Download Developing Conclusions About Different Modes of Inheritance

Misr University for Science and Technology
Faculty of Biotechnology
Principles of Genetics 102
"Laboratory manual"
Instructor
Dr. Mona Ibrahim
Demonstrators
Nermin Gamal
Heba Radwan
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Contents
 Chapter 1: Introduction to Drosophila melanogaster

Introduction of Drosophila melanogaster
o Development of Flies.
o Culturing.
o Handling of the Flies
o Distinguish sex
o Phenotype characterization.
o Securing virgin females.
 Chapter 2: 1st Mendelian law

Drosophila and maize experiments in genetics :Monohybrid crosses
o The first experiment in Drosophila( Monohybrid cross)
o ♂ EE x ♀ ee
o ♀ EE x ♂ ee
o The first experiment in Maize (Monohybrid cross)
 Chapter 3: Chi square test
 The Chi Square Test
o Analysis of monohybrid's offspring in Drosophila.
o Analysis of monohybrid's offspring crosses in maize.
 Chapter 4: 2nd Mendelian law

Drosophila and maize experiments in genetics: Dihybrid crosses
o The second experiment of drosophila (dihybrids crosses)
o ♂eevgvg x ♀EEVgVg
o ♀ eevgvg x ♂EEVgVg
o Analyzing the offspring (F2) of the dihybrid cross in
drosophila(Chi-square test)
o The second experiment of maize ( Dihybrid cross )
o Analyzing the offspring (F2) of the dihybrid cross in maiz(Chisquare test)
 Chapter 5: Gene interaction and epistasis
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Gene interaction and epistasis
o Gene interaction in maize
o Analysis of gene interaction ratios in maize by (Chi-square test
)
 Chapter 6: sex linkage

Drosophila experiments in genetics: sex - linkage
o Introduction of sex chromosomes and gene transition
o Sex linkage in Drosophila
o ♂RГx♀ww
o ♀RRx♂wГ.
o Analysing the offspring (F2) of the sex linkage experiment in
drosophila
 Chapter 7: Pedigree Analysis
 Glossary
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Chapter one: Introduction to Drosophila
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Introduction to Drosophila melanogaster
The purpose of this chapter is to introduce students to the equipment and
techniques used to handle “fruit flies" in the laboratory. You will learn about
the life cycle of the Dipteran Drosophila melanogaster. You will also learn to
differentiate between male and female flies and how to identify aberrant
phenotypes that are attributable to specific mutations, many of which will
be studied during the semester.
 Objectives of this investigation
Upon completion of this exercise you should be able to
Distinguish between male and female D. melanogaster
Categorize mutant flies based on aberrant phenotypes
 Prepare controlled genetic crosses of D. melanogaster

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Background
The fruit fly Drosophila melanogaster is a very useful organism for genetic
research and has probably been used to define more fundamental genetic
principles than any other ulticellular eukaryote. One person responsible for
development of D. melanogaster into a model genetic system was named
Thomas Hunt Morgan who, in 1910, published one of the first descriptions of
sex linkage
Morgan and his students elucidated many basic principles of heredity,
including sex-linked inheritance, epistasis, multiple alleles, and gene mapping.
Thomas Hunt Morgan and colleagues extended Mendel's work by describing
X-linked inheritance and by showing that genes located on the same
chromosome do not show independent assortment. Studies of X-linked traits
helped confirm that genes are found on chromosomes, while studies of linked
traits led to the first maps showing the locations of genetic loci on
chromosomes"
Due to its simple culturing requirements, short generation time, copious
offspring, and well-defined genetics, this diminutive organism has become a
versatile model system that is routinely used for inquiries into the genetics
of eukaryotic development, behavior, and population dynamics. Since the
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arrival of recombinant DNA technology, much has been learned about the
biology of Drosophila melanogaster.
 Development of Flies
The Life Cycle of D. melanogaster
D. melanogaster progress through four stages during their life cycle: egg,
larva, pupae, and adult. Fertilization of eggs is internal, and females deposit
fertilized eggs on the surface of the culture medium. Usually within one day,
the eggs develop into larvae, which burrow into the nutritive medium. Over a
period of 4-7 days, the larvae pass through three stages, or instars, and
eventually crawl onto a firm surface to pupate.
During the pupal stage, which usually lasts from 4-6 days, metamorphosis
occurs and the adult form develops. The adult fly emerges (ecloses) as an
imago, which is slender, elongated, and light in color, with rumpled and
unexpanded wings. Within a few hours, the adult matures, becoming darker
and more rotund, with fully expanded wings. Adult flies may live for a month
or more.
The rate of D. melanogaster development is greatly influenced by
temperature. When propagated at 21o C (69.8o F), the progression from egg
to adult usually takes about two weeks However, if temperature is
maintained at 25o C (77.7o F), this time can be shortened to about 10 days.
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Working with D. melanogaster
 Culturing of the flies
 Medium
For our purposes, D. melanogaster will be raised in plastic culture vials, which
may contain a small piece of plastic netting to increase surface area for
attachment of pupae. The culture medium is a complex mixture of agar,
sugars, other nutritional supplements, and mold inhibitors.
 Material needed for each student for this investigation
 Medium
Stereo dissecting microscope
Bottles of culture
Gummed labels
Etherizer
Re-etherizer (Petri dish)
Ether in dropping bottle
Dropper
Teasing needle or fine camel's hair brush
Morgue containing 70% ethyl alcohol
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Handling the Flies
 To etherize and examine adult flies, proceed as follows:
1- Place a few drop of ether on the absorbent material of etherizer
2- Strike the base of bottle lightly on the palm of the hand so the flies
will drop to the bottom.
3- Remove the culture bottle, plug quickly replace it with the mouth of
the etherizer, invert on the bottle over the etherizer, and shake the
flies into the etherizer.
4- Subject the flies to ether for about 30seconds after they cease
moving. Avoid over etherization if they are to be used in further
mating the flies will die if left in etherizer too long .over etherized
flies hold their wings vertically over their body (in contrast to the
normal at-rest position).
5- Transfer the etherized flies to clean white card. A 3-by- 5 in .file
card is ideal.
6- Examine the etherized flies with a dissecting microscope at 10 x to
25x magnification. Use a soft brush or teasing needle for moving the
flies about on the stage of the microscope.
7- If the flies revive before finish examine them ,add a few drops of
ether to the absorbent pad on the re-etherizer and cover flies on the
microscope stage for a few seconds .
8- If the flies are not needed after observation they may be discarded
in the morgue. Etherized flies to be used for further mating should be
permitted to recover in a dry vial or on a dry surface in the culture
bottle before they come into contact with moist medium.
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 Distinguishing sex
Female
Size
Shape
Color
Sex combs
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Male
Larger than the male
The caudal extremity is Sharpe and
protruding. The abdomen of the female is
distended and appears spherical or ovate.
Black pigment is more than male and
Uniformly distributed
Where as in male is round and blunt
the abdomen of the male is relatively
narrow and cylindrical .
Black pigment is more extensive on
the caudal extremity of the male than
on that of the female
The marking or pigment occurs only in the
dorsal region
On male the marking extend
completely around the abdomen and
meet on venteral side.
Only male have small tuft of black
bristles called a sex comb on the
anterior margin at the basal tarsal joint
of each front leg
Examination of the external genitalia under magnification is the best means
of distinguishing the sex of flies. Only males flies exhibit darkly colored
external genitalia which are visible on the ventral side of the tip of the
abdomen following characteristics, illustrated in figure (1.2) can also help to
distinguish males from females
Female genitalia
Male genitalia
Female flies have anal plates and very dark ovipositor Males have anal plates, a dark colored genital
plates on their ventral side.
arch and a penis.
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Scoring Phenotypes
It will often be necessary to accurately identify phenotypic characteristics
each as eye color or shape, body color, and bristle or wing morphology.
In most cases, you will be using a stereo microscope (dissection microscope)
to examine various phenotypes or determine the sex of an individual fly. To
accurately score flies for phenotype differences that are difficult to
distinguish, it is easiest to make direct comparisons by having a wild type fly
and the mutant strain both in the field of view of the stereo microscope.
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 Securing virgin females
To begin across between two varieties of flies, you must secure a virgin
female. Once inseminated; females retain viable sperm for several days.
Thus, the only way to ensure a controlled mating between different genetic
stocks is to use virgin females.
The most common method for obtaining virgin females is to select those
that have recently emerged from their pupa cases. These flies
(D.melanogaster) do not mate for at least 8 hours (probably not for 10
hours) after emerging from the pupa case. Thus, 8-hour-old females will still
be virgins even if male flies are present. To secure virgin females, follow this
procedure
1- Empty the appropriate stock bottle of all adult flies .Record the time
on the bottle
2- Within 8 hours of removing the adult flies, etherize any newly
emerged adults. Such newly emerged flies are distinguished by their
pale body color and a characteristics dark spot on the ventral side of
the abdomen, slightly to the left of the midline.
3- Females among those newly emerged adults will be virgins that can be
used in the appropriate cross
Another method for obtaining virgin females is to isolate pupae from which
the adult flies are about to emerge .Single pupa cases should be placed in
small vials containing a narrow strip of moistened paper towel. The vial
should be stoppered with cotton. Of necessity, any females hatching alone in
a vial will be a virgin. This method of obtaining virgin females is laborious and
can prove unwieldy because it requires many vials.
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Exercise
You will be given flies with “unknown” mutations to identify and complete the
table
Trait
Wild Unknown
Type 1
Body
color
Eye
color
Eye
shape
Wing
shape
and size
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Unknown2
Uknown3
Unknown4
Unknown5
Unknown
6
Notes
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Chapter two: 1st Mendelian law
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Drosophila and maize experiments in genetics
Monohybrid crosses
In this exercise, you will investigate the inheritance of traits in D.
melanogaster, and maize using the Mendelian model to develop your
hypotheses. You will work with monohybridd crosses to determine the
genotypes of the F1 and F2 generations and the phenotypes.
Objectives of this investigation
Upon completion of this lab, you should be able to:

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
Predict the offspring of a monohybrid, cross.
Determine the parents of a designed cross from the ratios of
observed offspring.
Calculate phenotypic ratios and carry out Chi-squared analyses for
each cross.
Background:
Each cross begins with true breeding (homozygous) parents. A homozygote
has two identical alleles for a given trait. If parents with different traits
are crossed (aa x AA), the offspring will be heterozygous (Aa), receiving one
allele from each parent. However, the offspring will express only one allele,
which Mendel described as dominant. The masked trait is recessive; two
alleles must be present for a recessive trait to be expressed. The parents in
this cross are referred to as the P generation, and the hybrid offspring, as
the F1 (first filial or hybrid) generation. Two F1 hybrids are crossed to
produce the F2 generation.
The predicted Mendelian results for the F2 generation from a monohybrid
cross (Aa x Aa) is a 1:2:1 genotypic ratio and a 3:1 phenotypic ratio. Mendel
formulated his first law of segregation to explain these results: allele pairs
separate during gamete formation and randomly return to the paired
condition during the fertilization to form a zygote
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First experiment
Monohybrid crosses in Drosophila
D. melanogaster has historically been the model system for research in
eukaryotic genetics and plays an important role in the development of our
knowledge of heredity. D. melanogaster have a low chromosome number (n=
4), referred to as X (1), 2, 3, 4 chromosomes. Chromosomes 2, 3, 4 are
autosomes (same in both sexes). X and Y are sex chromosomes. Females are
XX and males are XY. Chromosomes 4 and Y contain few genes and for
practical purposed can be ignored. Almost the entire genetic content of the
D. melanogaster genome resides on only three chromosomes: X, 2, and 3.
Making crosses:
1- When making crossing between two varieties, For example, in crossing
flies having ebony body color with those having white eyes, only body
color and eye color need to be observed carefully.
2- Secure etherized male flies of one variety and etherized virgin
females of the other. While holding the culture bottle on its side,
place these flies in the bottle. Be sure to add some dry yeast granules
or yeast suspension to the medium before introducing the flies.
3- Keep the bottle on its side until the flies have recovered from
etherization. This position will prevent their becoming stuck in the
medium.
4- Label the bottle and record data in table 1.2 after 7 or 8 days
remove the parent flies (p1) to prevent their being confused with or
mated with their offspring.
5- Flies of the first filial generation (F1) will soon being to emerge.
After several F1 flies have appeared, etherized and examine them,
especially with regard to the characters by which the P1 flies
differed.
6- Record in table 1.2 the phenotypes of the F1 flies of each sex; place
these F1 flies in a fresh bottle of medium. (Be sure to label the
bottle) This mating will allow for a production of the second filial
generation (F2). It is not necessary that the F1 female flies are virgins
for this mating.
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Exercise
Why the instructor may wish to have you test cross an F1 female flies?
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In this case you should use a virgin female?
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Why a virgin female is needed in this case?
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What is test cross?
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Laboratory record
Mush of success in elucidating the basic principle of inheritance depend on
his keeping accurate records about his experiments .keeping accurate and
detailed notes is also necessary to the successful completion of your
laboratory work .To this end ,you are provided with Table 1.2 in which to
records as you do the various steps of this investigation.
Table 1.2 Record of a drosophila Experiment (1)
Experiment number……………. .Name………………………………………………………….
1. Cross………………………….female x …………………………………..male.
2. Date P1 mated………………………………………………………………………………………….
3. Date P1 s removed…………………………………………………………………………………..
4. Date F1 first appeared ………………………………………………………………………....
5. Phenotype of F1 males ……………………………………………………………………….....
6. Phenotype of F1 females ……………………………………………………………………….
7. Date F1 male and female placed in fresh bottle (or date F1 virgin female
test crossed)……….
8. Date F1 flies removed ………………………………………………………………………….
9. Date F2 appeared…………………………………………………………………………….......
10. Record F2 in the following table:
Phenotype
a ………………….
Male
Number
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b. …………………
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c. ………………….
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Phenotype
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Female
Number
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Total
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D …………………….
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e. ……………………
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f. …………………
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g. …………………
h. ……………….
Total ………………..
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11. Indicate the genotypes and phenotypes of the flies in each of three
generations
Male
Female
Genotype of P1
Genotype of F1
Phenotype of F1
Genotype of F2
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What F2 phenotypic ratio do you expect to obtain?
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What F2 genotypic ratio is expected?
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Which trait is dominant in your experiment?
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How do you know?
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Second Experiment
Monohybrid cross in Maize
The extensive genetic variability in maize also obeys mendelian principles;
both seedling (shape) and aleurone (endosperm) (color) characters lend
themselves to classroom use. Figure (1.3) shows an ear of F2 kernels of corn
that are segregating in a ratio of 3 colored: 1colorless aleuronic.
figure 1.3
You probably will not be able to conduct actual experimental mating with it,
Because maize has relatively long life cycle (3 months or more to complete)
Your instructor might provide you with ears of corn or flats of F 2 seedlings
and request that you determine the number of individuals having the various
phenotypes and then to interpret the data in terms of mendelian principles.
For example, suppose you counted 40 green and 12 albino seedlings in an F 2
population.
How would you interpret these data?
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Ears bearing F2 kernels can be purchased from a number of biological supply
companies. Without removing the kernels from the ears, you can count the
number of kernels in the different phenotypes, record data in table 1.3, and
then formulate hypotheses to explain the data collected. Two possible
illustrations of endosperm traits follow.
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Parents
1- SuSu
Starchy
(smooth)
x susu
sweet
(wrinkled)
Parents
2- CC
x
Pigmented
(Purple)
cc
non pigmented
(White)
F1
Phenotype…………………………………
Genotype……………………………….…
F1
Phenotype………………………………
Genotype……………………………….
F2
Phenotype…………………………………
Genotype………………………………….
F2
Phenotype……………………………….
Genotype…………………………...........
What F2 phenotypic ratio do you expect to obtain?
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What F2 genotypic ratio is expected?
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Which trait is dominant in your experiment?
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How do you know?
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Chapter three
Chi square test
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Chi square test
2
The purpose of the chi square X test is to determine whether
experimentally obtained data constitute a good fit to a theoretical,
2
expected ratio. In other words, the X test enables one to determine
whether it is reasonable to attribute deviations from an expected value to
chance. Obviously, if deviations are small then they can be more reasonably
attributed to chance.
The question is how small must deviations are in order to be attributed to
chance?
2
The formula for X is as follows:
∑ (O-E)2 /E
Where O = the observed number of individuals of a particular phenotype, E =
the expected number in that phenotype, and ∑ = the summation of all
possible values of (O-E)2 /E for the various phenotypic categories.
2
The following is an example of how one might apply the X test to genetics.
In a cross of tall tomato plants to dwarf ones, the F1 consisted entirely of
tall plants, and the F2 consisted of 102 tall and 44 dwarf plants. Does this
2
data fit a 3:1 ratio? To answer this question, X value can be calculated (see
table 1.4).
2
The calculated X value is 2.0548. What does this mean? If the observed
2
had equaled the expected, the value would have been zero. Thus, a small X
value indicates that the observed and expected ratios are in close
agreement. However, are the observed deviations within the limits expected
by chance? In order to determine this, one must look up the X2value on a chi
square table (see table 1.5). Statisticians have generally agreed on the
arbitrary limits of odd of 1 chance in 20 (probability = .05) for drawing the
line between acceptance and rejection of the hypothesis as a satisfactory
explanation of the data tested. A X2value of 3.841 for a two-term ratio
corresponds to a probability of .05. That means that one would obtain X2
value of 3.841 due to chance alone on only about 5% of similar trials.
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When X2exceeds 3.841 for a two term ratio, the probability that the
deviations can be attributed to chance alone is less than 1 in 20.
Thus, the hypothesis of compatibility between the observed and expected
ratios must be rejected. In the example given the X2 value is much less than
3.841. Therefore, one can attribute the deviations to chance alone.
Notice that across table 1.5 are probability (P) values and down the side are
degrees of freedom (df) values. The number of degrees of freedom is one
minus the number of terms in the ratio. In the example above (3:1) there are
two terms. Therefore, the degrees of freedom is 2 - 1 = 1. Thus, on the 1
degree of freedom row and under the .05 column the X2value of 3.841 is
found. In the example given, the X2 value obtained is 2.0548. Looking on the
1 degree of freedom row, one finds that value between the .05 (X2= 3.841)
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and the .20 (X2= 1.642) probability columns. This means that the probability
that the deviations occurred by chance alone is between 5% and 20%. In
other words, if you were to repeat this experiment 100 times, you would
expect to observe deviations at least as large as you obtained in between 5
and 20 of the 100 experiments due to chance alone.
Important points of chi-square
It must be used only on the numerical data itself, never on percentage or
ratios derived from the data. Two mathematical chick points are always
percent in chi-square calculations:
(1) The total of the expected column must equal the total observations and
(2) The sum of the deviations should equal 0 the squaring of negative
deviations convert all values to a positive scale.
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Application of the Chi Square Test
 First experiment
Transfer the F2 data obtained in the first experiment (see Table 1.2) to
table 1.6.Calculate X2for the total of males and females based on the
hypothesis that the classical monohybride F2 ratio has been obtained.
Table 1.6 Calculation of x2 on Data from experiment one.
Phenotypes
Ratio%
O
E
O-E
(O-E)2
(O-E)2/E
Totals
X2=……
1. How many degrees, of freedom do you have in the interruption of the
X2value?
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2. Using Table 1.5 determine what X2values lie on either side of the X2
calculated in Table 1.6 record these values here……………………………………………...
3. What probability values are associated with these table values of X2
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4. In this case, do you accept or reject the hypothesis that these data
approximate a 3:1 ratio?
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 Second experiment
Transfer the F2 data obtained in the second experiment to table
1.3.Calculate X2for the hypothesis that the classical monohybride F2 ratio
has been obtained.
Table 1.3 Calculation of x2 on Data from experiment two.
Phenotypes
Ratio %
O
E
O-E
(O-E)2
(O-E)2/E
Totals
X2=……….
1. How many degrees, of freedom do you have in the interruption of the
X2value?
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2. Using Table 1.5 determine what X2values lie on either side of the X2
calculated in Table 1.3 record these values here………………………………………………
3. What probability values are associated with these table values of X2
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
4. In this case, do you accept or reject the hypothesis that these data
approximate a 3:1 ratio?
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
28 | P a g e
Chapter four: 2nd Mendelian law
29 | P a g e
Drosophila and Maize Experiments in Genetics:
Dihybrid Crosses
In this exercise, you will investigate the inheritance of traits in D.
melanogaster, and maize using the Mendelian model to develop your
hypotheses. You will work with dihybridd crosses to determine the
genotypes of the F1 and F2 generations and the phenotypes.
 Objectives of this investigation
Upon completion of this lab, you should be able to:



Predict the offspring of a dihybrid, cross.
Determine the parents of a designed cross from the ratios of
observed offspring.
Calculate phenotypic ratios and carry out Chi-squared analyses for
each cross.
Background
In considering two traits simultaneously, Mendel postulated second law of
independent assortment of the genes. Mendel studied seven different traits
in the garden pea, and, in the entire situation investigated, he obtained
dihybride F2 ratio of 9:3:3:3:1and dihybride test cross ratio of 1:1:1:1:1. This
law states that allele pairs separate independently during the formation of
gametes. Therefore, traits are transmitted to offspring independently of
one another.
What is the mechanism of independent assortment?
Genes located on separate pairs of chromosomes assort independently of one
another .Thus ,one might assume that the seven different traits studied by
Mendel were controlled by seven gene loci located –one locus per
chromosome pair-on the seven chromosome pairs of the garden pea .This ,in
fact, is not the case ,as we documented by Novitski and Blixt
(1978) .Because of recombination .genes located on the same chromosome
pair but at great distance from one another ,will also assort indecently of
one another .
30 | P a g e
Investigation of other organism other than garden pea revealed that
independent assortment, occurs in them also,Figure (2.1) shows an ear of F2
Kernels of corn :the Kernel are distributed according to the law of
independent assortment ,with (A)9/16 of the Kernels being colored with
starchy (smooth)endosperm, (B)3/16 being colored with sugary (wrinkled)
endosperm .(C)3/16 being colorless starchy and,(D)1/16 being colorless
sugary.
Figure 2.1
Third Experiment:
Dihydrid cross in Drosophila
1.The gene for vestigial wings (vg) is located on the second chromosome,
and the gene for ebony body color (e) is located on the third
chromosome. Mating may be made between flies from this two stocks .Be
certain to use virgin females flies for this cross. Reciprocal crosses ,that
is , after about 8 days remove and discard the parent flies record all
data relative to thisexperiment
2. Predict the results in the F1 and F2generations from the mating
prepared
F1 phenotype(s): ………………………………………………………………………………………………...
F2 phenotypes and frequencies
………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………
31 | P a g e
3. When the F1 flies appear in the culture bottles, check to see if your
expectations have been obtained .Record data relative to the F1 flies in
Table 2.1. Mate F1 males and F1 females in new culture bottles. After
about 8 days remove and discard the F1 flies
4. When the F2 adult flies emerge in the culture bottle, etherize and
classify them for the various characteristics involved in this cross
.Record data in Table 2.1
Table 2.1 Record of drosophila Experiment
Record both the phenotype and stock number of the parent flies.
Experiment number …………………………….. Name……………..
1. Cross …………………………… Female x ………………………
2. Date P1 s mated …………………………………………………..
3. Date P1 s removed…………………………………………………
4. Date F1 s first appeared ……………………………………………
5. Phenotype of F1 males …………………………………………….
6. Phenotype of F1 females …………………………………………..
7. Date of male and female placed in fresh bottle ……………………..
8. Date F1 flies removed …………………………………………
9. Date F2 progeny appeared …………………………………
10. Record F2 in the following table
32 | P a g e
Male
Phenotype
a.
b.
c.
d.
e.
f.
g.
h.
Female
Number
………….
........
........
........
........
........
........
…………..
……………..
..........
.........
.........
.........
.........
.........
…………….
Totals
Phenotype
Number
Total
…………….
.........
.........
.........
.........
.........
…………….
…………….
……………
.......
.......
.......
.......
.......
………….
……………
…………
......
.......
.......
......
......
……….
…….
……………….
…………..
………..
4.-Indicate the genotype and phenotype of the flies in each of the three
generations.
P1…………………………………………………………………………
F1…………………………………………………………………………
F2………………………………………………………………………..
5-Calculate the expected number of individuals in each of the phenotypic
categories and calculate the deviation (differences) between observed
(o) and expected (E) numbers. Record this data in Table 2.2
Table 2.2 Summary of Data and Calculation for the second drosophila
Experiment
Phenotypes
Totals
33 | P a g e
Ratio%
O
E
O-E
(O-E)2
(O-E)2/E
X2=……
How many degrees, of freedom do you have in the interruption of the
X2value?
……………………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………………..
2. Using Table 1.5 determine what X2values lie on either side of the X2
calculated in Table 1.3 record these values here………………………………………………
3. What probability values are associated with these table values of X2
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
4. In this case, do you accept or reject the hypothesis that these data
approximate a 3:1 ratio?
………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………
34 | P a g e
Fourth Experiment:
Dihydrid cross in maize
You will be given an envelope of F2 maize kernels or an intact ear selected by
your instructor from one of the alternative crosses listed in the instructor's
manual.
1-Classify the kernels into four different phenotypes.
2-Select gene symbol to represent the two alleles of each gene
studied in this example
Gene symbol
Phenotype
…………………………………………..
……………………………………………………
………………………………………….
……………………………………………………
…………………………………………..
……………………………………………………
…………………………………………..
……………………………………………………
A. Which characteristics are dependents on dominant alleles?
……………………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………………..
3. Now, using the gene symbol you have chosen, give the possible genotype of
the original homozygous parents (P1) of this F2 generation.
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
a. What are the phenotypes of these parents' individuals?
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
b. Assuming the parents to be homozygous, could more than one original
cross have been used to produce this F2 generation? If so, what crosses
could have been made? (Give genotype and phenotype)
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
35 | P a g e
4. In any case, what must have been the genotype and the phenotype of the
F1 generation that subsequently was self-pollinated to produce the F2
kernels with you have been working?
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………………….
5. Calculate the number of kernels expected in each phenotypic category and
the deviation between observed and expected (O-E). Record this data in
Table 2.3
Table 2.3 Summary of Data and Calculation for the second Maize
Experiment
Phenotypes
Totals
Ratio%
O
E
O-E
(O-E)2
(O-E)2/E
X2=………
In this case, do you accept or reject the hypothesis that these data
approximate a 9:3:3:1 ratio?
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
36 | P a g e
Chapter five: Gene interaction and epistasis
37 | P a g e
Gene interaction in maize
In this exercise, you will investigate the genes interaction in maize. You will
work with dihybrid crosses to determine the genotypes of the F1 and F2
generations and the phenotypes
Objectives of this investigation
Upon completion of this lab, you should be able to:
 Determine how some of these genes interact to determine endosperm
color.
 Calculate phenotypic ratios and carry out Chi-squared analyses for
each cross.
Background
Numerous gene loci scattered over the 10 pairs of chromosomes of maize
determine endosperm color as well as other endosperm characterics . The
interaction between two or more genes to control a single phenotype in which
one allele of the first gene becomes (epistatic) (i.e. the gene or locus that
suppressed or masked the action of a gene at the second locus) the gene or
locus suppressed was called (hypostatic)
The classical phenotype ratio of 9:3:3:1 observed in progeny of dihybride
parents becomes modified by epistasis into ratio that are various
combination of the 9:3:3:1 groupings s seen in the table below
Type of
epistasis
Genotype
aabb
aaB-
None
1
Recessive
4
Dominant
1
Duplicate
recessive
7
Duplicate
dominant
1
15
Incomplete
duplicate
dominant
1
6
38 | P a g e
3
3
Types of epitasis Ratio
A-bb
A-B-
3
9
9:3:3:1
3
9
9:3:4
12
12:3:1
9
9:7
15:1
9
9:6:1
Dominant
and
recessive
39 | P a g e
3
13
13:3
Fifth Experiment:
Gene interaction in maize
You will be given an ear of corn containing F2 kernels of two or more colors.
Count the number of F2 kernels of each color. From the F2 data and from the
information provided about the P1 and F1 phenotypes, determine the
mechanisms of inheritance that account for the data.
Among the possible kinds of ears of corn your instructor may ask you to
study the following characters
Phenotype of the truebreeding parents
Purple x White
Yellow x Yellow
White x white
Purple x yellow
…………………………
Phenotype of the F1
kernels
purple
yellow
purple
purple
………………………
Phenotype of the F2
kernels
Purple, red, white
Yellow, purple
Purple, white
Purple, yellow ,whit
………………………………
Your instructor will give you an ear of corn and tell you which of the four
kinds it is.
1. Record the number of the ear here
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
2. Record the phenotypes of the true-breeding parental generation here.
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
3. Record the phenotype of the F1 kernels here.
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
40 | P a g e
4. Count the F2 kernels on the ear and record their phenotypes, numbers,
and ratios in the following spaces
Phenotype
Number
Ratio suggested by F2
data
Total
5. How do genes interact to produce such epistasic ratio as 9:3:4? The
following account explains the inheritance of mouse coat color: a truebreeding agouti (wild-type) mouse genotype CCAA, and a true breeding albino
may have the genotype ccaa. If mated, two such animals would produce F1
offspring having the genotype CcAa and the agouti phenotype .In the F2
generation produced from crossing the F1 among themselves, one would
expect 9 C - A- : 3C- aa : 3cc A- : 1ccaa. The 9/16 that are C-A- will be
agouti, and the 1/16 that is ccaa will be albino, but what about 3/16 C-aa and
the 3/16 ccA-? Thos having the ccA- genotype will also be albino because
they are homozygous recessive cc, a genotype that prevents them from
producing any pigment whatsoever .Those individuals constituting the 3/16
C-aa will be able to make pigment because they have the dominant C allele
,but the pigment that they produce will be black because they are
homozygous recessive aa .In summary ,9/ 16 C-A- can make agouti
pigment,3/16 C-aa make black pigment ,and 4/16 (=3/16ccA-plus 1/16
ccaa)can make no pigment at all and are thus albinos.
6. Now ,attempt to devise a hypothesis to explain the mechanism of
inheritance of endosperm color for the ear of corn you have been studying
.In doing so ,answer the following questions
a) What ratio do the F2 data approximate?
…................................................................................................................................
...................................................................................................................................
..................................................................................................................................
41 | P a g e
b) How many gene loci appear to determine the F2 phenotypes?
………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………
c) Suggest gene symbols to be used in explaining the data .Indicate how
each gene and its alleles are functioning to regulate endosperm color by
completing the following table:
Gene symbol
Gene effect
……………………….
…………………………..
……………………….
……………..……………
……………………….
…………….…………….
……………….………
…………………………..
d) Now, using the Parental gene symbols you have devised, write the
following genotypes
Parental generation
Phenotype
Parent no.1
………………………
Parent no .2
………………………
F1 generation
Phenotype
………………………
F2 generation
Phenotype
…………………………
…………………………
…………………………
Genotype
………………….
…………………..
Genotype
…………………..
Genotype
…………………..
…………………..
…………………..
e) You have already suggested a ratio based on your F2 data. How closely
do the data approximate the ratio you have hypothesized? To answer the
42 | P a g e
question, calculate the expected number (E) for each F2 phenotype based
on the ratio you have hypothesized and the total number of F2 kernels
you have counted. Record your calculations in the following table.
F2Phenotypes
Observed nos. (O)
Expected nos.(E)
……………….
……………………….
……………….
………………………..
……………………….
……………….
………………………..
………………………..
Totals
……………………….
………………………..
………………………….
If the expected numbers closely approximate the observed numbers,
we conclude that the data to be satisfactorily explained by the
proposed hypothesis.
Phenotypes
Ratio%
Totals
O
E
O-E
(O-E)2
(O-E)2/E
X2=……….
Your comment
………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………….
43 | P a g e
Chapter sex: sex linkage
44 | P a g e
Sex chromosomes and Gene transmission
Cytology came to be better understood in the late nineteenth and early
twentieth centuries, the role that chromosomes play in sex determination
became more apparent .The presence of a particular nuclear structure (the
x body) was detected in half the sperms of certain insects as early as 1891.
Later (1902), C.E. McClung studied grasshopper species and demonstrated
the difference between chromosome constitution of males and females:
males – always have one less chromosome than females. Grasshopper sperm
cells were shown to be of two types, those having darkly staining
(heterochromatic) chromosome (the X body) and those not having such a
chromosome. This heterochromatic chromosome came to be known as the X
(sex-determining) chromosome.
Finally, it became clear that female grasshoppers have two X chromosomes
per somatic cell (XX), whereas males have only one (XO). The sex of the
offspring is determined by the kind of the sperm that fertilizes the ovum;
X-bearing sperm cell produce female offspring, and the sperm lacking X
chromosome result in male progeny. Subsequent research revealed that this
XO system of sex determination is, in fact, less common than the XY method
of sex determination. The latter process involves two heteromorphic, not
completely homologous, sex chromosomes. Half the sperm cells produced per
the X chromosome are female determining and half bear the Y and are male
determining. Both Drosphila and
humans
have
the
XY method of
determination. Research, however, has shown some fundamental differences
in the basic genetics of sex determination in these two species despite their
similar chromosomal basis for sex differences. In 1916, C.B Bridges
reported that sex in Drosophila is determined by a balance between femaledetermining genes located on the X chromosome and male determining gene
located on the various autosomes. In humans and perhaps in mammals
generally, the male-determining genes are located on the Y chromosome.
In 1910, Thomas Hunt Morgan studied the white eye mutant gene in
Drosophila in the first investigation to provide extensive experimental
45 | P a g e
evidence for X-linked inheritance. Prior to that time, relationships between
sex and gene transmission (e.g., the relationship noted since biblical times
for hemophilia) had been observed, but no specific mechanism explaining the
relationship had been set forth. Classical hemophilia, of course, is now
understood to be due to an X-linked recessive gene. X-linked gene exhibit a
crisscross pattern of inheritance in which the gene is transmitted by a male
to all of his daughters and from them to approximately on-half of his
grandsons.
Objectives of the investigation
Open completion of this investigation, the student is able to:


Diagram how an x-linked is transmitted from parents to F1 and F2
generations in an experimental mating of Drosophila.
Trace the transmission of an x-linked gene in a human pedigree.
………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………….
………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………….
………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………….
46 | P a g e
Sixth Experiment:
Sex linkage in Drosophila
For this part of the investigation you may use any one of many sex-linked
mutants of Drosophila crosses so following discussion is based on the
assumption that you will use the x-linked white eye (w) mutant.
Different members of the class will be asked to do reciprocal crosses so
that the results can be compared .Review the procedures for making
Drosophila mating (see investigation1)
A. Cross
Some of you will mate a white-eyed virgin female fly (Xw Xw ) to wild
type male (X+Y).Produce F1 and F2 generations .Record data in Table
9.1
B. Reciprocal cross
Others will mate awild-type virgin female (X+X+) To white eyed male
(XwY).Produce F1 and F2 generations. Record data in Table 9.1.
Now, compare the F1 and F2 phenotypes and genotypes of the
reciprocal crosses by completing Table 9.2.
Write a brief summary paragraph describing how the trait you have been
studying (white eyes) would be inherited in reciprocal crosses if it were
controlled not by a gene that is X-linked but rather by one that is
autosomes .
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………………….
47 | P a g e
Table 9.1 Records of Drosophila Experiment
Record data of cross involving a sex-linked gene
Experiment number…………………..Name……………………………………
1. Cross………………….…………..Female X……………………………Male.
2. Date P1s mated………………………………………………………………..
3. Date P1 s removed…………………………………………………………….
4. Date F1 s First appeared………………………………………………………
5. Phenotype of F1 female ………………………………………………………
6. Phenotype of F1 male………………………………………………………….
7. Date F1 male and female placed in fresh bottle ……………………………….
8. Date F1 flies removed …………………………………………………………
9. Date F2 progeny appeared…………………………………………………….
10. Record F2 data in the following table:
Males
Phenotype
a………………………..
b………………………..
c………………………..
d………………………..
Totals
48 | P a g e
Females
Number
……………………..
………………………
………………………
………………………
……………………..
Phenotype
Number
……………
……….
……………. ……….
…………… ……….
…………. ……….
……............
Total
……...
………
………
………
………
Table 9.2 A comparison of the results of reciprocal Crosses involving an Xlinked Gene
Cross (A)
Reciprocal Cross (B)
Phenotype
P1 female
P1male
F1 female
F1 male
F2 female
F2 male
49 | P a g e
Genotype
………….
………….
………….
………….
………….
………….
……………...
……………...
……………...
……………...
……………...
……………...
Phenotype
...…………...
...…………...
...…………...
...…………...
...…………...
...…………...
Genotype
……………..
……………..
……………..
……………..
……………..
……………..
Chapter seven: Pedigree Analysis
50 | P a g e
Pedigree Analysis
Introduction
A pedigree is a diagram of family relationships that uses symbols to represent
people and lines to represent genetic relationships. These diagrams make it easier
to visualize relationships within families, particularly large extended families.
Pedigrees are often used to determine the mode of inheritance (dominant,
recessive, etc.) of genetic diseases. A sample pedigree is below.
In a pedigree, squares represent males and circles represent females. Horizontal
lines connecting a male and female represent mating. Vertical lines extending
downward from a couple represent their children. Subsequent generations are
therefore written underneath the parental generations and the oldest individuals
are found at the top of the pedigree.
If the purpose of a pedigree is to analyze the pattern of inheritance of a
particular trait, it is customary to shade in the symbol of all individuals that
possess this trait.
In the pedigree above, the grandparents had two children, a son and a daughter.
The son had the trait in question. One of his four children also had the trait.
In the exercises below, assume that the trait in question is a genetic disease or
abnormality. We will learn patterns of inheritance that have the following modes of
inheritance:



Autosomal dominant.
Autosomal recessive.
X-linked recessive.
51 | P a g e
Developing Conclusions About Different Modes of Inheritance
Autosomal Dominant
1. The pedigree below is for a genetic disease or abnormality. We do not yet know
if it is dominant or recessive. We will determine if it is possible that the trait is
autosomal dominant. If the trait were dominant, we would use the following
designations:
A = the trait (a genetic disease or abnormality, dominant)
a = normal (recessive)
If the trait were recessive, we would use the following designations:
A = normal (dominant)
a = the trait (a genetic disease or abnormality, recessive)
a) Assume for the moment that the trait is dominant (we don't know yet). The
pedigree shows that three of the individuals have the recessive (normal) phenotype
and one individual has the dominant (abnormal) phenotype. Write the genotype of
the affected (abnormal) individual next to her symbol in the pedigree below. If you
only know one of the genes (letters), use a ?" for the unknown letter. Write the
genotype of the three recessive individuals next to their symbols. As you write the
genotypes, keep in mind that the pedigree may not be possible for a dominant trait.
b) Is it possible that the pedigree above is for an autosomal dominant trait?
……………………………………………………………………………………………………………………………………………
c) Write the genotypes next to the symbol for each person in the pedigree below
assuming that it is for a dominant trait.
…………………………………………………………………………………………………………………………………………………….
52 | P a g e
d) Is it possible that this pedigree is for an autosomal dominant trait?
…………………………………………………………………………………………………………………………………………………..
e) What can you conclude from these two examples about the parents of a child
that has a dominant characteristic?
............................................................................................................................................................
............................................................................................................................................................
............................................................................................................................................................
2. We will determine if the pedigree below can be for a trait that is autosomal
dominant. Use "A" and "a" as you did for the pedigrees above.
a) Write the genotype of each individual next to the symbol.
………………………………………………………………………………………………………………………………………………………..
b) Is it possible that this pedigree is for an autosomal dominant trait?
………………………………………………………………………………………………………………………………………………….
c) In conclusion, can two individuals that have an autosomal dominant trait have
unaffected children?
………………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………………
53 | P a g e
Autosomal Recessive
3. We will determine if the pedigree below can be for a trait that is autosomal
recessive. Use the following designations:
A = normal
a = the trait (a genetic disease or abnormality)
a) Assuming that the trait is recessive, write the genotype of each individual next
to the symbol.
………………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………………….
b) Is it possible that the pedigree above is for an autosomal recessive trait?
………………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………………
c) Assuming that the pedigree below is for a recessive trait, write the genotype
next to the symbol for each person.
…………………………………………………………………………………………………........................................................
............................................................................................................................................................
d) Is it possible that this pedigree is for an autosomal recessive trait?
………………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………………
54 | P a g e
e) If a trait is autosomal recessive, what can you conclude about the children if
both parents are affected?
………………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………………
4. We will determine if the pedigree below can be for a trait that is autosomal
recessive. Use "A" and "a" as you did for the previous example.
a) Write the genotype of each individual next to the symbol.
………………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………………
b) Is it possible that this pedigree is for an autosomal recessive trait?
………………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………………….
c) If a trait is autosomal recessive, what can you conclude about the children of
two parents that are not affected?
………………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………………..
5. We will determine if the pedigree below can be for a trait that is autosomal
recessive.
a) Write the genotype of each individual next to the symbol.
………………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………………….
55 | P a g e
b) Is it possible that this pedigree is for an autosomal recessive trait?
………………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………………
c) In this pedigree, two generations have been skipped. What can you conclude
about recessive traits skipping generations?
………………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………………..
X-Linked Recessive
The conclusions that you made for autosomal recessive traits apply to X-linked
traits. In this exercise, we will work on some additional conclusions because males
have only one X chromosome and females have two.
6. We will determine if the pedigrees below can be for a trait that is X-linked
recessive. Use the following designations:
XA = normal
Xa = the trait (a genetic disease or abnormality)
Y = Y chromosome (males only)
a) Write the genotype of each individual next to the symbol.
………………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………………..
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b) Is it possible that the pedigree above is for an X-linked recessive trait?
………………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………………
c) Write the genotype next to the symbol for each person in the pedigree below.
………………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………………..
d) Is it possible that this pedigree is for an X-linked recessive trait?
………………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………………….
e) Write the genotype next to the symbol for each person in the pedigree below.
………………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………………
f) Is it possible that this pedigree is for an X-linked recessive trait?
………………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………………...
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g) Write the genotype next to the symbol for each person in the pedigree below.
………………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………………….
h) Is it possible that this pedigree is for an X-linked recessive trait?
………………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………………….
i) What can you conclude about the children of mothers affected with an X-linked
recessive characteristic?
………………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………… ……………………..
7. We will determine if the pedigree below can be for a trait that is X-linked
recessive. We will continue to use the designations "XA and Xa".
a) Write the genotype of each individual next to the symbol.
………………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………………
b) Is it possible that this pedigree is for an X-linked recessive trait?
………………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………………
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c) Which parent did the son get the Xa gene from?
………………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………………..
d) What can you conclude about father-to-son transmission of X-linked traits?
………………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………………..
8. We will determine if the pedigree below can be for a trait that is X-linked
recessive.
a) Write the genotype of each individual next to the symbol.
………………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………………...
b) Is it possible that this pedigree is for an X-linked recessive trait?
………………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………………...
c) What can you conclude about the children if both parents are affected with an
X-linked recessive trait?
………………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………………...
d) How does this conclusion compare with the one you made earlier if about both
parents being affected by an autosomal recessive trait?
………………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………………
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e) Do the conclusions that you made for autosomal recessive traits apply to Xlinked recessive traits?
………………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………………
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Glossary
A dihybrid cross -- is a breeding experiment between P generation (parental
generation) organisms that differ in two traits.
A monohybrid cross -- is a breeding experiment between P generation (parental
generation) organisms that differ in one trait.
An allele -- is an alternative form of a gene (one member of a pair) that is located
at a specific position on a specific chromosome.
Autosome -- a nuclear chromosome other than the X- and Y-chromosomes.
Centromere -- a region of a chromosome to which spindle traction fibers attach
during mitosis and meiosis; the position of the centromere determines whether the
chromosome is considered an acrocentric, metacentric or telomeric chromosome.
Chromosome -- in the eukaryotic nucleus, one of the threadlike structures
consisting of chromatin and carry genetic information arranged in a linear sequence.
Dominance -- An allele or corresponding phenotypic trait that is expressed in
heterozygotes .
Dominant -- alleles that determine the phenotype displayed in a heterozygote with
another (recessive) allele.
Gene -- The fundamental physical and functional unit of heredity, which carries
information from one generation to the next; a segment of DNA, composed of a
transcribed region and a regulatory sequence that makes possible transcription.
Genotype -- genetic constitution of an organism.
Hemizyg -- A male organism in which a mutant trait is present at the X chromosome
(i.e. Drosophila sp).
Heterozygote -- having two alleles that are different for a given gene.
Homologous chromosomes -- chromosomes that pair during meiosis; each homologue
is a duplicate of one chromosome from each parent.
Homozygote -- having identical alleles at one or more loci in homologous
chromosome segments.
Hybrid -- An offspring resulting from mating between individuals of different
genetic constitution.
Incomplete dominance -- is a form of intermediate inheritance in which one allele
for a specific trait is not completely dominant over the other allele.
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Linkage -- analysis of pedigree the tracking of a gene through a family by following
the inheritance of a (closely associated) gene or trait and a DNA marker.
Linkage -- the greater association in inheritance of two or more nonallelic genes
than is to be expected from independent assortment; genes are linked because they
reside on the same chromosome.
Locus --The particular physical location on the chromosome of which the gene for a
given trait occurs.
Pedigree -- a diagram of the heredity of a particular trait through many
generations of a family.
Phenotype -- observable characteristics of an organism produced by the organism's
genotype interacting with the environment.
Recessive -- a gene that is phenotypically manifest in the homozygous state but is
masked in the presence of a dominant allele.
X-Chromosome -- The sex chromosome found in two doses in female mammals and
many other species.
Y-Chromosome -- The sex chromosome found in a single dose in male mammals and
many other species.
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