Exam 3 Genetics

Practice questions for BIO 340 (Dr. Holechek’s Exam 3) Note: We will be discussing these questions during office hours on Tuesday, February 20 from 9:30 am - 11:00 am . To join the meeting: https://asu.zoom.us/j/85451466686 1. Which activity of DNA polymerase is responsible for its ability to proofread newly synthesized DNA? Exonuclease 2. How many mutations would you expect to see in the human genome (6 x 10 9 ) if mismatches could not be repaired? Hint: Because of tautomeric shift-induced mismatches, DNA replication inherently can be no more accurate than 10 -4 mutations/base pair. 6 x10^-4 x 10^9 = 600,000. 9-4=5 3. Proofreading removes approximately 99% of mismatches before DNA replication moves on. After proofreading what is the mutation rate for DNA replication in humans? 10^-6 mutations/base pairs 4. Each of the five DNA strands below (A-E) are slowly heated. Which strand will melt first? Hint: Think about the number of hydrogen bonds between A and T vs G and C. A. 5’-ATATCGATTA-3’ less G’s-C’s 3’-TATAGCTAAT-5’ B. 5’-ACGGATGCAC-3’ 3’-TGCCTACGTG-5’ C. 5’-GCGCCCGGCG-3’ 3’-CGCGGGCCGC-5’ D. 5’-TTCGAGTGAA-3’ 3’-AAGCTCACTT-5’ E. 5’-CGCCGGATGC-3’ 3’-GCGGCCTACG-5’

5. What is the sequence of the right hand strand of the DNA shown below? Hint: The figure below can help, think about where the 5’ phosphate and the 3’ hydroxyl groups are located with respect to the pentose sugar. A. 5’ ACTG 3’ B. 5’ TGAC 3’ C. 3’ CAGT 5’ D. 3’ TGAC 5’ E. 5’ TGAC 3’ 6. The two oligonucleotides below are allowed to anneal and a set of sequencing reactions are performed. What will the sequencing gel of the first 6 nucleotides added to the DNA look like? Hint: Think about directionality and the concept that DNA strands are antiparallel. 5’ ATCCTGGACACTGTACCATCGGTACCAATCACAGGTCCTTACAGT 3’ 5’ CAGTGTC 3’

11. What would the consequences be of having DNA polymerase without 5’->3’ exonuclease activity? The consequences of having DNA polymerase without 5' to 3' exonuclease activity would include a higher error rate, an increased risk of mutations, reduced DNA fidelity, impaired DNA repair mechanisms, and potentially reduced viability for the organism or cell. The exonuclease activity of DNA polymerase is critical for maintaining the integrity of the genetic material during DNA replication . 12. What is the correct order of enzyme action during DNA replication? A. Ligation, priming, DNA polymerization, helicase/topoisomerase activity, proofreading, 5’-3’ exonuclease activity/DNA polymerization. B. Priming, helicase/topoisomerase activity, proofreading, DNA polymerization, 5’-3’ exonuclease activity/DNA polymerization, ligation. C. Helicase/topoisomerase activity, priming, 5’-3’ exonuclease activity/DNA polymerization, ligation, DNA polymerization. D. Helicase/topoisomerase activity, priming, DNA polymerization/proofreading, 5’-3’ exonuclease activity/DNA polymerization, ligation. E. Priming, DNA polymerization, helicase/ligase, proofreading, ligation, 5’-3’ exonuclease activity/DNA polymerization. 13. What is the sequence of the DNA strand being synthesized during sequencing ? 14. Why is telomerase important? It allows for telomere length and equilibrium maintenance by adding on repeats to the end of the chromosome . 15. For the 3 genes shown below, what are the template strands?

A. Lower, lower, upper. Read Template 3’ to 5’ B. Upper, upper, lower. C. Lower, upper, lower. D. Upper, lower, lower. E. Lower, upper, upper. F. Upper, lower, lower. Use the DNA sequence below, which encodes a prokaryotic gene to answer the next 3 questions. 1 11 21 31 41 ATGAGGAGTT GACACACAAG AGGAGGTAGC AGTATGGGTA TAATCTAATG 51 61 71 81 91 CGTAATTGAG GAGGTAGTTG ACGTATGAAT AGTTAACGTA CGGGGGGGAA 101 111 121 131 141 ACCCCCCCTT TTTTTTTTTC GAGCAATAAA AGGGTTACAG ATTGCATGCT 16. What region of this prokaryotic DNA sequence will be transcribed into mRNA? A. 1-131 B. 71-119 C. 74-149 D. 54-119 E. 74-131 17. What will the sequence be for the protein translated from this mRNA? A. N-Met-Glu-Ser-C B. N-Met-Asn-Ser-C C. N-Met-Gly-Ile-Ile-C D. N-Met-Arg-Asn-C E. N-Met-Thr-Gly-C 18. Where is the 3’UTR? A. 9-131 B. 41-119 (F) (L) (L) (I) (M) (V) (S) (P) (T) (A) (Y) (H) (Q) (N) (K) (D) (E) (C) (W) (R) (S) (R) (G)

N-Met-Gln-Cys-Lys-Thr-C What mutation can give you this mutant protein? A. An A>U transition B. Deletion of a C C. Deletion of a U D. An A>G transition E. A C>U transition 23. Which of the following statements is false ? A. The Trp Operon is under negative control B. The Trp Operon is under positive control C. The Lac operon is under positive control D. The Lac operon is under negative control E. Both B and C are both incorrect 24. What is the phenotype of the following cell? Hint: X-gal is an analog of lactose, and therefore hydrolyzed by the - β galactosidase enzyme. lacI lacI- O+lacZ+ LacY+ lacA+ A. Always blue , even without lactose present. B. Blue when lactose is present, clear when lactose is absent. C. Clear when lactose is present, blue when lactose is absent D. Always clear, even without lactose present. E. Clear when lactose is present, blue when lactose is present. 25. What is the phenotype of the following cell?

O+ lacI+O C lacZ+ LacY+ lacA+ A. Always blue , even without lactose present. B. Blue when lactose is present, clear when lactose is absent. C. Clear when lactose is present, blue when lactose is absent D. Always clear, even without lactose present. E. Clear when lactose is present, blue when lactose is present. 26. Under which of the following conditions will E. coli make large amounts of  - galactosidase? A. I + O + Z + , glucose absent, lactose present. B. I - O + Z + /F’I S , glucose absent, lactose present. C. I + O + Z - , glucose absent, lactose present. D. I - O C Z + /F’O+, glucose absent, lactose absent. E. A and D will both give high levels of  - galactosidase gene expression. 27. Under which conditions will E. coli make the least amount of  -galactosidase? A. I + O + Z + , glucose absent, lactose present. B. I S O C Z + glucose present, lactose absent. C. I - O + Z + , glucose absent, lactose absent. D. I S O + Z + , glucose absent, lactose present. E. B and D will both make low levels of  -galactosidase. 28. Design primers that amplify the open reading frame (Start to Stop codon) and allow its insertion into a plasmid vector pBIO340 which has a multi-cloning site (MCS) that includes restriction sites for Bam HI and Hind III restriction enzymes. You want to add the Bam HI site (GGATCC) to your forward primer and the Hind III (AAGCTT) to your reverse primer. Your primers will be 15 bp long each and at the 3’ end of each you will have either the start codon ( Bam HI Forward primer) or the stop codon ( Hind III Reverse primer).g