Lab 3 Report Checklist – Inheritance Patterns and Linkage in Drosophila
Title Page (1 mark)
-
descriptive title
name
lab section
due date
Abstract (5 marks) – no references required
-
maximum of 250 words
purpose of the experiment – why is what you studied important?
short summary of main methodology
most important findings/results
main conclusions
Introduction (20 marks) – references required
-
maximum of 750 words
introductory paragraph providing support for why the experiment was
performed (what research question(s) are you answering?)
explanation of Drosophila as a good model organism in context to this
experiment
theory of alleles and dominant/recessive expression
expected distribution of genotypes & phenotypes when observing autosomal
alleles
expected distribution of genotypes & phenotypes when observing sex-linked
alleles (role of paternal origin of alleles)
theory of linkage and how this alters expected genotypic and phenotypic
ratios
concluding paragraph which briefly outlines methods and their relation to
your study objectives, and states your hypothesis (what are you expecting
the outcome(s) of the experiment to be?)
Methods (7 marks) – must reference lab manual
-
-
outline the following methods: preparation of vials for breeding, setting up of
crosses for observation, procedures for removing parents, counting and
recording phenotypes of progeny flies, setting up testcrosses for linkage
experiment, disposing of flies, how data was analyzed for each part of the
experiment
include specific details regarding number of flies used, incubation settings
(time/temperature), number of trials, breeding time periods, etc.
Results (20 marks) – no references required
-
-
-
descriptive text outlining the suggested inheritance patterns for each
mutation (sepia, white, and ebony) as indicated by total cross counts in the
F2 generations
tables displaying total cross counts and expected numbers for the F2
generations of each mutation. Include expected numbers for each
inheritance pattern
descriptive text outlining the results of chi-square analyses for F2 crosses of
each mutation
tables displaying the chi-square analysis for each cross analyzed
descriptive text outlining the determination of expected results for
determining linkage
table displaying the total counts of the linkage experimen
contingency table for determining expected values
text describing the results of the chi-square test for independence (including
hypothesis, degrees of freedom, chi-square value, p-value range, and
acceptance or rejection)
Discussion (25 marks) – references required
-
maximum of 750 words
explanation of results for each of the 3 mutations – were the results
statistically significant? (as indicated by the chi-square analysis)
support findings for mutation inheritance patterns with literature references
explain linkage results (is there or isn’t there linkage)
explain why your results do or do not align with the expected results based
on the known location of the genes (include a reference to literature!)
why might results deviate from expected numbers, even when statistically
significant?
what errors may have occurred during the experiment?
how would you improve this experiment and build on your findings by
answering new research questions?
concluding paragraph which relates the results back to the original study
objectives/hypotheses and summarizes the main report findings
References (2 marks)
-
references must be numbered in order of appearance within the text of the
report
in this section, references should appear in numerical order
3 references minimum + the lab manual
BIOL2301L Report Requirements for Lab 3:
Inheritance Patterns and Linkage in Drosophila
Submission: All reports are due by 11:59 PM on the published date to the lab Nexus site in
the appropriate Assignments folder. Assignments should be submitted as a pdf document.
A late penalty of 5% / day will apply to assignments received after the deadline. NO
assignments will be accepted after one week past the due date.
Guidelines:
The report assignment in this lab is designed to allow you to gain experiencing in
preparing scientific articles for publication. As the ability to effectively communicate in a
written format is essential to various fields, the development of these skills will be invaluable
to you regardless of your final career goals. For this assignment, you are expected to
demonstrate your understanding of the technical aspects of these experiments, as well as
thetheory associated with the processes studied. It is strongly advised that you start working
on your report as early as possible. Excerpts from Writing Papers in the Biological Sciences
by
V. E. McMillan is provided on Nexus for you, and contains useful suggestions for how to
approach scientific writing. If you have any questions about the expectations for your report,
please contact your instructor. REMEMBER that ALL parts of a lab report should be written in
a formal style using the third person. Your report should be left aligned and include
indentations. Text should be double- or one-and-a-half-spaced, except for figure and table
captions, which should be single-spaced. Where a word range is indicated below, marks will
be deducted based upon the amount of words over that maximum. Inappropriate or
insufficient referencing will result in the deduction of marks and/or referral for Departmental
Review in the case of serious infractions. The specific sections found within a research
articleare as follows:
Total Marks (80 marks)
Title Page (1 mark)
Your title should be on a separate page at the front of your assignment. Please
includethe title, your name, lab section, and the due date of the assignment. The title of your
report needs to be complete and provide a good summary of what was investigated.
“Drosophila Lab” or “Inheritance and Linkage” are NOT adequate titles.
Abstract (Max 250 words; 5 marks)
This is a summary of your work and includes a brief summation of your ENTIRE
reportand should include the following:
- introductory sentence that highlights the purpose/importance of your work (1 mark)
- briefly describe the methods employed (1 mark)
- be sure to include specific, key results/observations (2 marks)
- provide your overall conclusions (1 mark)
You do not need to provide references within your abstract, as the same information will be
provided in further detail within the report.
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Introduction (Max 750 words; 20 marks)
You must summarize the relevant background information that relates to the
investigation being undertaken. Include an introductory paragraph which highlights why your
report is of interest (2 marks). You must state your general hypothesis/purpose in the
introductory paragraph (1 mark).
Explain, with references, the general theory of your topic, focusing on the specific
processes you are investigating. Depending on the focus of your work, basic information
about the techniques employed may also be included here. Remember to include
appropriatereferences to relevant literature. DO NOT summarize your findings within the
introduction. It should only include the required background information. Items to be covered
here include:
- Drosophila as a model organism: why are we using them? (2 marks)
- Concepts of alleles and dominance (2 marks)
- normal distribution of genotypes and phenotypes for autosomal alleles dependent
ondominance (4 marks)
- altered ratios observed for sex-linked genes dependent on dominance and
paternalorigin of alleles (4 marks)
- concept of linkage and its alteration of expected ratios (2 marks)
Conclude your introduction with a brief summary of the methods you will employ and
how they relate to addressing your specific hypotheses to be tested (2 marks). Do not
forgetto state what your specific hypotheses are in this paragraph (1 mark).
Deductions will be made in this section for poor grammar/spelling, organization, and
overall writing quality up to a maximum of 5 marks.
Methods (No word limit; 7 marks)
Summarize the methods used in your experiments and reference them to the lab
manual. They should be written in proper paragraphs, with each procedure clearly identified.
Never use point-form notation, bullet-point lists, or numbered lists of instructions. The use of
headings within your methods section is optional, but unnecessary. Here (and throughout
your report) do not refer to your experiments by the numbers used in the lab manual, but by
the method used/experimental questions. Methods must include all key details, such as fly
numbers for matings, time of incubation, etc. You do not need to include basic procedures,
such as how to determine the sex of Drosophila, as you can assume that your reader has a
basic understanding of laboratory procedures. These basic procedures can simply be
referenced to the lab manual and the stepwise procedure does not need to be explained.
You will receive a 1 mark deduction if the lab manual is insufficiently referenced here.
Inappropriate formatting will result in a 1 mark deduction as well. Any omissions or errors in
methods will result in 0.5 mark deductions for each instance.
Results (No word limit; 20 marks)
Your results section must be organized in a logical manner and provide context for
your results through the inclusion of written text. Your text is NOT optional. The text within
the results section is supported by your tables/figures, not the other way around. All of the
information that you include in your tables/figures must be presented within the text and
referred to by table/figure number. You should not list all of the values you obtain within your
text, but you need to clearly indicate what was measured, how you did any calculations, and
the general trends/minima/maxima observed. DO NOT discuss the accuracy or validity of
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your results or interpret their meaning in this section.
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Tables and figures should be inserted in your report immediately following the
paragraph in which they were first mentioned. Tables and figures must be numbered and
have TITLES. The title and captions must be informative, providing your reader with
information as to the general procedures used and what is depicted in that table or figure.
Appropriate units must be included, and formatting should ensure easy interpretation by your
reader. Key observations should be stated. This additional information is provided in the
caption following your title. The titles/captions for tables go ABOVE, and for figures go
BELOW.
Include the following in your results:
- tables containing the total counts and expected numbers for each of your crosses
forthe F2 generations. For these tables, be sure to include the numbers expected for
each of the different forms of inheritance. (3 marks)
-within your text state what these numbers suggest as to the inheritance pattern for
each mutation: autosomal or sex-linked, dominant or recessive? (2 marks)
-based on this determination, include a table with Chi-squared results for the F2
generations. You will want to indicate the allele, hypothesized form of inheritance,
degrees of freedom, Chi-square value, p-value range, and acceptance or rejection (3
marks)
-in your text be certain to state your hypotheses and describe your Chi-squared
analyses, including what classes are being analyzed. Be certain to state the final result
of each analysis. (3 marks)
-table of total counts for your linkage cross (1 mark)
-contingency table to determine expected numbers for your linkage test (1 mark)
-explain the associated Chi-squared test for independence you performed for linkage.
A table of this data is unnecessary, but you will want to include your hypothesis and
other information as stated above for the inheritance patterns. (2 marks)
-in addition, you will receive 1 mark for good organization, 1 mark for spelling/grammar,
and 3 marks for the overall quality of your text:
1 = poor: minimal text with lack of context provided
2 = good: writing style and quality meets expectations for a second-year class
3 = excellent: writing style and quality exceed expectations
There are no strict rules as to which data should be presented first, but keep in mind
that you must present it in a way which makes logical sense and allows you to provide
somewhat of a narrative as you progress through this section.
Discussion (Max 750 words; 25 marks)
This section should be a comprehensive analysis and discussion of the implications
ofthe experiments performed. The discussion should put your results within the context of
established scientific knowledge. This should include referenced information from the lab
manual and other published sources. Do not simply repeat your results, but discuss them
along with new information or considerations. Interpret the validity of your results, and draw
conclusions as to whether they agree or disagree with previous research. Do not forget to
discuss potential sources of error or limitations of the procedures employed. Also, make
suggestions for means of improving your work and potential future inquiries to be made.
Yourfinal 1-2 paragraphs must specifically address your hypotheses and state the overall
conclusions to be drawn from your report. Conclude with a final statement which stresses
theimportance of your work.
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You will be graded as follows:
- Explain your results for each of the three mutations. Were your results statistically
significant? Did they match with the available literature? (3 marks/mutation = 9
markstotal)
-Explain your linkage results. Were you able to determine whether linkage exists
between the two genes? Do your results make sense based on the known location of
these genes? (4 marks)
-Why would your results, even though statistically significant, deviate from the
expected numbers? What errors may exist in your experiments? (2 marks)
-Are there means by which you can improve this experiment? Are there
otherquestions you would like to investigate? (2 mark)
-Overall conclusions given in final paragraph with hypotheses addressed and strong
final statement (3 marks)
-in addition, you will receive 1 mark for good organization, 1 mark for spelling/grammar,
and 3 marks for overall writing quality:
1 = poor: explanation of reasoning is lacking, overall communication is lacking in
cohesion
2 = good: reasoning behind conclusions is presented clearly, sufficient critical
analysis for second-year level
3 = excellent: exceptional use of literature and critical analysis, clearly explained
in a cohesive and concise manner
References (2 marks)
References will be needed throughout the report whenever you present information
from an outside source (journal article, textbook, lab manual, websites, etc.). You must
include a minimum of 3 references in addition to the lab manual. Some references are
viewedas being more reputable than others. Peer-reviewed literature from scientific journals
are viewed as the most reputable, followed by edited books, etc.
You will need to use the CSE Citation-Sequence system described on pages 130-131
of Writing Papers in the Biological Sciences. We use the Number Systems Style, which
requires superscript numbers in order of appearance following your cited statement or at the
end of the sentence, with references listed by number at the end of your report. If you cite the
same source more than once, reuse the previously assigned number. However, when citing
from books and the lab manual, you must include a page number or range, with each having
its own number. For examples of proper citation using this system, please refer to: https://
www.scientificstyleandformat.org/Tools/SSF-Citation-Quick-Guide.html
For each reference under the minimum four (lab manual plus three others), you will be
deducted 0.5 marks. Improper formatting will result in a deduction of 1 mark. In the case of
minor infractions, deductions for a lack of or improper referencing will be applied in the
section with the error. Extensive lack of appropriate referencing will result in your report
beingsent to the Departmental Review Committee to determine whether Academic
Misconduct has occurred.
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MAKE SURE YOU UNDERSTAND THE FOLLOWING:
This course has a zero tolerance policy for academic misconduct / plagiarism. Failure
to include or to properly cite external references will be penalized. Ongoing failure to
reference material will be treated as academic misconduct and will be subject to
departmental review.
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LAB 3A: Determining Inheritance Patterns of Mutant Alleles in Drosophila
Objectives
1. To become familiar with the use of Drosophila melanogaster as an organism for genetic
research.
2. To be able to determine the sex and identify specific mutant phenotypes of Drosophila.
3. To be able to design and perform genetic crosses in order to determine:
the dominant or recessive nature of the mutant allele
whether the gene is located on an autosomal or sex chromosome
Drosophila melanogaster, the fruit fly, is one of the most powerful tools that geneticists
use to examine the relationship between genotype and phenotype. The species is useful
because there is a series of easily recognised mutant forms, it has a short two-week life cycle, a
high reproductive capacity and readily grows and breeds in the laboratory on a variety of foods.
It has been used as a genetic test organism since the beginning of the twentieth century when
Morgan established several genetic relationships, including sex linkage, using Drosophila.
In this experiment, you will be provided with data from a number of crosses between two
stocks of flies; wild type flies and a mutant strain with either altered eye colour (sepia or white)
or body colour (ebony). Each stock is true breeding (i.e. homozygous) for the trait you are
examining. The wild type flies are homozygous wild type for all characteristics and have redcoloured eyes, and tan coloured bodies. These phenotypes are shown in Figure 1. Once
we have carried our crosses through to the F2 generation, we will determine the phenotypic
ratios for each cross. These experimental results are then compared to the expected ratios
through the use of Chi-Squared statistical analysis (discussed previously in Lab 2).
Figure 1: Drosophila melanogaster phenotypes. Eye colour phenotypes include the wild-type red eye (a) and the
sepia (b) and white (c) mutants. The wild-type flies have tan bodies (d), which can be easily differentiated from the
ebony (e) mutants. Images obtained from https://www.biologie.uni-halle.de/entwicklungsgenetik/lehre/studenten/
drosophila/mutanten/?lang=en
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In addition to being able to distinguish the different phenotypes, it is essential that you
learn to identify male vs. female flies (see images in Figure 2). Mature flies are easily
distinguished, as females are larger in size, and contain banding along the length of the
abdomen, while males are only striped on the anterior portion. In addition, the posterior end of
the abdomen is pointed in females and rounded in males. Younger flies can be more difficult to
sex, and determining the presence or absence of the male sex combs (bristles on the lower
portion of the anterior set of legs) is often necessary.
Figure 2. Distinguishing features of female and male Drosophila. Mature flies in both (a) and (b) have visible
differences in the shape and banding of the abdomen. Females are also generally larger in size. The male sex
combs appear as dark “spots” on the anterior legs at low magnification (indicated with a red arrow in c). Images
obtained from (a) https://www.berkeley.edu/news/media/releases/2002/07/03_paras.html, (b) Pulver SR, Berni J.
The fundamentals of flying: simple and inexpensive strategies for employing Drosophila genetics in
neuroscience teaching laboratories. J Undergrad Neurosci Educ. 2012 Fall;11(1):A139-48. Epub 2012 Oct 15.
PMID: 23493248; PMCID: PMC3592735, and (c) https://researchguides.library.vanderbilt.edu/c.php?
g=156859&p=1515661
Procedures
Materials
True breeding stocks of wild type and mutant (sepia, white, and ebony) Drosophila
Immobilisation apparatus, paintbrush, file cards, dissecting microscope, jar of ethanol (fly
morgue)
Food vials, Drosophila food, and scoops.
Incubators
Step 1: Immobilisation of Drosophila
Insects become immobile but are usually not killed by short exposures to low temperatures.
The low temperature is achieved through temporarily housing the flies in ice-filled cooler
boxes. Additionally, an ice/water slush mixture is prepared in a small plastic container and a
petri dish is placed on top of the slush to provide a chilled viewing platform for microscope
manipulation. The immobilised flies can be sexed and phenotyped while they are immobilized
in the petri dish.
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1. Place a vial of flies horizonatlly in an ice-filled cooler box for 10 to 15 minutes. This will
slow the flies down, however some may still be mobile. Be sure to place the vials of flies
on their sides so that the immobilised flies do not become stuck to the moist food at the
bottom of the vial. Be sure to maintain the horizontal position at all times until you
have dumped the flies into the petri dish so that the flies do not get stuck in the
food.
Note: Avoid extremely long periods of cold exposure as you may kill the flies or render
them infertile.
2. Prepare an ice/water slurry in the plastic containers provided. Be sure to have enough
water to ensure complete contact between the bottom of the petri dish and the
ice/water surface.
3. When your flies are sluggish remove the plug from the vial and then pour the flies into
the petri dish bottom on the ice/water slurry. Cover them with the petri dish lid until
they have stopped moving.
4. You can put the plastic container, plus the petri dish of flies, on top on the stage of a
dissecting microscope. Remember to move the lens assembly up the column before
doing so. Manipulate the flies with a gentle sweeping motion with a small paintbrush.
Be careful - they are fragile!
5. As the ice melts, the petri dish may become unstable. Drain off some of the water and
add more ice as required.
6. When you are done viewing and/or counting your flies, place all unused flies in the Fly
Morgue (container of 90% ethanol). Transfer selected flies into the prepared food vial.
Note: The low temperature causes condensation. You may need to wipe the petri dish
dry with a Kleenex. Also take care that the flies do not stick to the petri dish bottom.
Step 2: Food Preparation
The food you will be using in this experiment is a commercial carbohydrate preparation to which
baker's yeast is added. The Drosophila feed on the dividing yeast cells. Prepare 12 food vials
using the method below.
1. Add one scoop of food flakes. Pour 16 ml of cold water into each food vial and allow the
food to absorb the water.
2. Set the vial aside and allow the food to finish thickening without disturbing it. Sprinkle a
pinch of dried yeast grains onto the surface of the food.
3. Plug the vials with the sponges immediately after the food is prepared to prevent stray
flies from entering your vials.
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Step 3: Initiation of Parental Cross
The female Drosophila stores sperm and can use the sperm from one insemination for most of
her reproductive life. In order for the sperm to be from the appropriate male fly, the females
must be less than 6 hours old to guarantee that the females have not been previously
inseminated. To ensure this is the case, the stock vials are cleared of all flies in advance of your
experiment, and only flies that emerge from the pupae in the following 6 hours are used to
initiate crosses. For each mutation, two different crosses are performed in duplicate:
Cross A: wild type female x mutant male
Cross B: mutant female x wild type male.
1. Immobilise the flies, sex them and check their phenotypes. Transfer 3 females and 3
males of the appropriate phenotypes into each vial. Label each vial with the cross it
contains (e.g., sepia female X wild type male), your name and lab section. Set up a
duplicate vial for each cross. You should have 12 vials when you are finished, each
containing six flies, three of each sex of different phenotypes.
2. Remember to leave the vials on their sides after transferring the flies so that the
immobilized flies do not become stuck in the food and die.
3. When the flies have recovered from the cold treatment, turn the vials upright and fasten
them together with an elastic band. Place the vials in the appropriate incubator. Record
the temperature of the incubator.
Step 4: Removal of Parental Flies
The parental flies must be removed from the vials before the emergence of the progeny flies to
prevent the parental male flies from inseminating newly hatching female progeny. This can be
done 7 days after the crosses were initiated since the parental females will have laid many eggs
by this time guaranteeing adequate numbers of progeny. Eggs (visible under the dissecting
microscope), larvae and perhaps pupae should be present, indicating successful matings.
1. Immobilise the parental flies and move them to a petri dish, working with each vial
individually. Determine the sex and phenotype of the parental flies and check these
against the crosses that were initiated. Note any discrepancies on the vial and in
your lab manual. Place these parental flies in the Fly Morgue. Why is it necessary to
check the phenotypes of the parental flies at this stage of the experiment?
2. Check the consistency of the food in all vials and add water or dry food as appropriate.
Return the vials to the incubator.
Step 5: Setting up F2 Crosses
Two weeks after the removal of the parental flies, numerous F1 progeny will have hatched. We
are now ready to cross these F1 flies in order to produce the F2 generation that we will use in
our analysis.
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1. Immobilise the F1 flies and quickly assess whether or not the progeny are following with
the expected pattern of inheritance.
2. For each original parental cross (2 vials), set up 2 new vials containing 3 F1 female flies
and 3 F1 male flies. Because these female flies are being mated with males from the
same vials, we do not need to clear the vials in advance of preparing our final cross.
3. Discard all extra flies in the morgue. Remove all labels from vials that you are finished
with and place these vials on the designated tray for disposal.
4. When the flies have recovered from the cold treatment, turn the vials upright and fasten
them together with an elastic band. Place the vials in the appropriate incubator. Record
the temperature of the incubator.
Step 6: Removal of F2 Parents
The parental flies from this cross are again discarded 7 days after initiation. These flies can be
immediately discarded. Check and adjust if necessary, the consistency of the food in all of your
vials.
Step 7: Analysis of F2 Generation
Normally you would immobilize your flies at this point in order to determine the sex and
phenotype of each of the flies from your crosses. As we are doing the lab online, you will be
utilizing images which have be uploaded to Nexus which represent the flies collected from
each vial (1 image = 1 vial). Figure 3 below provides examples of how these images will
appear.
Figure 3: Drosophila images for Nexus data analysis. Illustrations of female (a) and male (b) wild-type flies.
Mutant phenotypes are sepia eyes (c), white eyes (d), and ebony body (e).
Score at least 100 flies from each cross using the data provided on Nexus and the tables on the
next page. In order to ensure you are not counting flies more than once, utilize the grid and start
at the top left square and move across or down systematically until you reach your total of 100
flies. Add your phenotype counts to the appropriate class data tables (available on Nexus) and
use these for performing your Chi-square analyses for your Lab 3 Report.
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Wild Type vs. Sepia
F2 Phenotypes for Cross A: Wild Type Female X Mutant Male
Wild Type Female
Wild Type Male
Mutant Female
Mutant Male
F2 Phenotypes for Cross B: Mutant Female X Wild Type Male
Wild Type Female
Wild Type Male
Mutant Female
Mutant Male
Wild Type vs. White
F2 Phenotypes for Cross A: Wild Type Female X Mutant Male
Wild Type Female
Wild Type Male
Mutant Female
Mutant Male
F2 Phenotypes for Cross B: Mutant Female X Wild Type Male
Wild Type Female
Wild Type Male
Mutant Female
Mutant Male
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Wild Type vs. Ebony
F2 Phenotypes for Cross A: Wild Type Female X Mutant Male
Wild Type Female
Wild Type Male
Mutant Female
Mutant Male
F2 Phenotypes for Cross B: Mutant Female X Wild Type Male
Wild Type Female
Wild Type Male
Mutant Female
Mutant Male
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Before the next lab:
Determine the expected ratios which would be obtained for each of the four inheritance patterns
by generating Punnett squares for each. You can use the space provided below for this
purpose. Compare your experimental results with the expected results. Generate hypotheses
as to the means of inheritance for each of the three mutations based upon the collected data.
1.
Autosomal dominant
2.
Autosomal recessive
3.
Sex-linked dominant
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4.
Sex-linked recessive
Hypotheses:
Sepia: ______________________________________________________________________
____________________________________________________________________________
____________________________________________________________________________
White: ______________________________________________________________________
____________________________________________________________________________
____________________________________________________________________________
Ebony: ______________________________________________________________________
____________________________________________________________________________
____________________________________________________________________________
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LAB 3B: Determination of Gene Linkage
Objectives
1. To understand the concept of genetic linkage.
2. To be able to calculate the recombination frequency.
3. To be able to apply the Chi-Square test to determine whether genes are linked.
4. To be able to determine the distance between genes through the calculation of the
recombination frequency.
Linkage and Recombination
The basic Mendelian laws of inheritance state that independent assortment will occur
during the formation of gametes, resulting in new combinations of traits, distinct from those
observed within the parental generation. The frequency at which these new combinations
occurred is determined using the formula
recombination frequency (%) = # recombinant progeny x 100
total progeny
As illustrated in Figure 1, the frequency of these new combinations (or recombinant
phenotypes) should be 50% if independent assortment occurs. However, while this holds
true for many of the traits we examine, this is not always the case.
Thomas Morgan was the first to describe instances where independent segregation did
not occur, resulting in phenotypic recombination occurring in less than 50% of the progeny.
These genes are said to be “linked” as the alleles obtained from a single parent will more
frequently be inherited together. Through further study, his lab determined that the cause of
this reduced recombination was due to these genes being located close together on the same
chromosome.
Figure 1: Two genes undergoing independent segregation. Genes on separate chromosomes can segregate in
one of two ways, leading to four different potential allelic combinations: nonrecombinant (left) or recombinant
(right).
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Crossing over between genes during prophase I is a random event, and as such,
crossing over will occur at different locations during different meiotic divisions. Unlinked
genes are those which are far apart and will likely have one or more crossovers occurring
between them, resulting in a 50% recombination frequency which is indistinguishable from
that observed for genes on separate chromosomes (see Figure 2). Conversely, some genes
are located so close to one another that crossing over “never” occurs, and these are said to
be completely linked. Those genes which demonstrate intermediate levels of recombination
(less than 50%, but greater than 0%) exhibit incomplete linkage. See Figure 3 on the next
page for a pictorial representation of complete and incomplete linkage.
Figure 2: Crossing over
between unlinked genes.
Genes which are far apart may
undergo a single or many
crossover events. An odd
number of events will result in
the production of both
nonrecombinant and
recombinant phenotypic
combinations (left), while even
numbers of crossovers will
result in only nonrecombinant
phenotypes being observed
(right).
Figure 3: Crossing over is reduced between linked genes. Genes which are completely linked are so close
together that crossovers do not occur between them, resulting in no progeny with recombinant phenotypes (left).
Incomplete linkage is observed when crossing over occurs less than 50% of the time due to the proximity of the
genes (right).
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The recombination frequency not only indicates whether the genes are linked, but is
also used as a measure of the genetic distance for mapping purposes. Gene distance is
provided in centiMorgans (cM) or map units (m.u.), both of which are equivalent to 1%
recombination. While this will not be done in this lab, the pair-wise comparison of multiple
linked genes can allow the mapping of an entire chromosome. Unlinked genes cannot be
used for this purpose, as they will always show 50% recombination, and therefore will always
appear to be 50 cM apart.
In order to determine the recombination frequency, we will perform a series of crosses
utilizing the same basic procedures employed in Lab 3A. Our first cross will utilize flies which
are homozygous for each of our two characteristics (wild type females and double mutant
males), thereby generating flies which are heterozygous for both genes. Females of this
generation are then used in a testcross with our double-mutant male flies, producing a
generation consisting of both nonrecombinant (wild type and double-mutant) and recombinant
(those possessing one or the other mutant phenotype) progeny (see Figure 4).
Figure 4: Diagram representing the theory of a testcross for linkage. Diagram depicts the phenotypes of the
flies (left) and their genotype (right). Hypothetical recessive mutations for purple eyes (p) and green bodies (g)
are shown. The phenotype of the testcross progeny are determined by the genetic input from the female (lefthand chromosomes) only, as the male contributes only recessive genes. Recombinant progeny are only
produced following crossing over between the linked genes.
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Chi-Square Test for Independence
While looking at recombination frequencies will suggest whether genes are linked, we
must utilize the Chi-Square Test for Independence in order to be able to determine whether
our results are statistically significant. Not only is this necessary to ensure that there is
sufficient evidence that our hypothesis is correct, but it will also allow us to state the distance
between the genes with greater certainty. Testing for linkage is performed in the same
manner as done in Lab 2, but with the following exceptions:
1. We must test the hypothesis that the genes are unlinked.
As we do not know the actual distance between the genes, there is no way to
accurately determine our expected numbers to test the hypothesis of linkage directly.
However, we can test the hypothesis that that genes are unlinked, as regardless of
how distant they are, all unlinked genes demonstrate 50% recombination.
2. The determination of the expected numbers for each phenotypic class requires the use
of a contingency table.
In order to minimize the effects of alterations in the observance of one trait on the
other, a contingency table is constructed using our observed values (see example
below).
Bb
bb
Column totals
Aa
15
6
21
aa
8
18
26
Row Totals
23
24
47
In this example, 15 progeny have both dominant phenotypes so would be found in the
intersecting square for the genotypes of Aa and Bb, 8 have the only the dominant B
phenotype and would be found in the intersecting square for aa and Bb, etc.
We use this table to calculate our expected numbers for each phenotypic class by
using the formula
Expected number = Row total x Column total
Grand total
In this example, the expected number of progeny exhibiting both dominant traits would
then be (23 x 21) / 47 = 10.28.
3. The degrees of freedom are now calculated using the following formula
d.o.f. = (# of rows - 1) (# of columns - 1)
In this case the d.o.f. is then (2-1)(2-1) = 1.
The remainder of the analysis proceeds with the use of the usual Chi-Square equation in
order to determine the p-value using the Table of Critical values on page 22. Remember
though that we are testing the hypothesis that the genes are not linked. Therefore, a p-value
less than 0.05 would indicate that the genes are indeed linked.
40
Procedures
In Lab 3A we examined the inheritance of three mutant Drosophila phenotypes: sepia
eyes, white eyes, and ebony body. This analysis allowed us to determine both the dominance
relationship between these alleles and the wild type alleles, as well as whether the genes for
these characteristics were autosomal or located on a sex chromosome. We are now going to
further analyze the two characteristics located on an autosome to determine whether or not
they exhibit linkage.
Materials
True breeding stocks of wild type and double mutant Drosophila
Immobilisation apparatus, paintbrush, file cards, dissecting microscope, jar of ethanol (fly
morgue)
Food vials, Drosophila food, and scoops
Incubators
All of the basic procedures for handling and crossing Drosophila were covered in Lab 3A.
This procedure only outlines the basic steps, indicating details that are specific to this
experiment.
Step 1: Generation of Heterozygotes
For this experiment, we will be crossing females that are homozygous for both wild type traits
with males that are homozygous for both mutant traits.
1. Prepare 2 food vials as previously described in Lab 3A.
2. Immobilize the flies and sex them. Transfer 3 wild type females and 3 double-mutant
males to each vial, leaving them on their side until the flies have recovered.
3. Place your vials into the incubator and record the temperature.
Step 2: Removal of Parental Flies
Parental flies are again removed after 7 days. Again, check flies for any discrepancies in
phenotype prior to their disposal in the Fly Morgue and ensure the food is of the proper
consistency prior to returning the vials to the incubator.
Step 3: Performing the Testcross
Two weeks following the removal of the parental flies, we will cross female progeny with the
double-mutant parental males. The vials were cleared 6 hours prior to this step in order to
ensure that the females have not been previously inseminated by the F1 males. The cross
will be performed in duplicate.
41
1. Prepare 2 food vials.
2. Immobilize progeny from the previous cross and sex them. Transfer 3 females into
each vial.
3. At the same time, immobilize double-mutant flies and sex them. Transfer 3 males into
each vial along with the females from step 2.
4. Once flies have recovered, place the vials in the incubator and record the temperature.
Step 4: Removal of Parental Flies
Parental flies are removed 7 days after the initiation of the testcross. Again check the
phenotypes and adjust food consistency if required.
Step 5: Analysis of Testcross Generation
1. Two images have been provided for you on Nexus. Count a minimum of 100 flies, using
both images if necessary. As we are dealing with autosomal genes, you do not need to
include the sex of the flies. Add your data to the class data table (available on Nexus)
and calculate the total observed for each phenotypic combination.
2. Identify the recombinant and nonrecombinant combinations and calculate the
recombination frequency.
3. Create a contingency table and perform the Chi-Square Test for Independence to
determine whether these genes are linked.
42
U2021F LAB 3A CLASS DATA
Enter your data from your individual counts in the highlighted cells
-Be sure to count both Vials 1 & 2.
Wild Female x Sepia Male
Wild Female Wild Male Mutant Female Mutant Male
35
37
15
13
37
36
11
16
39
35
16
10
40
44
7
9
49
51
0
0
42
47
4
7
36
40
13
11
Vial 1
Vial 2
Total
278
290
66
66
Wild Female x White Male
Wild Female Wild Male Mutant Female Mutant Male
54
25
0
21
51
22
0
27
55
20
0
25
53
27
0
20
50
24
0
26
51
23
0
26
53
23
0
24
Vial 1
Vial 2
Total
367
164
0
169
Wild Female x Ebony Male
Wild Female Wild Male Mutant Female Mutant Male
34
35
17
14
36
38
13
13
41
41
15
3
47
36
10
7
35
30
16
19
38
36
16
10
42
38
19
1
Vial 1
Vial 2
Total
273
254
106
67
Total
100
100
100
100
100
100
100
0
0
700
Sepia Female x Wild Male
Wild Female
42
37
38
34
45
39
36
Vial 1
Vial 2
Total
271
Total
100
100
100
100
100
100
100
0
0
700
White Female x Wild Male
Wild Female
31
24
22
28
27
24
21
Vial 1
Vial 2
Total
177
Total
100
100
100
100
100
100
100
0
0
700
Ebony Female x Wild Male
Wild Female
38
34
39
32
39
35
32
Vial 1
Vial 2
Total
249
e x Wild Male
Wild Male Mutant Female Mutant Male
29
18
11
42
15
6
37
19
6
40
18
8
32
14
9
33
19
9
38
16
10
251
119
59
e x Wild Male
Wild Male Mutant Female Mutant Male
20
28
21
23
28
25
23
25
30
21
27
24
28
19
26
29
28
19
24
26
29
168
181
174
le x Wild Male
Wild Male Mutant Female Mutant Male
38
11
13
35
7
24
37
9
15
41
18
9
32
12
17
34
13
18
39
12
17
256
82
113
Total
100
100
100
100
100
100
100
0
0
700
Total
100
100
100
100
100
100
100
0
0
700
Total
100
100
100
100
100
100
100
0
0
700
1
2
3
4
5
B
C
D
Wild-type Female X Sepia Male
(Cross A) Vial 1
A
E
1
2
3
4
5
B
C
D
Wild-type Female X Sepia Male
(Cross A) Vial 2
A
E
1
2
3
4
5
B
C
D
Sepia Female X Wild-type Male
(Cross B) Vial 1
A
E
1
2
3
4
5
B
C
D
Sepia Female X Wild-type Male
(Cross B) Vial 2
A
E
1
2
3
4
5
B
C
D
Wild-type Female X White Male
(Cross A) Vial 1
A
E
1
2
3
4
5
B
C
D
Wild-type Female X White Male
(Cross A) Vial 2
A
E
1
2
3
4
5
B
C
D
White Female X Wild-type Male
(Cross B) Vial 1
A
E
1
2
3
4
5
B
C
D
White Female X Wild-type Male
(Cross B) Vial 2
A
E
1
2
3
4
5
B
C
D
Wild-type Female X Ebony Male
(Cross A) Vial 1
A
E
1
2
3
4
5
B
C
D
Wild-type Female X Ebony Male
(Cross A) Vial 2
A
E
1
2
3
4
5
B
C
D
Ebony Female X Wild-type Male
(Cross B) Vial 1
A
E
1
2
3
4
5
B
C
D
Ebony Female X Wild-type Male
(Cross B) Vial 2
A
E
1
2
3
4
5
B
C
D
Vial 1: F1 Tan, Red Female X
Ebony, Sepia Male Testcross
A
E
1
2
3
4
5
B
C
D
Vial 2: F1 Tan, Red Female X
Ebony, Sepia Male Testcross
A
E
U2021F LAB 3B CLASS DATA
Enter your data from your individual counts in the highlighted cells
-Be sure to count both Vials 1 & 2.
F1 Tan, Red Female x Ebony, Sepia Male Testcross
Tan, Red
Tan, Sepia
Ebony, Red
26
19
27
35
23
21
32
19
15
36
22
15
38
17
14
33
22
19
27
23
21
Vial 1
Vial 2
Total
227
145
132
Ebony, Sepia
28
21
34
27
31
26
29
196
100
100
100
100
100
100
100
0
0
700
7:40
a
GENETICS
LAB MANUAL
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LAB 2: Genetic Inheritance: Probability & Significance
Objectives
1. To be able to apply the multiplication and addition rules to determine the probability of
a particular combination of events occurring.
2. To be able to use the Chi-Square (x²) Test to analyze the observed ratios and to
determine if they indicate the acceptance or rejection of the hypothesis.
3. To be able to apply these concepts to basic genetic inheritance patterns.
Determining the Probability of Events
<
Understanding how to determine the probability of a particular set of events occurring
is essential to the study of genetic inheritance. For the purposes of our labs, it is sufficient
that you can determine the probability of a given event, as well as being able to apply the two
most basic rules dictating the probability of multiple events: the multiplication and addition
rules.
The multiplication rule is used to determine the probability of two independent
events occurring together. These events are those whereby the fact that one has occurred
does not affect the probability of the other occurring. For example, having obtained heads on
a single toss of a coin does not alter the probability of obtaining heads on a second toss. As
the name of the rule implies, we would simply multiply the individual probabilities to obtain our
answer as follows:
As there are only two possible outcomes for a coin toss (heads or tails), and each is
just as likely to occur, the probability of obtaining heads is 2 and the probability of tails
is also ¹12.
• Therefore, the probability of obtaining heads on each of two tosses of the coin
becomes 1/2 x ¹/2 = 14.
The addition rule is used to determine the probability that one of two or more
mutually exclusive events will occur. These can be described as either/or events. For
example, if we wanted to know the probability of obtaining two heads or two tails if we tossed
a coin twice, we would need to add the probabilities of each possibility, as you could only
achieve either two heads or two tails if performing only two tosses of a coin, not both. In this
case, we can obtain the probability of either two heads or two tails as follows:
We know from above that the probability of two heads is 14.
We would determine the probability of two tails in the same manner, so it is also 14.
Therefore, the probability of either two heads or two tails becomes 14/4 + ¹/4 = ¹/2.
Remember also, that when dealing with probability, the sum of all possible events will equal 1.
Can you determine the other possibilities for our coin toss example in order to demonstrate
that this rule is true?
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Determining the Probability of Genetic Events: Monohybrid and Dihybrid Crosses
The inheritance of genetic information follows these same basic concepts of probability.
As you will recall from Lab 1, each somatic cell contains two copies of each chromosome, and
following meiosis each gamete contains only a single copy of each. You will also recall that
each gene is found at a specific locus of a particular chromosome. Therefore, each gamete
contains a single copy of each gene.
1/2 A
The simplest way of looking at the probability of genetic inheritance is through
performing a monohybrid cross. That is, a cross which examines the inheritance of a single
gene. As an example, let's assume we are interested in looking at the gene which encodes
protein A. There are two different alleles (versions) of this gene within the population: a
dominant allele (A) and a recessive allele (a). If an individual was heterozygous for this gene,
then half of their gametes would contain A and the other half would contain a. This means
the probability of an offspring inheriting A is 1/2 (as is the probability of inheriting a).
During the process of fertilization, a random gamete from the female parent is joined
with a random gamete from the male parent. As the allele inherited from one parent does not
impact which allele is inherited from the other, this is an independent event. As such, the
probability of each possible combination of alleles is determined through the application of the
multiplication rule. We can see the results of this applied to the completion of the Punnett
square in Figure 1, with each possibility having a probability of 14.
This simple example also demonstrates the use of the addition rule for mutually
exclusive genetic events in order to determine the probability of a particular genotype or
phenotype. As seen in Figure 1, there are two ways in which a heterozygous progeny can be
produced. They can inherit the dominant allele from their mother and the recessive allele
from their father or the can inherit the recessive allele from their mother and the dominant
allele from their father. Therefore, through the application of the addition rule, the probability
of an individual being heterozygous becomes 14 + ¹/4 = ¹/2.
1/2 Ax
1/2 a
nexus.uwinnipeg.ca
1/2 A
1/2 A 1/4 AA
1/2 a x
1/2 a
1/2 a
x
1/2 A
1/2 a
1/2 A
1/2 a
1/2 A
1/4 AA
1/4 Aa
1/2 A
1/4 AA
1/4 Aa
Genotypic probabilities:
1/4 AA + (1/4 Aa + 1/4 Aa) + 1/4 aa = 1
1/4 AA + 1/2 Aa + 1/4 aa = 1
1/2 a
1/4 Aa
18
<
1/2 a
1/4 Aa
1/4 aa
Each gamete randomly
chosen from each parent =
independent events =
multiplication rule
Random fertilization means
each possible combination
has an equal probability of
occurring = 1/4
Equivalent combinations are either/or for a given individual progeny =
mutually exclusive = addition rule
Figure 1: Demonstration of the application of the probability rules to a monohybrid cross
Phenotypic probabilities:
(1/4 AA + 1/4 Aa + 1/4 Aa) + 1/4 aa = 1
314 Dominant + 1/4 recessive = 18:01
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<
Similarly, again referring to the example in Figure 1, when determining the probability
of a progeny having a dominant phenotype, it is the sum of all progeny carrying one or more
copies of A. Thus, this probability is determined to be 4 AA + 1/2 Aa = 34.
We can extend this further by examining two genes simultaneously, which is called a
dihybrid cross. The application of the probability rules to a dihybrid cross will be discussed
during the lab tutorial, where we will again analyze the inheritance of the gene encoding
protein A, but also simultaneously analyze the inheritance of the gene for protein B. In this
example, we will again be dealing with two alleles: the dominant B allele and the recessive b
allele.
Determining the Probability of Genetic Events: Inheritance Patterns
The example provided above is one where one allele demonstrates complete
dominance over the other. However, this is not always the case, as alleles may also
demonstrate incomplete dominance or codominance. However, regardless of what form of
dominance exists, the probability of genotypic combinations will remain the same. What
differs will be the probability of a particular phenotype being expressed. In both incomplete
dominance and co-dominance, the heterozygous individuals express a different phenotype
than either type of homozygous individual. In this case, rather than having a 3 dominant :1
recessive ratio (or 34 and 14 probabilities), we now have a 1:2:1 ratio (or ½, ½, and
probabilities). In the case of incomplete dominance, ½ of the progeny will express a
phenotype that is intermediate to that of the homozygous phenotypes. Conversely, in
codominance, the phenotype of the heterozygous individuals is a combination of the
homozygous phenotypes.
In addition to phenotypic probabilities being determined by dominance relationships,
they are also affected by the location of the gene. Genes which are located on the sex
chromosomes (X and Y in both humans and Drosophila) are not inherited in the same manner
between the sexe Females receive two copi of genes located on the X chromosome and
inheritance patterns will be the same as those observed for autosomal genes. However,
males receive only a single X chromosome and are therefore hemizygous for these genes.
In addition, only males receive a Y chromosome and the genes it carries. Thus, the sex of the
individual is always included in both the genotype and phenotype.
Statistical Analysis of Data
As you are aware, the segregation and the independent assortment of allele pairs,
together with the random nature of fertilization, determine genetic ratios. All of these events
are affected by chance, and it is rare that the observed ratios are an exact match to the
expected ratios. We can see the operation of chance if we toss a coin and observe the
numbers of heads and tails in 10 throws. Our expectation, or Null hypothesis, is that since a
coin can fall in one of two ways, the chance of either occurance is 2. Yet we would be
surprised if we got 5 heads and 5 tails, and it is possible to get 10 heads or tails. We would
expect that we would get closer to our predicted 1:1 ratio if we made 100 or 1000 tosses, and
to get all heads or all tails in this instant would be highly unlikely. In this lab, we will utilize the
Chi-Square Analysis to determine whether our observed data significantly agrees with our
expectations based upon our null hypothesis.
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<
Statistical Analysis of Genetic Data
Formulating a null hypothesis requires the ability to determine the expected results.
When focusing on genetics, we will use our understanding of the patterns of inheritance to
determine the null hypothesis. For example, if we want to form a hypothesis about a genetic
trait we believe to be autosomal dominant, we would utilize the basic Mendelian phenotypic
ratio for a monohybrid cross. As such, our null hypothesis would be that we would expect ¾
of the progeny to exhibit our trait of interest, and 14 to exhibit the wild type phenotype. This
statement would of course be much more specific when performing a specific analysis, as you
will observe in our example below.
Similar to the examination of a single (hypothesized) autosomal dominant trait, we would
determine our hypothesis for alternative forms of inheritance, such as recessive alleles or sex-
linked characteristics (both of which will be considered in this lab). More complex analyses can
also be performed, such as looking at dihybrid crosses, or even epistatic gene interactions.
Finally, the Chi-square analysis can also be used to determine whether or not genes are located
near one another on the same chromosome: a phenomenon termed genetic linkage (to be
discussed in detail in Lab 3B).
Chi-Square Analysis
Using the chi-square (x²) test, we can analyze experimental data to determine whether
it matches our expected ratios. This test takes into account both the sample size and the
number of classes (or groups) which exist within our experiment. For example, when
tossing a coin there are two classes: heads or tails. Conversely, when tossing a die there are
six classes: 1, 2, 3, 4, 5, or 6. Each of these classes are equivalent to one possible result for
each event. This test will compare the observed and expected numbers for a particular data
set in order to calculate a probability value which will indicate whether we should accept or
reject our null hypothesis. Please note that you must have at least two observed classes and
you must be analyzing counted data (not fractions, frequencies, or percentages). Utilize the
following procedure in order to perform the chi-square test.
1. Determine your Null Hypothesis utilizing your expected ratios. You must write your
hypothesis in a proper sentence(s).
2. Determine the number of classes (n) in your experiment and use this to calculate the
degrees of freedom (d.o.f.) for your analysis. This value represents how many classes
are able to vary within an individual experiment. As one class is always constrained by
the others, d.o.f = n - 1.
For example, in a coin tossing experiment the two classes are heads and tails,
giving us a d.o.f. of 2 - 1 = 1. This is due to the fact that if we toss a heads, the
only other option must therefore be tails. So the number of tails must be our total
tosses - the number of heads.
• Another example would be the tossing of a die, where we would have six classes,
for a d.o.f. of 6-1=5. The number of 6's rolled must be the total rolls - the
number of 1's - the number of 2's - the number of 3's -- the number of 4's - the
number of 5's. That is, if the number rolled is not a 1, 2, 3, 4, or 5, it must
necessarily be a 6.
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<
3. Perform your experiment and record the data. These values will be your observed (0)
numbers for each class.
4. Determine the expected (E) numbers for each class utilizing the total number of
observed and your null hypothesis. Always include the decimal places (up to three)
within your expected numbers. Do not round this numbers up or down.
5. Calculate the deviation (D) between the observed and expected values using the
formula: D=O-E
6. Calculate the chi-square (x²) value by determining the sum (2) of the squared
deviations of each class divided by its expected values, as given by the formula:
χλ = Σ D? = Σ (O-E)2
E
E
7. Utilize the Table of Critical Values of the Chi-Square Distribution (Table 1 on the next
page). This table will allow you to determine the probability of having obtained this
degree of deviation if your null hypothesis is correct. To use the table follow the
following procedure:
Find the d.o.f. for your analysis, which will be listed on the left hand side.
Move along the row until you reach the two values which "bracket" your Chi-Square
value. In other words, find the neighboring values that are just above and just
below your calculated value.
Move up to find your probability (p-value) range. You will notice that these values
range from 0.995 to 0.005 (or 99.5% to 0.5% as to their likeliness of occurring).
This range could (for example) be 0.01 < p < 0.05, which means that you would
expect to see the deviation observed in your experiment 1-5% of the time if your
hypothesis is true.
●
8. Use this range to determine whether your hypothesis should be accepted by using the
following guidelines:
If the p-value is > 0.05, you would accept your hypothesis. This value means that if
your hypothesis is true, the deviation observed in your data would be expected in
5% or more trials. In other words, there is up to 95% probability that the observed
deviation is due to chance.
• If p < 0.01, then the deviation from the expected is significant, and the Null
hypothesis would be rejected. You should be aware however, that this does not
mean that your hypothesis is definitively incorrect, as a correct hypothesis could still
exhibit this degree of variation in up to 1% of trials. At this point, you would want to
reexamine your hypothesis and consider alternatives.
If 0.01 < p < 0.05, then we would state that we have reason to doubt our Null
hypothesis. Ideally, you would repeat any experiment which falls in this region.8:01
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Table 1: Critical Values of the Chi-Square Distribution. Utilize this chart for determining the p-value range of your
Chi-square analyses for deciding whether to accept or reject your hypotheses.
PROBABILITY (p)
<
-ACCEPT ???? REJECT>
22
D.O.F. 0.995 0.975 0.900
0.500
0.100 0.050 0.025
2.706 3.841 5.024
1
5
6
7
8
9
10
2.156 3.247
0.000 0.000 0.016 0.445
2 0.010 0.051 0.211 1.386 4.605 5.991 7.378
3 0.072 0.216 0.584 2.366 6.251 7.815 9.348
4 0.207 0.484 1.064 3.357 7.779 9.488 11.143
0.412 0.831 1.610 4.351 9.236 11.070 12.832
0.676 1.237 2.204 5.348 10.645 12.592 14.449
0.989 1.690 2.833 6.346 12.017 14.067 16.013
1.344 2.180 3.490 7.344 13.362 15.507 17.535
1.735 2.700 4.168 8.343 14.684 16.919 19.023
4.865 9.342 15.987 18.307 20.483
2.603 3.816 5.578 10.341 17.275 19.675 21.920
12 3.074 4.404 6.304
11.340 18.549 21.026 23.337
13 3.565 5.009 7.042 12.340
4.075 5.629 7.790 13.339
15 4.601 6.262 8.547 14.339
16 5.142 6.908 9.312 15.338
17 5.697 7.564 10.085
18 6.265 8.231 10.865
19 6.844 8.907 11.651
7.434 9.591 12.443
11
19.812 22.362 24.736
14
21.064 23.685 26.119
22.307 24.996 27.488
23.542 26.296 28.845
16.338 24.769
17.338
27.587 30.191
25.989
28.869 31.526
18.338
27.204
30.144 32.852
36.191
20
19.337
28.412
31.410 34.170
37.566
21
29.615 32.670 35.479
38.932
8.034 10.283 13.240
14.042
14.848 22.337
20.337
21.337 30.183 33.924 36.781
22
8.643 10.982
40.289
23
9.260 11.688
41.638
32.007 35.172 38.076
33.196 36.415 39.364
24
9.886 12.401
15.659 23.337
42.980
25
44.314
10.520 13.120 16.473 24.337 34.382 37.652 40.646
11.160 13.844 17.292 25.336 35.563
26
38.885 41.923
45.642
27
14.573 18.114 26.336 36.741 40.113 43.194
46.963
11.808
12.461 15.308 18.939
28
48.278
29
13.121 16.047 19.768
27.336 37.916 41.337 44.461
28.336 39.088 42.557 45.722
29.336 40.256 43.773 46.969
49.588
30
13.787 16.791 20.599
50.892
31
44.985 48.232
52.192
32
46.194 49.481
53.486
33
47.400 50.725
54.776
34
56.061
14.458 17.539 31.434 30.336 41.422
15.135 18.291 22.271 31.336 42.585
15.816 10.047 23.110 32.336 43.745
16.502 19.806 23.952 33.336 44.903 48.602 51.966
35 17.192 20.570 24.797 34.336 46.059 49.802 53.203
36 17.887 21.336 25.643 35.336 47.212 50.998 54.437 58.619
Values from 1 to 30 degrees of freedom from Thompson, Biometrika 32:188 189 (1941); values from 31 to 36
degrees of freedom from Statistical Tables, 2nd edition, Rohlf and Sokal (1981) W. H. Freeman & Co.
57.342
0.010
6.636
9.210
11.345
13.277
15.086
16.812
18.475
20.090
21.666
23.209
24.725
26.217
27.688
29.141
30.578
32.000
33.409
34.8058:01
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Chi-Square Example
As an example, we will be analzying data from a simple monohybrid cross examining the
inheritance of green versus yellow leaves.
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1. Determine your Null Hypothesis. In this example, the experimenter observed 85 green
plants and 25 yellow plants when determining the phenotype of the progeny in the F2
generation. This would suggest that green leaves are dominant to yellow leaves, so
our Null Hypothesis is:
The expected ratio is 3 green plants : 1 yellow plant.
2. Determine the degrees of freedom. As there are only two observed phenotypes (green
and yellow) there are two classes, so n = 2, and d.o.f. = 2 - 1 = 1
3. Calculate the Chi-Square Value by setting up a table as follows:
a) Include the expected ratio and observed data.
Fraction Observed Expected Deviation
Expected (0) (E) (D)
3/4
85
1/4
25
1
110
Your fractions of expected progeny should always equal 1.
Phenotype
Green
Yellow
Total
Phenotype
b) Calculate the expected values by multiplying the total number of observed progeny
by the fraction expected for each class, keeping decimal places.
Fraction Observed Expected Deviation
Expected (0)
(E)
(D)
Green
3/4
85
82.5
2.5
Yellow
1/4
27.5
Total
1
110
* Your total deviation should always equal 0.
Phenotype
Fraction Observed Expected Deviation
Expected (0)
(D)
(E)
Green
3/4
85
82.5
Yellow
1/4
25
27.5
Total
1
110
110
* The total expected should be the same as the total observed.
c) Complete the remainder of the table. The sum of D²/E is the Chi-Square (x²) value.
25
110
<
D²
23
-2.5
0
D²
D²/E
D²
6.25
6.25
D²/E
D²/E
0.076
0.230
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021F
4. Use the Table of Critical Values on page 22 to determine your p-value.
With a d.o.f. of 1, then our Chi-Square value of 0.306 gives the following p-value range:
0.50 < p < 0.900
This means that this occur in 50% to 90% of trials due to chance alone if our
hypothesis is correct.
5. Decide whether to accept or reject your hypothesis.
As p > 0.05, we would accept our hypothesis.
Laboratory Exercises
Materials
F2 corn cob from dizygotic cross of purple, smooth and yellow, wrinkled P generations
Corn cob from testcross of F1 cob from the above cross
Two dice, tally counter
A. Analysis of Genetic Ratios in Corn
We will be analyzing the inheritance patterns for two corn characteristics: kernel color (yellow
or purple) and kernel texture (smooth or wrinkled). For both characteristics, one of the alleles
is completely dominant to the other. The information completed below will be used to
complete your Lab 2 assignment.
Kernel color:
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1. Using the images of the results on monohybrid crosses shown in Figure 2 on the next
page, define your alleles to indicate dominance and provide the expected phenotypic
ratios for each of the two characteristics in the form of a Null hypothesis.
Define alleles:
Null hypothesis:
Kernel texture:
Define alleles:
<
Null hypothesis:
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021F
0000000000010
Kernel color: Yellow x Purple
Kernel texture: Smooth x Wrinkled
Figure 2: F1 generation of monohybrid crosses for corn kernel characteristics. Crosses
between homozygous plants were performed for yellow and purple kernels (left) and smooth
and wrinkled kernels (right). Note that each kernel has a distinct genotype/phenotype on the
F1 cobs. Images obtained from https://www.carolina.com/life-science/genetics/plant-
Small
genetics/10422.ct?N=2287668716&Nr=&mCat=10337&sCat=10415&nore=y
2. Using your null hypothesis for kernel color, complete table 2 below.
each trait were
Table 2: F1 generation of monohybrid crosses for kernel color. Plants homozygous for
crossed (yellow x purple), and the resultant progeny were analyzed for kernel color. Samples of two
different sizes are included below.
Sample size
Phenotype
Expected number
Purple
Large
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Yellow
Total
Purple
Yellow
Total
Expected
frequency
Observed
number
371
25
118
489
4688
<
1685
6373
3. Perform a Chi-square analysis for these data sets. Does the sample size make a
difference as to whether you would accept or reject your hypothesis?8:01
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netics Lab Manual
021F
4. Use the below results of a dihybrid cross between pure-breeding purple, smooth and
yellow, wrinkled plants when completing your Lab 2 Assignment.
FX-
F
20040
<
Figure 3: Diagram depicting the results of a dihybrid cross for kernel color and kernel texture. Parental crosses
were performed with plants which were homozygous for both characteristics as indicated. Images obtained from
https://www.carolina.com/life-science/genetics/plant-genetics/10422.ct?
N=2287668716&Nr=&mCat=10337&sCat=10415&nore=y
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021F
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B. Investigation of Probability by Tossing Dice
This is a practice exercise that can be attempted at the end of the lab or at home. An answer
key will be posted following the lab tutorial.
1. Determine the probability of each sum of numbers on two dice (2-12). Enter this into
row 2 of Table 3 below.
Observed
number
2. Using the probabilities determined for step 1, calculate the expected number in 1000
tosses (row 3) for each combination by multiplying row 1 by 1000.
3. Use this data to perform a Chi-Square analysis.
Table 3: Class data table for sum of two dices tossed. Dice were tossed a total of 1000 times.
Sum of
2
3
4
5
6
7
8
9
10 11 12
24
48
116 144 171 137 112 86
54 27
Expected
frequency
<
Expected
number
27
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