Exam 3 End of Chapter Questions

How 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.

The gene encoding the E. coli enzyme β-galactosidase begins with 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.

The sequence of the consensus −10 region is TATAAT. If two 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 more efficiently transcribed, and why?

Assuming the two genes use similar transcription initiation modes, tesA will be more efficiently transcribed. The tesB −10 sequence deviates more from the consensus sequence, and its higher G≡C content will be more difficult to melt.

Activator proteins A and B are required to express gene X. Analysis of the DNA upstream from the gene X promoter identified an 18 bp sequence with near twofold symmetry that is required for activation. Purification of the gene A and gene B products showed 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 palindrome, the activator probably functions as a dimer. 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 follow 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 with 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 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 most heavily utilized by the cell.

Perhaps 3,000 or more transcription factors participate in the activation of human genes. However, this is far fewer than the number of genes in the human genome (∼20,000 to 25,000). Explain how specific gene activation is achieved when genes outnumber gene activators by 10:1.

First, most genes are regulated by multiple transcription 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 make a total of 10 different dimeric species that can recognize 10 different DNA sequences.

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 low but constant level. If an essential housekeeping gene were experimentally moved from euchromatin to a region of heterochromatin, what would be the likely effect on the cell?

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

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 chromosomal region. The acetyl groups are removed by histone deacetylases (HDACs).

Briefly describe the relationship between chromatin structure and transcription 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.

Within the histone structure, do protein modifications occur primarily near the N-terminus or near the Cterminus? 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.

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 helix-turn-helix motif. This mutant repressor does not repress transcription from gene X. When the mutant repressor is expressed at high levels on a plasmid that is introduced into the bacterial cell, transcription of X is increased even though the wild-type repressor (capable of binding its normal DNA binding site and shutting down transcription) is present in the same cell. Explain.

Most repressors with helix-turn-helix motifs function as oligomers, many as homodimers. When 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.

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

No. The two 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.

Steroid hormone receptors are located in the cytoplasm, where they can interact with incoming hormones. However, 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.

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.

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

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.

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.

In bacteria, there are many examples of two (or even more) genes being transcribed from one promoter—for example, the promoter is followed by gene A and then gene B, with both genes transcribed into a single mRNA. In some cases, the first gene in the linear sequence is transcribed at much higher levels than the second gene (i.e., many but not all of the mRNAs do not include gene B). What kind of DNA sequences might be present between the first and second genes to account for the lower level of transcription of gene B?

There may be a palindromic sequence that allows formation of a hairpin structure during transcription, creating aρo-independent terminator that is not perfectly efficient, or an imperfect rut sequence to guide the loading of the ρ protein. In both cases, the sequences would have to be imperfect so that some transcripts could be elongated through gene B.

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 import signal in the receptor protein

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