Recombination January 2009

1. RecombinationJanuary 2009

2. Orientation of alleles on a chromosome

3. T t A A a a t T A man is a carrier for sickle cell anemia (Aa) and Tay-Sachs disease (Tt). Let’s say the genes involved in both of these disorders are on the same chromosome. Which of the following are possible representations of his genotype: #1 #2 #1 only #2 only #1 and #2

4. R r Y Y y y r R Yellow pea color (Y) is dominant to green pea color (y). Round seed shape (R) is dominant to oval pea shape (r). Let’s say the genes responsible for both of these phenotypes are on the same chromosome. Which of the following genotypes should result in a yellow pea plant with round seeds? A. YyRr B. C. D. Both A and B E. All of the above

5. T t A A a a t T A man is a carrier for sickle cell anemia (Aa) and Tay-Sachs disease (Tt). Let’s say the genes involved in both of these disorders are on the same chromosome. Which of the following are possible representations of his genotype: #1 #2 #3 AaTt #1 only #1 and #2 #1 and #3 #2 and #3 #1, #2, and #3

6. vg tz tz+ vg+ vg+ tz tz+ vg Wild-type female Muscle defects, shriveled wing male tz vg X tz vg 155 Wild-type 330 muscle defects, normal wings 165 muscle defects, shriveled wings 350 normal muscles, shriveled wings What is the genotype of the wild-type parental female fly? A. B. C. There’s not enough information tz = mutant version of tafazzin tz+ = normal version of tafazzin vg = mutant version of vestigial vg+= normal version of vestigial Normal dominant over mutant for both

7. wg p p+ wg+ wg+ p p+ wg Wild-type female Pink eye, shriveled wing male p wg X p wg 155 Wild-type 330 pink eye, normal wings 165 pink eye, shriveled wings 350 normal eyes, shriveled wings What do the chromosomes of the wild-type parental female fly look like? A. B. C. There’s not enough information p = mutant version of pink p+ = normal version of pink wg = mutant version of wing wg+ = normal version of wing Normal dominant over mutant for both

8. A b a b a b a b A B a b a b A B a b a B a b a b In the test cross above what is the arrangement of the alleles? A. A. X D. A or B E. A,B, or C B X C. X

9. 2 and 3 point crosses, calculating recombination frequencies

10. Adrenoleukodystrophy (ALD) Accumulation of long fatty acid chains in the body Neurodegeneration, problems with vision, hearing, and motor coordination In Drosophila bubblegum gene is similar to the human gene that when mutant causes ALD. You want to study the linkage between bubblegumand another gene, curly

11. 99 Vision defects, normal wings 395 Wild-type 405 curly wings, vision defects 101 curly wings, normal vision Curly wings, vision defects male Wild-type female bgm cu bgm+ cu+ bgm cu bgm cu X Which classes of progeny arose from parental type gametes? Wild-type & Vision defects, normal wings Wild-type & Curly wings, vision defects Vision defects, normal wings & Curly wings, normal vision Curly wings, vision defects & Curly wings, normal vision

12. Let’s look at the linkage between tafazzin and vestigial genes A Drosophila gene called tafazzin is similar to the human gene that when mutant causes Barth syndrome. Mutations in the gene vestigial prevents flies from normally spreading out their wings. In humans: problems with the heart, susceptibility to infections, and muscle weakening tz = mutant version of tafazzin tz+ = normal version of tafazzin vg = mutant version of vestigial vg+= normal version of vestigial Normal dominant over mutant for both Mutations in tafazzin cause muscle weakness in flies.

13. Wild-type female c+ c, pr+ pr Curly wings, purple eyes c c, pr pr X Progeny: 155 Wild-type 330 normal wings, purple eyes 165 curly wings, purple eyes 350 curly wings, normal eyes 1,000 total progeny What is the recombination frequency between curly and purple? A. 0.16 B. 0.32 C. 0.49 D. 0.52 E. 0.68 c = mutant version of curly c+ = normal version of curly pr = mutant version of purple pr+ = normal version of purple Normal dominant over mutant for both

14. Wild-type female b+ b, g+ g Black, green eye male b b, g g X Progeny: 36 wild-type 15 gray, green eyes 36 black, green eyes 13 black, white eyes 100 total progeny What is the recombination frequency between black and green? A. 0.15 B. 0.13 C. 0.28 D. 0.51 E. 0.72 b = mutant, black body b+ = normal, gray body g = mutant, green eye g+= normal, white eye Normal dominant over mutant for both

15. Wild-type female b+b g+g Black, green eye male bb gg X Progeny: 15 wild-type 36 gray, green eyes 13 black, green eyes 36 black, white eyes 100 total progeny What is the recombination frequency between black and green? A. 0.15 B. 0.13 C. 0.28 D. 0.49 E. 0.72 b = mutant, black body b+ = normal, gray body g = mutant, green eye g+= normal, white eye Normal dominant over mutant for both

16. 150 Wild-type 340 normal wings, orange eyes 160 curly wings, orange eyes 350 curly wings, normal eyes Wild-type female c+c o+o Curly wings, orange eyes c c o o X Progeny: 1,000 total progeny What is the recombination frequency between curly and orange A. 0.16 B. 0.31 C. 0.49 D. 0.50 E. 0.69 c = mutant version of curly c+ = normal version of curly o = mutant version of orange o+= normal version of orange Normal dominant over mutant for both

17. Given 3 genes, how many gene orders are possible? A) 1 B) 3 C) 6 D) 9

18. Let’s look at the y w m cross again 758 y+ w+ m+ 700 y w m 401 y+ w+ m 317 y w m+ y+ w m 12 y w+ m+ y+ w m+ 0 y w+ m What are the NCO chromosomes? A. w+ y+ m+ & w y m B. w+ y+ m & w y m+ C. w+ y m+ & w y+ m D. w+ y m & w y+ m+

19. Let’s look at the y w m cross again 758 y+ w+ m+ 700 y w m 401 y+ w+ m 317 y w m+ y+ w m 12 y w+ m+ y+ w m+ 0 y w+ m What is the order of the genes? A. w - y - m B. y - w - m C. w - m - y

20. Female flies phenotypically wild-type and heterozygous for each of three different mutations [curly wings (c), short bristles (b), and sepia eyes (s)] were crossed to male flies that have curly wings, are short bristles, and have sepia eyes. The number of progeny in eight different phenotypic classes is: Which of the three genes (c,b,s) is in the middle? A. Curly B. Short Bristles C. Sepia

21. Female flies phenotypically wild-type and heterozygous for each of three different mutations [purple eyes (pr), dumpy wings (dp), and hairy (h)] were crossed to male flies that have purple eyes, have dumpy wings, and are hairy. The number of progeny in eight different phenotypic classes is: Which of the three genes (c,b,s) is in the middle? A. hairy B. purple C. dumpy

22. Before answering “Why?,” let’s do a linkage analysis (based on Fig. 5.9 in the book): Female flies who appear wild-type are heterozygous ebony body (e) and short bristles (s) (Gray body (+) and long bristles (+) are wild-type and dominant) are crossed to ebony body, short bristle males. The progeny are 537 wild-type flies, 76 flies with wild-type bristles and ebony bodies, 75 flies with wild-type body color and short bristles and 542 flies with ebony body and short bristles. What are the parental (NCO) chromosomes? +,+ & +, s +,+ & e, + +,+ & e, s e,s & e, + e,s & +, s e, + & +, s

23. Before answering “Why?,” let’s do a linkage analysis (based on Fig. 5.9 in the book): Female flies who appear wild-type are heterozygous ebony body (e) and short bristles (s) (Gray body (+) and long bristles (+) are wild-type and dominant) are crossed to ebony body, short bristle males. The progeny are 537 wild-type flies, 76 flies with wild-type bristles and ebony bodies, 75 flies with wild-type body color and short bristles and 542 flies with ebony body and short bristles. What are the recombinant (SCO) chromosomes? +,+ & +, s +,+ & e, + +,+ & e, s e,s & e, + e,s & +, s e, + & +, s

24. Before answering “Why?,” let’s do a linkage analysis (based on Fig. 5.9 in the book): Female flies who appear wild-type are heterozygous ebony body (e) and short bristles (s) (Gray body (+) and long bristles (+) are wild-type and dominant) are crossed to ebony body, short bristle males. The progeny are 537 wild-type flies, 76 flies with wild-type bristles and ebony bodies, 75 flies with wild-type body color and short bristles and 542 flies with ebony body and short bristles. What is the map distance between ebony and short? A. 6.9 B. 10.9 C. 12.3 D. 14.0 E. 27.9

25. Let’s go back to the white, yellow, miniature cross and check the gene order. Here’s the data for crossing a phenotyically wild-type female by a white eyed (w), yellow bodied (y), miniature winged (m) male: w+ y+ m What are the NCO chromosomes? 2157 w y m+ A. w+ y+ m+ & w y m 1203 w y m B. w+ y+ m & w y m+ 1092 w+ y+ m+ C. w+ y m+ & w y+ m 49 w+ y m D. w+ y m & w y+ m+ 41 w y+ m+ 2 w+ y m+ 1 w y+ m

26. Let’s go back to the white, yellow, miniature cross and check the gene order. Here’s the data for crossing a phenotyically wild-type female by a white eyed (w), yellow bodied (y), miniature winged (m) male: w+ y+ m What are the DCO chromosomes? 2157 w y m+ A. w+ y+ m+ & w y m 1203 w y m B. w+ y+ m & w y m+ 1092 w+ y+ m+ C. w+ y m+ & w y+ m 49 w+ y m D. w+ y m & w y+ m+ 41 w y+ m+ 2 w+ y m+ 1 w y+ m

27. Let’s go back to the white, yellow, miniature cross and check the gene order. Here’s the data for crossing a phenotyically wild-type female by a white eyed (w), yellow bodied (y), miniature winged (m) male: w+ y+ m What is the order of the genes? 2157 w y m+ A. w - y - m 1203 w y m B. y - w - m 1092 w+ y+ m+ C. w - m - y 49 w+ y m 41 w y+ m+ 2 w+ y m+ 1 w y+ m

28. Let’s go back to the white, yellow, miniature cross and check the gene order. Here’s the data for crossing a phenotyically wild-type female by a white eyed (w), yellow bodied (y), miniature winged (m) male: w+ y+ m 2157 w y m+ What is the map distance for w to m? w y m A. 1.4 1092 w+ y+ m+ B. 33.6 49 w+ y m C. 33.7 41 w y+ m+ D. 34.1 2 w+ y m+ E. 35.0 1 w y+ m

29. Wild-type female fly, heterozygous for +/a, +/b and +/c is crossed to an a, b, c male. The progeny include: NCOs: abc and +++ DCOs: ab+ and ++c What is the map?: A. a - b - c B. a - c - b C. b - a - c

30. Wild-type female fly, heterozygous for +/a, +/b and +/c is crossed to an a, b, c male. The progeny include: NCOs: a+c and +b+ DCOs: ++c and ab+ What is the map?: A. a - b - c B. a - c - b C. b - a - c

31. Mitotic Recombination

32. s + + y A recessive mutation in the X-linked yellow gene of Drosophila confers a yellow body color to the flies, and a recessive mutation in another X-linked gene, singed, gives singed hairs. Let’s say you are working with a fly that has the genotype: y sn+/ y+sn, which of the possible phenotypes could you see as a result of mitotic recombination if a single crossover occurs between the sn gene and the centromere? A. a spot that is both yellow and singed B. a yellow spot C. a singed spot D. a twin spot

33. A smurf with a wild type phenotype is heterozygous for the red gene (r+r) and the hairy gene (h+h). His chromosomes are shown below: r h+ r+ h If mitotic crossing over occurs between the r and h genes, what phenotypes are possible?A) a red spot B) a hairy spot C) a red and hairy spot D) a twin spot (a red spot next to a hairy spot) E) only wild type