1. Recombination January 2009

2. Orientation of alleles on a chromosome

3. 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: A a T t A a t T A.#1 only B.#2 only C.#1 and #2 #1#2

4. 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 Y y R r Y y r R

5. 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 A a T t A a t T A.#1 only B.#1 and #2 C.#1 and #3 D.#2 and #3 E.#1, #2, and #3

6. 155 Wild-type 165 muscle defects, shriveled wings 330 muscle defects, normal wings 350 normal muscles, shriveled wings Wild-type femaleMuscle defects, shriveled wing male 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 tz vg What is the genotype of the wild-type parental female fly? A. B. C. There’s not enough information tz tz + vg vg + tz tz + vg + vg X

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

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

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 bubblegum and another gene, curly

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

12. In humans: problems with the heart, susceptibility to infections, and muscle weakening Mutations in tafazzin cause muscle weakness in flies. 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. 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 Let’s look at the linkage between tafazzin and vestigial genes

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

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

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

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

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

18. 758 y+ w+ m+ 700 y w m 401 y+ w+ m 317 y w m+ 16y+ w m 12 y w+ m+ 1y+ w m+ 0 y w+ m Let’s look at the y w m cross again 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. 758 y+ w+ m+ 700 y w m 401 y+ w+ m 317 y w m+ 16y+ w m 12 y w+ m+ 1y+ w m+ 0 y w+ m Let’s look at the y w m cross again 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 Wild-type = 446Sepia eyes & Curly wings= 10 Curly wings = 42Sepia eyes & Short bristles = 40 Sepia eyes = 2Short bristles & Curly wings = 1 Short bristles = 10Sepia eyes, Curly Wings, & Short bristle = 449

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 Wild-type = 298Dumpy wings & Hairy = 30 Hairy = 8Vision defects & Purple eyes = 165 Purple eyes = 28Dumpy wings & Purple eyes = 10 Dumpy wings = 161Dumpy wings, Purple eyes, Hairy = 300

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? A.+,+ & +, s B.+,+ & e, + C.+,+ & e, s D.e,s & e, + E.e,s & +, s F.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? A.+,+ & +, s B.+,+ & e, + C.+,+ & e, s D.e,s & e, + E.e,s & +, s F.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. 6.9B. 10.9C. 12.3 D. 14.0E. 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: 2278 w + y + mWhat are the NCO chromosomes? 2157w y m + A. w + y + m + & w y m 1203w y mB. w + y + m & w y m + 1092w + y + m + C. w + y m + & w y + m 49w + y mD. w + y m & w y + m + 41w y + m + 2 w + y m + 1w 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: 2278 w + y + mWhat are the DCO chromosomes? 2157w y m + A. w + y + m + & w y m 1203w y mB. w + y + m & w y m + 1092w + y + m + C. w + y m + & w y + m 49w + y mD. w + y m & w y + m + 41w y + m + 2 w + y m + 1w 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: 2278 w + y + mWhat is the order of the genes? 2157w y m + A. w - y - m 1203w y mB. y - w - m 1092w + y + m + C. w - m - y 49w + y m 41w y + m + 2 w + y m + 1w 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: 2278 w + y + m 2157w y m + What is the map distance for w to m? 1203w y mA. 1.4 1092w + y + m + B. 33.6 49w + y mC. 33.7 41w y + m + D. 34.1 2 w + y m + E. 35.0 1w 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. 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 s + + y

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