Adam Capriola
Biology Lab Section 156
Dr. Lee
4/1/07
Fly Lab
Title: A Study on Inheritance Mechanisms of Physical Traits Found in Drosophila Melanogaster
Abstract
This was a study on determining the inheritance mechanisms for different physical traits
found in Drosophila melanogaster. In order to accomplish this, a reciprocal cross was performed
by mating wild type male flies with mutant female flies and then wild type female flies with
mutant male flies. After two weeks, the counts and phenotypes of the resulting F1’s were
recorded, and then another reciprocal cross was performed by mating the F1’s from each of the
parental crosses. Predictions were then made for the phenotypic ratios of the F2’s by following a
believed inheritance mechanism. After another two weeks, the actual counts and phenotypes of
the F2’s were recorded. The predicted ratios were compared to the actual ratios using a chisquared test to see if the predicted inheritances mechanisms were correct. The F2 ratios for
mutant A were predicted using the autosomal inheritance pattern. Both chi-squared values
failed, but this is believed to be because of human error or too small of a sample size. The F2
ratios for mutant D were predicted using the X-linked inheritance pattern. Both chi-squared
values these ratios passed, which means that the trait for mutant D is passed on by X-linked
inheritance. In conclusion, the inheritance mechanisms were determined, but there is some doubt
for the inheritance mechanism of mutant A.
Introduction
The genes that code for specific physical traits found in species are passed on from parent
to progeny by different inheritance mechanisms. One mechanism, autosomal inheritance,
involves the passing of a gene found on an X chromosome coded for by two alleles. In most
cases, there is one dominant allele and one recessive allele, in which trait coded for on the
dominant allele shows over the recessive trait. There can however, be more than just two
different types of alleles on a gene. Often, the dominant allele codes for the wild type trait of a
species. Wild type is a term which refers to the most common physical trait found in a species.
This does not mean that the allele coding for this trait is a dominant allele, however this is often
the case. When the gene is passed to the progeny, one allele from each parent is contributed to
make the gene for the progeny. Physical traits can sometimes be coded for by two different
genes, or there can be alleles that are not completely dominant or recessive to each other. This
allows room for many different variations and ratios of physical traits.
Another inheritance mechanism is X-linked inheritance. In this type of inheritance, the
gene for the physical trait is encoded for on the X sex chromosome. The gene can also have
different variations of alleles, just as in autosomal inheritance, but in this case the male will only
have one allele. This is because males only have one X sex chromosome. Females have two X
sex chromosomes, so they will have two alleles that work much like they do in autosomal
inheritance. Because the males only have one X chromosome, this means they only pass one
allele on to their female progeny and none to their male progeny (Goodnenough, 1984). Females
have two alleles to pass on, so they pass one allele to the male progeny and one to the female
progeny.
A way to find the inheritance mechanism for a certain trait is by crossing a male wild
type with a female that has the trait, and then doing the reciprocal cross, female wild type with
male with the trait (Greenspan, 1997). After performing this, the F1’s can then be used for
another reciprocal cross. By this time, it should be possible to analyze the counts of the progeny
and map out the inheritance pattern. In this experiment, the inheritance patterns of different fly
traits found in Drosophila melanogaster were determined though reciprocal crosses of parental
flies and their F1 progeny.
Materials and Methods
Six different variations of drosophila melanogaster were distributed to teams in vials
containing about 10 adult flies each. The six variations of drosophila were labeled wild type,
mutant A, mutant B, mutant C, mutant D, and mutant E. Each species of fly was observed under
a microscope. In order to view the flies under the microscope without them flying away, a
cooler with ice was first obtained and an empty vial was placed into the ice. The flies in the first
vial were knocked down from the top by tapping the vial on the table. While the flies were still
down, the foam cork was removed and the vial was quickly flipped upside-down on top of the
empty vial in the cooler. In a few moments, the flies fell asleep from the coldness of the ice and
fell into the empty vial. The dormant flies were then transferred from the new vial to a metal
outlet cover placed on a plate of ice. The flies were now observed under a 10x lensed light
microscope. Fine paint brushes and toothpicks were used to manipulate the flies to better view
them. The sex was determined by looking for sex combs, which are only on males, or by
looking for darker tails, which are also found only on males. The phenotype was determined by
eye color, eye size, or wing size. Once observations were finished being recorded, the flies were
discarded into the fly morgue, which was a can with about one inch of vegetable oil in the
bottom.
After observing all six species of drosophila, teams then picked a mutant to mate with the
wild type. A reciprocal cross was performed, mating the wild type male with the mutant female
and the wild type female with the mutant male. First, two empty vials were filled about a third
of the way up with instant blue drosophila food and 0.5% propionic acid, which was needed to
feed the larvae. Six to eight grains of yeast were also put in each vial to feed the adults. Then,
vials containing wild type males and females and mutant males and females were distributed to
each team. The flies were “knocked out” in the same manner as they were before, and were
observed under the microscope in order to determine phenotypes and sex them. Once teams
became better at discerning fly characteristics, the flies could be viewed simply with the naked
eye. These figures were recorded, then about six mutant females and wild type females were put
into one of the new vials with food, and about six mutant males and six wild type females were
put into the other new vial. These vials were stored in a 20º C lab room.
A week later, the adult flies were cleared (put into the morgue) and observations were
recorded about the appearance of the vial. After another week, some F1 progeny had emerged.
These flies were knocked out and observed. Their sex and phenotypes were recorded. During
the next two weeks, the F1’s were observed and cleared four more times. These counts were
recorded on a sheet containing all of the class’s information. The F1’s were then crossed in the
same manner the parentals were crossed. About six female and six male progeny from the
mutant males by wild type female cross were put in a vial together, and about six female and six
male progeny from the mutant female by wild type male cross were put in a vial together. These
flies were observed and their F2 progeny counts were recorded after two weeks. This data was
also recorded on a sheet along with the rest of the class’s data. All of the class’s data was
recorded into a notebook to take home and analyze.
Results
When observing our mutant fly (mutant D), we noticed it had eyes which were the same
size as the wild type flies, but they were white in color (or this could be described as being
colorless). Other than this trait, it appeared the same as the wild type. It seemed to be the same
width and length, have the same sized wings, and have the same beige body color. The eye color
is what distinguished it as being different than the wild type.
For the cross between mutant A males and wild type females, the resulting F1 progeny
were 87 wild type males and 103 wild type females (Table I). The reciprocal of this cross,
mutant A female by wild type male, yielded 75 wild type males and 88 wild type females (Table
I). For the crosses between these F1’s, the progeny of the mutant A males by wild type female
cross produced 40 mutant A males, 35 mutant A females, 131 wild type males, and 177 wild type
females (Table II). These F2’s were compared to an expected ratio of 1 mutant A male to 1
mutant A female to 3 wild type males to 3 wild type females (Table III). The resulting chisquared value of 13.62 was compared the 95% confidence value of 7.82 for 3 degrees of freedom
(Table III, IX). Chi-squared was greater than this value, so that meant the proposed ratio had to
be rejected. The progeny of the mutant A female by wild type male cross yielded 7 mutant A
males, 16 mutant A females, 53 wild type males, and 71 wild type females (Table II). These
F2’s were also compared to an expected ratio of 1 mutant A male to 1 mutant A female to 3 wild
type males to 3 wild type females (Table IV). The resulting chi-squared value of 12.01 was
compared the 95% confidence value of 7.82 for 3 degrees of freedom (Table IV, IX). Chisquared was greater than this value, so that meant the proposed ratio had to be rejected.
For the cross between mutant D males and wild type females, the resulting F1 progeny
were 25 wild type males and 25 wild type females (Table V). The reciprocal of this cross,
mutant D female by wild type male, yielded 78 mutant D males and 87 wild type females (Table
V). For the crosses between these F1’s, the progeny of the mutant D males by wild type female
cross produced 58 mutant D males, 66 wild type males, and 159 wild type females (Table VI).
These F2’s were compared to an expected ratio of 1 mutant D male to 1 wild type male to 2 wild
type females (Table VII). The resulting chi-squared value of 4.78 was compared the 95%
confidence value of 5.99 for 2 degrees of freedom (Table VII, IX). Chi-squared was less than
this value, so that meant the proposed ratio could not be rejected. The progeny of the mutant D
female by wild type male cross yielded 26 mutant D males, 31 mutant D females, 27 wild type
males, and 36 wild type females (Table VI). These F2’s were compared to an expected ratio of 1
mutant D male to 1 mutant A female to 1 wild type male to 1 wild type female (Table VIII). The
resulting chi-squared value of 2.06 was compared the 95% confidence value of 5.99 for 2
degrees of freedom (Table VIII, IX). Chi-squared was less than this value, so that meant the
proposed ratio could not be rejected.
Discussion
After analyzing the counts of the F1’s, I was able to determine the mechanism by which
the specified trait was passed on. For mutant D, it appeared that the trait for colorless eyes is an
X-linked recessive trait. This is the only model of inheritance that seemed to fit with the
resulting progeny. I denoted the wild type gene as X+ and the mutant gene as XD. The first
parental cross was mutant D male by wild type female (XD/Y x X+/ X+), which hypothetically
should have resulted in ½ wild type females (X+/ XD) and ½ wild type males (X+/Y). The second
parental cross was wild type male by mutant D female (X+/Y x XD/ XD), which hypothetically
should have resulted in ½ wild type females (X+/ XD) and ½ mutant D males (XD/Y). The F1
data seemed to follow these numbers. Following the X-linked recessive inheritance pattern, I
predicted what the F2’s should be. With the use of a Punnett square, I found that the cross
between F1 wild type males and F1 wild type females (X+/Y x X+/ XD) should result in ½ wild
type females, ¼ wild type males, and ¼ mutant males. The genotypes of these progeny would be
¼ X+/ X+, ¼ X+/ XD, ¼ X+/Y, and ¼ XD/Y. In order to determine whether my predicted
phenotypic ratio matched the observed ratio, I used a chi squared test.
A chi-squared test takes a null hypothesis, which says that there is no difference between
observed data and predicted data, and determines whether the null hypothesis is valid (Russell,
2003). The observed data is compared to expected data using a mathematical formula to produce
a chi-squared value. That value is then compared to a probability value from a table, which is
obtained depending on the degrees of freedom. Degrees of freedom is found by taking the total
number of classes and subtracting one. The most common probability value is 95% confidence.
If the chi-squared value is greater than the probability value, the null hypothesis is rejected, but if
it is less than the probability value, the null hypothesis cannot be rejected.
I compared my predicted phenotypic ratio to the actual F2 counts and my chi-squared
value of 4.78 was less than the 95% confidence value of 5.99 for 2 degrees of freedom, which
means my predicted ratio could not be rejected (Table VII). I also used a Punnett square to find
the hypothetical F2’s between the F1 mutant D males and F1 wild type females (XD/Y x X+/ XD).
The Punnett square predicted ¼ wild type females, ¼ wild type males, ¼ mutant D females, and
¼ mutant D males. The genotypes of these progeny would be ¼ X+/ XD, ¼ X+/Y, ¼ XD/ XD, and
¼ XD/Y. I also compared the phenotypic ratio to the actual F2 counts and my chi-squared value
of 2.06 was less than the 95% confidence value of 7.82 for 3 degrees of freedom, which means
my predicted ratio could not be rejected (Table VIII). This means that I am most likely correct
in my assumption of colorless eyes being an X-linked recessive trait. Also, whenever the
reciprocal crosses give significantly different F1 and F2 phenotypic ratios, the trait is usually Xlinked, which was the case here (Goodenough, 1984). This also supports the hypothesis of Xlinked inheritance for white eyes.
In concern to mutant A, after analyzing the results of the F1’s, it seemed that having
small eyes is an autosomal recessive trait. I denoted the wild type gene as X+ and the mutant
gene as XA. The first parental cross was mutant A male by wild type female (XA/XA x X+/ X+),
which hypothetically should have resulted in ½ wild type females (X+/ XA) and ½ wild type
males (X+/ XA). The second parental cross was wild type male by mutant A female (X+/ X+ x
XA/ XA), which also hypothetically should have resulted in ½ wild type females (X+/ XA) and ½
wild type males (X+/ XA). The F1 data seemed to come close to these numbers. Using a Punnett
square, I predicted what the F2 progeny should be after crossing the F1’s. I found that the cross
between F1 wild type males and F1 wild type females (X+/ XA x X+/ XA) should result in 1/8
mutant A males, 1/8 mutant A females, 3/8 wild type males, and 3/8 wild type females. The
genotypes of these progeny would be ¼ X+/ X+, ½ X+/ XA, and ¼ XA/ XA. I compared the
phenotypic ratio to the actual F2 counts and my chi-squared value of 13.62 was greater than the
95% confidence value of 7.82 for 3 degrees of freedom, which means my predicted ratio had to
be rejected (Table III). I also used a Punnett square to find the hypothetical F2’s between the
other cross for mutant A, which was also F1 wild type males and F1 wild type females (X+/ XA x
X+/ XA). The Punnett square again predicted 1/8 mutant A males, 1/8 mutant A females, 3/8
wild type males, and 3/8 wild type females. The genotypes were the same as before, being ¼ X+/
X+, ½ X+/ XA, and ¼ XA/ XA. I also compared the phenotypic ratio to the actual F2 counts and
my chi-squared value of 12.01 was greater than the 95% confidence value of 7.82 for 3 degrees
of freedom, which means my predicted ratio could had to be rejected (Table IV).
Even though my proposed ratios were rejected for mutant A, I still believe that the trait
for small eyes is an autosomal recessive trait. There is no other method of inheritance that would
fit with the phenotypes found in the F1’s and F2’s. If more flies were counted, I feel that the chisquared value may be lower, and thus my proposed ratios would be accepted. However, there
were less mutant D flies counted than mutant A flies and my proposed ratios for mutant D passed
the chi-squared test. This could mean that there was error in counting the mutant A flies. The
trait of colorless eyes is easier to discern than small eyes, so I feel the results for mutant D are
more accurate than the results for mutant A.
The genes for white eyes (colorless eyes) and bar eyes (small eyes) are both found on
chromosome 1 of drosophila melanogaster (Russell, 1994). This is an X chromosome, which
rules out either of the traits being passed on by Y-linked inheritance. Only genes on the Y
chromosome as passed on by Y-linked inheritance. This proves that my assumptions of
autosomal inheritance for mutant A and X-linked inheritance for mutant D could be correct.
Those are the only two inheritance mechanisms that involve the X chromosome. The traits for
white eyes and bar eyes are also recessive (Russell, 1994). This goes along with my prediction
of both traits being recessive. The only thing I do not know for certain is if the specific traits are
either autosomal or X-linked.
Literature Cited
Goodenough, Ursula. 1984. Genetics (3rd edition). (Saunders College Publishing, Philadelphia
PA).
Greenspan, Ralph J. 1997. Fly Pushing: The Theory and Practice of Drosophila Genetics. (Cold
Spring Harbor Laboratory Press, Plainview NY).
Russell, Peter. 1994. Fundamentals of Genetics. (HarperCollins, New York NY).
Russell, Peter J. 2003. Essential iGenetics. (Benjamin Cummings, San Francisco).
Tables
Table I:
Mutant A Males
Mutant A Females
Wild Type Males
Wild Type Females
Mutant A Male x Wild Type
Female
0
0
87
103
Mutant A Female x Wild Type
Male
0
0
75
88
F1 Wild Type Males x F1 Wild
Type Female
40
35
131
177
F1 Wild Type Female x F1 Wild
Type Male
7
16
53
71
Table II:
Mutant A Males
Mutant A Females
Wild Type Males
Wild Type Females
Table III:
F1 Wild Type
Males x F1
Wild Type
Female
Mutant A
Males
Mutant A
Females
Wild Type
Males
Wild Type
Females
Total
Table IV:
F1 Wild Type
Female x F1
Wild Type
Male
Observed
Count
Observed
Ratio
Expected
Ratio
Expected
Count
(Obs-Exp)2/Exp
40
1.14
1
47.88
1.30
35
1
1
47.88
3.46
131
3.74
3
143.63
1.11
177
5.06
3
143.63
7.75
χ2 = 13.62
383
Observed
Count
Observed
Ratio
Expected
Ratio
Expected
Count
(Obs-Exp)2/Exp
Mutant A
Males
Mutant A
Females
Wild Type
Males
Wild Type
Females
Total
7
1
1
18.38
7.05
16
2.29
1
18.38
0.31
53
7.57
3
55.13
0.08
71
10.14
3
55.13
4.57
χ2 = 12.01
147
Table V:
Mutant D Males
Mutant D Females
Wild Type Males
Wild Type Females
Mutant D Male x Wild Type
Female
0
0
25
25
Mutant D Female x Wild Type
Male
78
0
0
87
F1 Wild Type Male x F1 Wild
Type Female
58
0
66
159
F1 Wild Type Female x F1
Mutant D Male
26
31
27
36
Table VI:
Mutant D Males
Mutant D Females
Wild Type Males
Wild Type Females
Table VII:
F1 Wild Type
Male x F1
Wild Type
Female
Mutant A
Males
Mutant A
Females
Wild Type
Males
Wild Type
Females
Total
Table VIII:
F1 Wild Type
Female x F1
Mutant D
Observed
Count
Observed
Ratio
Expected
Ratio
Expected
Count
(Obs-Exp)2/Exp
58
1
1
70.75
2.30
0
0
0
0
0
66
1.14
1
70.75
0.32
159
2.74
2
141.50
2.16
χ2 = 4.78
283
Observed
Count
Observed
Ratio
Expected
Ratio
Expected
Count
(Obs-Exp)2/Exp
Male
Mutant A
Males
Mutant A
Females
Wild Type
Males
Wild Type
Females
Total
26
1
1
30
0.53
31
1.19
1
30
0.03
27
1.04
1
30
0.30
36
1.38
1
30
1.20
χ2 = 2.06
120
Table IX:
Degrees of Freedom
2
3
95% Confidence Value
5.99
7.82