Cell Bio - Smartwork for Test 2

Auxin is a plant hormone that regulates cell division and stem and root growth. Arabidopsis thaliana is a small plant used as a model plant organism. A genetic screen in Arabidopsis identified plants that were resistant to auxin and no longer responded to auxin treatment. When the mutant gene was identified, it coded for a tRNA with a mutation in the anticodon. The normal anticodon 5′UGC3′ was mutated to 5′UAC3′. Consider that the tRNA would still carry the same amino acid but would bind the mRNA differently. How would this mutation alter the translation of genes?

Alanine would be added in place of valine.

Condensins are used to condense chromatin in the early stages of mitosis. Their localization within the cell may help regulate their function. Condensin I is found in the cytoplasm and condensin II is found in the nucleus. Suppose the cell cycle is arrested in late prophase before the nuclear envelope is broken down in prometaphase. What will be the state of chromatin in the arrested cell?

Condensin II will have formed large loops of chromatin but there would be no loops within loops.

Another step in PCR requires small single-stranded DNA primers to anneal to a target sequence on denatured DNA. Two primers are used. Ideally these primers will have similar melting temperatures. The melting temperature is defined as the temperature at which 50% of the DNA is in the single-stranded form. Which of the following primers will have the highest melting temperature?

G G G G A A A T T T C C C C

Many antibiotics work by inhibiting bacterial protein synthesis. Investigators have isolated a promising new compound and wish to determine its mechanism of action. Using a cell-free translation system similar to the ones originally used to deduce the genetic code, the researchers incubate their drug with the synthetic polynucleotide 5'-AUGUUUUUUUUU. In the absence of the drug, this polynucleotide directs the synthesis of the peptide Met-Phe-Phe-Phe. When the drug is added, only the peptide Met-Phe is produced. Based on this observation, which is most likely the mechanism of action of this potential new antibiotic?

It blocks translocation of the large ribosomal subunit, preventing the movement of peptidyl-tRNA from the A site to the P site of the ribosome.

Identify which statement(s) is/are true regarding telomere characteristics.

Only statements C and D are true regarding telomere characteristics. Telomeres cap the ends of linear chromosomes and prevent them from being recognized by the cell as broken DNA in need of repair. Telomeres contain repeated nucleotide sequences that are required to replicate the ends of linear chromosomes.

Suppose the 3' splice site is mutated from AGG to ACG. Predict the consequence of this mutation.

Splicing would begin but not be properly completed if the 3' splice site is altered.

Suppose a mutant RNA polymerase II is used that lacks one of the phosphorylation sites on the C-terminal tail. The mRNA synthesized using the mutant RNA polymerase lacks a poly-A tail. Inside cells, no protein is synthesized from this mRNA. How could this be explained?

The mRNA is rapidly degraded. The mRNA is not transported out of the nucleus.

Once heterochromatin has been established, it will often spread until it encounters which of the following?

a barrier DNA sequence

The response to chaetocin is partially mediated by changes in the p15INK4B gene expression. p15 is a tumor suppressor that regulates the cell cycle, and its expression is commonly altered in cancer contributing to tumor growth. Which of the following graphs shows the expected changes to p15 expression after chaetocin treatment inhibits SUV39H1?

graph C where as chaetocin increases , p15 increases

In the absence of repair, what would the replication of a double helix containing a mismatch yield?

one DNA molecule with the normal sequence and one DNA molecule with a mutated sequence

The regulation of chromatin is involved in inflammation. Chronic low levels of inflammation can contribute to cardiovascular disease, cancer, diabetes, and other conditions. Suppose you want to develop a new drug that could lessen inflammation by blocking expression of the inflammatory gene IL-6. You have identified a candidate drug that alters the expression of IL-6. The following northern blot shows the mRNA levels of IL-6 in untreated cells (0 u) and cells treated with increasing units of the drug (1 u, 3 u, and 10 u). Units (u) of drug added are shown on the figure. Actin expression is shown as a control. Does your new drug increase or decrease expression of IL-6? Which of the following mechanisms of action could have led to the results seen in the northern blot?

- Decreases IL-6 - Deactivation of a chromatin-remodeling complex - Deactivation of a histone acetyltransferase enzyme

Place the following steps of RNA splicing in the correct order from first step to last.

1. BBP and U2AF bind to the A in the branch-point site in the intron, while the U1 snRNP binds the 5' splice site 2. the U2 snRNP binds the branch-point site and the U4/U6 and U5 snRNPs bind to the 5' splice site 3. a conserved adenine in the branch-point site attacks the 5' splice site, cutting the RNA at the 5' splice site 4. the 5' end of the intron binds the branch-point adenine, forming a lariat 5. the 3' end of the first exon interacts with the 3' splice site forming a new bond in the sugar-phosphate backbone 6. the lariat is released and degraded

Modification of chromatin can lead to increases or decreases of local or global gene expression. Choose whether each of the following modifications increase transcription, decrease transcription, or could increase or decrease transcription depending on the context. 1. Chromatin-remodeling complex uses ATP energy to alter chromatin. 2. Histone acetyltransferase adds acetyl groups to histone tails. 3. Histone deacetylase removes acetyl groups from histone tails. 4. Methyltransferases add methyl groups to histone tails.

1. increase Or decrease transcription 2. increase transcription 3. decrease transcription 4. increase OR decrease transcription

A primary transcript (immature, non-processed) single-stranded RNA molecule has the following nucleotide composition: 30% A, 20% G, 24% C, and 26% U. What is the nucleotide composition of the double-stranded DNA molecule from which it was transcribed?

28% A, 22% G, 22% C, and 28% T

In 1952, Alfred Hershey and Martha Chase, working with a virus called T2, conducted what is now considered a landmark experiment to determine whether genes are made of DNA or protein. When this virus, which is made entirely of DNA and protein, infects E. coli, it injects its genetic material into the bacterial cell, leaving the empty virus head stuck to the cell surface. To determine whether it was DNA or protein that enters the infected bacterial cell, the researchers radioactively labeled one batch of T2 with the isotope 35S, which resulted in radiolabeled viral proteins, and a second batch of T2 with 32P, which resulted in radiolabeled DNA. They then incubated the radioactive viruses with E. coli. After allowing a few minutes for the viruses to transfer their genetic material to the bacterial cells, the researchers used a blender to shear the empty virus heads from the bacterial cell surface. They then used a centrifuge to separate the infected bacteria from the empty virus heads: spinning the sample at high speed caused the heavier, infected bacteria to pellet at the bottom of the centrifuge tube, while the lighter, empty virus heads remained in solution. Using this protocol, what

32P in the pellet, 35S in the solution Correct. Their hypothesis was that either DNA or protein was injected during infection. Because DNA is the genetic material and the viruses inject their genetic material into the bacterial cells, after infection most of the DNA should have been in the bacterial cells in the pellet. Thus, the 32P that was used to label the DNA should have been in the pellet. The head of the virus is composed mainly of protein, so the 35S should have remained with the empty virus heads in the solution.

The nucleotide sequence of one DNA strand in a DNA double helix is 5'-CATTGCCAGAAAAAT-3'. What is the sequence of the complementary strand produced during replication?

5'-ATTTTTCTGGCAATG-3'

The sequence of the template strand of a DNA molecule is 5'-ACTGGCAATG-3'. What is the sequence of the RNA transcribed from this DNA?

5'-CAUUGCCAGU-3'

The sequence of the coding strand of a DNA molecule (that is, the DNA strand that contains the codons specifying the protein sequence) is 5'-CGGATGCTTA-3'. What is the sequence of the RNA made from this DNA?

5'-CGGAUGCUUA-3'

Synthetic biologists are trying to create cells from raw material. One step in the process is encapsulating genetic material into a compartment. Researchers of the origin of life think that the earliest cells on Earth may have used RNA as their genetic material instead of DNA. As biologists consider which genetic material to use in creating their synthetic cells, which of the following characteristics of a single-stranded RNA genome should they keep in mind?

A newly synthesized RNA strand is not identical to the template strand.

What evidence suggests that the large amount of excess "junk" DNA in a genome may serve an important function?

A portion of "junk" DNA is highly conserved in its DNA sequence among many different eukaryotic species.

Consider the following image of multiple genes (a and b) on a chromosome and then answer the question. Based on the figure, which of these statements is/are correct?

A. For different genes, opposite strands of DNA can serve as a template. B. RNA is always polymerized in the 5'-to-3' direction.

You have joined a lab studying DNA replication in E. coli. The graduate student you are working with has identified a mutation in primase that makes primase very inefficient. Your project is to characterize the cells with this mutation. Predict the defects you would most likely see in the mutant E. coli cells.

A. a delay in DNA polymerase beginning synthesis B. a longer total time to replicate DNA

Acyclovir is an antiviral drug used to treat shingles, chickenpox, cold sores, and genital herpes caused by the herpes family of viruses. Acyclovir treatment prevents viral replication to treat the infection. Acyclovir is phosphorylated by the viral thymidine kinase and host cell kinases to form the active drug acyclovir triphosphate. Compare the structure of the normal nucleotide deoxyguanosine triphosphate and acyclovir triphosphate. What is the most likely mechanism of action for how acyclovir limits viral replication?

Acyclovir blocks further nucleotide addition by DNA polymerase thereby inhibiting DNA replication.

The SMC ring complexes, including cohesin and condensin I and II, help organize the structure of interphase and mitotic chromosomes. Imagine that you are studying the function of these complexes in regulating chromatin structure by using microscopy to visualize the formation of DNA loops in in vitro reactions. Your reactions contain a DNA template, ATP, sequence-specific clamp proteins, and extracts from interphase or mitotic cells. You then add either cohesin or condensin to your reactions. You can measure the loops forming in the microscope. You then make each of the following alterations to your reaction. Match each of the alterations to the reaction with your hypothesis of how DNA loop formation would be affected by the alteration. (Note: Each answer will only be used once.) 1. Add a nonhydrolyzable analog of ATP <_________> 2. Add more binding sites for sequence-specific clamp proteins to the DNA template. <_________> 3. Add cohesin with a defect in the ATP binding domain. <____________> 4. Add condensin with a defect in the ATP binding domain. <____________> 5. Add sequence-specific clamp proteins with mutations so they no longer bind DNA. <____________>

Add a nonhydrolyzable analog of ATP. >> block DNA loop information 2. Add more binding sites for sequence-specific clamp proteins to the DNA template. >> smaller DNA loops form 3. Add cohesin with a defect in the ATP binding domain. >> DNA loops do not form in interphase. 4. Add condensin with a defect in the ATP binding domain >> DNA loops do not form in mitotic extract. 5. Add sequence-specific clamp proteins with mutations so they no longer bind DNA. >> Larger DNA loops form

In the original Meselson-Stahl experiment shown above, bacteria were cultured in 15N-containing "heavy" medium and then transferred to 14N-containing "light" medium for one cell division. The results of this experiment ruled out the "conservative" model of DNA replication, but to distinguish between the "semiconservative" model and the "dispersive" model, the DNA had to be heated to separate the two strands. You are repeating the experiment, but are unable to heat the DNA to perform the second part. However, you realize that allowing the bacteria to undergo a second round of division would also distinguish between the different models. Which of the following would you expect to see as a result after the second division in "light" medium?

After the second division, one population of DNA would run at the same position as DNA from bacteria grown fully in "light" medium, and one population would run at a position halfway between fully "heavy" and fully "light" DNA.

The following is a portion of the DNA sequence from the beginning of a gene that codes for a protein. The bottom strand is the coding strand. What is the sequence of the mRNA produced from this region of the gene? 5′ CCTATGTACTTCCAGGTACATCGC 3′ 3′ GGATACATGAAGCTCCATGTAGCG 5′ Using the mRNA from Part 1, what is the sequence of the beginning of this protein? Write your answer using the one letter code for amino acids. (For example: THNP).

B. 5′ GCGAUGUACCUCGAAGUACAUAGG 3′ MYLEVHR

The thermophilic bacterium Thermus aquaticus (T. aquaticus), originally isolated from hot springs in Yellowstone National Park, has a DNA polymerase (Taq polymerase) that lacks the 3'-to-5' proofreading function found in most DNA polymerases. T. aquaticus has a chromosome that is approximately 2 Mb in size. Assuming that the error rate of nucleotide misincorporation of Taq polymerase is similar to other bacterial DNA polymerases, and the mismatch repair rate is also comparable to other bacteria as shown in the table above, which of the following statements most accurately describes DNA replication in T. aquaticus?

C. On average, there will be a point mutation introduced to the T. aquaticus chromosome approximately once every 5 replication cycles.

During DNA replication, an incorrect nucleotide is incorporated into the newly synthesized strand of DNA approximately 1 in 107 nucleotides. In order for this error to be corrected, the DNA mismatch repair machinery must be able to distinguish between the parental strand with the correct DNA sequence and the daughter strand with the incorrect DNA sequence. Which of the following would represent a possible way to determine which strand is which?

C. The parental strand will be continuous (unbroken), while the daughter strand will have gaps and/or nicks, allowing the repair machinery to determine which is which.

You are working in a synthetic biology lab, developing new and unique biological systems. You are designing a DNA polymerase that is able to synthesize DNA in the 3'-to-5' direction, and you want to incorporate proofreading activity in this enzyme. You will need several different types of enzymatic activities in your polymerase to be able to accomplish this. Aside from the actual 5'-to-3' proofreading exonuclease, which of the following activities might allow for the design of a proofreading 3'-to-5' DNA polymerase?

DNA ligase activity

Duplicated chromosomes are separated during mitosis of the cell cycle. Which of the following chromosomal alterations would prevent proper chromosome segregation during mitosis?

Deletion of the centromere sequence

In their 1953 paper on the double-helical structure of DNA, Watson and Crick famously wrote: "It has not escaped our notice that the specific pairing we have postulated immediately suggests a possible copying mechanism for the genetic material." What did they mean?

Each strand in a DNA double helix contains all the information needed to produce a complementary partner strand. Correct. Because each DNA strand in a double helix contains a sequence of nucleotides that is complementary to the sequence of its partner strand, each strand can serve as a template to direct the synthesis of a new strand identical in sequence to its former partner.

If DNA replication were conservative (although we know it is not), what would Meselson and Stahl have seen following the first round of replication in E. coli that had been switched from a heavy (15N-containing) nutrient medium to a light (14N-containing) nutrient medium?

Half the DNA would be heavy, and the other half would be light.

AMP-PNP is a non-hydrolyzable analog of ATP that can bind to proteins in a similar manner as ATP but is no longer hydrolyzed. Predict what would happen to helicase activity if AMP-PNP were added to a DNA replication reaction.

Helicase would no longer function, since the AMP-PNP is not hydrolyzed. ATP binding and hydrolysis induce the conformational changes that facilitate DNA unwinding by helicase.

Which statement is true about the association of histone proteins and DNA?

Histone proteins have a high proportion of positively charged amino acids, which bind tightly to the negatively charged DNA backbone.

ATPγS is a non-hydrolyzable analog of ATP that can be used to determine if a process is dependent on ATP. You are studying the ATP dependence of DNA replication and treat cultured cells with ATPγS. Sort the following enzymes into those that would be inhibited by ATPγS treatment and those that would be unaffected.

Inhibited by ATPγS: - Clamp loader - Primase - DNA helicase - DNA ligase Unaffected by ATPγS: - Telomerase - Topoisomerase - DNA polymerase

The animated figure shows the normal splicing process used to remove introns from eukaryotic mRNA. The XPC gene codes for a protein that functions in DNA repair. This gene contains 16 exons with the first 5 shown in Figure A. Defects in splicing can lead to human disease. Individuals with xeroderma pigmentosum have high levels of skin cancer due to defects in the nucleotide excision DNA repair process because of nonfunctional XPC protein. DNA damage caused by exposure to the UV rays in sunlight is not repaired leading to mutations in critical genes that control the cell cycle leading to high rates of skin cancer. Several members of a particular family were found to have a mutation in intron 3 of the XPC DNA repair gene (Figure B). The mutation disrupted normal splicing leading to skipping of exon 4 and little to no functional XPC protein in their cells. Figure A. The pre-mRNA transcript spanning the first 5 exons of the XPC gene is shown with introns (yellow) between the exons. Splicing removes the introns, generating the functional mRNA. Figure B. The intron 3 mutation leads to skipping of exon 4 and nonfunctional XPC protein. Studying the gene furtherin vitroshowed that the U2 snRNP

Mutation of lariat branch point

Everyone is exposed regularly to ionizing radiation found in the soil, water, and air and from cosmic rays. In fact, 80% of the ionizing radiation people are exposed to comes from naturally occurring sources. Ionizing radiation can cause double-strand breaks in the DNA. Often, the DNA breaks have missing nucleotides at the broken ends. What type of repair would likely be used, and what would be the result of repairing this type of damage?

Nonhomologous end joining would be used to join the DNA, but errors would still remain.

Which of the following statements about nucleosomes is false?

Nucleosomes are found only in mitotic chromosomes.

In the late 1950s and early 1960s, Leonard Hayflick performed experiments demonstrating that vertebrate cells had limited proliferative capacity, and most cells were limited to approximately 50 rounds of division. This limit, called the "Hayflick limit," is explained by the fact that most normal body cells do not express the enzyme telomerase. With each cell division, the telomere shortens, and when it reaches a critical length, cells can no longer divide. Only cells that are able to express telomerase could continue to divide beyond the Hayflick limit. Which of the following statements is a conclusion that follows from the telomerase explanation for the Hayflick limit?

One change that is likely to occur when cells become cancer cells is the re-expression of telomerase.

In vertebrate cells, ~5% of cytidine in DNA is methylated to produce 5-methylcytidine. This mainly occurs in the promoter regions of genes, and is used to silence transcription. Like cytidine, 5-methylcytidine can undergo spontaneous deamination. Deamination of cytidine produces uridine, while deamination of 5-methylcytidine produces thymidine, both of which would be mismatched with the guanosine that would have been paired with the cytidine/5-methylcytidine. Based on this, which of the following statements would most likely be correct?

Over evolutionary time, one would expect to see a high rate of C-to-T mutations in the promoter regions of vertebrate genes.

Rifampin, along with several other antibiotics, is administered to people with tuberculosis caused by Mycobacterium tuberculosis infection. The following figure shows a transcription reaction containing a DNA template, sigma, and RNA polymerase from a Mycobacterium species. Lane 1 shows that in the absence of rifampin, mostly long mRNA products are produced. Rifampin treatment causes the production of short RNA products seen at the bottom of the gel in Lanes 2 and 3 (Figure 1). Consider how prokaryotic transcription occurs. Which of the following possible mechanisms of rifampin action is consistent with the data in Figure 1?

Rifampin blocks the exit of mRNA from the channel on RNA polymerase.

Telomerase was first described in the ciliate Tetrahymena thermophila by Elizabeth Blackburn and her student Carol Greider. They, along with Jack Szostak, subsequently won a Nobel Prize for this discovery. As the animation shows, the template RNA sequence in Tetrahymena is 3'-ACCCCAAC-5'. The telomerase protein and RNA template together extend the 3' end of the chromosome by adding 5'-GGGTTG-3' repeats to the chromosome. The complementary strand is then synthesized by DNA polymerase α.Blackburn's lab altered the sequence of the telomerase RNA. Predict what might happen to telomeres in Tetrahymena cells if the Tetrahymena template RNA were mutated to 3'-ACCCCGAC-5'.

Telomere sequence would be altered to 5'-GGGCTG-3'.

Predict what might happen to telomeres in Tetrahymena cells if the Tetrahymena template RNA were mutated from 3'-ACCCCAAC-5' to 3'-AGCCCAAC-5'.

Telomeres would become shorter in every generation compared to normal cells.

Which of the following occurs when a cell repairs a double-strand DNA break by the process of nonhomologous end joining?

The DNA sequence at the site of repair is altered by a short deletion.

The following GIF demonstrates how chromatin-remodeling complexes can slide nucleosomes along the DNA. The motor subunit (purple) uses the energy of ATP to slide the nucleosome down the DNA. The chromosome (red) contains a short region marked in blue that can be seen changing location in relation to the nucleosome (yellow) as the chromatin-remodeling complex functions. Nucleosome sliding plays important roles in regulating chromatin packing and consequently gene expression. It also plays a role in providing access to the DNA for other proteins like DNA repair enzymes. The nucleosome sliding activity of chromatin-remodeling complexes (CRC) can be studied in an in vitro electrophoretic mobility shift assay (EMSA) containing a DNA template, histones, chromatin-remodeling complexes, and ATP. When the nucleosomes are preloaded at the end of the DNA template and run on a polyacrylamide gel, they move through the gel. Added chromatin-remodeling complexes (CRC) can move the nucleosome to the middle of the DNA template. This slows down the electrophoretic mobility of the nucleosome and DNA through the gel allowing the researchers to measure chromatin-remodeling complex activity (Figure 1). DNA c

The chromatin-remodeling complex (CRC) moves the nucleosomes to the center of the DNA when ATP is added. The added CR5538 drug activates the chromatin-remodeling complex (CRC) motor subunit.

Some applications in biology, such as polymerase chain reaction (PCR), require melting the DNA double helix into single strands of DNA. This can be accomplished by heating the DNA. As DNA is heated, why does the double helix structure denature into single strands of DNA but not into individual nucleotides? In other words, why do the single strands remain intact even though the double helix does not?

The double helix is held together with hydrogen bonds, while the single strands are linked by phosphodiester bonds.

In the 1920s, bacteriologist Fred Griffith demonstrated that a heat-killed, infectious pneumococcus produced a substance that could convert a harmless form of the bacterium into a lethal one. Fifteen years later, researchers prepared an extract from the disease-causing S strain of pneumococci and showed that this material could transform the harmless R-strain pneumococci cells into the infectious S-strain form. This change to the bacteria was both permanent and heritable, suggesting that this "transforming principle" represents the elusive genetic material of the cells. The researchers subjected their extract to a variety of tests to determine the chemical identity of the "transforming principle." In one experiment, they treated the material with enzymes that destroy all proteins. This treatment did not affect the ability of the extract to transform harmless bacteria into an infectious form. From this result, what could the researchers conclude?

The genetic material is not protein. Correct. From the results, the researchers could conclude that the genetic material is not protein. The results ruled out protein as the genetic material because experimentally destroying the proteins had no effect on the ability of the extract to transform harmless bacteria into an infectious form.

Telomerase was first described in the ciliate Tetrahymena thermophila by Elizabeth Blackburn and her student Carol Greider. They, along with Jack Szostak, subsequently won a Nobel Prize for this discovery. As the animation shows, the template RNA sequence in Tetrahymena is 3'-ACCCCAAC-5'. The telomerase protein and RNA template together extend the 3' end of the chromosome by adding 5'-GGGTTG-3' repeats to the chromosome. The complementary strand is then synthesized by DNA polymerase α. Blackburn's lab altered the sequence of the telomerase RNA. Predict what might happen to telomeres in Tetrahymena cells if the Tetrahymena template RNA were mutated to 3'-ACCCCGAC-5'. What problem with replication of linear chromosomes does telomerase address?

The lagging strand stops short of the 3' end during replication, so chromosomes would shorten in each replication cycle without telomerase

Seed dormancy and seed germination are regulated in plants including Arabidopsis. Seed dormancy inhibits expression of seed germination genes and, under the right conditions, switching to germination inhibits seed dormancy gene expression. Plants also contain many reader and writer regulatory proteins that recognize and add methyl groups to lysines on histone H3 tails to maintain chromatin in daughter cells. The Arabidopsis protein SUVH4 is expressed during seed germination but not during seed dormancy. When expressed, SUVH4 adds methyl groups to the histone H3 tails on lysine 9 near genes involved in seed dormancy. Seed germination then occurs. The chromatin state near the seed dormancy genes is then inherited in daughter cells formed during seed germination. Which of the following hypotheses fits with the above information?

The lysine 9 methylation leads to heterochromatin formation and reduced expression of genes involved in seed dormancy.

In an experiment conducted in 1962, investigators took tRNAs bearing cysteine and chemically converted the charged amino acid to an alanine. They then introduced these "hybrid" alanine-bearing tRNAs into a cell-free translation system from which they removed all of the normal, cysteine-bearing tRNAs. How did this chemical manipulation affect the proteins produced by this altered system?

The proteins contained alanines where cysteines were supposed to be.

To crack the genetic code, researchers introduced synthetic messenger RNAs into in vitro translation systems and determined which proteins were produced from these synthetic mRNAs. mRNAs consisting of poly-UUC led to production of three different proteins: poly-Phe, poly-Ser, and poly-Leu. What best explains this result?

The synthetic mRNA was read in all three reading frames.

The Encyclopedia of DNA Elements (ENCODE) project aims to discover functional elements in DNA. One technique employed is DNase-seq. This technique employs a DNase enzyme that digests accessible regions of chromatin, while inaccessible regions remain undigested. The undigested DNA is then sequenced. Which of the following is likely true of the DNase-seq results?

Unsequenced DNA is likely part of euchromatin. Centromeric DNA is likely to be sequenced in all samples.

Investigators treat cells with a chemical that introduces random mutations into the DNA, including single-nucleotide changes that turn one base into another. They then isolate two mutants: one produces a protein that carries an alanine at a site that normally contains a valine; the other produces a protein that carries a methionine instead of the valine. When these mutant cells are subjected to the same mutagenic treatment, they both produce proteins that contain a threonine at the site of the original valine. Assuming that the mutations causing these alterations are single-nucleotide changes, what were the codons that specified each of the amino acids discussed?

Val, GUG; Ala, GCG; Met, AUG; Thr, ACG

The genetic code was originally deciphered, in part, by experiments in which synthetic polynucleotides with repeating sequences were used as mRNAs to direct protein synthesis in cell-free extracts. Under these conditions, ribosomes could be made to start translation anywhere within the RNA molecules, with no start codon necessary. What peptide would be made by translation from a synthetic mRNA made of the repeating trinucleotide UCGUCG...?

a polymer of serine (Ser-Ser-Ser...), a polymer of arginine (Arg-Arg-Arg...), and a polymer of valine (Val-Val-Val...)

When does homologous recombination most likely occur in order to flawlessly repair double-stranded DNA breaks?

after the cell's DNA has been replicated

Choose all of the following that correctly describe a characteristic of mismatch repair.

all A.DNA polymerase and ligase fill in the gap. B.Regions of improper base-pairing between parent and daughter strand are detected and repaired. C.Helicase unwinds the DNA in the mismatched area. D.Exonuclease removes the newly synthesized DNA in the mismatched area.

Where is heterochromatin not commonly located?

chromosomal regions carrying genes that encode ribosomal proteins

The DNA helicase animation shows the bacteriophage T7 helicase unwinding DNA. Which of the following are critical components of the helicase mechanism of action necessary to unwind DNA?

conformational changes of subunits oscillating loops pulling the single-stranded DNA through a central hole ATP binding and hydrolysis

Position effect variegation in Drosophila eyes occurs when a chromosomal inversion moves an eye color gene near a region of heterochromatin. The eye color gene is sometimes then silenced leading to white sectors in the eye. The proteins involved in heterochromatin regulation are also found in humans and alterations in the system can lead to disease including cancer. Chaetocin is produced by the Chaetomium species of fungi. This metabolite possesses several antitumor properties both in vivo and in vitro. One chaetocin activity inhibits SUV39H1, the human homolog of Su(var) 3-9. Sort the following events into those that would increase and those that would decrease after chaetocin treatment inhibits SUV39H1

decreased: - recruitment of HP1 - methylation of histone H3 at position 9 - formation of heterochromatin - spreading of lysine 9 methylation along the chromosome increased: - expression of the genes at the location where SUV39H1 is inhibited

Researchers often want to isolate a certain type of RNA. For some RNA species, this can be accomplished via affinity chromatography, using beads coated with chains of poly-deoxythymidine (poly-dT). The desired RNA will stick to the beads while unwanted RNAs will flow through the column. The retained RNA can then be eluted. What RNA species can be purified using this method?

eukaryotic mRNA

Ionizing radiation can also cause damage to the nitrogenous bases of DNA. Repair of the damaged bases takes several steps. Which of the following repair mechanisms can be used to repair the nitrogenous bases of DNA damaged by ionizing radiation?

excision repair

One approach to determining the function of a protein is to inhibit protein function by mutating the gene coding for the protein and observing the resulting phenotype. Cleves et al. sought to uncover the role of fibroblast growth factor 1a (FGF1a) in the coral Acropora millepora by mutating the FGF1a gene via CRISPR technology (Cleves et al., 2018 CRISPR/Cas9-mediated genome editing in a reef-building coral PNAS 115 (20) 5235-5240). The CRISPR process uses the Cas9 endonuclease to cut DNA at a site targeted by a guide RNA (see image). What happens next to induce gene mutations?

nonhomologous end joining

Which macromolecule(s) is/are critical in the active site of the ribosome for catalysis of peptide bond formation? Translation proceeds in a series of steps in the active site of the ribosome. Which of the following are important steps in polypeptide formation?

ribosomal RNA - The RNA in the P site makes hydrogen bonds with the 3' end of the aminoacyl-tRNA. - The hydrogen bonds formed between ribosomal RNA and tRNAs position the aminoacyl-tRNAs to catalyze peptide bond formation.

It has been proposed that the first cells used RNA for both information storage and catalysis and that DNA and proteins evolved later. Which modern macromolecules may be relics of the hypothesized RNA world?

spliceosome ribosome

The telomerase enzyme is unusual in having both protein and RNA components, but it belongs to the larger family of nucleic acid polymerases, all of which share a common structure. Family members with the most similar function are likely to have sequences and structures that are also most similar to each other. Given this, rank the following DNA polymerases based on predicted sequence similarity to telomerase, from most similar to least similar.

the ranking from most similar to least similar is as follows: 1. reverse transcriptase encoded by LINE-1 retrotransposon 2. DNA polymerase alpha 3. terminal deoxynucleotidyl transferase

Several organisms have a homologous protein (inherited from a common ancestor) that is highly similar at the amino acid level. You are comparing the genes that code for these proteins in the different organisms when you note that one of the codon nucleotide positions shows more nucleotide variation than the other nucleotide positions. In which codon nucleotide position do you expect to see the most variability among species?

third nucleotide position

An electromobility shift assay (EMSA) can detect protein binding to DNA. Proteins are preincubated with labeled DNA and then run on a native polyacrylamide gel that leaves protein-protein and protein-nucleic acid interactions intact. Protein binding to DNA slows the movement of the labeled DNA through the gel. Lane 1 of Figure 1 shows the labeled prokaryotic DNA without added proteins. Lane 2 shows bacterial sigma and RNA polymerase bound to the prokaryotic DNA. Figure 2 shows an EMSA assay using a eukaryotic promoter DNA, RNA polymerase II (RNA pol II), and several general transcription factors. Match each of the reactions with the correct lane on Figure 2.

this one was so ASS B C A

The mismatch repair system recognizes mismatched base pairs, removes a portion of the DNA strand containing the error, and then resynthesizes the missing DNA using the correct sequence as a template. But what if the mismatch repair system instead removed a piece of the DNA strand that contained the correct sequence? What would replication of this improperly repaired sequence produce?

two DNA molecules bearing the same mutation