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BIO 181 Lab Exercise
Patterns of Inheritance
John Nagy & Dennis Massion
Preparation
‹ Read Chapter 14, sections 14.1 through 14.3.
‹ Familiarize yourself with the concepts in Table 14.1 of the textbook.
‹ Read Sections 1 and 2 in this protocol.
‹ Launch Lab 5: Patterns of Inheritance in Canvas.
‹ Download, either from the ‘Labs’ page of the course website or Canvas, and launch
the Mendelian Genetics Data Frame.
1 Introduction
Recall that in one of his experiments, Mendel mated true-breeding, round-seeded pea plants
with true breeding, wrinkled-seeded plants. His results and explanation are represented in Fig.
1. Mendel’s hypothesis predicts a 3:1 ratio in the F2 generation—on average 75% should be
round while 25% should be wrinkled. In the actual experiment, however, Mendel counted 5474
round and 1850 wrinkled in the F2—only 74.74% were round while 25.26% were wrinkled. The
numbers were close to 3:1, but not exactly the same. Why not?
The reason is randomness. Since the F1 parents are both heterozygous, they can contribute
either a R or r allele to the ofspring. Alleles, according to Mendel, segregate independently,
which means that the probability the child inherits the R allele is 0.5 (and therefore the proba- bility of inheriting the r allele is 1 − 0.5 = 0.5). It’s like fipping a fair coin to determine which
allele is inherited—a heads means the child inherits R while tails means it inherits r. It also
means that the probability of inheriting the R allele remains 0.5 no matter how many children
the parents have and how many of those got R. (Here one has to beware of the gambler’s
fallacy. Have you ever heard anyone reason along these lines: “I’ve played the lottery a bunch
of times and never won—my luck is bound to change soon,” or “I’ve played this slot machine
(or card game or any other gambling game) and lost almost every time, so sooner or later the
law of averages will catch up and I’ll start winning?” That’s the gambler’s fallacy. It doesn’t
matter how many times the player played the game; the probability of winning doesn’t change.)
Because the allele that a child inherits is randomly determined, results of a dihybrid (e.g.,
Rr×Rr) cross are usually not 3:1. But that creates a problem for scientists. Suppose Mendel’s
results were actually 72% round and 28% wrinkled. Then someone could argue, “look, if we
round to the nearest 5% then Mendel actually got closer to 70% round seeds instead of 75%.
Therefore, I claim Mendel’s hypothesis is wrong.” So, the question is, if we are testing to see if
a given trait is Mendelian, what results would support or refute the Mendelian hypothesis?
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Figure 1: Results and explanation of Mendel’s experiments on seed shape in garden peas.
Table 1: Goodness of ft table for Mendel’s experiment on seed shape.
Phenotypes Obs Exp Obs − Exp (Obs − Exp)2 (Obs − Exp)2/Exp
Round 5474 5493 −19 361 0.0657
Wrinkled 1850 1831 19 361 0.1972
2 Goodness of Fit
The way forward here is a statistical test called goodness of ft. Such tests measure how well
data ft a particular hypothesis. In our case, we will use a statistic called χ2.
2.1 Calculating χ2
We want to know if the data support or contradict the idea that this trait obeys Mendel’s
principle. Either it does (the null hypothesis) or it doesn’t (the alternative hypothesis). We
test the null hypothesis with the following procedure.
1. List the relevant phenotypes. These are spherical (round) and wrinkled (Table 1).
2. Record the observed data. This is listed in Table 1 under “Obs.”
3. Calculate and record the expected frequencies: Mendel counted 7324 F2 seeds in
this experiment. If his hypothesis were correct, then we’d expect his results to be about
75% round and 25% wrinkled, in which case he would have observered 0.75(7324) = 5493
round and 0.25(7324) = 1831 wrinkled. These results are listed in the “Exp” column of
Table 1.
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4. Complete the calculations of the goodness of ft table: In the fourth column we
calculate the diference between the observed and expected, Obs − Exp. In the ffth
column we square the values in the previous column, (Obs − Exp)2. Finally, in the last
column we divide the previous column by the expected frequencies.
5. Calculate χ2: The χ2 statistic is simply the sum of values in the fnal column of the
goodness of ft table.
Example 1: Calculating χ2
Let’s recalculate table 1 to see how it’s done.
1. Fill in phenotypes, observed and expected values. These can be obtained in the
preceding text.
2. Calculate the diference between observed and expected frequencies. In the frst row
we have
5474 − 5493 = −19,
while in the second we get
1850 − 1831 = 19.
3. Square these diferences. That is, calculate −192 = 192 = 361.
4. Divide this square by expected frequency. In this example we have
361
5493 = 0.0657
for row 1 and
361
1831 = 0.1972.
for row 2.
5. Therefore, χ2 = 0.0657 + 0.1972 = 0.2629.
2.2 Interpreting χ2
The χ2 statistic measures ft between hypothesis and experiment. The larger the χ2, the less
the two agree. In practice, we compare our experiment’s χ2 to a critical value, χ2 (see crit
footnote1 for more information).
1For those interested in where this number comes from, I’ll try to give a quick motivation. We started by
assuming that Mendel’s theory was correct, just for the sake of argument. That assumption is called the null
hypothesis. If the null hypothesis is true, then mathematicians can show that the probability one would get
a χ2 value equal to or greater than the critical value is less than a predetermined level of signifcance. In
this example I set this level of signifcance at 0.05, meaning that if we were to get an experimental result with
a probability less than 5% under the null hypothesis, I’ll conclude that the null hypothesis is not likely to be
true.
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Observed Results
Expected Results
Figure 2: Expected and observed results of Mendel’s experiments on 2 traits—seed shape and
color—in garden peas.
Our decision rule is the following:
1. If χ2 > χcrit
2 , then the experimental result is signifcantly diferent from the expectations
of the hypothesis; i.e., the hypothesis is not supported.
2. if χ2 ≤ χ2
crit, then there is no strong evidence from the data that the hypothesis is false.
Example 2: Interpreting χ2
In this case, the critical value is 3.841. Our χ2 = 0.2629. Since 0.2629 < 3.841, we
conclude that the experimental results do not contradict the hypothesis.
2.3 Mendel’s Principle of Independent Assortment
In addition to his studies on single traits, Mendel also studied how pairs of traits were passed
from parent to ofspring. Recall that in one experiment he crossed a true-breeding variety
of pea that had round seeds and a yellow seed leaf with another true-breeding variety with
wrinkled seeds and a green seed leaf. All the F1 ofspring were round/yellow (Fig. 2). He
then self-fertilized the F1 plants. If the 2 traits were passed independently, then Mendel would
expect to see an approximate (3 : 1)2 = 9 : 3 : 3 : 1 ratio of phenotypes in the F2, which he did
(Fig. 2). This observation led him to conclude that these 2 traits are inherited independently.
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Table 2: Goodness of ft table for Mendel’s experiment on seed shape and color.
Phenotypes Obs Exp Obs − Exp (Obs − Exp)2 (Obs − Exp)2/Exp
Round/Yellow 315 312.75 2.25 5.0625 0.0162
Round/Green 108 104.25 3.75 14.0625 0.1344
Wrinkled/Yellow 101 104.25 −3.25 10.5625 0.1013
Wrinkled/Green 32 34.75 −2.75 7.5625 0.2176
Example 3: Testing Mendel’s second conclusion
Mendel’s observations are given in Fig. 2. He counted a total of 556 F2 plants; therefore,
we expect 9
16 (556) = 312.75 round/yellow, 3
16 (556) = 104.25 round/green and wrin- kled/yellow, and 1
16 (556) = 34.75 wrinkled/green. From here we complete Table 2, sum
the last column and get χ2 = 0.4695. In this case the critical value has risen to 7.815,
which is well above Mendel’s χ2 value, so his hypothesis is not contradicted.
3 Procedure
Corn (scientifc name Zea mays) can vary in the color of its seeds or kernels. Like most traits,
many genes contribute to the trait, so corn can range from purple to red to yellow to white.
Here we study one particular variation—purple vs. yellow kernel color. Against a fxed genetic
background, cross-breeding a true-breeding purple with a true-breeding yellow plant will always
produce purple F1 ofspring. However, cross-breeding two purple F1s can yield both purple and
yellow ofspring. (See Fig. 3 on page 6 for details.)
So, superfcially this looks Mendelian. However, we should not assume that it is because
most traits are not. Our goal today will be to test the hypothesis that this trait (kernel color)
is Mendelian. If so, then kernel color would be determined by a single gene with two alleles: say
A (purple) and a (yellow). Yellow kernels would have genotype aa and purple kernels would
be either AA or Aa. (How do we know that purple would be the dominant trait? You can tell
just from the information already given.)
3.1 Data Gathering
3.1.1 Dihybrid cross
1. In Canvas, fnd the image of the dihybrid (F2) cob. Note that each kernel is an individual.
2. Note that the kernels run more-or-less in rows down the cob. Starting with any initial ker- nel, count the number of purple/starchy, purple/sweet, yellow/starchy and yellow/sweet
kernels in that kernel’s row. If you get to a point where you can’t tell where the row
goes just make your best guess and move on. If you get to the end of a row, move to
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