1. Obtain a vial of flies.
2. Gently tap the vial to knock flies away from the top of the foam stopper.
3. Soak the wand with fly nap solution and gently place it in the vial
5. After the flies stop moving, gently, tap the vial of flies onto a piece of paper
Step 3: Sexing and Counting Flies
1. Turn on the top light. Make sure the bottom light is OFF.
2. Using the small paint brush, sex and sort the flies on to different halves of a piece of paper
3. Male flies have sex combs on their forelimbs and have dark, blunt abdomens while females have a stripped abdomen and lighter, pointed abdominals.
Step 4: Crossing Flies
1. After counting and recording the sex and phenotype of your F1 flies, remove the foam
stopper and place the vial on its side next to the edge of the index card containing the flies.
2. Gently, brush the 3-4 female and 2-3 male flies into the vial and replace the foam stopper.
5. Use a dissecting microscope to check for eggs and larva. Eggs are white and are sometimes hard to distinguish, so look carefully.
6. Remove the adults by anesthetization and place them in the fly morgue containing alcohol solution
7. About a week later, anesthetize, sort count and record the data for the F2 generation.
8. Repeat step 1 to 4 for the F3 Generation
Independent variable: Phenotypes and genders of the parents
Dependent variables: Phenotypes and genders of the offspring
Original Cross
Cross 1 Results: F2 Generation
Cross 2 Results: F2 Generation
Cross 3 Results: F2 Generation
Cross 4 Results: F2 Generation
Cross of Apterous from F2 Generation
Cross 5 Results: F3 Generation
The data that I have collected is not very reliable. My results indicate that apterous is a recessive trait while the wild type is dominant according to offspring. However, my group made a huge mistake in the F1 generation. For cross three and cross four of the F1 generation, the offspring phenotype should not have contained any wild type fruit flies because apterous winged is a recessive phenotype. Instead of crossing apterous fruit flies, my group mistakenly crossed a different vial of flies. Instead of waiting for the original vial of (parent) fruit flies to lay offspring for the F1 generation, my group crossed the parent fruit flies.
From the data that I have collected, I noticed that there are a significantly higher number of fruit flies with red eyes and wild type. There is not a 1:1 ratio of males and females from the data. Also, apterous flies seem to be recessive because there are a significantly lower number of flies with those phenotypes than the wild type. Also, sepia eyes seem to have a recessive trait because there is a considerably lower number of flies with sepia than red eyes.
Conclusion:
This investigation’s purpose was to observe the effects of the offspring from a parent with altered genes. With the results of all four crosses of the F2 generation, the number of wild fruit flies with red eyes is capacious compared to the number of all the other fruit flies; thus wild type and red eyes are dominant sex-linked traits. With the results of all wild flies, the number of sepia and white eyes is significantly less than that of the red eyes, concluding that sepia and white eyes are recessive traits. Contrasting the number of wild fruit flies with red eyes, the number of apterous fruit flies is accordingly less, which signifies that the apterous wing is a recessive trait. Even with a significantly less number of offspring sexed, the conclusions can still be drawn that red eyes and wild type is dominant (indicated by Cross 4). From the cross of apterous flies of the F1 generation, the offspring (F2 generation) denotes that red eyes are dominant and sepia eyes are recessive. It also shows that with a cross of the same apterous phenotype, the ability to not have wings is a hereditary trait, and as long as one parent had no wings, the offspring would not have wings. Thus it can be concluded that the wild type, which has wings, is dominant to apterous (lacks wings) and that red eyes is sex linked dominant to sepia/white eyes. Many mutant characteristics have been selected in Drosophila. The mutants which I found in this exercise were eye color. White eyes indicated that it was a sex-linked, recessive mutant eye color. This gene locus is found on chromosome 1 - the sex chromosome, and so is found in diploid status in females, but only a single copy in males (the Y chromosome which determines maleness does not carry this locus.) The wild-type allele is "W" and the mutant is "w". Sepia eyes are a recessive mutant eye color, with the eyes appearing brownish-red as the individuals emerge from the pupa, and darkening to nearly black in older adults. Wild-type allele is "Se" or "S" and mutant is "se" or "s". Thus, my hypothesis is correct. The offspring of a pair of reciprocal Drosophila crosses is determined by which the mutation is dominant or recessive and whether it is x-linked or autosomal.
Results from this lab could have been affected by many things. Constantly knocking out the fruit flies could have caused some of the larvae to not hatch therefore affecting the data recorded for the offspring. Also some of the traits of the fruit flies are difficult to distinguish, especially if they are newly hatched. Thus, incorrectly identifying the characteristics of the flies could have also greatly affected the results received. Also, instead of waiting for the original vial of (parent) fruit flies to lay offspring for the F1 generation, my group crossed the parent fruit flies. Hence, improper calculation of numbers could have also caused inaccurate results. Finally, some flies could have gotten stuck in the medium and could have been identified. Improvements can be made if the investigation crossed more fruit flies, thus giving a more veritable number of fruit flies for each phenotype and hence a more reliable data. Finally, crossing the F1 generation instead of the parent generation would be more veritable for the data collected.
Apterous x sepia X Apterous x wild (ww x rr) x (Ww x Rr)
Red x Apterous = 7/28
Wild x Wild = 18/28
Sepia x wild = (3/18)
The Chi square (χ2) test is used for discontinuous values.
χ2 = Σ ((O-E)2/E)
Where E = expected values
O = observed values
Σ= sum of
Table to show Chi square analysis and probability of F1 generation for monohybrid cross
Degrees of freedom = number genotypes – 1 2 genotypes - 1 = 1 degree of freedom Critical value = 3.84