Wi09 Exam 3 key

Wi09 Exam 3 key - BICD 100 Genetics Exam 3 March 4, 2009...

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Unformatted text preview: BICD 100 Genetics Exam 3 March 4, 2009 Prof. Soowal Be sure to put all answers on your scantron. No electronics (calculators, phones, MP3 players, etc.) permitted during exam. BICD 100 Exam 3 Page 2 Questions 1‐4. In Drosophila, recessive mutations brown (br) causes brown eyes, rudimental (ru) causes rudimentary‐shaped wings, and yellow (y) causes yellow body color. A true‐breeding brown‐eyed male is crossed to a true‐breeding rudimentary‐winged, yellow‐bodied female. The resulting F1 females are testcrossed to triple‐homozygous recessive males to generate the following progeny: Phenotype Number Wild type 40 Yellow body 10 Brown eyed 365 Rudimentary wings 110 Yellow, rudimentary 345 Brown, yellow 80 Brown, yellow, rudimentary 50 TOTAL 1000 1. Which gene is in the middle? a. brown d. These genes are not linked. b. rudimental e. Cannot determine from c. yellow information given. 2. What is the map distance between brown and yellow? a. 18 mu d. 30 mu b. 26 mu e. These genes are not linked. c. 28 mu 3. What is the map distance between rudimental and yellow? a. 10 mu d. 30 mu b. 16 mu e. These genes are not linked. c. 20 mu 4. Which phenotypic class(es) represent the double crossovers? a. Wild type & yellow body b. Brown eyed, yellow body & brown eyed, yellow body, rudimentary wings c. Wild type & brown eyed, yellow body, rudimentary wings d. Yellow body e. These genes are not linked. BICD 100 Exam 3 Page 3 Questions 5‐7. Three linked genes in tobacco control leaf morphology (M), leaf color (C) and leaf size (S). In the pure breeding “Carolina” strain of tobacco, all recessive alleles are present. In the “Virginia” strain, all of these alleles are dominant. These loci are arranged as follows: M‐‐‐‐5 mu‐‐‐‐C‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐20 mu‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐S There is no interference in this region. A cross is made between the Virginia and Carolina strains, and the F1 is backcrossed to the Carolina strain. 5. What proportion of the backcross progeny will resemble the Virginia strain? a. 76% b. 74% c. 50% d. 38% e. 37% 6. What proportion of the backcross progeny will have the McS phenotype? a. 4% b. 2.5% c. 2% d. 1% e. 0.5% 7. What proportion of the backcross progeny will have the MCs phenotype? a. 20% b. 9.5% c. 10% d. 2.5% e. 2% Questions 8‐9. You are examining the relative positions of 5 genes in the newly discovered bacterial species, E. geneticophilus. Those genes are exm, wnt, qtr, fnl, and bio. • It takes 16 minutes to transfer the entire chromosome. • The F factor in Hfr‐1 is integrated into the chromosome at a site that is 1 minute away from the integration site of the F factor in Hfr‐2. • In Hfr‐1, wnt is the first gene transferred after 1 minute, with fnl coming 4 minutes later, followed by exm 1 minute later. • In Hfr‐2, qtr is the first gene transferred after 1 minute, followed 6 minutes later by bio. 8. What is the distance (in minutes) between bio and exm? a. 1 d. 4 b. 2 e. None of the above. c. 3 9. What is the distance (in minutes) between fnl and qtr? a. 1 d. 4 b. 2 e. None of the above. c. 3 BICD 100 Exam 3 Page 4 Questions 10‐14. You’ve just begun work as a lab assistant with Dr. Vy Rus, whose lab studies bacteriophages. Wild type bacteriophages (T4) can infect E. coli strains B and K, while mutant forms can infect B but not K. Dr. Rus identified 9 mutants (shown below). She coinfects E. coli with these phage in pairwise combinations, and determines whether a plaque can grow (+) or not (0) on strain K. A few combinations have not yet been tested (blank). 1 2 3 4 5 6 7 8 9 0 0 + + 0 0 0 + + 1 0 + + 0 0 0 + + 2 0 0 0 + 0 0 3 0 + 0 + 0 0 4 0 0 + + 5 0 0 0 0 6 0 + + 7 0 0 8 0 9 10. What is the experiment type performed above? a. Complementation test d. Continental test b. Conjugation test e. Cabotage test c. Compilation test 11. How many cistrons can be identified from the given data? a. 1 e. Unable to determine b. 2 with the current gaps in c. 3 the chart. d. 4 12. Which mutations are in the largest cistron? a. 1,2,5,6,7 d. A and B b. 3,4,6,8,9 e. A and C c. 3,4,8 13. Your first project in the lab is to fill in the rest of the table, but you accidentally drop the two plates, 5+3 and 5+7. Hoping to avoid the wrath of Dr. Rus, you fill in the above table without getting experimental results. Given the data, what do you expect to be the results of the plates? a. 5 would not form plaques with either 3 or 7. b. 5 should form plaques with 7 but not with 3. c. 5 should form plaques with both 3 and 7. d. 5 should form plaques with 3 but not with 7. e. Don’t guess, Dr. Rus is way too smart and you’re going to get chewed out! 14. What is the nature of mutation 6? a. It is an example of interference. b. It coincidentally failed to complement any situation. c. It is a point mutation. d. It is a defective mutant. e. It is a deletion that spans every cistron. BICD 100 Exam 3 Page 5 For questions 15‐17, use one of the following choices. Each answer may be used once, more than once, or not at all. a. Statement I is true, Statement II is true. b. Statement I is true, Statement II is false. c. Statement I is false, Statement II is true. d. Statement I is false, Statement II is false. 15. (I) Transduction requires cell contact. (II) Transduction is sensitive to Dnase. D 16. (I) Transduction occurs when bacterial genes are carried from a donor cell to a recipient cell. (II) Transduction is carried out by bacteriophage. A 17. (I) Gene transfer in bacteria is bidirectional. (II) Auxotrophic strains of bacteria can grow on minimal media. D 18. A donor strain of bacteria with genotype leu+ gal­ pro+ is infected with phages. The phage lysate from the bacterial cells is collected and used to infect a second strain of bacteria that are leu­ gal+ pro­. The second strain is selected for leu+. Testing for the other genes shows that 47% of the leu+ are also pro+, and 74% of the leu+ are also gal+. Which gene is closest to leu? a. gal b. pro c. Either is theoretically possible. d. Cannot determine from the information given. 19. L and M are 65 map units apart on the same chromosome. What fraction of the progeny of a cross between LM/lm and lm/lm would be expected to be Lm/lm? a. 10% d. 50% b. 25% e. None of the above. c. 32.5% BICD 100 Exam 3 Page 6 Questions 20‐23. Wild‐type unicorns have long horns (L), blue eyes (B), and a white coat (W). As a unicorn breeder, you are very excited to discover a new (if not so photogenic) strain of unicorn that has short horns (l), brown eyes (b), and a patched coat (w). After a series of crosses between the two different strains of unicorns, you obtain several true‐breeding lines. You cross a true‐breeding short horned, blue eyed, white coat unicorn with a true‐breeding long horned, brown eyed, patched coat unicorn. You cross the F1 progeny to the original newly discovered strain and obtain the following progeny: Phenotype brown, short, patched blue, short, patched blue, long, patched brown, long, patched brown, short, white blue, long, white blue, short, white brown, long, white TOTAL Number 111 13 14 119 11 114 106 12 500 20. What is the genotype of the F1 progeny? (Disregard gene order here.) a. LBW/lbw d. LbW/lBw b. Lbw/lBW e. None of the above. c. LBw/lbW 21. Which gene is in the middle? a. L b. B c. W d. The genes are unlinked. e. Cannot determine from the information given. 22. What is the map distance between L and W? a. 5 mu b. 8 mu c. 10 mu d. 40 mu e. The genes are unlinked 23. What is the map distance between B and W? a. 5 mu d. 40 mu b. 8 mu e. The genes are unlinked c. 10 mu 24. Which of the following is/are true of a negative interference value? a. A crossover decreases the probability of a subsequent crossover event. b. A crossover increases the probability of a subsequent crossover event. c. The coefficient of coincidence is less than 1. d. The number of observed crossovers is less than expected. e. More than one of the above is true. BICD 100 Exam 3 Page 7 Questions 25‐27. Three genes, X, Y, and Z, are arranged in that order (XYZ) on a chromosome. A cross between a triple heterozygote with the three genes in the cis configuration and a triple homozygous recessive individual yields the following progeny: Phenotype Number XYZ 419 xyz 411 Xyz 44 xYZ 46 XYz 33 xyZ 37 XyZ 6 xYz 4 TOTAL 1000 25. What are the map distances between X and Y, and Y and Z, respectively? a. 5 mu; 4 mu d. 6 mu; 10 mu b. 4 mu; 5 mu e. None of the above. c. 10 mu; 6 mu 26. What is the coefficient of coincidence in this cross? a. 0.0 b. 0.5 c. 1.0 d. 1.25 e. 1.67 27. What can you conclude from this information? a. There is no interference in this region. b. There is complete interference in this region. c. There is positive interference in this region. d. There is negative interference in this region. e. We cannot make any conclusions without additional information. Questions 28‐30. True/False. A=True; B=False. 28. Plasmids do not have to integrate into the host cell chromosome in order to be replicated. A 29. Interrupted conjugation results in the production of Hfr strains. B 30. The process of transferring DNA from one bacterium to another through a bacteriophage is transformation. B ...
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This note was uploaded on 09/01/2011 for the course BICD 100 taught by Professor Nehring during the Winter '08 term at UCSD.

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