GENERAL BIOLOGY LAB 1 (BSC1010L)
Lab #8: Mendelian Genetics
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OBJECTIVES:
 Understand Mendel’s laws of segregation and independent assortment.
 Differentiate between an organism’s genotype and phenotype.
 Recognize different patterns of inheritance.
 Perform monohybrid and dihybrid crosses.
 Use pedigree analysis to identify inheritance patterns.
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INTRODUCTION:
Through his studies of the inheritance patterns of the garden pea, Pisum sativum, Gregor
Mendel changed our understanding of heredity. Mendel studied characters/traits that differed
between plants and designed cross-fertilization experiments to understand how these characters
transmit to the next generation. The results of Mendel’s work refuted the prevailing hypothesis
of blending inheritance and provided a new framework for understanding genetics. Ultimately,
Mendel postulated two laws to explain heredity: (1) the law of segregation and (2) the law of
independent assortment.
Monohybrid crosses and the law of segregation
The law of segregation, also termed the “first law,” states that during gamete formation
the alternate forms of a gene (i.e. alleles) on a pair of chromosomes segregate randomly so that
each allele in the pair is received by a different gamete. For example, if you were to examine the
gene responsible for petal color, you may discover that the gene can be expressed as either
yellow or white flowers. In this scenario, the gene is petal color, while the alleles are yellow and
white. Depending on which allele is expressed, petal color will vary. Examine Figure 1 below
making sure that you can follow the path of each allele from parent to offspring.
Figure 1. Schematic of Mendel’s law of segregation
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In diploid organisms, all alleles exist in pairs; identical alleles within a pair are
homozygous, while different alleles are heterozygous. Allele forms are represented by a single
letter that explains whether a particular trait is dominant or recessive. Dominant alleles are
assigned an uppercase letter (E), while recessive alleles are lowercase (e).In general, a dominant
trait is expressed when at least one of the alleles present in the resulting allelic pair is dominant
(EE or Ee). In contrast, for a recessive trait to be expressed, both alleles within the pair must be
recessive (ee). For example, when considering ear lobe shape, two forms (attached and
unattached) are apparent (Fig. 2). This trait is regulated by a single gene where unattached ear
lobes are dominant (E) while attached ear lobes (e) are recessive.
Figure 2. (a) Unattached (EE or Ee) vs. (b) attached earlobes (ee)
An organism’s genotype (EE, Ee, ee) is the combination of alleles present whereas the
phenotype is the physical expression of the genotype. In the earlobe shape example above, an
individual can have a genotype of EE, Ee or ee. People with EE or Ee genotypes have the
unattached earlobe phenotype (Fig 2a), while those with an ee genotype express the attached
earlobe form (Fig 2b). Note that dominant traits can be either homozygous (EE) or heterozygous
(Ee) while recessive traits are always homozygous (ee).
Question:
Given that the allele for brown eyes (B) is dominant and the allele for blue eyes (b) is recessive,
which of the following genotypes would result in individuals with brown eyes? Which
genotype(s) is/are homozygous and which is/are heterozygous?
BB: __________________________________
Bb: ___________________________________
bb: ___________________________________
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In today’s lab you will use the concepts of Mendelian Genetics to solve problems regarding
inheritance.
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TASK 1 – Patterns of Inheritance I: Simple Dominance
Simple dominance is the term used to describe a common outcome of allelic
combinations, where one allele, if present, will dominate over the other and will be expressed.
Information about alleles present in a parental population can be used to determine the
probability of different genotypic and phenotypic ratios for a variety of traits in the offspring. In
instances when only 1 or 2 traits are being considered the Punnett square (Fig. 3) approach is
used to predict the possible outcomes of the parental cross. When only one trait is being
considered the cross is monohybrid while a dihybrid cross involves 2 traits.
General instructions on how to perform a cross using the Punnett square approach:
1.
2.
3.
4.
Write down the genotypes of the parents
Note the gametes that each parent can contribute
Draw a Punnett Square
Across the top write the gametes that one parent contributes and along the side write
the gametes contributed by the other parent
5. Perform the cross
6. Determine the genotypic and phenotypic ratios
Figure 3. Example of a Monohybrid cross
In the example above (Fig. 3), the genotypic ratio is 1:2:1 (1: CC, 2: Cc, 1: cc) while the
phenotypic ratio is 3:1. Since C = curly hair and c = straight hair, ¾ of the possible offspring will
have curly hair while only ¼ will have straight hair.
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Procedure:
1. You will now simulate a cross between two heterozygous individuals, Tt and Tt. Each
group should obtain two coins from your TA. You will flip the coins simultaneously to
represent the potential outcomes of a cross between two Tt individuals. A head represents
the dominant tall allele (T) while a tail symbolizes the recessive dwarf allele (t). Before
you begin flipping the coins, perform the Tt x Tt cross in the Punnett square below to
estimate the expected genotypic and phenotypic ratios.
Parent 2
Parent 1
Based on this cross, what do you anticipate the genotypic and phenotypic ratios to be?
Write your hypotheses (Ho and Ha) in Table 1.
Table 1:
Expected Genotypic Ratio
Expected Phenotypic Ratio
2. Begin flipping the two coins simultaneously for a total of 64 times. Record your results in
Table 2.
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Table 2:
Response
Number
TT
Tt
tt
Questions:
a. What ratio of allele combinations did you observe?
b. What genotypes and phenotypes result from these crosses?
c. What are the genotypic and phenotypic ratios?
d. How did your results compare to your expectations? Do your results support or reject
your null hypothesis?
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e. Do you think your results would have been closer if you flipped the coins 6400 times
instead of just 64? Why or why not?
3. Albinism, a recessively inherited trait, results in organisms that lack pigment in the skin,
hair or eyes. A female with normal pigmentation, but who had an albino mother, mates
with an albino male. They have one child. Using the information you have learned so far
complete Table 3.
Table 3:
Genotype of child’s mother
Genotype of child’s father
Possible gametes of mother
Possible gametes of father
Possible genotype and phenotype of the offspring
Genotypic ratio of children
Phenotypic ratio of children
TASK 2 - Patterns of inheritance II: Incomplete vs. Complete Dominance & Codominance
Inheritance of traits can occur in multiple forms. So far you have considered complete
dominance, where a homozygous dominant or a heterozygous individual expresses the dominant
phenotype, while an individual that is homozygous recessive expresses the recessive phenotype.
However, in certain cases a cross between two different allele forms results in a
phenotypic expression that combines the two allelic traits. This type of inheritance is known as
incomplete dominance. For example, if an offspring resulting from a cross between a red (RR)
and a white (rr) snapdragon plant receives the dominant allele for red flower color (R) from one
parent and the allele for white flower color (r) from the other, the resulting genotype will be Rr.
The heterozygous form (Rr) of the plant will bear pink flowers since neither allele is completely
dominant over the other (Fig. 4).
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Figure 4. Pink snapdragons are an example of incomplete dominance
1. Determine the possible phenotypes of the F1 offspring when two pink snapdragons are
crossed. Show your work in the space provided below.
Parent 2
Parent 1
2. What would be the resulting genotypes of a cross between a pink and a white
snapdragon? Show your work in the space provided below.
Parent 2
Parent 1
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Expression of both alleles of a particular gene is known as codominance. When alleles
are inherited codominantly, both phenotypes are expressed at the same time in the heterozygous
condition in contrast to incomplete dominance where the heterozygote is an intermediate
between the two homozygotes (Fig. 5).
Figure 5. Different types of inheritance
The ABO blood type system is an excellent example of codominance. Humans have four
blood types, namely, A, B, AB and O. All individuals carry two alleles, one from each parent. In
this system, both alleles inherited determine one’s blood type, where a person with Type AB
blood possesses phenotypic traits of both A and B blood types (Table 4).
Blood type (phenotype)
Type A
Type B
Type AB
Type O
Genotype
IAIA or IAi,
IBIB or IBi,
IAIB
ii
Table 4. Relationship between blood type and genotype
For example, an individual with Type B blood can have two possible genotypes, IBIB or
I i, where I (dominant) and i (recessive) represent an allele from each parent (Table 4). The
different blood types are characterized by the presence of a particular sugar molecule attached to
the proteins on the surface of red blood cells (Fig 4). In Type A blood, the attached sugar
molecule is galactosamine, while in Type B blood it is galactose. In contrast, individuals with
Type O blood, have no sugars present on the surface of their red blood cells. These protein-sugar
complexes are antigens that act as recognition markers for the immune system. The immune
system is tolerant to its own antigens but produces antibodies against antigens that differ from
B
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its own. The antibodies formed bind to the antigens causing agglutination (clumping) and lysis of
the foreign red blood cells. Therefore, an individual with Type A blood could not receive a blood
transfusion from a Type B blood donor because the antigens on the donor’s red blood cells will
trigger an immune response from the recipient’s antibodies. Thus, the Type A recipient will
produce antibodies against the donor’s Type B antigens.
Figure 6. Human ABO blood types
When the wrong blood type is given to a patient, agglutination of the blood occurs and
this can ultimately lead to death. Table 5 provides a quick overview of which blood types can
donate to which, and which types can receive from which. A plus indicates that mixing of the
donor and recipient blood types results in agglutination whereas a blank cell means that no
agglutination occurs when the blood types are mixed.
Recipient (Antibodies)
Donor
(Antigens)
A
A
B
+
AB
+
B
+
AB
O
+
+
+
+
O
Table 5. Effects of mixing different blood types
Questions:
1. Given the information above, why do you think that Type O is the universal donor
and AB is the universal recipient?
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2. There was a mix of up children in the maternity ward of a hospital. The children
in question and their blood types are listed below.
Child 1: type A (genotype IAIA or IAi)
Child 2: type B (genotype IBIB or IBi)
Child 3: type AB (genotype IAIB )
Child 4: type O (genotype ii)
Which child or children could belong to a couple having AB and O blood types?
3. Based on the previous question, is it possible to prove paternity based on blood types?
Explain.
4. A woman with Type O blood has a child with the same blood type. Can the child’s
father have Type AB blood? Why or why not?
Another trait involved in blood typing is the Rh factor. The Rh factor works along the
principle of simple dominance instead of codominance. An individual who is Rh positive
possesses the Rh antigen on his/her blood cells while someone who is Rh negative lacks Rh
antigens. Generally, the Rh status of an individual is always included with the blood type. For
example, a person that is AB+ has Type AB blood and is Rh positive. This information is
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particularly important during pregnancy since Rh incompatibility can develop in women that are
Rh- and who have an Rh+ baby. Mixing of maternal and fetal blood through the placenta can
cause the mother to develop antibodies against the Rh antigens from the baby. This condition is
usually not harmful to the first child but may cause mild to severe symptoms (depending on the
amount of Rh antibodies created) during subsequent pregnancies since the mother’s Rh
antibodies attack the Rh antigens of the developing fetus.
Determining blood type
Procedure:
1. Prepare your station by obtaining the following supplies:
a. 4 small plastic blood typing trays
b. Toothpicks for mixing
c. Four bottles of blood (on your table) representing four different
individuals
d. One bottle representing A antibodies
e. One bottle representing B antibodies
f. One bottle representing Rh antibodies
2. You will use each blood typing tray to determine the blood type of a particular
individual. Note that each tray contains 3 wells, labeled A, B and Rh.
3. Add 3 drops of blood from individual 1 to every well in a blood typing tray.
4. Add 3 drops from the bottle labeled A antibodies to the well labeled A.
5. Add 3 drops of B antibodies to the well labeled B.
6. Add 3 drops of Rh factor solution to the well labeled Rh.
7. Mix each well with a toothpick. Note: Use a different toothpick for each well
and tray.
8. After 1 min, examine the tray for the presence of crystals. Presence of
crystals, indicates agglutination and a positive test* for a particular blood type
and Rh factor.
9. Repeat steps 3-8 for the remaining individuals.
*Important note: In this particular experiment, agglutination indicates a positive test
for a particular blood type. For example, if agglutination occurs in well A, then the
individual has blood type A. However, when working with real blood (i.e. for the
purposes of transfusions), agglutination would be a negative result. With regular
blood, agglutination would indicate that the antibodies of one’s blood detected a
foreign substance (an antigen), causing an immune response and cell lysis.
10. Record your results in Table 6. Note which wells agglutination occurs in each
tray. Based on your results, determine the blood type of the four individuals
examined.
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Table 6:
Agglutination (Yes or No)
Individual Well A Well B
Rh factor (+/-)
Blood type
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TASK 3 - Patterns of Inheritance III: Lethal Inheritance
Lethal inheritance, the last pattern to be examined, occurs when an offspring dies after
acquiring a particular trait. An example of such a scenario would be albinism in plants.
1. Why is albinism fatal in plants?
2. Using the tray of plants in the front of the lab, count the number of green and albino plants
you observe. Using this information, determine the ratio of green to albino plants.
Green ________: Albino ________
3. Based on the above ratio, predict the genotypes of the parental generation.
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Dihybrid crosses and the law of independent assortment
Mendel’s “second law” or the law of independent assortment states that alleles from
different genes assort independently from one another during meiosis if they are located on
separate chromosomes (Fig. 7). For instance, alleles for the attached earlobe gene will assort
independently from those that determine height as long as they are located on different
chromosomes. This seems easy enough to grasp but is often the bane of biology students. Think
of it like this - if you are flipping a coin and recording the results of each toss (heads or tails),
and your friend is doing the same thing, do your tosses have any effect on the results of his? No,
because your two actions are not linked. The chance of getting any two events to happen at the
same time is simply the product of the chance of each event happening at all. For example, the
chance of getting a “heads” in a coin flip is 50%. However, the chance that you and your friend
get heads on the same toss is .5 x .5 = .25 or 25%.
Figure 7. Schematic of Mendel’s law of independent assortment
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TASK 4 - Drosophila Virtual Genetics Lab
Procedure:
1. Click to open the Drosophila Genetics Lab Program on the desktop.
2. Each group will perform one of the Single Gene Inheritance experiments (1, 2 or 3) per
group.
3. Follow the instructions listed in the pdf under "Procedure" and then complete the
"Results" and "Questions" sections.
4. Once completed, perform one of the Double Gene Inheritance experiments (4 or 5) and
repeat step 3.
5. Lastly, perform experiment 15 on Lethal Inheritance and repeat step 3.
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TASK 5 - Testing Mendel’s laws of inheritance in Brassica rapa
Throughout the semester you have been growing the rapid-cycling Brassica rapa plants.
As you noted in previous labs, your plants have different phenotypes, i.e., some have purple
stems and green leaves while others have green stems and yellow-green leaves. In fast plants, the
presence of the anthocyanin pigment produces a purple color (P) that is apparent on the stems
and leaf tips of the developing plant. In the absence of this pigment, the stems and leaves are
bright green (p) in color. The second trait, yellow-green leaf color (y), being observed by some
of you is the result of a recessive gene, that when present in the homozygous condition, produces
leaves, stems and seed pods that are yellow-green in color. In contrast, plants that are
homozygous dominant (YY) or heterozygous (Yy) for this trait produce leaves and stems that are
green in color. The seeds you planted during the last lab period are the F2 generation that
resulted from a cross of your F1 plants (see Fig. 8 and 9).
The objective of this task is to investigate the pattern of gene inheritance (i.e. dominant
vs. recessive) as well as Mendel’s laws of segregation and independent assortment. You will also
learn how to use a statistical test to examine how well the inheritance patterns observed in class
agree with expected statistical ratios.
Note: Before beginning your analyses, you will need to determine whether you are investigating
single-gene (Fig. 8) or double gene (Fig. 9) inheritance patterns. This semester two separate
types of F1 plants were grown and since both inheritance patterns can produce plants that bear
purple stems and green leaves, you will need to check for the presence of trichomes (hairs) on
the stem. Since the number of hairs is controlled by more than one gene, we will not consider
this trait in the current exercise. Therefore if your plant’s stem is hairy, then your investigation
will be of monohybrid inheritance patterns since only the anthocyanin gene will be considered
(see Fig 8).
Figure 8. Possible F2 Genotypes and phenotypes for the Anthocyanin gene
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Figure 9. Possible F2 Genotypes and phenotypes for the Anthocyanin and YellowGreen Leaf genes
1. Based on your records, what were the parental phenotypes (F1) of your group’s plants?
i. What are the possible genotypes that produced these phenotypes?
2. Examine the seedlings in your group and record the following information in Table 7.
i. phenotypes of the plants in your group
ii. genotype for each phenotype observed (see Fig. 8 and 9)
iii. quantity of each genotype
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Table 7:
Group
#
# seedlings
Phenotypes present
(can be > 1)
Genotype
Quantity of
Genotype
1
2
3
4
5
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3. Combine your results with the data from the rest of the class in Table 8.
Table 8:
Phenotype
Quantity
Purple stem, green leaf
Purple stem, yellow-green leaf
Non-purple stem, green leaf
Non-purple stem, yellow-green leaf
Total number of plants
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4. Depending on which inheritance pattern you are investigating, calculate the expected
phenotypic ratios of the F2 population resulting from the cross of F1 individuals, either Pp
x Pp or Ppyy x PpYy.
Parent 2
Parent 1
Parent 2
Parent 1
5. Based on your expected ratios formulate null and alternate hypotheses (Ho and Ha)
regarding the phenotypic pattern you should observe in your F2 plants.
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6. Chi-Square analyses: You will use this statistical test to examine how the predicted
phenotypic ratios of Brassica rapa compare to your observed data.
A test that is often applied to determine how well observed ratios fit expected statistical
ratios is the chi-square (χ2) or “goodness of fit” test. This test calculates the deviations of
observed numbers from expected numbers into a single numerical value called χ2. The difference
between the number observed and the number expected for a particular phenotype is squared and
then divided by the number expected. This is repeated for each phenotype class. The χ2 value
consists of the summation of these values for all classes. The formula for χ2 (Suzuki, et al., 1986)
is:
χ2 = ∑ (observed - expected) 2
expected
Degrees of
Freedom (df)
1
2
3
4
5
6
7
8
9
90%
0.016
0.211
0.584
1.064
1.610
2.204
2.833
3.490
4.168
Possibility of Chance Occurrence in Percentage
(5% or Less Considered Significant)
80% 70% 50%
30%
20%
10% 5% (sig.)
0.064 0.148 0.455 1.074 1.642 2.706
3.841
0.446 0.713 1.386 2.408 3.219 4.605
5.991
1.005 1.424 2.366 3.665 4.642 6.251
7.815
1.649 2.195 3.357 4.878 5.989 7.779
9.488
2.343 3.000 4.351 6.064 7.289 9.236 11.070
3.070 3.828 5.348 7.231 8.558 10.645 12.592
3.822 4.671 6.346 8.383 9.083 12.017 14.067
4.594 5.527 7.344 9.524 11.030 13.362 15.507
5.380 6.393 8.343 10.656 12.242 14.684 16.919
1%
6.635
9.210
11.341
13.277
15.086
16.812
18.475
20.090
21.666
Table 9. χ2 Values and Probabilities
Associated with each χ2 value is a probability that indicates the chance that, in repeated
experiments, deviations from the expected would be as large as or even larger than the ones
observed in this experiment. The calculated χ2 value can be used to obtain probabilities, or p
values, from a chi square table (Table 9). These probabilities indicate the likelihood that the
observed deviations are due to random chance alone. The most widely accepted p value is 0.05,
which means that in cases where the p < 0.05, the chance of obtaining the observed results by
chance alone is 5% or less. In general, low χ2 values indicate a high probability that the observed
deviations could be due to random chance alone, while high χ2 values indicate a low probability
that the observed deviations can be explained by chance events and the deviations are likely the
result of the experimental treatment. If the χ2 value results in p < 0.05, there is less than a 5%
chance that the observed results occurred by chance alone and the observed results are
significantly different from the expected (i.e. you reject the null hypothesis). If p≥ 0.05, then 5%
of the time or more, any deviation from the expected results is due to chance only (i.e. you fail to
reject the null hypothesis). The χ2 test does not prove that a hypothesis is correct but evaluates
whether or not the data and the hypothesis have a “good fit.”
In Table 9, which lists probabilities and χ2 values, note the “Degrees of Freedom”
column. In any experiment, the degree of freedom (df) is a measure of the number of categories
that are independent of each other. The df is one less than the number of different phenotypes
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possible. In the current experiment we have four possible phenotypes so there are 4 - 1 = 3 df.
Example:
If, in an F2 population of 100 plants, results are 60 wild type: 40 mutant (expected ratio would
be 75 wild type : 25 mutant), then:
χ2 = (60 - 75)2 + (40 – 25)2
75
25
=3+9
= 12.0
Because we have two possible phenotypes (mutant and wild), we have 2-1 df. Looking in the χ2
table for χ2 = 12 with 1 degree of freedom, probability is < 0.01; therefore, these results are not
supportive of a 3:1 ratio since the probability is less than 5% (0.05).
To read the χ2 table, locate the df specific to your experiment in the first column. Now look
across the row to find the probability column that best matches your calculated χ2 value. The row
matching 1 degree of freedom is marked by an orange square in Table 9 . You will notice that
our calculated χ2 value is too high and the closest number to that value is 6.635 in the last column
(circled in green). This means that the probability of obtaining the expected 3:1 ratio simply by
chance is less than 1% therefore the null hypothesis must be rejected.
7. Now calculate the χ2 based on your class data. Show all of your calculations in the space
provided below.
8. Note the following information:
Degrees of freedom: __________
χ2:________________________
Probability of chance occurrence (from Table 9):________
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9. Based on your results, what can you conclude about your Ho?
10. How do your observed ratios compare to those expected? Explain.
11. What do your class results suggest about the inheritance patterns of Brassica rapa?
TASK 6 - Analyzing Pedigrees
A pedigree is a map of relatives that is used to determine the inheritance pattern of a
particular disease or trait. This map usually includes the gender of each family member, how
each is related (through lines connecting individuals) and also provides information about
genetic traits. Certain symbols are used to indicate these variables (see Fig. 10).
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Figure 10. Symbols used in pedigree analysis
Questions:
1. What would you look for in a pedigree to determine if a trait was dominant?
2. What would you look for in a pedigree to determine if a trait was recessive?
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3. The following is a pedigree for albinism. Note: the half-filled circles and squares in this
pedigree represent carriers but in other pedigrees carriers may not be clearly denoted as in
this case. Can you determine if this trait is dominant or recessive? Explain.
4. Below is a pedigree where multiple individuals are afflicted with Brachydactyly or
shortening of the digits. Can you determine if this trait is dominant or recessive? Explain.
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LOOK AHEAD:
Before coming to lab next week, make sure to read the Molecular Biology task sheet as well as
Chapter 7 (pgs. 71-74) in your lab manual.
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REFERENCES:
http://www.fastplants.org/
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