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MCB 121 Spring 2012 Midterm 1 KEY (Test Form A) Question Form A Form B Form C 1.

hi can you explain question 6-9. Answers are attached

MCB 121 Spring 2012 Midterm 1 KEY (Test Form A) Question Form A Form B Form C 1. D B D 2. B A B 3. C A C 4. A B A 5. B D B 6. B B B 7. A C A 8. A A A 9. B B B 10. C E C 11. D B D 12. B B B 13. B A B 14. C D C 15. E E E 16. C C C 17. A A A 18. A B A 19. E C E 20. B B B 21. B C B 22. A D A 23. D B D 24. E B E 25. C C C 26. A E A 27. B C B 28. C A C 29. B A B 30. E E E 31. B B B 32. C C C 33. B B B Exam Statistics #Students Hi Low Mean Test Form A 86 100 24 76 Test Form B 85 100 28 74 Test Form C 82 100 30 75 Combined A-C 253 100 24 75
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2 I) (Problems 1-5) (16 points) 1 (4 points) Consider the following (very simplified) pathways for head and tail formation for a newly discovered bacteriophage T45: Three steps have been identified that are important for the conversion of tail and head components into complete heads and tails necessary for the formation of a viable virus. Three genes have been identified that carry out these different steps. To determine which gene is responsible for a given step, a number of temperature sensitive (ts) and cold sensitive (cs) mutations have been isolated in each gene and double mutant viruses have been created. Below are the results of reciprocal temperature shift experiments using these double mutant viruses to infect E. coli. Remember that ts mutants produce viable phage at 20°C but not 40°C and cs mutants produce viable phage at 40°C but not 20°C. For comparison, some infections are shown using either wild type virus or single mutant viruses. PFU stands for “Plaque Forming Units” and is a measure of the success of the infection. A PFU above 95 is considered a productive infection and a PFU below 5 is considered unproductive. Genotype of Phage T4 20°C shifted to 40°C (PFU) 40°C shifted to 20°C (PFU) Wild type 100 101 Gene 1 (ts) 99 100 Gene 1 (ts) + Gene 2 (cs) 100 99 Gene 1 (ts) + Gene 3 (cs) 3 98 Gene 2 (cs) + Gene 3 (ts) 97 99 Based on the data in the chart, which of the following best describes the function of each gene? a) STEP A = Gene 1; STEP B = Gene 2; STEP C = Gene 3 b) STEP A = Gene 1; STEP B = Gene 3; STEP C = Gene 2 c) STEP A = Gene 2; STEP B = Gene 1; STEP C = Gene 3 d) STEP A = Gene 3; STEP B = Gene 1; STEP C = Gene 2 e) STEP A = Gene 3; STEP B = Gene 2; STEP C = Gene 1
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Biology - 8230367.doc

In class we discussed the idea that the 5’-CAP and Poly A tail of a eukaryotic mRNA associate
with a set of proteins that causes the mRNA to circularize, in particular eIF4E, Poly A Binding...

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