BIS101_Exam2_PracticeExam_v3_KEY

BIS101 - Exam 2 – Practice Exam 1 Directions: Answer all questions as best you can. Partial credit is possible if your work and/or reasoning is clear. Keep answers brief – unnecessary information may result in reduced points. No regrades solely due to issues of legibility, so make sure we can read your responses. Question 1: 28 pts Your research project is to engineer a mutation in human cells using Cas9/CRISPR to heart disorders. You want to target the protein coding region of the GATA4 gene, with the coding strand shown below and a gRNA (underlined) and PAM (bold) sequence you identified indicated. 5’-GCCTAGGCATGATCGTGCATGCGAA CGG CA-3’ 1.1 (6 pts): Complete the human gene dsDNA molecule by giving the sequence of the template strand. Include representation of the polarity and bases (no need to represent hydrogen bonds). 3'-CGGATCCGTACTAGCACGTACGCTTGCCGT-5' 5'-TGCCGTTCGCATGCACGATCATGCCTAGGC- 3' Partial credit possible depending on polarity and sequence match. Credit given for just gRNA portion and for single base errors, with the assumption this is a typo. 1.2 (6 pts): Give the RNA sequence transcribed from the GATA4 gene DNA sequence, show bases and polarity. 5'-GCCUAGGCAUGAUCGUGCAUGCGAACGGCA-3' Partial credit possible for polarity and mRNA molecule or coding/template issues. Credit given for gRNA portion only if otherwise correct. 1.3 (12 pts) You are comparing the human and mouse versions of the full GATA4 gene. 1. Give the missing percentages for each nucleotide for the human and mouse genes 2. Which gene version (human or mouse) will have a higher melting temperature? Why? 3. Briefly explain why the information in the DNA table is not enough to know what the percent RNA bases will be for the GATA4 mRNA. (Note 2 versions of table) The higher melting temperature will be the sequence with higher percent GC. This is because more GC bonds means increased proportion of base pairs with 3 hydrogen bonds vs. 2 for AT pairs. Requires more energy/heat to denature/melt GC vs. AT. Two reasons: 1. dsDNA base composition doesn’t indicate specific coding strand sequence, so we don’t know the bases in the mRNA. 2. Since both human and mouse are eukaryotes, there will be introns that are spliced out of the mature mRNA. Partial credit for other explanations DNA Base Human Base % Mouse Base % DNA Base Human Base % Mouse Base % A 36 18 A 14 32 T 36 18 T 14 32 C 14 32 C 36 18 G 14 32 G 36 18

BIS101 - Exam 2 – Practice Exam 2 1.4 (4 pts): You use the gRNA in your experiment, delivering the gRNA and Cas9 to human heart cells. When you examine which sites in the human genome were cut, you find that there are 4 different site that Cas9 cut with this gRNA. Cutting occurred at the intended GATA4 gene target and 1 other site with similar high frequency. The other 2 sites were cut at much lower rates. Give a brief explanation for these results based on the likely DNA sequences present at the 3 sites that are not in the GATA4 gene. . The results are consistent with there being a second site in the genome with the exact same (or 1bp different) DNA sequence that is targeted by the gRNA - this would be the second high frequency cut site. In addition, there are two other sites that are close to the same sequence as targeted by the gRNA (perhaps 1 to 3 mismatched bases, more than the high frequency site), which are much less frequently cut but still occur every so often – these are the two sites cut at much lower rates. Question 2: 20 pts Fill in the table with an “X” for whether the process is related to Replication, Transcription, Translation and whether applies to Prokaryotes or Eukaryotes. A process may be involved in multiple central dogma steps and may be in one, the other, or both Prokaryotes and Eukaryotes. Process Replica-on Transcrip-on Transla-on Prokaryotes Eukaryotes Nucleo’de added to 3' end of growing strand X X X X Binding to Shine Dalgarno sequence X X Interac’on of tRNA an’codon with mRNA codon X X X Synthesis of Okazaki fragments X X X Rho-dependent termina’on X X Telomere lengthening X X Removal of intronic sequence X X AUG defines start of synthesis X X X Produce a molecule that can have enzyma’c ac’vity in the cell X X X X Ac’vity of RNA primase X X X

BIS101 - Exam 2 – Practice Exam 4 CAG to TAG: This introduces a new stop codon (nonsense mutation), protein truncated at this point Likely phenotype: full LFS as likely to result in non-functional protein. Mutation 2: GAT to GAC: This is a synonymous/silent mutation, as the new codon codes for the same AA (Asp) Likely phenotype: healthy as no impact on protein Mutation 3: GAG to ---: Three base deletion will result in loss of one (Glu) AA, otherwise same protein Likely phenotype: milder symptoms as protein changed, but only one AA. Could be either full LFS or healthy if no other information, but compared to other mutations, most likely to be milder symptoms. Mutation 4: GTTGAT to new TGA: Two base deletion creates frameshift that produces a new stop codon Likely phenotype: full LFS as likely to result in non-functional protein. Mutation 5: AAA to ACA: Substitution changes codon (missense) to go from Lys to Thr, so one changed AA Likely phenotype: milder symptoms as protein changed, but one one AA. Could be either full LFS or healthy if no other information, but compared to other mutations, most likely to be milder symptons. Mutation 6: gtg to ---: Deletion of the first three bases of the intron. Likely phenotype: full LFS, as this will impact splicing of the mRNA. In this case, the splicesosome will fail to recognize the start of the intron, thus will not be excised. The protein will contain AA sequence corresponding to the intron. This is likely to have a more significant impact on protein function than a single AA change or single AA deletion. Note: 2pts given for answers that communicated “no impact on protein due to intron mutation” as this is a generally logical line of thinking for intronic mutations

BIS101 - Exam 2 – Practice Exam 5 4 (14 pts) What are the similarities and differences in non-coding function between an origin of replication, promoter and ribomosomal binding site (RBS) with regard to step in the central dogma (replication, transcription, translation), specific role in the relevant central dogma process, and molecular interaction partner? Similarities: (5 pts) All represent non-coding DNA sequences that are recognized by enzymes (rRNA and proteins) that can bind to a specific DNA/RNA sequence and each is critical for starting a central dogma step . All are present in prokaryotes, but no deduction if not stated or eukaryotes stated. Differences: (9 pts) Origen of Replication (Ori): involved in replication , DNA bound by proteins (e.g. DNAa) during initiation step of replication by binding at Ori site. Promoter: involved in transcription , DNA bound by proteins (e.g. Sigma factor) critical in the recruitment RNA Polymerase to the start of the gene. Ribosomal binding site (AKA Shine Dalgarno sequence): involved in translation , mRNA bound by small ribosomal subunit, required for initiation of translation by mediating mRNA binding (start codon recognition next part of initiation). For full credit, need to specify somewhere that these are all sequences that are encoded as noncoding DNA information, that each sequence is bound by an enzyme that can recognize a specific DNA/RNA sequence, and need to state the specific step in the relevant central dogma process. 5. (6 pts) Why does a drug targeting a replication enzyme like topoisomerase have a generally stronger impact on cancerous cells vs. non-cancerous human cells? In contrast, why does the amanita mushroom toxin that inhibits RNA polymerase impact all human cells? 1. Cancerous cells undergo continuous or high rates of replication and are frequently more sensitive to disruption of this process (i.e. cannot recover from things that block or disrupt DNA replication). As most human cells are not actively replicating, this makes drugs that inhibit replication more specific to cancer cells. 2. In contrast, all cells require mRNA transcription, so amanita toxin has a non-specific impact on all human cells over time as new mRNA can’t be produced. Full credit requires correctly addressing both chemotherapy/replication and amanita/transcription.