Patterns of Inheritence in Drosophila Melanogaster

The Fruit Flies of Melanogaster

 
Introduction:  
Many simple patterns of inheritance follow the laws of Mendel. Dominant traits will always be expressed when present, and recessive traits will only be expressed when two recessive alleles are present. When crossing a pure homozygous dominant trait with a pure homozygous recessive trait as the P generation, it is expected that all the offspring in the F1 generation will express the dominant trait, since every offspring will receive one copy of the dominant allele from one parent and one copy of the recessive allele from the other. In the F2 generation, the expected outcome will be a 3:1 phenotypic ratio of dominant to recessive, and a 1:2:1 genotypic ratio of homozygous dominant to heterozygous to homozygous recessive (Campbell et al. 268).

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This simple inheritance pattern explains many of the inheritance phenomena exhibited in nature, but some inheritance patterns go beyond Mendel’s laws of genetics. In incomplete dominance, neither allele is dominant over the other so the outcome is a blend of both traits. In codominance, both traits are expressed separately. In mitochondrial inheritance, all offspring will receive specific genes from the mother. In X-linked recessive traits, the alleles are located on the X chromosome, and these conditions frequently appear in males because they only have one copy of the X chromosome (“Inheritance Patterns”). When the exact inheritance pattern is unknown in a cross, the ratios of each type of offspring help to determine if the inheritance pattern follows Mendel’s laws or if it is one of the above varieties.
In fruit flies, the red and brown genes for eye color are located on autosomes. However, a mutation on the white gene in fruit flies on the X chromosome prevents any eye color from developing at all (“The Genetics of Eye Color”). The gene for white eye color is epistatic to the red and white eye genes. This is how fruit flies are able to have three different eye colors – when the white mutation is not present, there will be a simple inheritance pattern between red and sepia eyes. When the mutation is present, the red or sepia eyes will not be expressed because they will be masked behind the white mutation.
Drosophila melanogaster were used in this procedure because they reproduce very quickly and are easily manageable. All their food and water needs are taken care of by the substance called media in the bottom of the vial. They are a convenient size because they are not too big, but they are small enough to easily distinguish traits under a microscope (“The Fruit Fly and Genetics”). The life cycle of the flies begins as eggs. From the eggs emerge the larvae, which look like tiny worms. The larvae grow through three stages until they reach the pupal stage. The pupae mature and darken in color for three to four days until they break forth from the pupal case to become adult flies (“Development”).
In this experiment, three crosses were performed between different varieties of the fruit fly Drosophila Melanogaster. Cross 1 was between a sepia eyed female and a wild type male, Cross 2 was between a white eyed female and a wild type male, and Cross 3 was between a red eyed, vestigial winged female with a sepia eyed, normal winged male.
In Cross 1, a simple pattern of Mendel’s laws is predicted to be expressed. Wild type flies with red eyes is the dominant phenotype over sepia colored eyes. Sepia colored eyes are a result from a recessive gene, and only result when two sepia-eyed flies mate or when two heterozygous flies mate. Furthermore, sepia colored eyes is not dependent on the sex of the fly, so in the case of this cross all flies in the F1 generation should have red eyes, but be carriers for the sepia colored eye trait. In the F2 generation when the heterozygous flies mate, the predicted phenotypic ratio will be 3:1, where for every three red eyed flies there would be one sepia colored fly. The related genotypic ratio of homozygous dominant to heterozygous to homozygous recessive will be 1:2:1. Our hypothesis for Cross 1 is if there are no mutations and the cross follows Mendel’s laws of independent assortment, then the ratio of red to sepia eyed flies will be 3:1 for the F2 generation.
In cross 2, sex linked inheritance plays a role. The mutation for white colored eyes is X-linked recessive. When the white eyed female is crossed with a red eyed male, all the males in the F1 generation should exhibit the mutation, and all the females should have red eyes. This is because the males can only accept a recessive allele from the mother and the Y chromosome from the father which does not carry the mutation for white eye color. The females will receive the red gene from the father’s X chromosome which will cover the white gene from the mother. The F2 generation produced by the white eyed male and heterozygous female will thus have a genotypic ratio of 1:1:1:1. Therefore, our hypothesis for Cross 2 for the F2 generation is that if the gene for the white eye mutation is located on the X chromosome, then the phenotypic ratio will be 1:1:1:1 if sex is considered.
In cross 3, the focus shifted from just looking at eye colors to looking at eye colors and wing type. The fruit flies could either have normal wings or exhibit vestigial wings, which are shortened. Flies with vestigial wings have a defect in their vestigial gene located on the second chromosome. So, a dihybrid cross will be used to determine the predicted phenotypes and genotypes of the F1 and F2 generation. A dihybrid cross uses two traits with two alleles each, and so two different aspects of an organism are crossed. Vestigial wings are a recessive trait, so two recessive alleles must be inherited in order to express the trait. This is also the case with sepia colored eyes. So when a parent generation of a red eyed vestigial winged female is crossed with a sepia eyed normal winged male, all of the offspring in the F1 generation should have red eyes and normal wings. The F2 generation, however, are produced by heterozygotes and thus four phenotypes should be seen: red eyed normal, sepia normal, sepia vestigial, and red eyed vestigial. Therefore, our hypothesis for the F2 generation in Cross 3 is that if both traits follow Mendel’s laws of independent assortment for the dihybrid cross, then the predicted phenotypic ratio will be 9:3:3:1.
Methodology:
Materials used include:

Vials
Microscopes
Fly nap and anesthesia wand
Paint brushes
Fly food
Fly netting
Cotton plugs
Plain white index card

Procedure:

First, prepare vials for the fruit flies to live in. Obtain three glass vials, and estimate a few centimeters of Carolina Instant Drosophila Medium in each. After, put a few drops of water in the culture and let it sit a few minutes to soak in the medium. At this point, also put in a fly net. Obtain F1 flies from instructor for the three crosses.
Check the vials for life. The flies need to be alive for active breeding purposes. However, ensure that there are no F2 larvae yet, as this could be misleading for the results. Check the food for moisture, and add water with a pipet if the food gets too dry.
Anesthetize the fruit flies. Put the vials of flies upside down in the refrigerator, as this forces the flies into a state of inactivity. After approximately ten minutes, take the vials out and transport each of the three tubes (for the separate three crosses) into three different vials. Mark the three vials with tape for either cross 1, cross 2, or cross 3. Tap the flies into the new vials, and close it with a cotton plug with an anesthesia wand connected to it soaked in Flynap. Wait a few minutes for the flies to stop moving or flying to begin the next procedure.
Shake the fruit flies onto a white index card and place the card under the dissecting microscope. Use the paintbrush to move the flies to the center of the viewing field in order to sex them and view them for the desired traits. Record the data in the data tables. Males have a solid black abdomen and sex combs on their forelegs, while females have a striped abdomen and no sex combs. Additionally, females are generally larger than males[1].
After recording the data for the flies, place them in the morgue. Record F1 data for three days, or until F2 larvae are seen.
Repeat the above procedure using the same vials, but this time using only F2 flies. Record data for three days.
Set the extra flies free, and clean out all the vials thoroughly.

Results:
Cross 1
Punnett square: Cross between heterozygous male and heterozygous female for eye color
Table 1: Lab Group data for Cross 1

Fasdfasdasdffasdfasdf

Day 1

Day 2

Day 3

Total

Red M

3

2

55

60

Red F

4

1

58

63

Sepia M

0

0

6

6

Sepia F

0

0

4

4

Fasdfasdasdffasdfasdf

Day 1

Day 2

Day 3

Total

Red M

2

3

1

6

Red F

3

2

0

5

Sepia M

0

0

0

0

Sepia F

2

0

0

2

Chi Square Analysis for Lab Group Data for Cross 1

Expected Totals:
Red: ¾ x x/13 = 9.75 red
(11-9.75)^2/9.75 = 1.160
Sepia: ¼ x x/13 = 3.25 sepia
(2-3.25)^2/(3.25) = .481
∑=1.641
Degrees of Freedom: 1
.20 > p > .10
Accept the null hypothesis
Table 2: AP Bio 2015 Class Data for Cross 1

Fasdfasdasdffasdfasdf

Pd 8

Pd 6

Total

Red M

39

58

97

Red F

46

64

110

Sepia M

12

7

19

Sepia F

11

4

15

F2

sex

1

2

3

4

5

6

7

8

9

10

11

total

RED M

M

19

13

8

35

44

21

6

8

36

48

22

260

RED F

F

37

12

12

38

51

36

5

4

36

50

29

310

SEPIA M

M

4

3

5

12

10

6

0

3

5

12

6

66

SEPIA F

F

5

5

3

16

13

7

2

1

13

14

9

88

Chi Square Analysis for AP Bio Group Data for Cross 1:

Expected Totals:
Red: ¾ x x/724 = 543
(570-543)^2/543 = 1.343
Sepia: ¼ x x/724 = 181
(154-181)^2/181 = 4.028
∑= 5.371
.01 > p > .001
Reject the null hypothesis
Cross 2:
Punnett Square: Cross between heterozygous red eyed female and hemizygous white eyed male

Table 3: Lab Group Data for Cross 2   

F1

day 1

day 2

day 3

total

RED M

0

0

15

15

RED F

24

1

32

57

WHITE M

24

3

10

37

WHITE F

0

0

13

13

 
 
 
 
 

F2

day 1

day 2

day 3

total

RED M

8

3

0

11

RED F

9

6

1

16

WHITE M

7

1

0

8

WHITE F

6

3

1

10

Chi Square Analysis for Lab Group Data for Cross 2
Expected Totals:
Red M: 1/4 x x/45 = 11.25
(11-11.25)^2/11.25 = .006
Red F: 1/4 x x/45 = 11.25
(16-11.25)^2/11.25 = 2.01
White M: 1/4 x x/45 = 11.25
(8-11.25)^2/11.25 = .939
White F: 1/4 x x/45 = 11.25
(10-11.25)^2/11.25 = .139
∑= 3.094
Degrees of freedom: 3
0.50 > p > .30
Accept the null hypothesis
Table 4: AP Bio 2015 Class Data for Cross 2

F1

1

2

total

 
 
 
 
 
 
 
 

RED M

60

15

75

 
 
 
 
 
 
 
 

RED F

87

58

145

 
 
 
 
 
 
 
 

WHITE M

50

35

85

 
 
 
 
 
 
 
 

WHITE F

10

13

23

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

 
 
 
 
 
 
 
 
 
 
 
 

F2

1

2

3

6

7

8

9

10

11

total

 
 
 

RED M

24

23

22

8

11

23

38

9

23

211

 
 
 

RED F

25

30

38

7

16

14

42

16

14

231

 
 
 

WHITE M

18

27

20

9

8

11

13

17

21

162

 
 
 

WHITE F

20

23

24

11

10

11

12

11

24

176

 
 
 

Chi Square Analysis for AP Bio Class Data for Cross 2:
Expected Totals:
Red M: 1/4 x x/780 = 195
(211-195)^2/195 = 1.312
Red F: 1/4 x x/780 = 195
(231-195)^2/195 = 6.646
White M: 1/4 x x/780 = 195
(162-195)^2/195 = 5.585
White F: 1/4 x x/780 = 195
(176-195)^2/195 = 1.851
∑=15.394
Degrees of freedom: 3
p > .001
Reject the null hypothesis
Cross 3:
Punnett Square: Cross between two flies heterozygous for both red eyes and normal wings

Table 5: Lab Group Data for Cross 3

F1

sex

day 2

day 3

total

RED / NORMAL

M

6

16

40

RED / NORMAL

F

12

17

62

RED / VESTIGAL

M

0

0

0

RED / VESTIGAL

F

0

0

0

SEPIA / NORMAL

M

0

5

5

SEPIA / NORMAL

F

0

1

1

SEPIA / VESTIGAL

M

0

0

0

SEPIA / VESTIGAL

F

0

0

0

 
 
 
 
 
 

F2

sex

day 1

day 2

day 3

 

RED / NORMAL

M

12

7

2

 

RED / NORMAL

F

13

13

10

 

RED / VESTIGAL

M

1

2

0

 

RED / VESTIGAL

F

3

3

1

 

SEPIA / NORMAL

M

8

4

1

 

SEPIA / NORMAL

F

4

3

4

 

SEPIA / VESTIGAL

M

0

0

0

 

SEPIA / VESTIGAL

F

1

0

1

 

Chi Square Analysis for Lab Group Data for Cross 3
Expected Totals:
Red normal: 9/16 x x/93 = 52.313
(57-52.313)^2/52.313 = .420
Sepia normal: 3/16 x x/93 = 17.438
(24-17.438)^2/17.438 = 2.470
Red vestigial: 3/16 x x/93 = 17.438
(10-17.438)^2/17.438 = 3.172
Sepia vestigial: 1/16 x x/93 = 5.813
(2-5.813)^2/(5.813) = 2.501
∑= 8.563
Degrees of freedom: 3
.05 > p >.01
Reject the null hypothesis
Table 6: AP Bio 2015 Group Data for Cross 3