molecular exam 3

ch 10 10. Haw does epigenetic inheritance differ from Mendelian inheritance?

"Epigenetic inheritance" refers to chromatin modifications (particularly histone modifications) that are retained in the chromatin after cell division and affect gene transcription. Such modifications are not encoded in the DNA and thus are not subject to Mendelian inheritance.

transcript for RNA polymerase 1

35s rRNA precursor

ch 15 3. The gene encoding the E. coli enzyme ~-galactosidase begins ,,vith the sequence ATGACCATGATTACG. What is the sequence of the mRNA transcript specified by this part of the gene?

5'-AUGACCAUGAUUACG. The sequence reported for a gene is, by convention, that of the coding strand, and sequences are always written in the 5'~3' direction.

RNA polymerase 3

5s rRNA

The sequence of the messenger RNA molecule (primary transcript) synthesized from a DNA molecule with a coding strand having the sequence 5′ ATGCTACCGTTA 3' is

5′ AUGCUACCGUUA 3'

Working in a research lab, you wish to examine the kinetics of the initiation phase of bacterial RNA polymerase, as a function of promoter sequence. You want to prevent the reaction from entering the elongation phase. How many nucleotides can be added to an RNA polymer in the initiation phase?

8-10 nucleotides

ch 15 4. The sequence of the consensus -10 region is TATAAT. If nvo genes, tesA and tesB, have identical promoter sequences except in the -10 region, where the tesA sequence is TAATAT and the tesB sequence is TGTCGA, ·which gene do you expect to be 1nore efficiently transcribed, and why?

At 50 to 90 nucleotides per second, the enzyme would take 34 to 61 seconds to transcribe the gene.

ch 19 2. Activator proteins A and B are required to express gene X. Analysis of the DNA upstream fro1n the gene X promoter identified an 18 bp sequence ·with near nvofold symmetry that is required for activation. Purification of the gene A and gene B products sho'\>ved that both proteins form homodimers, but neither the A nor the B homodimer binds the 18 bp site. What are the possible functions of the A and B activators with respect to the 18 bp site? Propose a test of one of your ideas.

Because the 18 bp site is a near palindro1ne, the activator probably functions as a di1ner. Given that both protein A and protein B are required for activating gene X, they may form a heterodimer that has specificity for the binding site. This could be tested in an electrophoretic mobility shift assay or any other assay that measures DNA binding. Alternatively, the 18 bp site might be bound by a third, unidentified protein, and proteins A and B might bind different DNA sites. To test this, the site could be used in a functional DNA-binding assay to follo'l-v the binding protein during purification, allowing identification of the correct protein. Footprinting could be used as an assay (see Chapter 20), and the DNA sequence could be used to make an affinity chromatography resin to aid purification (see Chapter 7). A final possibility is that protein A and/or protein B interact ,.vith a different protein to bind the 18 bp site. This could be tested by purification using a functional assay such as footprinting. The purified active protein would reveal the additional protein.

In which of the following sentences is the word constitutive used correctly, and, the statement is most likely true?

Enzymes of the central metabolic pathways are expressed at a nearly constant, or constitutive, level.

ch 15 13. In most organisms, specialized DNA repair systems are closely linked to transcription. Suggest a biological rationale for this close relationship.

Errors in the genetic information in actively transcribed genes are of greater immediate importance to the cell than errors in silent genes. Transcription-coupled DNA repair focuses repair on those DNA sequences 1nost heavily utilized by the cell.

ch 21 6. Perhaps 3,000 or more transc1iption factors participate in the activation of human genes. Ho,,vever, this is far fe,,ver than the number of genes in the human genome (- 20,000 to 25,000). Explain ho,,v specific gene activation is achieved ,,vhen genes outnumber gene activators by 10:1.

First, most genes are regulated by multiple transc1iption factors (activators), and different (often unique) combinations of factors are used at different genes. Second, families of activators form heterodimers, such that a family of four related proteins can 1nake a total of 10 different dimeric species that can recognize 10 different DNA sequences.

ch 21 9. Housekeeping genes are those that must be expressed at all times, providing a protein or RNA that is essential for general cellular metabolism. They are often expressed at a lo"v but constant level. If an essential housekeeping gene '\>Vere experimentally moved from euchromatin to a region of heterocl1romatin, what would be the likely effect on the cell?

Gene expression is generally silenced in regions of heterochromatin. If the genes were essential, as most housekeeping genes are, moving it into heterochromatin would be lethal for the cell.

ch 21 3. A histone acetyltransferase (HAT) is activated, transferring acetyl groups to histones in a particular region of the genome. What amino acid residues in histones are generally modified by HATs? What is the likely effect of the modifications on the transcription levels of genes in that region? What enzymes reverse the effects of HATs?

HATs generally modify Lys residues in the C-terminal tails of histones. Histone acetylation reduces the affinity of nucleosomes for DNA and can increase the transcriptional activity of a chro1noso1nal region. The acetyl groups are removed by histone deacetylases (HDACs).

ch 19 4. Briefly describe the relationship between chromatin structure and transc1iption in eukaryotes.

Heterochromatin is highly condensed and transcriptionally inert, because the histone proteins make promoters inaccessible. The less-condensed euchromatin has undergone a structural remodeling, allowing some regions to be transcribed. The alterations include covalent modification (such as acetylation) of histones and displacement of nucleosomes, creating exposed regions of DNA that are probably binding sites for regulatory proteins.

ch 19 9. A repressor protein effectively blocks transcription from bacterial gene X. A mutant form of the repressor is engineered with an altered DNA-binding site in the helixturn-helix motif. This mutant repressor does not repress transcription from gene X. vVhen the 1nutant repressor is expressed at high levels on a plasmid that is introduced into the bacterial cell, transcription of Xis increased even though the '\>vild-type repressor (capable of binding its normal DNA binding site and shutting do'\>Vl1 transcription) is present in the same cell. Explain.

Most repressors with helix-turn-helix motifs function as oligo1ners, many as homodimers. vVhen the plasmid-encoded mutant repressor is synthesized at high levels in the cell, most of the "wild-type repressor molecules synthesized are incorporated into less functional heterodimers with a mutant repressor subunit.

ch 15 12. Gene A encodes protein A. A genetic engineer excises a pro1noter sequence for gene A from tl1e DNA and reinserts it at the other end of gene A, oriented so that an RNA polymerase binding at the pro1noter ,,vill transcribe across gene A. Will the 1nRNA synthesized by the RNA polymerase still possess a sequence that produces a functional protein A? Why or why not?

No. The t,,vo strands of a DNA molecule are antiparallel and complementary (not identical). when the promoter is inverted, it will direct RNA synthesis that uses what was originally the coding strand as the template strand. The mRNA sequence, derived from a different DNA strand and synthesized in the opposite direction, would be very different from the mRNA produced by the original gene and might not even contain an open reading frame.

ch 19 13. Steroid hormone receptors are located in the cytoplasm, where they can interact ,,vith incoming hormones. Ho,.vever, steroid hormones act by regulating gene function, and genes are in the nucleus. How is this regulation achieved?

On binding a hormone molecule, the steroid hormone receptor dimerizes and the hormone-receptor complex is transported into the nucleus.

differences between RNA and DNA polymerases

RNA polymerases is directed to specific stretches of the DNA as templates. They dont use the entire genome as templates. RNA polymerases do not require a primer to initiate synthesis. DNA polymerases use both DNA strands.

Which of the following transcription factors is utilized by all three eukaryotic RNA polymerases?

TBP

Which protein would be most likely found interacting with DNA at a class II promoter?

TBP

What is the CORRECT order of general transcription factor assembly at an RNA polymerase II promoter?

TFIID, TFIIB, TFIIF, TFIIH

By what mechanism is the protein bound to the distant "Regulatory binding site" most likely to interact with the RNA polymerase bound to the promoter sequence?

The DNA bends or loops bringing the binding site and promoter site into close proximity.

ch 15 2. The -10 and -35 sequences in bacterial promoters are separated by about two turns of the DNA double helix. How· would transcription be affected if a deletion ·were introduced in the promoter region that moved the -35 sequence to the -29 position?

The deletion would move the - 35 sequence closer to the - 10 sequence by half a helical turn of the DNA, putting the two elements on opposite faces of the DNA duplex. This would dramatically reduce binding of sigma factor to the promoter, thereby decreasing transcription efficiency.

ch 15 1. The sequences of promoters tend to be rich in A and T residues. Suggest why this is so.

The promoter is a site for RNA polymerase loading and initiation of transcription. To accomplish this, the RNA polymerase must form an open complex in which the two DNA strands are separated over a short distance. Due to the effects of AT-rich versus GC-rich DNA on the overall thermal stability of DNA, the strand separation is more readily accomplished in sequences that are AT-rich

Researchers artificially mutated a serine to an alanine in the C-terminal domain (CTD) of RNA polymerase II. In their Xenopus transcription system, they found that this mutated polymerase could no longer transcribe. What is the best hypothesis to explain this observation?

The serine is important for the phosphorylation of CTD necessary for transcription elongation; without it no transcription could occur.

ch 19 3. Most proteins that regulate gene expression bind at specific DNA sequences, recognizing those sequences primarily through protein-DNA interactions within in the major groove of the DNA. Why is the major groove used for sequence recognition more often than the minor groove or the phosphoribose backbone?

There are more hydrogen-bond donor and acceptor groups on the nucleotide bases in the major groove than on those in the minor groove, providing much better discrimination between bases.

ch 19 14. Expression of the CRP transcription activator in E. coli readily leads to transcription of the lactose metabolism genes when lactose is present and glucose is not. If a particular eukaryotic activator is expressed in the appropriate eukaryotic cell, introduced on an engineered virus or plasmid, it often does not trigger transcription of its target gene. Explain.

There are several possible explanations. Many eukaryotic genes are regulated by more than one activator protein, and another activator may be needed. Many eukaryotic genes are encapsulated (and silenced) in heterochromatin, and remodeling of the chromatin may be required in the region where the gene is located to allow activation. The protein may need to be modified, and the modifying enzyme ( e.g., kinase) may not be present in the cell. Finally, perhaps the activator cannot be transported into the nucleus.

ch 21 11a. A scientist is studying the function of a type of nuclear steroid receptor protein in mouse cells. She introduces various mutations into the gene encoding the receptor protein and transfers the genes into mice. If mutations are introduced that (a) eliminate the nuclear i1nport signal in the receptor protein, how will the molecular pathway of hormone-receptor interaction be altered?

With elimination of the nuclear import signal, the receptor could bind the hormone in the cytoplasm, but the co1nplex would not be imported into the nucleus to activate gene expression.

In ChIP-seq, an antibody is used to purify _____________________ in order to identify bound ______________________.

a protein of interest; DNA sequences

general transcription factor PIC component

bind to the core promoter

TATA binding protein PIC component

binding of this transcription factor bends the DNA

repressor PIC component

binds to a silencer

What is the CORRECT order of transcriptional events occurring after RNA polymerase binds to the promoter?

closed complex formation, open complex formation, start of RNA synthesis, promoter clearance

weakest promoter 1

contains -35 with one base pair DOWN mutation, -10 box

strongest promoter 3

contains UP element, -35 box and -10 box

mild promoter 2

contains Up elements, -35 box with a down mutation, -10 box

mediator PIC component

coordinates binding of activators to general transcription factors

The analogous structure of the -10 and -35 boxes in prokaryotes is the ________________ in eukaryotes.

core promoter

In a DNA footprinting experiment, what are the bands on a gel?

fragments of DNA

activator PIC component

has a DNA binding domain and an activating domain

promoter

include DNA sequences bound by proteins that can either recruit or block RNA polymerase

transcript for RNA polymerase 2

mRNA

repressor

molecular that turns off transcription

activator

molecule that turns on transcription

ch 10 4. Within the histone structure, do protein modifications occur primarily near the N-terminus or near the C terminus? what features distinguish the structure ·where modifications occur?

most histone modifications occur near the N-terminus. This part of the histone molecule is a relatively unstructured tail that extends out from the nucleosome core

TFIIH PIC component

phosphorylates the CTD of RNA polymerase II

polycistronic mRNA

product of operon transcription

DNA damage

prokaryotic RNA polymerase pausing is thought to be due to

All of the following are characteristics of RNA polymerase III promoters EXCEPT

within a species, all class III core promoters are the same.