Reebops Lab Report
I. INTRODUCTION
For this report, four different crosses were completed: two monohybrid crosses and two
dihybrid crosses. A hypothesis was formed for each cross:
1. Monohybrid Cross #1:The traits being studied are autosomal and they undergo
independent assortment.
2.Monohybrid Cross #2: The traits being studied are autosomal and they undergo
independent assortment.
3. Dihybrid Cross #1: The traits being studied are autosomal and they undergo
independent assortment.
4. Dihybrid Cross #2: The traits being studied are autosomal and they undergo
independent assortment.
Gregor Mendel allowed me to form the basis for my hypothesis. Through the selective
cross-breeding of common pea plants over many generations, Mendel discovered that certain
traits show up in offspring without any blending of parent characteristics. Therefore, I knew the
Reebops would have both autosomal and sex linked chromosomes. I first assumed that the traits
were autosomal chromosomes (chromosomes that are not sex chromosomes) because most
chromosomes in an organism are autosomal. In humans, there are 22 pairs of autosomal
chromosomes. The 23rd pair of chromosomes is the sex chromosomes, and they differ between
males and females. In all of my hypotheses, I assumed that the traits being studied were
autosomal. If my hypothesis was incorrect, I knew the trait or one of the traits would be sex
linked. I knew the chromosomes would undergo independent assortment because all
chromosomes that undergo meiosis undergo independent assortment. This occurs when pairs of
homologous chromosomes line up and separate independently of one another.
We are doing this lab to better understand how genetics work. Specifically, in Reebops
Part 1, we developed an understanding of the processes of meiosis, fertilization, independent
assortment, and segregation. We also learned about alleles, genotypes, phenotypes,
chromosomes, genes, gametes, zygotes, and genotypes. Through this lab, we learned the genetic
basis for the variability within a species. In Parts 2 and 3, we learned how to use Punnett squares
and chi square tests in order to test a hypothesis. In Part four, we learned about sex-linked traits.
II. METHODS
In order to complete this lab, one needs to understand how to complete a monohybrid
cross, a dihybrid cross, and chi-squares. Monohybrid crosses are fairly simple. First, you name
the trait that is being discussed. There will only be one. Next, you list the possible alleles (there
will be two). Then, you will list the phenotypes of the parents, and then the genotypes of the
parents. This will allow you to make a list of the possible gametes. With both the possible female
and male gametes, you can complete a Punnett square. The Punnett square will show the
offspring genotypes, offspring phenotypes, the expected ratios of phenotypes, and the genotype.
After you find this information, you can do the same thing for the F1 generation, or the parent’s
offspring. This will show you the expected genotypes and phenotypes of the F2 generation.
Dihybrid crosses are similar, but are a little more complicated because two traits are
involved. First, list the two traits and the four possible alleles. Then, list the phenotypes and
genotypes of the parents. Also list the possible alleles. Then complete the Punnett Square. This
will give you the offspring genotypes, phenotypes, expected ratios phenotype, and genotype. Do
the same thing with the F1 generation in order to find the expected phenotypes and genotypes of
the F2 generation.
After you have completed your crosses, you must complete a chi square worksheet to test
your hypothesis. If you follow the worksheet, you will find a chi square number. You will also
find your degrees of freedom by subtracting your number of classes by 1. Then refer to the Chisquare table, and if your p value is greater than 0.05, your hypothesis is accepted. If not, the
hypothesis is rejected, and at least one of your traits is sex linked. If that is the case, you need to
redo your cross-using sex linked traits. Your hypothesis should then be correct.
In order to determine which trait is sex linked in a dihybrid cross, add the similar traits
together and compare them. The two traits with a ratio of 3:1 are the autosomal traits. The others
are the sex-linked traits.
RESULTS:
For Monohybrid Cross 1, I crossed a P1 homozygous recessive female who had green
antennae with a P1 male who had silver antennae (which are dominant), and got the following F2
offspring:
434 Females who had silver antennae
448 females who had green antennae
440 males who had silver antennae
437 males who had green antennae
For the Monohybrid Cross 2, I crossed a P1 homozygous female who had 3 body
segments with a P1 homozygous male who had only 2. I got he following F2 offspring from the
cross:
348 females with 3 body segments
123 females with 2 body segments
368 males with 3 body segments
115 males with 2 body segments
For the first dihybrid cross , I crossed a P1 female who was homozygous recessive for the
traits of leg color and number of eyes, with a P1 male who was homozygous dominant for leg
color and number of eyes. I got the following F2 offspring:
494 females with blue legs and two eyes
158 females with blue legs and three eyes
156 females with red legs and two eyes
54 females with red legs and three eyes
508 males with blue legs and two eyes
143 males with blue legs and three eyes
160 males with red legs and two eyes
51 males with red legs and three eyes
For the second dihybrid cross, I crossed a P1 homozygous recessive female who had a
straight tail and smooth legs with a P1 male who had a curly tail and hairy legs. The F2
offspring that I got are:
256 females with curly tails and hairy legs
245 females with curly tails and smooth legs
80 females with straight tails and hairy legs
87 females with straight tails and smooth legs
240 males with curly tails and hairy legs
257 males with curly tails and smooth legs
79 males with straight tails and hairy legs
85 males with straight tails and smooth legs
Summaries:
For the first monohybrid cross, the expected phenotypic ratio of the F1 generation is 4/4
silver. The expected phenotypic ratio of the F2 generation is 3:1 Silver: Green. The Chi square
value was 601.1. The p value was much less than 0.01, so the hypothesis was rejected. So then, I
redid the cross, using sex-linked traits. The expected phenotypic ratio of the F1 generation is 2:2,
Silver: Green. The expected phenotypic ratio of the F2 generation is 1:1:1:1, female, green:
female, silver: male, green: male, silver. The new chi square value is 0.2501 and the p value is
greater than .30 so the hypothesis is accepted.
For the second monohybrid cross, the expected phenotypic ratio of the F1 generation is
4/4 3 body segments. The expected phenotypic ratio of the F2 generation is 3:1, body segments:
2 body segments. The Chi square value was 0. The p value was greater than 0.3 so the hypothesis
was accepted.
For the first dihybrid cross, the expected phenotypic ratio of the F1 generation is 16/16
blue, 2 eyes. The expected phenotypic ratio of the F2 generation is 9:3:3:1 blue 2: blue 3: red 2:
red 3. The Chi square value was 2.83. The p value was greater than 0.3 so the hypothesis was
accepted.
For the second dihybrid cross, the expected phenotypic ratio of the of the F1 generation is
16/16 curly, hairy. The expected phenotypic ratio of the F2 generation is 9:3:3:1 curly, hairy:
curly, smooth: straight, hairy: straight, smooth. The Chi square value was 469.01. The p value
was much less than 0.01, so the hypothesis was rejected. So then, I redid the cross-using sex
linked traits. The expected phenotypic ratio of the F1 generation is 8:8 female, curly, hairy: male,
curly, smooth. The expected phenotypic ratio of the F2 generation is 3:3:1:1:3:3:1:1, female,
curly, hairy: female, curly, smooth: female, straight, hairy: female, straight, smooth: male, curly,
hairy: male, curly, smooth: male, straight, hairy: male, straight, smooth. The new chi square
value is 1.38, and the p value is greater than 0.3. The hypothesis is now accepted.
Key:
1 antenna = AA
2 antenna = Aa
No antenna=aa
Red nose=QQ
Orange nose = Qq
Yellow nose =qq
2 eyes = EE or Ee
3 eyes= ee
3 body segments=DD or Dd
2 body segments= dd
Blunt antennae=BB or Bb
Sharp antennae=bb
1 green hump=MM
2 green humps= Mm
3 green humps=mm
Curly tail = TT or Tt
Straight tail= tt
Blue legs = LL of Ll
Red legs = ll
Curly proboscis =RR or Rr
Straight proboscis =rr
G= silver antennae
g= green antennae
CONCLUSIONS:
For this report, four different crosses were completed: two monohybrid crosses and two
dihybrid crosses. A hypothesis was formed for each cross:
1. Monohybrid Cross #1:The traits being studied are autosomal and they undergo
independent assortment.
2.Monohybrid Cross #2: The traits being studied are autosomal and they undergo
independent assortment.
3. Dihybrid Cross #1: The traits being studied are autosomal and they undergo
independent assortment.
4. Dihybrid Cross #2: The traits being studied are autosomal and they undergo
independent assortment.
The first monohybrid cross was not supported. I know this because after completing my
chi square worksheet, The Chi square value was 601.1. The p value was much less than 0.01, so
the hypothesis was rejected. The F2 data does not makes sense because it does not coincide with
the observed phenotype.
The second monohybrid cross was supported. I know this because after competing my chi
square worksheet, The Chi square value was 0. The p value was greater than 0.3 so the
hypothesis was accepted. The F2 data makes sense because it coincides with the observed
phenotype.
The first dihybrid cross was supported. I know this because after competing my chi
square worksheet, The Chi square value was 2.83. The p value was greater than 0.3 so the
hypothesis was accepted. The F2 data makes sense because it coincides with the observed
phenotype.
The second dihybrid cross was not supported. I know this because after completing my
chi square worksheet, The Chi square value was 469.01. The p value was much less than 0.01, so
the hypothesis was rejected. The F2 data does not make sense because it does not coincide with
the observed phenotype.
Revised Hypothesis:
The first monohybrid cross was not supported. I know this because after completing my
chi square worksheet, The Chi square value was 601.1. The p value was much less than 0.01, so
the hypothesis was rejected. The F2 data does not make sense because it does not coincide with
the observed phenotype. Therefore, the trait is most likely sex linked.
New Hypothesis: The antennae color trait is sex linked and undergoes independent
assortment.
My new hypothesis was supported because the new chi square value is 0.2501 and the p
value is greater than .30. The F2 data now coincides with the observed phenotypes.
The second dihybrid cross was not supported. I know this because after completing my
chi square worksheet, The Chi square value was 469.01. The p value was much less than 0.01, so
the hypothesis was rejected. The F2 data does not make sense because it does not coincide with
the observed phenotype. Therefore, the trait is most likely sex linked.
New Hypothesis: The tail trait is autosomal, and the leg trait is sex linked, and both traits
undergo independent assortment.
The new chi square value is 1.38, and the p value is greater than 0.3. The hypothesis is
now accepted.
Error Analysis:
When filling out the chi square chart, I rounded to two decimal places. This could result in slight
error. Also, while making the many Punnet Squares, I could have accidentally wrote the wrong
genotypes. This could greatly alter results from a simple error.
Discussion Questions:
1. I believe there is inheritance mechanism so there is genetic diversity. Therefore, species can
have a more diverse gene pool and can withstand disease that may wipe out others that don’t
have the same gene pool. There was genetic diversity present when we created the Reebops in
class. Everyone had the same parents, (each parent had the same genotype) and we all ended up
with different phenotypes for our offspring. The Punnett Squares show all the options for genetic
diversity with those traits.
2. The inheritance of Reebop traits does not involve the blending of physical traits. The offspring
inherit one trait or another. For example, if one parent has a curly proboscis and another has a
straight one, the child will inherit one or the other, not a mixture of both. This is because the
offspring can inherit three different genotypes, RR, Rr, or rr. The first two will give the offspring
a curly proboscis and the rr will give it a straight one. The offspring will always inherit one of
the parents phenotypes, not something completely different.
3.The alleles are stable from each generation because they are constant; they don’t change what
they represent. If RR represents a curly proboscis in the P1 generation, it will in the F2
generation as well.
4. Segregation is the separation of pairs of homologous chromosomes. It takes place during
Anaphase I. Independent Assortment occurs when pairs of homologous chromosomes line up
and separate independently of one another. It occurs at the metaphase plate and begins during
Prophase I, continues through Metaphase I, and ends in Anaphase I. All of the Reebops undergo
meiosis so they undergo both of these processes.
5. Both parents contribute to offspring. Each parent donates their gametes to create the zygotes.
If sex linked chromosomes are involved, the mother will donate X’s and a father can donate a Y.
If a Y is donated, a male zygote is created. If two X’s are donated by a mother, a female zygote
is created. Both parents must contribute if an organism reproduces sexually.
6. Sex-linked chromosomes are an exception to the basic rules of inheritance. According to page
53 in the lab Manuel, in humans there are 22 pairs of autosomal chromosomes. The 23rd pair are
sex chromosomes and are called X and Y chromosomes. They differ between males and females.
The X and Y chromosomes line up and segregate like homologous pairs during meiosis, but they
are not a homologous pair. They are different sizes and code for different genes.