#17: gene regulation in eukaryotes

0.0(0)
studied byStudied by 0 people
learnLearn
examPractice Test
spaced repetitionSpaced Repetition
heart puzzleMatch
flashcardsFlashcards
Card Sorting

1/192

encourage image

There's no tags or description

Looks like no tags are added yet.

Study Analytics
Name
Mastery
Learn
Test
Matching
Spaced

No study sessions yet.

193 Terms

1
New cards

Where is the TATA box typically located in eukaryotic genes?
A) 100-150 bp upstream of the transcription start site
B) 25-30 bp upstream of the transcription start site
C) At the transcription start site
D) Within the first exon of the gene

25-30 bp upstream of the transcription start site

2
New cards

Which of the following is TRUE about promoters in eukaryotic genes compared to bacterial genes?
A) Eukaryotic genes often produce polycistronic mRNAs
B) Each eukaryotic gene has its own promoter
C) Bacterial genes do not have promoters
D) Bacterial genes always have TATA boxes

Each eukaryotic gene has its own promoter

3
New cards

The __________ box is a key promoter element found about 25-30 base pairs upstream of the transcription start site in many eukaryotic genes and is essential for transcription initiation.

TATA

4
New cards

Unlike bacteria, eukaryotic cells do not typically produce __________ mRNAs; instead, each gene has its own promoter and is transcribed separately.

polycistronic

5
New cards

True or False: The TATA box is found in both bacterial and human genes and serves the same function in transcription initiation.

false

6
New cards

True or False: In eukaryotic cells, one promoter can control transcription of multiple genes like in bacterial operons.

false

7
New cards

What is the function of the 5’ cap added to eukaryotic mRNA?
A) Protects mRNA from degradation and is important for ribosome binding
B) Signals the start of transcription
C) Marks the mRNA for export from the nucleus only
D) Acts as a polyadenylation signal

Protects mRNA from degradation and is important for ribosome binding

8
New cards

Where does the ribonuclease cleave the pre-mRNA during the addition of the poly-A tail?
A) At the 5’ end of the mRNA
B) Immediately after the start codon
C) 11-30 nucleotides downstream of the AAUAAA sequence
D) At the exon-intron junction

11-30 nucleotides downstream of the AAUAAA sequence

9
New cards

Which enzyme adds the poly-A tail to the 3’ end of the mRNA?
A) RNA polymerase II
B) Poly-A polymerase
C) Ribonuclease
D) Spliceosome

Poly-A polymerase

10
New cards

Eukaryotic genes contain _________, which are removed during RNA processing, and ________, which are expressed regions coding for proteins.

introns; exons

11
New cards

The sequence ________ acts as a polyadenylation signal in the pre-mRNA that guides cleavage and polyadenylation.

AAUAAA

12
New cards

True or False: The poly-A tail is encoded directly in the DNA sequence of the gene.

false

13
New cards

Which cellular structure is primarily responsible for separating transcription and translation in eukaryotic cells?
A) Ribosome
B) Nuclear membrane
C) Golgi apparatus
D) Endoplasmic reticulum

Nuclear membrane

14
New cards

Which of the following is NOT a common point at which eukaryotic cells regulate gene expression?
A) Transcription initiation
B) mRNA transcript processing
C) mRNA translation
D) DNA replication

DNA replication

15
New cards

Why is mRNA export from the nucleus a critical regulatory step in eukaryotes?
A) It allows mRNA to be spliced properly
B) It controls when mRNA becomes available for translation in the cytoplasm
C) It increases mRNA stability
D) It enhances transcription initiation

It controls when mRNA becomes available for translation in the cytoplasm

16
New cards

What is the primary role of the promoter in gene transcription?
A) Site where ribosomes bind
B) Site where RNA polymerase binds
C) Site of mRNA splicing
D) Site where tRNA binds

Site where ribosomes bind

17
New cards

Which protein specifically recognizes the TATA box during transcription initiation in eukaryotes?
A) Sigma factor
B) RNA polymerase II
C) TATA-binding protein (TBP)
D) Enhancer-binding protein

TATA-binding protein (TBP)

18
New cards

Basal transcription factors are necessary because:
A) They enhance mRNA stability
B) They help recruit RNA polymerase to the promoter
C) They process introns from the mRNA
D) They initiate translation

They help recruit RNA polymerase to the promoter

19
New cards

Basal gene expression refers to a __________, __________ level of transcription that occurs even in the absence of additional regulatory signals.

low; consistent

20
New cards

True or False: The promoter region includes binding sites for both RNA polymerase and basal transcription factors.

true

21
New cards

Enhancer sequences are best described as:
A) Coding DNA sequences close to the promoter
B) DNA sequences that must be immediately upstream of the gene
C) Non-coding DNA sequences that regulate transcription and can be located far from the gene
D) RNA sequences that bind to repressors

Non-coding DNA sequences that regulate transcription and can be located far from the gene

22
New cards

Which of the following is TRUE about enhancer sequences?
A) They only activate transcription, never repress it
B) They function only when located upstream of the transcription start site
C) Their function depends on their DNA-protein interactions, not their position relative to the gene
D) They must be within 50 base pairs of the promoter to work

Their function depends on their DNA-protein interactions, not their position relative to the gene

23
New cards

Enhancers can act as __________ or __________ of transcription by interacting with proteins that influence RNA polymerase activity.

activators, repressors

24
New cards

Enhancer sequences are found in __________ regions of DNA and can be located either __________ or __________ the transcription start site.

non-coding, 5’, 3’

25
New cards

True or False: Enhancer sequences lose their function if they are moved from their original position near the gene.

false

26
New cards

True or False: Enhancers are part of euchromatic gene regulation because they influence the accessibility and activity of genes in loosely packed chromatin.

true

27
New cards

Transcription factors regulate gene expression by:
A) Binding only to promoter sequences
B) Binding only to enhancer sequences
C) Binding to both promoters and enhancers and recruiting other proteins to influence transcription
D) Binding randomly to DNA without sequence specificity

Binding to both promoters and enhancers and recruiting other proteins to influence transcription

28
New cards

Which feature allows transcription factors to regulate only specific genes?
A) Their ability to bind RNA polymerase directly
B) Their recognition of specific DNA sequences in enhancers or promoters
C) Their high abundance in the cell
D) Their capacity to bind to any DNA sequence

Their recognition of specific DNA sequences in enhancers or promoters

29
New cards

Which of the following is NOT a type of transcription factor?
A) Basal factors
B) Activators
C) Repressors
D) Ribosomes

Ribosomes

30
New cards

Transcription factors bind to __________ sequences to either induce or silence gene expression by recruiting other proteins that influence __________.

enhancer, transcription

31
New cards

The human genome encodes approximately __________ transcription factors that regulate gene expression.

2000

32
New cards

True or False: Transcription factors have no role in silencing gene expression.

false

33
New cards

True or False: Basal transcription factors are a class of transcription factors involved in general transcription initiation.

true

34
New cards

How do transcriptional activators enhance gene transcription?
A) By degrading RNA polymerase
B) By recruiting RNA polymerase to the promoter
C) By blocking enhancer sequences
D) By preventing DNA looping

By recruiting RNA polymerase to the promoter

35
New cards

What mechanism allows enhancer sequences located far from a gene’s promoter to influence transcription?
A) DNA methylation
B) RNA splicing
C) DNA looping
D) Protein degradation

DNA looping

36
New cards

Transcription factors bound to enhancer sequences can physically interact with the promoter through __________, which brings distant regions of DNA close together.

DNA looping

37
New cards

DNA looping allows enhancer-bound transcription factors to interact with the transcription machinery at the gene’s __________ site.

promoter

38
New cards

True or False: Enhancer sequences must be immediately adjacent to the promoter to influence transcription.

false

39
New cards

How do transcriptional activators enhance transcription initiation?
A) By degrading basal transcription factors
B) By directly interacting with basal transcription factors or via mediator proteins
C) By blocking RNA polymerase binding
D) By binding to the coding region of genes

By directly interacting with basal transcription factors or via mediator proteins

40
New cards

What is the role of mediator proteins in transcription activation?
A) To prevent RNA polymerase from binding to DNA
B) To facilitate interactions between activators and the basal transcription machinery
C) To degrade mRNA after transcription
D) To inhibit enhancer sequences

To facilitate interactions between activators and the basal transcription machinery

41
New cards

Transcriptional activators help stabilize the __________ at the promoter to increase the rate of transcription initiation.

transcription machinery

42
New cards

Cooperation between different transcription factors increases the recruitment of __________ to the promoter.

RNA polymerase

43
New cards

True or False: Transcriptional activators only interact directly with RNA polymerase to enhance transcription.

false

44
New cards

True or False: Mediator proteins act as bridges between transcriptional activators and basal transcription factors.

true

45
New cards

What is the purpose of the chemical crosslinking step in ChIP-Seq?
A) To break DNA into small fragments
B) To create covalent bonds between DNA and proteins, preserving their interaction
C) To sequence DNA fragments
D) To remove unbound proteins from DNA

To create covalent bonds between DNA and proteins, preserving their interaction

46
New cards

In ChIP-Seq, the antibody used is specific for:
A) DNA sequences near promoters
B) Any chromatin-associated protein
C) The transcription factor of interest
D) RNA polymerase only

The transcription factor of interest

47
New cards

Which of the following best describes why DNA is fragmented in ChIP-Seq?
A) To enable the transcription factors to bind DNA
B) To create smaller, manageable pieces for immunoprecipitation and sequencing
C) To remove unbound DNA
D) To denature proteins

To create smaller, manageable pieces for immunoprecipitation and sequencin

48
New cards

n ChIP-Seq, after the DNA-protein complexes are isolated using antibodies, the crosslinks are _________ to release the DNA for sequencing.

reversed

49
New cards

The sequencing step in ChIP-Seq helps identify the exact _________ across the genome where the transcription factor binds.

DNA binding sites (or sequences)

50
New cards

True or False: ChIP-Seq can only identify binding sites that are close to gene promoters, not distant enhancers.

false

51
New cards

True or False: The antibody in ChIP-Seq pulls down all DNA fragments from the genome, regardless of whether they are bound to the transcription factor.

false

52
New cards

Which of the following best explains the role of antibodies in ChIP-Seq?
A) Antibodies digest the DNA not bound to transcription factors
B) Antibodies specifically bind to transcription factors to pull down DNA-protein complexes
C) Antibodies sequence the DNA fragments
D) Antibodies fragment the chromatin into smaller pieces

Antibodies specifically bind to transcription factors to pull down DNA-protein complexes

53
New cards

Crosslinking in ChIP-Seq is typically performed using:
A) Formaldehyde
B) DNase
C) Protease
D) Ethanol

Formaldehyde

54
New cards

The DNA sequences identified by ChIP-Seq represent the _________ of transcription factors across the genome.

binding sites

55
New cards

True or False: ChIP-Seq results can help identify enhancer sequences that regulate gene expression from a distance

true

56
New cards

True or False: In ChIP-Seq, the specificity of the antibody is crucial to avoid pulling down unrelated DNA fragments.

true

57
New cards

What is the main advantage of using ChIP-Seq over whole-genome sequencing when studying transcription factor binding?
A) ChIP-Seq sequences the entire genome faster
B) ChIP-Seq targets only DNA bound by specific proteins, reducing data complexity
C) Whole-genome sequencing identifies binding sites more precisely
D) ChIP-Seq does not require DNA fragmentation

ChIP-Seq targets only DNA bound by specific proteins, reducing data complexity

58
New cards

In ChIP-Seq data analysis, what do the “peaks” represent when mapped to the genome?
A) Regions of high DNA methylation
B) Transcription factor binding sites
C) Areas of DNA mutation
D) RNA polymerase binding sites

Transcription factor binding sites

59
New cards

Why are many short DNA reads generated in ChIP-Seq?
A) Because the DNA is sequenced as a continuous long molecule
B) Due to fragmentation of DNA before sequencing
C) Because only RNA is sequenced
D) To remove unbound DNA

Due to fragmentation of DNA before sequencing

60
New cards

In ChIP-Seq, after sequencing, the short DNA reads are ________ back to the reference genome to locate transcription factor binding sites.

mapped

61
New cards

The regions with the highest number of aligned reads in ChIP-Seq data are called ________, indicating protein-DNA interactions.

peaks

62
New cards

True or False: ChIP-Seq provides data on all DNA in the genome, including regions not bound by proteins.

false

63
New cards

True or False: The Illumina platform is commonly used in ChIP-Seq to generate short DNA reads for high-throughput sequencing.

true

64
New cards

In yeast, the genes GAL1, GAL7, and GAL10 are inducible. What does “inducible” mean in this context?
A) The genes are always transcribed at a constant level
B) The genes are transcribed only when galactose is present
C) The genes are permanently turned off
D) The genes are transcribed only during cell division

The genes are transcribed only when galactose is present

65
New cards

Which of the following is true about GAL1, GAL7, and GAL10 genes in yeast?
A) They are transcribed from a single promoter producing polycistronic mRNA
B) They are tightly linked but have separate promoters
C) They are located on different chromosomes
D) They are constitutively expressed regardless of galactose

They are tightly linked but have separate promoters

66
New cards

Why might yeast tightly regulate the expression of galactose metabolism genes?
A) To conserve energy by producing enzymes only when galactose is available
B) Because galactose is toxic to yeast cells
C) Because the enzymes are needed at all times
D) To prevent glucose metabolism

To conserve energy by producing enzymes only when galactose is available

67
New cards

The genes GAL1, GAL7, and GAL10 in yeast are examples of _______ gene expression because they are only turned on in the presence of ________.

inducible; galactose

68
New cards

Although GAL1, GAL7, and GAL10 are tightly linked, each gene has its own ________ which controls transcription initiation.

promoter

69
New cards

What role does GAL4 play in the regulation of galactose metabolism genes in yeast?
A) It represses transcription by binding to the operator
B) It acts as a transcriptional activator by binding to upstream activating sequences (enhancers)
C) It degrades galactose molecules
D) It transports galactose into the cell

It acts as a transcriptional activator by binding to upstream activating sequences (enhancers)

70
New cards

How does GAL80 regulate GAL4 activity in the absence of galactose?
A) GAL80 activates GAL4 to turn on transcription
B) GAL80 binds GAL4 and inhibits its ability to activate transcription
C) GAL80 degrades GAL4 protein
D) GAL80 binds DNA and blocks RNA polymerase directly

GAL80 binds GAL4 and inhibits its ability to activate transcription

71
New cards

What happens to GAL80 when galactose is present in the cell?
A) GAL80 binds more tightly to GAL4 to enhance activation
B) GAL3 binds GAL80, preventing GAL80 from entering the nucleus and inhibiting GAL4
C) GAL80 is degraded immediately
D) GAL80 binds DNA and represses GAL genes

GAL3 binds GAL80, preventing GAL80 from entering the nucleus and inhibiting GAL4

72
New cards

The transcriptional activator ________ binds to upstream activating sequences (enhancers) near the GAL genes to stimulate transcription.

GAL4

73
New cards

In the absence of galactose, the protein ________ binds to GAL4 and inhibits its ability to activate transcription.

GAL80

74
New cards

When galactose is present, ________ binds to GAL80, preventing GAL80 from entering the nucleus and thus allowing GAL4 to activate transcription.

GAL3

75
New cards

What happens to GAL80 protein in the absence of galactose?
A) GAL80 is degraded rapidly
B) GAL80 enters the nucleus and binds GAL4 to repress transcription
C) GAL80 remains in the cytoplasm and cannot repress transcription
D) GAL80 binds DNA directly to activate transcription

GAL80 enters the nucleus and binds GAL4 to repress transcription

76
New cards

Why does GAL80’s location within the cell influence transcriptional regulation?
A) Because GAL80 can only bind DNA when in the cytoplasm
B) Because transcription occurs inside the nucleus, so GAL80 must be nuclear to repress transcription
C) Because GAL80 can only be degraded in the cytoplasm
D) Because GAL80 activates GAL4 when in the cytoplasm

Because transcription occurs inside the nucleus, so GAL80 must be nuclear to repress transcription

77
New cards

What causes GAL80 to be sequestered in the cytoplasm?
A) Presence of galactose, via interaction with GAL3
B) Absence of galactose
C) Phosphorylation of GAL80 by kinases
D) Degradation of GAL4

Presence of galactose, via interaction with GAL3

78
New cards

True or False: The chemical modification and localization of regulatory proteins like GAL80 can control gene expression

true

79
New cards

true or False: Transcription of GAL genes occurs in the cytoplasm, so GAL80 represses gene expression only when in the cytoplasm.

false (Transcription occurs in the nucleus, so GAL80 must be nuclear to repress it.)

80
New cards

Explain how GAL80 regulates transcription differently in the presence versus absence of galactose.

In the absence of galactose, GAL80 enters the nucleus and binds to GAL4, blocking recruitment of the transcription initiation complex, repressing GAL gene expression. When galactose is present, GAL3 binds GAL80, sequestering it in the cytoplasm, preventing it from repressing GAL4. This allows GAL4 to activate transcription.

81
New cards

What is the purpose of reverse transcription in RNA sequencing?
A) To degrade RNA before sequencing
B) To convert RNA into DNA for more stable sequencing
C) To fragment DNA into smaller pieces
D) To add adaptors to DNA fragments

To convert RNA into DNA for more stable sequencing

82
New cards

Adaptors attached to DNA fragments during RNA-Seq serve all EXCEPT:
A) Helping primers bind during PCR
B) Acting as unique identifiers for samples
C) Preventing DNA from entering the sequencing machine
D) Allowing hybridization to the sequencing flow cell

Preventing DNA from entering the sequencing machine

83
New cards

True or False: RNA sequencing only tells us whether a gene is turned on or off, but not how much it is expressed.

false

84
New cards

True or False: Fragmentation of cDNA is an important step to create small pieces suitable for sequencing.

true

85
New cards

Why is RNA sequencing important?

  • tells us how many copies of each gene’s mRNA are present in a cell or tissue  

  • This shows us which genes are active and how active they are  

86
New cards

Purpose of DNA adaptors?

  • For hybridization purposes 

  • As primers for PCR 

  • As primers for sequencing reaction  

  • As identifiers in subsequent bioinformatic analyses (data analysis to know which fragment came from which sample) 

87
New cards

What is the role of the flowcell in sequencing by synthesis?
A) To amplify DNA fragments by PCR
B) To hold DNA fragments in place for sequencing
C) To chemically label nucleotides
D) To fragment DNA into smaller pieces

To hold DNA fragments in place for sequencing

88
New cards

Adaptors on DNA fragments are designed to:
A) Help the DNA fragments bind to the flowcell
B) Break the DNA into pieces
C) Convert RNA to DNA
D) Stop PCR amplification

Help the DNA fragments bind to the flowcell

89
New cards

True or False: Sequencing by synthesis adds nucleotides randomly to the DNA fragment.

false (Nucleotides are added one at a time, and the sequence is determined by detecting each added base.)

90
New cards

True or False: PCR amplification happens after the DNA fragments are attached to the flowcell.

false (PCR amplification happens before attaching fragments to the flowcell.)

91
New cards

What is the primary reason we use normalization like RPM in RNA-Seq data analysis?
A) To remove sequencing errors
B) To make sure reads are from exons only
C) To account for differences in total sequencing depth
D) To identify mutations in RNA

To account for differences in total sequencing depth

92
New cards

Which type of gene region do RNA-Seq reads most commonly align to?
A) Introns
B) Promoters
C) Exons and UTRs
D) Telomeres

Exons and UTRs

93
New cards

True or False: If one RNA-seq sample has twice the total number of reads as another, it will always show twice the expression for every gene.

false (Raw counts can be misleading if not normalized; expression depends on RPM or other normalized measures.)

94
New cards

True or False: RPM allows fair comparisons of gene expression across samples with different total read counts.

true

95
New cards

In RNA-seq, if a sample has 100 reads for a gene and the total number of reads in the library is 5 million, the RPM is: ___

20 RMP (100/5)

96
New cards

Sample A has 250 reads for Gene B and 10 million total reads. Sample B has 500 reads for Gene B and 25 million total reads.
What is the RPM of Gene B in each sample, and which sample has higher normalized expression?

  • Sample A: 250/10=25 RPM

  • Sample B: 500/25​=20 RPM

  • Sample A has higher normalized expression.

97
New cards

You have two RNA-seq samples. Sample 1 has 300 reads for gene X out of 15 million total reads. Sample 2 has 600 reads for gene X out of 30 million total reads. What can you conclude about gene X after RPM normalization?

A. It is more active in Sample 1
B. It is more active in Sample 2
C. It is equally active in both samples
D. There is insufficient data to determine

It is equally active in both samples

98
New cards

RPM normalization divides raw read counts for a gene by:

A. The gene's length in kilobases
B. The total number of reads in the sample (in millions)
C. The number of exons in the gene
D. The number of samples in the experiment

The total number of reads in the sample (in millions)

99
New cards

Which of the following is not a limitation of RPM normalization?

A. It doesn’t account for gene length
B. It fails to adjust for library size differences
C. It doesn't handle compositional biases between samples
D. It may not distinguish between biological and technical variation

It fails to adjust for library size differences

100
New cards

In RNA-seq, a higher number of reads mapping to a gene generally indicates higher __________ of that gene.

expression