The document summarizes Gregor Mendel's experiments with pea plants and how they laid the foundation for genetics. It then describes how the author will simulate Mendel's experiments virtually using fruit flies. The author performs a series of crosses between fruit flies with different traits and uses a chi-square analysis to determine if the results match expected Mendelian ratios. The crosses are designed to show traits behaving as dominant, recessive, recessive sex-linked, and dihybrid traits.
Ethical and bio-safety issues related to GM cropsMahammed Faizan
a seminar presentation on ethical and bio-safety issues related GM crops.
impact of gm crops on human, animal and environmental health.
safety measure related transgenic crops.
international governmental bodies
Ethical and bio-safety issues related to GM cropsMahammed Faizan
a seminar presentation on ethical and bio-safety issues related GM crops.
impact of gm crops on human, animal and environmental health.
safety measure related transgenic crops.
international governmental bodies
Role of secondary metabolites in insect pest managementMohd Irshad
SECONDARY METABOLITES ARE THOSE COMPOUNDS WHICH ARE DIRECTLY INVOLVED IN PLANT DIFFENCE MECHANISM SO HERE I ADDED SOME SLIDES WITH KNOWLEDGABLE INFORMATION AND CITED SOME CLEAR CUT EXAMPLES.
Its about how fruit ripening occurs and how we can manipulate ripening process by using biotechnology to delay ripening and to reduce postharvest losses
LINKAGE AND CROSSING-OVER SMG
A brief description of Linkage - Bateson and Punnett's Experiment on Sweet pea, Lathyrus odoratus, Coupling and Repulsion Theory, Complete and Incomplete Linkage, Significance of Linkage, Crossing-over: Cytological basis, Types, Factors influencing the frequency , Significance, Mitotic crossing-over
Insect-resistant transgenic crops were first commercialized in the mid-1990s with the introduction of GM corn (maize), potato and cotton plants expressing genes encoding the entomocidal δ-endotoxin from Bacillus thuringiensis (Bt; also known as Cry proteins). In 2010, 148 million ha of biotech crops were grown in 29 countries, representing 10% of all 1.5 billion hectares of cropland in the world. The global value of this seed alone was valued at US $11.2 billion in 2010, with commercial biotech maize, soybean grain and cotton valued at approximately US $150 billion per year. In recent years, it has become evident that Bt-expressing crops have made a significant beneficial impact on global agriculture, not least in terms of pest reduction and improved quality. However, because of the potential for pest populations to evolve resistance, and owing to lack of effective control of homopteran pests, alternative strategies are being developed. Some of these are based on Bacillus spp., e.g. vegetative insecticidal proteins (VIPs) or other insect pathogens.
Role of secondary metabolites in insect pest managementMohd Irshad
SECONDARY METABOLITES ARE THOSE COMPOUNDS WHICH ARE DIRECTLY INVOLVED IN PLANT DIFFENCE MECHANISM SO HERE I ADDED SOME SLIDES WITH KNOWLEDGABLE INFORMATION AND CITED SOME CLEAR CUT EXAMPLES.
Its about how fruit ripening occurs and how we can manipulate ripening process by using biotechnology to delay ripening and to reduce postharvest losses
LINKAGE AND CROSSING-OVER SMG
A brief description of Linkage - Bateson and Punnett's Experiment on Sweet pea, Lathyrus odoratus, Coupling and Repulsion Theory, Complete and Incomplete Linkage, Significance of Linkage, Crossing-over: Cytological basis, Types, Factors influencing the frequency , Significance, Mitotic crossing-over
Insect-resistant transgenic crops were first commercialized in the mid-1990s with the introduction of GM corn (maize), potato and cotton plants expressing genes encoding the entomocidal δ-endotoxin from Bacillus thuringiensis (Bt; also known as Cry proteins). In 2010, 148 million ha of biotech crops were grown in 29 countries, representing 10% of all 1.5 billion hectares of cropland in the world. The global value of this seed alone was valued at US $11.2 billion in 2010, with commercial biotech maize, soybean grain and cotton valued at approximately US $150 billion per year. In recent years, it has become evident that Bt-expressing crops have made a significant beneficial impact on global agriculture, not least in terms of pest reduction and improved quality. However, because of the potential for pest populations to evolve resistance, and owing to lack of effective control of homopteran pests, alternative strategies are being developed. Some of these are based on Bacillus spp., e.g. vegetative insecticidal proteins (VIPs) or other insect pathogens.
Tutorial and investigation for DrosophiLab, a free, downloadable fruit fly editor and genetics simulator. Written to satisfy the ICT component for simulations in the IB Biology course, and differentiated to suit standard level (monohybrid and codominant crosses) and higher level students (dihybrid and linked genes).
For my IB Biology Class - 5.3 Populations. To be used in conjunction with the Excel StatBook. (http://i-biology.net/ia/statexcel/).
A series of hemacytometer slides to allow students to practice counting populations and plotting their growth.
Figure 3- Ventral view of the female and male abdomen Virgin females-.docxSUKHI5
Figure 3: Ventral view of the female and male abdomen Virgin females: Difference between a virgin and a mated female fruit fly. Virgin (left): pale in colour, black spot in abdomen (meconium) and often bloated. Mated (right): more pigmented, no meconium. (Source: Twitter) Crosses: For this lab we are observing and crossing the F 1 offspring. The parents and some individuals from the offspring will show the phenotypic traits for the mutations, some will show the wildtype phenotype. The parental cross and F 1 are showed below: Dihybrid cross 1: P l = apterous (vestigial wings showed) x sepia Dihybrid cross 2 : P l = wildtype x white Cross 1: sepia male x apterous virgin female 9 Cross 2: white virgin female 9 x wildtype male So, what is your observed ratio? To determine the proportion each class represents of the total number of offspring, divide \# observed/total offspring: Proportion of wildtype 171/220 = 0.78 , which can also be shown as 78% (multiply by 100 ) Proportion of apterous 49/220 = 0.22 , which can also be shown as 22% To calculate the observed ratio in the offspring among the phenotypic classes: 1. Write the classes with the corresponding numbers as a ratio: 171 wildtype: 49 apterous. 2. Simplify the ratio by dividing each number by the smallest number in the ratio. So, 171/49 = 3.5 and 49/49 = 1 . So, the simplified ratio is 3.5 wildtype: 1 apterous. This is obviously close to a 3:1 ratio, but is it close enough to support your hypothesis? This is the point at which you would do your Chi-Square calculations. Review questions (5 pts) 1) What is the Parental cross genotype and phenotype? What would you expect for the genotype and phenotype about the F 1 and F 2 ? 2) What are the expected rations for a monohybrid and for a dihybrid cross?
.
Problems2. This is a map for a diploid plantR--------35-------.pdfFootageetoffe16
Problems:
2. This is a map for a diploid plant:
R--------35--------E--------17---------F
One parental plant is R e f/r E F. It is mated with a homozygous recessive plant: r e f/r e f. Give
the proportion of the offspring assuming there is both no interference, and complete interference.
3. In a newly discovered insect, the three genes R, A, and P located on the same chromosome
(R= red body, A= spotted coloring, P=purple eyes, r=white body, a=solid coloring, p=white
eyes). A heterozygous parent(R A P/r a p) is mated with a homozygous recessive parent (r a p/ r
a p).
The following offspring are shown below:
(210) dominant for all traits
(16) red body, solid coloring, purple eyes
(70) red body, spotted coloring, white eyes
(50) red body, solid coloring, white eyes
(200) recessive for all traits
(55) white body, spotted coloring, purple eyes
(13) white body, spotted coloring, white eyes
(72) white body, solid coloring, purple eyes
Perform a three-point linkage to determine the gene order and relative distance between the
genes. How did the distance between the 2 genes affect the frequency of crossing-over events?
4. Here are four recombinant sequences that came from four Hfr matings during an interrupted
mating experiment. Determine the order and relative distance by setting A to 0/100 minutes.
Hfr1) A T1 H T2
8 26 33 53
Hfr2) E1 T1 E2 E3
4 12 24 54
Hfr3) T2 S T3 A
7 37 52 62
Hfr4) T3 C E1 E3
2 17 22 72
5. In a u-tube experiment, a researcher placed a kanamycin resistant strain of E. coli on one side
of a membrane impervious to the cells, and a wild-type strain on the other side. After a day, he
then plated a sample from both sides of the tube on plates containing kanamycin and observed
growth on both plates. He then repeated this experiment, but this time he added DNase to the
tube and observed the same results when he plated both cell types. What happened inside the
tube?
6. Pretend the following is a chromosome that underwent some changes. What were they? (the
dot is the centromere).
A B C D E F . G H I J A B C G . F E H I J
a. pericentric inversion, deletion
b. paracentric inversion, deletion
c. deletion, translocation
d. Robertsonian translocation,
pericntric inversion
7. Given:
Type of mutation Cistron A or B
1. point A
2. point A
3. point B
4. point B
5. deletion A
Results from 5 crosses:
1x2 (plaque) 2x3 (plaque) 3x4 (plaque) 4x5 (plaque)
1x3 (plaque) 2x4 (plaque) 3x5 (plaque)
1x4 (plaque) 2x5 (no plaque)
1x5 (plaque)
8. The following nucleotide sequence is found in a piece of DNA
5’-TTAACA-3’
3’-AATTGT-5’
if the cytosince undergoes a tautameric shift for one replication cycle draw two round of
replication and explain what type of mutation occurred (Transition or Transversion)?
Bonus:
Explain the relationship between mosaicism and x-inactivation using an example.
9. In a diploid fly, the three loci R E and D are linked as follows
R--------------22cm------------E----------16cm-------------D
One fly is available to you and its genotype is R.
BIOL 105Experiment to determine the Drosophila genome .docxAASTHA76
BIOL 105
Experiment to determine the Drosophila genome and Inheritance of both the red-eyed and white-eyed traits (word count 1000)
Name: Mai Alqattan
Instructor: Timothy McClure
Experiment to determine the Drosophila genome and Inheritance of both the red-eyed and white-eyed traits
Mai Alqattan
Abstract Comment by Tim I McClure: The abstract is supposed to have portions of each aspect of the text. You basically have no introduction or discussion, and your results are interpreted incorrectly Also, you need to clearly say what you mean – for example instead of “the values obtained…” you should just tell us that those are the p-values. A lot of this is fluff that isn’t actually telling the reader anything new.
This experiment was aimed at demonstrating that Red and white eye colors areeye color in Drosophila melanogaster is s sex linked trait cs hence carried onin the female X-chromosome. The results matched both the expected, observed and predicted values. The experiment involved a cross-link carried between a red-eyed and white fruit fly. The values obtained were less than 1%. This means that there was no significance difference in the transmission of the red-eyed and white-eyed characteristics in the fly Drosophila melanogaster. The characteristics shown in both the F1 and F2 was due to cross link and inheritance of the traits from the parental forms. Comment by Tim I McClure: We’re going to need a bit more background in the intro portion of the abstract. For example, why do we care about model organisms? Comment by Tim I McClure: How did we get those expected ratios? Why do we care? The point of this experiment was to test whether eye color was a sex-linked trait or not, so our hypothesis was that it was sex linked, and we figured out the expected values from sex-linked heritability. So, our null hypothesis (stasticical hypothesis) was that there would be no difference between the observed and expected (from sex-linkage) phenotypic ratios. The biological interpretation of rejecting the null would be support for our hypothesis that eye color is sex-linked. Comment by Tim I McClure: What is this? Comment by Tim I McClure: What values are you talking about here? Comment by Tim I McClure: Okay, you’re talking about your p-values? If the p-values we obtained were less than 0.01 (1%), we would definitely reject our null hypothesis. This is unclear, and wrong even if it were.
You will need to revisit your hypotheses in order to correctly interpret the statistical results. You also have no discussion here.
Introduction
Drosophila melanogaster is a three millimetersmall species of fruit fly which is commonly found in ripe fruits. It is mostly applied in research particular developmental biology and genetics. It is also applied in reproduction in studying the life cycle, embryonic development, understanding the complexity of living organisms and development of organs in complex organism. This is because it has ...
2024.06.01 Introducing a competency framework for languag learning materials ...Sandy Millin
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
For more information, visit-www.vavaclasses.com
Instructions for Submissions thorugh G- Classroom.pptxJheel Barad
This presentation provides a briefing on how to upload submissions and documents in Google Classroom. It was prepared as part of an orientation for new Sainik School in-service teacher trainees. As a training officer, my goal is to ensure that you are comfortable and proficient with this essential tool for managing assignments and fostering student engagement.
The Roman Empire A Historical Colossus.pdfkaushalkr1407
The Roman Empire, a vast and enduring power, stands as one of history's most remarkable civilizations, leaving an indelible imprint on the world. It emerged from the Roman Republic, transitioning into an imperial powerhouse under the leadership of Augustus Caesar in 27 BCE. This transformation marked the beginning of an era defined by unprecedented territorial expansion, architectural marvels, and profound cultural influence.
The empire's roots lie in the city of Rome, founded, according to legend, by Romulus in 753 BCE. Over centuries, Rome evolved from a small settlement to a formidable republic, characterized by a complex political system with elected officials and checks on power. However, internal strife, class conflicts, and military ambitions paved the way for the end of the Republic. Julius Caesar’s dictatorship and subsequent assassination in 44 BCE created a power vacuum, leading to a civil war. Octavian, later Augustus, emerged victorious, heralding the Roman Empire’s birth.
Under Augustus, the empire experienced the Pax Romana, a 200-year period of relative peace and stability. Augustus reformed the military, established efficient administrative systems, and initiated grand construction projects. The empire's borders expanded, encompassing territories from Britain to Egypt and from Spain to the Euphrates. Roman legions, renowned for their discipline and engineering prowess, secured and maintained these vast territories, building roads, fortifications, and cities that facilitated control and integration.
The Roman Empire’s society was hierarchical, with a rigid class system. At the top were the patricians, wealthy elites who held significant political power. Below them were the plebeians, free citizens with limited political influence, and the vast numbers of slaves who formed the backbone of the economy. The family unit was central, governed by the paterfamilias, the male head who held absolute authority.
Culturally, the Romans were eclectic, absorbing and adapting elements from the civilizations they encountered, particularly the Greeks. Roman art, literature, and philosophy reflected this synthesis, creating a rich cultural tapestry. Latin, the Roman language, became the lingua franca of the Western world, influencing numerous modern languages.
Roman architecture and engineering achievements were monumental. They perfected the arch, vault, and dome, constructing enduring structures like the Colosseum, Pantheon, and aqueducts. These engineering marvels not only showcased Roman ingenuity but also served practical purposes, from public entertainment to water supply.
Unit 8 - Information and Communication Technology (Paper I).pdfThiyagu K
This slides describes the basic concepts of ICT, basics of Email, Emerging Technology and Digital Initiatives in Education. This presentations aligns with the UGC Paper I syllabus.
Biological screening of herbal drugs: Introduction and Need for
Phyto-Pharmacological Screening, New Strategies for evaluating
Natural Products, In vitro evaluation techniques for Antioxidants, Antimicrobial and Anticancer drugs. In vivo evaluation techniques
for Anti-inflammatory, Antiulcer, Anticancer, Wound healing, Antidiabetic, Hepatoprotective, Cardio protective, Diuretics and
Antifertility, Toxicity studies as per OECD guidelines
1. Hyohyun Lee
IB Biology HL: Period 5
April 7th, 2010
1
Virtual Fly Lab
Introduction
Gregor Johann Mendel’s experiments with garden peas dramatically influenced the field of
biology. Mendel’s results became the foundation for the discipline known as genetics, which is the
study of variation and inheritance. Mendel’s experiments were done by using varieties of pea plant.
He crossed those varieties of pea together by transferring the male pollen from one variety to the
female parts in flowers of another variety artificially. He considered seven different characters
including flower color, flower position, seed color, seed shape, pod color, pod shape, and stem length.
Variations of a given character are known as traits. Then, he collected the seeds and grew them to
find out what their characteristics looked like. Later his results were surprising; characteristics from
the original parents had disappeared or reappeared in their offspring. During his experiments, he
discovered that there were particles of inheritance factors, which are now called genes. Specifically,
genes are heritable factors that control a specific characteristic. Mendel also thought that were
alternative forms of a gene which were responsible for variations in inherited characters. Those
forms of a gene are now called alleles. After he got surprising results, he observed that the ratio of
pea plants was always close to 3:1, dominant alleles : recessive alleles. For instance, in the pea plants
that Mendel used have two alleles for stem length, a tall stem and a dwarf stem. With true-breeding
homozygous parents called the P generation, he first crossed a tall plant with a dwarf plant. The first
offspring, called the F1 generation, were all tall. This is because a tall stem is a dominant character in
pea plant and a dwarf stem is a recessive character. In F2 generation from the self-pollination of F1
plants, the phenotype ratio of tall plants to dwarf plants was approximately 3:1. Here, Mendel
discovered that this ratio was made from a cross between heterozygotes. He also used dihybrid
crosses in order to explain his law of independent assortment, which is that alleles for different
characters segregate into each gamete independently. In a dihybrid cross, the phenotype ratio of
heterozygous F1 cross is approximately 9:3:3:1 in F2 generation. Furthermore, Punnett square is
useful to predict the genotypes and phenotypes from a genetic cross.
Chi-square analysis is a statistical test that makes a comparison between the data collected in
an experiment and expected data. In genetics, the Chi-square analysis used to evaluate data from
experimental crosses to determine if the assumed genetic explanation is supported by the data.
The person who originally proposed the use of the fruit fly was Thomas Hunt Morgan while
he worked at Columbia University in the early 1900’s. For his genetics work with fruit fly, he
received the Nobel Prize in Medicine in 1933. In this virtual Fly Lab, fruit flies, also called
2. Hyohyun Lee
IB Biology HL: Period 5
April 7th, 2010
2
Drosophila melanogaster, will be used. In addition, with Punnett square, Chi-square analysis, and
other basic knowledge, it will be easy to simulate basic principles of genetic inheritance based on
Mendel’s genetics by performing crosses between fruit flies. The Chi-square analysis will be used so
that I can accept or reject my hypothesis for the expected phenotype ratio of offspring for each cross.
The tasks of this lab are to investigate traits of fruit flies and to show proof that I have found traits
which are inherited in the following five manners: a dominant allele, a recessive allele, a recessive
sex-linked allele, a dominant lethal allele, and dihybrid cross.
In this lab, there will be also used some genetic abbreviations. The normal “wild type”
version will be represented as “+”. A table of the genetic abbreviations used in this lab appears below.
Table 1) The Genetic Abbreviations
Abbreviation Phenotype Abbreviation Phenotype Abbreviation Phenotype
AP Apterous Wings EY Eyeless Eyes SE Sepia Eyes
AR Aristapedia
Antennae
DP Dumpy Wings SN Singed
Bristles
B Bar Eyes F Forked
Bristles
SS Spineless
Bristles
BL Black Body L Lobe Eyes ST Star Eyes
BW Brown Eyes M Miniature
Wings
SV Shaven
Bristles
C Curved Wings PR Purple Eyes T Tan Body
CV Crossveinless
Wings
RI Radius
Incompletus
Wings
VG Vestigial
Wings
CY Curly Wings S Sable Body W White Eyes
D Dichaete
Wings
SB Stubble
Bristles
Y Yellow Body
E Ebony Body SD Scalloped Wings
3. Hyohyun Lee
IB Biology HL: Period 5
April 7th, 2010
3
A dominant allele (Eye shape Bar Female x Wild type Male)
A dominant allele is an allele that has the same effect on the phenotype whether it is present
in the homozygous or the heterozygous state. To test for a dominant allele, the character of lobe eyes
is used because this character is assumed as a dominant allele. In this cross, one of the parents must
have identical lobe eyes allele and another must be a wild fruit fly.
F1
--------------------------------------------------------------------------
Results of Cross #27
Ignoring Sex
Parents
(Female: B) x (Male: +)
Offspring
Phenotype Number Proportion Ratio
B 1016 1.0000 1.000
Total 1016
--------------------------------------------------------------------------
Chi Square Hypothesis Using Cross #27
Ignoring Sex
Phenotype Observed Hypothesis Expected Chi-Square Term
B 1016 1.0000 1016.0000 0.0000
Total 1016 1.0000 1016.0000 0.0000
Chi-Squared Test Statistic = 0.0000
Degrees of Freedom = 0
Level of Significance = 1.0000
Recommendation: Do not reject your hypothesis
--------------------------------------------------------------------------
Explanation:
From this data, it is shown that when a fly with homozygous bar eyes mates with a fly with
wild type eyes, the resulting offspring only has bar type eyes. This evidence can suggest that bar eye
shape is the dominant allele over wild type because in order for the offspring to have only have bar
eye shape from parents that have different homozygous allele, the bar eye shape must be the
dominant allele. This is shown in the Punnett square below:
B = Bar Eyes + = Wild Type
Table 2) Punnett Square for a Cross between Bar Eyes Female and Wild Eyes Male
B B
+ B+ B+
+ B+ B+
Genotype of the offspring = B+
Phenotype of the offspring = All Bar eyes (supported by the result of cross #27)
4. Hyohyun Lee
IB Biology HL: Period 5
April 7th, 2010
4
F2
--------------------------------------------------------------------------
Results of Cross #28
Ignoring Sex
Parents
(Female: B) x (Male: B)
Offspring
Phenotype Number Proportion Ratio
+ 251 0.2505 1.000
B 751 0.7495 2.992
Total 1002
--------------------------------------------------------------------------
Chi Square Hypothesis Using Cross #28
Ignoring Sex
Phenotype Observed Hypothesis Expected Chi-Square Term
+ 251 1.0000 250.5000 0.0010
B 751 3.0000 751.5000 0.0003
Total 1002 4.0000 1002.0000 0.0013
Chi-Squared Test Statistic = 0.0013
Degrees of Freedom = 1
Level of Significance = 0.9709
Recommendation: Do not reject your hypothesis
----------------------------------------------------------------------------------------------------------------
Explanation:
From Cross #28, it further supports that bar eye is the dominant allele for eye shape. As
suggested from the previous Punnett Square, the F1 generation only has a bar eye shape phenotype
with genotype of B+. When these heterozygotes cross together, they are expected to produce F2
generation with a phenotype ratio, 3:1 for bar eye shape to wild eye shape, as shown in the Chi-
Square #28 above. According the Chi-squared test, the result is not statistically significant because
the result is nearly accurate compared to the expected data. This can only suggest that bar eye shape
is the dominant allele and this is shown in the Punnett square below:
B = Bar Eyes + = Wild Type
Table 3) Punnett Square for a Cross between Heterozygous Bar Eye Parents
B +
B BB B+
+ B+ ++
Genotype ratio of the offspring – BB : B+ : ++ = 1:2:1
Phenotype ratio of the offspring – Bar Eyes : Wild Eyes = 3:1
5. Hyohyun Lee
IB Biology HL: Period 5
April 7th, 2010
5
A recessive allele (Wing shape Dumpy Female x Wild type Male)
A recessive allele is an allele that only has an effect on the phenotype when it is present in
the homozygous state. In this case, dumpy wing shape is assumed as a recessive allele. Therefore, if
dumpy wing shape female and wild type male cross together, they will breed offspring that has all
phenotype with wild wing shape.
F1
--------------------------------------------------------------------------
Results of Cross #3
Ignoring Sex
Parents
(Female: DP) x (Male: +)
Offspring
Phenotype Number Proportion Ratio
+ 1004 1.0000 1.000
Total 1004
--------------------------------------------------------------------------
Chi Square Hypothesis Using Cross #3
Ignoring Sex
Phenotype Observed Hypothesis Expected Chi-Square Term
+ 1004 1.0000 1004.0000 0.0000
Total 1004 1.0000 1004.0000 0.0000
Chi-Squared Test Statistic = 0.0000
Degrees of Freedom = 0
Level of Significance = 1.0000
Recommendation: Do not reject your hypothesis
--------------------------------------------------------------------------
Explanation:
From this data, it is shown that when a fly with homozygous dumpy wings mates with a fly
with wild type wings, the resulting offspring only has wild type wings. This evidence can suggest
that dumpy wing shape is the recessive allele over wild type wing because in order for the offspring
to have only wild wing shape from parents that have different homozygous allele, the dumpy wing
shape must be the recessive allele. This is shown in the Punnett square below:
DP = Dumpy Wings + = Wild Type
Table 4) Punnett Square for a Cross between Dumpy Wings Female and Wild Wings Male
DP DP
+ DP+ DP+
+ DP+ DP+
Genotype of the offspring = DP+
Phenotype of the offspring = All Wild Wings (supported by the result of cross #3)
6. Hyohyun Lee
IB Biology HL: Period 5
April 7th, 2010
6
F2
--------------------------------------------------------------------------
Results of Cross #4
Ignoring Sex
Parents
(Female: +) x (Male: +)
Offspring
Phenotype Number Proportion Ratio
+ 741 0.7647 3.250
DP 228 0.2353 1.000
Total 969
--------------------------------------------------------------------------
Chi Square Hypothesis Using Cross #4
Ignoring Sex
Phenotype Observed Hypothesis Expected Chi-Square Term
+ 741 3.0000 726.7500 0.2794
DP 228 1.0000 242.2500 0.8382
Total 969 4.0000 969.0000 1.1176
Chi-Squared Test Statistic = 1.1176
Degrees of Freedom = 1
Level of Significance = 0.2904
Recommendations: Do not reject your hypothesis
--------------------------------------------------------------------------
Explanation:
From Cross #4, it further supports that dumpy wing is the recessive allele for wing shape. As
suggested from the previous Punnett Square, the F1 generation only has wild wing shape phenotype
with genotype of DP+. When these heterozygotes cross together, they are expected to produce F2
generation with a phenotype ratio, 3:1 for wild wing shape to dumpy wing shape, as shown in the
Chi-Square #4 above. According the Chi-squared test, the result is statistically significant because
the result is not that accurate compared to the expected data and the hypothesized rate. This can only
suggest that dumpy wing shape is the recessive allele over wild wing shape and this is shown in the
Punnett square below:
DP = Dumpy Wings + = Wild Type
Table 5) Punnett Square for a Cross between Heterozygous Wild Wing Parents
DP +
DP DPDP DP+
+ DP+ ++
Genotype ratio of the offspring – DPDP : DP+ : ++ = 1:2:1
Phenotype ratio of the offspring – Wild Wings : Dumpy Wings = 3:1
7. Hyohyun Lee
IB Biology HL: Period 5
April 7th, 2010
7
A recessive sex-linkedallele (Wild type Female x Body color Sable Male)
A recessive sex-linked allele is an allele which appears in sex chromosome, X chromosome.
Crossing a wild type female with a sable color male is the first step to find a recessive sex-liked
allele.
F1
--------------------------------------------------------------------------
Results of Cross #5
Parents
(Female: +) x (Male: S)
Offspring
Phenotype Number Proportion Ratio
Female: + 488 0.4905 1.000
Male: + 507 0.5095 1.039
Total 995
--------------------------------------------------------------------------
Chi Square Hypothesis Using Cross #5
Phenotype Observed Hypothesis Expected Chi-Square Term
Female: + 488 1.0000 497.5000 0.1814
Male: + 507 1.0000 497.5000 0.1814
Total 995 2.0000 995.0000 0.3628
Chi-Squared Test Statistic = 0.3628
Degrees of Freedom = 1
Level of Significance = 0.5469
Recommendation: Do not reject your hypothesis
--------------------------------------------------------------------------
Explanation:
From this data, it is shown that when a fly with normal color crosses with a fly having
homozygous sable color, the resulting offspring only has wild body color. This evidence can suggest
that sable body color is included in the recessive sex-linked allele over the wild body color because
in order for the offspring to have only wild body color from parents that have different homozygous
allele, the sable body color must be the recessive sex-linked allele. This is shown in the Punnett
square below:
S = Sable Body + = Wild Body
Table 6) Punnett Square for a Cross between Wild Body Color Female and Sable Color Male
XS Y
X+ XSX+ X+Y
X+ XSX+ X+Y
Genotype of the offspring = XSX+, X+Y
Phenotype of the offspring = All Wild Body Color (supported by the result of cross #5)
8. Hyohyun Lee
IB Biology HL: Period 5
April 7th, 2010
8
F2
--------------------------------------------------------------------------
Results of Cross #6
Parents
(Female: +) x (Male: +)
Offspring
Phenotype Number Proportion Ratio
Female: + 509 0.4947 2.004
Male: + 254 0.2468 1.000
Male: S 266 0.2585 1.047
Total 1029
--------------------------------------------------------------------------
Chi Square Hypothesis Using Cross #6
Phenotype Observed Hypothesis Expected Chi-Square Term
Female: + 509 2.0000 514.5000 0.0588
Male: + 254 1.0000 257.2500 0.0411
Male: S 266 1.0000 257.2500 0.2976
Total 1029 4.0000 1029.0000 0.3975
Chi-Squared Test Statistic = 0.3975
Degrees of Freedom = 2
Level of Significance = 0.8198
Recommendation: Do not reject your hypothesis
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Chi Square Hypothesis Using Cross #6
Ignoring Sex
Phenotype Observed Hypothesis Expected Chi-Square Term
+ 763 3.0000 771.7500 0.0992
S 266 1.0000 257.2500 0.2976
Total 1029 4.0000 1029.0000 0.3968
Chi-Squared Test Statistic = 0.3968
Degrees of Freedom = 1
Level of Significance = 0.5287
Recommendation: Do not reject your hypothesis
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Explanation:
From Cross #6, it further supports that sable body color is the recessive sex-linked allele for
body color. As suggested from the previous Punnett Square, the F1 generation only has wild body
color phenotype with genotype of S+. When these heterozygotes cross together, they are expected to
produce F2 generation with a phenotype ratio, 3:1 for wild body color to sable body color, as shown
in the Chi-Square #6 above. According the Chi-squared test, the result is not statistically significant
because the result is nearly accurate compared to the expected data and the hypothesized rate. This
can suggest that sable body color is the recessive sex-linked allele over wild body color and this is
shown in the Punnett square in next page:
9. Hyohyun Lee
IB Biology HL: Period 5
April 7th, 2010
9
S = Sable Body + = Wild Body
Table 7) Punnett Square for a Cross between Heterozygous Wild Body Color Parents
X+ Y
XS XSX+ XSY
X+ X+X+ X+Y
Female genotype ratio of the offspring – XSX+ : X+X+ = 1:1
Female phenotype ratio of the offspring – All Wild Body Color
Male genotype ratio of the offspring – XSY : X+Y = 1:1
Male phenotype ratio of the offspring – Sable Body Color : Wild Body Color = 1:1
A dominant lethal allele (Wing angle Dichaete Female x Wing angle Dichaete Male)
A lethal allele is an allele that can cause death if organisms have homozygous dominant
allele. For fruit flies, dichaete wing angle is one of the lethal alleles. The reason is that when two
fruit flies with the same trait, dichaete wing angle, are mated, the ratio is 2:1, dichaete wing angle :
wild wing angle. Here, the one that survives and that will be used for mating has to be carrier
because homozygous dichaete wing angle fly cannot live anymore.
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Results of Cross #13
Ignoring Sex
Parents
(Female: D) x (Male: D)
Offspring
Phenotype Number Proportion Ratio
+ 350 0.3418 1.000
D 674 0.6582 1.926
Total 1024
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Chi Square Hypothesis Using Cross #13
Ignoring Sex
Phenotype Observed Hypothesis Expected Chi-Square Term
+ 350 1.0000 341.3333 0.2201
D 674 2.0000 682.6667 0.1100
Total 1024 3.0000 1024.0000 0.3301
Chi-Squared Test Statistic = 0.3301
Degrees of Freedom = 1
Level of Significance = 0.5656
Recommendation: Do not reject your hypothesis
--------------------------------------------------------------------------
10. Hyohyun Lee
IB Biology HL: Period 5
April 7th, 2010
10
Explanation:
From this data, it is shown that when two flies with the same trait, dichaete wing angle, the
resulting offspring only has wild wing angle and dichaete wing angle. However, we can know that by
drawing a Punnett square, the homozygous dichaete wing angle flies are dead because the trait is a
lethal allele. This evidence can suggest that dichaete wing angle is included in the lethal allele
because in order for the offspring to have the ratio of 2:1, dichaete wing angle : wild wing angle, the
dichaete wing angle must be the lethal allele. This is shown in the Punnett square below:
D = Dichaete Wings + = Wild Wings
Table 8) Punnett Square for a Cross between two dichaete wing angle parents
D +
D DD (died) D+
+ D+ ++
Genotype of the offspring = DD (died), D+, ++
Phenotype of the offspring = Dichaete Wing, Wild Wing (supported by the result of cross #13)
Genotype ratio of the offspring – DD (died) : D+ : ++ = 1:2:1
Phenotype ratio of the offspring – Dichaete Wing : Wild Wing = 2:1 (because one dichaete wing fly
is dead)
11. Hyohyun Lee
IB Biology HL: Period 5
April 7th, 2010
11
Dihybrid Cross (Wild type Female x Wing size Apterous & Eye color Sepia Male)
The 9:3:3:1 ratio is often found when parents that are heterozygous for two genes are
crossed together. The dihybrid cross follows Mendel’s Law of Independent Assortment because the
genes are unlinked.
F1
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Results of Cross #7
Ignoring Sex
Parents
(Female: +) x (Male: SE;VG)
Offspring
Phenotype Number Proportion Ratio
+ 1011 1.0000 1.000
Total 1011
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Chi Square Hypothesis Using Cross #7
Ignoring Sex
Phenotype Observed Hypothesis Expected Chi-Square Term
+ 1011 1.0000 1011.0000 0.0000
Total 1011 1.0000 1011.0000 0.0000
Chi-Squared Test Statistic = 0.0000
Degrees of Freedom = 0
Level of Significance = 1.0000
Recommendation: Do not reject your hypothesis
--------------------------------------------------------------------------
Explanation:
From this data, it is shown that when a fly with wild type crosses with a fly having sepia eye
and vestigial wings, the resulting offspring only has wild type. This evidence can suggest that this
cross is dihybrid cross because in order for the all offspring to have only wild type from parents that
have different characteristics, the cross between the wild type and sepia eye and apterous wing must
be the dihybrid cross. This is shown in the Punnett square below:
SE = Sepia Eyes + = Wild Type VG = Vestigial Wings
Table 9) Punnett Square for a Cross between a wild type female and a vestigial wing / sepia
eye male
SE VG
+ SE+ VG+
+ SE+ VG+
Genotype of the offspring = SE+, VG+
Phenotype of the offspring = All Wild Eye Color / Wild Wing (supported by the result of cross #7)
12. Hyohyun Lee
IB Biology HL: Period 5
April 7th, 2010
12
F2
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Results of Cross #8
Ignoring Sex
Parents
(Female: +) x (Male: +)
Offspring
Phenotype Number Proportion Ratio
+ 561 0.5571 9.508
SE 195 0.1936 3.305
VG 192 0.1907 3.254
SE;VG 59 0.0586 1.000
Total 1007
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Chi Square Hypothesis Using Cross #8
Ignoring Sex
Phenotype Observed Hypothesis Expected Chi-Square Term
+ 561 9.0000 566.4375 0.0522
SE 195 3.0000 188.8125 0.2028
VG 192 3.0000 188.8125 0.0538
SE;VG 59 1.0000 62.9375 0.2463
Total 1007 16.0000 1007.0000 0.5551
Chi-Squared Test Statistic = 0.5551
Degrees of Freedom = 3
Level of Significance = 0.9066
Recommendation: Do not reject your hypothesis
--------------------------------------------------------------------------
Explanation:
From Cross #8, it further supports that a cross between a wild type and sepia eye / apterous
wing is a dihybrid cross. As suggested in previous Punnett square, the F1 generation only has wild
heterozygous type. When these heterozygotes cross together, they are expected to produce F2
generation with a phenotype ratio, 9:3:3:1 for wild body color to sepia eye color to vestigial wing
shape to mixture of sepia eye and vestigial wing, as shown in the Chi-Square #8 above. According
the Chi-squared test, the result is not statistically significant because the result is nearly accurate
compared to the expected data and the hypothesized rate. This can suggest that this cross is a
dihybrid cross and this is shown in the Punnett square in next page:
13. Hyohyun Lee
IB Biology HL: Period 5
April 7th, 2010
13
SE = Sepia Eyes + = Wild Type VG = Vestigial Wings
Table 10) Punnett Square for a Cross between two heterozygous flies
++ VG+ SE+ VGSE
++
VG+
++ ++ VG+ ++ SE+ ++ VGSE ++
VG+ ++ VGVG ++ VGSE ++ VGVG SE+
SE+
VGSE
SE+ ++ VGSE ++ SESE ++ VG+ SESE
VGSE ++ VGVG SE+ VG+ SESE VGVG SESE
Genotype ratio of the offspring – ++ ++ : VG+ ++ : SE+ ++ : VGSE ++ : VGVG ++ : VGVG SE+ :
SESE ++ : VG+ SESE : VGVG SESE = 1:2:2:4:1:2:1:2:1
Phenotype ratio of the offspring – 9 Wild type : 3 vestigial-winged : 3 sepia-eyed : 1 vestigial-
winged sepia eyed