Exam4_S2024 and Key

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BIOL 2323 Exam 4 (Spring 2024) 1 EXAM 4 GENERAL GENETICS BIOL 2323 – SPRING 2024 • Mark your answers to the exam questions on a blue ANSWER SHEET NO. 4521 using a pencil No. 2. It is essential that you enter your name and university ID on the form by filling in the bubbles on the sheet!! • Return both the answer sheet and the exam. Be aware that only entries on the scantron answer sheet will count for grade assignment! If your answer on the exam is correct, but not on the answer sheet, you will (for obvious reasons) not get credit for answering the questions correctly. • Read each question and the selection of answers completely and carefully before choosing the best answer . Mark the correct answers on your exam by circling the corresponding letter. It is recommended that you transfer the results to the answer sheet after you have answered all questions and you are sure that your answers are final. • Trying to correct entries on the answer sheet may lead to errors when the sheet is read. It you are unable to erase completely, it is much safer to fill in a new sheet. • You can earn a maximum of 36 points on this exam by answering all 35 questions and the bonus question (on page 11) correctly. GOOD LUCK! ……………………………………………………………………………………………………………….

BIOL 2323 Exam 4 (Spring 2024) 2 The following table may be useful for answering some of the questions: 1. The following is the sequence of a very small eukaryotic protein: N-Met-Ser-Trp-His-Tyr-C Which of the following is a plausible mRNA sequence for translating the above protein? (Hint: *think* like a eukaryotic ribosome initiating translation) A. 5’ -ATGAGTTGGCACTAT- 3’ B. 5’ - ACAGCAUGUCUUGGCAUAUACCGGAUU3’ C. 5’ -GACGCAUAGUUGGCCUUCGAAAUAC- 3’ D. 5’ -UCCACGUGUGACCUUAAACCAAAAAA- 3’ E. None of the above. 2. As hinted at in question above, there can be several possible mRNA sequences that are translated into an identical protein. Why is this the case? A. The genetic code is unambiguous. B. The genetic code is degenerate. C. The genetic code is a triplet. D. The genetic code is nearly universal. E. The genetic code is non-overlapping.

BIOL 2323 Exam 4 (Spring 2024) 3 3. What was the significance of Sir Archibald Garrod ’s research on alkaptonuria (black urine disease)? A. He was the first to sequence DNA to discover that alkaptonuria patients lacked a key metabolic gene. B. He was the first to show that alkaptonuria patients had a hypermorphic mutation leading to excessive production of a metabolite that turns urine black. C. He was the first to connect a heritable disease to a specific metabolic defect. D. He was the first to visualize intragenic recombination. E. He was the first to discover why asparagus makes pee smell funny. 4. The following figure shows a simplified pathway for methionine biosynthesis in E. coli . If you generated a mutant defective only in producing functional Enzyme 4, which compound would you expect to accumulate inside of the cell? A. serine B. O -acetyl serine C. cysteine D. cystathione E. homocysteine 5. Mutations in the gene encoding isocitrate dehydrogenase (IDH) are associated with several different types of cancers. Wild-type IDH converts isocitrate to alpha-ketoglutarate + CO 2 , while the mutant IDH instead produces 2- hydroxyglutarate (an “oncometabolite” that promotes cancer progression). This is what type of mutation? A. null. B. hypomorphic. C. hypermorphic. D. neomorphic. E. mighty morphin. 6. Still considering the IDH allele above, which description best matches your predictions for the properties of the disease allele? A. The disease allele is recessive to the wild-type allele, and results from a deletion of the IDH gene. B. The disease allele is recessive to the wild-type allele, and results from a missense mutation in the IDH gene. C. The disease allele is dominant to the wild-type allele, and results from a deletion of the IDH gene. D. The disease allele is dominant to the wild-type allele, and results from a duplication of the IDH gene. E. The disease allele is dominant to the wild-type allele, and results from a single amino acid substitution of IDH. Serine O- acetyl serine cysteine cystathione homocysteine methionine Enzyme 1 Enzyme 2 Enzy me 3 Enzyme 4 Enzyme 5

BIOL 2323 Exam 4 (Spring 2024) 4 7. All of the following were key observations in Neurospora crassa that contributed to Beadle and Tatum devising the “one gene - one enzyme” hypothesis except… A. the observation that UV could induce auxotrophic mutations in haploid conidia (spores). B. the observation that auxotrophic mutations showed at least dihybrid inheritance patterns when backcrossed to the wild-type parental strain. C. the observation that auxotrophic mutants required a single type of nutrient. D. the observation that auxotrophic mutants were blocked at a single biochemical steps. E. All of the above were key observations. 8. The tertiary structure of a protein… A. is determined by its primary amino acid sequence. B. refers to the amino acid sequence of a protein. C. can include α - helices or β -pleated sheets. D. refers to the oligomeric state of a protein. E. can include stem loops and hairpins. 9. Your friend Logan has discovered a bacterium that biosynthesizes Adamantium and requires it for growth. Being something of a scientist yourself, you perform a feeding experiment with four known intermediates. Based on the results of the feeding experiment below, what is the biochemical pathway for Adamantium biosynthesis. (Note: +/- refers to growth or no growth, respectively) A. Compound A → Compound B → Compound C → Compound D → Adamantium B. Compound C → Compound D → Compound B → Compound A → Adamantium C. Compound D → Compound A → Compound B → Compound C → Adamantium D. Compound D → Compound C → Compound B → Compound A → Adamantium E. Compound B → Compound A → Compound D → Compound C → Adamantium 10. Based on the results of the feeding experiment above, which enzyme is responsible the reaction that ultimately produces Adamantium? (Note: Mutant 1 corresponds to a defect in Enzyme 1, etc.) A. Enzyme 1 B. Enzyme 2 C. Enzyme 3 D. Enzyme 4 E. Wild Type

BIOL 2323 Exam 4 (Spring 2024) 5 11. Which of the following events signals the termination of translation? A. The ribosome runs out of tRNAs. B. The polypeptide folds into a protein. C. The ribosome reaches the end of the mRNA. D. The ribosome reaches a start codon. E. The ribosome reaches a stop codon. 12. You have isolated 2 new rII - T4 phage mutants and wish to determine whether they are located in either the rII A gene or rII B gene. Fortunately, you have previously characterized a mutant with a mutation in the rII A gene (Mutant A1). You performed complementation analysis by co- infecting mutant A1 with either Mutant 2 or Mutant 3. The results are shown below. Note that lysis only occurs if wild-type rII + phage are recovered from co-infection. Which genes are likely mutated in Mutants 2 and 3, respectively? A. gene A, gene A B. gene B, gene B C. gene A, gene B D. gene B, gene A E. gene C, gene C 13. If you were designing an experiment to calculate the recombination frequency between two different T4 rII - mutations, which scheme would you use to most easily identify rare recombinant phage? A. A selection for phage able to infect the E. coli K12 strain. B. A selection for phage unable to infect the E. coli K12 strain. C. A screen for large plaques on the E. coli B strain. D. A screen for normal plaques on the E. coli B strain. E. A screen based on microscopic inspection of individual phage. 14. When studying proflavin-induced mutants, Crick and colleagues observed that combining 3 single base-pair deletions or 3 single base-pair insertions restored the phage to Wild Type. However, wild-type phage were only recovered if the 3 deletions or insertions mapped very close together. What is the most likely explanation for this observation? A. Proflavin induced mutations naturally cluster next to each other. B. Proflavin induced mutations always occur in threes. C. Cellular repair machinery prevents insertions or deletions from occurring at great distances. D. Recombination reduces the distance between insertions or deletions. E. The further the mutations are separated from each other, the more of the open reading frame remains frameshifted. Co-infected Mutants Results (+/- lysis) A1 + 2 + A1 + 3 -

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