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Mendelian Genetics
Objectives
1. Use “genetic corn” to measure and interpret the results of crosses that illustrate Mendel’s laws of segregation and
independent assortment.
2. Learn the Chi-square procedure for testing the agreement of observations and
expectations, and use this procedure to
evaluate the genetic corn results.
3.
Demonstrate the principles of ABO and
Rho blood typing as an example of
codominant inheritance.
Procedures
1- Mendelian Genetics of Corn
Domestic corn (Zea mays) is ideal for demonstrating the principles of inheritance. An
ear of corn is a stem with multiple female
flowers arranged along its length. Each
flower produces an egg, which will develop
into a seed (kernel). The male flowers are
on the tassels, which bear anthers and produce pollen. If a young ear of corn on one
plant is fertilized with pollen from the tassels of another plant, each kernel of corn
that forms on that ear is an individual offspring of the mating. Each ear has hundreds
of kernels (offspring), providing the statistical sample that is so important for demonstrating the principles of inheritance.
An advantage of domestic corn is that the
kernels remain attached to the cob, which
makes them easy to observe and handle.
Incidentally, the Native Americans that domesticated the corn plant selected an allele
that causes the seeds to remain attached to
the cob. This is a key mutation that made
corn easy to harvest. Respect their accomplishment and DO NOT pick the kernels off
the ears!
We will examine the phenotypes of the corn
kernels and try to explain the observed ratios of phenotypes using Mendel’s principles. You will also perform statistical tests
of the results to see if they support the predictions made from Mendel’s principles.
First, though, we need to consider just what
a corn kernel is and what genes are involved
in the tests.
Corn kernels consist of several parts (see
Figure 1). The outer layer or shell is called
the pericarp. The pericarp is not part of the
embryo- it is “packaging” from the parent
plant. Inside the pericarp is a layer of cells
called the aleurone, which can be colored,
and the endosperm, which is the bulk of the
seed– the starchy part. Also inside is the
embryo. We’ll assume that the aleurone,
endosperm and the embryo all have the
same genotype, which we’ll call this the
embryo genotype. We will look at two phenotypic traits- kernel color and kernel
shape. In our corn, these characters depend
on the embryo genotype.
Figure 1– Anatomy of a corn seed (kernel)
endosperm
Aleurone
Pericarp
embryo
Seed shape. There are two phenotypes in
our corn– smooth and wrinkled– and they
are controlled by the Su locus. The Su gene
has two alleles- these are Su and su. (Try
not to be confused by the use of a 2-letter
symbol for one allele). Possible diploid
genotypes are SuSu, Susu, and susu. Bad
enough to write it, but try to say it and make
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Mendelian Genetics
any sense! Lets call Su starch and su sugar. These alleles control the amount of sugar in the
endosperm. The dominant allele Su causes starchy endosperm (little sugar). Homozygous
susu produces sugary endosperm (sweet corn) and causes the kernel to be wrinkled and
somewhat translucent when it dries.
Seed color: The locus that we will consider affects aleurone color and is symbolized Pr.
There are two alleles represented in our corn. The allele Pr is dominant and causes the seeds
to be dark purple. The allele pr is recessive and causes the kernels to be yellow. Possible
genotypes are PrPr, Prpr, and prpr. These two-letter symbols get awkward when you write
a genotype for both loci, for example SusuPrPr. Ask your instructor if you don't understand.
Table 1. Genetic corn loci and alleles
Locus
Dominant
allele
Recessive
allele
Dominant Effect
Su
Su
su
Enzyme converts sugar to starchmakes smooth kernels
Pr
Pr
pr
Purple pigment colors kernels
(otherwise they are yellow)
Count the phenotypes: Work in pairs. The instructor will provide ears resulting from different crosses involving the color and shape genes. Note the code number of each ear and report this number with your results. Examine the ears and be sure that you can distinguish all
the phenotypes (smooth vs wrinkled, and purple vs yellow). Use good judgment. For example, a little dimple on the end is not wrinkled- they should be really shriveled.
For each ear, count and record the phenotypes of the kernels. Count at least 400 kernels on
each ear, or all the kernels, whichever comes first. OK, it is tedious. Doing science is usually 10% inspiration and 90% perspiration. But just think of the glory that was heaped on
Mendel (mostly posthumously, of course). Classify each kernel in whole rows, one after the
other. When you come to the end of a row, drop down to the next and then work your way
back to the other end. As you call out the phenotypes, your partner can keep tally on a piece
of scratch paper. Add up the tallies and enter them in Table 2. Note: some ears will have all
4 phenotypes, others only 2. Leave boxes empty as needed.
Table 2. Corn phenotype counts
ear #
Phenotype counts
smooth red
wrinkled red
smooth yellow
wrinkled red
Total counted
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Mendelian Genetics
For each ear, calculate the percentages of
each phenotype represented (divide the
number of kernels X 100 in each category
by the total number counted on that ear).
Enter the results in Table 3.
Table 3. Corn phenotype percentages.
ear #
Phenotype percentages
smooth wrin- smooth
red
kled red yellow
wrinkled
yellow
2- Chi-Square Test
Remember that inheritance is a matter of
probability. How do we decide if our observations fit our predictions? If you were
flipping a coin, you would not conclude that
the coin was unbalanced just because it
came up heads the first two times you
flipped it. However, if you flipped heads 9
of 10 times, you would probably be suspicious!
Statisticians have quantified these judgments. The Chi-square test is commonly
used to compare observed results with predictions according to a specific hypothesis.
For example, if you expected 10 of 30 offspring to have the recessive phenotype, and
the observed number was 14 of 30, you
might want to know how likely it is that the
difference between observed and expected
result was caused by something other than
chance. How different do the observed and
expected results have to be before you decide that the prediction is wrong?
The formula for calculating the chi-square
statistic (X2) is:
X 2= Σ[(o-e)2/e]
The formula can be stated as follows: chisquare is the sum of the squared differences
between observed (o) and expected (e) results in all categories. The symbol Σ means
“sum”. What are you summing? The term
(o-e)2/e is calculated for each category of
the result. For example, the categories in a
genetic cross would be each phenotype.
For example, let’s say that you expect a 3:1
ratio of phenotypes in a monohybrid test
cross. Suppose that the actual results are
638 purple kernels and 240 white, of 878
total. Your hypothesis is the 3:1 ratio, so
your prediction is (878*0.75) = 659 purple
and (878*0.25) = 219 white. The calculation of X2 for this example is shown in Table 4.
Table 4. Calculation of Chi square for the
example discussed above.
Phenotypes
Purple
White
Observed (o)
638
240
Expected (e)
659
219
(o - e)
-21
21
(o-e)2
441
441
(o-e)2/e
0.668
2
X2 = Σ[(o-e)2/e] = 2.668
We get a value of 2.668 for X2. The more
the results differ from the predictions, the
bigger 2 will be. Statisticians have determined exactly how probable it is to obtain
each value of X2 through chance. We can
compare our result with a table of X2 values
to find its probability. If our results are
very unlikely, we can conclude that the prediction was probably wrong.
Mendelian Genetics
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Of course, we have to pick a criterion for
how unlikely a result can be before we reject our hypothesis. The usual standard in
natural sciences is 5% (p < 0.05). The criterion is the probability (p) that the result
could occur by chance, if the hypothesis is
true. If the result would occur less than 5%
of the time by chance alone, its fair to guess
that a X2 so large is not due to chance*.
When we say a result is “statistically significant”, we mean that it is very unlikely to
occur by chance alone (< 5% of the time) .
Step-by-Step Procedure for Testing Your
Hypothesis and Calculating Chi-Square
The value of X2 also depends on the number
of categories being compared. In this simple example, there are only two phenotypes.
Therefore there are only two categories in
the calculation of X2. If you think about it,
it seems a bit redundant to include both
categories in the calculation. After all, if
the kernels don’t have the one phenotype,
they must have the other. This consideration is called the “degrees of freedom” (df).
Degrees of freedom is the number of categories in the data minus 1. In our example,
there are two categories (purple and white);
therefore, there is 1 degree of freedom.
4. Compare your value of X2 to the chisquare distribution table to determine
how likely that value is:
Now we are ready to see how unlikely our
result really is. Refer to the chi-square distribution table (Table 5 below). Using the
appropriate degrees of freedom, locate the
value nearest to the calculated value of chisquare. For our example (X2 = 2.67) you
should find that the p value is about 0.10.
This means that there is a 10% probability
that you would see results this far from predictions (or further) due to chance. In terms
of your criterion, the results do not differ
significantly from expected.
*Keep in mind that unlikely results can happen by chance. If you were a real stickler
for proof, you might set your criterion at
p=0.01 rather than 0.05.
1. State the hypothesis and the predicted
results.
2. Based on the hypothesis, predict the expected numbers for each observational
class. Remember to use numbers, not
percentages.
3. Calculate X2. Round your answer to
two decimal places.
A. Determine df and locate the appropriate row.
B. Locate the value closest to your calculated 2 on that row.
C. Read the p value from the head of
that column.
5. State your conclusion in terms of your
hypothesis.
A. If the p value for the calculated X 2
is p > 0.05, accept your hypothesis.
'The deviation is small enough that
chance alone accounts for it. A p value
of 0.6, for example, means that there is
a 60% probability that this X2 or a larger
one could occur by chance. The difference between observation and prediction is not significant.
B. If the p value for the calculated X2 is
p < 0.05, reject your hypothesis, and
conclude that something other than
chance is operating for the difference to
be so large.
Chi-square requires that you use numerical values, not percentages or ratios. Also note
that Chi-square is not reliable if the expected value in any category is less than 5.
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Mendelian Genetics
Chi-Square Distribution. Compare values of X2 to this table to determine how likely
they are to be due to chance alone. You won’t probably won’t find your exact value of
X2. Just check to see if your value is larger or smaller than the X2 for the critical value
of p (0.05) and the appropriate degrees of freedom.
(df)
Probability (p)
0.95
0.90
0.80
0.70
0.50
0.30
0.20
0.10
0.05
0.01
0.001
1
0.004
0.02
0.06
0.15
0.46
1.07
1.64
2.71
3.84
6.64
10.83
2
0.10
0.21
0.45
0.71
1.39
2.41
3.22
4.60
5.99
9.21
13.82
3
0.35
0.58
1.01
1.42
2.37
3.66
4.64
6.25
7.82
11.34
16.27
4
0.71
1.06
1.65
2.20
3.36
4.88
5.99
7.78
9.49
13.28
18.47
5
1.14
1.61
2.34
3.00
4.35
6.06
7.29
9.24
11.07
15.09
20.52
6
1.63
2.20
3.07
3.83
5.35
7.23
8.56
10.64
12.59
16.81
22.46
7
2.17
2.83
3.82
4.67
6.35
8.38
9.80
12.02
14.07
18.48
24.32
8
2.73
3.49
4.59
5.53
7.34
9.52
11.03
13.36
15.51
20.09
26.12
9
3.32
4.17
5.38
6.39
8.34
10.66
12.24
14.68
16.92
21.67
27.88
10
3.94
4.86
6.18
7.27
9.34
11.78
13.44
15.99
18.31
23.21
29.59
“Not significant”
“Significant”
3– Blood Typing
The ABO and Rh factor blood groups are a
good example of codominance– each of two
alleles having simultaneous effects on the
phenotype regardless of the other’s presence. There are 3 common alleles of the
ABO gene locus. They are IA, IB, and i.
The codominant IA and IB alleles cause proteins (form A and form B) to be present on
the red blood cells. The recessive i allele
does not produce a protein. Depending on
which combination of these alleles a person
inherits, their blood cells may have the following phenotypes: A, B, AB or O.
Foreign proteins that enter your body act as
antigens, molecules that cause the production of antibodies. Antibodies are proteins
made by the immune system that selectively
bind to the antigen that induced them. Each
antibody molecule bears two binding sites
for its antigen, so antibodies can act as a
kind of molecular glue that will cause antigen molecules to stick together.
All of this becomes an issue if you receive a
blood transfusion of cells that bear a protein
form that you do not have on your own
cells. In that case you produce antibodies
that cause the foreign cells to agglutinate,
or stick together.
A convenient way to determine blood type
is to expose a sample of the blood cells to
solutions containing the antibodies, and see
which antibodies cause agglutination. The
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Mendelian Genetics
antibody solutions are called antisera because they are the serum portion of the
blood of animals that were sensitized by injections of that particular antigen.
In the “old days” we used to test our own
blood with antisera in the classroom. However, after AIDS became widespread in the
1980’s, we began using synthetic substitutes
for blood, out of the prevailing fear of infection. These substitutes work reasonably
well. They are particles coated with certain
antigens, and they will agglutinate in the
presence of antibody like blood cells would.
Sure do miss poking fingers to draw blood
though...
Procedure:
You will be provided with several samples
of synthetic blood representing different
blood types. Obtain a glass slide and lable
the ends A and B. Then place a small drop
of antiserum A near the A end and a drop of
antiserum B at the B end. Next, place
small drops of the “blood” that you are testing near to, but not touching, each of the
drops of antisera. Use a clean toothpick to
mix each pair of drops together. Be sure to
use a different toothpick for each pair.
Observe the mixed drops for agglutination.
The agglutinating “blood” will take on a
grainy appearance as the particles clump
together. Determine and note the blood
type of each of the “persons” whose blood
you are testing. Have your instructor check
your results and interpretation. For purposes of the lab final, you may be asked to
determine the blood type of a sample using
this procedure.
Assignment:
Prepare the following to hand in next time:
Genetic Corn & Chi Square statistics
1. For each of the four ears that you examined, hypothesize about the genotypes
of the parents. What parental genotypes
would you predict would give the ratios
of offspring phenotypes that you saw?
2. Test each of your hypotheses using the
Chi square test. For each ear, prepare a
table in Excel showing the predicted
and observed numbers for each phenotype, and the value of X2. Give the approximate p value (from Table 5), and
state whether or not the results are consistent with your hypothesis about parental genotypes (i.e., is the p value
greater than or less than 0.05?).
Blood typing questions
1. Which blood type shows codominance?
Explain.
2. Which blood type is the universal donor?
3. Which blood type is the universal acceptor?
4. What is Rh factor? Research this blood
type and write a brief essay in your own
words explaining its significance.
Compare with the ABO blood groups.
Why are women who are Rh negative
and marry Rh positive men concerned
about effects on the health of their second child, but not their first?
Genetics jokes
Genetics explain why you look like your father and if you don't why you should. Did you know that
sterility is hereditary? If your grandfather didn't have children and your father didn't have children,
you won't have children either. And by the way– how do you tell the difference between a male chromosome and a female chromosome? You must take down their genes .