What are transcription factors?
Regulatory proteins that bind to specific DNA sequences.
Which of the following elements is not a DNA binding domain?
Alternative splicing is an example of gene regulation that occurs after the synthesis of mRNA.
Which of the following best describes the mechanism by which steroid hormones control gene expression?
Steroid hormones that enter the cell activate receptors. These hormone-receptor complexes then bind HREs and influence gene expression.
Which of the following best describes the role of chaperone proteins in the regulation of gene expression by steroid hormones?
Chaperone proteins maintain functionality of the receptor.
The reason some cells respond to the presence of a steroid hormone while others do not is that _______.
the receptors necessary for regulation differ among cells of various types.
Which statements about the modification of chromatin structure in eukaryotes are true?
Some forms of chromatin modification can be passed on to future generations of cells.
Acetylation of histone tails is a reversible process.
DNA is not transcribed when chromatin is packaged tightly in a condensed form.
Acetylation of histone tails in chromatin allows access to DNA for transcription.
Methylation of histone tails in chromatin can promote condensation of the chromatin.
Which statements about the regulation of transcription initiation in these genes are true?
Control elements C, D, and E are distal control elements for the imaginin gene.
The fantasin gene will be transcribed at a high level when activators specific for control elements A, B, and C are present in the cell.
Both the fantasin gene and the imaginin gene will be transcribed at high levels when activators specific for control elements A, B, C, D, and E are present in the cell.
Which of the following choices represent mRNA molecules that could be produced from the primary RNA transcript by alternative RNA splicing? (In each choice, the yellow part on the left represents the 5′ cap, and the yellow part on the right represents the poly-A tail.)
The diagram below shows a length of DNA containing a bacterial gene.
a. same sequence as RNA transcript (except for having U instead of T).
b. produces stem loop structure in RNA transcript.
c. recognized by sigma subunit of RNA polymerase.
d. complementary to RNA transcript.
e. leads to an unstable RNA-DNA duplex.
Promoter recognition:
RNA polymerase is a holoenzyme composed of a five-subunit core enzyme and a sigma (?) subunit. Different types of ? subunits aid in the recognition of different forms of bacterial promoters. The bacterial promoter is located immediately upstream of the starting point of transcription (identified as the +1 nucleotide of the gene). The promoter includes two short sequences, the -10 and -35 consensus sequences, which are recognized by the ? subunit.
Chain initiation:
The RNA polymerase holoenzyme first binds loosely to the promoter sequence and then binds tightly to it to form the closed promoter complex. An open promoter complex is formed once approximately 18 bp of DNA around the -10 consensus sequence are unwound. The holoenzyme then initiates RNA synthesis at the +1 nucleotide of the template strand.
Chain elongation:
The RNA-coding region is the portion of the gene that is transcribed into RNA. RNA polymerase synthesizes RNA in the 5? > 3? direction as it moves along the template strand of DNA. The nucleotide sequence of the RNA transcript is complementary to that of the template strand and the same as that of the coding (nontemplate) strand, except that the transcript contains U instead of T.
Chain termination:
Most bacterial genes have a pair of inverted repeats and a polyadenine sequence located downstream of the RNA-coding region. Transcription of the inverted repeats produces an RNA transcript that folds into a stem-loop structure. Transcription of the polyadenine sequence produces a poly-U sequence in the RNA transcript, which facilitates release of the transcript from the DNA.
Drag the labels into the flowchart to show the order of events as they are thought to occur during eukaryotic transcription involving RNA polymerase II (RNA pol II).
1. TFIID binds to the TATA box.
2. TFIIA, TFIIB, TFIIF, and RNA pol II bind.
3. TFIIE and TFIIH bind.
4. Synthesis of the pre-mRNA begins at the +1 nucleotide.
5. A 5′ cap is added to the pre-mRNA.
6. Spliceosome complexes carry out intron splicing.
7. A poly-A tail is added to the pre-mRNA.
Exon 1. Exon 2. Intron 2. Exon 3.
mutation in a splicing signal sequence in intron 2.
No mRNA produced.
mutation in the gene’s promoter sequence.
Exon 1. Intron 1. Exon 2. Intron 2. Exon 3.
mutation in a splicing sequences in introns 1 ; 2.
Exon 1. Exon 2. Exon 3.
no mutation in any splicing signal sequence.
Why is gene regulation more complex in a multicellular eukaryote than in a prokaryote?
Eukaryotic cells contain greater amounts of DNA and this DNA is associated with various proteins.
The diversity of cells in a multicellular eukaryote suggests that certain genes are active in some cells but not in others.
Eukaryotes have many chromosomes and those chromosomes are enclosed in a nuclear envelope.
Why is the study of this phenomenon more difficult in eukaryotes?
There are different levels at which regulation can occur in eukaryotes.
In eukaryotes, it is difficult to isolate certain molecular species.
In eukaryotes, it is difficult to interpret the behavior of isolated molecules in an artificial environment.
Gene expression in eukaryotes is tissue specific .
How might one determine the influence of CAAT boxes on the transcription rate of a given gene?
Delete the CAAT box sequence and measure the transcription rate.
Make mutations in the CAAT box sequence and measure the transcription rate.
Introduce extra CAAT box sequences and measure the transcription rate.
A deletion within the GAL4 gene that removes the region encoding amino acids 1 to 100.
A deletion within the GAL4 gene that removes amino acids 1-100 would remove the DNA-binding section and not allow transcriptional activation.
A deletion of the entire GAL3 gene.
Without the product of the GAL3 gene, there would be no disruption of the Gal4p/Gal80p complex and therefore no transcription of the GAL1 gene.
A mutation within the GAL80 gene that blocks the ability of Gal80 protein to interact with Gal3p.
If the GAL80 gene product can’t interact with Gal3p, there can be no interaction with the Gal4p/Gal80p complex and therefore no GAL1 transcription.
A deletion of one of the four UASG elements upstream from the GAL1 gene.
A deletion of one of the four UASG elements would reduce transcription of the GAL1 gene.
A point mutation in the GAL1 core promoter that alters the sequence of the TATA box.
Generally, mutations in the TATA box of a promoter reduce transcription of the relevant gene.
Required for basal-level transcription.
TATA, CAAT, GC boxes.
Always upstream of the gene within 100 bases of the transcription initiation site.
Not required for basal level transcription.
Responsible for tissue- and time- specific gene expression.
Position can be upstream, downstream, or with the gene.
May influence the expression of more than one gene.
Chromatin remodeling
Refers to changes in DNA/chromosome structure that can influence overall gene output.
Promoters 2
are sequences in DNA to which transcription factors and RNA polymerase bind to initiate transcription.
Enhancers 2
are cis-acting DNA sequences to which transcription factors bind to regulate transcription.
They can act over distances of thousands of base pairs and can be upstream, downstream, or internal to the gene they affect.
Transcription factors
bind DNA and regulate transcription.
The addition of a 5′ cap and poly-A tail, and removal of introns are all steps in_________.
RNA processing and transport.
RNA processing contributes to _______, which regulates translation.
mRNA stability

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