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Drosophila - Laboratory Report 2
1. FACULTY OF ENGINEERING AND NATURAL SCIENCES
DEPARTMENT OF GENETICS AND BIOENGINEERING
BIO 323: GENETIC ENGINEERING-1
LABORATORY REPORT
Experiments Name: Crossing of Drosophila F1 Phenotypes
Experiments Date and Submission Date: (29.11.2016)&(06.01.2017)
Students Name / Surname: Necla YÜCEL
Students ID Number: 116201132
Evaluator Name: Asst. Prof. Dr. Özgür Gül
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CONTENTS
List of Tables……...……………………………………………………………………………………..3
List of Figures..………………………………………………………………………………………….3
List of Graph…………………………………………………………………………………………….3
Aim……………………………………………………………………………………………………...4
Introduction…………………………………………………………………………………………...4-6
Materials………………………………………………………………………………………………...6
Method………………………………………………………………………………………….……..6-8
Result……………………………………………………………………………………………….....8-9
Discussion……………………………………………………………………………………………9-10
Conclusion…………………………………………………………………………………………..…10
References………………………………………………………………………………………….10-11
Appendix……………………………………………………………………………………………….11
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LIST OF TABLES
Table 1: Phenotypes of the F1 Generation
Table 2: Calculation of chi squared value for a cross between F1 and F1 flies
Table 3: Chi-Square Table
LIST OF FIGURES
Figure 1: F1 x F1 Drosophila
Figure 2: F2 phenotypes in the pupa stage
Figure 3: Drosophila separated by genotype and phenotypes
LİST OF GRAPH
Graph 1: Chi-Square Distribution Graph
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CROSSING OF DROSOPHILA F1 PHENOTYPES
AIM
In this study, the genetic variability of wild-living flies and their laboratory progeny was
investigated over several generations. All of the parent generation, or flies we started with, were
wild-types phenotypically, but their genotypes were unknown,in this experiment we were
explored the basic principles of genetics such as Mendelian genetics using the model organism,
Drosophila melanogaster.In the previous experiment were obtained F1 phenotypes, we were
utilized them. We put the flies under a Magnifying Glass to determine which phenotypes they
exhibited, recorded the phenotypes in a table, used the data to determine the chi squared value,
and compared our chi squared value to that of a table to determine if it actually fit the expected
ratio.
INTRODUCTION
Gregor Mendel was born in 1822, is now known as Father of Genetics. He initially studies
inheritance of just one pair of contrasting traits. Mendel begins his experiment with garden pea
plant. Mendel recognized two principles that were later call Principle of Mendelian Inheritance:
Principle of Mendelian Inheritance
1. Law of Segregation (The "First Law")
The Law of Segregation states that when any individual produces gametes, the copies of a gene
separate so that each gamete receives only one copy. A gamete will receive one allele or the
other. The direct proof of this was later found when the process of meiosis came to be known.
In meiosis, the paternal and maternal chromosomes are separated and the alleles with the traits
of a character are segregated into two different gametes.
2. Law of Independent Assortment (The "Second Law")
The Law of Independent Assortment, also known as "Inheritance Law", states that alleles of
different genes assort independently of one another during gamete formation. While Mendel's
experiments with mixing one trait always resulted in a 3:1 ratio between dominant and recessive
phenotypes, his experiments with mixing two traits (dihybrid cross) showed 9:3:3:1 ratios.
Mendel concluded that different traits are inherited independently of each other, so that there is
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no relation, for example, between a cat's colour and tail length. This is actually only true for
genes that are not linked to each other.1
Hereditary Traits of Drosophila
Before one observes their mutants, one needs to be familiar with the appearance of the wild
type Drosophila, the type found most often in natural populations of the organisms. Although
thousands of mutations in Drosophila are known, only those which are relevant to these
exercises are listed.
1. Eyes
Wild type: red, oval in shape and many-faceted
Mutants: white, black, apricot, scarlet red, pink, or brown; changes in shape and number of
facets
2. Wings
Wild type: smooth edges, uniform venation, extend beyond the abdomen
Mutants: changes in size and shape; absence of specific veins; changes in position in which
wings are held when at rest
3. Bristles
Wild type: fairly long and smooth (note distribution on head and thorax)
Mutants: shortened, thickened, or deformed bristles changes in patterns of distribution
4. Body colour
Wild type: basically gray, with pattern of light and dark areas
Mutants: black (in varying degrees), yellow, in doubtful cases, color can often be determined
most clearly on wing veins and legs
Mutants’ traits can be assumed as recessive to the wild type.2
Chi-Square Test
The chi-square test is a statistical tool that compares experiment results with an accepted set of
data to determine how much the experimental values deviated from the accepted ones and
whether or not that deviation can be explained solely by chance. The square of the difference
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between the observed and expected values (O – E)2 for each data point (phenotype category in
this case) is calculated. Then, by dividing this by the expected value, the amount of deviation
between the experiment data and the accepted value for that data point can be determined.
Adding the statistic for each data point yields a value known as the X2 (chi-square) statistic.
X2 = ∑ (O – E)2 /E
Because all the values for each category, or data point, are being added together, the value of
X2 will rise as the number of data points used increases. For this reason, “degrees of freedom”
must be included in the parameters of the analysis. The number of degrees of freedom (v) for a
chisquare test is equal to the number of data points minus one. The number of degrees of
freedom does not have an impact on the value of X2 itself, but rather is used in the interpretation
of the importance of the value, as shown in the chi-square table, below. The numbers get larger
as you go down and the value for degrees of freedom increases.3
MATERIAL
- Drosophila Melanogaster (F1 phenoype)
- Drosophila Medium (Each vial contain 10 mL Media and 10 mL Distilled water)
- Anesthefly Solution
- Vial Tube With Sponge Cover
- Soft Paint Brush
- Marker (pen)
- Magnifying Glass
- q-tips
METHOD
In the second phase of the experiment, the F1 phenotypes obtained, we were crossed between
themselves because of to get F2 phenotypes. F1 phenotypes obtained in the previous step were
used. To accomplish this crossing, we were used eight female Drosophila F1 phenotypes and
eight male Drosophila F1 phenotypes. We prepared medium culture for put Drosophila into it.
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To prepare medium culture we were used 10 mg media and 10 mL distilled water, mixed in
vial. Flies were stun with Anesthefly Solution and they were put medium culture. But the vial
was held horizontally because the flies would cling to the medium culture and maybe they die.
Like this,
Figure 1: F1 x F1 Drosophila
Until the flies were wake up, we waited. Then,we were bring the tube to vertical position. We
waited during a week for the Drosophila to form the F2 phenotypes.
Figure 2: F2 phenotypes in the pupa stage
After this step, everyday flies were checked. For this, flies were stun with Anesthefly Solution.
Then, separated in accordance with gender and phenotypes. Such as:
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Figure 3: Drosophila separated by genotype and phenotypes
Drosophila separated by sex and phenotypes were counted and it was noted. Subsequently,
these flies were put on a morgue. In this way underwent checks for 10 days.
RESULT
Table 1: Phenotypes of the F1 Generation
Phenotypes Number of progeny
Males Females Total
Wild type 26 27 53
Sepia 4 15 19
Table 2: Calculation of chi squared value for a cross between F1 and F1 flies.
Class Observed Expected (O-E)2 (O-E)2/
Expected
Wild type 53 54 ( 53-54)2
= 1
1/ 54
=0.018
Sepia 19 18 ( 19-18)2
=1
1/18
=0.055
Totals 72 72 X2
= 0.073
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Graph 1:Chi-Square Distribution Graph
If X is a random variable having a χ2 distribution with ν = 1 degrees of freedom, then p = Pr[X
≥ 0.074] = 0.7856. By conventional criteria, this difference is considered to be not statistically
significant.
DISCUSSION
Chi Square is based on whether the difference between observed and expected frequencies is
significant. It is used in the analysis of qualitatively stated data.
The calculation needs to be performed only twice - once for each phenotype. Adding the results
together provides the value for X2.
Total number of flies observed: 72
Flies with Wild type: 53
Flies with Sepia: 19
Expected ratio: 3:1
Expected number of flies with Wild type: 54
Expected number of flies with Sepia: 18
X2 = [(53–54)2 /54]+ [(19–18)2 /18]=0.01851+0.05555=0.074
Degrees of freedom [df = (# rows - 1) * (# columns - 1)] : 1
P value:0.7854 ; P > 0.005
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Table 3: Chi-Square Table4
Since only two phenotypes are used, there will be only one degree of freedom, as the number
of degrees of freedom is one less than the number of phenotypic categories. According to the
chi square table given above, the X2 value (0.074) is between 0.00393 and 0.45500, the values
for 50% and 80% confidence respectively, there is an 50 to 80% confidence that the deviations
from the expected experiment values are due solely to chance. To test the result that we
obtained, we use the Chi square statistic to test hypotheses concerning expected and observed
ratios. If the p value is greater than 0.005, so the hypothesis can not be rejected.
CONCLUSION
I expect from this experiment, if the mutation vestigial is autosomal recessive in the P
generation, then by the F2 generation the expected Phenotype Ratio is 75% wild : 25% vestigial.
The Hypothesis was supported based on data and percent error.
REFERENCES
1) Retrieved on 8 April 2010 at
http://www.dreamessays.com/customessays/Science%20Research%20Papers/11452.ht
m
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2) Basic Genetics: Thomas Hunt Morgan and Sex-linked Traits. Retrieved April 6, 2010
from http://library.thinkquest.org/20465/sexlinked.html
3) http://www2.vernier.com/sample_labs/BIO-A-07-COMP-genetics_of_drosophila.pdf
4) Krieger, K. and U. Pott. 2002. Bio 1 Laboratory Manual. Genes and Hypothesis, pp. 7-
1 – 7-10. UWGB, WI.
APPENDIX
Chi-Square equation:
Degrees of freedom formula:
DF = (# rows - 1) * (# columns - 1)