Genetics Final Dr. Konopka

Let's suppose a mutation in the glucocorticoid receptor does not prevent the binding of the glucocorticoid hormone to the protein but prevents the ability of the receptor to activate transcription. Make a list of all the possible defects that may explain why transcription cannot be activated.

1. It could be in the DNA-binding domain, so that the receptor would not recognize a GRE. 2. It could be in the HSP90 domain, so that HSP90 would not be released when the hormone binds. 3. It could be in the dimerization domain, so that the receptor would not dimerize. 4. It could be in the nuclear localization domain, so that the receptor would not travel into the nucleus. 5. It could be in the domain that activates RNA polymerase, so that the receptor would not activate transcription, even though it could bind to GREs.

Discuss the common points of control in eukaryotic gene regulation.

1. Transcription. This includes regulatory transcription factors; DNA methylation (the attachment of methyl groups, which usually inhibits transcription); and changes in the arrangements and composition of histones and nucleosomes. 2. RNA level. This includes RNA processing, regulation of alternative splicing via SR proteins; RNA stability, regulation of RNA half-life and RNA translation via RNA binding proteins; regulation via miRNA and siRNA. 3. Protein level. This includes feedback inhibition, small molecules that modulate enzyme activity; and posttranslational modification-covalent changes to protein structure that affect protein activity.

What is a CpG island? Where would you expect one to be located? How does the methylation of CpG islands affect gene expression?

A CpG island is a stretch of 1,000 to 2,000 base pairs in length that contains a high number of CpG sites. CpG islands are often located near promoters. When the island is methylated, this inhibits transcription. This inhibition may be the result of the inability of the transcriptional activators to recognize the methylated promoter and/or the effects of methyl-CpG-binding proteins, which may promote a closed chromatin conformation.

In response to potentially toxic substances (e.g., high levels of iron), eukaryotic cells often use translational or posttranslational regulatory mechanisms to prevent cell death, rather than using transcriptional regulatory mechanisms. Explain why.

A cell may need to respond rather quickly to a toxic substance in order to avoid cell death. Translational and posttranslational mechanisms are much faster than transcriptional activation, in which it is necessary to up regulate the gene, synthesize the mRNA, and then translate the mRNA to make a functional protein.

What is the meaning of the term consensus sequence? Give an example. Describe the locations of consensus sequences within bacterial promoters. What are their functions?

A consensus sequence is the most common nucleotide sequence that is found within a group of related sequences. An example is the -35 and -10 consensus sequences found in bacterial promoters. At -35, it is TTGACA, but it can differ by one or two nucleotides and still function efficiently as a promoter. In the consensus sequences within bacterial promoters, the -35 site is primarily for recognition by sigma factor. The -10 site, also known as the Pribnow box, is the site where the double-stranded DNA will begin to unwind to allow transcription to occur.

What is the difference between a constitutive gene and a regulated gene?

A constitutive gene is unregulated, which means that its expression level is relatively constant. In contrast, the expression of a regulated gene varies under different conditions. In bacteria, the regulation of genes often occurs at the level of transcription by combinations of regulatory proteins and small effector molecules. In addition, gene expression can be regulated at the level of translation or the function of a protein can be regulated after translation is completed.

When you would you expect gene transcription and DNA replication to take place, during M phase or interphase? Explain your answer

A high level of compaction prevents gene transcription and DNA replication from taking place. Therefore, these events occur during interphase.

What is a histone variant?

A histone variant is a histone with an amino acid sequence that is slightly different from a core histone. Histone variants play specialized roles with regard to chromatin structure and function

What chemical group (phosphate, hydroxyl, amino, carboxyl or methyl) is found at the 3' end of DNA and the 5' end of DNA?

A hydroxyl group is at the 3' end and a phosphate group is at the 5' end

Some mutations have a cis-effect on gene expression, whereas others have a trans-effect. Explain the molecular differences between cis- and trans-mutations.

A mutation that has a cis-effect is within a genetic regulatory sequence, such as an operator site, that affects the binding of a genetic regulatory protein. A cis-effect mutation affects only the adjacent genes that the genetic regulatory sequence controls. A mutation having a trans-effect is usually in a gene that encodes a genetic regulatory protein.

Would a mutation that inactivated the lac repressor and prevented it from binding to the lac operator site result in the constitutive expression of the lac operon under all conditions? Explain. What is the disadvantage to the bacterium of having a constitutive lac operon?

A mutation that prevented the lac repressor from binding to the operator would make the lac operon constitutive only in the absence of glucose. However, this mutation would not be entirely constitutive because transcription would be inhibited in the presence of glucose. The disadvantage of constitutive expression of the lac operon is that the bacterial cell would waste a lot of energy transcribing the genes and translating the mRNA when lactose was not present.

What is a nucleosome-free region? Where are they typically found in a genome? How are nucleosome-free regions thought to be functionally important?

A nucleosome-free region (NFR) is a location in the genome where nucleosomes are missing. They are typically found at the beginning and ends of genes. An NFR at the beginning of a gene is thought to be important so that genes can be activated. The NFR at the end of a gene may be important for its proper termination.

What is the unique feature of ribozyme function? Give two examples described in this chapter.

A ribozyme is catalyst that is composed of RNA. Examples are RNase P and self-splicing group I and II introns. It is thought that the spliceosome may contain catalytic RNAs as well.

In which of the ribosomal sites, the A site, P site, and/or E site, could the following be found? A tRNA without an amino acid attached A tRNA with a polypeptide attached

A tRNA without an amino acid attached The E site and P sites. (Note: A tRNA without an amino acid attached is only briefly found in the P site, just before translocation occurs.) A tRNA with a polypeptide attached P site and A site (Note: A tRNA with a polypeptide chain attached is only briefly found in the A site, just before translocation occurs.) A tRNA with a single amino acid attached Usually the A site, except the initiator tRNA, which can be found in the P site.

Briefly describe three ways that ATP-dependent chromatin- remodeling complexes may change chromatin structure.

ATP-dependent chromatin remodeling complexes may change the positions of nucleosomes, evict histones, and/or replace histones with histone variants.

What is an Okazaki fragment? In which strand of DNA are Okazaki fragments found? Based on the properties of DNA polymerase, why is it necessary to make these fragments?

An Okazaki fragment is a short segment of newly made DNA in the lagging strand. It is necessary to make these short fragments because in the lagging strand, the replication fork is exposing nucleotides in a 5 to 3 direction, but DNA polymerase is sliding along the template strand in a 3 to 5 direction away from the replication fork. Therefore, the newly made lagging strand is synthesized in short pieces that are eventually attached to each other.

What is an insulator?

An insulator is a segment of DNA that functions as a boundary between two adjacent genes. An insulator may act as a barrier to changes in chromatin structure or block the effects of a neighboring enhancer

What is antisense RNA? How does it affect the translation of a complementary mRNA?

Antisense RNA is RNA that is complementary to a functional RNA such as mRNA. The binding of antisense RNA to mRNA inhibits translation.

When connecting two Okazaki fragments, DNA ligase needs to use ATP as a source of energy to catalyze this reaction. Explain why DNA ligase needs another source of energy to connect two nucleotides, but DNA polymerase needs nothing more than the incoming nucleotide and the existing DNA strand. Note: You may want to refer to Figure 13.13 to answer this question.

As mentioned in the answer to part C, the 5 end of the DNA in the middle Okazaki fragment is a monophosphate. It is a monophosphate because it was previously connected to the RNA primer by a phosphoester bond. At the location of the right arrow, there was only one phosphate connecting this deoxyribonucleotide to the last ribonucleotide in the RNA primer. For DNA polymerase to function, the energy to connect two nucleotides comes from the hydrolysis of the incoming triphosphate. In this location shown at the right arrow, however, the nucleotide is already present at the 5 end of the DNA, and it is a monophosphate. DNA ligase needs energy to connect this nucleotide with the left Okazaki fragment. It obtains energy from the hydrolysis of ATP or NAD+ .

What is meant by the term attenuation? Is it an example of gene regulation at the level of transcription or translation? Explain your answer

Attenuation means that transcription is ended before it has reached the end of an operon. Because it causes an end to transcription, it is a form of transcriptional regulation even though the translation of the trpL region plays a key role in the attenuation mechanism.

Which RNA primer would be the first one to be removed by DNA polymerase I, the primer on the left or the primer on the right? For this primer to be removed by DNA polymerase I and for the gap to be filled in, is it necessary for the Okazaki fragment in the middle to have already been synthesized? Explain.

B. The RNA primer in the right Okazaki fragment would be removed first. DNA polymerase would begin by elongating the DNA strand of the middle Okazaki fragment and remove the right RNA primer with its 5 to 3 exonuclease activity. DNA polymerase I would use the 3 end of the DNA of the middle Okazaki fragment as a primer to synthesize DNA in the region where the right RNA primer is removed. If the middle fragment was not present, DNA polymerase could not fill in this DNA (because it needs a primer).

Describe the sequence in bacterial mRNA that promotes recognition by the 30S subunit.

Bacterial mRNAs contain a Shine-Dalgarno sequence, which is necessary for the binding of the mRNA to the small ribosomal subunit. This sequence, UUAGGAGGU, is complementary to a sequence in the 16S rRNA. Due to this complementarity, these sequences will hydrogen bond to each other during the initiation stage of translation.

How does the use of this template result in a telomere sequence that is tandemly repetitive?

Because it uses this sequence many times in row, it produces a tandemly repeated sequence in the telomere at the 3′ ends of linear chromosomes.

Unlike DNA polymerase, reverse transcriptase does not have a proofreading function. How might this affect the proliferation and evolution of the virus?

Because reverse transcriptase does not have a proofreading function, it is more likely for mistakes to occur. This creates many mutant strains of the virus. Some mutations might prevent the virus from proliferating. However, other mutations might prevent the immune system from battling the virus. These kinds of mutations would enhance the proliferation of the virus.

How is the function of reverse transcriptase similar to the function of telomerase?

Both reverse transcriptase and telomerase use an RNA template to make a complementary strand of DNA.

What is DNA methylation? When we say that DNA methylation is heritable, what do we mean? How is it passed from a mother to a daughter cell?

DNA methylation is the attachment of a methyl group to a base within the DNA. In many eukaryotic species, this occurs on cytosine at a CG sequence. After de novo methylation has occurred, it is passed from mother to daughter cell. Because DNA replication is semiconservative, the newly made DNA contains one strand that is methylated and one that is not. DNA methyltransferase recognizes this hemimethylated DNA and methylates the cytosine in the unmethylated DNA strand; this event is called maintenance methylation.

Name the 7 the enzymes required for DNA replication in bacterial cells and for each, briefly describe their job.

DNA polymerase III: catalyzes the dehydration reaction of adding a deoxyribonucleotide onto the growing DNA chain (requires energy), b. Helicase: Unwinds and breaks the hydrogen bonds between DNA strands (requires energy), c. Primase: A type of RNA polymerase that uses a DNA template to make a small RNA molecule, a primer. DNA polymerase adds the first deoxyribonucleotide onto the this primer. The RNA primer is eventually replaced with DNA nucleotides by DNA polymerase I; d. DNA ligase: Ligates or "sews" the Okazaki fragments on the lagging strand together; e. Topoisomerase: Relieves the stress of unwinding the DNA helix by breaking, swiveling and rejoining DNA strands downstream of the replication fork; f. DNA polymerase I: replaces the RNA primer bases with DNA bases; g. Single-stranded DNA binding proteins: Not really an enzyme. Binds to and stabilized single-stranded DNA after strand separation until replication occurs.

A DNA strand has the following sequence: 5'-GATCCCGATCCGCATACATTTACCAGATCACCACC-3' In which direction would DNA polymerase slide along this strand (from left to right or from right to left)?

DNA polymerase would slide from right to left

Discuss the differences in the compaction levels of metaphase chromosomes and interphase chromosomes.

During interphase, much of the chromosomal DNA is in the form of the 30 nm fiber, and some of it is more highly compacted heterochromatin. During metaphase, all of the DNA is highly compacted, as shown in Figure 12.15d.

What sequence elements are found within the core promoter of protein-encoding genes in eukaryotes? Describe their locations and specific functions.

Eukaryotic promoters are somewhat variable with regard to the pattern of sequence elements that may be found. In the case of protein-encoding genes that are transcribed by RNA polymerase II, it is common to have a TATA box, which is about 25 bp upstream from a transcriptional start site. The TATA box is important in the identification of the transcriptional start site and the assembly of RNA polymerase and various transcription factors. The transcriptional start site defines where transcription actually begins.

The glucocorticoid receptor is an example of a transcriptional activator that binds to response elements and activates transcription. (Note: The answer to this question is not directly described in this chapter. You have to rely on your understanding of the functioning of other proteins that are modulated by the binding of effector molecules, such as the lac repressor.) How would the function of the glucocorticoid receptor be shut off?

Eventually, the glucocorticoid hormone will be degraded by the cell. The glucocorticoid receptor binds the hormone with a certain affinity. The binding is a reversible process. Once the concentration of the hormone falls below the affinity of the hormone for the receptor, the receptor will no longer have the glucocorticoid hormone bound to it. When the hormone is released, the glucocorticoid receptor will change its conformation, and it will no longer bind to the DNA.

If a eukaryotic chromosome has 25 origins of replication, how many replication forks does it have at the beginning of DNA replication?

Fifty, because two replication forks emanate from each origin of replication. DNA replication is bidirectional.

An organism has a G + C content of 64% in its DNA. What are the percentages of A, C, T and G?

G = 32%, C = 32%, A = 18%, T = 18%

Write out a sequence of an RNA molecule that would form a stem-loop with 24 nucleotides in the stem and 16 nucleotides in the loop.

Here is an example of an RNA molecule that could form a hairpin that contains 24 nucleotides in the stem and 16 nucleotides in the loop. 5-GAUCCCUAAACGGAUCCCAGGACUCCCACGUUUAGGGAUC-3 The complementary stem regions are underlined.

What are the structural and functional differences between heterochromatin and euchromatin?

Heterochromatin is more tightly packed than euchromatin. This is due to a greater compaction of the radial loop domains. Functionally, euchromatin can be transcribed into RNA, whereas heterochromatin is inactive. Heterochromatin is most abundant in the centromeric regions of the chromosomes and in the telomeric regions.

Consider how histone proteins bind to DNA and then explain why a high salt concentration can remove histones from DNA.

Histones are positively charged and DNA is negatively charged. They bind to each other by these ionic interactions. Salt is composed of positively charged ions and negatively charged ions. For example, when dissolved in water, NaCl becomes individual ions of Na+ and Cl− . When chromatin is exposed to a salt such as NaCl, the positively charged Na+ ions could bind to the DNA and the negatively charged Cl− ions could bind to the histones. This would prevent the histones and DNA from binding to each other.

The initiation phase of eukaryotic transcription via RNA polymerase II is considered an assembly and disassembly process. Which types of biochemical interactions—hydrogen bonding, ionic bonding, covalent bonding, and/or hydrophobic interactions—would you expect to drive the assembly and disassembly process? How would temperature and salt concentration affect assembly and disassembly?

Hydrogen bonding is usually the predominant type of interaction when proteins and DNA follow an assembly and disassembly process. In addition, ionic bonding and hydrophobic interactions could occur. Covalent interactions would not occur. High temperature and high salt concentrations tend to break hydrogen bonds. Therefore, high temperature and high salt would inhibit assembly and stimulate disassembly.

If a DNA-binding protein does not recognize a specific nucleotide sequence, what part of the DNA helix, would the protein recognize? Explain your answer.

If a DNA-binding protein does not recognize a nucleotide sequence, it probably is not binding in the grooves, but instead is binding to the DNA backbone (i.e., sugar-phosphate sequence). DNA binding proteins that recognize a base sequence must bind into a major or minor groove of the DNA, which is where the bases would be accessible to a DNA-binding protein. Most DNA-binding proteins that recognize a base sequence fit into the major groove. By comparison, other DNA binding proteins such as histones, which do not recognize a base sequence, bind to the DNA backbone.

What is the relationship between mRNA stability and mRNA concentration? What factors affect mRNA stability?

If mRNA stability is low, this means that it is degraded more rapidly. Therefore, low stability results in a low mRNA concentration. The length of the polyA tail is one factor that affects stability. A longer tail makes mRNA more stable. Certain mRNAs have sequences that affect their half-lives. For example, AU-rich elements (AREs) are found in many short-lived mRNAs. The AREs are recognized by cellular proteins that cause the mRNAs to be rapidly degraded.

In eukaryotes, what type of modifications occur to pre-mRNAs?

In eukaryotes, pre-mRNA can be capped, tailed, spliced, edited, and then exported out of the nucleus.

If a gene is repressible and under positive control, describe what kind of effector molecule and regulatory protein are involved. Explain how the binding of the effector molecule affects the regulatory protein.

In this case, an inhibitor molecule and an activator protein are involved. The binding of the inhibitor molecule to the activator protein would prevent it from binding to the DNA and thereby inhibit its ability to activate transcription.

Discuss the differences between ρ-dependent and ρ-independent termination.

In ρ-dependent termination, the ρ protein binds to the rut site in the RNA transcript after the ρ site has been transcribed. Eventually, an RNA sequence forms a stem-loop structure that will cause RNA polymerase to pause in the transcription process. As it is pausing, the ρ protein, which functions as a helicase, will catch up to RNA polymerase and break the hydrogen bonds between the DNA and RNA within the open complex. When this occurs, the completed RNA strand is separated from the DNA along with RNA polymerase. In ρ-independent transcription, there is no ρ protein. The RNA forms a stem-loop complex that causes RNA polymerase to pause. However, when it pauses, a uracil-rich region is bound to the DNA template strand in the open complex. Because this is holding on by fewer hydrogen bonds, it is rather unstable. Therefore, it tends to dissociate from the open complex and thereby end transcription

What are the three stages of translation? Discuss the main events that occur during these three stages.

Initiation: The mRNA, initiator tRNA, and initiation factors associate with the small ribosomal subunit; then the large subunit associates. Elongation: The ribosome moves one codon at a time down the mRNA, adding one amino acid at a time to the growing polypeptide. Three sites on the ribosome, the A, P, and E sites, are important in this process. The A site is where the tRNA (except for the initiator tRNA) binds to the ribosome and recognizes the codon in the mRNA. The growing polypeptide is then transferred to the amino acid attached to this tRNA. The ribosome then translocates so that this tRNA is now moved to the P site. The empty tRNA that was in the P site is moved into the E site. This empty tRNA in the E site is then expelled and the next charged tRNA can bind to the A site. Termination: A stop codon is reached and a termination factor binds to the A site. The hydrolysis of GTP initiates a series of events that leads to the disassembly of the ribosomal subunits and the release of the completed polypeptide.

A gene mutation changes an AT base pair to a GC pair. This causes a gene to encode a truncated protein that is nonfunctional. An organism that carries this mutation cannot survive at high temperatures. Make a list of all the genetic terms that could be used to describe this type of mutation.

It is a gene mutation, a point mutation, a base substitution, a transition mutation, a deleterious mutation, a mutant allele, a nonsense mutation, a conditional mutation, and a temperature-sensitive lethal mutation

With regard to cellular efficiency, why do you think the cell has several mechanisms to regulate gene expression at the level of translation?

It takes a lot of cellular energy to translate mRNA into a protein. A cell wastes less energy if it prevents the initiation of translation rather than a later stage such as elongation or termination

If an abnormal repressor protein could still bind allolactose, but the binding of allolactose did not alter the conformation of the repressor protein, how would this affect the expression of the lac operon?

It would be impossible to turn the lac operon on even in the presence of lactose because the repressor protein would remain bound to the operator site.

If you were given a sample of chromosomal DNA and asked to determine if it is bacterial or eukaryotic, what experiment would you perform, and what would be your expected results?

Lots of possibilities. You could digest it with DNase I and see if it gives multiples of 200 bp or so. You could try to purify proteins from the sample and see if eukaryotic proteins or bacterial proteins are present.

Within a protein, certain amino acids are positively charges, some are negatively charged, some are polar but uncharged and some are nonpolar. If you knew that a DNA-binding protein was recognizing the DNA backbone, which amino acids in the protein would be good candidates for interacting with the DNA?

Lysines and arginines, and also polar amino acids.

Mutations in bacterial promoters may increase or decrease gene transcription. Promoter mutations that increase transcription are termed up-promoter mutations, and those that decrease transcription are termed down-promoter mutations. The sequence of the −10 region of the promoter for the lac operon is TATGTT (see Figure 14.5). Would you expect the following mutations to be up-promoter or down-promoter mutations?

Mutations that make a sequence more like the consensus sequence are likely to be up promoter mutations, whereas mutations that cause the promoter to deviate from the consensus sequence are likely to be down promoter mutations. Also, in the -10 region, AT pairs are favored over GC pairs, because the role of this region is to form the open complex. AT pairs are more easily separated, because they form only two hydrogen bonds compared to GC pairs, which form three hydrogen bonds.

A species of bacteria can synthesize the amino acid histidine so it does not require histidine in its growth medium. A key enzyme, which we will call histidine synthetase, is necessary for histidine biosynthesis. When these bacteria are given histidine in their growth medium, they stop synthesizing histidine intracellularly. Based on this observation alone, propose three different regulatory mechanisms to explain why histidine biosynthesis ceases when histidine is in the growth medium. To explore this phenomenon further, you measure the amount of intracellular histidine synthetase protein when cells are grown in the presence and absence of histidine. In both conditions, the amount of this protein is identical. Which mechanism of regulation would be consistent with this observation?

One mechanism is that histidine could act as corepressor that shuts down the transcription of the histidine synthetase gene. A second mechanism would be that histidine could act as an inhibitor via feedback inhibition. A third possibility is that histidine inhibits the ability of the mRNA encoding histidine synthetase to be translated. Perhaps it induces a gene that encodes an antisense RNA. If the amount of histidine synthetase protein was identical in the presence and absence of extracellular histidine, a feedback inhibition mechanism is favored, because this affects only the activity of the histidine synthetase enzyme, not the amount of the enzyme. The other two mechanisms would diminish the amount of this protein

Define epigenetics. Are all epigenetic changes passed from parent to offspring? Explain.

One way to define epigenetics is the study of mechanisms that lead to changes in gene expression that can be passed from cell to cell and are reversible, but do not involve a change in the sequence of DNA. Not all epigenetic changes are passed from parent to offspring. For example, those that occur in somatic cells, such as lung cells, would not be passed to offspring.

A eukaryotic protein-encoding gene contains two introns and three exons: exon 1-intron 1-exon 2-intron 2-exon 3. The 5′ splice site at the boundary between exon 2 and intron 2 has been eliminated by a small deletion in the gene. Describe how the pre-mRNA encoded by this mutant gene would be spliced. Indicate which introns and exons would be found in the mRNA after splicing occurs.

Only the first intron would be spliced out. The mature RNA would contain: exon 1-exon 2-intron 2-exon 3.

Let's suppose that a vertebrate organism carries a mutation that causes some cells that normally differentiate into nerve cells to differentiate into muscle cells. A molecular analysis of this mutation revealed that it was in a gene that encodes a methyltransferase. Explain how an alteration in a methyltransferase could produce this phenotype.

Perhaps the methyltransferase is responsible for methylating and inhibiting a gene that causes a cell to become a muscle cell. The methyltransferase is inactivated by the mutation.

What parts of the nucleotide make up the backbone?

Phosphate and sugar are found in the backbone.

Two circular DNA molecules (A and B) are topoisomers of each other. When viewed under the microscope, molecule A appears more compact than molecule B while the level of gene transcription is much lower for molecule A. Which of the following 3 possibilities could account for these observations? Molecule A has 3 positive supercoils and molecule B has 3 negative supercoils Molecule A has four positive supercoils and molecule B has 1 negative supercoil Molecule A has zero supercoils and B has 3 negative supercoils.

Possibility b, in which molecule A has 4 positive supercoils and molecule B has 1 negative supercoil fits these data. Molecule A would be more compacted because it has more supercoils. Also, molecule B would be more transcriptionally active, because it is more negatively supercoiled. The first possibility does not fit the data because both molecules have the same level of supercoiling so molecule A and molecule B would have the same level of compaction. The third possibility does not fit the data because molecule B would be more compact.

Discuss the similarities and differences between RNA polymerase and DNA polymerase (described in Chapter 13).

RNA and DNA polymerase are similar in the following ways: 1. They both use a template strand. 2. They both synthesize in the 5 to 3 direction. 3. The chemistry of synthesis is very similar in that they use incoming nucleoside triphosphates and make a phosphoester bond between the previous nucleotide and the incoming nucleotide. 4. They are both processive enzymes that slide along a template strand of DNA. RNA and DNA polymerase are different in the following ways: 1. RNA polymerase makes RNA and DNA polymerase makes DNA. 2. RNA polymerase does not require a primer. 3. RNA polymerase does not have a proofreading function or exonuclease activity.

What is the phenomenon of RNA interference (RNAi)?

RNA interference refers to the phenomenon in which the presence of a double-stranded RNA molecule leads to the silencing of a complementary mRNA.

What is the subunit composition of bacterial RNA polymerase holoenzyme? What are the functional roles of the subunits?

RNA polymerase holoenzyme consists of sigma factor plus the core enzyme, which is composed of five subunits, α2ββ′ω. The role of sigma factor (σ) is to recognize the promoter sequence. The α subunits are necessary for the assembly of the core enzyme and for loose DNA binding. The β and β subunits catalyze the covalent linkages between adjacent ribonucleotides. The ω subunit is important for the proper assembly of the core enzyme.

An RNA molecules has the following sequence: Region 1 Region 2 Region 3 5'-CAUCCAUCCAUUCCCCAUCCGAUAAGGGGAAUGGAUCCGAAUGGAUAAC-3' Parts of region 1 can form a stem-loop with region 2 and with region 3. Can region 1 form a stem loop with region 2 and region 3 at the same time? Why or why not? Which stem-loop would you predict to be more stand: between 1 and 2 OR between 1 and 3. Explain your choice.

Region 1 cannot form a stem-loop with region 2 and region 3 at the same time. Complementary regions of RNA form base pairs, not base triplets. The region 1/region 2 interaction would be slightly more stable than the region 1/region 3 interaction because it is one nucleotide longer, and it has a higher amount of GC base pairs. Remember that GC base pairs form three hydrogen bonds compared to AU base pairs, which form two hydrogen bonds. Therefore, helices with a higher GC content are more stable.

Discuss the structure and function of regulatory elements. Where are they located relative to the core promoter?

Regulatory elements are relatively short genetic sequences that are recognized by regulatory transcription factors. After the regulatory transcription factor has bound to the regulatory element, it will affect the rate of transcription, either activating it or repressing it, depending on the action of the regulatory protein. Regulatory elements are typically located in the upstream region near the promoter, but they can be located almost anywhere (i.e., upstream and downstream) and even quite far from the promoter

Suppose that a cell had a mutation in a gene that encodes for single-stranded binding proteins such the cell cannot make functional SSBPs. Would DNA replication occur faster, slower or not all? Explain your answer.

Single-stranded DNA binding proteins bind to and stabilized single-stranded DNA after strand separation until replication occurs. If there were no SSBPs, DNA replication would mostly likely not happen al all. To re-separate the strands, a helicase would need to rebind at the origin. If there was any section that was able to complete DNA replication, the helicase may separate this region as well.

For each of the following transcription factors, how would eukaryotic transcriptional initiation be affected if it were missing? TFIIB TFIID TFIIH

TFIIB RNA polymerase would not bind to the core promoter. TFIID contains the TATA-binding protein. If it were missing, RNA polymerase would not bind to the core promoter. TFIIH The formation of the open complex would not take place.

Which eukaryotic transcription factor(s) shown in Figure 14.14 plays an equivalent role to σ factor found in bacterial cells?

TFIID and TFIIB would play equivalent roles to sigma factor. Sigma factor does two things: it recognizes the promoter (as does TFIID), and it recruits RNA polymerase to the promoter (as does TFIIB).

Compared with DNA polymerase, how is telomerase different in its ability to synthesize a DNA strand?

Telomerase is different than DNA polymerase in that it uses a short RNA sequence, which is part of its structure, as a template for DNA synthesis.

A lariat contains a closed loop and a linear end. An intron has the following sequence: 5′-GUPuAGUA-60 nucleotides-UACUUAUCC-100 nucleotides-Py12NPyAG-3′. Which sequence would be found within the closed loop of the lariat, the 60-nucleotide sequence or the 100-nucleotide sequence?

The 60-nucleotide sequence would be found in the closed loop. The closed loop is the region between the 5 splice site and the branch site.

Explain the proofreading function of DNA polymerase.

The active site of DNA polymerase has the ability to recognize a mismatched nucleotide in the newly made strand and remove it by exonuclease cleavage. Proofreading occurs in a 3 to 5 direction. After the mistake is removed, DNA polymerase resumes DNA synthesis in the 5 to 3 direction.

Explain how the acetylation of core histones may loosen chromatin packing

The attraction between DNA and histones occurs because the histones are positively charged and the DNA is negatively charged. The covalent attachment of acetyl groups decreases the amount of positive charge on the histone proteins and thereby may decrease the binding of the DNA. In addition, histone acetylation may attract proteins to the region that loosen chromatin compaction.

What parts of the nucleotide occupy the major and minor grooves of double-stranded DNA?

The bases occupy the major and minor grooves.

What are the 3 components of a nucleotide? When regard to the 5' and 3' positions on a sugar molecule, how are nucleotides linked together to form a nucleic acid?

The building blocks of a nucleotide are a sugar (ribose or deoxyribose), a nitrogenous base, and a phosphate group. In a nucleotide, the phosphate is already linked to the 5 position on the sugar. When two nucleotides are hooked together, a phosphate on one nucleotide forms a covalent bond with the 3 hyrdroxyl group on another nucleotide.

On rare occasions, a chromosome can suffer a small deletion that removes and centromere. When this occurs, the chromosome usually is not found within subsequent daughter cells. Explain why a chromosome without a centromere is not transmitted very efficiently from mother to daughter cells (Note: a chromosome that is located outside the nucleus after telophase, it is degraded).

The centromere is the attachment site for the kinetochore, which attaches to the spindle. If a chromosome is not attached to the spindle, it is free to "float around" within the cell, and it may not be near a pole when the nuclear membrane re-forms during telophase. If a chromosome is left outside of the nucleus, it is degraded during interphase. That is why the chromosome without a centromere may not be found in daughter cells.

A diagram of a linear chromosome is shown here. 5'-A-------------------------------------------------------------B-3' 3'-C-------------------------------------------------------------D-5' The end of each strand is labeled with an A, B, C, or D. Which ends could not be replicated by DNA polymerase? Why not?

The ends labeled B and C could not be replicated by DNA polymerase. DNA polymerase makes a strand in the 5 to 3 direction using a template strand that is running in the 3 to 5 direction. Also, DNA polymerase requires a primer. At the ends labeled B and C, there is no place (upstream) for a primer to be made.

What is the function of a splicing factor? Explain how splicing factors can regulate the tissue-specific splicing of mRNAs.

The function of splicing factors is to influence the selection of splice sites in RNA. In certain cell types, the concentration of particular splicing factors is higher than in other tissues. The high concentration of particular splicing factors, and the regulation of their activities, may promote the selection of particular splice sites and thereby lead to tissue-specific splicing.

What enzymatic features of DNA polymerase prevent it from replicating one of the DNA strands at the ends of linear chromosomes?

The inability to synthesize DNA in the 3 to 5 direction and the need for a primer prevent replication at the 3′ end of the DNA strands.

What types of genetic activities occur during interphase? Explain why these activities cannot occur during M phase.

The main activities that can occur during interphase are transcription and DNA replication. For these activities to occur, the DNA must be in a relatively loose conformation. During M phase, there is relatively little genetic activity, although there is evidence that a few genes are transcribed. However, during M phase most genes are transcriptionally inactive

What is the function of the nucleolus?

The nucleolus is a region inside the eukaryotic nucleus where the assembly of ribosomal subunits occurs.

How does a eukaryotic ribosome select its start codon? Describe the sequences in eukaryotic mRNA that provide an optimal context for a start codon.

The ribosome binds at the 5 end of the mRNA and then scans in the 3 direction in search of an AUG start codon. If it finds one that reasonably obeys Kozak's rules, it will begin translation at that site. Aside from an AUG start codon, two other important features are a guanosine at the +4 position and a purine at the -3 position.

What is the role of aminoacyl-tRNA synthetase?

The role of aminoacyl-tRNA synthetase is to specifically recognize tRNA molecules and attach the correct amino acid to them. This ability is sometimes described as the second genetic code because the specificity of the attachment is a critical step in deciphering the genetic code. For example, if a tRNA has a 3-GGG-5 anticodon, it will recognize a 5-CCC-3 codon, which should specify proline. It is essential that the aminoacyl-tRNA synthetase known as prolyl-tRNA-synthetase recognizes this tRNA and attaches proline to the 3 end. The other aminoacyl-tRNA synthetases should not recognize this tRNA.

What re the roles of the core histone proteins compared with the role of histone H1 in the compaction of eukaryotic DNA?

The role of the core histones is to form the nucleosomes. In a nucleosome, the DNA is wrapped 1.65 times around the core histones. Histone H1 binds to the linker region. It may play a role in compacting the DNA into a 30 nm fiber.

List the structural differences between DNA and RNA.

The sugar in DNA is deoxyribose; in RNA it is ribose. DNA contains the base thymine, while RNA has uracil. DNA is a double helical structure. RNA is single stranded, although parts of it may form double-stranded regions.

For protein-encoding genes, one DNA strand is called the template strand, and the complementary strand is called the coding strand. Are these two terms appropriate for nonprotein-encoding genes? Explain.

The term template strand is still appropriate because one of the DNA strands is used as a template to make the RNA. The term coding strand is not appropriate because the RNA made from nonprotein-encoding genes does not code for an amino acid sequence of a polypeptide.

In what ways are the actions of the lac repressor and trp repressor similar and how are they different with regard to their binding to operator sites, their effects on transcription, and the influences of small effector molecules?

The two proteins are similar in that both bind to a segment of DNA and repress transcription. They are different in three ways. (1) They recognize different effector molecules (i.e., the lac repressor recognizes allolactose, and the trp repressor recognizes tryptophan. (2) Allolactose causes the lac repressor to release from the operator, whereas tryptophan causes the trp repressor to bind to its operator. (3) The sequences of the operator sites that these two proteins recognize are different from each other. Otherwise, the lac repressor could bind to the trp operator, and the trp repressor could bind to the lac operator.

Let's assume the linker DNA averages 54 bp in length. How many molecules of H2A would you expect to find in a DNA sample that is 46,000 bp in length?

There are 146 bp around the core histones. If the linker region is 54 bp, we expect 200 bp of DNA (i.e., 146 + 54) for each nucleosome and linker region. If we divide 46,000 bp by 200 bp, we get 230. Because there are two molecules of H2A for each nucleosome, there would be 460 molecules of H2A in a 46,000-bp sample of DNA.

Coumarins and quinolones are two classes of drugs that inhibit bacterial growth by directly inhibiting DNA gyrase. Discuss two reasons why inhibiting DNA gyrase might inhibit bacterial growth.

These drugs would diminish the amount of negative supercoiling in DNA. Negative supercoiling is needed to compact the chromosomal DNA, and it also aids in strand separation. Bacteria might not be able to survive and/or transmit their chromosomes to daughter cells if their DNA was not compacted properly. Also, because negative supercoiling aids in strand separation, these drugs would make it more difficult for the DNA strands to separate. Therefore, the bacteria would have a difficult time transcribing their genes and replicating their DNA, because both processes require strand separation. As discussed in Chapter 13, DNA replication is needed to make new copies of the genetic material to transmit from mother to daughter cells. If DNA replication was inhibited, the bacteria could not grow and divide into new daughter cells. As discussed in Chapters 14-16, gene transcription is necessary for bacterial cells to make proteins. If gene transcription was inhibited, the bacteria could not make many proteins that are necessary for survival.

What is one reason why an amphibian, such as a salamander, might have a larger genome than a mammal, such as a chimpanzee?

They are both eukaryotes and so would both have introns in their genes. A chimpanzee is more complex and so would expected to have more coding DNA. Therefore, the salamander would need more non-coding DNA, which is typically repetitive sequences.

In the tertiary structure of tRNA, where is the anticodon region relative to the attachment site for the amino acid?

They are far apart, at opposite ends of the molecule.

Discuss and make a list of the similarities and differences in the events that occur during the initiation, elongation, and termination stages of transcription (see Chapter 14) and translation (Chapter 15).

This could be a very long list. There are similarities along several lines: 1. There is a lot of molecular recognition going on, either between two nucleic acid molecules or between proteins and nucleic acid molecules. Students may see these as similarities or differences, depending on their point of view. 2. There is biosynthesis going on in both processes. Small building blocks are being connected together. This requires an input of energy. 3. There are genetic signals that determine the beginning and ending of these processes. There are also many differences: 1. Transcription produces an RNA molecule with a similar structure to the DNA whereas translation produces a polypeptide with a structure that is very different from RNA. 2. Depending on your point of view, it seems that translation is more biochemically complex, requiring more proteins and RNA molecules to accomplish the task.

Explain what is meant by the "coupling of transcription and translation." Does coupling occur in bacterial and/or eukaryotic cells? Explain

This means that translation can begin before transcription of the mRNA is completed. This cannot occur in eukaryotic cells because transcription and translation occur in different cellular compartments. In eukaryotic cells, transcription occurs in the nucleus, whereas translation occurs in the cytosol

A mutation within a gene changes the start codon to a stop codon. How will this mutation affect the transcription of this gene?

This will not affect transcription. However, it will affect translation by preventing the initiation of polypeptide synthesis.

How can environmental agents that do not cause gene mutations contribute to cancer? Would these epigenetic changes be passed to offspring?

Though they don't change the DNA sequence, epigenetic modifications can affect gene expression. Such changes could increase gene expression and thereby result in oncogenes or they could inhibit the expression of tumor suppressor genes. Either type of change could contribute to cancer. For example, DNA methylation of a tumor suppressor gene could promote cancer.

What are the functions of transcriptional activator proteins and repressor proteins? Explain how they work at the molecular level

Transcriptional activation occurs when a regulatory transcription factor binds to a regulatory element and activates transcription. Such proteins, called activators, may interact with TFIID and/or mediator to promote the assembly of RNA polymerase and general transcription factors at the promoter region. They also could alter the structure of chromatin so that RNA polymerase and transcription factors are able to gain access to the promoter. Transcriptional inhibition occurs when a regulatory transcription factor inhibits transcription. Such repressors may interact with TFIID and/or mediator to inhibit RNA polymerase.

List the components required for translation. Describe the relative sizes of these different components. In other words, which components are small molecules, macromolecules, or assemblies of macromolecules?

Translation requires mRNA, tRNAs, ribosomes, proteins such as initiation, elongation, and termination factors, and many small molecules. ATP and GTP are small molecules that contain high-energy bonds. The mRNA, tRNAs, and proteins are macromolecules. The ribosomes are a large complex of macromolecules.

According to the adaptor hypothesis, are the following statements true or false? The sequence of an anticodon in a tRNA directly recognizes a codon sequence in mRNA, with some room for wobble. The amino acid attached to the tRNA directly recognizes a codon sequence in mRNA. The amino acid attached to the tRNA affects the binding of the tRNA to a codon sequence in mRNA.

True, false, false

What are the components of a nucleosome? Pick all the correct components: Two turns of DNA. 2 copies each of H2A, H2B, H3, and H4 - This makes up the histone octet 1 copy of H1. A histone octet 146 bases - This makes up two turns of DNA around the octet 200 bases

Two turns of DNA. 2 copies each of H2A, H2B, H3, and H4 - This makes up the histone octet A histone octet 146 bases - This makes up two turns of DNA around the octet

When chromatin is treated with a moderate salt concentration, the linker histone H1 is removed. Higher salt concentration removes the rest of the histone proteins. If the experiment of Figure 12.10 were carried out after the DNA was treated with moderate or high salt, what would be the expected results?

With a moderate salt concentration, the nucleosome structure is still preserved so the same pattern of results would be observed. DNase I would cut the linker region and produce fragments of DNA that would be in multiples of 200 bp. However, with a high salt concentration, the core histones would be lost, and DNase I could cut anywhere. On the gel, you would see fragments of almost any size. Because there would be a continuum of fragments of many different sizes, the lane on the gel would probably look like a smear rather than having a few prominent bands of DNA

Let's consider how DNA ligase connects the left Okazaki fragment with the middle Okazaki fragment. After DNA polymerase I removes the middle RNA primer and fills in the gap with DNA, where does DNA ligase function? See the arrows on either side of the middle RNA primer. Is ligase needed at the left arrow, at the right arrow, or both?

You need DNA ligase only at the right arrow. DNA polymerase I begins at the end of the left Okazaki fragment and synthesizes DNA to fill in the region as it removes the middle RNA primer. At the left arrow, DNA polymerase I is simply extending the length of the left Okazaki fragment. No ligase is needed here. When DNA polymerase I has extended the left Okazaki fragment through the entire region where the RNA primer has been removed, it hits the DNA of the middle Okazaki fragment. This occurs at the right arrow. At this point, the DNA of the middle Okazaki fragment has a 5 end that is a monophosphate. DNA ligase is needed to connect this monophosphate with the 3 end of the region where the middle RNA primer has been removed.

Viral genomes can either be composed of RNA or DNA and can either be single- or double-stranded. The base composition of a specific virus was analyzed and found to be 14.1% A, 14% U, 36,2% G and 35.7% C. Is the genome of the virus most likely to be single-stranded DNA, single-stranded RNA, double-stranded DNA or double-stranded RNA?

You would conclude that the virus is an RNA virus since it has Uracil in its genome, which is not found in DNA. You would also conclude it is probably double-stranded RNA because the amount of A equals U and the amount of G equals C. Therefore, this molecule could be double stranded and obey the AU/GC rule. However, it is also possible that it is merely a coincidence that A happens to equal U and G happens to equal C, and the genetic material is really single stranded

What are the advantages and disadvantages of mRNAs with a short half-life compared with mRNAs with a long half-life?

a disadvantage of mRNAs with a short half-life is that the cells probably waste a lot of energy making them. If a cell needs the protein encoded by a short-lived mRNA, the cell has to keep transcribing the gene that encodes the mRNA because the mRNAs are quickly degraded. An advantage of short-lived mRNAs is that the cell can rapidly turn off protein synthesis. If a cell no longer needs the polypeptide encoded by a short-lived mRNA, it can stop transcribing the gene, and the mRNA will be quickly degraded. This will shut off the synthesis of more proteins rather quickly. With most long-lived mRNAs, it will take much longer to shut off protein synthesis after transcription has been terminated.

What does telomerase use as its template for the synthesis of a DNA strand?

a short RNA sequence is part of the enzyme

As described in Figure 16.12, four regions within the trpL mRNA can form stem-loops. Let's suppose that mutations have been previously identified that prevent the ability of a particular region to form a stem-loop with a complementary region. For example, a region 1 mutant cannot form a 1-2 stem-loop, but it can still form a 2-3 or 3-4 stem-loop. Likewise, a region 4 mutant can form a 1-2 or 2-3 stem-loop but not a 3-4 stem-loop. Under the following conditions, would attenuation occur? a) Region 1 is mutant, tryptophan is high, and translation is not occurring. b) Region 2 is mutant, tryptophan is low, and translation is occurring. c) Region 3 is mutant, tryptophan is high, and translation is not occurring. d) Region 4 is mutant, tryptophan is low, and translation is not occurring.

a) Attenuation will not occur because loop 2-3 will form. b) Attenuation will occur because 2-3 cannot form, so 3-4 will form. c) Attenuation will not occur because 3-4 cannot form d) Attenuation will not occur because 3-4 cannot form.

Describe three examples of genes that are not protein-encoding genes.

tRNA genes encode tRNA molecules, and rRNA genes encode the rRNAs found in ribosomes. There are also nonprotein-encoding genes for the RNAs found in snRNPs and so on.