Chapter 16

RECAP 16.1

  1. The cell could reduce the transcription of the Xase gene, hydrolyze Xase mRNA, prevent Xase RNA translation, hydrolyze Xase protein, or inhibit Xase protein.

  2. In the presence of lactose, the promoter is exposed and RNA polymerase binds to begin transcription. In the absence of lactose, the promoter is occupied by the repressor and there is no transcription.

  3. Sigma factors direct RNA polymerase to bind to recognition sequences at certain promoters. If several promoters of genes with related functions (e.g., heat shock response) have the same recognition sequence, transcription of those genes will occur at the same time and rate.

  4. To keep a constant, low-level expression of repressor protein, the regulatory gene would have to have an inefficient promoter, and synthesis of the repressor would be constitutive.

RECAP 16.2

  1. General transcription factors bind to the promoter and to RNA polymerase in a complex to direct RNA polymerase to the promoter to initiate transcription and locally denature DNA so that the template strand is available for base pairing during RNA synthesis. Specific transcription factors bind to specific promoters or promoters with recognition sequences. Other transcription factors bind to enhancer sequences that can be far from the actual promoter and induce DNA to bend to attract the rest of the initiation complex for transcription.

  2. Proteins such as transcription factors fit into the DNA double helix by structural motifs, and their amino acids may form hydrogen bonds with bases on the interior of the double helix. The sequences of amino acids (proteins) and of bases (DNA) are specific so that only certain proteins bind to certain DNA sequences.

  3. A-17

    In response to a signal from the environment or within the cell, a specific transcription factor is made and/or translocated into the nucleus where it binds to a recognition sequence in DNA at the promoters of target genes. This binding attracts the initiation complex for transcription, so multiple genes are activated.

RECAP 16.3

  1. In the bacteriophage: Phage DNA is injected into the host cell. Early phase genes are transcribed at phage promoters by using DNA sequences similar to those of the host cell (positive regulation). This leads to early protein that binds to host promoters to shut them down (negative regulation). Other early proteins lead host RNA polymerase to transcribe middle and late phage genes (positive regulation).

    In HIV: HIV RNA is injected into the host cell. HIV reverse transcriptase is activated to make cDNA and integrase to splice cDNA into the host chromosome (positive regulation). Later, host RNA polymerase binds to HIV promoters to make HIV mRNAs (positive regulation). HIV tat protein acts as an anti-terminator for the transcription of HIV genes integrated into the host genome.

  2. HIV infection, cDNA formation and integration would occur as normal. Viral genes would be expressed and a large precursor protein would be made. But it would not be cut into separate viral proteins. HIVE particles would not be packaged or released from the cell.

RECAP 16.4

  1. Histone proteins are positively charged and bind to DNA, generally blocking transcription. Acetylation of histones neutralizes the positive charge and thus the histones do not bind to DNA as tightly, which opens up the chromatin structure for transcription. By contrast, histone deacetylation removes acetyl groups, restoring positive charges on histones so transcription is repressed.

  2. See the opening story and Investigating Life: Gene Expression and Behavior (p. 351). The behavioral environment appears to alter DNA epigenetically by increasing DNA methylation at promoters in genes in brain tissues, thus changing the rate of gene transcription.

  3. X chromosome inactivation is shown in Figure 16.16. The Xist gene on the X chromosome is transcribed to make a short RNA that binds to the rest of the X chromosome, inhibiting transcription of the other genes. Chromosome proteins bind to the inactive X chromosome, causing heterochromatin to form and inhibiting gene expression. X chromosome inactivation is believed to occur in order to balance the expression of X-linked genes between males (XY) and females (XX) since the Y chromosome does not usually contain X-linked genes.

  4. You could sequence the relevant genes of colorectal cancer cells and look for mutations that lead to aberrant function, then isolate the proteins involved and determine that their functions are indeed abnormal. To show epigenetic silencing, you might sequence the promoters of the genes and look for epigenetic changes (e.g., cytosine methylation, which would be increased if there were transcriptional silencing). Then you could examine the tumor cells to see if the active proteins were there but in small amounts.

RECAP 16.5

  1. miRNAs and siRNAs can bind by base pairing to target mRNAs and prevent their translation because tRNA cannot bind; or the inhibitor RNAs can bind to pre-mRNA in the nucleus and lead to its hydrolysis by RNase; or the inhibitor RNAs can bind to DNA at the transcription site and block RNA polymerase from working for transcription.

  2. mRNA can fold back on itself by hydrogen bonding of complementary bases, forming looped structures. These structures can bind to proteins that then inhibit translation at the ribosome.

  3. The proteasome binds to proteins that are targeted with ubiquitin for breakdown. Within the proteasome are proteases that hydrolyze targeted proteins.

  4. On average, a human gene can form at least four different mRNAs by alternative splicing. Each of these mRNAs is translated to a unique protein. So the number of different proteins is much greater than the number of genes.

  5. miRNAs targeted to activated oncogenes will block the translation of target mRNAs that would make proteins that otherwise stimulate cell division.

WORK WITH THE DATA, P. 351

    1. mRNA was measured in the heads because gene expression in the brain determines whether a honey bee will be a worker or a queen.

    2. The level of inhibition of DNMT mRNA was about 60 percent after 48 h. This is good but not perfect inhibition. So some DNMT mRNA probably remained.

  1. The reduction of DNA methylation was about 20 percent. So some 5-methylcytosine remained.

  2. In the controls, about 23 percent of the larvae developed into queens. But in the larvae that had reduced DNA methylation, about 72 percent developed into queens. This is a remarkable shift given that DNMT mRNA and DNA methylation were reduced but not as much. There must be a threshold for methylation, above which gene expression is affected.

FIGURE QUESTIONS

Figure 16.1 Yes. There can be multiple binding sites for regulatory proteins at the promoter region.

Figure 16.7 An advantage of the nucleus is compartmentation—the separation of transcription/processing and translation. This allows for finer regulation of gene expression. Also, the nucleus is protected from nucleases in the cytoplasm, which might hydrolyze mRNA and reduce its lifetime as it is made.

Figure 16.13 Lysogeny is a term used for bacteriophage to describe the integration of the viral genome into the host cell for a period of time. Although the term is not used for eukaryotic viruses, HIV does integrate its genome as cDNA and so could be called lysogenic.

Figure 16.14 5-methylcytosine mutations (deamination) are not repaired. So locations with 5-methylcytosine will tend to accumulate mutations and their potential to regulate transcription will vary.

Figure 16.17 7

APPLY WHAT YOU’VE LEARNED

  1. An inhibitor of DNA methylation restores tumor suppressor activity in the cancer cells. The same effect is observed with an inhibitor of histone deacetylation. These results suggest that both DNA methylation and histone deacetylation operate to shut down expression of genes that suppress tumor growth.

  2. The data show that inhibitors of both histone deacetylation and DNA methylation act together with greater effect than either inhibitor alone. This suggests that some gene expression is inhibited by DNA methylation, some by histone deacetylation, and some by both processes. In this last group of genes, both inhibitors must be required to release the genes from being silenced.)

  3. Gene 1 can be reactivated by either inhibitor, but Gene 2 can be reactivated only by a combination of the two inhibitors. For Gene 1, the extent of DNA methylation and histone deacetylation may not be great. In this case, either DNA demethylation or histone acetylation may tip the balance just enough to alter the nucleosome to allow access of RNA polymerase to bind to the promoter and express the gene. In the case of Gene 2, RNA polymerase may be prevented from binding to the promoter for two reasons: the promoter site could be heavily methylated and histones could also be deacetylated, thus preventing loosening of the nucleosome to expose the promoter region. Therefore, for Gene 2 both DNA methylation and histone deacetylation must be inhibited for expression to be restored.