Biochemistry Exam 4

What four amino acid side chains are integral to the DNA binding function of the TATA binding protein? Choose one: A. four phenylalanine residue side chains B. four histidine residue side chains C. four tyrosine residue side chains D. four tryptophan residue side chains

A. four phenylalanine residue side chains

There are several similarities between the miRNA and RNAi pathways. Which of the following are common features of both pathways? A. proteins required for processing of the primary transcript B. source of the primary ncRNA transcript C. complementarity of the processed transcript to mRNA D. cellular role of the pathway

A. proteins required for processing of the primary transcript

Prokaryotic and eukaryotic DNA polymerases rarely make mismatched base pairs because ____________. -the mismatched pairs make covalent bonds instead of hydrogen bonds. -the active site does not fit mismatches well. -there are no hydrogen bonds that line up between the mismatches. -the DNA primer does not allow for mismatched pairs.

-the active site does not fit mismatches well.

14. What are iPS cells? How are they produced?

14. iPS cells are induced pluripotent stem cells that have the potential to develop into almost any cell type, given the right conditions and stimuli. They can be produced from many cell types by adding the genes for four transcription factors, Sox2, Oct4, Klf4, and c-Myc, either as free DNA or as viruses. A small fraction of cells treated in this way are converted to iPS cells

15. Explain how the spliceosome recognizes introns.

15. The U1 and U2 snRNAs in the U1 and U2 snRNPs recognize and bind the 5′ splice site and branch point through base pairing

15. What are the major differences between meiotic homologous recombination by double-strand break and homologous recombination for the repair of double-strand breaks caused by DNA damaging events?

15. The meiotic homologous recombination (MHR) process exerts control over where strand breaks occur, whereas DNA damageinduced, double-strand breaks might occur anywhere within the chromosome. During MHR, the break is catalyzed by Spo11, with ends that are fairly easily repaired using Spo11, Rad51, and Dmc1. DNA-damaging breaks yield DNA ends that must be extensively processed, often involving resection of DNA to produce repairable ends, repaired by Rad51, BRCA1, and BRCA2.

16. Describe what occurs when bacteriophage λ invades an E. coli cell and establishes lysogeny. What would happen if the E. coli cell were exposed to a high dose of UV light?

16. Initially the phage binds the cell and injects its DNA genome into the cell. The DNA is circularized and integrated into the host chromosome using integrase and integration host factor. Once integrated, bacterial replication passively replicates the phage DNA as well. If the cell is exposed to a high dose of UV, a lytic cycle is initiated due to the DNA damage: The prophage excises from the host DNA using integrase, integration host factor, and Xis (an excision-specific phage protein), then replicates many times and produces phage proteins. Phage virions assemble and cell lysis releases the virions.

16. Describe two ways in which a point mutation in a gene can result in alternative splicing.

16. Point mutations can cause alternative splicing through elimination of a splice site (mutation in one of the conserved U1 or U2 recognition sequences) or introduction of a cryptic 5′ or 3′ splice site, leading to splicing within an exon.

4. Why does the human mitochondrion have a slightly different genetic code than that found in the cytosol?

4. This is probably due to its ancestral origin. It is widely believed that mitochondria evolved from bacteria taken up by early nucleated cells. Because mitochondrial DNA has different evolutionary origins from chromosomal DNA, it contains a separate protein translation system, including tRNA genes

4. You are studying the regulation of a Drosophila gene, and you have available the genome sequences of several related Drosophila species. By comparing these sequences in the vicinity of your gene, you identify several regions that appear to be conserved. Moreover, you identify potential binding sites for known regulatory proteins. You speculate that this region is an enhancer. Suggest how you might test this idea. Assume that you are competent at recombinant DNA work and that you have a method to introduce desired constructs into Drosophila. (Hint: One of the defining properties of an enhancer is that it works from variable distances and in both orientations.)

4. To validate your hypothesis, you would use recombinant DNA to place the proposed enhancer upstream of a reporter gene in Drosophila embryos. Several constructs should be prepared, with the enhancer in either orientation and at a variable distance from the promoter for the reporter. If the DNA region includes an enhancer, you should see expression of the reporter. See Levine and colleagues' study to analyze the eve stripe 2 enhancer (see Figure 23.54).

4. Describe the secondary and tertiary structure of tRNA.

4. tRNA forms a cloverleaf secondary structure with three arms: D, TΨC, and anticodon. The top stem of tRNA is the acceptor site for amino acid attachment. The D and TΨC loops further interact to form the L-shaped tertiary structure.

All DNA is synthesized in which direction? Question options: 5'->3' 5'->1' 1'->5' 3'->5'

5'->3'

Introducing Oct4, Sox2, c-Myc, or Klf4 into a differentiated cell causes Question options: cell death. a pluripotent state. no change. cell meiosis.

a pluripotent state.

The enzyme responsible for generating siRNA from double strand RNA is ____________. a.snoRNA particle b. RISC c. Sigma d. Dicer

d. Dicer

Aminoacyl-tRNA synthetases carry out editing _______________. a. by binding to a separate enditing subunit associated with the enzyme b. before activation of the amino acid with ATP c. after the aminoacyl-tRNA has been released by the enzyme d. at a site that is distinct from the active site

d. at a site that is distinct from the active site

The Trp repressor protein is bound to ____ in the ____ of tryptophan.

double-stranded DNA presence Correct; in this molecular structure, the Trp repressor protein can be seen associating with double-stranded DNA. Correct; the Trp repressor protein regulates the process of transcription. In the presence of tryptophan, the Trp repressor will associate with nucleic acid and halt transcription.

A nucleophilic amino group of the amino acid bound to the terminus of A-site tRNA attacks the electrophilic carbonyl carbon in the ester bond between the terminus of P-site tRNA and its bound amino acid during the __________ step of translation. Question 3 options: elongation initiation translocation termination

elongation

The initiation, elongation, and termination of eukaryotic transcription is controlled by the Question options: extent of phosphorylation and dephosphorylation of the RNA polymerase CTD. presence of initiation, elongation, and termination factors binding to RNA polymerase. unwinding the double-stranded DNA. length of the RNA transcript.

extent of phosphorylation and dephosphorylation of the RNA polymerase CTD.

Producing an iPS cell is remarkable because the pathway Question options: produces a very unstable cell. is a highly irreversible reaction. for conversion is very specific and limited. for conversion is general, not specific.

for conversion is general, not specific.

A protein that has a weak affinity for a DNA site and is at ___________ concentrations of protein will bind to DNA __________. low; indiscriminately low; strongly high; strongly high; indiscriminately

high; strongly

One of the most common binding interactions between proteins and DNA are __________ bonds. Question options: covalent ionic hydrogen polar

hydrogen

The __________ strand of DNA is transcribed into mRNA. Question 2 options: leading template lagging coding

template

Why are mutations that result in decreased cell viability are most likely to be lost ?

because they affect only one cell before it is able to divide and produce daughter cells.

The roles of U1, U2, U4, U5, and U6 in the spliceosome complex are to Question options: bind mRNA and facilitate the splicing reaction. bind the small nuclear RNA. carry the products from the nucleus to the cytoplasm. bind proteins and hold the complex together.

bind mRNA and facilitate the splicing reaction.

The best description of how allosteric regulation works is that a ligand __________ the transcriptional regulatory protein, which __________ the protein's affinity for DNA. Question options: binds to; causes conformational changes that can only decrease covalently modifies; increases covalently modifies; decreases binds to; causes conformational changes that affect

binds to; causes conformational changes that affect

In an iteration of the Nirenberg-Leder experiment to assign triplet codons to specific amino acids, radioactively labeled aminoacyl-tRNA with the anticodon of 3'-CUG-5' was retained on the filter at the end of the experiment. Which mRNA was used in this iteration of the experiment? a. 5'-CAG-3' b. 5'-GTC-3' c. 5'-GAC-3' d. 5'-CUG-3'

c. 5'-GAC-3'

The DNA sequence of prokaryotic gene promotors were found to be Question 1 options: largely conserved. distinct for each gene. capable of binding different promoters. strong binders of DNA polymerase.

largely conserved.

The information gained from the DNA footprinting technique is the Question options: location of a DNA binding protein on DNA. DNA sequence. location of a gene in DNA. promotor region of a gene.

location of a DNA binding protein on DNA.

Transcriptional elongation is favored by the trp operon when tryptophan levels are Question options: low. high. constant. The trp operon is not affected by tryptophan levels.

low.

The bacteriophage λ life cycle events outlined in Part 1 represent the lytic/lysogenic cycle of phage replication.

lysogenic Correct; the bacteriophage λ life cycle events outlined in Part 1 represent the lysogenic cycle of phage replication. DNA integration takes place in this form of phage replication. Phage DNA is replicated along with cellular DNA during normal DNA replication that precedes binary fission.

Of the following protein synthesizing RNA molecules, which codes for a protein? Question options: tRNA siRNA rRNA mRNA

mRNA

What kind of control mechanism is ligand-induced binding of an activator protein? Question options: negative covalent positive neutral

positive

RNA is typically more susceptible to backbone hydrolysis than DNA because of the Question options: presence of simpler nucleotides. lack of thymine nucleotides. presence of a 2' OH group. presence of a 3' OH group.

presence of a 2' OH group.

Excess exposure to ultraviolet light can cause DNA damage because of Question options: causes protein misfolding. produces photoproducts. produces methylated thymidine. causes an abasic site to form. causes double strand breaks in DNA.

produces photoproducts.

One function of eukaryotic transcriptional activator proteins is to __________. phosphorylate DNA cis-acting sequences to alter DNA structure initiate RNA synthesis by synthesizing the first RNA primer insulate chromatin so that gene transcription is unregulated recruit chromatin modifying proteins such as HAT and HDAC enzymes

recruit chromatin modifying proteins such as HAT and HDAC enzymes

A function of transcriptional activator proteins is to Question options: initiate DNA synthesis. recruit other transcription factors. recruit gene promotors. initiate RNA synthesis.

recruit other transcription factors.

The spliceosome is composed of five ribonucleoproteins, each of which has a specific role in the splicing process. Different combinations of these proteins are associated with different stages of the splicing reaction. Place the snRNP complexes listed into their appropriate stage in the splicing reaction. Each stage should contain only a single combination of proteins.

Pre-spliceosome: U1 and U2 Precatalytic: U1, U2, U4, U5, U6 Active spliceosome: U2, U5, and U6

The two key elements of the oriC include Question options: three 13-bp repeats and four 9-bp repeats with enriched G-C base pairs. 20 subunits of A-T pairs. three 13-bp repeats and four 9-bp repeats with enriched A-T pairs. the RNA primer.

three 13-bp repeats and four 9-bp repeats with enriched A-T pairs

When there are high levels of tryptophan in the bacterial cell, the effect it has on the Trp repressor is _______________ and the effect on the structure of the Trp operon mRNA is _____________. to inhibit operon binding; to promote formation of the regions 3-4 stem loop to stimulate operon binding; to promote formation of the regions 2-3 stem loop to inhibit operon binding; to promote formation of the regions 2-3 stem loop to stimulate operon binding; to promote formation of the regions 3-4 stem loop

to stimulate operon binding; to promote formation of the regions 3-4 stem loop

12. What are the three types of mutations that result from single point mutations? What is the potential severity of each?

. (1) Silent mutations do not change the coded amino acid due to degeneracy of the genetic code; (2) nonsense mutations result in premature stop codons and generally yield nonfunctional proteins; (3) missense mutations incorporate incorrect amino acids into the protein. Both nonsense and missense mutations can be detrimental to a cell, depending upon the protein altered and the degree of alteration

1. Why was Francis Crick dissatisfied with protein synthesis models that brought amino acids into direct contact with DNA or mRNA?

1 Crick did not support these models because in eukaryotes, DNA and protein synthesis occur in different cellular compartments. Also, on a physicochemical basis, nucleotide structure seemed insufficiently complex to act directly as a template for 20 amino acids.

23r1. Regulation can occur at many steps in the pathway of gene expression. Give at least one example of a regulator at the level of (a) transcription initiation, (b) protein stability, and (c) protein posttranslational modification.

1. (a) lac operon, yeast galactose genes, Drosophila eve gene; (b) LexA protein; (c) metabolic regulation through phosphorylation, methylation, or acetylation.

C1. Why did George Gamow state that at least three nucleotides must be required to code for each amino acid?

1. Gamow concluded that three nucleotides were the minimum number required for each codon because a two-nucleotide codon would result in only 42 , or 16, possible combinations. Because there are 20 standard amino acids, a two-nucleotide codon would have required more than one amino acid to be associated with the same codon. However, a codon of three nucleotides gives 43 , or 64, possible combinations.

C1. Consider the regulation of the trp operon in E. coli. If this system were present in yeast instead of E. coli, would you expect it to function properly? Explain

1. It would not work in eukaryotes because this mechanism depends on transcription and translation occurring simultaneously; in eukaryotes these processes occur in different cellular compartments.

Order these steps (from first to last) of the Nirenberg-Leder experiment, which was used to determine which codons encoded which amino acids.

1. purify protein synthesis machinery from cells 2. Add radioactively labeled amino acid 3. Add a trinucleotide RNA sequence 4. Allow proteins, especially ribosomes, to bind to a filter, while other components pass through 5. Determine if radioactive material is bound to filter

10. From the following sequence, locate the sites of potential photoproduct formation. Indicate what photoproducts will most likely form and what the potential effect on the DNA would be if left unrepaired. 5′-ACGTCAGTTACGTACTGACGT

10. 5′-ACGTCAGTTACGTACTGACGT. The TC photoproduct is likely to be a (6-4) photoproduct, which can stall replication forks if not repaired. The TT and CT photoproducts would most likely be pyrimidine dimers and, if unrepaired, would most likely result in adenine on the daughter strand. The TT photoproduct would not result in a mutation, but CT would result in a mismatch, with high potential for mutation, depending upon when repair occurs. If the daughter strand were used to correct the lesion, then the C would be converted to a T.

10. In bacteria, the same DNA strand is often a template for both replication and transcription. Why are codirectional collisions a common occurrence?

10. Both DNA and RNA polymerase read 3′ to 5′, resulting in a 5′ to 3′ polymerization reaction. When both replication and transcription complexes proceed down the same region of DNA, the DNA polymerase can overtake RNA polymerase because transcription is much slower than replication; this results in a codirectional collision.

10. What happens during cytosine deamination, and how is it repaired?

10. Cytosine can be spontaneously deaminated, producing uracil. Uracil is removed by uracil DNA glycosylase to produce an abasic site that is repaired by base excision repair, where a nick is created in the DNA by AP endonuclease. The nucleotide is replaced by short-patch repair or a short section of nucleotides is replaced by long-patch repair

10. In the galactose regulatory system of yeast, what would be the behavior of mutants that lack one of the following components: GAL3, GAL4, or GAL80? Consider the expression of the target genes with and without the presence of galactose. What would be the behavior of a GAL3-GAL80 double mutant?

10. GAL3 and GAL4 mutants will not express the target genes in the presence of galactose. A GAL80 mutant and a GAL3- GAL80 double mutant will express the target genes in the absence of galactose.

10. What is an epigenetic state?

10. It is a long-lasting pattern of gene expression, almost always defined as passing from one generation to the next. It does not involve a change in DNA sequence, but may involve modification of cytosine to form 5-methylcytosine.

10. Why is the ribosome considered a ribozyme?

10. It is a ribozyme because it contains a catalytic RNA (23S rRNA in E. coli) that catalyzes the formation of a peptide bond between amino acids.

10. Why do acetylation and methylation not occur simultaneously on the same lysine residue on histones?

10. Methylation and acetylation are not found on the same regions of histones because they have opposite effects on histones. Methylation leads to tighter compaction of histones, often found in transcriptionally inactive regions. Acetylation is often found in histones that are not as tightly condensed and are therefore more accessible by the transcriptional machinery.

10. Why is the phosphorylation state of the RNA polymerase II CTD an important component of transcription? Describe the changes that the CTD undergoes from initiation to termination.

10. Phosphorylation of the conserved heptapeptide repeat in the CTD is important because it mediates interaction of capping and splicing factors and polymerase activity. Certain serine residues are phosphorylated during initiation and elongation, but upon termination the CTD is dephosphorylated, facilitating release of the RNA polymerase II complex from the nascent transcript.

11. What would happen if a mutation in the λ phage Xis gene occurred such that the resulting protein was not functional?

11. Because Xis is required for λ phage DNA excision, a mutation rendering Xis nonfunctional would prevent excision and prevent initiation of a lytic cycle.

11. If unrepaired, what effect can cyclobutane pyrimidine dimers or cross-linked guanine residues have on cellular processes?

11. Both of these can distort the DNA helix and present a roadblock for access to individual bases. The replication fork stalls at such lesions or skips over them and attempts to place an incorrect nucleotide. RNA polymerase either stalls or skips over lesions, putting a potentially incorrect nucleotide at the site.

11. Why are multiple lipid modifications often found on membrane-associated proteins?

11. Lipid modifications can be used for strong association of a protein with a membrane. As soluble proteins fold, the hydrophobic regions are mostly sequestered in the interior of the protein, while hydrophilic regions remain exposed to the polar, aqueous environment of the cell. Membranes have hydrophobic fatty acid tails arranged in the center of the membrane. The addition of lipids to the exterior of a protein adds hydrophobic regions that can insert within the membrane and therefore firmly associate with the membrane.

11. The paper by Balbo and Bohm [Balbo, P. B., and Bohm, A. (2007). Mechanism of poly(A) polymerase: structure of the enzyme-MgATP-RNA ternary complex and kinetic analysis. Structure, 15, 1117-1131] illustrates the interactions between poly(A) polymerase, the 3′ end of mRNA, and ATP. Figures 2 and 3 therein detail the interactions between the protein and the two substrates. Why is single-stranded DNA a poor substrate for this reaction?

11. Poly(A) polymerase interacts with the 2′-OH in the three adenines at the end of the existing poly(A) tail (positions -1, -2, and -3). Lack of 2′-OH in a single-stranded DNA template would decrease the affinity for the substrate.

11. Explain why decapping and deadenylation are common mechanisms involved in RNA decay.

11. The 5′ m7 G cap and poly(A) tail protect mRNA from digestion. During mRNA decay, removal of either (or both) allows exonucleases to completely digest the mRNA. Other RNA decay mechanisms exist in which the transcript is first cleaved within the sequence, but exonuclease digestion is more common.

11. What are the three types of coatomer proteins that assist with vesicle formation, and with what type of transport is each associated?

11. The three types are COPI, COPII, and clathrin. COPI vesicles mediate transport within the Golgi apparatus; COPII vesicles mediate transport between the endoplasmic reticulum and the Golgi apparatus; and clathrin vesicles mediate transport between the Golgi apparatus and plasma membrane.

11. What is cooperative DNA binding? Why might it be useful in stabilizing epigenetic states?

11. When the binding of one regulatory protein at its site increases the occupancy of another molecule at a different site, usually close to the first one. This usually involves direct interaction between the two bound molecules but it can involve dimerization in solution, followed by dimer binding, or more indirectly by one protein making the binding site for the second one available (e.g., fluctuations "breathing" in nucleosome structure). In cooperative DNA binding, occupancy of the second site goes from near zero to nearly 100% over a very narrow range of concentrations of the DNA binding molecules. This confers switch-like behavior on the regulatory circuit, and can help stabilize epigenetic states.

12. How does the existence of ribozymes like the hammerhead ribozyme and group I introns provide support for the RNA world hypothesis?

12. Most biological reactions require a catalyst to occur in a physiologic time frame suitable for the organism. The RNA world hypothesis relies on the catalytic ability of RNA. Hammerhead ribozyme and group I introns are examples of catalytic RNA that can function without the assistance of any protein components, suggesting that life is possible without proteins.

12. Formation of the 5′ m7 G cap requires that the 5′-nucleotide be a triphosphate. Explain why only this nucleotide retains all three phosphoryl groups.

12. RNA synthesis requires NTPs; the first nucleoside added has three phosphoryl groups. The reaction to add each subsequent nucleotide is initiated by a nucleophilic attack on the α phosphate, resulting in cleavage and release of PPi .

12. Regulatory systems that control anabolic or catabolic pathways operate in ways that conserve energy for the cell. Give an example of each type of pathway, and indicate how the regulation serves this purpose.

12. Regulatory systems save energy by ensuring that gene products are made only when needed. If the product of the biosynthetic pathway is present, there is no need to make more; if a metabolite is not available, there is no need to make enzymes to catabolize it. Examples: Anabolic pathway, tryptophan or histidine biosynthesis. Catabolic pathway: lactose metabolism in E. coli, galactose metabolism in yeast.

12. What effect would an RNase inhibitor found in a human cell have on HIV? What problems might be associated with such an agent?

12. This would prevent degradation of the RNA-DNA hybrid formed by reverse transcriptase, which would prevent the production of the double-stranded DNA that is required for insertion into the host genome and thus prevent an infection by HIV. However, an RNase inhibitor might also prevent degradation of human RNA, reducing the recycling of RNA nucleotides that provides a cell with building blocks for RNA. This would place tremendous energy demands upon the cell

13. How does the mechanism of group I introns differ from that of group II introns? Which is more similar to the spliceosome-catalyzed reaction?

13. In both types of introns, splicing is initiated by a nucleophilic attack on the 3′ splice site. In group I introns this attack occurs from an essential guanosine cofactor, whereas in group II introns it is initiated from a conserved adenosine in the branch site between the 5′ and 3′ splice sites. The released group II intron has a lariat structure, whereas the released group I intron is linear. The group II intron mechanism closely resembles a spliceosomecatalyzed reaction, and there are also sequence similarities between the splice and branch sites.

13. What are eukaryotic pioneer factors, and why is their action important? How are they able to operate? Give an example

13. Pioneer factors are the first trans-acting factors to bind to an enhancer or promoter, even in the presence of nucleosomes in these regions. Binding can lead to availability of other cisacting sites thereby setting in motion a long chain of events that assemble a pre-initiation complex and subsequent transcription. They may bind because their binding site is not covered by a nucleosome or because the nucleosomes have variable positions and the site is sometimes free. Alternatively, they bind to DNA while it is wrapped around the histone core, or when their site becomes transiently available due to fluctuations in nucleosome structure. Pioneer factors include yeast PHO4, and Oct4 and Sox2, which can convert a range of cell types to iPS cells.

13. What is the primary difference between strong and weak bacterial promoters? How does transcription differ from these two types of promoters?

13. Strong bacterial promoters differ only slightly from the conserved sequences at the -35 and -10 boxes, whereas weak promoters can contain several base pair changes. These are the sites of σ-factor binding, so the closer the -35 and -10 boxes are to the conserved sequences, the stronger the association between RNA polymerase and the promoter. Therefore, strong promoters are more likely to be bound by σ factor and transcribed than weak promoters.

13. Show the product of O6 -methylation of guanine, and explain how it is repaired.

13. The methylation product is shown in Figure 20.42. O6 - methylguanine-DNA methyltransferase (MGMT) is required to remove O6 -methylguanine. It is not a true enzyme, as it is irreversibly modified during the process. A cysteine in the active site performs a nucleophilic attack upon the methyl group, removing it from the guanine and converting the cysteine to homocysteine. The protein is then degraded to metabolize homocysteine.

14. Why does tRNA contain more types of base modifications than rRNA?

14. Base modifications in tRNA make the correct tertiary structure possible and enable specific binding between tRNAs and their aminoacyl-tRNA synthetases, as primary sequence and secondary structure are conserved in all tRNAs and cannot be used for specific interactions. By contrast, rRNA sequence and structure vary enough to be used for specific interactions.

14. What are the two roles of methylation by Dam methylase— one in the replication process, the other in repair?

14. The oriC region has 11 sites methylated by Dam methylase to distinguish the template strand from the newly synthesized strand. This highly methylated region is a SeqA binding site. The SeqA-oriC complex is sequestered at the membrane to prevent further initiation until the current replication process is at least a third completed. In DNA repair, methylation sites allow recognition of the template strand for mismatch repair by MutS-MutL-MutH. These proteins locate mismatches and use the methylation site to determine the template strand.

3. What would be the effect of a mutation in the HIV-1 reverse transcriptase gene that decreased the rate of mismatches?

3. Any decrease in the error rate of HIVRT would result in an overall decrease in the rate of viral mutations. This would probably not be an advantage for the virus, as the relatively high error rate often helps to thwart the use of antiviral drugs.

17. What are the differences in the RNAi and miRNA pathways?

17. RNAi and miRNA differ primarily in the structure of the initiating double-stranded RNA and in the mechanism of translational repression. The siRNAs in RNAi are perfectly matched doublestranded RNA fragments, whereas miRNAs may contain loops and bulges. Thus, siRNAs are generally a perfect match to their target, leading to mRNA cleavage, but miRNAs are often an imperfect match with mRNA, resulting in translational repression.

17. Describe the events in V(D)J recombination and explain the importance of this process.

17. V(D)J recombination produces specific antibody-heavy chains. With 51 V segments, 27 D segments, and 6 J segments, more than 8,000 different heavy chains are possible by combining one of each segment. Pairing with a light chain produces even greater potential for variety. The process involves first combining D and J segments, with all intervening segments removed. Next, the V segment is combined with the DJ product, once again removing intervening segments. Removal of the remaining V and J segments is not done on a DNA level, but rather by splicing the mRNA.

For a human chromosome that has a central centromere (metacentric), the single (non-replicated) metacentric chromosome is expected to contain ____ telomeres.

2 Correct; a metacentric human chromosome is expected to contain two telomeres. One telomere is expected to be at each end of the linear chromosome.

2. The modified base 5-methylcytosine is used in higher eukaryotes as an epigenetic mark of particular chromatin states. But 5-methylcytosine is also known to be a "hot spot" for mutation; that is, it has a higher rate of mutation from C to T than most C residues. Explain (on the basis of your understanding of DNA repair) why this is the case.

2. Cytosine residues are deaminated at a very low rate to form uracil, which will lead to a uracil-guanine (U-G) base pair. Because U is not a normal component of DNA, this is quickly repaired to a C-G base pair by uracil N-glycosylase and subsequent events. Deamination of 5-MeC would lead to formation of a T-G base pair, and the cell would not rapidly repair this, as there is no way to be sure which base is the correct base. If DNA replication occurs before the mismatch is repaired, it fixes a C-to-T transition mutation in one of the progeny duplexes.

2. What are Okazaki fragments?

2. DNA segments produced by discontinuous replication of the lagging strand. One is started each time a new β clamp loads onto the lagging strand template and is completed when DNA polymerase encounters the preceding Okazaki fragment.

2. In the experiment by Jacob and Meselson, bacteria were grown for several generations in a medium containing heavy isotopes (15NH4Cl and 13C-glucose). The bacteria used these isotopes to synthesize all cellular components, resulting in their incorporation into all of the bacterial nucleic acids and proteins. The bacteria were switched to a normal medium after infection with bacteriophage, and the ribosomes were isolated. Jacob and Meselson discovered that the isolated ribosomes contained the heavy isotopes. Why did this result indicate that ribosomes are not carriers of genetic information?

2. If ribosomes were the source of genetic material, the bacteriophage would synthesize its own ribosomes soon after infection as a way of propagating itself. This would occur in the normal medium, and so these ribosomes would not contain heavy isotopes. However, if ribosomes were only sites of protein synthesis, then the viral mRNA would be translated by bacterial ribosomes, which were synthesized while the heavy isotopes were present. The presence of isotopes in ribosomes enabled Jacob and Meselson to confirm that ribosomes were not the viral genetic material

2. Why was Khorana's method of nucleotide synthesis so essential to the discovery of the genetic code?

2. Khorana's method of nucleotide synthesis was the first procedure that allowed directed synthesis, not simply random combinations of nucleotides. This was essential to cracking the genetic code, as it allowed Khorana and others to test specific triplet combinations in filter-binding and in vitro translation experiments.

2. Why is it beneficial for organisms to have more than one copy of rRNA genes in the genome? Why do more complex organisms tend to have more copies of these genes?

2. Ribosomes are essential for protein synthesis, and rRNA is a structural and catalytic component. Each ribosome translates only one mRNA at a time; so the more ribosomes present, the more protein synthesis. More complex organisms (that is, yeast, humans) have a greater number of proteins expressed and so require more active ribosomes than organisms such as bacteria

C2. Explain the events that would need to occur at the replication fork if conservative replication occurred.

2. The difference would basically occur following the actual polymerization process. Following the core polymerase, a pair of helicases would be required to separate the template and daughter strands, followed by a protein with the ability to rejoin the template strands and move the daughter strands together for double-strand formation. Mismatch repair would need to occur before this separation of the template-daughter strands. Thus, conservative replication would require increased complexity over the semiconservative process.

2. What are the three possible amino acids resulting from the translation of the RNA sequence CCAAA?

2. They are Pro (CCA), Gln (CAA), or Lys (AAA).

2. What are trans-acting factors and cis-acting sites?

2. Trans-acting factors can operate on DNA sites anywhere in the genome; not limited to the chromosome that encodes them. Cisacting sites are binding sites for trans-acting factors, generally located at or near the genes on which the trans-acting factors act. The actions of trans-acting factors are highly specific because of these sites.

3. In the trp attenuation system of E. coli, predict the effects of the following mutations on the operation of the regulatory system, and give your reasons. To simplify matters, assume that the Trp repressor is not active. Three broad possibilities exist for whether the operon is transcribed in the absence or presence of charged tRNATrp: behavior is normal, transcription is always low ("noninducible"), or transcription is always high ("constitutive"). a. Mutation in region 4 so it cannot pair with region 3. b. Mutation in region 3 so it can pair with region 4 but not with region 2. c. Change of the two Trp codons to glycine codons. d. Change of the AUG codon of the leader peptide to AUA. e. Combination of the first mutation with each of the other three.

3. (a) Mutation in region 4 so it cannot pair with region 3: would be constitutive, because no terminator is formed. (b) Mutation in region 3 so it can pair with region 4 but not with region 2: would be noninducible; antiterminator cannot form in the absence of Trp-tRNATrp, so terminator would always form. (c) Change of the two Trp codons to glycine codons: noninducible, because the ribosome would not pause at the Gly codons in the presence of low Trp-tRNATrp. It might pause if glycine levels are low, but glycine is rarely low. (d) Change of the AUG codon of the leader peptide to AUA: noninducible, because the ribosome would not be able to stabilize the antiterminator. (e) Combination of the first mutation (a) with each of the other three (b, c, or d). Constitutive, because if the terminator cannot form, it does not matter what happens upstream. For the mutation (a) and (d) combined: pausing at the 1-2 stem-loop is too brief to reduce the rate of transcription significantly. This latter is an "epistasis" test, often used to order the events in a regulatory pathway.

3. In bacterial transcription, the σ factor dissociates during elongation. How would the activity of the polymerase and transcription elongation be affected if the σ factor remained bound to the holoenzyme after initiation?

3. Continued association with σ factor would make RNA polymerase unable to continue down the template. Dissociation of σ factor increases the affinity of the holoenzyme for general DNA, enabling the enzyme to release from the -35 and -10 boxes and move along the template.

3. The mRNA sequence AUGCACAGU codes for the first three amino acids of a particular protein. Which nucleotides can be changed without modifying the amino acid sequence that will result after translation?

3. Normally, AUGCACAGU is translated to produce the tripeptide Met-Thr-Gly. The sixth or ninth nucleotides can be changed without changing the amino acid sequence, as there are four codons for Thr and Gly that only differ at the third position. It would not be possible to change the third nucleotide because there is only one Met codon

3. How are DNA binding proteins able to recognize a specific double-stranded DNA sequence?

3. Specific DNA binding usually involves interaction of the protein with the major groove of the DNA at the site. The edges of the base pairs present different patterns of hydrogen-bond donors and acceptors between the major and minor grooves. In the major groove, all four base pairs have distinct patterns; in the minor groove, AT and TA patterns cannot be distinguished, nor can GC and CG. Hence the major groove is more suitable for sequencespecific interactions. Additionally, protein α helices fit into the major groove, but are too large for the minor groove. This means that some side chains projecting from α helices can make specific contacts with the edges of base pairs in the major groove. Some protein-DNA interactions involve less direct contact, such as a well-positioned water molecule forming hydrogen-bond networks between the protein and the DNA, or even van der Waals interactions, but the latter are usually less specific.

3. What is the nucleophile in the reaction catalyzed by DNA polymerase? What is the significance of this in terms of initiating DNA synthesis?

3. The 3′-hydroxyl of the previously added nucleotide is the nucleophile for DNA polymerase. Initiation requires a primer, a short stretch of RNA produced by primase, to provide the nucleophile.

3. The DNA sequence of a gene follows. If this is the sense strand, write the sequence of the mRNA transcript formed. CGCGGATCCTTGAATTCTAAATAAACCATTT ACCACCATGACC

3. The antisense strand is used as the template for transcription to make a complementary strand. The mRNA sequence would be CGCGGAUCCUUGAAUUCUAAAUAAACC AUUUACCACCAUGACC

3. What is the wobble hypothesis, and how does it explain the degeneracy of the genetic code?

3. The wobble hypothesis states that noncanonical base pairs can exist between the third (3′) position in the codon and the first (5′) position in the anticodon. Wobble base pairing explains how some mRNA codons that differ by the nucleotide at the third position can bind to the same tRNA, facilitated by the presence of modified nucleotides such as inosine at the wobble position. Inosine can form hydrogen bonds with A, C, or U.

What is the correct order of the following steps to synthesize new DNA? 1. Addition of an RNA primer 2. Extension of the RNA primer 3. Conversion of double-stranded DNA to single-stranded DNA 4. Synthesize new DNA Question options: 3; 1; 2; 4 4; 1; 3; 2 2; 3; 4; 1 1; 2; 3; 4

3; 1; 2; 4

4. The activity of regulatory proteins can be modulated in many different ways. Give at least one example of regulation by each mechanism. a. Protein is blocked by an inhibitor. b. Protein requires a binding partner. c. Active protein is unmodified. d. Active protein is a ligand-bound form. e. Active protein is a form not bound to ligand.

4. (a) Yeast GAL4; (b) mammalian Oct4 and Sox2; (c) yeast PHO4; (d) CRP, Trp repressor; (e) lac repressor.

4. What feature of rRNA makes co-transcriptional folding necessary for correct assembly of the ribosome? How do ribosomal proteins contribute to the formation of the conserved secondary structure of rRNA?

4. Co-transcriptional folding is necessary because of the size of the transcripts; folding of the 5′ end as soon as it exits the polymerase decreases the number of structural variations possible making it less likely that the RNA would be trapped in an unfavorable conformation. The secondary structure also generates binding sites for ribosomal proteins, and such interactions during transcription help "lock" the structures into place and protect the RNA.

4. What effect would an absence of Mg2+ have on the activity of DNA polymerase?

4. DNA polymerase has two Mg2+ ions complexed to active site aspartate residues. When a dNTP binds, the Mg2+ withdraws electron density from phosphoryl groups, making the α phosphate more susceptible to nucleophilic attack by the 3′-hydroxyl. Electron density is also withdrawn from the β and γ phosphoryl groups, making the pyrophosphate a better leaving group. Absence of Mg2+ would significantly decrease DNA polymerase activity.

4. What are the primary replicative polymerases in prokaryotic and eukaryotic cells?

4. In prokaryotes: DNA polymerase III; in eukaryotes: DNA polymerase δ (lagging strand synthesis) and DNA polymerase ε (leading strand synthesis).

4. RNaseP can cleave both pre-tRNA and pre-rRNA sequences. If the primary sequences of these RNAs are not always similar, how the does the enzyme recognize and bind its substrates?

4. RNaseP recognizes the secondary structure of tRNA and rRNA rather than the primary structure. Such structures form cotranscriptionally, so the primary transcript is cleaved on the basis of the presence of these structures and not any individual sequences.

5. Bacteriophages are often studied by the following approach. First, cells are mixed with the bacteriophages; after a period of time to allow the virus to bind the cells, the cells are mixed with a small amount of liquid agar, and the mixture is poured onto the surface of a plate containing solid agar. The top agar quickly hardens. Eventually, the cells grow and form a dense layer of cells in the top agar layer. Because the viruses multiply much more rapidly than the cells, however, they eat a hole, or "plaque," in the bacterial layer. Each plaque develops from a single infected cell, so this is a way to count the number of viruses. It is also a way to do genetics, as was done in the early days of studying bacteriophage λ. Plaques of bacteriophage λ differ from those of most viruses. Most viruses kill all the cells in the plaque, so there are no surviving cells and the plaques are "clear"; you can see through them. In contrast, plaques of bacteriophage λ are cloudy or "turbid" because lysogens (bacterial cells in which the virus is in the lysogenic stage) arise in the plaque and can grow. a. In the plaque, lysogens are often infected by other λ viruses present in the plaque. Why don't these newly infecting viruses grow lytically and kill the cells? b. Mutants of λ can arise that form clear plaques. Which virus genes are likely to be mutated?

5. (a) A lysogen in a λ plaque can be superinfected by other λ virions, but the lysogen makes CI, which will bind to the lytic promoters of the incoming viral DNA, repressing them and preventing them from growing lytically. This is termed "immunity." (b) Clear plaques have mutations in the cI gene; they cannot make CI and cannot establish or maintain the lysogenic state. Clear plaques led to the discovery of this gene and two others; hence the name of the gene cI as clear plaque, class I.

5. Why is it important that all tRNA molecules adopt a similar tertiary structure?

5. All tRNA molecules must have similar structure in the region where they bind to elongation factors and the A, P, and E sites on the ribosome. These interactions are not dictated by the particular amino acid charged to the tRNA, so all tRNAs have similar structures.

5. Describe the basic aminoacylation reaction. How does this reaction differ in class I and class II aminoacyl-tRNA synthetases?

5. Aminoacylation is a two-stage reaction: First the amino acid is activated through the formation of an aminoacyl-AMP moiety; second, the amino acid is transferred to the acceptor stem of tRNA. Class I aminoacyl-tRNA synthetases transfer the amino acid to the 2′-hydroxyl on the last conserved adenosine residue in the acceptor stem. Class II enzymes transfer the amino acid to the 3′-hydroxyl of this nucleotide residue.

5. Describe the mechanism by which HIV-1 reverse transcriptase (HIVRT) produces DNA for chromosomal integration.

5. HIVRT uses the single-stranded RNA of the HIV genome as a template to produce a DNA-RNA duplex, using the 3′ end of the host cell Lys-tRNA as a primer. HIVRT has RNase activity and degrades the RNA strand once DNA is produced. RNA fragments serve to prime synthesis of the complementary DNA strand to generate double-stranded DNA.

5. What are positive regulation and negative regulation? Give at least two examples of each for both prokaryotes and eukaryotes.

5. Positive regulation: gene expression is stimulated by the action of regulatory proteins; prokaryotes: cAMP-CRP in lac operon, CI activation of λ cI; eukaryotes: GAL4 activation of GAL genes, Bicoid and Hunchback activation of eve. Negative regulation: gene expression is inhibited by the action of regulatory proteins; prokaryotes: lac repressor in lac operon, Trp repressor in trp operon, LexA in SOS regulon; eukaryotes: TR inhibition in absence of ligand by recruited SMRT complex, HPI inhibition of DNA, Giant and Krüppel inhibition of eve stripe 2.

5. The following sequence is from a region of the M13 bacteriophage genome. Identify and label the promoter elements that would be recognized by the bacterial RNA polymerase. Where would transcription begin? CAGGCGATGATCAAATCTCCGTTGTACTTT GTTTCGCGCGTTGGTATAATCGCTGGGGTCAA GATGAGT

5. The -35 and -10 boxes are underlined. Transcription starts at the ATG after the -10 box (in gray). CAGGCGATGATCAAATCTCCGTTGTACTT TGTTTCGCGCGTTGGTATAATCGCT GGGGTCAAGATGAGT

5. RNAi-based modification of amylopectin content in wheat occurs through decreased expression of starch-branching enzymes. If the siRNA used to knock down plant starch-branching enzymes was transmitted to humans, what enzyme might it affect, and why would this be harmful?

5. The structure of amylopectin is similar to that of glycogen, as both contain an α(1→4) chain of glucose molecules with α(1→6) branches. The spacing between branches varies, so it is reasonable to conclude that starch-branching enzymes and glycogen-branching enzymes that catalyze formation of the same bond are similar. If siRNA against a starch-branching enzyme was transmitted to mammals, it could result in repression of a glycogen-branching enzyme, and the cell would store only straight-chain glycogen, reducing the rate of glycogen breakdown and the capacity to store glycogen. These effects would negatively affect ATP production and could lead to symptoms similar to those observed for glycogen storage disorders

6. Describe the typical major elements of a replication fork.

6. Ahead of the replication fork, topoisomerase adds negative supercoils to relieve the positive supercoiling caused by separation of strands. Helicase and primase are coupled at the front of the replication fork to separate the strands and synthesize RNA primers, respectively. Single-stranded binding proteins maintain single-stranded DNA and protect it from damage. DNA polymerase III, tethered to helicase, consists of a core enzyme and a β clamp, which increases processivity of the polymerase. DNA polymerase I and ligase follow the replication fork to remove RNA primers and seal nicks in the newly formed DNA.

6. Your colleague is studying the regulation of an E. coli gene, and she purifies a protein that stimulates expression of the gene in an in vitro transcription system. In this system, the gene is expressed at a very low level in the absence of this protein and at a high level in its presence. She interprets these data to mean that it is an activator protein. Can you suggest another possibility, along with one or more experiments that would distinguish between the two models?

6. Another possible explanation is that the purified protein is an alternate σ factor. Sigma determines the promoter binding specificity, and there are multiple σ factors. To distinguish these models, the in vitro transcription system could use an RNA polymerase without the usual σ70 factor, called the "core" RNA polymerase. The new-sigma model predicts that the target gene would be transcribed; the activator model predicts no transcription.

6. Many antibacterial agents work by inhibiting RNA polymerase. These agents are typically molecules that interact with various regions of the catalytic center to prevent DNA binding or transcript synthesis. Why are many of these considered "broad-spectrum antibiotics"; that is, antibacterial agents that act against a large number of different bacterial species? Why are these broad-spectrum antibiotics advantageous for both physicians and their patients? These agents, however, do not affect eukaryotic RNA Pol I, II, or III. What does this tell you about the sequence conservation between the bacterial and eukaryotic enzymes?

6. Bacterial RNA polymerase is highly conserved; therefore, an inhibitor of its activity would likely affect all bacteria, meaning that the antibiotic can be used without knowing the species of bacteria causing the infection, an advantage because most bacterial infections require treatment as soon as possible. The structure of RNA polymerase is conserved from bacteria to eukaryotes, but these enzymes do not share a high degree of sequence conservation. The interaction between the antibacterial molecule and RNA polymerase depends on specific base contacts, and differences at the primary sequence level means that the eukaryotic RNA polymerases do not form the same contacts, and therefore the inhibitor has no effect.

6. Why do bacteria contain multiple σ factors?

6. Bacterial σ factors are transcription factors that direct RNA polymerase to specific gene promoters. The holoenzyme can interact with any DNA strand, but σ increases holoenzyme affinity for -10 and -35 boxes. Because bacteria use different genes to respond to changing nutrient and environmental conditions, they have σ factors that preferentially increase transcription at specific sets of these genes as needed.

6. What are the functions of the three tRNA binding sites on the ribosome?

6. The aminoacyl (A) site is the entry point during elongation for aminoacylated tRNAs in the ribosome. The peptidyl (P) site is where peptide bond formation occurs between the tRNA-bound amino acid and the growing polypeptide chain attached to the P-site tRNA. The uncharged tRNA leaves the ribosome from the exit (E) site.

6. Why does the initiator tRNA bind in the P site rather than the A site of the ribosome?

6. The initiator tRNA charged with methionine begins the protein sequence. Because Met is the first amino acid, it must bind directly to the P site so that a second tRNA can bind to the A site and the first peptide bond can be formed.

7. Why do some bacterial mRNA transcripts contain more than one Shine-Dalgarno sequence?

7. A bacterial mRNA may encode more than one polypeptide; each translational start site has a corresponding Shine-Dalgarno sequence.

7. What are the major events that occur at oriC to allow initiation of DNA synthesis?

7. Binding of DnaA begins DNA melting and guides helicase to single-stranded DNA (with the help of DnaC). Gyrase allows further opening of the double helix by relieving torsion created by the single-strand region, and single-stranded DNA binding proteins maintain the single-stranded state. DNA Pol III binds to release DnaA and start the replication process.

7. A nucleosome is positioned as shown in the diagram here, with two cis-acting sites lying on the DNA close to the exit point of the DNA. At a low frequency, the two sites are transiently exposed by dissociation of a segment of DNA from the histones. If the second site (in green) is exposed, the green factor can bind to it, as shown. What effect will this have on the ability of the red trans-acting factor to bind to its site? Can this same effect on binding occur by other mechanisms?

7. Binding of the green factor would enhance the ability of the red factor to bind, as its binding site is now accessible, not bound up in the nucleosome. This is a form of cooperativity, which can also involve protein-protein interactions between the two proteins bound to DNA or in solution.

7. Describe the similarities and differences between snRNA and snoRNA, including cellular location and function.

7. Both are small noncoding RNAs with roles in RNA processing found in protein complexes in which they are critical for activity and for binding to the target RNA. They differ in their cellular location and RNA target. Small nuclear RNAs are part of the snRNP complexes of the spliceosome, located in the nucleus. Each spliceosome subunit (that is, U1, U2, and so forth) is named after the snRNA bound within the complex. The U1 and U2 snRNAs are responsible for binding the intron sequences. The snoRNAs are located in the nucleolus: The two main classes of snoRNAs facilitate base modification of rRNAs and bind to the ribonucleoprotein complexes that assemble on rRNA as it is transcribed.

7. Compare and contrast the action of transcriptional activators in eukaryotes and prokaryotes.

7. In prokaryotes, activators generally interact directly with RNA polymerase, recruiting it to a promoter or increasing the rate of open complex formation. In eukaryotes, activators act on other transcription machinery components, nearly always recruiting proteins to particular genes where they open up the chromatin, or recruit components of the general transcription machinery.

7. Why is Mg2+ essential for the activity of RNA polymerase?

7. There are two Mg2+ ions in the active site of RNA polymerase that interact with the negatively charged oxygens on the α and β phosphoryl groups of the incoming NTP and that lower the pKa of the 3′-OH of the last ribose in the existing transcript. These interactions facilitate nucleophilic attack by positioning the NTP and assisting deprotonation of the 3′-OH.

7. How does translation termination occur?

7. When a termination codon reaches the A site, a release factor binds the A site, preventing further tRNA binding and leading to dissociation of the ribosomal subunits.

8. What is unique about the structure of telomerase that allows it to extend the ends of chromosomes?

8. Human telomerase contains a 451-nucleotide RNA with a tandem RNA at the 3′ end that is complementary to the chromosome end. After association of the RNA with the end of the chromosome, telomerase extends the chromosome using the RNA as template. When the end of the template is reached, telomerase releases from the chromosome.

8. Compare and contrast the action of transcriptional repressors in eukaryotes and prokaryotes.

8. In prokaryotes, repressors affect RNA polymerase directly; usually they block access to the promoter or affect later steps in transcription. Repressors in eukaryotes work indirectly to set up repressive chromatin states. They recruit HDACs to deacetylate histones, or they recruit histone methyltransferases, which in turn recruit HP1

8. Assume that you have identified all the cis-acting sites in a prokaryotic genome for a particular gene regulatory protein. Can you predict the consequences of the regulatory protein binding to these sites? Give your reasons, with examples.

8. No, you cannot be certain what the effect of protein binding will be. Numerous examples indicate that the effect of a bound trans-acting factor depends on its context. Bound bacteriophage λ CI protein and CRP-cAMP both can act as either activators or repressors. The mammalian Oct4-Sox2 heterodimer activates several ES-specific genes and works to repress developmental genes.

8. Why is puromycin not useful as an antibiotic to treat bacterial infections in humans?

8. Puromycin is not a useful antibiotic because it affects both bacterial and eukaryotic protein synthesis. It would lead to cell death in humans as it would in bacteria.

8. Compare and contrast Rho-dependent and Rhoindependent termination. What is a common feature in both mechanisms?

8. Rho-dependent termination involves destabilization of the transcription bubble by Rho binding to a C-rich sequence in mRNA, resulting in transcriptional pausing and termination. Rho-independent termination also destabilizes the interaction between RNA polymerase and the transcript, however structural elements within the mRNA interfere with the transcription bubble: Conserved sequences at the 3′ end of the transcript fold into a hairpin loop leading to release of the polymerase.

8. Open the PDB file 3RTX of mouse guanylyltransferase (Mce1) in a molecular viewer and read the associated paper in Molecular Cell (2011) describing the crystal structure. What role does phosphorylation at Ser5 play in the interaction between mammalian RNA guanylyltransferase and the RNA polymerase II CTD? Is phosphorylation at this residue essential for the interaction? Why is Ser2 phosphorylation not required for efficient capping?

8. The crystal structure of mouse guanylyltransferase (Mce1) bound to an 18-AA CTD phosphopeptide shows that the phosphoryl group bound to Ser5 interacts with Arg386, Arg330, and Lys331. The presence of these amino acid contacts increases the affinity of Mce1 for the RNA polymerase II CTD. Elimination of these contacts would not completely prevent interaction because Mce1 contains additional residues (Cys383, Val372, and Cys375) whose interaction does not depend on the phosphorylation state of Ser5. In the crystal structure, the Ser2 phosphoryl group is pointed away from the binding pocket and cannot contribute to binding. But if pointed toward the binding site, it still would not increase the affinity because interactions between the Ser2 phosphoryl group and Cys375 would replace existing van der Waals interactions between Cys375 and Cα of Ser2 and Cδ of Pro3, not add new ones.

8. An Ames test of a suspected mutagen was examined both before and after incubation with rat liver extract, giving the following results. What can you conclude about the suspected mutagen?

8. The results suggest the suspected mutagen is mutagenic without liver metabolism, and even more mutagenic after metabolism by liver enzymes.

8. When comparing the three phases of translation in prokaryotes and eukaryotes, why is initiation the most different?

8. Translation initiation is different in prokaryotes and eukaryotes because the structure and organization of mRNA is different. Transcription and translation occur simultaneously in prokaryotes, and the mRNA generally does not contain significant secondary structure. Eukaryotic mRNA travels to the cytosol from the nucleus, so it is modified to ensure its stability [5′ cap and 3′ poly(A) tail] and may contain secondary structures. The size of the 5′ untranslated region can vary because there are additional factors in eukaryotic systems that bind the 5′UTR in the preinitiation complex

9. How are proteins directed to the endoplasmic reticulum during protein synthesis?

9. Proteins that contain a specific N-terminal signal sequence are directed to the endoplasmic reticulum during translation, when the N-terminal signal sequence is recognized and bound by the signal recognition particle. Translation halts until the ribosomal complex is associated with the endoplasmic reticulum. Signal recognition particle binding to a receptor on the endoplasmic reticulum membrane allows the nascent polypeptide to enter and translation to resume.

9. TATA-less promoters do not contain a recognizable TATA box but are still associated with TBP. What features are often found in these promoters?

9. TATA-less promoters often contain the downstream promoter element that may interact with TBP.

9. Describe the Ames test and explain its importance.

9. The Ames test uses a mutant strain of Salmonella that cannot produce histidine to test the mutagenicity of various molecules. The test molecule is incubated with Salmonella and liver extract to simulate mammalian metabolism. If colonies are produced on minimal-histidine media, then a back mutation has occurred in the histidine synthesis genes so the pathway is functional. Substances that are mutagenic to bacteria can be identified, which are potentially carcinogenic in humans.

9. What are the two primary roles for activators in eukaryotes? Give several different examples of each

9. The first role is to antagonize repressive chromatin states; for example, by recruiting HATs to acetylate histones and open up chromatin, enabling acetylated histones to recruit more components. Activators can also recruit chromatin remodeling complexes to move or remove nucleosomes, exposing other cisacting sites for additional factors. The second role is to recruit components of transcription machinery, including the large protein complexes mediator and TFIID. Function of these complexes requires the promoter region to be relatively free of nucleosomes.

9. Why must proteins be completely translated before nuclear transport can occur?

9. There are no nuclear transport complexes analogous to the ER signal recognition particle-receptor complex that work cotranslationally.

9. Histone acetylation is generally associated with promoter regions for genes that are actively being transcribed. Suggest several mechanistic reasons why this might be the case.

9. Three models have been suggested for this correlation, which are not mutually exclusive. First, acetylation neutralizes the charge on the lysine side chain, which may weaken interaction of the histone tails with the DNA in a nucleosome. Second, charge neutralization may weaken interaction between nucleosomes, disrupting higher-order structure. Third, acetylated histones serve as docking sites for other proteins and complexes. Although these interactions do not necessarily favor activation, this is found to be true in many cases.

9. The review by Valadkhan [Valadkhan, S. (2010). Role of the snRNAs in the spliceosomal active site. RNA Biology, 7, 345-353] describes the catalytic activity of the spliceosome complex, and Figure 1 therein illustrates interactions between U2 and U6 snRNA. Describe these interactions, and explain why they are necessary for correct positioning of the intron and for catalysis.

9. U2 and U6 snRNA base pair, leaving a bulge in the center of the structure. Residues in this bulge are then free to base pair with the mRNA intron. U2 snRNA binds the conserved RY sequence in the branch point, leaving the catalytic adenine unpaired and extruding from the structure. The 5′ splice site is stabilized through base pairing with the conserved U6 snRNA sequence ACAGAGA, bringing the branch-point adenine in close proximity to the 5′ splice site for the first transesterification

Briefly describe the structure and function of the 1) one key E. coli protein that are each needed for high fidelity termination of DNA replication and 2) one key human protein and the one key. A. 1) Tus protein prevents fork replication past termination, 2) Telomerase prevents lagging strand deletion. B. 1) Rho protein functions in E. coli., 2) HDAC/HAT proteins function in humans to stimulate termination. C. 1) Ter protein prevents fork replication past termination, 2) Telomerase prevents leading strand deletion. D. 1) Telomerase binds to E. coli Tus sequence, 2) Ter protein works in humans to prevent over-replication. E. 1) Primase functions in E. coli, 2) Gyrase functions in humans to prevent early termination.

A. 1) Tus protein prevents fork replication past termination, 2) Telomerase prevents lagging strand deletion.

Which of the following statements is FALSE concerning translation initiation in prokaryotes? A. ATP hydrolysis provides energy necessary for formation of the initiation complex. B. The 70S ribosomal complex is made up of the 30S and the 50S subunits. C. The ribosome binding site is known as the Shine-Dalgarno sequence. D. Three initiation factors (IFs) are necessary.

A. ATP hydrolysis provides energy necessary for formation of the initiation complex.

Francis Crick hypothesized that the Genetic Code must require 3 nucleotides in the mRNA to specify each amino acid based on a simple calculation. What was the logic behind this calculation? A. Based on the number of known amino acids and the number of known nucleotide bases. B. The square root of 16 equals 4 and the square root of 9 equals 3. C. 4 x 4 = 16, whereas 4 x 4 x 3 = 48; since there are 20 amino acids, it must require 3 codons (48 > 16). D. Based on the number of known amino acids and the number of known tRNA molecules. E. Based on the fact that thymine is present in DNA and uracil in RNA; both base pair with adenine.

A. Based on the number of known amino acids and the number of known nucleotide bases.

Use the nine nucleotide DNA base pairs below to determine the mRNA and protein sequences in Q #2 and Q #3. Section of DNA sequence shown is located somewhere in the middle of the gene not at the start site. PROMOTER 5'....... CCA TAC CGG .......3' TERMINATION 3'....... GGT ATG GCC........5' 2. mRNA 5'..... _C_ _C_ __A__ codon 1 __U__ __A__ __C__ codon 2 __C__ __G__ _G__...3' codon 3 The correct order for codons 1, 2, and 3 written 5' to 3' on the mRNA strand is: A. CCA, UAC, CGG B. GGU, AUG, GCC C. GGC, CAU, ACC D. CCG, GUA, UGG E. GGG, CAU, CCC

A. CCA, UAC, CGG

Which of the following statements is FALSE concerning tRNA synthetases? Choose one: A. Each tRNA-amino acid codon combination has its own designated aminoacyl-tRNA synthetase. B. The amino acid is ultimately linked to the 3′ OH of the terminal adenosine of the tRNA. C. The tRNA synthetases use two proofreading steps to ensure accurate charging of tRNAs. D. There are two classes of aminoacyl-tRNA synthetases.

A. Each tRNA-amino acid codon combination has its own designated aminoacyl-tRNA synthetase.

Why does the initiator fMet-tRNAfMet bind to the P site rather than the A site of the ribosome? A. It needs to bind to the P site so AA2-tRNAAA2 can bind to the A site and a peptide bond can be formed. B. It needs to bind to the P site because the fMet chemical structure only fits the P site ribosomal active site. C. It binds to the A site as do all incoming AA-tRNAAA molecules so that a peptide bond can be formed. D. It needs to bind to the P site first and then it moves to the A site so that a peptide bond can be formed. E. It needs to bind to the P site first so that the E site is open for the incoming AA-tRNA^AA molecules.

A. It needs to bind to the P site so AA2-tRNAAA2 can bind to the A site and a peptide bond can be formed.

Which of the following statements is NOT correct about the structure of the nucleosome? Choose one: A. Histone tails are rich in negatively charged amino acids. B. The core is a histone octamer. C. Approximately 147 bp of DNA is part of each nucleosome. D. The histones commonly have posttranslational modifications.

A. Histone tails are rich in negatively charged amino acids.

In the wobble position, inosine binds to adenine, cytosine, and uracil, but not guanine. Why not? A. Inosine and guanine are chemically incompatible to form hydrogen bonds. B. Cytosine is created by chemical modification of adenine. C. Guanine is a pyrimidine, and therefore too large to bind to inosine. D. Guanine is a pyrimidine and inosine is a purine, whereas cytosine is a purine. E. Inosine is created by chemical modification of guanine.

A. Inosine and guanine are chemically incompatible to form hydrogen bonds.

What accounts for the accuracy of DNA synthesis at the replication fork? A. Proofreading function of DNA polymerase that removes incorrect nucleotides in 3' to 5' direction. B. The ability to replicate one DNA strand twice (lagging + lagging) in one round of replication. C. Sieve function in the editing site that removes incorrect amino acids from the replication fork. D. DNA polymerase fidelity is ensured by the G-A and C-T base pairs forming between two DNA strands. E. Mg2+ in the active site provides proofreading functions through ionic bonds with only correct nucleotides

A. Proofreading function of DNA polymerase that removes incorrect nucleotides in 3' to 5' direction.

Observe the nucleic acid to which the Trp repressor protein is bound. Based on your observation of the molecular structure, which statement best describes the interaction between the Trp repressor protein and nucleic acid? Choose one: A. The Trp repressor binds upstream of the first gene of the trp operon, and the protein interaction spans two major grooves and one minor groove along the longitudinal axis of the binding site. B. The Trp repressor binds downstream of the first gene of the trp operon, and the protein interaction spans two major grooves and one minor groove along the longitudinal axis of the binding site. C. The Trp repressor binds upstream of the first gene of the trp operon, and the protein interaction spans two minor grooves and one major groove along the longitudinal axis of the binding site. D. The Trp repressor binds downstream of the first gene of the trp operon, and the protein interaction spans two minor grooves and one major groove along the longitudinal axis of the binding site.

A. The Trp repressor binds upstream of the first gene of the trp operon, and the protein interaction spans two major grooves and one minor groove along the longitudinal axis of the binding site. Correct; the Trp repressor binds upstream of the first gene of the trp operon, and the protein interaction spans two major grooves and one minor groove along the longitudinal axis of the binding site. This best describes the interaction between the Trp repressor protein and nucleic acid.

Which of the following is not a mechanism that enables the binding of a pioneer factor to a cis-acting sequence? Choose one: A. The pioneer factor binds to the DNA with the help of the mediator complex. B. The pioneer factor binds a site in the linker between nucleosomes. C. The pioneer factor binds to DNA that has transiently dissociated from the nucleosome core. D. The pioneer factor binds to DNA that is bound to a nucleosome but still accessible.

A. The pioneer factor binds to the DNA with the help of the mediator complex. Correct; the pioneer factor is a pioneer because it binds first. The mediator complex does not interact until other proteins are already bound to the DNA.

Why is the trp operon regulated in this manner? In other words, why would a mechanism, like the one observed in this molecular structure, be evolutionarily favorable? Choose one: A. The production of the tryptophan biosynthetic enzymes is a waste of metabolic energy if tryptophan is available to the bacterial cell from the environment. B. High concentration of tryptophan is toxic to the cell, so if too much tryptophan is produced, the cell will not survive and will not reproduce. C. Interaction between tryptophan and the Trp repressor protein will cause the activated complex to engage transcription of genes involved in DNA replication. D. All of these reasons describe why the mechanism of transcriptional control, like the one observed in this molecular structure, would be evolutionarily favorable and therefore selected for.

A. The production of the tryptophan biosynthetic enzymes is a waste of metabolic energy if tryptophan is available to the bacterial cell from the environment. Correct; a mechanism of transcriptional control, like the one observed in this molecular structure, would be evolutionarily favorable and therefore selected for because the production of the tryptophan biosynthetic enzymes is a waste of metabolic energy if tryptophan is available to the bacterial cell from the environment. The cell only needs to have access to enough tryptophan to maintain a critical concentration of charged Trp-tRNA that can support protein synthesis.

The figure shows that the distance between a start codon and the Shine-Dalgarno sequence is not the same in every gene. Which of the following hypotheses could explain how a ribosome deals with this to always start translation at the start codon? A. The ribosome moves along the mRNA starting at the Shine-Dalgarno sequence and uses the first start codon it finds to initiate translation, so the length does not matter. B. The amino acids encoded by the mRNA between the start codon and the Shine-Dalgarno sequence are cleaved off the final protein, so the length does not matter. C. The sequence between the start codon and the Shine-Dalgarno sequence forms a stem loop, so the length does not matter. D. The sequence between the start codon and the Shine-Dalgarno sequence gets spliced out, so the length does not matter.

A. The ribosome moves along the mRNA starting at the Shine-Dalgarno sequence and uses the first start codon it finds to initiate translation, so the length does not matter.

A ribozyme is an RNA enzyme that functions as a catalyst to mediate substrate reactions. Which of the following is true regarding ribozyme-mediated cleavage? A. The ribozyme reaction involves a phosphodiester cleavage through acid-base and metal ion catalysis B. Ribozymes decrease the standard change in free energy of the reaction but not the reaction rate. C. The ribozyme reaction involves a transacetylation reaction which removes exons from mRNA. D. Ribozymes protect cells against gene silencing by blocking the ability of Dicer to cleave dsRNA E. Ribozymes are required for use of alternate promoters, which gives rise to multiple proteins.

A. The ribozyme reaction involves a phosphodiester cleavage through acid-base and metal ion catalysis

What is the main purpose of the Ames test? A. To determine if chemicals mutate DNA by quantifying their effect on reverting a His- phenotype. B. It allows researchers to study salmonella digestion of liver cell extracts as a function of mutations. C. To determine if chemicals mutate DNA by quantifying their effect on causing a His- phenotype. D. It allows researchers to find mutations that cause bacteria to die when grown on Histidine plates. E. It allows researchers to study antibiotic resistance as a function of DNA mutations.

A. To determine if chemicals mutate DNA by quantifying their effect on reverting a His- phenotype.

A single strand of RNA could potentially interact with which of the following molecules? Choose one or more: A.metabolic intermediates B.mRNA C.proteins D.DNA

A.metabolic intermediates B.mRNA C.proteins D.DNA

Name two reasons why the regulation mechanisms of trp biosynthetic enzymes in bacterial cells are not found in yeast cells, i.e., how are yeast different? A. 1) hairpin loops do not form in eukaryotic mRNA; 2) bacterial ribosomes are 70S eukaryotic are 80S B. 1) trp is an essential amino acid:no enzymes 2) transcription/translation are uncoupled:no attenuation; C. 1) chromatin packaging of mRNA blocks hairpin loops; 2) trp synthesis in yeast is blocked by glyphosate D. 1) bacterial genes are in operons: not so in yeast; 2) yeast have more than one trp biosynthetic pathway E. 1) trp is a nonessential amino acid:no enzymes; 2) attenuation would not work because of RNA splicing

B. 1) trp is an essential amino acid:no enzymes 2) transcription/translation are uncoupled:no attenuation;

Number the steps from start of RNA synthesis to start of protein synthesis in eukaryotes (a->g). (a)__3__ U1/U2 form a complex at the 5' end of introns (b) _6___Poly(A) polymerase adds ~200 adenine residues (c)__2__Guanine-N7-methyltransferase adds the m7 G cap (d)__5__ CStF protein binds to AAUAAA sequence at 3' end (e)__1__ RNA Pol II CTD is hyperphosphorylated by TFIIH (f)__4___U6/U5/U2 complex cleaves the 3' end of introns (g)__7__mRNA is exported through the nuclear pore complex A. 2, 6, 7, 3, 5, 1, 4 B. 3, 6, 2, 5, 1, 4, 7 C. 7, 3, 6, 1, 5, 2, 4 D. 3, 5, 2, 6, 1, 4, 7 E. 4, 6, 2, 5, 1, 3, 7

B. 3, 6, 2, 5, 1, 4, 7

When the signal recognition particle (SRP) binds to the endoplasmic reticulum (ER) signal peptide sequence on the amino terminal end of the polypeptide in the early stages of translation, the ribosome pauses on the mRNA and stops translation. What is the purpose of this ribosome pausing? A. Permits SRP dephosphorylation and ribosome disassembly. B. Allows time for the SRP-Ribosome complex to bind the SRP receptor C. Permits the translocon protein to be exported to the nucleus. D. The signal sequence gets hyperacetylated and bound by histones. E. The mRNA gets cleaved by an endonuclease and releases the m7 G cap

B. Allows time for the SRP-Ribosome complex to bind the SRP receptor

Which is an effect of acetylation of lysines in the histone tails? Choose one: A. Acetylation recruits proteins containing chromodomains. B. Interactions between DNA and the histone tail are decreased. C. Interactions between nucleosomes is increased. D. Acetylated tails prevent binding of additional transcription factors.

B. Interactions between DNA and the histone tail are decreased

Which is an effect of acetylation of lysines in the histone tails? Choose one: A. Acetylated tails prevent binding of additional transcription factors B. Interactions between DNA and the histone tail are decreased C. Interactions between nucleosomes is increased D. Acetylation recruits proteins containing chromodomains

B. Interactions between DNA and the histone tail are decreased Acetylation of lysine leads to a loss of a positive charge. DNA is negatively charged and interacts strongly with the positively charged lysine. A loss of this positive charge decreases the interactions and weakens the nucleosome. Additionally, interactions between nucleosomes are weakened and other transcription factors are able to bind. Proteins with bromodomains bind to these acetylated histones, while proteins with chromodomains bind to methylated histones.

Which of the following statements is NOT correct concerning lysine methylation? Choose one: A. There is no change in the positive charge on lysine. B. Methylated lysines bind to bromodomains of proteins .C. Methylation lasts longer than acetylation. D. Up to three methyl groups can be added to the lysine amino group.

B. Methylated lysines bind to bromodomains of proteins Methylation affects lysine differently than acetylation. Lysine maintains its positive charge and can accept up to three methyl groups. The number of groups added can affect the function of the modified side chain. Methylation also lasts longer than acetylation. Methylated lysines bind to chromodomains in proteins.

What is the function of the resolvase enzyme in DNA recombination? A. Base excision repair of DNA mutations. B. Repair of Holliday junctions. C. Adenylation of DNA parental strands. D. Methylation of DNA daughter strands. E. Mismatch repair of DNA mutations.

B. Repair of Holliday junctions.

As stated in the text, Marshall Nirenberg and Heinrich Matthaei discovered that addition of rRNA increased the amount of protein synthesized in vitro by a mixture of lysed bacterial cells, which contained all of the biomolecules needed to synthesize proteins. Given what you know about protein synthesis, what can you conclude from this experiment? Choose one: A. rRNA can act as a template for translation. B. The extract contains ribosomal proteins that are not bound to rRNA, so additional ribosomes form. C. rRNA does not require ribosomal proteins for protein synthesis. D. rRNA encodes ribosomal proteins.

B. The extract contains ribosomal proteins that are not bound to rRNA, so additional ribosomes form.

Reconsider the image in Part 1, and determine what would likely happen first during DNA replication if the temperature were elevated by a few degrees Celsius (normal human temperature is 37°C).(Note: This question is not asking about proteins, which might denature at increased temperatures, so focus on the nucleic acid.) Choose one: A. The new polynucleotide of the replicating leading strand will become denatured from the template strand. B. The new polynucleotides of the replicating lagging strand will become denatured from the template strand. C. The new polynucleotides of the replicating lagging strand will become covalently linked to the template strand. D. The new polynucleotide of the replicating leading strand will become covalently linked to the template strand.

B. The new polynucleotides of the replicating lagging strand will become denatured from the template strand. Correct; if the temperature is elevated by a few degrees Celsius, then the weakest of the hydrogen-bonded systems will denature first. In this circumstance, the new polynucleotides of the replicating lagging strand will become denatured from the template strand. These are the Okazaki fragments, and because these are of limited length, they are more susceptible to denaturation compared to longer double-stranded DNA.

What cell is not expected to contain an active telomerase enzyme? Choose one: A. primary spermatocytes within a male testis B. alveolar epithelial cells within deep lung tissue C. B cells that have undergone a clonal selection D. embryonic stem cells within the blastula

B. alveolar epithelial cells within deep lung tissue Correct; an alveolar epithelial cell within deep lung tissue does not need to normally give rise to a significant amount of cellular descendents. Therefore, it is highly unlikely that this cell possesses an active telomerase.

Which of the following components of eukaryotic mRNA are removed by RNA processing? A. poly (A) sequences B. intronic sequences C. the m7 G cap D. exonic sequences E. all hairpin loops

B. intronic sequences

The resolvase active site contains an amino acid with which of the following? Choose one: A. side-chain thiol (sulfhydryl) that reacts with a phosphate group of the substrate B. side-chain hydroxyl that reacts with a phosphate group of the substrate C. side-chain thiol (sulfhydryl) that reacts with a nitrogenous base of the substrate D. side-chain hydroxyl that reacts with a nitrogenous base of the substrate

B. side-chain hydroxyl that reacts with a phosphate group of the substrate Correct; the resolvase active site contains an amino acid with a side-chain hydroxyl that reacts with a phosphate group of the substrate. One strand of duplex is cleaved to yield a 3′-phosphotyrosine and a 5′-hydroxyl, thereby cleaving a phosphodiester bond.

The siRNA that mediate RNAi are a focus of many studies into ways to use gene silencing to treat or prevent disease. One example is the use of siRNA to reduce expression of a host gene required for viral production under normal conditions. This reduction could prevent a virus like HIV from replicating inside cells by inhibiting the production of human proteins that replicate the virus. siRNA has also been explored as a way to reduce the expression of proteins that promote the growth and spread of cancerous cells. While there have been some positive results in experimental studies using siRNA to target the expression of specific genes in humans, many issues still need to be addressed before the use of siRNA to treat disease becomes widespread. What would be the most difficult part of using a specific siRNA to reduce expression of a human gene involved in the development of cancer or whose product allows for the expression of viral-encoded proteins? A.determining the effectiveness of the siRNA in cells B.reducing potentially harmful side effects of siRNA activity C.identifying an siRNA that is effective in targeting a specific mRNA D.delivering siRNA to cells in the body

B.reducing potentially harmful side effects of siRNA activity D.delivering siRNA to cells in the body

Place the following steps in proper order for the translation of a membrane-bound protein.A. GTP binds to SRP.B. Protein synthesis occurs on free ribosome.C. Protein synthesis halts.D. SRP binds to the signal peptide sequence. Question options: A; B; D; C B; D; C; A D; C; B; A A; B; C; D

B; D; C; A

Why do single stranded DNA binding proteins (SSBs) bind to the single stranded DNA after the helicase action to prevent reassociation of the 2 DNA strands?

Because single stranded DNA will quickly form double stranded DNA whenever possible

Why is the dissociation of the enzyme from the DNA molecule a wasteful process?

Because the product of one round of polymerization becomes the substrate for the next round instead, enzyme remains bound to DNA, allowing reaction to proceed

Number the steps from start to finish for prokaryotic mRNA translation (a->e). (a)__4__Translational elongation proceeds in the 3' direction along the mRNA strand (b) __3_ Binding of 50S subunit, GTP hydrolysis, release of IFs, 70S ribosome forms (c)__1__Initiation factors (IFs) bind to the 30S ribosome along with GTP (d)__5__Stop codon enters A site, binding of RF protein, disassembly of ribosome. (e)_2__ mRNA Shine-Dalgarno sequence base-pairs with 16S rRNA in 30S subunit A. 3, 4, 1, 5, 2 B. 4, 1, 3, 5, 2 C. 4, 3, 1, 5, 2 D. 4, 3, 5, 1, 2 E. 4, 3, 1, 2, 5

C. 4, 3, 1, 5, 2

Which of the following does NOT activate the lytic pathway of bacteriophage λ? A. Cro binding to OR3 B. DNA damage in the host C. CI binding to OR1 and OR2 D. Activation of RecA*

C. CI binding to OR1 and OR2 CI cooperatively binding to OR1 and OR2 promotes expression of more CI protein and favors the lysogenic pathway. When Cro is expressed, it shuts off CI expression and favors the lytic pathway. DNA damage to the host turns on the SOS pathway, which activates RecA*. RecA* induces self-cleavage and inactivation of CI, which also favors the lytic pathway.

Match the bacterial DNA replication component on the left with its primary function on the right. (a)__D__Clamp protein (b) _E__ Topoisomerase (c)__B__DNA helicase (d)__F__Single-strand binding protein (e)__C__Primase (f)__A__ DNA A protein A. Unwind the DNA at OriC B. Unwind the DNA double helix C. Synthesize an RNA primer D. Maximize processivity of DNA Pol E. Remove supercoils ahead of fork F. Prevent DNA strand reannealing A. E, D, B, F, C, A B. D, E, C, F, B, A C. D, E, B, F, C, A D. A, E, B, F, C, D E. D, F, B, E, C, A

C. D, E, B, F, C, A

Lynch syndrome is marked by which of the following characteristics? A. Cancer resistance due to autosomal dominant mutations in the enzyme telomerase that prevents the shortening of telomere fibers. B. Higher cancer rates due to X-linked inherited mutations in DNA ligase, which prevents the repair of single strand breaks in DNA. C. Higher cancer rates due to autosomal dominant mutations in genes encoding mismatch repair enzymes hMLH1 and hMSH2. D. Cancer resistance due to autosomal recessive mutations in MutS/MutL, which prevent the deamination of cytosine to uracil. E. The inability to excrete uric acid due to a deficiency in the HGPRT enzyme as a result of X-linked mutations, resulting in severe gout and colon cancer.

C. Higher cancer rates due to autosomal dominant mutations in genes encoding mismatch repair enzymes hMLH1 and hMSH2

Which of the following statements is NOT correct about the structure of the nucleosome? Choose one: A. The histones commonly have posttranslational modifications. B. Approximately 147 bp of DNA is part of each nucleosome. C. Histone tails are rich in negatively charged amino acids. D. The core is a histone octamer.

C. Histone tails are rich in negatively charged amino acids.

Where are pioneer factors least likely to bind? Choose one: A. Cis-acting sites transiently exposed by nucleosome movement B. Cis-acting sites between nucleosomes C. Linker DNA between nucleosomes D. Cis-acting sites on the outside of nucleosomes

C. Linker DNA between nucleosomes Pioneer factors are the first transcription factors to bind to cis-acting sites, such as promoters or enhancers. These regions may be between nucleosomes and already openly exposed. In other cases, these cis-acting sites are transiently exposed by nucleosome movement. These pioneer factors can also bind to cis-acting elements on the outside of nucleosomes and lead to chromatin remodeling.

What is the type of molecular defect that has been linked to the disease retinitis pigmentosa? A. Mutation in the mRNA sequence causing an alternate splicing reaction to occur. B. Mutation in the mRNA pathway owing to deadenylation of poly(A) tail. C. Mutation in a splicing protein causing defects in the spliceosome machinery D. Mutation causing mRNA redirection during nuclear export owing to loss of m7 G cap E. Mutation causing insertion of a lariat structure into the middle of an exon sequence

C. Mutation in a splicing protein causing defects in the spliceosome machinery

Record T (true) or F (false) for statements below regarding protein synthesis. T F a. Prokaryotic and eukaryotic ribosomes both contain RNA that functions as a catalytic enzyme. T F b. Accurate protein synthesis requires tRNA synthetases to link GTP to 3' end of tRNA molecules. T F c. Ribosomal E, P, and A sites align tRNA anti-codons with the corresponding mRNA codons. T F d. Cells contain only ~30 tRNA molecules, of which 3 correspond to termination tRNA molecules. T F e. The wobble position in tRNA is 5' nucleotide in the anti-codon and 3' nucleotide in mRNA codon. A. T, T, T, F, T B. T, F, F, F, T C. T, F, T, F, T D. T, F, T, T, T E. T, F, T, F, F

C. T, F, T, F, T

Which of statement is most correct regarding the "trombone model" of DNA synthesis? A. The Pol III core on the lagging strand stays bound to the DNA and only releases at termination. B. The Pol III core on the lagging strand synthesizes Okazaki fragments in the 3'→5' direction. C. The lagging strand template is released by Pol III when it reaches the 5' end of the RNA primer. D. DNA Pol II mediates leading strand synthesis in the 5'→3' direction and then reverses to 3' to 5'. E. The leading and lagging strands are synthesized by different types of DNA polymerase enzymes.

C. The lagging strand template is released by Pol III when it reaches the 5' end of the RNA primer.

Which of the following is not a mechanism that enables the binding of a pioneer factor to a cis-acting sequence? Choose one: A. The pioneer factor binds a site in the linker between nucleosomes. B. The pioneer factor binds to DNA that is bound to a nucleosome but still accessible. C. The pioneer factor binds to the DNA with the help of the mediator complex. D. The pioneer factor binds to DNA that has transiently dissociated from the nucleosome core.

C. The pioneer factor binds to the DNA with the help of the mediator complex. Correct; the pioneer factor is a pioneer because it binds first. The mediator complex does not interact until other proteins are already bound to the DNA.

Resolvase activity will act upon Holliday junctions and will generate which of the following? A. only nonrecombinant nucleic acid B. only recombinant nucleic acid C. either recombinant or nonrecombinant nucleic acid

C. either recombinant or nonrecombinant nucleic acid

Histone acetylation leads to transcriptional activation. Which of the following is not a mechanism that leads to an increase in transcription? Choose one: A. weakening interactions with other nucleosomes B. weakening electrostatic interactions with the DNA backbone C. facilitating binding of histone deacetylases D. facilitating binding of additional transcription factors

C. facilitating binding of histone deacetylases+ Correct; acetylated histones do bind to histone deacetylases. While histone deacetylases can bind acetylated histones, they do not increase transcriptional activity. Instead, they remove the acetyl group and end up reducing transcription.

Glutamine has hydrogen-bond acceptors and a hydrogen-bond donor in its side chain, allowing it to make bidentate interactions with DNA. A bidentate interaction is one that makes two or more hydrogen bonds with a base or base pair.Where could glutamine make a bidentate interaction with this G-C base pair? A. in the major groove B. in the minor groove C. in the major groove or the minor groove D. spanning both the major groove and the minor groove E. in the major groove or the minor groove, or spanning both the major groove and the minor groove

C. in the major groove or the minor groove

What is the BEST explanation for why there is redundancy in the genetic code? A. tRNA wobble position at 3'-position of tRNA anticodon and 5'-position of mRNA codon allows non canonical base pairing to occur B. Aminoacyl tRNA synthetase charges tRNAs with amino acids with low specificity C. tRNA wobble position at 5'-position of tRNA anticodon and 3'-position of mRNA codon allows non canonical base pairing to occur D. tRNA lacks the high-fidelity 3'→5' exonuclease activity, so reading errors are common. E. Cytosine deamination to uracil is more common at the 3' position on the mRNA codon.

C. tRNA wobble position at 5'-position of tRNA anticodon and 3'-position of mRNA codon allows non canonical base pairing to occu

Which proteins are present in the cells of eve stripe 2? Choose one or more: A.Krüppel B.Giant C.Bicoid D.even-skipped E.Hunchback

C.Bicoid D.even-skipped E.Hunchback Transcription of eve is activated by Bicoid and Hunchback binding to the eve stripe 2 enhancer. These two transcription factors must therefore be present. If Giant and Krüppel were to bind to the eve stripe 2 enhancer, they would repress transcription. These two transcription factors must therefore be absent. Expression leads to formation of even-skipped protein.

If you modified the trp operon leader region to replace the following regions with nonfunctional versions (just one at a time), which versions would prevent termination of translation via attenuation? Choose one or more: A.modifying region 1 B.modifying region 2 C.modifying region 3 D.modifying region 4

C.modifying region 3 D.modifying region 4

Which of the following are examples of transcriptional activators? Choose one or more: A.acetylated histone B.TFIID C.thyroid hormone receptor D.RecA protein E.CRP F.GAL4

C.thyroid hormone receptor E.CRP F.GAL4 Correct; transcriptional activators either recruit chromatin modifiers to genes, or recruit transcriptional machinery to promoters.

1. A scientist wishes to express a eukaryotic protein in bacterial cells. The gene is cloned along with its promoter region and is inserted into a plasmid. After transforming the plasmid into bacterial cells, protein expression is initiated, but no protein is observed after the cells are lysed. Why? How could this problem be fixed?

C1. Eukaryotic and prokaryotic promoters have different sequences for RNA polymerase, and so the bacterial σ factor would most likely not bind the eukaryotic promoter sequence, thus RNA polymerase would not associate with the DNA, preventing transcription and thus protein expression. Cloning the gene behind a bacterial promoter sequence could solve this problem

Which types of posttranslational modifications can be attached to which amino acids on proteins?

Cysteine: farnesyl, geranylgeranyl, palmitoylate asparagine: n-linked glycan glycine: myristoylate

Circle the underlined words describing the process of DNA synthesis and record the number. DNA replication is conservative (1) / semiconservative (2), which requires that helicase (3) / gyrase (4) unwinds the DNA double helix. Next, DNA Pol III (5) / DNA Pol II (6) binds and forms the replisome and moves along the template strand in a 5'--3'direction (7) / 3'--5'direction (8). If a mistake is made on the newly synthesized strand of DNA, DNA Pol III corrects the error in a 5'--3'direction (9) / 3'--5'direction (10). After DNA replication is complete, the double stranded DNA is a left (11) / right (12) -handed DNA helix. A. 1, 3, 5, 8, 10, 12 B. 2, 4, 5, 7, 9, 12 C. 2, 3, 6, 7, 10, 11 D. 2, 3, 5, 7, 10, 12 E. 1, 4, 6, 8, 9, 11

D. 2, 3, 5, 7, 10, 12

Which of the following statements is not a true description of both RNA and protein? Choose one: A. Both can interact with other RNA and protein molecules. B. Both can be found in the nucleus and cytoplasm. C. Both can fold into tertiary structures that are important for correct function of the molecule. D. Both require the same amount of time for synthesis.

D. Both require the same amount of time for synthesis.

Which is NOT a class of nucleosome modification? Choose one: A. Chromatin remodeling B. Histone variant C. Histone tail posttranslational modification D. DNA acetylation

D. DNA acetylation Nucleosome modifications focus on the histone core. The tails of the histones can be posttranslationally modified or a histone could be replaced with a variant that alters its properties. Moving nucleosomes to different locations relative to the DNA sequence or complete removal of a histone can expose different regulatory sites.

Which of the following transcription factors has the highest protein levels in the presence of eve stripe 2 enhancer? Choose one: A. Bicoid B. Giant C. Krüppel D. Hunchback

D. Hunchback The eve stripe 2 enhancer has multiple cis-acting binding sites and at least four transcription factors. The relevant occupancy of these transcription factors on the enhancer affects the expression levels of the enhancer.

Are iPS cells likely to revert back to their previous differentiated state? Choose one: A. Yes, because their epigenetic state never changed .B. Yes, once the added transcription factors are used up. C. No, because pluripotency is irreversible. D. No, because Oct4, Sox2, and Nanog create a positive feedback loop to maintain their expression.

D. No, because Oct4, Sox2, and Nanog create a positive feedback loop to maintain their expression.

All three eukaryotic polymerases contain subunits homologous to subunits that make up the bacterial RNA polymerase (α2ββ′ω). However, the eukaryotic polymerases contain additional subunits that are not homologous to these proteins. Which of the following provides one explanation for the role of these subunits in transcription? A. The eukaryotic RNA polymerases synthesize RNA using a different reaction mechanism that requires these subunits. B. Eukaryotic RNA polymerase complexes are larger and these additional subunits are required only to provide the complex with more stable protein-protein interactions. C. Eukaryotic RNA transcripts are longer, so transcription requires the activities of additional subunits. D. Promoter regions and regulatory elements differ for all three eukaryotic RNA polymerases. Additional subunits are required to facilitate the interaction between the polymerase complex and DNA.

D. Promoter regions and regulatory elements differ for all three eukaryotic RNA polymerases. Additional subunits are required to facilitate the interaction between the polymerase complex and DNA.

How might the even-skipped stripe 3 enhancer differ from the stripe 2 enhancer? Choose one: A. The stripe 2 enhancer would have a binding site for Giant. B. The stripe 3 enhancer causes higher levels of even-skipped expression than the stripe 2 enhancer. C. The stripe 3 enhancer is active a day after the stripe 2 enhancer is active. D. The stripe 3 enhancer would not have a cis-acting site for Krüppel. E. They both direct expression of even-skipped, so they do not differ.

D. The stripe 3 enhancer would not have a cis-acting site for Krüppel Correct; Krüppel is a repressor that is present at high levels in the posterior parts of the embryo. If the stripe 3 had binding sites from Krüppel, it would not direct expression in the stripe 3 region.

Which statement below is most correct regarding the function of a) mRNA and b) rRNA? A. a) mRNA is the major RNA constituent of ribosomes, b) rRNA is an adaptor molecule in protein synthesis B. a) mRNA is the carrier of genetic information, b) rRNA is an adaptor molecule in protein synthesis. C. a) mRNA is the major RNA constituent of ribosomes, b) rRNA is the carrier of genetic information D. a) mRNA is the carrier of genetic information, b) rRNA is the major RNA constituent of ribosomes. E. a) mRNA is the carrier of genetic information, b) rRNA is the major RNA constituent of RNA polymerase

D. a) mRNA is the carrier of genetic information, b) rRNA is the major RNA constituent of ribosomes.

What property of the hammerhead ribozyme allows this molecule to serve in a ribozymal capacity? A. reacting with substrate in a manner such that the ribozyme is never used again B. 15-nucleotide variable sequence that could possibly fold into thousands of base-paired stems C. randomly shaped catalytic core that is formed through interactions between nonconserved nucleotides D. binding of substrate and orienting that substrate in a more reactive conformation

D. binding of substrate and orienting that substrate in a more reactive conformation

Eukaryotic chromosomes present a special challenge for replication in terms of nucleotide erosion over time. This is true because eukaryotic chromosomes are Choose one: A. circular and when mutations are removed, the nucleotide sequence is shortened. B. circular and become catenated once replication is completed. C. linear and when mutations are removed, the nucleotide sequence is shortened. D. linear and there are RNA primers that will be removed at each end.

D. linear and there are RNA primers that will be removed at each end. Correct; eukaryotic chromosomes present a special challenge for replication at the ends of the chromosomes. This is because eukaryotic chromosomes are linear and there are RNA primers that will be degraded at each end. When this takes place during each round of the cell cycle, telomeres shorten.

The trp operon Choose one: A. relies on a Trp activator protein. B. is only regulated at the level of transcriptional elongation. C. contains enzymes that are involved in tryptophan degradation. D. uses attenuation as a regulatory mechanism.

D. uses attenuation as a regulatory mechanism. The trp operon is composed of five genes involved in tryptophan biosynthesis, so the level of tryptophan in a cell affects activation of the operon. Just like the lac operon, the trp operon uses a repressor protein to control transcription initiation. The operon also uses a second mechanism, called attenuation, which regulates transcriptional elongation.

Which of the following are consequences of bee larvae eating royal jelly? Choose one or more: A.More histones are methylated. B.Fewer histones are methylated. C.More DNA is methylated. D.Less DNA is methylated. E.Expression of many other bee genes is increased. F.Expression of many other bee genes is reduced.

D.Less DNA is methylated. E.Expression of many other bee genes is increased. Correct; cytosine methyltransferases do not methylate histones, but instead methylate cytosine bases. Without Dnmt3, less DNA is methylated. Since methylation leads to local chromatin condensation, Dnmt3 normally reduces expression of the genes it methylates. Royal jelly prevents the methylation, and therefore increases gene expression.

Which of the following would cause a genetic mutation? Question options: DNA makes too many copies in the cell. DNA damage is not corrected. Death of a cell occurs. DNA synthesis does not occur.

DNA damage is not corrected.

Order the following events in the life cycle of bacteriophage λ starting with the initial infection of the host cell.

DNA is injected in a linear form DNA circularizes within the cell insertion into the host chromosome Replication of λ DNA by host cell polymerase Correct; DNA is injected in a linear form and immediately circularizes within the cell, using the complementary overhangs on the ends of the DNA to form the circular structure. The ends are then ligated by a host ligase that prepares the DNA for integration into the host chromosome. Once integration has taken place, the bacteriophage λ genetics are replicated along with the host chromosome during normal DNA replication.

Which of statement is most correct regarding characteristics of codons in the Genetic Code? A. Genetic code is 54 codons with 51 corresponding to the 20 amino acids and 3 that are stop codons B. Genetic code is 64 codons with 60 corresponding to the 20 amino acids and 4 that are stop codons C. Genetic code is 54 codons with 50 corresponding to the 20 amino acids and 4 that are stop codons D. Genetic code is 64 codons with 62 corresponding to the 20 amino acids and 2 that are stop codons E. Genetic code is 64 codons with 61 corresponding to the 20 amino acids and 3 that are stop codons

E. Genetic code is 64 codons with 61 corresponding to the 20 amino acids and 3 that are stop codons

What function do HAT and HDAC enzymes perform in the chromatin-modifying process? A. HAT acetylates and represses the gene. HDAC deacetylates and activates the gene. B. HAT acetylates the gene and HDAC deacetylates the gene. Both repress the gene. C. HAT enzymes methylate DNA, HDAC enzymes do not. D. HAT acetylates the gene and HDAC deacetylates the gene. Both activate the gene. E. HAT acetylates and activates the gene. HDAC deacetylates and represses the gene

E. HAT acetylates and activates the gene. HDAC deacetylates and represses the gene

What is the amino acid sequence (start at N-term) encoded by these three mRNA codons? A. Pro-Val-Gln B. Gly-Met-Glu C. Pro-Val-Trp D. Gly-His-Ile E. Pro-Tyr-Arg

E. Pro-Tyr-Arg

What is the difference between models for semiconservative and dispersive DNA replication? A. Semiconservative replication yields two DNA molecules: one that is all template DNA and one that is all new DNA; dispersive replication yields two DNA molecules that are mixed DNA. B. Semiconservative replication yields two DNA molecules; where one is all new DNA strands and the other is mixed DNA strands; dispersive replication yields two DNA molecules that are both new DNA. C. Semiconservative replication yields two DNA molecules where each contains one template and one new DNA strand; dispersive replication yields two DNA molecules with one mixed strand and one new strand. D. Semiconservative replication yields two DNA molecules that are both made of a mixed DNA; dispersive replication yields two new DNA molecules and two old DNA molecules, so four total DNA molecules. E. Semiconservative replication yields two DNA molecules where each contains one template strand and one new DNA strand; dispersive replication yields two DNA molecules that a both mixed DNA.

E. Semiconservative replication yields two DNA molecules where each contains one template strand and one new DNA strand; dispersive replication yields two DNA molecules that a both mixed DNA.

Choose the one correct statement regarding tRNA synthetases below. A. Hydrolysis of GTP drives conformational changes resulting in formation of the tRNA-amino acid complex. B. tRNA synthetase is a ribozyme that uses RNA to catalyze the formation of the tRNA-amino acid complex. C. tRNA synthetases use small RNA molecules to correctly recognize the anticodon of the tRNA molecules. D. The amino acid is linked to the 3'OH of the ATP before formation of the tRNA-amino acid complex. E. The editing site hydrolyzes an amino acid-AMP or a tRNA-amino acid if the amino acid is not correct.

E. The editing site hydrolyzes an amino acid-AMP or a tRNA-amino acid if the amino acid is not correct.

What happens during protein synthesis if a tRNA is charged with the wrong amino acid? A. The protein will not be synthesized because the incorrectly charged tRNA will be stuck in the P site. B. Protein synthesis will continue, however the RNA splicing reaction will be blocked by the amino acid. C. The tRNA will be excluded from the A site of the ribosome but not from the P site. D. The amino acid will be removed by a ribozyme that cleaves tRNA molecules at the CCA end. E. The protein will contain the wrong amino acid in the corresponding codon position.

E. The protein will contain the wrong amino acid in the corresponding codon position.

This DNA binding protein associates with _____ DNA sequence _____ a protein-coding sequence of DNA. If the TATAAA sequence were altered to AGGCTA then that change would likely result in _________ in the rate of transcription at this gene locus

EUKARYOTIC UPSTREAM OF A DECREASE

Gene regulation of the lactose operon in E. coli is controlled by two transcription factors, CRP and the lac repressor. Beta-galatosidase is one product of the lab operon. Why does adding glucose to a bacterial culture containing lactose inhibit beta-galactosidase expression? Choose the ONE best answer. Question options: The presence of lactose is more inhibitory that the presence of glucose. Therefore if lactose is present, beta-galactosidase is not expressed because lactose is tightly bound to the lac repressor, inhibiting expression. When lactose is present, the lac repressor binds cAMP and therefore represses transcription of beta-galactosidase. Beta-galactosidase enzyme expression levels are not actually decrease, but the activity of beta-galactosidase decreases because at high levels of lactose, enzyme activity decreases. Even though lactose is present, there is no beta-galatosidase expression because the high levels of glucose mean that there are low levels of cAMP. If cAMP levels are low, CRP does not bind and activate transcription.

Even though lactose is present, there is no beta-galatosidase expression because the high levels of glucose mean that there are low levels of cAMP. If cAMP levels are low, CRP does not bind and activate transcription.

In an iteration of the Nirenberg-Leder experiment to assign triplet codons to specific amino acids, radioactively labeled aminoacyl-tRNA with the anticodon of CUG was used. The radioactivity was retained on the filter at the end of the experiment. Which mRNA was used in this iteration of the experiment? Question options: CAG GAC CTG GTC

GAC

An exciting potential application for induced pluripotent stem cells (iPS cells) is to replace cells that have been damaged by disease. One of the first human diseases doctors have tried to treat with iPS cells is macular degeneration, an age-related condition that causes death of a group of cells in the retina. Macular degeneration has been considered a useful test case for the use of iPS cells. In fact, protocols for reprogramming iPS cells into retinal pigment epithelial (RPE) cells are well established and the cells can be implanted into the patient's retina without invasive surgery.

Isolate cells from patients skinCause expression of genes required for pluripotencyCause expression of genes required fro RPE differentiationImplant cells into patient's retina Correct; isolated cells must first be converted to iPS cells by making them pluripotent. Then they are differentiated into retinal cells and implanted into the patient to replace the degenerated cells.

Consider the chemical composition of the hammerhead ribozyme. Rotate the molecule and zoom in to observe the atoms as well as the overall chemical structure of each monomer.Which of the following is true of this molecule? It is composed of both RNA and amino acids and the complex does not possess catalytic activity. It is composed entirely of RNA and possesses catalytic activity. It is composed entirely of amino acids and does not possess catalytic activity. It is composed of both RNA and amino acids and the complex possesses catalytic activity. It is composed entirely of amino acids and possesses catalytic activity. It is composed entirely of RNA and does not possess catalytic activity.

It is composed entirely of RNA and possesses catalytic activity.

where does replication initiate?

Origin of replication A DNA sequence at which DNA replication initiates.

6. List at least six different ways in which eukaryotes and prokaryotes differ, with emphasis on issues bearing on transcription and gene regulation.

Prokaryotes: -no nucleus -transcription and translation can be coupled -one RNA polymerase -RNA polymerase binds promoter differently -very little RNA processing -no chromatin -activators and repressors work directly on RNA polymerase -enhancers rare -default state: genes are accessible Eukaryotes: -have nucleus -transcription and translation in separate compartments -3 types of RNA polymerase -transcription factors bind promoters (usually) and recruit RNA polymerase II -extensive RNA processing and splicing -chromatin -activators and repressors work indirectly by effects on chromatin and recruiting other factors -enhancers very common, esp. in higher eukaryotes -default state: are inaccessible due to chromatin

1. What three forms of RNA are found in both prokaryotes and eukaryotes? Why are these three common to all organisms?

R1. They are mRNA, tRNA, and rRNA, essential in all organisms for converting the genetic code into proteins.

: Describe the biochemical mechanism by which RNA splicing occurs precisely at the exon-intron borders considering that a single nucleotide offset by a mistake in splicing will reset the open reading frame register for the encoded protein.

RNA splicing involves small nucleoprotein RNA (snRNP) complexes that contain "guide" RNA molecules and proteins. Sequential loading of spliceosome complexes onto the RNA provide alignment and catalysis to excise lariat structure. U1 and U2 snRNA form specific base pairs with regions of the target intron to ensure specific splicing

Telomerase contains a(n) ____ sequence that binds complementary to and extends the length of the _______ strand after DNA replication.

RNA, template Correct; telomerase contains the RNA sequence 3'-CCCAAUCCC-5' that binds complementary to DNA, which will be extended in length. Correct; telomerase binds complementary to and extends the length of the template strand.

1. What is meant by semiconservative replication? How is this different from conservative or dispersive replication?

Semiconservative replication creates two daughter DNA molecules, each containing one template strand and one newly synthesized strand. Conservative replication would yield one DNA duplex made up of both templates and one made of two newly synthesized strands. Dispersive replication would result in both DNA molecules made of mixed portions of template and newly synthesized DNA, template paired with template, and new DNA paired with new DNA.

Identify one way in which RNA splicing can give rise to multiple mRNA transcripts from a gene containing 3 exons and 2 introns. (check all that apply) Splicing between the 3' end of exon 1 and 5' end of exon 3. Splicing between the 3' end of intron 1 and the 3' end of intron 2 Splicing between 3' end of intron 2 and the 3' end of exon 3. Splicing between the 5' end of exon 2 and the 5' end of exon 3 Splicing between the 5' end of exon 1 and the 3' end of exon 3

Splicing between the 3' end of exon 1 and 5' end of exon 3.

The mRNA codon 5'-UGU-3' corresponds to 5'-TGT-3' in the DNA coding strand. Name each type of mutation - and the functional consequence - of the 3' thymidine deoxynucleotide being mutated to C, A, and G with regard to the encoded protein

The UGU codon corresponds to Cys amino acid. If the 3' deoxynucleotide is mutated to a C, then the codon is UGC and it is a silent mutation (still Cys). If T is mutated to A, then codon is UGA and it is a nonsense mutation (Stop codon). If T is mutated to a G, then codon is Trp and it is a missense mutation (pg 1033)

9. If the sequence 5′-AACGC-3′ were damaged by reactive oxygen species, what would be the most prevalent product, and what would be the result of replication (show both strands after replication)?

The guanine would become 8-hydroxyguanine and tautomer, 8-oxoguanine. These base pair with adenine, resulting in a G-A mismatch that is not likely to be repaired by DNA repair mechanisms. After a second round of replication, the result is a G→C substitution in one of the daughter strands. First round of replication: 5′-AACGC-3′ 5′-AACGC-3′ (8-hydroxyguanine = G): 3′-TTGAG-5′ 3′-TTGCG-5′ Second round of replication: 5′-AACCC-3′ 5′-AACGC-3′ 3′-TTGAG-5′ 3′-TTGCG-5′ 5′-AACGC-3′ 5′-AACGC-3′ 3′-TTGAG-5′ 3′-TTGCG-5′

Termination of DNA synthesis in E. coli and humans differ significantly because of the genomic structures involved, namely a circular genome in bacteria and linear chromosomes in humans. Briefly describe structure and function of the ONE key E. coli protein and the ONE key human protein that are each needed for high fidelity termination of DNA replication.

The key protein in high fidelity termination in E. coli is the Tus protein, which binds to Ter sequences to prevent replication forks from moving through the termination region. The Tus protein blocks replication by forming a tight association with a specific G-C base pair in the Ter sequences. The key protein in human cells is Telomerase, which contains a RNA primer and is needed to extend lagging strand synthesis at the telomeres of linear chromosomes

Considering that binding of the correct tRNA to the aminoacyl-tRNA synthetase is the critical first step (amino acid proofreading occurs after tRNA binding), how is specific tRNA recognition and binding accomplished?

The specificity for tRNA recognition by aminoacyl-tRNA synthetases is not limited to the anticodon sequence as it is insufficient (2 of 3 nucleotides). Instead, a number of modified nucleotide bases scattered throughout the tRNA structure provide a recognition signature, many of which are in the D loop and acceptor stem

Which of the following is correct concerning spliceosome-mediated trans splicing? Question 4 options: The spliceosome mechanism involves two transesterification reactions. The spliceosome is composed only of proteins. The snRNPs involved in splicing are not recycled. The spliceosome mechanism closely resembles the mechanism of group I introns.

The spliceosome mechanism involves two transesterification reactions.

Describe the THREE experimental approaches used to decipher the Genetic Code as illustrated in figures 22.5, 22.6, and 22.7 in the textbook

The three experimental approaches were 1) the Nirenberg-Matthaei experiment in which poly RNA molecules were incubated with each of 20 amino acids in a protein synthesis extract, 2) the Khorana experiment using DNA templates to synthesize RNA molecules of defined sequences that have three open reading frames, and 3) the NirenbergLeder experiment using ribosome complexes and a filter assay to match triplet codons with amino acids

Transcriptional initiation at defined sequences in DNA is required to ensure that the 5' end of the RNA transcript is correct. Describe the two types of experimental approaches that have been used to identify regulatory DNA sequences located near gene start sites.

The two types of experimental approaches to mapping binding sites of transcription factors are 1) DNaseI footprinting, which physical maps protected areas of DNA in a protein binding assay using a "protection" assay with radioactive DNA, and 2) bioinformatic analysis of known promoter regions searching for conserved sequences (pgs 1067-1068)

How could both DNA strands be synthesized simultaneously?

While one strand is synthesized continuously, the other is synthesized in fragments, aka Okazaki fragments continuously synthesized strand = leading strand stand containing Okazaki fragments = lagging strand -synthesized discontinuously

What explains the observation that a single eukaryotic protein coding gene can give rise to multiple different proteins, i.e., why are there only ~25,000 human genes in the genome, but ~150,000 different proteins in the human "proteome?" Choose the TWO best correct answers. Question 5 options: a. Genes can contain more than one polyadenylation site, which alters the 3' of the mRNA transcript and the inclusion/exclusion of exons. b. Some genes are transcribed from both strands of the DNA and this generates isozymes of the same protein. c. mRNA transcripts from the same gene can be differentially spliced to include/exclude exons. d. Serine phosphorylation of a polypeptide changes its amino acid sequence, and thereby generates multiple proteins from the same gene. e. There are ~6 times more proteins than genes because each gene makes 6 different proteins. f. Polypeptide subunits of the same protein complexes are often differentially cleaved by protease enzymes, resulting in loss of some subunits and duplication of others.

a. Genes can contain more than one polyadenylation site, which alters the 3' of the mRNA transcript and the inclusion/exclusion of exons. c. mRNA transcripts from the same gene can be differentially spliced to include/exclude exons.

The Tus-Ter complex terminates E. coli DNA synthesis by ____________. a. blocking the opening of the dna helix at the fork by helicase b. blocking dna methylation on the nascent dna strand c. addition of dNTPs to the lagging dna strand at the fork d. stimulating the dissociation of RNA primase from the lagging strand

a. blocking the opening of the dna helix at the fork by helicase

In the ER, prenylation can occur. Prenylation is the attachment of an isoprenoid group to a _____________ residue via a(n) _____________. a. cysteine thioester b. serine; ester c. threonine; ester d. lysine; amide

a. cysteine thioester

Individuals with mutations of BRCA have increased incidence of cancer because of their___________. a. decreased ability to repair double-strand dna breaks b. increased ability to form pyrimidine dimers c. decreased ability to repair single-strand dna breaks d. increased ability to form methylated guanine

a. decreased ability to repair double-strand dna breaks

Why do helicase and gyrase need to work together to keep the replication fork moving forward? a. helicase unwinds dna and gyrase relieves the torsional strain created by the undwinding b. helicase unwinds the dna in front of the fork and gyrase unwinds the dna behind the fork c. helicase synthesizes dna and gyrase prevents helicase from dissociating from the dna at the fork d. gyrase adds the rna primer and helicase removes the rna primer after the fork passes thru it

a. helicase unwinds dna and gyrase relieves the torsional strain created by the undwinding

Lysogeny by bacteriophages causes what result to the host DNA? a. integration of virus Dna into host dna b. formation of holliday junctions c. increased transcription of virus genes d. production of new bacteriophage particles

a. integration of virus Dna into host dna

The function of the lac operon is to provide a. the enzymes needed to utilize the disaccharide lactose. b. the enzymes needed to produce lactose. c. lactose. d. glucose and galactose to make lactose.

a. the enzymes needed to utilize the disaccharide lactose.

The Ames test is designed to identify mutation rates using a strategy based on ___________. a. the freqency of new mutations that compensate for an existing mutation b. the frequency by which a dna damaging agent blocks histidine biosynthesis c. the frequency by which a carcinogen causes uncontrolled bacterial cell growth d. the frequency that a new mutation blocks amino acid biosynthesis

a. the freqency of new mutations that compensate for an existing mutation

In which direction is mRNA synthesized by prokaryotic and eukaryotic RNA polymerases? a. the mrna is synthesized with 5' to 3' phosphodiester linkages b. the mrna is synthesized in the antiparallel direction c. the mrna is synthesized with 3' to 5' phosphodiester linkages d. it could be 5'-3' or 3'-5' depending on the template or coding strands

a. the mrna is synthesized with 5' to 3' phosphodiester linkages

When do somatic mutations occur? Question options: S phase G1 phase after zygote formation G2 phase

after zygote formation

Which component of the Nirenberg-Leder experiment, that assigned triplet codons to specific amino acids, was radioactively labeled? Question options: mRNA ribosome tRNA aminoacyl-tRNA

aminoacyl-tRNA

What role do BRCA1 and BRCA2 play in DNA repair? Question options: repair pyrimidine dimers synthesize new DNA bases find the double-strand break assist with strand invasion

assist with strand invasion

Which of the following statements best describes how the terms trans-acting and cis-acting apply to gene regulation? a. Cis-acting factors can bind to specific DNA sequences whereas trans-acting sites are DNA sequences. b. Trans-acting factors are proteins that bind to specific DNA sequences, which are themselves cis-acting sites in DNA. b. Trans- and cis-acting factors can both only bind to DNA elements to which they are physically linked. d. Trans- and cis-acting factors can both bind to specific DNA sequences.

b. Trans-acting factors are proteins that bind to specific DNA sequences, which are themselves cis-acting sites in DNA.

A Holliday junction can best be defined as a region of ________________. a. a pyrimidine dimer b. a quadruplex dna structure c. a region of dna damage d. a duplex dna structure

b. a quadruplex dna structure

How do bacterial sigma factor proteins promote RNA synthesis? a. they function as rna primase proteins that synthesize rna primers b. they recruit rna polyerase to specific sequences at the 5' end of genes c. they bind to rna polymerase and alter its conformation by phosphorylation d. they bind to dna and induce structural changes that open the dna helix

b. they recruit rna polyerase to specific sequences at the 5' end of genes

What is the mechanism by which replication errors are fixed in E. coli? Question options: degrading of the entire DNA strand to start over base excision mismatch repair abasic repair

base excision

How is the Ames test used to quantitate mutagenic potential of chemical compounds? Select the TWO best answers. Question options: a. Bruce Ames invented the Ames test to determine which chemical compounds cause human cancer, however bacterial cells are not liver cells and so it is not very useful to prevent gout. b. Histone deficient plates are used to calculate the background mutation rate, which is the control plate for the filter disk experiment. The more colonies growing on the plate, the better the mutagen. c. The chemical compound to be tested is added to a filter paper disk and placed on agar plates containing the tester Salmonella strain; potent mutagens result in MORE colonies forming on histidine free plates as compared to control plates. d. The Ames test uses filter disks containing chemical compounds to quantitate the level at which the compound is lethal to the Salmonella tester strain; the FEWER colonies on the plate, the more toxic is the chemical regardless of mutagenic potential. e. The ratio of bacterial colonies to drug concentration (bc:dc) is used to calculate the mutagenic index of a chemical compound; high bc:dc ratios means the drug functions as a growth factor for the cells. f. The Salmonella test strain used in the Ames test contains a mutation in the histidine biosynthetic pathway, which is used to identify compounds that induce back mutations to permit bacterial growth on histidine free plates. g. The Salmonella tester strain is grown in histidine containing media since it lacks the ability to synthesize its own histidine for protein synthesis; chemical compounds that mimic histidine lead to MORE bacterial growth on the plates.

c. The chemical compound to be tested is added to a filter paper disk and placed on agar plates containing the tester Salmonella strain; potent mutagens result in MORE colonies forming on histidine free plates as compared to control plates. f. The Salmonella test strain used in the Ames test contains a mutation in the histidine biosynthetic pathway, which is used to identify compounds that induce back mutations to permit bacterial growth on histidine free plates.

How are RNA tertiary structures different from protein tertiary structures? a. protein structures are hydrophobic and only found in membranes, RNA structures are soluble and always only found in the nucleus b. rna structures are more chemically complex than protein structures c. rna adopts multiple structures,whereas protein structures are restricted d. protein structures have covalent bonds, whereas rna structures have noncovalent bonds

c. rna adopts multiple structures,whereas protein structures are restricted

What step in the process of translation of a nascent protein into the lumen of the ER requires hydrolysis of GTP? a. binding of SRP to the SRP receptor b. binding of SRP to the signal peptide sequence c. transfer of the signal peptide sequence into the translocon d. release of SRP from the ribosome

c. transfer of the signal peptide sequence into the translocon

If 15N DNA (heavy) replicated using conservative replication in 14N (light) media, the outcome would be different than semi-conservative replication, and in this case, the two daughter DNA double strands would be ______________after one generation. a. light for both daughter dna products b. both daughter dna molecules would have one heavy and one light dna strand c. heavy for both daughter dna products d. heavy only for one daughter dna and light only for the other daughter dna

d. heavy only for one daughter dna and light only for the other daughter dna

A common feature of protein synthesis in both eukaryotes and prokaryotes is _____________. a. the E site that can bind a tRNA covalently bound to the new protein b. hydrolysis of ATP to facilitate translocation of the ribosome c. the P site in the ribosome can bind an uncharged tRNA d. hydrolysis of GTP to facilitate binding of the AA-tRNA

d. hydrolysis of GTP to facilitate binding of the AA-tRNA

A common chemical modification found in RNA is __________________. a. carboxylation of trna in the region of the anti-codon b. phophorylation of the C-terminal domain of rRNA c. alkylation of mRNA at the intron-exon border d. methylation of nucleotide bases in tRNA

d. methylation of nucleotide bases in tRNA

The spliceosome ___________. a. performs the same function in prokaryotes and eukaryotes b. is a cis acting ribozyme that autocleaves mrna c. is a protein catalyst that functions to ligate exons to introns d. resembles group II introns in its mechanism and product

d. resembles group II introns in its mechanism and product

Long noncoding RNA (lncRNA) is generated from _____________. a. viral rfna that did not integrate into host dna b. telomere sequences at the ends of chromosomes c. intronic sequences spliced out of mrna d. transcription of genomic dna

d. transcription of genomic dna

iPS cells can be defined as __________ cells. Question options: somatic stem dedifferentiated embryonic stem differentiated

dedifferentiated

If a protein is covalently modified by ubiquitin, it will be: targeted to the nucleus. anchored to the plasma membrane. degraded. secreted.

degraded.

A common RNA base modification is Question options: hydroxylation. methylation. amination. carboxylation.

methylation.

Describe the structure and function of miRNA, snRNA, and lncRNA, which represent the three major types of noncoding RNAs; what are four mechanisms by which lncRNA molecules are thought to regulate cellular processes?

miRNA is ~18-32 nucleotides short noncoding RNA that functions in translational regulation; snRNA ~70- 200 nucleotide small noncoding RNA that functions in RNA splicing, and lncRNA is ~200-1000 nucleotide long noncoding RNA is that functions in gene expression (pg 1057). The four mechanisms by which lncRNA is thought to regulate cellular processes are 1) binding to mRNA to regulate protein synthesis, 2) binding to DNA to regulate transcription, 3) binding to proteins to regulate protein functions, and 4) ligand-induced riboswitches to regulate signaling pathways

What is the function of the MutS-MutL-MutH protein complex? Question options: DNA adenylation DNA methylation resolution of Holliday junctions. base excision mismatch DNA repair

mismatch DNA repair

Together, the eight histone molecules are called the histone Question options: octamer. octane. dimer. tetramer.

octamer.

If 15N DNA replicated using conservative replication in 14N media, the outcome would be that new DNA has Question options: only the higher density. one high density and one low density strand. a lower density. a single intermediate density.

one high density and one low density strand.

A protein is targeted to the plasma membrane. Its final functional location is as a subunit of a transmembrane protein that interacts with the central subdomain of the membrane. Analysis of this protein would most likely show modification with a Question options: isoprenoid. palmitoylate. phosphate. myristoylate.

palmitoylate.

What is the function of telomerase in termination of DNA synthesis? Question options: remove the telomeres reverse transcription of the telomeres bind to single-strand DNA to prevent refolding shorten the DNA strand after each replication

reverse transcription of the telomeres

Which type of RNA facilitates RNA interference by resulting in degraded mRNA? Question options: snoRNA siRNA miRNA rRNA

siRNA

What are the three molecular processes that are required to convert a skin cell into an induced pluripotent stem cell after introduction of the four transcription factor genes? (check all that apply) replication of DNA squences encoding c-Myc, Klf4, Sox2, and Oct4 genes formation of the c-Myc/Klf4/Sox2/Oct4 autoregulatory complex activation of the Nanog gene promoter by the Sox2/Oct4 complex converting iPS cells into differentiated skins cells to then make muscle cells the Sox2/Oct4/Nanog complex activates expression of stem cell genes the Sox2/Oct4/Nanog complex inhibits expression of cell differentiating genes formation of tumor cells in skin biopsies of the patient, which are then removed the Sox2/Oct4/Nanog complex activates a positive autoregulatory loop

the Sox2/Oct4/Nanog complex activates a positive autoregulatory loop the Sox2/Oct4/Nanog complex activates expression of stem cell genes the Sox2/Oct4/Nanog complex inhibits expression of cell differentiating genes

Long noncoding RNA (lncRNA) is generated from Question options: an infection of viral RNA. excision of intron lariat segments. the splicing of used mRNA. the tips of chromosomes. the transcripti.on of genomic DNA

the transcripti.on of genomic DNA