1. INHERITANCE OF TWO OR MORE INDEPENDENT GENES

2. In Drosophila melanogaster, body colour is determined by the e gene: the recessive allele is responsible for the black colour of the body, the dominant allele e+ is responsible for the grey body. Vestigial wings are determined by the recessive allele vg, normal wings are determined by the dominant allele vg+. These two genes are independent. If dihybrid flies for these two genes are crossed and resulting progeny is composed by 368 individuals, how many individuals are present in every phenotypic class? e+ e+e+ e+ e e vg + vg + vg vg Parental Genotypes: e + e vg + vg X e + e vg + vg Gametes e (1/2) vg + (1/2) vg (1/2) vg + (1/2) vg (1/2) e + (1/2) e vg (1/4) e + vg + (1/4) e + vg (1/4) e vg + (1/4)

3. Phenotypes for gene e Phenotypes for gene vg Phenotipical classes (cross e + e X e + e) (cross vg + vg X vg + vg) vg + (3/4) e (1/4) e + (3/4) vg (1/4) e + vg + (¾ X ¾= 9/16) e + vg (¾ X ¼ = 3/16) vg + (3/4) vg (1/4) e vg + (¾ X ¼ = 3/16) e vg (¼ X ¼=1/16) 9/16 di 368 = 207 individuals e + vg + brown body and normal wings 3/16 di 368 = 69 individuals e + vg brown body and vestigial wings 3/16 di 368 = 69 individuals e vg + black body and normal wings 1/16 di 368 = 23 individuals e vg black body and vestigial wings

4. Phenotype of individuals used for the cross Black Rough Black Smooth Brown Rough Brown Smooth a) Black Rough X Brown Smooth50000 b) Black Rough X Black Rough c) Black Rough X Brown Smooth d) Brown Rough X Black Rough a)The parents have different phenotypes and the progeny is all Black and Rough: -Black and Rough are dominant characters (B, black; R, rough) -The parents are homozygous: BB RR X bb rr We do not perfomed statistic test because we do not have differences between expected progeny and observed progeny. n° of individuals in the progeny In guinea pigs, the R gene determinates fur rough or smooth, while the B gene controls fur color. Determine the genotype of individuals used for the crosses and verify your hypothesis with χ2 test.

5. Phenotype of individuals used for the cross Black Rough Black Smooth Brown Rough Brown Smooth a) Black Rough X Brown Smooth50000 b) Black Rough X Black Rough185605718 c) Black Rough X Brown Smooth d) Brown Rough X Black Rough b) We know that Black and Rough are dominant traits. The crossed individuals have the same phenotypes but in the progeny we have the recessive characters, thus both parents are double heterozygous: Bb Rr X Bb Rr In the progeny we excpect: 9/16 BR (Black rough): 3/16 Br (black smooth): 3/16 bR (brown rough): 1/16 br (brown smooth) n° of individuals in the progeny In guinea pigs, the R gene determinates fur rough or smooth, while the B gene controls fur color. Determine the genotype of individuals used for the crosses and verify your hypothesis with χ2 test.

6. PhenotypesXo (observed individuals) H (hypothesi s) Xa Expected individuals (Xo –Xa) 2 Xa Black Rough 1859/16180(185-180) 2 = 25 25/180= 0,14 Black Smooth 603/166000/60= 0 Brown Rough 573/1660(57-60) 2 = 9 9/60=0,15 Brown Smooth 181/1620(18-20) 2 = 4 4/20= 0,20 Total32016/16320  =0,49  2 = 0,49 Freedom Degree= 4 – 1 = 3 22

7. Accepted Rejected P >90%, we accept the hypotesis

8. Phenotype of individuals used for the cross Black Rough Black Smooth Brown Rough Brown Smooth a) Black Rough X Brown Smooth50000 b) Black Rough X Black Rough185605718 c) Black Rough X Brown Smooth1051009897 d) Brown Rough X Black Rough c) The parents have different phenotypes. The second parent is homozygous recessive for both genes (bb rr). The first parent has the dominant phenotypes for both characters B - R -. In the progeny we observe classes of individuals with recessive phenotypes thus the first indiviudual id double heterozygous: Bb Rr. In the progeny we expect 4 phenotypic classes with ratio: n° of individuals in the progeny In guinea pigs, the R gene determinates fur rough or smooth, while the B gene controls fur color. Determine the genotype of individuals used for the crosses and verify your hypothesis with χ2 test.

9. PhenotypesXo (observed individuals) H (hypothes is) Xa Expected individuals (Xo –Xa) 2 Xa Black Rough 105¼100(105-100) 2 = 25 25/100= 0,25 Black Smooth 100¼ 00/60= 0 Brown Rough 98¼100(98-100) 2 = 4 4/100=0,04 Brown Smooth 97¼100(97-100) 2 = 9 9/100= 0,09 Total4004/4400  =0,38  2 = 0,38 Freedom Degree = 4 – 1 = 3 22 P =>90%, we accept the hypothesis

10. Phenotype of individuals used for the cross Black Rough Black Smooth Brown Rough Brown Smooth a) Black Rough X Brown Smooth50000 b) Black Rough X Black Rough185605718 c) Black Rough X Brown Smooth1051009897 d) Brown Rough X Black Rough63175822 d) The first individual is homozygous recessive for color (bb) and he has the dominant phenotype for Rough gene (R -). The second individual is dominant for both characters (B - R -). In the progeny we observe the classes with recessive phenotypes for brown ans smooth hair, thus the original individuals have a recessive allele to donate to the progeny: b b R r X B b R r n° of individuals in the progeny In guinea pigs, the R gene determinates fur rough or smooth, while the B gene controls fur color. Determine the genotype of individuals used for the crosses and verify your hypothesis with χ2 test.

11. Gene BGene Rphenotypic Classes bb X BbRr x Rr R (3/4)B R (1/2 x 3/4 = 3/8)black rough B (1/2) r (1/4)B r (1/2 x 1/4 = 1/8)black smooth R (3/4)b R (1/2 x 3/4 = 3/8)brown rough b (1/2) r (1/4)b r (1/2 x 1/4 = 1/8)brown smooth With branch diagram determine expected phenotypic classes:

12. Phenotyp es Xo (observed individual s) H (hypoth esis) Xa Expected individuals (Xo –Xa) 2 Xa Nero arruffato 633/860(63-60) 2 = 9 9/60= 0,15 Nero liscio 171/820(17-20) 2 = 9 9/20= 0,45 Bruno arruffato 583/860(58-60) 2 = 4 4/60=0,07 Bruno liscio 221/820(22-20) 2 = 4 4/20= 0,2 Totale1608/8160  =0,87  2 = 0,87 Degrees of freedom = 4 – 1 = 3 2 2 P =80-90%, we accept the hypothesis

13. Determine the genotype of individuals used for the crosses and verify your hypothesis with χ2 test. For each cross create a table, according to the scheme reported below. Gene W determines Red color, gene D determines the plant height Phenotype of individuals used for the cross Red tall Red short White tall White short a) red tall x red tall 1200450 b) red tall x white short c) red tall x white short d) white tall x red tall e) red tall x red tall a)Color: the crossed individuals have the same phenotypes but in the progeny we have also recessive phenotype individuals: both parents are heterozygous (Ww) and Red (W) is dominant over white (w). b)Height: we cannot establish wich is the dominant allele since the character does not segregate. All the progeny is tall. Both parents could be recessive (dd) or D D x D - o or D - x D D Numero di individui della progenie

14. PhenotypesXo (observed individuals) H (hypothes is) Xa Expected individuals (Xo –Xa) 2 Xa Red tall1203/4123,75(120-123,75) 2 = 14 14/123,75= 0,11 White tall451/441,25(45-41,25) 2 = 14 14/41,25= 0,33 Total1654/4165  =0,44  2 = 0,44 Degrees of freedom= 2 – 1 = 1 2 2 P =50-70%, we can accept the hypothesis

15. Phenotype of individuals used for the cross Red tall Red short White tall White short a) red tall x red tall 1200450 b) red tall x white short 10001050 c) red tall x white short d) white tall x red tall e) red tall x red tall b) The crossing plants have two different phenotypes. In the progeny we observe plants red and white: we know that red is dominant over white thus the genotypes are W w X w w Height: since we have only tall plants we estabish that tall is dominant over short. In the progeny there isn’t segregation of character then the tall plant is homozygous. The genotypes are: D D X d d n° of individuals in the progeny Determine the genotype of individuals used for the crosses and verify your hypothesis with χ2 test. For each cross create a table, according to the scheme reported below. Gene W determines Red color, gene D determines the plant height

16.  2 = 0,12 Degrees of freedom = 2 – 1 = 1 P = 70-80%, we accept the hypothesis Ww DD X ww dd We expect: ½ WD (red and tall) and ½ wD (white and tall). Verifichiamo con il test del  2 se possiamo accettare l’ipotesi. Phenotype s Xo (observed individuals) H (hypothes is) Xa Expected individuals (Xo –Xa) 2 Xa Red tall100½102,5(100-102,5) 2 = 6,25 6,25/102,5= 0,06 White tall105½102,5(105-102,5) 2 = 6,25 6,25/102,5= 0,06 Total2052/2205  =0,12

17. Phenotype of individuals used for the cross Red tall Red short White tall White short a) red tall x red tall 1200450 b) red tall x white short 10001050 c) red tall x white short 45434844 d) white tall x red tall e) red tall x red tall c) The crossed plants have different phenotypes. In the progeny we observed segregation of the character: we know that red is dominant over white and we assess that the cross is between a heterozygous and a homozygous recessive: W w X w w Height: Tall is dominant over short thus also for height the cross is between a heterozygous and a homozygous recessive D d X d d n° of individuals in the progeny Determine the genotype of individuals used for the crosses and verify your hypothesis with χ2 test. For each cross create a table, according to the scheme reported below. Gene W determines Red color, gene D determines the plant height

18. Phenotype s Xo (observed individuals) H (hypothes is) Xa Expected individuals (Xo –Xa) 2 Xa Red tall45¼ (45-45) 2 = 0 0/45= 0 Red short43¼45(43-45) 2 = 4 4/45= 0,09 White tall48¼45(48-45) 2 = 9 9/45=0,2 White short 44¼45(44-45) 2 = 1 1/45= 0,02 Total1804/4180  =0,31  2 = 0,31 Degrees of freedom = 4 – 1 = 3 test  2 P =>90%, we can accept the hypothesis W w D d X w w d d We expect 4 phenotypic classes with ratio 1 : 1 : 1 : 1.

19. Phenotype of individuals used for the cross Red tall Red short White tall White short a) red tall x red tall 1200450 b) red tall x white short 10001050 c) red tall x white short 45434844 d) white tall x red tall 1756718258 e) red tall x red tall d) In the progeny we have recessive phenotypic classes. We can suppose that the cross is between: w w D d X W w D d n° of individuals in the progeny Determine the genotype of individuals used for the crosses and verify your hypothesis with χ2 test. For each cross create a table, according to the scheme reported below. Gene W determines Red color, gene D determines the plant height

20. We expect 4 phenotypic classes: 3/8 : 1/8 : 3/8 : 1/8 Gene WGene DPhenotypic classes ww X WwDd x Dd D (3/4)W D (1/2 x 3/4 = 3/8)Red Tall W (1/2) d(1/4)W d (1/2 x 1/4 = 1/8)Red short D (3/4)w D (1/2 x 3/4 = 3/8)White Tall w (1/2) d (1/4)w d (1/2 x 1/4 = 1/8)White short What are phenotypic classes expected ?

21. Phenotype s Xo (observed individuals) H (hypothes is) Xa Expected individuals (Xo –Xa) 2 Xa Red tall1753/8180,75(175-180,75) 2 = 33,06 /180,75= 0,18 Red short671/860,25(67-60,25) 2 = 45,56 /60,25= 0,75 White tall1823/8180,75(182-180,75) 2 = 1,56 1,56/180,75=0, 1 White short 581/860,25(58-60,25) 2 = 5,06 5,06/60,25= 0,08 Total4828/8482  =1,11  2 = 0,31 Degrees of freedom = 4 – 1 = 3 P =70-80%, Hypothesis accepted

22. Phenotype of individuals used for the cross Red tall Red short White tall White short a) red tall x red tall 1200450 b) red tall x white short 10001050 c) red tall x white short 45434844 d) white tall x red tall 1756718258 e) red tall x red tall 265929328 e)In the progeny we observe recessive phenotypes. The individuals must be double heterozygous: W w D d X W w D d From a dihybrid cross we expect 4 phenotipic classes: 9/16 WD (Red tall) : 3/16 Wd (Red short) : 3/16 wD (white tall) : 1/16 wd (white short). n° of individuals in the progeny Determine the genotype of individuals used for the crosses and verify your hypothesis with χ2 test. For each cross create a table, according to the scheme reported below. Gene W determines Red color, gene D determines the plant height

23. Phenotype s Xo (observed individuals) H (hypothes is) Xa Expected individuals (Xo –Xa) 2 Xa Red tall2659/16268,87(265-268,87) 2 = 14,97 /268,87= 0,05 Red short923/1689,62(92-89,62) 2 = 5,66 /89,62= 0,06 White tall933/1689,62(93-89,62) 2 = 11,42 11,42/89,62=0, 12 White short 281/1629,87(28-29,87) 2 = 3,49 3,49/29,87= 0,11 Total47816/16478  =0,34  2 = 0,34 Degrees of freedom = 4 – 1 = 3 P >90%, hypothesis accepted

24. Crossing a pure line of melons whit white and spherical fruits with an other pure line of melons with yellow and flat fruits the f1 progeny is melons with white and spherical fruits. Crossing two plants of F1 we have: Determine the genotypes of individuals of P1, P2 e F1, make the hypothesis of traits segregation and verify the results with X2 P1P2 X F1 148 Plants with white and spherical fruits 52 Plants with yellow and spherical fruits 49 Plants with white and flat fruits 23 Plants with yellow and flat fruits F2 From first cross we know that spherical is dominant over flat and white is dominant over yellow. WE decide to name R the gene that control the shape melon and G the gene that control the color melon: the first parents is homozygous dominant and the second parent is homozygous recessive. The progeny will be heterozygous for both genes. RR GGrr gg Rr Gg

25. Phenotype s Xo (observed individuals) H (hypothes is) Xa Expected individuals (Xo –Xa) 2 Xa White Spherical 1489/16153(148-153) 2 = 25 /153= 0,16 Yellow Spherical 523/1651(52-51) 2 = 1 /51= 0,02 White Flat 493/1651(49-51) 2 = 4 4/51=0,08 Yellow Flat 231/1617(23-17) 2 = 36 36/17= 2,12 Total27216/16272  =2,38  2 = 2,38 Degrees fo freedom = 4 – 1 = 3 P=30-50%, hypothesis accepted In F2 we have 4 phenotipic classes. This is a dihybrid cross and we expect frequencies 9 : 3 : 3 : 1. test  2.

26. Gene A Aa A (1/2) a (1/2) Gene B Bb B (1/2) b (1/2) B (1/2) b (1/2) Gene C CC C (1) Gametes ABC (½ X ½ X 1= 1/4) AbC (½ X ½ X 1= 1/4) aBC (½ X ½ X 1= 1/4) abC (½ X ½ X 1= 1/4) Using the branch diagram, determine the gametes produced by individuals with the following genotype (A, B and C genes are independent): a) Aa Bb CC

27. b) Aa bb Cc Dd Gene A Aa A (1/2) a (1/2) Gene B bb b (1) Gene C Cc C (1/2) c (1/2) C (1/2) c (1/2) Gene D Dd D (1/2) d (1/2) D (1/2) d (1/2) D (1/2) d (1/2) D (1/2) d (1/2) Gametes AbCD (½ X 1 X ½ X ½ = 1/8) AbCd (½ X 1 X ½ X ½ = 1/8) AbcD (½ X 1 X ½ X ½ = 1/8) Abcd (½ X 1 X ½ X ½ = 1/8) abCD (½ X 1 X ½ X ½ = 1/8) abCd (½ X 1 X ½ X ½ = 1/8) abcD (½ X 1 X ½ X ½ = 1/8) abcd (½ X 1 X ½ X ½ = 1/8)

28. Gene A Aa A (1/2) a (1/2) Gene B Bb B (1/2) b (1/2) B (1/2) b (1/2) Gene C Cc C (1/2) c (1/2) C (1/2) c (1/2) C (1/2) c (1/2) C (1/2) c (1/2) c) Aa Bb Cc Gametes ABC (½ X ½ X ½ = 1/8) ABc (½ X ½ X ½ = 1/8) AbC (½ X ½ X ½ = 1/8) Abc (½ X ½ X ½ = 1/8) aBC (½ X ½ X ½ = 1/8) aBc (½ X ½ X ½ = 1/8) abC (½ X ½ X ½ = 1/8) abc (½ X ½ X ½ = 1/8)

29. Phenotypes for Gene A Aa X aa A (1/2) a (1/2) Phenotypes for Gene B Bb X bb B (1/2) b (1/2) B (1/2) b (1/2) Phenotypes for Gene C Cc X cc C (1/2) c (1/2) C (1/2) c (1/2) C (1/2) c (1/2) C (1/2) c (1/2) Final Phenotypes ABC (½ X ½ X ½ = 1/8) ABc (½ X ½ X ½ = 1/8) AbC (½ X ½ X ½ = 1/8) Abc (½ X ½ X ½ = 1/8) aBC (½ X ½ X ½ = 1/8) aBc (½ X ½ X ½ = 1/8) AbC (½ X ½ X ½ = 1/8) Abc (½ X ½ X ½ = 1/8) For the following cross, determine the phenotypic classes expected in the progeny and their frequencies (A, B and C genes are independent) AaBbCc x aabbcc