BIO 181 Laboratory Exercise Online ProtocolMendelian GeneticsJohn Nagy and Dennis Massion

Preparation

• Read Chapter 14, sections 14.1 through 14.3 (pgs. 289-299).

• Familiarize yourself with the concepts in Table 14.1 of the textbook.

• Read Sections 1 and 2 in this protocol.

• Launch Lab 5: Mendelian Genetics in Canvas.

• Download, either from the ‘Labs’ page of the course website or Canvas, and launchthe Mendelian Genetics Data Frame.

1 Introduction

Recall that in one of his experiments, Mendel mated true-breeding, round-seeded pea plants withtrue 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 roundwhile 25% should be wrinkled. In the actual experiment, however, Mendel counted 5474 roundand 1850 wrinkled in the F2, which is not exactly 3:1—only 74.74% were round while 25.26% werewrinkled. The numbers were close, but not exactly the same. Why not?

The reason is randomness. Since the F1 parents are both heterozygous, they can contributeeither a R or r allele to the offspring. Alleles, according to Mendel, segregate independently, whichmeans that the probability the child inherits the R allele is 0.5 (and therefore the probability ofinheriting the r allele is 1 − 0.5 = 0.5). It’s like flipping a fair coin to determine which allele isinherited—a heads means the child inherits R allele while tails means it inherits the r allele. It alsomeans that the probability of inheriting the R allele remains 0.5 no matter how many children theparents 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 andnever won—my luck is bound to change soon,” or “I’ve played this slot machine (or card game orany other gambling game) and lost almost every time, so sooner or later the law of averages willcatch up and I’ll start winning?” That’s the gambler’s fallacy. It doesn’t matter how many timesthe player played the game; the probability 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’sresults were actually 72% round and 28% wrinkled. Then someone could argue, “look, if we roundto 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 traitis 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 fit 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 0.0657Wrinkled 1850 1831 19 361 0.1972

2 Goodness of Fit

The way forward here is a statistical test called a goodness of fit test. We will measure howwell our data fit a particular hypothesis using 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 nullhypothesis 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 thisexperiment. 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 roundand 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 fit table: In the fourth column wecalculate the difference between the observed and expected, Obs−Exp. In the fifth columnwe square the values in the previous column, (Obs− Exp)2. Finally, in the last column wedivide the previous column by the expected frequencies.

5. Calculate χ2: The χ2 statistic is simply the sum of values in the final column of the goodnessof fit 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 pre-ceding text.

2. Calculate the difference between observed and expected frequencies. In the first row wehave

5474 − 5493 = −19,

while in the second we get1850 − 1831 = 19.

3. Square these differences. 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 and361

1831= 0.1972.

for row 2.

5. Therefore, χ2 = 0.0657 + 0.1972 = 0.2629.

2.2 Interpreting χ2

The χ2 statistic measures fit between hypothesis and experiment. The larger the χ2, the less thetwo agree. In practice, we compare our experiment’s χ2 to a critical value, χ2

crit (see footnote1

for more information). Our decision rule is the following:

1For those interested in where this number comes from, I’ll try to give a quick motivation. We started byassuming that Mendel’s theory was correct, just for the sake of argument. That assumption is called the nullhypothesis. 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 significance. In this exampleI set this level of significance at 0.05, meaning that if we were to get an experimental result with a probability lessthan 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 andcolor—in garden peas.

1. If χ2 > χ2crit, then the experimental result is significantly different from the expectations of

the hypothesis; i.e., the hypothesis is not supported.

2. if χ2 ≤ χ2crit, 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 concludethat 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 fromparent to offspring. Recall that in one experiment he crossed a true-breeding variety of pea thathad round seeds and a yellow seed leaf with another true-breeding variety with wrinkled seeds anda green seed leaf. All the F1 offspring 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 ledhim to conclude that these 2 traits are inherited independently.

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Table 2: Goodness of fit 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 0.0162Round/Green 108 104.25 3.75 14.0625 0.1344Wrinkled/Yellow 101 104.25 −3.25 10.5625 0.1013Wrinkled/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, weexpect 9

16(556) = 312.75 round/yellow, 3

16(556) = 104.25 round/green and wrinkled/yellow,

and 116

(556) = 34.75 wrinkled/green. From here we complete Table 2, sum the last columnand get χ2 = 0.4695. In this case the critical value has risen to 7.815, which is well aboveMendel’s χ2 value, so his hypothesis is not contradicted.

3 Procedure

Corn (scientific 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 fixed geneticbackground, cross-breeding a true-breeding purple with a true-breeding yellow plant will alwaysproduce purple F1 offspring. However, cross-breeding two purple F1s can yield both purple andyellow offspring. (See Fig. 3 on page 6 for details.)

So, superficially this looks Mendelian. However, we should not assume that it is because mosttraits are not. Our goal today will be to test the hypothesis that this trait (kernel color) isMendelian. If so, then kernel color would be determined by a single gene with two alleles: sayA (purple) and a (yellow). Yellow kernels would have genotype aa and purple kernels would beeither AA or Aa. (How do we know that purple would be the dominant trait? You can tell justfrom the information already given.)

3.1 Data Gathering

3.1.1 Dihybrid cross

1. In Canvas, find 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 kernel,count the number of purple/starchy, purple/sweet, yellow/starchy and yellow/sweet kernelsin that kernel’s row. If you get to a point where you can’t tell where the row goes just makeyour best guess and move on. If you get to the end of a row, move to an adjacent row andcontinue. It really doesn’t matter as long as you don’t count any kernel twice. Keepcareful track of your tally on a separate sheet of paper.

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Figure 3: Summary of the experiment analyzed in this protocol. Two traits are shown: seed color(purple or yellow) and carbohydrate content (starchy or sweet). Here we test if these traits areMendelian. (A) To produce the F2 cob, true-breeding purple/starchy plants were bred with true-breeding yellow/sweet (generation P ). The offspring (generation F1 were all purple/starchy. Thesewere bred together to make the F2 cob. The test cross cob was produced by backcrossing the F1

with its yellow/sweet parent. (B) Close-up views of the 4 possible phenotypes in this experiment.Section 3.2 looks only at seed color. Section ?? looks at both seed color and carbohydrate content.

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3. Continue in this fashion until you have counted at least 100 kernels.

4. Record your final tally in the “F2 Tally” column of the table labeled, ”Main Data Frame”in the “Mendelian Genetics Data Frame” spreadsheet.

3.1.2 Test Cross

1. In Canvas, find the image of the test cross cob.

2. Tally the phenotypes of 100 kernels as you did with the dihybrid cob.

3. Record your observations in the column labeled, “Test Cross Tally” of the table labeled,”Main Data Frame” in the “Mendelian Genetics Data Frame” spreadsheet.

3.2 Data Analysis for Seed Color

Here we will calculate a goodness of fit test to measure how well the data fit the null hypothesisthat seed color in corn is a Mendelian trait.

3.2.1 Hybrid analysis

First we look at the F2 cob to see if how well its data fit the hypothesis.

1. In the “F2 Tally” data from the main data frame, count separately the total number ofpurple and yellow kernels (starchy and sweet combined). [Example: Suppose you count 37purple/starchy and 22 purple/sweet. That would give you 59 purple.]

2. Place these data in the “Observed” column of the “Goodness of Fit Table—Single Trait F2.”

3. Use the procedure described in Secction 2.1 to calculate the expected frequencies based onthe hypothesis that this is a Mendelian trait. Record your results in the “Expected” column.Recall that, if these traits are Mendelian, then we expect 75% of the kernels to be purple and25% to be yellow. So, to find the expected frequency of purple, multiply the total numberof kernels you counted by 0.75. Calculate expected frequency of yellow by multiplying thetotal by 0.25.

4. Use your data and the procedures described in Box 1 to complete the goodness of fit tablein the spreadsheet.

5. Calculate χ2 from that table as explained in Box 1 and record your answer in the properlocation of the data frame (directly below the table you just filled in).

6. State the null and alternative hypotheses of this experiment. Record your response in Canvas.

7. The critical value is 3.841. Do your data suggest that Mendel’s principle of segregationshould be abandoned for corn kernel color? Explain why or why not in detail in the spaceprovided on Canvas.

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3.2.2 Test cross analysis

Recall that Mendel tested his hypothesis of segregation by predicting the outcome of a novelexperiment, in particular a test cross. Here, Mendel backcrossed F1 plants with the recessiveparents (see Fig. 3). The procedures below will walk you through analysis of a similar test crossfor kernel color in corn.

1. As with the F2 cob, count the number of purple and yellow kernels.

2. Record your observations in the “Observed” column of table, “Goodness of Fit Table–SingleTrait Test Cross” of the data frame.

3. If Mendel’s hypothesis is correct, then this test cross would be between individuals of geno-types Aa× aa. Use that information to calculate the expected proportions of offspring thatare purple and record your results in Canvas.

4. Complete the goodness of fit table.

5. Use the goodness of fit table to calculate χ2 for this test. Record your result in the properlocation directly below the goodness of fit table.

6. Interpret the χ2 statistic for your results using a critical value of 3.841. Place your responsein Canvas.

3.3 Data Analysis for Seed Color and Carbohydrate Content

Now that we have explored the single trait—seed color—in corn, we now ask a follow-up question:Are seed color and carbohydrate content (starch or sugar) independent? As before, we will testthis hypothesis using the F2 and test cross cobs.

3.3.1 Dihybrid Analysis

1. Copy your observed data for the F2 cob to the “Observed” column of table “Goodness ofFit Table–Two-Trait F2” in the data frame.

2. Calculate the expected frequencies under the null hypothesis that these traits (color andcarbohydrate content) are Mendelian and independent using the information in Section 2.3.

3. Record these calculations in the “Expected” column. Note: Keep the decimal fractions; donot round.

4. Complete the goodness of fit table.

5. Calculate χ2 and record your result on the spreadsheet just below the goodness of fit table.

6. The critical value is 7.815. Use that information to interpret χ2 and explain the result indetail in the space provided in Canvas.

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3.3.2 Test cross analysis

As with the single trait, Mendel performed a test cross for the two trait case. That is, he back-crossed the purple/round F1 with its yellow/wrinkled parent. Here, we analyze a similar cross incorn.

1. Calculate the expected phenotypes and their frequencies of such a test cross for this case.Submit your answers in Canvas.

2. Copy the test cross data from the main data frame to the “Observed” column of the “Good-ness of Fit Table–Two-Trait Test Cross” table as you did before.

3. Fill in the expected number of offpring under the null hypothesis in the “Expected” column.

4. Calculate χ2 from the χ2 table and record your result in the proper location below thegoodness of fit table of the spreadsheet.

5. As before, interpret the χ2 statistic for your results using a critical value of 7.815. Recordyour response in Canvas.

4 Exercises

Complete the exercises in Canvas.

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