BIOLOGY 103 FALL, 2002 LAB 9

For our experiment, we crossed a long-antennaed (Ar) female with an Eyeless (Ey) male.

Cross #1: Their Children

264 Wild-type (+) female, 256 (+) male
257 (Ar) female, 249 (Ar) male

The eyeless trait did not appear in the children.

Now, we'll cross the children with each other.

Cross #2a: (+) female crossed with (+) male

Results: 369 (+) female, 382 (+) male
134 (Ey) female, 140 (Ey) male

In this cross, the (Ar) trait did not show up

Cross #2b: (Ar) female crossed with (Ar) male

Results: 132 (+) female, 147 (+) male
265 (Ar) female, 248 (Ar) male
42 (Ey) female, 35 (Ey) male
103 (Ar/Ey) female, 87 (Ar/Ey) male

All traits appeared in the children, with a 3:3:1:1 ratio

Cross #2c: (Ar) female crossed with (+) male

Results: 171 (+) female, 200 (+) male
187 (Ar) female, 207 (Ar) male
63 (Ey) female, 65 (Ey) male
59 (Ar/Ey) female, 51 (Ar/Ey) male

Once again, all traits showed with a 3:3:1:1 ratio

Cross #2d: (+) female crossed with an (Ar) male

Results: 187 (+) female, 174 (+) male
175 (Ar) female, 192 (Ar) male
53 (Ey) female, 77 (Ey) male
62 (Ar/Ey) female, 80 (Ar/Ey) male

Again, all traits appear with a 3:3:1:1 ratio

First we established that the
purple eye gene is an example of true breeding, and recessive...
then that the lobed eye gene is also and example of true breeding,
but it is dominant.

We hypothesized that breeding a purple, lobed female with a wild type male would produce a
9 lobed m/f : 3 wild type m/f : 3 purple, lobed m/f : 1 purple m/f.

We tested the hypothesis by breeding a purple, lobed female with a wild type male.
They produced=> lobed females + lobed males (1:1) ratio
the lobed males and females produced=> 1 wild m/f : 2 lobed m/f : 1 purple, lobed m/f

Our hypothesis was disproven since tour results were in a
1 wild: 2 lobed: 1 purple, lobed : 0 purple ratio
INSTEAD OF A 3 wild: 9: lobed: 3 purple, lobed: 1 purple ratio

So because our hypothesis was disproven, we concluded that there must be some linkage between the genes. Instead of each gene being carried by a different chromosome (which would result in our hypothesized Punett Square with 16 combinations), the genes are linked and share the same chromosome. The lobed and the purple eyes were carried on one chromosome and the wild gene was carried on a chromosome. This would then give us 4 possible results when doing the Squares and explain the 2:1:1 0 ratio that resulted in our experiment.

Kate Amlin, Katie Campbell, Stephanie Lane

(e)PR & (e) PR = PR

* PR is true breeding; Phenotype PR, Genotype PR PR

2) Next we tested (e)PR and (a) AR

PR AR & PR AR = (1) PR + : (2) PR AR

* PR again was homozygous, the AR seemed to be heterozygous

3) We examined (a) AR further
AR & AR = (2) AR: (1) +
* As with (2) the part of the model of genes where the AR AR should have existed was absent. The only way to obtain the ratio is if the phenotype AR is genotypically AR +

4) In order to test this further, we bred (e)PR (a) AR and an (e)+ (a) +

PR AR & + + = (1) (e) + (a) AR : (e) + (a) +
* This shows that the (e) + is dominant over the (e) PR and that the AR AR combination does not exist and is always heterozygous

We decided to test the wing vein trait of crossveinlessness.

When both male and female were crossveinless (CV), the kids were both CV and there was an equal number of males and females. When we mated the kids, the grandkids were also both CV, and there was also an equal number of males and females.

When the female is CV and the male is wild (+), the female kids were all + and the males were all CV, with equal amounts of male and female. When the kids were mated together, there were equal amounts of male and female. Within the females, there were equal amounts of + and CV, and in the male there were also equal amounts of + and CV.

Now...
When the female is + and the male is CV, the kids were ALL +, with equal amounts of male and female. In the grandkids, there were equal amounts of male and female, but there were ONLY female +. Within the males, there were equal amounts of + and CV.

And then...
When we mated the grandkids of a female + and a male CV, we got equal amounts of male and female and within each gender, equal amounts of + and CV.

So how do we account for the reappearance of the female CV when it seemed to have disappeared in both the female kids and grandkids generations? Our theory is that CV is a recessive gene that exists on the sex chromosome. This follows the same type of genetic pattern as recessive genes in humans, where recessive genes on the X chromosome do not show up in the female unless it is on both of her X chromosome. However, because males only have one X chromosome, when they have a recessive gene on the X, it will show up no matter what. Therefore, it is more unlikely for a recessive gene to affect the phenotype of the female than in the male.

We decided to test the wing vein trait of crossveinlessness.

When both male and female were crossveinless (CV), the kids were both CV and there was an equal number of males and females. When we mated the kids, the grandkids were also both CV, and there was also an equal number of males and females.

When the female is CV and the male is wild (+), the female kids were all + and the males were all CV, with equal amounts of male and female. When the kids were mated together, there were equal amounts of male and female. Within the females, there were equal amounts of + and CV, and in the male there were also equal amounts of + and CV.

Now...
When the female is + and the male is CV, the kids were ALL +, with equal amounts of male and female. In the grandkids, there were equal amounts of male and female, but there were ONLY female +. Within the males, there were equal amounts of + and CV.

And then...
When we mated the grandkids of a female + and a male CV, we got equal amounts of male and female and within each gender, equal amounts of + and CV.

So how do we account for the reappearance of the female CV when it seemed to have disappeared in both the female kids and grandkids generations? Our theory is that CV is a recessive gene that exists on the sex chromosome. This follows the same type of genetic pattern as recessive genes in humans, where recessive genes on the X chromosome do not show up in the female unless it is on both of her X chromosome. However, because males only have one X chromosome, when they have a recessive gene on the X, it will show up no matter what. Therefore, it is more unlikely for a recessive gene to affect the phenotype of the female than in the male.

We crossed a Purple-eyed Female with a Wild-eyed male.
The offspring had a 3:1 ratio for wild-eye color to purple. And after hand graphing the crossing of a PR PR female and a + + male, it was confirmed that the PR gene is recessive.

We then crossed a Curley-winged Female to a normal (wild type) male. For that, we had a two to one ratio. When hand-graphing the results of mating a CY CY female and a + + male, we realized we should have had all offspring be the same if that was to work. Therefore, (because we know that the wild type is + + from a previous experiment) the curly eyed gene cannot be CY CY. Upon graphing the results of a CY+ female mating with a ++ male, the correct ratio of two to one appeared. In order for the 2:1 to hold, we must assume that the CY gene is dominant over + gene. Also, the CYCY gene does not exist.

Finally, we paired a purple eyed, curly winged female with a wild-type male. The first generation offspring have two wild type offspring(both eyes and wings) and two wild type eyes with curly wings. When hand graphing the mating of a CY+, PRPR female with the ++,++ male, we realize there is a 50% chance of finding an offspring with curly wings (CY+) and no chance of the PR gene appearing. This holds, since the CY gene is dominant over the wild type and the wild type is dominant over the Purple eye gene.

Cross 1:
Purple-Eyed Female x Purple-Eyed Male
result - 492f and 499m ALL PURPLE

1b:
offspring of cross 1
result - 505f and 513m ALL PURPLE

CONCLUSION - purple-eyed trait is true breeding

Cross 2:
Eyeless Female x Eyeless Male
result - 511f and 475m ALL EYELESS

2b:
offspring of cross 2
result - 498f and 506m ALL EYELESS

CONCLUSION - eyeless trait is true breeding

Cross 3:
Purple, Eyeless Female x Purple, Eyeless Male
result - 528f and 506m ALL EYELESS

3b:
offspring of cross 3
result - 498f and 493m ALL EYELESS

CONCLUSION - because both parents were homozygous for eyeless, there was no chance for the eye color trait to be displayed phenotypically since they could not have eyes

Cross 4:
Purple, Eyeless Female x Wild Eyed Male
result - 518f and 471m ALL HAD WILD EYES

4b:
offspring of cross 4
result - 297f and 280m WILD EYES
101f and 84m WILD-SHAPED, PURPLE EYES
129f and 133m EYELESS

CONCLUSION - for eye shape, parents in the second generation were heterozygous for wild eye and eyeless. as a result, some of the offspring in the 3rd generation showed the wild-eye trait and others were eyeless.

for eye color, parents in the second generation were heterozygous for wild eye and purple. as a result, some offspring showed the purple eye trait and others were wild.

the ratio for this double-trait cross was 3:1:1 instead of the regular 9:3:3:1 ratio. when more eye-color traits would have been displayed, some offspring turned out eyeless. this is because the 3rd gen. the genotypes that would have shown purple eyeless and wild-colored eyeless could not be distinguished as two different phenotypes. we could not see the eye color trait if they did not have eyes.

We were curious about the inheritance of the yellow gene and whether it was passed on specifically by the male or the female and if it was possibly passed on to a specific sex.

Our first experiment was conducted with a yellow male and a "true" wild female. The offspring were all phenotype wild.
Once we continued the experiment and bred both phenotype wild sexes they yielded all phenotype wild flies again.
Our first hypothesis was that the male fly could not pass on yellow to either sex offspring.

For our second experiment we mated a yellow female with a "true" wild male fly. The offspring were about half phenotype wild female and half yellow male. We had no idea then how female yellow flies could possibly exist. Next we bred the offspring (wild phenotype females with yellow males) and this produced both wild phenotype males and females, and yellow phenotype males and females.
After this occured we started pulling our hair out, we knew something was wrong with our original hypothesis but we couldn't figure out how yellow males and females had been produced or if one sex had passed it on to its offspring.

After strenuous thinking aided by our lab companion Will, we came to the conclusion that the yellow allele is a recessive trait on the X chromosome. Thus, if a male has a yellow allele on his X chromosome he will be phenotype yellow. In order for a female fly to be phenotype yellow, she needs both of her X chromosomes to have yellow alleles. We wondered if this trait was similar to that of color blindness in humans, since colorblindness is more common in males.

Hypothesis: Both male and female fruit flies have two genes for each characteristic and the characteristics which are exhibited in inheritence will reflect this fact, with the wild type of each trait being dominant.

Methods and Results: We mated male and female brown-eyed fruit flies and found them to be "true breeding" (ie: their off-spring were all brown-eyed). Then, we mated a brown-eyed female with a wild-type male, producing offspring which all exhibited the wild-type phenotype. These offspring, when mated, produced offspring some of which are wild-type, some brown-eyed in an approximately 3:1 ratio.

We then examined wing angle, first mating a male and female both with the dichaete wing angle. This trait did not appear to be "true-breeding" as some of the offspring produced had a wildtype wing angle and some had a dichaete in an approximately 1:2 ratio.

Experimenting with a combination of these traits (eye color and wing angle) produced other interesting results. We mated a brown-eyed female with a dichaete wing angle with a wild type male, resulting in offspring which all had wild-type eyes and some of which had a wild-type wing angle and some a dichaete wing angle. Mating two of these offspring (a female wild-type and a male with wild-type eyes and dichaete wing angle) produced offspring in a 3:3:1:1 ratio (+, dichaete, brown-eyed, brown-eyed dichaete).

Conclusion:
The results of breeding brown-eyed fruit flies supports our hypothesis, as the presence of two genes for every trait would lead to the 3:1 ratio which was present in our results. Experimenting with wing angle, however, suggested that our hypothesis (wild type is always dominant) was incorrect, as it suggests that dichaete is a dominant trait and, in fact, lethal in a double dose. Also, the experiment combining brown eyes and dichaete wing angle further supported these findings because the ratio produced was what was expected in accordance with the punnett square examining that combination.

Curly Wings:

First generation
Curly Fem. X Curly Male
353 Curly Fem.
168 Wild Fem
349 Curly Male
156 Wild Male

2:1 (ignoring sex)

Second generation
Wild Fem x Father (Curly Male)
257 Wild Fem
241 Curly Fem
235 Wild Male
255 Curly Male

1:1 (ignoring sex)

Hypothesis: Curly wings are a heterozygous genotype. Wild type wings are homozygous dominant. A homozygous recessive genotype is fatal.

First generation

	MOTHER C c
F C A T H c E R	CC (wild)	Cc (curly)
	Cc (curly)	cc (fatal)

Second generation

	MOTHER C C
F C A T H c E R	CC (wild)	CC (wild)
	Cc (curly)	Cc (curly)

White eyes:

First generation
White Fem. X Wild Male
495 Wild Fem.
502 White Male

Ratio 1:1

Second generation
Wild Fem. X White Male
266 White Fem.
259 White Male
252 Wild Fem.
247 Wild Male

Ratio: 1:1:1:1

Third generation
First generation Wild Female X Wild Male
487 Wild Fem.
267WildMale
250 White Male

Ratio: 2:1:1

Hypothesis: White eyes are a X-chromosome linked characteristic. A female that carries only one chromosome associated with the white eye gene has wild eyes. A female must have both chromosomes associated with the white eye gene to have white eyes. Because males only have one X chromosome, if they carry the white eyed gene they have white eyes.

First Generation

	MOTHER X Xw
F X A T H Y E R	XwX (wild)	XwX (wild)
	XwY (white)	XwY (white)

Second Generation

	MOTHER Xw X
F Xw A T H Y E R	XwXw (white)	XwX (wild)
	XwY (white)	XY (wild)

Test Cross

	MOTHER Xw X
F X A T H Y E R	XwX (wild)	XX (wild)
	XwY (white)	XY (wild)

Our next experiment examines double trait flies in comparison to the wild type mate. We first bred a purple-eyed, cossveinless female with a wildtype male. Our results: 2 images; 528 wild type females and 493 CV males. Purple eyes is a recessive trait and CV is carried on the x chromosome of the female. We cross that generation and we got a 3:1:3:1 ratio of wild type to purple eyes to cross veined to purple eyed cross veined.

The female will always contribute the following traits: ++, +CV, p+, p CV. The male will contribute one of two sets of genes depending on whether the offspring is male or female. Is the offspring is female, he will either contribute +CV, or pCV. If the offspring is male, he will give either +Y, or pY. The Y trait here is essentially "empty" since it cannot carry the CV trait.

Scalloped winged: True Breeding
Wild Winged: True Breeding

Hypothesis:

Either wild winged or scalloped winged will be dominant. (regardless of sex).

One female Scalloped winged and One male Wild winged: all female offspring are wild winged and all males are scalloped. In the second generation males and females both had equal proportions of wild and scalloped winged.

One male Scalloped and One female wild winged: all offpring are wild winged in the first generation and in the second generation all females are wild winged and all males are equally split between scalloped and wild winged.

Conclusion:
Dominance is sex related in this case of scalloped versus wild winged fruit flies. This is due to the fact that some genes are located exclusively on the X chromosome and there is no counterdominant gene on the Y chromosome. Because females have 2 X chromosomes and males one X and one Y, then whatever gene is on the X (scalloped) will be dominant in males because there is no counter gene on the Y chromosome. Also, the female flies with wild prove that scalloped is recessive, since they have both scalloped and wild genes. Finally, a female with two scalloped is not lethal, merely guarentees that her son will be scalloped.

we tested aristapedia antennae and dichaete wing alignment. we have been testing both of these characteristics together and separately in males and/or females for the last two hours. we have figured out that both traits are lethal homozygos, meaning that they cannot truebreed because there can never be a homozygos aristopedia or a homozygos dichaete. any fruit flie that is aristapedia or dichaete is heterozygos. the problem is that these traits are only lethal homozygos in females. because they are only lethal in females, the genes for the aristapedia and dichaete are in the x chromosome. because the genes are both on the x chromosome, they cannot be separated, except in the case of cross-over when the chromosome splits.

our results were as follows:

m(ArD) & f(++):
1:1
of Ar:D

f(ArD) & m(++):
1:15-20:15-20:1
of ++:Ar:D:ArD