Genetics Exam III
How do bases pair?
%A = %T %C = %G
During transcription in ........(1)......... RNA Polymerase produces RNA in .........(2).......... orientation. Transcription is initiated at the .......(3)......... which consists of -35 and -10 conserved sequence elements. -10 sequence is AT rich because .........(4).......... .Transcription proceeds until the RNA Polymerase reaches terminator sequence that contains ........(5)........
(1) Prokaryotes (2) 5' to 3' (3) promoter (4) A binds T with two hydrogen bonds (5) hairpin forming sequence, rho binding site
Which of the following enzymatic activities is involved in proofreading during replication? 1. 5'-3' exonuclease 2. 3'-5' exonuclease 3. DNA ligase 4. DNA glycosylase 5. All of the above
2. 3'-5' exonuclease
A mutation that produces a mutant protein that difers from the wild-type protein at one amino acid position.
A mutation that changes only one amino acid in a protein is a missense mutation.
A mutation that changes several amino acids in a protein and results in a protein that is shorter than the wild-type product.
A mutation that changes several amino acids in a protein and results in a protein being shorter than must have changed more than one codon and create a premature, in-frame stop codon. This could be due to a frameshift mutation or a splice site mutation.
A mutation that produces a protein that is shorter than the wild-type protein but does not have any amino acid changes in the portion produced.
A mutation that produces a protein that is shorter than wild type but does not contain any amino acid changes in the protein produced would be caused by a nonsense mutation.
A mutation that produces about 5% of the wild-type amount of an mRNA.
A mutation that results in only 5% of the wild-type mRNA changes the amount of mRNA but not its sequence. This could be cause by a promoter mutation.
A null mutation that does not produce any functional protein product.
A null mutation that does not produce any function protein could be due to any of the mutation types.
What are the differences between a synonymous, missense, and nonsense mutation?
A synonymous mutation converts one codon into a different codon that codes for the same amino acid. A missense mutation converts one codon that codes for an amino acid into a different codon that codes for a different amino acid. A nonsense mutation converts a codon that codes for an amino acid into a stop codon.
Your goal is to obtain cDNA of ACTIN 2 from the plant Arabidopsis thaliana and insert it into a plasmid. You found the sequence of this cDNA in a sequence database. What procedures do you have to perform afterwards? A. 1. Design two primers, one matching the beginning of the cDNA and the other matching the end of the cDNA 2. Amplify the cDNA using RT-PCR (using total RNA as a template) 3. Ligate the PCR product into a plasmid B. 1. Design two primers, one matching the beginning of the cDNA and the other matching the end of the cDNA 2. Amplify the cDNA using PCR (using genomic DNA as a template) 3. Ligate the PCR product into a plasmid C. 1. Design one primer matching the beginning of the cDNA 2. Amplify the cDNA using RT-PCR (using total RNA as a template) 3. Ligate the PCR product into a plasmid D. 1. Design one primer matching the beginning of the cDNA 2. Amplify the cDNA using PCR (using genomic DNA as a template) 3. Ligate the PCR product into a plasmid E. None of the above
A. 1. Design two primers, one matching the beginning of the cDNA and the other matching the end of the cDNA 2. Amplify the cDNA using RT-PCR (using total RNA as a template) 3. Ligate the PCR product into a plasmid
In 2010 NASA astrobiologists discovered a new bacterial strain, which (as they claimed) has phosphorus in its DNA replaced with arsenic. This finding was later shown to be wrong but if it was true these bacteria would have arsenic in a building block of DNA which performs the following function: A. Connects two sugar groups together B. Connects two phosphate groups and a nitrogenous base together C. Connects two nitrogenous bases together D. Forms hydrogen bonds with the antiparallel strand E. Forms hydrogen bonds with RNA strand
A. Connects two sugar groups together
How would you design a genetic test capable of giving the most accurate estimate if a certain person's children may have an increased risk of developing Huntington's disease? A. Extract genomic DNA, PCR amplify the huntingtin gene and sequence the repeat region using Sanger sequencing B. Extract total RNA and run a northern blot with a probe complementary to the repeats C. Extract proteins and run a western blot with an anti-huntingtin antibody D. Generate a genomic library and screen for the huntingtin gene E. None of the above would be an informative test
A. Extract genomic DNA, PCR amplify the huntingtin gene and sequence the repeat region using Sanger sequencing
Alternative splicing means that there may be A. More proteins than protein-coding genes B. More protein-coding genes than proteins C. Multiple caps on a single mRNA D. Multiple poly A tails on a single mRNA E. None of the answers above is correct
A. More proteins than protein-coding genes
Which building block of DNA carries genetic information? A. Nitrogenous bases B. Sugars C. Phosphate groups D. Hydrogen bonds E. None of the above
A. Nitrogenous bases
Eukaryotic mRNA is capped: A. On 5' ends cotranscriptionally B. On 3' ends cotranscriptionally C. On 5' ends cotranslationally D. On 3' ends cotranslationally E. None of the answers above is correct
A. On 5' ends cotranscriptionally
Imagine you want to perform PCR in your garage. You found ways to extract template DNA, obtain primers and analyze PCR products on a gel. The only missing component is a thermostable DNA polymerase. Instead you have a DNA polymerase, which is active at 70 ˚C but inactivated by temperatures above 80 ˚C. Which protocol is the best way of overcoming this limitation?
A. Place in water bath at 95 ˚C for 1 minute Add DNA polymerase Place in water bath at 50 ˚C for 1 minute Place in water bath at 70 ˚C for 1 minute Repeat all steps 30 times
Which class of genomic sequences is most likely to contribute to the creation of chromosomal rearrangements? A. Repetitive DNA B. Introns C. Exons D. Existing balanced translocations E. Telomeres
A. Repetitive DNA Repetitive DNA (such as transposable elements which are DNA segments that can move within the genome) has the ability to amplify itself which then leads to these chromosomal rearrangements (inversion/duplication of DNA, etc).
Imagine you want to reproduce the experiment which demonstrated that DNA is the genetic material using transformation of the bacterium Streptococcus pneumoniae with purified cellular extracts. Having access to all modern methods, you want to make the conclusion of this experiment stronger. Which is the most important additional step you should add to the experiment? A. Test if the sample where DNA has been destroyed really has no DNA using PCR B. Test if the sample where RNA has been destroyed really has no RNA using western blot C. Test if the sample where polysaccharides have been destroyed really has no polysaccharides using northern blot D. Test if the sample where lipids have been destroyed really has no lipids using Sanger sequencing E. Test if the sample where proteins have been destroyed really has no proteins using complementation test
A. Test if the sample where DNA has been destroyed really has no DNA using PCR
Imagine that scientists discovered a previously unknown form of life. To test if the new organism is prokaryotic or eukaryotic they purified RNA Polymerase under conditions which preserve nascent transcripts. They analyzed the purified samples using western blot and detected ribosomal proteins. How should they interpret this result? A. This organism is prokaryotic B. This organism is eukaryotic C. This is something entirely new different from both prokaryotes and eukaryotes D. This organism is actively translating E. This experiment is inconclusive
A. This organism is prokaryotic [because RNA Polymerase and ribosomes bind to an RNA at the same time]
While investigating a new mutant in the HOXD gene, you found that a fragment of the coding sequence of the gene is changed in the following way: wild type allele: UAUAAUGCGCCCAUC mutant allele: UAUAAUGAGCCCAUC Assuming that the sequence above is in-frame, how would you describe the detected mutation? A. Transversion, missense B. Transversion, nonsense C. Transversion, frameshift D. Transition, nonsense E. Transition, frameshift
A. Transversion, missense
A mutant strain of Salmonella carries a mutation of the rho protein that has full activity at 37C but is completely inactivated when the mutant strain is grown at 40C. Speculate about the kinds of differences if you compared mRNAs at each temperature.
At 37C, all of the mRNAs normally present in Salmonella cells would be present in the mutant. At 40C, only the mRNAs that are coded by the genes using the intrinsic termination would be normal; those that use the rho-dependent termination would likely be longer than normal. Some or all of these transcripts would terminate downstream of their normal termination site.
How many anticodons are possible for Phe? A. 1 B. 2 C. 3 D. 4 E. 5
B. 2
Which change in gene sequence is NOT necessary to express a human gene in bacteria? A. Replace promoter with a prokaryotic counterpart B. Add 5' cap [because it is not in DNA sequence] C. Remove introns D. Replace polyadenylation site with a terminator E. All listed changes are necessary for expression
B. Add 5' cap [because it is not in DNA sequence] A gene expressed in bacteria (ie, transcription occurs) will be translated, regardless of a 5' cap. A cap is necessary for mRNA in human cells, since the mRNA transcript needs to be shipped out of the nucleus, into the cytoplasm, then into a ribosome for translation. And because of the path of the mRNA, eukaryotic degradation mechanisms can shutdown or degrade an uncapped mRNA transcript. The 5' cap, therefore, is not necessary in bacterial cells (bacterial DNA is not encapsulated in a nucleus and is in the cytoplasm). It's also necessary in eukaryotes, as it's important in the initiation of translation. Prokaryotes have other mechanisms of translation initiation. Specifically, the 5' cap is used to help bring the mRNA to ribosomes in eukaryotes, but prokaryotes have Shine-Dalgarno sequences instead, and don't need the cap for to bring mRNA to ribosomes.
You extract mature mRNA from an individual frog with a diploid chromosome number of 18. After reverse transcribing the mRNA, you sequence the cDNA library you have created. When you examine the sequences you notice there are numerous cases where cDNAs of substantially different lengths show 100% alignment in shared regions. A most likely explanation for this pattern is: A. Errors during transcription by RNA polymerase B. Alternative splicing of some mRNA C. Different alleles of the same gene D. Deletions during somatic growth E. Differential locations of promoter sites
B. Alternative splicing of some mRNA
What property of replication allows us to amplify a specific sequence of DNA using PCR? A. DNA polymerase requires dNTPs B. DNA polymerase requires existing 3' ends to elongate C. DNA polymerase requires magnesium for activity D. All of the above E. None of the above
B. DNA polymerase requires existing 3' ends to elongate
What is genetic information? A. Genetic code of an organism B. Information about all heritable traits of an organism C. All nucleic acids within an organism D. All proteins within an organism E. All of the above
B. Information about all heritable traits of an organism
DNA supercoiling is important for A. Maintaining genetic information B. Replication C. Restriction digestion D. Answers A and B are correct E. Answers B and C are correct
B. Replication
Which of the following features is NOT similar between replication and transcription in eukaryotes? A. Synthesis of nucleic acids in 5' to 3' orientation B. Requirement of a primer C. Initiation determined by DNA sequence D. Unwinding of DNA E. Involvement of RNase activity
B. Requirement of a primer
Which of the following features is NOT similar between replication and transcription in eukaryotes? A. Synthesis of a nucleic acid in 5' to 3' direction B. Requirement of a primer C. Initiation determined by DNA sequence D. Unwinding of DNA E. Involvement of an RNase activity
B. Requirement of a primer - transcription is primer independent
Which of the listed features of translation is different between prokaryotes and eukaryotes? A. Translation initiation at an AUG start codon B. The way the small subunit of the ribosome finds translation initiation site C. Catalytic activity in the large subunit of the ribosome D. Primary functions of the ribosomes performed by RNA E. None of the above
B. The way the small subunit of the ribosome finds translation initiation site [using Shine-Dalgarno sequence or 5' cap]
Imagine that for your honors thesis you are studying the Actin protein in flies. A collaborating lab sent you a plasmid containing the gene encoding actin. What is the best way to obtain large amount of full length circular plasmid DNA for further analysis? A. PCR amplify the plasmid B. Transform the plasmid into bacteria, grow the bacterial culture and extract the plasmid from bacteria C. Use Northern blot D. Use Western blot E. None of the answers above is correct
B. Transform the plasmid into bacteria, grow the bacterial culture and extract the plasmid from bacteria
Which of the listed features is different between DNA and RNA? A. Ability to contain genetic information B. Presence of nitrogenous bases C. Ability to have enzymatic activity D. Presence of phosphate groups E. None of the above
C. Ability to have enzymatic activity [RNA may form ribozymes]
Hershey and Chase's experiment in 1952 studied which part of a phage is injected into bacterial cells during infection. By tracking radioactively labeled DNA and proteins they found that phages inject DNA into their host cells. Why was that treated as evidence that DNA contains genetic information? A. Because DNA contains phosphorus but not sulphur B. Because proteins contain sulphur but not phosphorus C. Because it was already known that phages insert their genetic information into bacteria D. Because it was already known that phages do NOT insert their genetic information into bacteria E. Answers A, B and D are all correct
C. Because it was already known that phages insert their genetic information into bacteria
Which of the following is NOT an essential feature of the genetic material? A. High information content B. Ability to mutate C. Double helix D. Ability to replicate E. All of the above ARE essential features of the genetic material
C. Double helix
What does it mean that the genetic material has four essential properties (high information content, ability to replicate, ability to mutate and regulated expression)? A. Genes are encoded in DNA. B. Only DNA has these properties. C. Even if genes were contained in a molecule other than DNA, this molecule would have all these properties. D. If genes were contained in a molecule other than DNA, this molecule would NOT have all of these properties. E. None of the above.
C. Even if genes were contained in a molecule other than DNA, this molecule would have all these properties.
Which repair mechanism could repair DNA double strand break in an error-free manner? A. Translesion synthesis B. Nonhomologous end joining C. Homologous recombination D. Both A and B E. Both B and C
C. Homologous recombination
Recombination in meiosis occurs when .......I......... and .........II......... accompanied by heteroduplex formation. A. I. Holliday junctions are resolved in the same plane II. is B. I. Holliday junctions are resolved in the same plane II. is not C. I. Holliday junctions are resolved in different planes II. is D. I. Holliday junctions are resolved in different planes II. is not E. None of the above is correct
C. I. Holliday junctions are resolved in different planes II. is
Although DNA is the primary carrier of genetic material, under certain circumstances other molecules also carry genetic information. Which molecule is least likely to contain some genetic information? A. RNA B. Proteins C. Lipids D. None of the above molecules is capable of containing some genetic information E. All of the above sequences are equally capable of containing some genetic information
C. Lipids
Translating ribosomes together with translated mRNAs may be purified from living cells using ultracentrifugation. This approach can be used to measure the intensity of gene expression. Which of the following methods could be used to analyze the purified ribosomes to determine the level of translation of a specific mRNA? A. Western blot with an antibody against the large subunit of the ribosome B. PCR with primers matching a specific tRNA C. Northern blot with a probe complementary to the gene of interest D. Southern blot with a probe complementary to the gene of interest E. All of the above
C. Northern blot with a probe complementary to the gene of interest
Which of the following statements about replication is NOT true? A. DNA strands are separated by a helicase B. Proofreading dramatically increases accuracy of replication C. Primers are removed in 3' to 5' orientation D. Gaps in the backbone are sealed by DNA ligase E. DNA Polymerase requires a primer
C. Primers are removed in 3' to 5' orientation
In the bacteria E. coli, the Shine-Dalgarno sequence is incorporated into to the mRNA: A. Only in some of the alternative splicing events of the mature mRNA B. Co-translationally C. Shortly after initiation of transcription for all of the mRNA D. Just before termination of transcription E. At a variable time during elongation of the transcript depending on the gene
C. Shortly after initiation of transcription for all of the mRNA
Which of the following is NOT a correct description of Sanger sequencing? A. Sanger sequencing uses all dNTPs and one type of ddNTP per reaction to synthesize the strand B. The sequence we get from Sanger sequencing is complementary to the template strand C. Similar to PCR, we need two primers for Sanger sequencing D. Before doing Sanger sequencing, we need to perform PCR or cloning to get large quantity of the target gene E. A 3'-Hydroxyl group is present in dNTPs but not in ddNTPs
C. Similar to PCR, we need two primers for Sanger sequencing -- only one primer is needed
Which of the following are error-prone DNA repair pathways? A. Base excision repair and Nucleotide excision repair B. Homologous recombination and Base excision repair C. Translesion DNA synthesis and Non- homologous end joining D. Non- homologous end joining and Mismatch repair E. None of the above
C. Translesion DNA synthesis and Non- homologous end joining
Which of the following statements is NOT correct? A. Two primers are needed to amplify a target sequence by PCR B. Antibody is used in Western blot to detect the target protein C. Two primers are needed to sequence your target sequence using Sanger sequencing D. Reverse transcriptase is used to make cDNA from mRNA E. Both western blot and northern blot are used to quantify gene expression
C. Two primers are needed to sequence your target sequence using Sanger sequencing
Genetically modified plants are a big hope for the future of agriculture, especially in areas where people suffer from malnutrition. The most common way of obtaining transgenic plants is: A. Watering plants with a solution of DNA B. Injecting DNA into plant tissues C. Using a bacterium, which has a natural ability to transfer DNA into plant cells D. Crossing with other species E. None of the above
C. Using a bacterium, which has a natural ability to transfer DNA into plant cells
A geneticist obtained transgenic bacteria carrying a plasmid with cDNA of human insulin gene. Which of the following tools is the best way to test if these bacteria produce insulin? A. Southern blot B. Northern blot C. Western blot D. PCR E. RT-PCR
C. Western Blot - the only method that detects proteins
Imagine that for your honors thesis you are studying the Actin protein in flies. A collaborating lab sent you a plasmid containing the gene encoding actin. Now assume that the plasmid has the correct sequence and that it contains a cDNA sequence of the Actin gene driven by the native fly Actin promoter. Which system is the best choice for producing the Actin protein? A. Bacteria B. Bacteriophage C. Yeast D. Answers A and B are equally good E. All of the above are equally good
C. Yeast
DNA Repair Pathway: a cytosine that has been deaminated to uracil
Conversion of cytosine to uracil by deamination creates G:U base pairs in DNA. Uracil is removed from DNA by the base excision repair (BER) pathway.
What procedures should be performed to obtain a large quantity of DNA encoding a particular, known gene? A. (1) Find the sequence of the gene in a database, (2) design primers corresponding to the ends of the sequence, (3) perform Southern blot. B. (1) Digest genomic DNA with a restriction enzyme, (2) separate DNA on the gel and transfer to the membrane, (3) hybridize a radioactive probe. C. (1) Find the sequence of the gene in a database, (2) design primers corresponding to the ends of the sequence, (3) ligate the primers into a plasmid. D. (1) Find the sequence of the gene in a database, (2) design primers corresponding to the ends of the sequence, (3) perform PCR. E. None of the above
D. (1) Find the sequence of the gene in a database, (2) design primers corresponding to the ends of the sequence, (3) perform PCR.
Imagine that scientists found evidence of multiple proteins being produced from a single eukaryotic gene. How would you explain this observation and how could this be tested? A. Polycistronic mRNA and Southern blot with a probe complementary to the mRNA B. Alternative splicing and PCR with primers complementary to the flanking regions of the gene C. Polycistronic mRNA and genomic library generation D. Alternative splicing and northern blot with a probe complementary to the mRNA E. None of the above
D. Alternative splicing and northern blot with a probe complementary to the mRNA Option D is correct because we can explain the observation that: in eukaryotic cells, a single gene can produce multiple proteins through alternative splicing. During alternative splicing, different combinations of exons are joined together, while introns (are removed from the pre-mRNA. This results in various mature mRNA molecules being produced from a single gene, which are then translated into different protein isoforms. We could test this observation by using northern blot with a probe complementary to the mRNA. A northern blot involves separating RNA molecules by size using gel electrophoresis, transferring them to a membrane, and then hybridizing them with a labeled probe complementary to the mRNA of interest. By using a probe complementary to a region shared by all the alternatively spliced mRNAs, you can detect the different mRNA isoforms generated by alternative splicing.
Except for a few model organisms, there are few genomic resources available for most species. However, tools and sequence data from model organisms can frequently be applied to research questions involving non-model organisms. For example, if you wanted to amplify a specific gene in bumblebees, you could use the sequences from the honeybee genome to design primers. Which portion(s) of the gene would be best for designing primers? A. Introns B. 10bp region that is 1kb upstream of the promoter C. The TATA box D. Exons E. The 5' UTR
D. Exons
Which feature of mRNA is the most similar in its function to telomeres? A. Exons B. Introns C. Polysomes D. Polyadenylation E. There is no similarity whatsoever between any of those features and telomeres
D. Polyadenylation [because it protects ends]
Imagine you engineered the translation termination factor in a way that it binds the GGA codon instead of a STOP codon. How would you test if your engineered factor effectively causes premature termination at GGA codons? A. Produce the factor using in vitro transcription and incubate it with cellular extract and synthetic mRNA containing the GGA codon before the STOP codon B. Produce the factor using in vitro transcription and incubate it with cellular extract and synthetic mRNA containing the GGA codon after the STOP codon C. Produce the factor using in vitro transcription and incubate it with cellular extract and synthetic mRNA containing no GGA codon D. Produce the factor using in vitro translation and incubate it with cellular extracts and synthetic mRNA containing the GGA codon before the STOP codon E. More than one of the above would be a conclusive test
D. Produce the factor using in vitro translation and incubate it with cellular extracts and synthetic mRNA containing the GGA codon before the STOP codon
Compare and contrast the properties of DNA polymerase w/ RNA polymerase, listing at least three similarities and at least three differences between the molecules.
DNA and RNA polymerase are similar in that they both (1) catalyze phosphodiester bond formation to polymerize nucleotides into nucleic acids, (2) polymerize in a 5' to 3' direction, (3) are dependent on a DNA sequence template. They are different on the basis that (1) RNA polymerase can initiate strand synthesis, whereas DNA polymerase can only extend an existing strand, (2) most DNA polymerases can proofread using a 3' to 5' exonuclease activity, whereas RNA polymerase cannot, and (3) DNA polymerases use deoxyribosenucelotide triphosphate as substrates, where RNA polymerase uses ribonucleotide triphosphate
DNA mismatch repair can accurately distinguish between the template strand and the newly replicated strand of a DNA duplex. What characteristic of DNA strands is used to make this distinction?
DNA is methylated sometime after replication, such that daughter DNA molecules produced by DNA replication initially contain a methylated strand and non-methylated strand. Mismatch repair recognizes the methylated strand as the template strand and preferentially repairs the non-methylated strand at the site of mismatches. Methylation is the distinguishing characteristic.
What general mechanism do DNA polymerases use to check the accuracy of DNA replication and identify errors during replication?
DNA polymerase contributes to the accuracy of replication in two ways. First, it ensures that the correct nucleotide is added to the growing DNA strand. Second, it removes incorrect nucleotides immediately after they are added.
If a DNA replication error is detected by DNA polymerase, how is it corrected?
DNA polymerase proofreads each nucleotide after it is added to the growing DNA strand. If it detects a mismatched base pair, it removes the last nucleotide added using its 3' to 5' exonuclease activity and then selectively adds the correct nucleotide in its place.
You are interested in the Ccl2 gene in mouse, which might contribute to inflammatory response against bacterial infection. You want to 1) amplify the target gene segment for further study; 2) describe tissue specific expression in mouse 3) find out whether an orthologous gene (a gene with similar sequence) exists in human and whether you research has implications for human health. Choose the correct order of methods you would use to accomplish these three goals. A. DNA Cloning; Northern Blot; Southern Blot B. DNA Cloning; Western Blot; Northern Blot C. PCR; Southern Blot; Northern Blot D. PCR; Western Blot; Genome sequencing E. A and D
E. A and D
What are the general features of genetic material? A. Sufficient information capacity B. Ability to mutate C. Ability to replicate D. Regulated expression E. All of the above
E. All of the above
Imagine that for your honors thesis you are studying the Actin protein in flies. A collaborating lab sent you a plasmid containing the gene encoding actin. You should expect the plasmid to contain not only the Actin gene but also some other sequences. Please choose the best answer. A. Antibiotic resistance B. Origin of replication C. Centromere D. Telomere E. Answers A and B are correct
E. Answers A and B are correct
Which of the following experiments is capable of demonstrating that DNA carries genetic information? A. Transformation of the bacterium Streptococcus pneumoniae with purified cellular extracts B. Tracking the phage DNA during infection of bacteria C. Discovering the DNA structure D. Discovery of Chargaff's rules E. Answers A and B are correct
E. Answers A and B are correct
Which of the following statements about telomerase is true? A. Telomerase extends the leading strand template B. Telomerase extends the lagging strand template C. Telomerase is required for protection of chromosome ends in prokaryotes D. Telomerase is required for protection of chromosome ends in eukaryotes E. Answers B and D are correct
E. Answers B and D are correct
Which of the building blocks of DNA is NOT involved in the formation of the DNA backbone? A. Phosphate B. Deoxyribose C. Adenine D. Guanine E. Answers C and D are both correct (are NOT components of the backbone)
E. Answers C and D are both correct (are NOT components of the backbone)
Both transcription and DNA repair rely on the recognition of one DNA strand versus the other. Which DNA repair pathway recognizes a specific DNA strand using the same mechanism as the transcriptional machinery? A. Mismatch repair B. Base-excision repair C. Nucleotide excision repair D. Repair by homologous recombination E. None of the above
E. None of the above [because transcription recognizes strands by binding to promoter sequences, which have a specific orientation]
For years researchers believed that neurons in adult human brain do not divide. You performed an experiment in which you treated mice with radioactive cytidine. You observed significant incorporation of the labeled nucleoside in brain neurons. Based on this result you may conclude that: A. This is evidence of replication and therefore neurons do divide B. This is evidence of translation and therefore neurons do divide C. This is evidence of meiotic recombination and therefore neurons do divide D. This is evidence of DNA methylation and therefore your experiment does not demonstrate cell divisions in neurons E. None of the conclusions above is correct
E. None of the conclusions above is correct Cytidine is distinct from cytosine in that it lacks a triphosphate group; it is a nucleoside, not a nucleotide. A would therefore be incorrect because it is not cytidine that is incorporated into DNA (or RNA) strands during replication (or transcription), but rather, cytosine. B is incorrect because translation is the process of converting an RNA transcript into a functional protein, which does not involve the corporation of additional nucleotides (or nucleosides, as is true for cytidine) into biomolecules. C is incorrect because meiotic recombination is a process of "rearranging" genetic information between parental chromosomes during meiosis; this process does not incorporate any additional nucleotides (or nucleosides) into the DNA. D would not be correct because DNA methylation does not incorporate new nucleotides (or nucleosides) into DNA; rather, it methylates nucleotides already present in the DNA. So, none of these conclusions are correct, and E is the right answer.
5'-bromouracil is an analog of thymine. After incorporated in DNA, however, it will easily switch to its isomer that could pair with guanine. If the scenario above occurs and there's no successful repair, what kind of point mutation could we observe? A. Adenine to cytosine B. Thymine to guanine C. Adenine to thymine D. Thymine to adenine E. Thymine to cytosine
E. Thymine to cytosine
Which of the following is NOT an appropriate application of a method? A. Use western blot to detect changes in expression level of a gene involved in immune response B. Use DNA cloning to obtain large amounts of a plasmid C. Use Sanger sequencing to confirm that you have the correct sequence obtained from PCR D. Use Next Generation Sequencing to determine the genome sequence of a newly discovered species E. Use northern blot to figure out whether a gene has been deleted from the genome
E. Use northern blot to figure out whether a gene has been deleted from the genome
For a eukaryotic gene whose transcription requires the activity of an enhancer sequence, explain how proteins bound at the enhancer interact with RNA pol II and transcription factors bound at the promoter.
Enhancers are positive, cis-acting DNA sequences that can be located at great distances from the promoter of the gene that they regulate. They function by binding to proteins which then bind to "recruit" complexes of other proteins that bend DNA and bring the enhancers and its associated proteins close to the promoter of the regulated gene. The associated proteins can then interact with RNA polymerase and its transcription factors to promote transcription of a gene.
Why do the genomes of eukaryotes need to have multiple origins of replication, whereas bacterial genomes have only a single origin?
Eukaryotic chromosomes are significantly larger than bacterial chromosome and the rate of DNA synthesis is slower in eukaryotes as compared to prokaryotes. Both types of cells must completely replicate their chromosome before they can divide. Given the rate of DNA synthesis by bacterial DNA polymerase, initiation of bidirectional replication at one original in an E. coli chromosome is sufficient to replicate the entire chromosome in about 30 minutes. For several reasons, eukaryotic genomes require multiple origins for timely replication. First, eukaryotic genomes are contained in multiple chromosomes and there must be at least one replication of origin per chromosome. Furthermore, eukaryotic DNA polymerases function about 10 times slower than E. Coli. A third rationale for multiple eukaryotic regions is presumed need for replication of different genomic regions at different times during the S phase.
Microbiologists describe the processes of transcription and translation as "coupled" in bacteria. This term indicates that a bacterial mRNA can be undergoing transcription at the same moment it is also undergoing translation. How is coupling of transcription and translation possible in bacteria?
First, there are no membranes separating the bacterial chromosomes from the bacterial ribosomes; therefore, as soon as an mRNA is transcribed, ribosomal subunits have physical access to it. Second, bacterial transcripts do not contain introns; therefore, the primary transcript contains the complete open reading frame, and bacterial ribosomes can translate mRNA without needing to wait for mRNA - splicing.
DNA Repair Pathway: heavily damaged bacterial DNA
Heavily damaged bacterial DNA induces the SOS repair system. SOS repair allows for error-prone repair in order to prevent cell death due to high levels of DNA damage.
Explain how the Hershey and Chase experiment identified DNA as the hereditary molecule.
Hershey and Chase found that essentially all the P32-labeled T2 DNA but little to none of the S35-labeled T2 protein was in the infected bacterial cells. Since T2 genetic material must be inside the infected cells in order to direct new virus particle synthesis, these results pointed to DNA as the genetic material of the phage T2.
DNA Repair Pathway: a double-strand break that occurs just after replication in an actively dividing cell
In cells that have completed DNA synthesis, double-stranded DNA breaks are repaired by homologous recombination in bacteria and archaea, and primarily by synthesis-dependent strand annealing (SDSA) in eukaryotes, although non-homologous end-joining (NHEJ) and homologous recombination may be used by eukaryotic cells.
A sample of double-stranded DNA is found to contain 20% cytosine. Determine the percentage of the other three DNA nucleotides in the sample.
In dsDNA, %C = %G and %A = %T and %C + %G + %A + %T = 100% Therefore, 20% G, 20% C, 30%A, and 30% T
DNA Repair Pathway: a double-stranded break that occurs during G1 and pre- vents completion of DNA replication
In eukaryotic cells that suffer a double-stranded DNA break in G1 phase, non-homologous end-joining (NHEJ) is the primary pathway for repair.
Describe the difference between introns and exons.
Introns and exons are part of the transcribed region of a gene and are both present in the primary transcript. Every intron is flanked by two exons. Introns are removed from the primary transcript and flanking exons are joined together to create a "spliced" RNA by a process called, pre-mRNA splicing. Transcription, which produces the primary transcript, and pre-mRNA splicing occur in the nucleus in eukaryotic cells.
Although DNA is the primary carrier of genetic material, under certain circumstances other molecules also carry genetic information. Which molecule is least likely to contain some genetic information? RNA Proteins Lipids None of the above All of the above
Lipids
DNA Repair Pathway: a change in DNA sequence caused by a mistake made by DNA polymerase during replication
Mistakes during DNA replication create mismatched based pairs. The replication error can be immediately corrected by the proofreading function of DNA polymerase or can be corrected later by the mismatch repair pathway.
If a replication error escapes detection and correction, what kind of abnormality is most likely to exist at the site of replication error?
Mistakes in DNA replication lead to the formation of incorrect or mismatched base pairs. The most likely abnormality is mismatched base pairs.
Which building block of DNA carries genetic information?
Nitrogenous bases
Identify two mechanisms that can correct mismatched pairs.
Proofreading by DNA polymerase and mismatch repair can correct this kind of abnormality. Mismatch base repairs are recognized by the DNA repair pathway called mismatch repair. The mismatch DNA repair pathway recognizes the mismatched base pair, determines which strand is the newly synthesize DNA strand, removes nucleotides on that strand, and then resynthesizes that portion of the DNA strand.
How does RNA participate in DNA replication?
RNA serves as a primer for elongation by DNA polymerase
What enzymatic activity is present in telomerase?
RNA-dependent DNA polymerase
How does the telomerase assemble telomeres?
Telomerase is a reverse transcriptase that includes an RNA molecule. The RNA acts as a template for synthesis of telomeric DNA. Telomerase uses a segment of its RNA as the template to add multiple copies of a simple sequence to the 3' end of each strand of DNA on a linear chromosome. This strand, which corresponds to the template for lagging strand synthesis, is copied by the normal mechanism of lagging strand synthesis after it is extended by telomerase.
What is the function of telomerase and how does it operate to synthesize telomeres?
Telomerase is an enzyme that increases the length of linear chromosome ends. End-lengthening by telomerase counteracts end-shortening due to lagging strand DNA synthesis, resulting in the maintenance of chromosome length in cells expressing telomerase. Telomerase is a reverse transcriptase that synthesizes DNA using an endogenous RNA as a template. Telomerase synthesizes DNA on the 3' end of both DNA strands of a chromosome, creating a long single stranded DNA that is then replicated by lagging strand synthesis. The DNA sequence of chromosome ends corresponds to the RNA sequence of the template portion of telomerase RNA.
Why is telomerase usually active in germ-line cells but not in somatic cells?
Telomerase is required to ensure complete chromosome replication in germ cells, thus ensuring that every mitosis produces two daughter cells with complete chromosomes. Telomerase is not required in somatic cells because they cannot divide enough times to result in the loss of important DNA at chromosome ends.
What is the sequence of composition of telomeres?
Telomeres are replicated by telomerase, which uses a portion of its RNA sequence as a template to repetitively add a short, six-nucleotide sequence to ends of chromosomes. In many organisms, the repeated sequence is 5' - TTAGGG - 3'.
What is the functional role of telomeres?
Telomeres are through to provide two functions, one in chromosome replication and the other in chromosome protection. Telomeres provide a mechanism for replication for the ends of linear chromosomes. Without telomeres, lagging strand synthesis would fail to extend the chromosome ends, leaving a gap at each end after each round of replication. This would shorten the chromosome and after many rounds of replication, would result in the loss of important DNA sequences. Telomeres are repetitive DNA, which prevents loss of important DNA sequencing if shortening occurs. Telomeres are also the binding site for telomerase, which extends the lagging strand template to compensate for sequences lost during incomplete lagging strand synthesis. Telomeres also provide a protective cap, on the ends of linear chromosomes, that distinguishes normal chromosome ends from ends generated by double-stranded chromosome breaks. Without telomeric DNA and the proteins that bind telomeric DNA, the ends of chromosomes are recognized as broken chromosomes and are fused together by DNA repair enzymes. Such breakage can create chromosome end-to-end fusions, which then create dicentric chromosomes that can be broken during the next cell division, thus creating new breaks, new fusions, in an endless cycle known as bridge-break-fusion cycle.
Matthew Meselson and Franklin Stahl demonstrated that DNA replication is semiconservative in bacteria. Briefly outline their experiment and its results for two DNA replication cycles, and identify how the alternative models of DNA replication were excluded.
The results showed that before transfer to the N14 medium, all of the DNA was N15/N15. After one round of replication in the N14 medium, all of the DNA was N14/N15; after two rounds of replication, half the DNA was N14/N15 and half was N14/N14; after three rounds of replication, one fourth was N14/N15 and three fourths was N14/N14. The conservative model predicted that the original N15/N15 DNA would remain throughout; therefore, the results ruled out the conservative model. The dispersive model predicted that after each round of replication, there would only be one form of DNA, which would become less and less dense. Although this model was not ruled out after one round of replication, the persistance of the N14/N15 DNA and the presence of two other classes of DNA after rounds two and three ruled out the dispersive model.
Explain why Avery, MacLeod, and McCarty's in vitro transformation experiment showed that DNA, but not RNA or protein, is the hereditary molecule.
The transforming principle was considered to be genetic material. The most reasonable interpretation of those results was that DNA was the only essential component of the transforming principle, and therefore, the genetic material.
Describe the two types of transcription termination found in bacterial genes. How does transcription termination differ for eukaryotic genes?
The two types of transcription termination in bacteria are intrinsic termination, which is dependent only on specific DNA sequences at the 3' end of genes, and rho-dependent termination, which requires specific DNA sequences and the rho protein. In intrinsic termination, transcription of an inverted sequence at the 3' end of genes results in an mRNA that folds into a hairpin structure. RNA polymerase pauses during transcription of the poly-A sequence that immediately follows the inverted repeat, which allows the polymerase to physically interact with the RNA hairpin. The combination of pausing and interaction with the RNA hairpin causes polymerases to release the mRNA and the gene, terminating transcription. For rho-dependent termination, transcription of the inverted repeats also generates a hairpin structure; however, sequences following hairpin RNA structures constitute binding site for rho, called a rho utilization site. Rho binds the rut site in the transcript and uses ATP hydrolysis to move along the RNA and melt the RNA-DNA hybrid. Disruption of the base-pairing between RNA and the template induces polymerase to release the transcript and the template, terminating transcription. Transcription termination in eukaryotes differs in that the 3' end of the RNA is not formed by the release of the RNA from polymerase, but rather caused by the cleavage of the primary transcript and the addition of the poly-A to the 3' end of the transcript.
DNA Repair Pathway: a thymine dimer induced as a result of UV exposure
Thymine dimers are repair by two pathways: nucleotide excision repair (NER) and photo reactive repair.
T/F: Viruses have to be single stranded.
True
Promotor Mutations
affect initiation of transcription and can increase or decrease the amount of transcript produced
Which of the features below is NOT similar between replication and translation? a.) Synthesis in the 5' to 3' direction b.) Proofreading c.) Synthesis of polymeric molecules d.) Involvement of RNA e.) All features are similar between replication and translation
b.) Proofreading
Splice Site Mutations
can prevent or alter pre-mRNA splicing, which alters the mRNA sequence at the site, introducing new codons and pre-mature, in-frame stop codon
What is the function of transcription initiation?
defining 5' end of mRNA
How is telomeric DNA converted from single-stranded to double-stranded form?
in the next cell cycle by Okazaki fragments - telomerase only produces single stranded DNA
Frameshift Mutations
insertions or deletions within the reading frame that change all the codons after the site of mutation and typically create a premature, in-frame stop codon
There is a problem completing the replication of linear chromosomes at their ends. Describe the problem and identify why telomeres shorten in each replication cycle.
linear chromosomal DNA is replicated by bidirectional replication, which involves leading and lagging strand synthesis. the problem replicating the ends of linear chromosomes is due to the lagging strand synthesis, which is responsible for replicating one DNA strand at each chromosome end. lagging strand synthesis initiates at an RNA primer and moves inward, away from the chromosome ends. because of this, one single strand at the end of each chromosome is not replicated all the way to the end. this happens every time a linear chromosome is replicated, which causes the ends of DNA to shorten with every round of replication.
What is maintaining genetic information?
maintaining genetic information means that genetic information is passed from generation to generation through inherited units of chemical information
Missense Mutations
point mutations that cause a single amino acid acid substitution in the encoded protein
Nonsense Mutations
point mutations that convert a codon that codes for an amino acid into a stop codon
What are the percentages for single stranded DNA?
random nucleotide base amounts, cannot be predicted
DNA supercoiling is important for
replication
