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Unformatted text preview: BICD 100 Soowal Fa08 Exam 2 Page 2 Questions 1‐6. True/False (A = True; B = False) 1. 2. 3. 4. 5. B Specialized transduction is used to map distances between phage genes. B Epistatic genes must be dominant. A Recombination frequencies can not be predicted by a chi‐square test. B All human A/B/O blood type alleles are codominant. A Plasmids do not have to integrate into the host chromosome in order to be replicated. 6. B The distance between loci on a chromosome does not affect the recombination frequency between them. Questions 7‐8. You are a geneticist studying the inheritance of kernel color in corn. You are having difficulty, however, in obtaining a true‐breeding line of pink corn. Whenever you cross any two pink corn plants, you obtain 1 white‐kernelled plant for every 2 pink ones. The white plants are true breeding, and only give other white plants when crossed among themselves. 7. What ratio of pink:white do you expect from a cross of a pink plant to a white plant? a. 1 pink:1 white d. all pink b. 2 pink:1 white e. more than one of the c. 3 pink:1 white above are possible 8. What is the best explanation for the results that you’re seeing? a. Epistasis d. Incomplete dominance b. Lethal allele e. Codominance c. Dominant/recessive 9. The figure below shows a partial chromosome map of an E. coli Hfr strain. Each mark equals 10 minutes. If transfer of genes begins at “*” and goes in the direction of the arrow, what is one of the predicted results from this map? ton lac * azi gal a. b. c. d. e. ton and lac will rarely be transferred together. lac and azi will rarely be transferred together. Ten minutes after transfer of ton, azi will be transferred. It would take 30 minutes to transfer all of the genes that are shown. None of the above. BICD 100 Soowal Fa08 Exam 2 Page 3 Questions 10‐12. A liger (the offspring of a lion and tiger) can be born with mutations in three separate genes concerning coat color, coat length, and eye color. Mutations in these traits results in orange fur (o), which is recessive to brown fur (O), long fur (l), which is recessive to short fur (L), and blue eyes (b), which is recessive to brown eyes (B). The map for these genes is as follows: ‐‐‐‐‐‐‐‐‐‐‐‐ L ‐‐‐‐‐‐‐‐5‐‐‐‐‐‐‐‐‐‐ O ‐‐‐‐‐‐‐‐‐‐‐‐10‐‐‐‐‐‐‐‐‐‐‐‐‐B‐‐‐‐‐‐‐‐‐‐‐‐‐‐ A purebred short‐ and orange‐furred, blue‐eyed male tiger mates with a purebred long‐ and brown‐furred, brown‐eyed female lion. The F1 ligers from this mating are mated to triply recessive homozygous ligers, and the resulting 400 offspring are below: 1 Blue eyes 34 Orange fur 1 Long fur, orange fur 34 Long fur, blue eyes 5 Normal 160 Orange fur, blue eyes 5 Blue eyes, orange fur, long fur 160 Long fur 10. How many ligers resulted from double crossovers? a. 2 b. 10 c. 34 d. 68 e. 320 11. Calculate the coefficient of coincidence for this cross. a. 0.25 d. 1 b. 0.5 e. None of the above c. 0.75 12. Calculate the interference for this cross. a. 0 d. 0.75 b. 0.25 e. None of the above c. 0.5 Questions 13‐18. Joe and Sally have two children. They know the blood types of their children, but not of themselves. Their first child, Billy, is type A, and their second child, June, is type B. 13. What are the possible blood types of Joe and Sally (in either order)? a. A and B d. B and AB b. AB and O e. All of the above c. A and AB 14. What are the possible genotypes of Joe and Sally (in either order)? a. IAIB; IAi d. IAIA; IAIB Ai; IAIA b. I e. Two of the above AIA; IBIB c. I 15. What are the possible genotypes of June and Billy? [Don’t make any assumptions about Joe and Sally here.] a. IBIB (June); IAIA (Billy) d. IBi (June); IAIA (Billy) b. IBIB (June); IAi (Billy) e. All of the above c. IBi (June); IAi (Billy) BICD 100 Soowal Fa08 Exam 2 Page 4 Joe is out drinking one night, and ends up in the hospital after attempting to drive home. Sally comes in and offers to donate blood for Joe, who needs a transfusion. When they are both typed, it is discovered that Joe is type O while Sally is type A. 16. What type of antibodies does Sally have? a. Anti‐A d. Anti‐O b. Anti‐B e. No antibodies are c. Both anti‐A and anti‐B present. 17. What is the most likely genotype of Joe? a. IAi, hh d. ii, H_ Bi, hh b. I e. a, b, or d are all possible. c. IAIB, H_ 18. What type of epistasis is exhibited with the ABO blood groups? a. Recessive epistasis b. Dominant epistasis c. Mutual complementation epistasis d. Duplicate recessive epistasis e. The blood groups do not exhibit epistasis Use the following list to answer questions 19‐22. Choose the answer that best fits each description about exchange of genetic material in bacteria. a. Conjugation b. Transformation c. Transduction d. Both B and C e. All of the above 19. A Requires physical contact between cells. 20. C Is mediated by viruses. 21. E Can result in the transfer of chromosomal genes. 22. A Allows Hfr mapping. Questions 23‐25. A series of two‐point crosses were carried out among five loci (a, c, d, f, i) in D. melanogaster producing the following recombination frequencies. Loci RF Loci RF a‐c 30% a‐f 50% d‐c 9% i‐d 4% a‐i 17% i‐f 50% 23. Which of the following statements are true? a. The distance between loci a and c is 30 m.u. b. The distance between loci i and c is 13 m.u. c. The distance between loci a and d is 21 m.u. d. All of the above are true. e. It is not possible to verify all these map distances. BICD 100 Soowal Fa08 Exam 2 Page 5 24. Based on the information given, what CAN you determine? a. Whether loci a and c are on the same chromosome. b. Whether loci f and c are on the same chromosome. c. Whether loci d and f are on the same chromosome. d. All of the above are able to determined from the data. e. None of the above can be determined from the data. 25. Which of the following statements is correct? a. The map distance between loci a and f is the same as the map distance between loci d and f. b. Crossing‐over happens just as frequently between loci i and f as between loci a and d. c. Crossing‐over happens just as frequently between loci i and a as between loci i and f. d. Loci c and f are in the same linkage group. e. Loci d and f are in different linkage groups. (either answer accepted.) Refer to the pedigree below to answer questions 26‐27. 26. In the pedigree above, can you conclude that the nail‐patella syndrome and blood type genes are linked? If so, which alleles are most likely to be in coupling? a. The dominant allele responsible for the nail‐patella disease is in coupling with the B blood type allele. b. The recessive allele responsible for the nail‐patella disease is in coupling with the B blood type allele. c. The dominant allele responsible for the nail‐patella disease is in coupling with the A blood type allele. d. The recessive allele responsible for the nail‐patella disease is in coupling with the O blood type allele. e. There is no linkage. BICD 100 Soowal Fa08 Exam 2 Page 6 27. If there is linkage, how many of the offspring are recombinants? a. 0 (no linkage) b. 1 c. 2 d. 3 e. 4 Questions 28‐30. A curious literary geneticist is characterizing loraxes from the Truffula tree forest. She finds out that a pleasantly‐plump body type of loraxes (P) is dominant to the mutant lean body type (p). She also finds out that bushy mustaches (B) are dominant to the mutant no mustache (b). A heterozygous plump, bushy‐ mustachioed lorax mates with a lean, non‐mustachioed lorax. The very adorable resulting F1 progeny are as follows: Plump, no mustache Plump, mustache Lean, mustache 69 12 71 8 Lean, no mustache 28. Choose the correct statement about the above progeny. a. The alleles of the heterozygous parent MUST be in cis configuration. b. The alleles of the heterozygous parent MUST be in trans configuration. c. Because of the “relative” 1:1 nature of the progeny, these genes exhibit independent assortment. d. There is not enough information to determine whether or not crossing‐over occurred. e. There is not enough information to determine whether the alleles of the heterozygous parent are in cis or trans coupling. 29. Assume that recombination occurs and that these two genes are on the same chromosome. Choose the statement below that BEST characterizes how to determine the recombination frequency. a. Add up the recombinant progeny (69 +71), and then divide by total number of progeny (160). Because that number is larger than 50%, divide by two. Multiply by 100%. b. Add up the recombinant progeny (69 +71); then divide by total number of progeny (160). Then subtract from 1. Multiply by 100%. c. Add up the recombinant progeny (12 +8), and then divide by the nonrecombinant number of progeny (140). Multiply by 100% d. Add up the nonrecombinant progeny (12 +8), and then divide by total number of progeny (160). Multiply by 100% e. None of the above is a valid method for finding the recombination frequency. BICD 100 Soowal Fa08 Exam 2 Page 7 The curious geneticist thinks that she had mislabeled the lorax progeny. The “correct” progeny statistics are now below. Plump, mustache 69 8 12 Plump, no‐mustache Lean, mustache Lean, no‐mustache 71 30. What does this data set say differently from the data set given earlier? a. Nothing is different as the numbers are the same. b. The cis/trans configuration is now changed. c. The recombination frequency is now changed. d. The gene order is now changed. e. It depends ‐ more information is needed. Questions 31‐33. A variety of Gerbera daisies are white, green, or mint (light green) in color. A series of crosses between the three different colors of snapdragons produce the following progeny: Cross Offspring 1. mint x white 7 mint, 5 white 2. mint x green 5 mint, 6 green 3. white x white 14 white 4. green x green 16 green 5. mint x mint 4 green, 10 mint, 5 white 6. green x white 19 mint 31. What is the most likely mode of inheritance in Gerbera daisies? a. Complete dominance d. Dominant epistasis b. Codominance e. Incomplete dominance c. Recessive epistasis 32. What color is the heterozygote? What cross(es) prove this? a. Green, Cross 2 d. Mint, Crosses 1 and 2 b. White, Cross 1 e. Mint, Crosses 1, 2 and 5 c. Mint, Cross 5 33. A green Gerbera daisy and a white Gerbera daisy mate and produced F1 progeny. The F1 progeny were then mated to each other. What are the expected phenotypic percentages of the F2 progeny? a. All mint d. 75% mint and 25% white b. All white e. 25% green, 50% mint, and c. 50% mint and 50% white 25% white BICD 100 Soowal Fa08 Exam 2 Page 8 Questions 34‐36. The mutant recessive traits of long‐body (l), wingless (meaning wings not present) (w), and hairy‐legs (h) all belong to the same chromosome of a certain fly. A heterozygous winged, short‐bodied, and smooth‐legged fly mates with a wingless, long‐bodied, hairy‐legged fly. The resulting progeny (500) are as follows: Long‐bodied, winged, hairy‐legged Short‐bodied, wingless, smooth‐legged Long‐bodied, winged, smooth‐legged Short‐bodied, wingless, hairy‐legged Long‐bodied, wingless, hairy‐legged Short‐bodied, winged, smooth‐legged Short‐bodied, winged, hairy‐legged Long‐bodied, wingless, smooth‐legged 163 137 52 50 38 41 8 11 34. What is the gene order of the chromosome? a. h w l b. l w h c. w h l d. h l w e. None of the above 35. Which of the following is a correct statement about the map distances? a. (8 + 11 + 50 + 52)/(total # of progeny) x 100 gives the map distance between the wing locus and hairy‐leg locus. b. (8 + 11 +50 + 52)/(total # of progeny) x 100 gives the map distance between the wing locus and the body‐length locus. c. (8 + 11 + 41 + 38)/(total # of progeny) x 100 gives the map distance between the hairy‐leg locus and the body‐length locus. d. (8 + 11 + 52 + 38)/(total # of progeny) gives the map distance between the wing locus and the body‐length locus. e. None of the above 36. Assume that the coefficient of coincidence is 1.1. Which of the following statements is most accurate? a. The inference for this cross is relatively high. b. A crossover between one pair of loci decreased the probability of another cross over taking place. c. We are observing 110% of the double crossovers that we expected on the basis of the single‐crossover frequencies. d. There is no interference for this cross. e. None of the above – a coefficient of coincidence must be less than 1. BICD 100 Soowal Fa08 Exam 2 Page 9 37. You are a TA on the genetics discussion board the night before a midterm, feverishly trying to edit out the bad advice from the good amongst the students. Of the 5 posts, which ONE contains the good advice that you would leave up? a. “If you’re given the gene order of three loci as x, y, and z, and given the mapping‐unit distance between x and y as 25, the mapping‐unit distance between y and z should be (50 m.u. – 25 m.u.,) or 25.” b. “In a three point cross: in order to find the gene order, you should compare the four classes of double‐crossover progeny with the nonrecombinant.” c. “In a two point cross: double‐crossover progeny should look different than the parents.” d. “If just given the interference and total numberof progeny, you could find the mapping‐unit distance between loci.” e. “In a three point cross: double‐crossover progeny should differ only at one loci when compared to nonrecombinant progeny.” Questions 38‐40. You have discovered a single‐celled organism that resembles E. coli. There are two phenotypes, solid green or purple with orange stars. Upon further investigation, you find that the green type can be converted to the purple/orange type after conjugation, since all 6 genes necessary for this conversion exist on the F factor. Although you know the enzymes (1‐6 given below) needed for the expression of the purple/orange phenotype, you don’t know the order of their genes on the F factor, nor do you know the distances between these genes. To find out, you perform interrupted conjugations (just like E. coli) with 2 purple/orange F+ strains and observe phenotypes. The results are shown below. Enzyme 1: synthesizes red pigment Enzyme 2: synthesizes blue pigment Enzyme 3: combines red & blue pigments to form purple pigment Enzyme 4: places purple pigment within the plasma membrane Enzyme 5: creates orange polka dots Enzyme 6: converts orange polka dots to stars Strains Time of Entry (minutes) Gene 1 Gene 2 Gene 3 Gene 4 Gene 5 1 25 40 5 30 55 2 31 46 11 36 51 Gene 6 60 1 38. In strain 1, what is the order of the genes on the F factor? a. 1‐2‐3‐4‐5‐6 d. 1‐4‐2‐6‐5‐3 b. 3‐1‐4‐2‐5‐6 e. 6‐5‐4‐3‐2‐1 c. 6‐5‐2‐4‐1‐3 BICD 100 Soowal Fa08 Exam 2 Page 10 39. What are the distances (in minutes) of the genes from strain 2 (in order from first transferred to last)? a. 11, 31, 36, 46, 51 d. 5, 10, 10, 15, 20 b. 10, 20, 5, 10, 15 e. 1, 11, 31, 36, 46 c. 15, 10, 5, 20, 10 40. How long does it take for the entire F factor to transfer from the purple/orange type to the green type of these bacteria you have just discovered? a. 55 b. 60 c. 65 d. 100 e. The F factor will never completely transfer over from the purple/orange type to the green type Questions 41‐45 In summer squash, the genotype ww allows colored fruit, while W fruit is white. Y fruit is yellow (if colored), while yy fruit is green. 41. What is the genotype of an F1 squash from a cross between a pure‐breeding white plant (that would have been green if it could express color) and a pure‐ breeding yellow plant? a. WWYY d. WWYy b. WwYy e. wwyy c. wwYy 42. What phenotypic ratio is expected in the F2 plants if the F1 from question 41 self‐pollinated? a. 9:3:4 d. 9:3:3:1 b. 9:7 e. 12:3:1 c. 15:1 43. What type of interaction is involved? a. Dominant Epistasis d. Duplicate Dominant Epistasis b. Recessive Epistasis e. Gene Linkage c. Duplicate Recessive Epistasis 44. A cross between a plant with yellow fruit and a plant with white fruit produce 58 white: 39 yellow: 16 green. What are the genotypes of the parents? a. WwYy and WwYy d. WwYY and wwYy b. wwYy and WwYy e. WWYY and wwyy c. WWYy and WwYy 45. Another trait in the summer squash involves the presence of seeds (S), which is dominant over the lack of seeds (s). What is the probability of getting a white seedless squash from the selfing of a triple heterozygous squash? a. 3/16 d. 9/16 b. 3/64 e. 27/64 c. 9/64 BICD 100 Soowal Fa08 Exam 2 Page 11 46. What is the difference between penetrance and expressivity? a. Penetrance is used to describe humans while expressivity is used to describe animals. b. Penetrance can be calculated while expressivity cannot. c. Expressivity is the percentage of individuals that express the phenotype while penetrance is the degree to which the trait is expressed. d. Expressivity can be used only to describe large populations while penetrance is used to describe individuals. e. Penetrance and expressivity are interchangeable. A bacterium that does not contain an F factor is considered to be (47)B . If it does contain an F factor, then there are 3 classes it can be categorized into. The F factor of a(n) (48)C exists as a separate circular DNA from the bacterial chromosome. The F factor of a(n) (49) D bacteria is integrated into the bacterial chromosome. The F factor of a(n) (50) A bacteria exists separately from the bacterial chromosome but does carry some chromosomal genes. a. F’ b. F‐ c. F+ d. Hfr e. Wildtype ...
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This note was uploaded on 01/03/2009 for the course BICD 100 taught by Professor Nehring during the Fall '08 term at UCSD.
- Fall '08