Practice_Exam_1

Exam Practice Questions (many of these are real questions from previous exams) 1. (3 points) Which of the following is TRUE about the Central Dogma of Molecular Biology? A. Information can only flow from DNA to RNA, this cannot happen in reverse. B. The genotype to phenotype transition occurs at translation. C. RNAs always encode for proteins. D. Proteins can be made directly from DNA. E. There is very little evidence to support the Central Dogma of Molecular Biology. 2. (3 points) Which of the following is the best definition of a gene? A. A gene is a unit of expression that always codes for a protein. B. A gene is all of the exons found in the mature mRNA, including the untranslated regions. C. A gene is a unit of expression, including the non-coding regions that are required for transcription and translation. D. A gene is any DNA sequence that has an open reading frame greater than 20 amino acids. E. A gene is an RNA sequence that has been exported from the nucleus. 3. (3 points) Which of the following is a TRUE statement about the human genome: A. Humans have more genes than all other multicellular organisms, which explains the increased complexity found in humans. B. Sequencing of the first human genome was done using Next Generation Sequencing and costs approximately $1000. C. The human genome encodes approximately 20,000 genes. D. The human genome has only 3% repetitive sequences. E. As compared to yeast, human genes are very dense and have less intergenic sequences. 4. (3 points) Which of the following is TRUE about the relationship between gene number and genome size? A. The complexity of an organism is dictated both by the gene number and the size of the genome. B. All organisms have about 20,000 genes, but the genome size increases with increasing organism complexity. C. There is a good correlation between the size of an organism’s genome and the number of genes found in that organism. D. Humans are the most complex animals because we have the most genes and the biggest genomes. E. An increase in intergenic and repetitive sequences leads to a decrease in gene density and an increase in genome size. 5. (3 points) Which of the following best describes the untranslated regions (UTRs) of mRNAs? A. UTRs are sequences located 5’ of the start codon and 3’ of the stop codon. These sequences are important for translation. B. UTRs are located in introns and are not included in the mature mRNA. C. UTRs are the noncoding regions that are required for transcription. D. UTRs are not located within genes. E. UTRs are sequences located between the start and stop codons and are required for both transcription and translation. Last

6. (3 points) The price of DNA sequencing has been falling rapidly since 2007 due to: A. increased funding available for sequencing genomes. B. the invention of new technologies such as Next Generation Sequencing. C. increased output from Sanger Sequencing technology due to automation. D. a decrease in the cost of reagents such as DNA polymerase. E. new and improved methods for collecting DNA samples. 7. (3 points) Consider the 5-carbon sugar of a DNA nucleotide. To which carbon is the base attached? A. 1’ B. 2’ C. 3’ D. 4’ E. 5’ 8. (3 points) At the 5’ terminus of an oligonucleotide, what is attached to the 5’ carbon of the 5- carbon sugar? A. Phosphodiester linkage B. Hydroxyl group C. Hydrogen D. Phosphate group E. A purine or pyrimidine base 9. (6 points) Numbers are used to label specific parts of the dinucleotide pictured below. Write the number to the left of each feature that corresponds to that feature in the picture. ____ Phosphodiester Bond ____ Purine Base –––– Pyrimidine Base ____ 5’ end of the dinucleotide ____ A hydroxyl (OH) is found in this location and is required for DNA synthesis ____ In RNA, a hydroxyl is found in this location, but a hydrogen is attached in DNA N N N N H N 2 P O O O O P O O O O 1 2 3 4 5 6 PO NB I 6 4 5

15. (3 points) Which of the following NOT TRUE about next generation sequencing (NGS)? A. NGS uses reversible chain terminating dNTPs. B. NGS is more accurate than Sanger sequencing. C. NGS is cheaper than Sanger sequencing when considering the per base cost. D. NGS was NOT used to sequence the human genome. E. Using NGS you can sequence millions of base pairs in one experiment. 16. (4 points) You have performed a Sanger sequencing experiment and produced the gel shown below. What was the sequence of the template strand? 17. (4 points) In the Sanger sequencing experiment described in question 13 above, imagine you forgot to add ddATPs to the reaction you loaded in lane #1. The reaction was otherwise assembled correctly and included DNA polymerase, the DNA template, and dNTPs. Explain the result shown in lane 1 of the gel below. 10 9 8 7 6 5 4 3 2 1 A C T G 10 9 8 7 6 5 4 3 2 1 A complementary 5 CT GAC GCT G A 3 temp 3 GA CT G C G ACT S 5 TCA GCGT CAG 3 DNA synthesis occurred as normal w o terminating

18. (3 points) What is the chemical difference between dNTPs and ddNTPs? A. dNTPs have a hydroxyl group attached to the 3’carbon and ddNTPs have a hydrogen attached to the 3’ carbon. B. dNTPs have a single phosphate group attached to the 5’ carbon and ddNTPs have three phosphate groups attached to the 5’ carbon C. dNTPs have purine bases attached to their 1’ carbon and ddNTPs have pyrimidine bases attached to their 1’ carbon D. dNTPs have a hydrogen attached to the 2’carbon and ddNTPs have a hydroxyl group attached to the 2’ carbon. E. dNTPs have a 5-carbon sugar and ddNTPs have a 6-carbon sugar 19. (3 points) Both Sanger Sequencing and Next Generation Sequencing (NGS) use DNA synthesis as a basis for sequencing DNA. Which of the following best describes a major difference between these two strategies: A. A primer to start synthesis is required for Sanger sequencing, but not NGS. B. A DNA template is required for Sanger sequencing, but not NGS. C. DNA polymerase is required for Sanger sequencing, but not NGS. D. In modern labs, Sanger sequencing is no longer used. All sequencing experiments are done with NGS. E. NGS chain terminating dNTPs have reversible blocks. By contrast, Sanger sequencing uses ddNTPs and chain termination cannot be reversed. 20. (3 points) Which of the following is NOT TRUE about histones? A. Post-translational modifications of histone tails affect gene expression. B. Histones bind DNA at the following consensus sequence in DNA: TTATAAT C. Histones are highly basic proteins because they have a high percentage of positively charged amino acids. D. The nucleosome core is made up of 8 histone proteins. E. Histones are highly conserved in eukaryotes. 21. (3 points) Which of the following combination of histones makes up the core of a nucleosome? A. Two molecules each of H2A and H2B and one molecule of H1 B. Two molecules each of H2A, H2B, H3 and H4 C. One molecule each of H2A, H2B, H3 and H4 D. Two molecules each of H3 and H4 and one molecule of H1 E. Two molecules of H1 22. (3 points) Which of the following statements best describes Histone H1? A. H1 is part of the nucleosome core and is found in all nucleosomes. B. H1 is specific to mammals where gene regulation is more complicated. C. H1 is the linking histone and is required to compact the DNA into the 30nm fiber. D. H1 is required for transcription. E. All of the above are true

27. (4 points) The results below show the melting curves for three pieces of DNA of the same size, but with different percentages of GC base pairs. The melting temperature (Tm) is defined as the temperature at which half of the DNA is single stranded (ssDNA) and half is double- stranded. What is the approximate Tm for the piece of DNA with 50% GC base pairs? Is the melting temperature for the piece of DNA with 80% GC base pairs higher or lower than this? Why? 28. (3 points) Why does a higher percentage of GC base pairs increase the melting temperature of double stranded DNA? A. GC base pairs have more hydrogen bonds than AT base pairs. B. GC base pairs are covalently linked, whereas AT base pairs are not. C. GC base pairs are less common than AT base pairs D. GC base pairs have 3 phosphodiester linkages, whereas AT base pairs have only 2. E. GC base pairs are more tightly bound by histones than AT base pairs because they are more negatively charged. 29. (3 points) You designed an RNA probe and used it to perform a Northern Blot. In your first attempt, you do not have any bands in your positive control. You suspect that the temperature you have used for hybridization is too high, thus not allowing for hybridization between the probe and the target DNA because the conditions are too stringent. Unfortunately, your incubator is broken and can only be used at one temperature. Which of the following might be a solution to this problem? A. Decrease the cation concentration in the hybridization buffer B. Increase the concentration of formamide in the hybridization buffer C. Decrease the concentration of formamide in the hybridization buffer 75 C higher because its farther to the right and has a higher GC content

30. (3 points) You have performed a Northern Blot and you find that your radioactive RNA probe is binding non-specifically to several species of RNA, thus giving you several unexpected bands in your negative control lane. This indicates your hybridization conditions are not stringent enough. Unfortunately, your hybridization oven is stuck at 65°C, so you cannot increase the temperature to reduce this nonspecific binding. Which of the following step(s) could you take to reduce non-specific binding and thus improve your results? A. Decrease the concentration of formamide in the solution. B. Increase the concentration of formamide in the solution. C. Decrease the cation concentration in your solution. D. Both A and C would improve specificity. E. Both B and C would improve specificity. 31. (3 points) Which of the following methods can be used to determine if a protein is expressed in the cytoplasm or the nucleus? A. Northern Blot B. Southern Blot C. Western Blot D. Immunofluorescence E. RNA in situ hybridization 32. (3 points) Which of the following reagents can be used to selectively detect proteins in the laboratory? A. Radioactive Oligonucleotides B. Antibodies C. Restriction Enzymes D. Plasmids E. Peptides 33. (3 points) What is the purpose of the secondary antibody in a western blot experiment? A. Secondary antibodies are fluorescently tagged and bind to the primary antibody. B. Secondary antibodies are fluorescently tagged and bind directly to your protein of interest. C. Secondary antibodies are fluorescently tagged and hybridize to specific pieces of DNA in your sample. D. Secondary antibodies disrupt the interaction between your protein of interest and the primary antibody. E. Secondary antibodies are only used in immunofluorescence experiments, not in western blot experiments.

37. (4 points) You are studying Very Mysterious Gene (VMG), which is expressed in cultured human fibroblast cells. Based on reading the literature, you hypothesize that VMG moves from the nucleus to the cytoplasm in response to heat stress. The cells are normally cultured at 37°C, but heat stress is induced by culturing the cells at 42°C for 30 minutes. Briefly outline an experimental design that uses a method discussed in class to test your hypothesis. Assume you have at your disposal an antibody that recognizes VMG, a culture of human fibroblast cells, and all of the standard equipment and reagents found in a molecular biology lab. If your hypothesis is correct what result do you expect to observe? (Note: Complete sentences are not necessary) 38. (3 points) What is the purpose of a selectable marker, such as antibiotic resistance, in a plasmid? A. The selectable marker ensures that the plasmid is copied during every cell cycle. B. The selectable marker allows you to select for the bacteria that carry the plasmid. C. The selectable marker contains several restriction enzyme sites. D. The selectable marker is used to ensure that your experiment is not contaminated by other bacteria in the environment. E. Both A and B are purposes of the selectable marker 39. (3 points) DNA ligase is an enzyme used by molecular biologists to ligate DNA together. In what biological process does a cell use this enzyme? A. Genome sequencing B. Transcription C. Translation D. Formation of Nucleosomes E. DNA Replication Immunofluorescence 1 Fix the cell permeabilize membrane 2 Wash w primary antibody secondary antibody wash excess away 3 Visualize the microscope MCS

Problems 40-42 You are studying a gene in the nematode worm C. elegans that is required for proper muscle function. You name the gene Super Cool Gene (SCG). After using BLAST and reading the literature, you discover a mammalian homolog of SCG that is associated with heart disease. You therefore become interested in studying the function of this gene (mSCG) in mice. To determine where the gene is expressed in mice, you extract mRNA from several different tissues. You run these mRNA samples on a gel and use a radioactive RNA probe that is antisense to mSCG mRNA to test for the presence of mSCG mRNA in each tissue. You use a radioactive probe that hybridizes to the ubiquitous metabolic enzyme GAPDH as a positive control. Here are the results of your experiment: 40. (3 points) What is the name of the experimental procedure performed above? A. Northern Blot B. Southern Blot C. Western Blot D. Immunofluorescence E. RNA in situ hybridization 41. (3 points) Which experimental procedure would you use to determine which cell types of the heart express mSCB protein ? A. Northern Blot B. Southern Blot C. Western Blot D. Immunofluorescence E. RNA in situ hybridization 42. (4 points) Based on the results of this experiment, which tissues can you conclude do not express mSCB? Which tissues do express mSCB? Kidney Heart Lungs Skeletal Muscle Ovary Testis Brain Liver mSCG GAPDH Kidney brain liver lungs ovary and testis dont express MSCB Skeletal muscle Heart express MSCB

Problems 46-50: In class we learned that there are five highly conserved histones that are found in all eukaryotic organisms. In addition to these, there are several histone variants that can be incorporated into nucleosomes under special circumstances or in specific cell types. Imagine that you are a scientist and you have just discovered a new variant of histone H2A, which you decide to call Histone H2A.V1. You perform both a northern blot and a western blot to discover which tissues express Histone H2A.V1 mRNA and protein. For both experiments you use GAPDH as a positive control because both the mRNA and protein of this metabolic enzyme should be expressed in all tissues. In both experiments you observed bands that you predict are H2A.V1 based on the expected size of H2A.V1 mRNA and protein. These bands are marked on the results presented below with a label at the appropriate size that says “H2A.V1.” However, in both experiments you also observe additional bands that are too large to be H2A.V1 mRNA or protein. 46. (3 points) In which tissue(s) can you reasonably conclude that H2A.V1 mRNA and protein are found? A. H2A.V1 mRNA is found in the Brain and the Testis, while H2A.V1 protein is only found in the Testis. B. H2A.V1 mRNA is found in all tissues, while H2A.V1 protein is found only in the Testis. C. H2A.V1 mRNA is found in all tissues except the Liver, while H2A.V1 protein is found only in the Testis. D. H2A.V1 mRNA is found in all tissues except the Liver, while H2A.V1 protein is found only in the Testis, Skeletal Muscle, and Heart. E. H2A.V1 mRNA is found in the Brain and the Testis, while H2A.V1 protein is found in the Testis, Skeletal Muscle, and Heart. 47. (4 points) Your results show that H2A.V1 mRNA and protein expression are not the same in each tissue. Thinking about the central dogma, what stage in the flow of genetic information is likely blocked? In what tissue does this block occur? Kidney Heart Lungs Skeletal Muscle Ovary Testis Brain Liver H2A.V1 GAPDH Northern Blot Kidney Heart Lungs Skeletal Muscle Ovary Testis Brain Liver GAPDH Western Blot H2A.V1 Transcription brain

48. (3 points) In your northern blot experiment above, you observe bands that are larger than the predicted size of H2A.V1 mRNA. These bands are likely background due to your probe binding non-specifically to another RNA. When you repeat the experiment, which of the following can you change to stop non-specific binding of your probe? A. Increase the temperature of hybridization. B. Increase the concentration of formamide in the hybridization solution. C. Increase the cation concentration in the hybridization solution. D. Both A and B would improve specificity. E. Both A and C would improve specificity. 49. (3 points) What experiment could you perform that would tell you which cell types of the testis express H2A.V1 protein? A. Northern Blot B. Southern Blot C. Western Blot D. Immunofluorescence E. RNA in situ hybridization 50. (4 points) In your western blot experiment above, you observe bands in the skeletal muscle, heart, and lungs that are larger than the predicted size of H2A.V1 protein. You hypothesize that the secondary antibody you are using is non-specifically binding to a protein in these tissues. In one sentence, what control experiment could you run to test this hypothesis? 51. (3 points) In genomics, what is the definition of a “DNA Contig?” A. A single DNA molecule containing genes and non-coding regions B. A set of genes with similar functions C. A contiguous sequence of DNA assembled from overlapping fragments D. The entire genome of an organism E. Short, non-contiguous sequences of DNA used for PCR amplification 52. (3 points) Which of the following is a feature of long read sequencing, but not short read NGS (e.g., Illumina)? A. Long read sequencing does not require DNA amplification. B. Long read sequencing is less susceptible to sequencing errors. C. Long read sequencing is faster and more efficient. D. Long read sequencing yields shorter read lengths. E. Long read sequencing uses fluorescently labeled ddNTPs. secondary antibody step only Repeat western blot but leave out primary antibody

58 . Mark the following statements about Northern Blot and RT-PCR as True or False: ____ Both techniques directly measure RNA. ____ Amplification of cDNA by PCR allows for use of lower concentration RNA samples in RT-PCR. ____ cDNA used in RT-PCR will contain intron sequences. ____ Northern blot uses PCR to amplify RNA. ____ Northern blot separates RNA by size and can detect multiple bands of different sizes within a sample. 59 . If RT-PCR primers were designed to bind within the first intron of a gene, would you expect to see a band on your gel for samples expressing the mRNA? 60. Which of the following techniques can be used to visually determine the specific cell type(s) in which an mRNA is expressed within a tissue? A. Northern blot B. Immunofluorescence C. RNA in situ hybridization D. Bulk RNA-sequencing E. A and C 61. Which of the following techniques preserves spatial information about expression? A. RT-PCR B. Immunofluorescence C. Western blot D. RNA in situ hybridization E. B and D 62. What is the purpose of an affinity tag in protein purification? No b c introns are spliced out in mature mRNA It allows us to separate our protein of interest by running all proteins from our sample in a column containing something like metal ions if using His tag that our tag has high affinity for This will separate our protein from the rest of the proteins due to its strong affinity for the column

63. Which of the following techniques use antibodies to recognize protein ? A. Western blot B. Northern blot C. RT-PCR D. Immunofluorescence E. A and D 64. Which of the following techniques DOES NOT employ nucleic acid hybridization in any aspect of the experiment? A. Northern blot B. Immunofluorescence C. RNA in situ hybridization D. RT-PCR 65. For the previous question, briefly explain the role of nucleic acid hybridization in the three techniques which you did not select as the answer in question 64 . 66. You want to clone Your Favorite Gene (YFG) into a plasmid so that you can make purified protein. If there is a BamHI restriction enzyme cut site in the middle of YFG, should you use this enzyme to clone this gene? Briefly explain why or why not. The role of Nucleic acid hybridization in the 3 techniques is that similar sequences will find each other No b c it will cut my gene so I will be unable to fully insert it into the plasmid maintain its parity