Objectives1. 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 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:

1

Lab #8: Mendelian Genetics

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:

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 inherited intact from parent to offspring (Mendel did not coin the term “gene”). Prediction: the “particle” determining white color is still present in the offspring.

EXPERIMENT Cross the offspring resulting from purple x white crosses, 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 that about 2/3 of the purple F2 would carry the “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

2

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 from the particle for purple color. He further proposed that these two genes are alternative states (alleles) for the same particle. We now know that 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.

II. The monohybrid crossThe 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. 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 they are on homologous 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: 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.

3

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.

Male

(g

enoty

pe

Ww

)

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 F2 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:

Observation Peas display four different phenotypes: (1) yellow and smooth, (2) yellow and wrinkled, (3) green and smooth, (4) green and wrinkled.

Question Does the inheritance of one trait influence the inheritance of another trait?

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

4

fashions. First, you can do a 4 x 4 Punnett square (you should finish filling in all possible genotypes):

Female: WwYy

Male

:

W

wYy

gametes: WY Wy wY wyWY WWYY WWYy WwYY WwYyWy WWYywY WwYYwy 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 multiple 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!)

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.

5

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.

IV. Today’s Lab with Corn.

For your experiments today, you will use what you have learned about the principles of independent assortment and segregation to test hypotheses concerning the mechanisms of inheritance for color and sugar content in 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.

For your experiments today, you will use what you have learned about the principles of independent assortment and segregation to infer the mechanisms of inheritance for color and sugar content in corn. You will pose questions concerning these traits, reformulate them as testable hypotheses, and collect data to test three of them by sampling kernels on the cob. Finally, you will evaluate your hypotheses by comparing your data with your predicted results using the 2 statistical test.

Forming Testable Hypotheses

Observation: Individual kernels of corn on a cob vary in color and shape.

6

Questions: What are four questions Mendel might have posed, had he been working with corn rather than garden peas?

1.

2.

3.

4.

Hypotheses: Working with your partner and the teaching fellow, reformulate three of your questions into hypotheses concerning the mechanisms of inheritance of these characteristics (color and shape). Then make 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.

hypotheses predictions

1.

2.

3.

Results: You will use the 2 goodness-of-fit test if your observed results are consistent with the predictions made based on Mendel’s principles.

The 2 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 method for calculating the 2 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/EA– 65 75 -10 1.333

7

aa 35 25 +10 4total 100 100 2 = 5.333

df=2-1 =1

The general equation for the chi=square value for a particular

experiment is: 2 =

€

O− E( )2

E∑

The last ingredient for the test is the “degrees of freedom”, d.f., which is calculated as the number of classes less one (in this example, df= 2-1 =1). The observed 2 value (calculated with the above equation) is compared to those in a statistical table (see section V, page 12) containing values from a 2 distribution for each number of degrees of freedom, and their associated probability of occurring at random. A 2 value of 0 (zero) indicates no deviation from the expected values. Therefore, any 2 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 (p) values. In our example, 0.05>0> 0.01, or the probability of finding these results purely by chance is less than 5% but greater than 1%.

Experimental Procedures

Methods: Obtain a cob of corn. These cobs are the results of F1 dihybrid crosses, in other words the kernels are the F2 generation. Before you start, determine a protocol that you will follow to test each of your hypotheses. For each hypothesis, decide how many kernels you will count (sample size), 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 1Your observation:

Your hypothesis:

Protocol: Number of kernels to be counted__________

Kernel characteristics (categories)______________

Predicted results: if your hypothesis is correct, what numerical results to you expect (how many kernels in each category)? These will be your Expected values for your 2 test.

8

Results:

Evaluation: do the 2 test.

Conclusion; do your results cause you to reject your hypothesis?

9

Experiment 2Your observation:

Your hypothesis:

Protocol: Number of kernels to be counted__________

Kernel characteristics (categories)______________

Predicted results: if your hypothesis is correct, what numerical results to you expect (how many kernels in each category)? These will be your Expected values for your 2 test.

Results:

Evaluation: do the 2 test.

Conclusion; do your results cause you to reject your hypothesis?

10

Experiment 3Your observation:

Your hypothesis:

Protocol: Number of kernels to be counted__________

Kernel characteristics (categories)______________

Predicted results: if your hypothesis is correct, what numerical results to you expect (how many kernels in each category)? These will be your Expected values for your 2 test.

Results:

Evaluation: do the 2 test.

Conclusion; do your results cause you to reject your hypothesis?

11

V.

VI. Questions and Interpretation of Data1. 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. You explored both monohybrid and dihybrid crosses, and the resultant phenotypic patterns exhibited by their offspring. How would you perform a trihybrid cross? If both parents breed true, what phenotypic patterns would be exhibited in the F1 offspring from this cross? What about the F2?

12