EXAM 2 Ext. Flashcards
Match the organism listed in the left column with the amount of DNA in its haploid genome from the right column.
(a) E. coli (bacterium) (1) 4600 kb
(b) S. cerevisiae (yeast) (2) 3000 Mb
(c) H. sapiens (human) (3) 12 Mb
(a) 1 (b) 3 (c) 2
Which of the following is not a characteristic of the epigenome?
(a) Differences in chromatin structure
(b) Covalent modifications of the DNA sequence
(c) Clustering of genes coding for steps in a given pathway
(d) Differential expression of genes based on cell type
(c) Clustering of genes coding for steps in a given pathway
What is a primary consequence of the folding of chromatin on gene regulation?
The tight folding of chromatin renders many DNA sites inaccessible to transcription machinery. Chromatin remodeling is required to make these sites available for gene expression.
Which of the following statements about histones are correct?
(a) They are highly basic because they contain many positively charged amino acid side chains
(b) They are extensively modified after their translation
(c) In combination with DNA, they are the primary constituents of chromatin
(d) They account for approximately one-fifth of the mass of a chromosome
a, b, c
Which of the following statements about nucleosomes are correct?
(a) They constitute the repeating units of a chromatin fiber
(b) Each contains a core of eight histones
(c) They contain DNA that is surrounded by a coating of histones
(d) They occur in chromatin in association with approximately 200 base pairs of DNA, on average
a, b, d
Describe the structure of the nucleosome.
A nucleosome consists of ~146 bp of DNA wrapped around an octamer of histone proteins (two copies each of H2A, H2B, H3, and H4) in 1.65 left-handed superhelical turns. The histone tails protrude outward from the core structure.
Does the formation of nucleosomes account for the observed packing ratio of human metaphase chromosomes? Explain.
No. Nucleosome formation provides only about 7-fold compaction. Higher-order chromatin structures including the 30-nm fiber, looping, and scaffolding are required to achieve the 10,000-fold compaction seen in metaphase chromosomes.
Specific combinatorial control of transcription
(a) is enabled by specific interactions between transcription factors and specific DNA sequences
(b) allows a given regulatory protein to have different effects depending upon the neighboring proteins with which it is associated
(c) is effected by transcription factors some of which do not themselves interact with DNA
(d) depends upon the assembly of multicomponent nucleoprotein complexes
(e) results from the ability of one protein to recruit another to a complex
a, b, c, d, e
Match each transcription factor or domain in the left column with its characteristics in the column on the right.
(a) Basic-leucine-zipper
(b) Homeodomain
(c) C₂-His₂ zinc finger
(1) α-helix inserts into major groove
(2) Dimers stabilized by hydrophobic interactions
(3) Helix-turn-helix motif
(4) Forms heterodimers
(5) Forms homodimers
(6) Occurs in sets
(7) Single, long α-helix
(8) Structure is a tandem set of a small α-helix and β-sheet domain
(a) 1,2,4,5,7 (b) 1,3,4 (c) 1,6,8
For each of the following, indicate whether it is a feature of an activator (A), an enhancer (E), or neither (N).
(a) They recruit other proteins to promote transcription
(b) They are often redundant; part of them can be deleted with little loss of function
(c) They are modular
(d) They can act synergistically
(e) They perturb local chromatin structure
(a) A,E (b) A (c) A,E (d) A,E (e) A
Which of the following statements about eukaryotic genes that are actively being transcribed are correct?
(a) They are cell-type specific
(b) They are highly condensed
(c) They are more susceptible to hydrolysis by DNAase 1 than are silent genes
(d) They are developmentally regulated
(e) They can be detected by chromatin immunoprecipitation
a, c, d, e
The DNA methylation involved in gene regulation
(a) requires S-adenosylmethionine as a source of methyl groups
(b) occurs at 5′-CpG-3′ sequences
(c) uses N⁵,N¹⁰-methylenetetrahydrofolate to form the 5-methyl group of thymine
(d) converts cytosine in DNA to 5-methylcytosine
(e) is less frequent at sites adjacent to actively transcribed genes
a, b, d, e
Which of the following statements about steroid hormones are correct?
(a) They bind to a seven-helix transmembrane receptor to initiate a series of phosphorylations that culminate in gene transcription
(b) Upon binding to their specific receptor proteins, they enable the receptors to bind specific DNA sequences
(c) They activate specific protein kinases and protein phosphatases
(d) They are recognized by members of the nuclear receptor superfamily of proteins
(e) They require plasma membrane transporters to go from the blood to the cytosol
b, d
Nuclear hormone receptors
(a) are dimers
(b) bind to response elements
(c) undergo conformational changes when they bind their ligand
(d) contain zinc-finger domains
(e) interact with coactivators in the presence of their ligands
a, b, c, d, e
The tails of histones
(a) when acetylated have lower affinity for DNA
(b) are involved in recruiting chromatin-remodeling engines that move nucleosomes
(c) when acetylated, serve as substrates for histone deacetylases
(d) have their positive charges reduced by acetylation
(e) when acetylated interact with the bromodomain of many eukaryotic transcription factors when that domain is brominated
a, b, c, d
What are the biochemical similarities and differences between an iron-response element (IRE) and an estrogen-response element (ERE)?
The IRE is an RNA sequence in the 5’UTR of ferritin mRNA that binds IRE-BP to block translation. The ERE is a DNA sequence that binds the estrogen receptor-estrogen complex to activate transcription.
Describe the structure and mechanism of the IRP and compare with that of the mitochondrial aconitase.
IRP is 30% identical to mitochondrial aconitase. When cellular iron is low, IRP loses its iron-sulfur cluster and binds IREs. When iron is abundant, it functions as an aconitase like its mitochondrial counterpart.
Which of the following is involved in the mechanism of microRNA posttranslational regulation?
(a) miRNAs bind to complementary mRNA sequences and prevent them from being translated
(b) Double-stranded RNAs are cleaved into 21-nucleotide fragments, the single-stranded components of which form RISC complexes that cleave complementary mRNAs
(c) The Argonaute family of proteins are RNases that cleave sequences complementary to the microRNA
(d) The single-stranded miRNAs are generated from larger genetically encoded precursors
a, c, d
Would you expect the interaction between protamines and DNA to be enhanced or diminished in solutions that have highly ionic strength, that is, high salt concentrations? Protamines, which are found in very high concentrations in sperm where they participate in condensation of the DNA, are low molecular weight compounds rich in groups with high pKa values, that is, they are basic compounds. Explain the basis for your answer. Does the action of histone acetyltransferases (HATs) use a similar principle? Explain.
The interaction would be diminished in high salt because salt disrupts ionic interactions between basic protamines and DNA phosphates. HATs similarly reduce positive charges on histones by acetylation, weakening DNA binding.
The differences in the amino acid sequences of histone H4 between calf thymus and pea seedlings are as follows:
AA position 60: Pea (Ile) vs Calf (Val)
AA position 77: Pea (Arg) vs Calf (Lys)
(a) What minimum number of mutational events accounts for these differences?
(b) Comment on the nature of these changes.
(a) Two mutations: A→G change at first position for Ile→Val, and G→A change at second position for Arg→Lys.
(b) These are conservative changes preserving charge (Arg/Lys) and hydrophobicity (Ile/Val), suggesting minimal functional impact.
Chromatin that is transcriptionally active (euchromatin) is disperse in structure, whereas chromatin that is transcriptionally inactive (heterochromatin) is compact. When the nuclei of chicken globin-producing cells were treated briefly with pancreatic DNase, the adult globin genes were selectively destroyed, but the genes for embryonic globins and ovalbumin remained intact. In contrast, when the nuclei of oviduct cells were treated with DNase, the ovalbumin genes were destroyed. Explain these results.
Active genes in dispersed chromatin are DNase-sensitive, while inactive genes in compact chromatin are protected. The results show tissue-specific gene activation (globin in blood cells, ovalbumin in oviduct).
The chromatin of globin-producing cells can be treated with micrococcal nuclease under conditions that cleave DNA almost exclusively in the linker regions between intact nucleosomes. When the resulting nucleosomes are isolated and their DNA is examined, it is found to contain DNA sequences for the synthesis of globin. Are these results consistent or inconsistent with the explanation for the results in problem 3? Explain.
Consistent. The globin genes are packaged in nucleosomes but remain accessible to transcription machinery in the linker regions, explaining both nuclease sensitivity and protection patterns.
Would you expect a zinc deficiency in eukaryotes to be associated with any sort of developmental abnormality? Explain.
Yes. Zinc is essential for zinc finger domains in many transcription factors (e.g., nuclear receptors) and enzymes. Deficiency would disrupt gene regulation and metabolism during development.
You have isolated the M gene involved in muscle cell differentiation. When grown in 5-azacytidine, the M gene is expressed, whereas cells grown normally do not express it. Explain these observations.
5-azacytidine inhibits DNA methyltransferases, causing demethylation of CG islands in the M gene promoter. This epigenetic change activates transcription in muscle cells.
Proteins that interact with specific DNA sequences usually remain bound under low ionic strength conditions during gel electrophoresis. Describe how you could use this property to isolate transcription factors from cell extracts.
Use gel shift assays: incubate labeled promoter DNA with cell extracts, then run on low-ionic-strength gels. Protein-DNA complexes migrate slower than free DNA. Specificity can be tested with competitor DNA.
Would you be surprised if your analysis of the gene regulatory machinery of an eukaryotic cell indicated that DNA sequences far removed (1 or more kbp) from the transcription start site were involved? Explain.
No. Eukaryotic genes often have enhancers that act at a distance by DNA looping to bring transcription factors into proximity with the promoter.
What would you predict would happen to a genetic male (XY) fetus whose testosterone receptor had a mutation in its C-terminal domain that rendered that domain resistant to binding the androgen?
The fetus would develop female characteristics (testicular feminization) because testosterone signaling required for male development would be disrupted despite normal XY genotype.
A group of molecular biologists showed that three different DNA sequences (X, Y, Z) could activate transcription. They purified three dimeric protein activators, but only two polypeptide monomers. How is this possible?
The two monomers can form both homodimers (binding X and Z) and a heterodimer (binding Y), allowing three distinct DNA-binding specificities from two proteins.
What are the advantages of positive regulation in the control of gene expression in eukaryotes compared to prokaryotes?
Positive regulation is more efficient for large genomes (one activator can regulate many genes) and prevents inappropriate activation of the many genes that must remain silent in specialized cells.
A human cell has approximately 1300 times more DNA than an E. coli cell. What biological process may render the eukaryotic genome functionally equivalent to the prokaryotic one with respect to the location of targeted nucleic acid sequences by specific-binding proteins?
Chromatin condensation limits DNA accessibility, effectively reducing the search space for DNA-binding proteins to active regions, making target finding comparable to prokaryotes.
Describe the experiment carried out by Shinya Yamanaka that demonstrated that four genes in embryonic stem cells could induce pluripotency (iPS). Comment on the potential therapeutic use of iPS cells.
Yamanaka introduced Oct4, Sox2, Klf4, and c-Myc into fibroblasts, converting them to iPS cells resembling embryonic stem cells. iPS cells offer potential for patient-specific regenerative medicine without embryo use.
