BIO 181 Laboratory Exercise Name:

Mendelian Genetics Instructor:John Nagy and Dennis Massion Lab partners:

1 Introduction

Preparation

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

• Familiarize yourself with the concepts in Table 14.1.

Recall that in one of his experiments, Mendel mated true-breeding, round-seeded pea plantswith 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 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?

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.

Rev: Fall 2019 1 Nagy

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

2.1 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.

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.

Rev: Fall 2019 2 Nagy

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:

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.

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.

Rev: Fall 2019 3 Nagy

Table 2: Goodness of fit table for the corn color analysis.

Phenotypes Obs Exp Obs− Exp (Obs− Exp)2 (Obs− Exp)2/Exp

Purple

Yellow

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.

3 Testing the Principle of Segregation in corn

Corn (scientific name Zea mays) can vary in the color of its seeds or kernels. They can be eitheryellow or purple. Cross-breeding a true-breeding purple with a true-breeding yellow plant willalways produce purple F1 offspring. However, cross-breeding two purple F1s can yield both purpleand yellow offspring.

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: A(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 Single trait hybrid cross procedure

1. Obtain a single ear of corn from the tray labeled “F2.” Note that each kernel is an individual.(It’s a seed capable of growing into an independent corn plant with a unique genotype.)

2. Note that the kernels run more-or-less in rows down the cob. Starting with any initial kernel,count the number of purple and yellow kernels in that kernel’s row. If you get to a pointwhere you can’t tell where the row goes just make your best guess and move on. It reallydoesn’t matter as long as you don’t count any kernel twice. Keep careful track of yourtally on a separate sheet of paper.

3. Continue in this fashion until you have counted at least 100 kernels. You may do more than100 if you wish.

4. Use your data to complete the relevant goodness of fit table (table 2). Recall that we expect75% of the kernels to be purple and 25% to be yellow. So, to find the expected frequencyof purple multiply the total number of kernels you counted by 0.75. Calculate expectedfrequency of yellow by multiplying the total by 0.25.

Rev: Fall 2019 4 Nagy

5. Calculate χ2 from table 2.

χ2 =

6. In the space below, state the null and alternative hypotheses of this experiment.

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.

3.2 Single trait test cross procedure

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. The procedures below will walk you through a similar test cross for kernel color in corn.

1. Use Mendel’s hypothesis to predict the results of a test cross between a purple F1 corn plantand its yellow parent. Place your prediction and show your work in the space below.

Rev: Fall 2019 5 Nagy

2. Return your F2 ear of corn to the supply table and obtain one from the pile labeled “testcross.” This cob was obtained by backcrossing purple F1 plants with their yellow parents.

3. As before, count at least 100 kernels on the test cross cob. Tally the number of purple andyellow kernels in your sample.

4. In the space below, construct and complete a χ2 table for this cross.

5. Calculate χ2 from the χ2 table above and record your result below:

χ2 =

6. In the space below, interpret the χ2 statistic for your results using a critical value of 3.841.

4 Testing the Principle of Independent Assortment in corn

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.

Rev: Fall 2019 6 Nagy

Observed Results

Expected Results

Figure 2: Expected and observed results of Mendel’s experiments on 2 traits—seed shape andcolor—in garden peas.

Table 3: 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 3, 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.

Now we test if kernal color and seed shape in corn obey Mendel’s Principle of IndependentAssortment. Note that some kernels are wrinkled and some are round. The F2 cob was producedby crossing purple/round F1 plants. So now we ask, are these traits passed independently?

Rev: Fall 2019 7 Nagy

Table 4: Goodness of fit table for test of Mendel’s second law.

Phenotypes Obs Exp Obs− Exp (Obs− Exp)2 (Obs− Exp)2/Exp

Purple/Round

Purple/Wrinkled

Yellow/Round

Yellow/Wrinkled

1. Obtain another F2 cob. As before, examine at least 100 kernels and note their phenotype(purple/round, purple/wrinkled, yellow/round, yellow/wrinkled).

2. Record your results in table 4 and then complete that goodness of fit table.

3. Calculate χ2 from table 4 and record your result below:

χ2 =

4. The critical value is 7.815. Do your data contradict Mendel’s hypothesis or not? Explain indetail in the space provided.

Rev: Fall 2019 8 Nagy

5. As with the single trait, Mendel performed a test cross for the two trait case. That is,he backcrossed the purple/round F1 with its yellow/wrinkled parent. In the space below,calculate the expected phenotypes and their frequencies of such a test cross. Show your work.

6. Return the F2 cob and obtain a test cross cob. Tally the phenotypes (for both traits)of at least 100 kernels as you did before. In the space below, construct and complete agoodness of fit table for this experiment.

7. Calculate χ2 from the χ2 table above and record your result below:

χ2 =

8. In the space below, interpret the χ2 statistic for your results using a critical value of 7.815.

Rev: Fall 2019 9 Nagy

5 Exercises

In each exercise below show your work and box, circle or underline your answer.

1. Early-onset Parkinson disease 6 (PARK-6), an autosomal recessive trait, causes progressivenervous degeneration throughout a person’s life. Arnold has been diagnosed with PARK-6,and his wife Nancy’s father died of it. Genetic tests show that Nancy will not get the disease.

(a) What are Nancy and Arnold’s genotypes?

Nancy Arnold

(b) Nancy is pregnant with Arnold’s child. What’s the probability that the child inheritedPARK-6?

2. A man heterozygous for PARK-6 marries a homozygous normal woman. Suppose they havethree children. What is the probability that any of their children will develop PARK-6?Explain your response.

3. Mid-chin fissure is a dominant trait that obeys Mendel’s first principle. Together, this traitand PARK-6 obey Mendel’s second principle. Suppose a man who is heterozygous for bothtraits marries a woman with a non-cleft chin who has PARK-6. What are the expectedrelative frequencies of phenotypes among their offspring, assuming they can have children?(NOTE: State the phenotypes associated with each frequency.)

Rev: Fall 2019 10 Nagy

Table 5: Goodness of fit table for T. H. Morgan’s fruit fly experiment.

Phenotypes Exp Obs(Obs− Exp)2

Exp

4. In the fruit fly Drosophila melanogaster, gray body color is dominant to black, and long wingsare dominant to short (called vestigial). Thomas Hunt Morgan, who studied these traits,performed test crosses of gray-bodied, long-winged dihybrids with black-bodied, vestigial-winged homozygotes (BbV v × bbvv), from which he obtained 2300 offspring.

(a) If these traits obey Mendel’s second principle, how many of each phenotype do we ex-pect to see among offspring?

(b) Morgan actually obtained the following results: 965 offspring were gray body/longwings; 944 were black body/vestigial wings, 206 were gray/vestigial and 185 wereblack/long. With this information fill in the goodness of fit table (table 5) and cal-culate χ2.

χ2 =

(c) The critical value for this test is still 7.815. Does Morgan’s study support or contradictthe claim that body color and wing length obey Mendel’s Principle of IndependentAssortment? Defend your answer.

Rev: Fall 2019 11 Nagy

5. Most Bengal tigers (Panthera tigris) are orange with black stripes. Occasionally, however,they can be white with black stripes. The white phenotype is recessive to orange. In onestudy,2 a white and an orange tiger were bred in a zoo. These 2 tigers eventually had 10offspring: 4 white and 6 orange. What are the genotypes of the two parents?

6. If the orange tiger from the previous question is crossed with one of its orange offspring,what is the probability of obtaining a white tiger from the cross?

7. In snapdragons, a species of plant, red flower color is incompletely dominant over white, theheterozygotes being pink. If you want to produce seed that will yield only pink floweringplants, what would you cross?

8. George has type A blood and his wife Harriet has type B blood. They have a daughter withtype O blood. What is the probability that their next child will have type AB blood?

2Xu et al. 2013. Curr. Biol. 23(11):1031.

Rev: Fall 2019 12 Nagy

9. Red-green colorblindness is a sex-linked trait. A normal woman whose father was red-greencolorblind marries a normal man. What is the expected proportion of color-blind children?Give both phenotypes and genotypes, including gender.

10. A woman and a man have a red-green colorblind daughter. What can you say about thegenotypes of the parents?

11. In cats, coat color is an X-linked trait. Black coat color is determined by allele XB, andyellow is generated by the Xb allele. Black and yellow are co-dominant; heterozygous catshave a tortoise shell coat. Suppose a female yellow cat is bred to a black tom. What willthe phenotype ratio of kittens be? (Include genders in your ratios.)

12. What is the expected phenotypic ratio of a cross between a tortoise shell female cat and ablack tom (male)?

Rev: Fall 2019 13 Nagy

13. A woman without red-green colorblindness and with a normal chin, whose mother was color-blind, marries a cleft-chinned, non-colorblind man whose mother had a normal chin. Recallthat red-green colorblindness is sex-linked recessive, but mid-chin fissure (cleft) is autosomaldominant. What are the expected phenotypic ratios of the children?

14. A man with a cleft chin marries a normal-chinned woman and they have 10 children, allwith cleft chins.

(a) Can you be sure the man is homozygous? Explain.

(b) If the 11th child has a cleft chin, will that prove that the father is homozygous? Explain.

(c) Assume that the father is actually Bb and has had 10 cleft-chinned children. What isthe probability that the next child will have a normal chin?

15. Rex has a cleft chin and lacks perfect pitch, both autosomal dominant traits. His wife, Mary,has a normal chin and perfect pitch. Rex’s father had a normal chin and Mary’s motherlacked perfect pitch. Rex and Mary plan to have three children. What’s the probability thatthe children all have normal chins and lack perfect pitch?

Rev: Fall 2019 14 Nagy

16. In the fruit fly Drosophila melanogaster, vestigial wings and hairy body are produced bytwo recessive genes located on different chromosomes. The wild-type (“normal”) alleles forlong wings and hairless bodies are both dominant. Suppose a vestigial-winged, hairy maleis crossed with a homozygous normal female.

(a) What phenotypes would be expected among the progeny, and what are the phenotypefrequencies?

(b) If the F1 from this cross are permitted to mate randomly among themselves, what arethe expected phenotypic frequencies in the F2?

17. In certain types of mice, coat color is determined by 2 epistatically interacting genes. Onegene determines if the mouse produces melanin, the pigment in hair. Mice with genotypecc cannot produce melanin and are always albino. Mice that are either CC or Cc producemelanin and can be either black or agouti (sort of brown). Which of these they end updepends on a second gene. At this second locus, melanin-producing mice that are either AAor Aa are agouti, while those that are aa are black. For example, a mouse with genotypeAaCc is agouti,, one that is aaCc is black, but both Aacc and aacc are albino. What is thephenotypic ratio of the following cross: aaCc× Aacc?

Rev: Fall 2019 15 Nagy

18. Let S and s be the alleles for seed shape (spherical and wrinkled, respectively) in gardenpeas. Similarly, let Y and y be alleles for cotyledon color (yellow and green), P and p bealleles for flower color (purple and white) and T and t be alleles for plant height (tall andshort). Consider the following cross:

SsY yPpT t× SsY Y PpTT.

What is the probability of obtaining an offspring of genotype SSY Y PpTt from this cross?

19. Sheala and Zoltan are thinking of having a child together, but they are worried. Both ofthem have siblings who died in their teens from a disease called cystic fibrosis, which is anautosomal (non-sex-linked) recessive disorder. Sheala also had a cousin who died at age10 from Duchenne muscular dystrophy (DMD), a recessive X-linked disease. On top of allof this, both Sheala and Zoltan have thalassemia minor. Given this unfortunate geneticbackground, both prospective parents go for genetic testing, the results of which indicatethat both are carriers for cystic fibrosis. In addition, Sheala is a carrier for DMD.

(a) What is the probability that a child born to Sheala and Zoltan would have thalassemiamajor, but neither of the other diseases, assuming you knew that the child was male?

(b) What is the probability that a child in this mating would have DMD and cystic fibrosis,but not thalessemia major, assuming you knew that the child was female?

(c) What is the probability that a child born to Sheala and Zoltan would have at least oneof the three diseases, cystic fibrosis, DMD or thalassemia major, assuming you don’tknow the sex of the child?

Rev: Fall 2019 16 Nagy