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Objectives 1. To understand the Principles of Segregation and Independent Assortment. 2. To understand how Mendel’s principles can explain transmission of characters from

one generation to the next. 3. To understand and use the χ

2 goodness-of-fit statistical test to evaluate hypotheses about mechanisms of inheritance.

I. Background Much of our understanding of genetics today is taken for granted, as the use of DNA has become common in forensics and medicine. However, compared to many fields in biology, genetics is relatively young. Mendel’s “Laws” of inheritance were rediscovered in the early 1900s, and Watson and Crick described the structure of DNA only about fifty years ago (if you are interested in exploring the history of genetics, you can find a list of good web sites at http://dmoz.org/Science/Biology/Genetics/History/) People have always been interested in explaining how traits were passed from one generation to the next. In the 19th century, the leading hypothesis of heritability was that offspring resembled their parents because parental traits blended. A prediction of this hypothesis is that offspring of a cross of purple-flowered and white-flowered peas will have pale purple flowers. We can sum up their position and the outcome of Mendel’s experiments in the following table:

OBSERVATION

The characters of the offspring appear to be a blend of the parental traits.

QUESTION How are traits inherited? HYPOTHESIS / PREDICTION

Blending of heritable information from both parents produces offspring with intermediate traits.

EXPERIMENT(S) Mendel’s pea experiments. Data: The offspring from certain crosses, such as purple x white-flowered peas, resemble only one parent, not a blend.

EVALUATION/ DRAW INFERENCES

The data do not support the hypothesis. Mendel proposed alternative models.

Mendel conducted these breeding experiments over and over again, with different traits; all of them with two phenotypes, and from his results developed his “laws” of segregation and independent assortment. We place Mendel’s experiments into our structure for the scientific method in the following table:

Mendelian Genetics

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OBSERVATION

Pea flower colour is either white or purple, but never a blend.

QUESTION

If blending does not explain how traits are inherited, then how are traits inherited?

HYPOTHESIS / PREDICTION

Some “particle” determining flower color is passed intact from parent to offspring (Mendel did not coin the term “gene”). Prediction: the “particle” determining white color is still present in the offspring, but its expression is masked.

EXPERIMENT

Cross the offspring resulting from purple x white crosses (F1), and see if some “grand-offspring” (F2) are white . Data: about 25% of the F2 are white, the remainder are purple

EVALUATE/DRAW INFERENCES

The data support the hypothesis. Mendel predicted (and then experimentally confirmed) that about 2/3 of the purple F2 would carry the “white” particle (i.e. had both a purple and a white particle), and that other characteristics such as pea color and plant height would show similar patterns of inheritance.

We can put this experiment into the context of modern genetics by using modern terminology. In the parental generation (P), plants are true-breeding: for the trait of interest such as flower color, crosses within the true-breeding group always produce offspring that resemble the parents. This is because the parents are homozygous (identical alleles) for either the purple or white color. When Mendel crossed the true-breeding purple and white flowers, he always obtained plants, the F1 generation, with purple flowers, but the allele for “white” is not lost: we refer to these individuals as heterozygous. When he then crossed two F1 heterozygotes and determined the flower color of the F2, he found 705 individuals with purple flowers and 224 individuals with white flowers. He referred to this as “Segregation” because he stated that the particle for white color segregated (separated) from the particle for purple color. He further proposed that these two “particles” are alternative states (alleles) for the same character (“gene”). In modern terms; these two states are two different alleles for the same gene, where we define a gene (or genetic locus) as a specific section of a chromosome coding for a particular protein. Different alleles of one genetic locus, code for different versions of the same protein.

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II. The monohybrid cross The experiment described above is a monohybrid cross: the true-breeding parents are homozygous for different flower colors, and the F1 are hybrids for that one trait. Our understanding of meiosis provides a mechanistic explanation for the pattern of segregation in monohybrid crosses. In meiosis, when the gametes are being formed, each gamete only gets one copy of each chromosome (one member of each chromosome “pair”). In the F1 plants, each gamete will either get the allele for purple or the allele for white flower color: no gametes can have both alleles because these alleles are on each of the homologous (paired) chromosomes. One gamete from one parent combines with one gamete from the other parent, but which gamete combines with which (i.e., purple with purple or with white) is random. For each gamete type (purple or white) from one parent, there is a 50:50 chance it will combine with either white or purple from the other parent. This will produce, in the F2 generation, purple/purple: purple/white (or white/purple): white/white genotypes in a 1:2:1 ratio. Since the purple allele “masks” or is dominant to the white allele, the phenotypes of the F2 generation will be 3 purple: 1 white. Female (genotype Ww) This is represented here in a Punnett

Square. P represents the allele for the dominant purple flower trait and p represents the allele for the recessive white flower trait.

Mal

e

(gen

otyp

e W

w) gamete

genotypes P p

P

PP

Pp

p

Pp

pp

III. Dihybrid Cross. Mendel also conducted experiments where he crossed true-breed plants that differed in two different traits, producing F1 offspring that were hybrids at two genetic loci, or dihybrids. Now it can be imagined that two traits affecting the same structure, pea color and shape for instance, might move jointly to the next generation: if the original homozygous parents have smooth, yellow peas then all descendents with smooth peas will have yellow peas. The alternative hypothesis is that the traits are separate, and that the inheritance of pea color will be independent of the inheritance of pea shape. These two alternative hypotheses can be put into our table of the scientific method, and a single experiment can test both of them:

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Observation Peas display four different phenotypes with respect to their seed characteristics: for color yellow or green, for shape smooth or wrinkled

Question Does the inheritance of one trait (shape) influence the inheritance of another trait (color)?

Hypothesis 1, Prediction 1

Yes: if the original pea parents have either smooth and yellow peas or green and wrinkled peas, all the F2 offspring will have one of these combinations.

Hypothesis 2, Prediction 2

No: no matter what combination of traits the original parents have, all possible combinations will be found in the F2 generation.

Experiment Cross true-breeding plants then cross the F1 offspring. Rear the resultant F2 plants and determine the frequencies of the different combinations of traits. Data: in the F2, all possible combinations of traits are found.

Evaluation, Inferences

Hypothesis 1 is rejected; there is no “linkage” of the two different traits as they are passed from generation to generation.

In modern terminology, the inference drawn from this result is that the two genetic loci, pea color and pea shape, are not “linked”. The predicted genotype and phenotype distribution in offspring of a dihybrid cross using unlinked genetic loci can be calculated in two fashions. First, you can do a 4 x 4 Punnett square (you should finish filling in all possible genotypes): Female: WwYy

Mal

e:

W

wY

y

gametes: WY Wy wY wy WY WWYY WWYy WwYY WwYy Wy WWYy wY WwYY wy WwYy

Since W- is always round and Y- is always green, we can determine the phenotypes of these offspring, and find a 9:3:3:1 ratio for W-Y- : wwY- : W-yy : wwyy. A second, and much quicker, way to predict phenotype frequencies is to recall that in a monohybrid cross, 3/4 of the offspring will have the dominant phenotype and 1/4 will have the recessive phenotype. To estimate how many offspring from a dihybrid cross will have, for instance, the phenotype “round green” (W-Y-) you simply multiply the proportion of W- by the proportion of Y-, or 3/4 x 3/4 = 9/16. Similarly, 1/4 x 3/4 = 3/16 will have the phenotype “wrinkled green” (wwY-). This approach has the benefit that it is easily expanded to many more genetic loci, when Punnett’s squares become awkward or impossible to draw (one needs an additional dimension for every additional genetic locus!)

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It is important to note that not all genetic loci assort independently. If two genetic loci are relatively close together on the same chromosome, they will not move independently into the gametes and hence will appear “linked” in experimental crosses.

Again, we can look to our understanding of meiosis for a mechanistic explanation of independent assortment. Remember that, during meiosis, homologous chromosomes pair along the center of the cell during metaphase I. When they are lined up, however, the orientation of each pair of chromosomes is independent of the orientation of every other pair. In other words, there is equal chance that the w-carrying chromosome will be aligned with the y-carrying chromosome or with the Y-carrying chromosome. Assigning letters to each allele (W:round, w:wrinkled, Y: green, y: yellow), we can diagram these possibilities in a drawing of metaphase I.

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IV. Today’s Lab with Corn. Corn Genetics: Kernel color and sugar content Ears of corn are good study systems for learning about genetics, because each kernel represents an independent union of gametes, and thus a cob has a population of genetically unique kernels. The cobs you will be studying are variable for two traits of interest, color and sugar content. You can detect sugar content differences because high-sugar kernels are wrinkled and low-sugar kernels are smooth. Each of these characteristics is determined by a single gene, with two alleles.

There are four kernel phenotypes in the above ear of corn: Purple & Smooth (A), Purple & Wrinkled (B), Yellow & Smooth (C) and Yellow & Wrinkled (D). For your experiments today, you will use what you have learned about the principles of segregation and independent assortment to infer the mechanisms of inheritance for color and sugar content in corn. You will collect data to test four hypotheses concerning these traits by sampling kernels on the cob. Finally, you will evaluate your hypotheses by comparing your data with your predicted results using the χ2 statistical test (Chi Square).

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Evaluating Hypotheses by Predicting results: Part of testing hypotheses is to describe what results you would expect to find if your hypothesis is correct. The more exact your predictions are, the easier it is to use your data to test the hypothesis. Today’s Observation: Individual kernels of corn on a cob vary in color and shape. Hypotheses and predictions: Below are four hypotheses regarding the inheritance of corn kernel color and shape. Working with your partner and the teaching fellow, develop predictions of what you would observe in the population of corn kernels if the hypothesis is correct. These will be the “working hypotheses” guiding your experiments. Hyp. 1. The genes for color and shape assort independently. Prediction 1 Hyp.2. Purple is dominant to yellow. (or use the reverse) Prediction 2 Hyp. 3. Sugar is dominant to starch (wrinkled dominant to smooth) (or the reverse) Prediction 3 Hyp. 4. The corncob you are examining represents the result of a dihybrid cross. Prediction 4

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Analysis of Experimental Results: You will use the χ2 goodness-of-fit test to determine if your observed results are consistent with the predictions made based on Mendel’s principles. The χ2 test allows you to determine whether your results could have happened by chance alone. Typically, the null hypothesis being tested is that the observed results (O) and the expected results (E) are not different (Ho: O = E). The alternative hypothesis is that the observed counts and the expected counts are different (Ha: O ≠ E). The general equation for the chi-square value for a particular experiment is: Chi-Square = the sum over all categories for (observed cell counts – expected cell

counts)2 /expected cell counts or χ2 =O − E( )2

E∑

The last ingredient needed for the test is the “degrees of freedom” (df), which is calculated as the number of classes, less one. Here, the “degrees of freedom” are related to the number of categories contained in the characteristic under study. For example if the characteristic under study is the sex ratio of a trait there are two possible classes (male or female) and one degree of freedom. This is the case since one class (e.g. female) is “free” to vary in % of the total, but once its value is known the % of the other class (male) is determined by default (no longer “free”). Likewise if there are three values that make up 100% of the characteristics (such as red, pink and white) there are two degrees of freedom, two could vary but the last will be determined by default, and so on for increasing numbers of classes. The observed χ2 value (calculated with the above equation) is compared to those in a statistical table (Table 1. below) containing values from a χ2 distribution for each number of degrees of freedom, and the associated probability “ρ” that the observed data match the null hypothesis (or conversely that any deviation from the expected was solely the result of random chance). A χ2 value of 0 (zero) indicates no deviation from the expected values. Therefore, any χ2 value greater than 0 represents some deviation from the expected values. The statistical table provides a basis for evaluating the probability that the deviation from 0 has occurred by chance. The results are expressed as a range of probability (ρ) values. Table 1. Chi-Square Probability Distribution ρρρρ 0.99 0.95 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.05 0.01 df 1 0.00 0.004 0.02 0.06 0.15 0.27 0.45 0.71 1.07 1.64 2.71 3.84 6.63 2 0.02 0.10 0.21 0.45 0.71 1.02 1.39 1.83 2.41 3.22 4.61 5.99 9.21 3 0.11 0.35 0.58 1.01 1.42 1.87 2.37 2.95 3.66 4.64 6.25 7.81 11.34 4 0.30 0.71 1.06 1.65 2.19 2.75 3.36 4.04 4.88 5.99 7.78 9.49 13.28 5 0.55 1.15 1.61 2.34 3.00 3.66 4.35 5.13 6.06 7.29 9.24 11.07 15.09

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The numbers in Table 1 represent χ2 values as calculated with the formula given above. This “χ2 statistic” reflects the magnitude of the discrepancies between observed and expected counts. In other words, the larger the difference between your observed counts, and the expected counts, the larger the χ2 value that results. Therefore, large values of χ2 suggest rejection of the null hypothesis Ho. As the magnitude of the difference between observed and expected values increases, the probability that large of a difference between observed and expected values would be found as a result of sampling error alone decreases. Similar to how it is unlikely, though possible, that you could toss 90 heads and 10 tails in 100 tosses of a coin. At some point we need to decide how unlikely something is to still be comfortable saying Ho: O = E. In practice biologists usually use the 5% probability of match (observed vs. expected) or less. Meaning that at a “ρ” (probability) of 5% (0.05) that we could get that sizeχ2 by sampling error (random chance) alone if O = E. Anything less likely than that, we reject our hypothesis that O=E, and accept the alternative hypothesis that O ≠ E. In Table 1 above the blue shaded region would be considered to be a poor match and lead to the rejection of the null hypothesis. So, if you have 1 DF and the χ2 value you calculate is larger than 3.84, you would reject your null hypothesis (O=E). The general method for calculating the χ2 value for a goodness-of-fit test is to fill in a table, shown as an example for a monohybrid cross: class observed expected* (O-E) (O-E)2/E A– 65 75 -10 1.333 aa 35 25 +10 4 total 100 100 χ2 = 5.333 df=2-1 =1 In this example, looking at Table 1, 5.33>3.84 and the probability of finding these (divergent) results purely by chance is less than 5% but greater than 1% (χ2 value 6.63) and we reject the hypothesis that O=E.

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Experimental Procedures Methods: Obtain a cob of corn. These cobs are the F2 generation of unknown crosses. Before you start, determine a protocol that you will follow to test each of the hypotheses you made predictions for earlier. For each hypothesis, decide how many kernels you will count (sample size – remember you will need a minimum of 5 counts for each “class” and preferably many more!), where on the cob you will start (we recommend a new start point for each sample), and how you will classify each kernel. Fill in your protocol for each experiment in the section below. Experiment 1 To support or reject the first hypothesis, no formal experimentation is required, only observation of the cob. Protocol: Study the specimen, and record the important observations. Predicted results: if your hypothesis is correct, what results do you expect? Results: Evaluation: Experiment to test hypothesis 2 Protocol: Number of kernels to be counted__________ Kernel characteristics (categories)______________ Predicted results: if your hypothesis is correct, what numerical results do you expect (how many kernels in each category)? These will be your Expected values for your χ2 test. Results: Evaluation: do the χ2 test.

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Conclusion; do your results cause you to reject your hypothesis? Experiment to test hypothesis 3 Protocol: Number of kernels to be counted__________ Kernel characteristics (categories)______________ Predicted results: if your hypothesis is correct, what numerical results do you expect (how many kernels in each category)? These will be your Expected values for your χ2 test. Results: Evaluation: do the χ2 test. Conclusion; do your results cause you to reject your hypothesis?

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Experiment to test hypothesis 4 Protocol: Number of kernels to be counted__________ Kernel characteristics (categories)______________ Predicted results: if your hypothesis is correct, what numerical results do you expect (how many kernels in each category)? These will be your Expected values for your χ2 test. (Hint: create a 4x4 Punnet Square for this cross as you did in the “background” section for Mendel’s Peas. Then the “expected” can be calculated from the total observations times the phenotype ratios -x/16- of the different classes.) Results: Evaluation: do the χ2 test. Conclusion; do your results cause you to reject your hypothesis?

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V. Questions and Interpretation of Data 1. Based on your results, what is the relationship between kernel color alleles? Support your answer. 2. Based on your results, what is the relationship between kernel sugar content alleles? Support your answer. 3. Do the genes for kernel color and sugar content assort independently among gametes? Support your answer. 4. Does the kernel phenotype distribution support the idea that the cob is the result of a dihybrid cross? What information supports your answer? If a dihybrid cross (i.e. F1 to F2 of standard Mendelian crosses) is not indicated what conditions might contribute to this finding.