes the entire housekeeping gene to undergo chromatin remodeling to a heterochromatin state.  The mutation event likely altered a ______________.

 

Regulatory element

 

Silencer

 

Enhancer

 

Insulator

Insulator

34
New cards

Downregulation of the glucocorticoid receptor leads to:

 

A greater response to the glucocorticoid hormone

 

Individuals who are less tolerant to stress

 

Imprinting of the glucocorticoid hormone

 

Fewer glucocorticoid response elements

Individuals who are less tolerant to stress

35
New cards

Which of the following would have the highest levels of gene expression?

 

A promoter that is methylated

 

A mutation that has rendered an enhancer inactive

 

HDAC activity on an enhancer

 

HAT activity on a promoter

HAT activity on a promoter

36
New cards

The inactive X chromosome would contain which histone variant?

 

H2AX

 

MacroH2A

 

H3.3

 

cenH3

MacroH2A

37
New cards

Which DNA Polymerase is responsible for most DNA replication?

DNA Pol III

38
New cards

What DNA Polymerase is responsible for removing and filling in primers?

DNA Pol I

39
New cards

A transition mutation results in:

A. A purine being replaced by a pyrimidine

B. A pyrimidine being replaced by a purine

C. A purine replaced by another purine

D. An insertion or deletion of one base

C

40
New cards

Which of the following describes a transversion?

A. G → A

B. A → G

C. T → G

D. C → T

C

41
New cards

A germ-line mutation differs from a somatic mutation because it:

A. Occurs in body tissues only

B. Cannot be passed to offspring

C. Occurs in gametes and is heritable

D. Always produces cancer

C

42
New cards

Huntington disease is caused by:

A. A deletion in exon 3 of the huntingtin gene

B. Expansion of a CAG trinucleotide repeat

C. A point mutation in the promoter

D. Loss of a start codon

B

43
New cards

Which of the following is a mutagen that inserts itself between base pairs?

A. Base analog

B. Deaminating agent

C. Intercalating agent

D. Alkylating agent

C

44
New cards

Photolyase repairs DNA damage caused by:

A. Oxidative stress

B. Ultraviolet light–induced thymine dimers

C. Depurination

D. Intercalating agents

B

45
New cards

Which of the following is a spontaneous mutation?

A. Deamination of cytosine to uracil

B. Incorporation of 5-bromouracil

C. Alkylation by EMS

D. Intercalation of ethidium bromide

A

46
New cards

A reactive oxygen species (ROS) can cause:

A. Base oxidation leading to mispairing

B. Insertion mutations

C. Frameshift suppression

D. Ribosomal stalling

A

47
New cards

A  trinucleotide repeat expansion leads to:

A. Gene deletion

B. Frameshift mutation

C. Hairpin formation and repeat instability

D. Chromosomal inversion

C

48
New cards

Photoreactivation directly reverses thymine dimers using:

A. ATP hydrolysis

B. Photolyase and visible light

C. Nucleotide excision repair

D. Homologous recombination

B

49
New cards

The mismatch repair system differs from nucleotide excision repair because it:

A. Removes bulky adducts

B. Recognizes distortions in the double helix

C. Detects incorrect base pairing after replication

D. Uses photolyase

C

50
New cards

Non-homologous end joining (NHEJ) differs from homologous recombination repair

(HRR) because:

A. NHEJ uses a homologous chromosome as a template

B. NHEJ is error-prone and can lead to indels

C. HRR joins blunt ends without loss

D. HRR acts only in G1 phase

B

51
New cards

The Pribnow box is located:

A. In prokaryotes

B. In eukaryotes

C. Upstream of the terminator

D. In the Shine-Dalgarno sequence

A

52
New cards

The transcription bubble forms when:

A. Ribosomes begin translation

B. RNA polymerase unwinds a small DNA region

C. DNA ligase seals Okazaki fragments

D. The rho protein binds to RNA

B

53
New cards

Which statement about the template strand is correct?

A. It has the same sequence as the RNA (except U for T)

B. It is read 5′ → 3′ by RNA polymerase

C. It is complementary to the mRNA

D. It always contains the start codon

C

54
New cards

Rho-independent termination involves:

A. Rho binding and ATP hydrolysis

B. Formation of a hairpin

C. The sigma factor

D. rho gene mutations

B

55
New cards

The TATA box is most similar in function to which bacterial element?

A. Operator

B. Pribnow box

C. Shine-Dalgarno sequence

D. Terminator

B

56
New cards

Enhancers function by:

A. Binding RNA polymerase II directly

B. Increasing transcription from distant sites via activator proteins

C. Repressing translation

D. Splicing introns

B

57
New cards

Which of the following is not a modification of pre-mRNA?

A. 5′ capping

B. Poly-A tail addition

C. RNA editing

D. DNA methylation

D

58
New cards

Introns are:

A. Coding sequences retained in mRNA

B. Noncoding regions removed during splicing

C. Sequences promoting transcription

D. Regions bound by repressors

59
New cards

Alternative splicing increases protein diversity by:

A. Changing promoter sequences

B. Producing multiple mRNAs from one gene

C. Modifying tRNA molecules

D. Altering ribosomal RNA

B

60
New cards

The sequence hypothesis proposes that:

A. DNA directly converts to protein

B. RNA sequence determines amino-acid sequence

C. Protein dictates mRNA structure

D. Ribosomes control gene expression

B

61
New cards

The start codon in most mRNAs is:

A. UAA

B. UAG

C. AUG

D. UGA

C

62
New cards

The anticodon of a tRNA pairs with:

A. The ribosome E site

B. The codon in mRNA

C. The Shine-Dalgarno sequence

D. The 3′ poly-A tail

B

63
New cards

The E, P, and A sites of the ribosome function respectively in:

A. Entry of tRNA, peptide bond formation, exit of empty tRNA

B. Exit, peptide-bond formation, aminoacyl-tRNA entry

C. Peptide bond formation, translocation, exit

D. Release, attachment, elongation

B

64
New cards

Ochratoxin A inhibits translation by:

A. Blocking the ribosomal A site

B. Degrading mRNA

C. Inhibiting RNA polymerase II

D. Causing tRNA mischarging

A

65
New cards

The lacZ gene encodes:

A. Lactose permease

B. β-galactosidase

C. CAP protein

D. Repressor

B

66
New cards

When lactose is present and glucose is absent:

A. The lac repressor binds the operator

B. cAMP–CAP activates transcription

C. The operon is repressed

D. No allolactose is formed

B

67
New cards

An effector molecule such as allolactose functions by:

A. Binding to RNA polymerase II

B. Changing the shape of a regulatory protein

C. Repressing CAP binding

D. Blocking the Shine-Dalgarno sequence

B

68
New cards

Housekeeping genes differ from regulated genes because they:

A. Are only expressed in response to signals

B. Encode proteins needed under all conditions

C. Are repressed by glucose

D. Lack promoters

B

69
New cards

The trp repressor binds to the operator when:

A. Tryptophan levels are high

B. Tryptophan levels are low

C. Attenuation occurs

D. Ribosomes stall at leader peptides

A

70
New cards

Attenuation in the trp operon depends on:

A. Repressor binding only

B. Formation of alternative hairpins in the leader mRNA

C. The CAP–cAMP complex

D. RNA polymerase pausing at the promoter

B

71
New cards

Riboswitches regulate gene expression by:

A. Changing mRNA structure upon ligand binding

B. Binding to DNA directly

C. Degrading RNA polymerase

D. Recruiting ribosomes

A

72
New cards

Combinatorial control refers to:

A. Genes controlled by a single activator

B. Integration of multiple transcription factors to regulate one gene

C. Translation control through ribosomes

D. Chromosomal deletions

B

73
New cards

General transcription factors (GTFs):

A. Bind enhancer regions only

B. Are required for RNA polymerase II to initiate transcription

C. Are unique to repressors

D. Mediate RNA splicing

B

74
New cards

The glucocorticoid receptor activates transcription when:

A. Bound by cortisol and dimerized

B. Unbound and cytoplasmic

C. Phosphorylated without ligand

D. Degraded by proteasomes

A

75
New cards

Upregulation of a gene results in:

A. Decreased mRNA transcription

B. Increased mRNA transcription

C. Loss of enhancer activity

D. Histone deacetylation

B