Mendel's Laws - Genetics and Analysis

1. Understand the highlighted terms and principals used in Mendelian genetics.
2. Know how to complete a Punnett square and estimate phenotypic and genotypic ratios.
3. Know how the results of monohybrid and dihybrid crosses support Mendel's first and second laws.

This lab will introduce you to the concepts and procedures of the monohybrid cross, dihybrid cross, and complete and incomplete dominance. You will also be introduced to the statistical analysis of data using the Chi-square test.

In the mid-1800's, an Austrian monk named Gregor Mendel conducted experimental crosses with garden peas in an attempt to understand the basic patterns of inheritance. Mendel's meticulous work provided the basis for modern genetics. Although Mendel used garden peas for his experimental crosses, we will use corn kernels and tobacco seedlings since it is easier to observe results. Today, you will examine two kinds of crosses performed by Mendel and see how he used the results from these crosses to formulate his 1st and 2nd laws.

MENDEL'S FOUR PRINCIPLES OF INHERITANCE

1. Genes In Pairs: Genetic characters are controlled by unit factors (genes) that exist in pairs in individual organisms.
2. Dominance and Recessiveness: When unit factor pairs for a character are unlike in a single individual, one is dominant (expresses) over the other and the other is recessive. 
3. The Principle of Segregation: The two alleles of a homologous pair of chromosomes separate (segregate) during gamete formation (Meiosis) such that each gamete receives only one allele.
4. The Principle of Independent Assortment: Alleles of a gene pair assort independently of other gene pairs on nonhomologous chromosomes.

• A gene is a unit of heredity on a chromosome and has alternate forms called alleles. In sexually reproducing organisms each parent contributes one allele to their offspring that may or may not be like the other parent's allele.
• Alleles for a particular gene occur in pairs. Alleles that mask expression of other alleles of a particular gene, but are themselves expressed are dominant, and are usually designated by a capital letter (for example, "T"). Alleles whose expression is masked by dominant alleles are recessive, and are designated by a lower case letter (for example, "t").
• The genotype of an organism includes all the alleles present in the cell, whether they are dominant or recessive. The physical appearance of the trait is the phenotype. Thus, if tallness (T) is dominant to dwarfism (t), a plant with a tall phenotype can have a genotype of (TT) or (Tt). A plant with a dwarf phenotype can only have genotype (tt).
• When the paired alleles are identical (TT or tt), the genotype is homozygous (homo = same). Heterozygous (hetero = different) refers to a pair of alleles that are different (Tt).

1. Data Gathering

A monohybrid cross begins with experimental breeding between two parents that breed true for different forms of a single trait. The trait you will investigate in this problem is plant color. The two forms of plant color are Green and White or Albino. The offspring that result from this cross are called hybrids and are the F₁ generation. When two individuals from the F₁ generation are crossed the offspring is called the F₂ generation.

Chlorophyll synthesis in tobacco is controlled by a dominant allele "C"; the recessive allele is designated by "c". As a result of this dominance, homozygous dominant plants (CC) and heterozygotes (Cc) will be green plants, while homozygous recessive plants (cc) will be albino plants.

In this experiment, student pairs will be given a petri dish containing about 70 young tobacco plants. The seeds were placed on moist filter paper in petri dishes about 12 days prior to the lab. A pair of heterozygous tobacco plants (Cc) parented the seeds.

Classify and count the seedlings in your petri dish and record the results in the table below.

In order to visualize a cross between these two parental tobacco plants and the F₁ generation (and any other diploid organism that reproduces sexually) you should to construct a Punnett square. Below is a Punnett square to help you calculate the genotypes of the F₁ individuals resulting from this parental corn cross and the F₂ individuals resulting from the the F₁ cross. To construct a Punnett square, alleles present in the gametes (N) of one parent are written along the top and those from the other parent are listed down the side. The combinations produced by filling in each box of the Punnett square show the possible genotypes of the offspring.

Parental Cross

As you can see all offspring from the parental cross are the green phenotype. However, if you cross two of these offspring you will get a different phenotypic ratio. The Punnett Square below shows the expected results of this cross. The only 2 possible gametes from each heterozygous parent in this cross are C and c.

Note the following with regard to monohybrid, dihybrid, trihybrid, etc. crosses:

n = haploid number of chromosomes involved in the cross

2n = number of possible gametes

(2n)² = number of possible combinations.

F₁ Cross

You would expect a phenotypic ratio of 3 green plants: 1 albino plant.

2. Analysis

Though the predicted ratio is 3 green : 1 albino, you will rarely find that exactly 3/4 of the seedlings are green and exactly 1/4 are albino. If the difference between the observed ("o") number of seedlings and the expected ("e") number of seedlings is not too large, you would probably not be surprised since you would expect some variation in any random process, such as the recombination of alleles that takes place in a monohybrid cross.

If there were a large deviation, you would probably suspect that something was wrong with the experiment or that your hypothesis was incorrect. In a cross of this type you are actually making the hypothesis, called the null hypothesis, that there will be no difference between the expected counts ("e") and the observed counts ("o"). A statistical test called the Chi-square test can be used to determine if deviations from expected genetic ratios are within an acceptable range of variation.

The formula for Chi-square (X²) is: X² = S (o - e)2/e where: "o" is the observed count, "e" is the expected count, and "S" means the sum of (o - e)2/e for each class of data.

The expression (o - e)2/e gives a Chi-square value for each class of data. In a monohybrid cross with complete dominance, such as our tobacco cross, there are 2 classes of data, green seedling and albino seedlings. In the Chi-square formula, the individual Chi-square values for each class are added to give the total chi-square value. This value is then compared to a table (Critical Values of the Chi-Square Distribution) which provides information about the probability of the deviations. An abbreviated Chi-square table can be found later in this lab write-up.

Note the following regarding the Chi-square formula:

The greater the deviation (o - e), the larger the Chi-square value for that particular class of data. The deviation is directly proportional to the Chi-square value for a class of data.

The larger the expected count (e), the smaller the Chi-square value for a particular class of data. The expected count is indirectly proportional to the Chi-square value for a class of data. The expected counts make up the sample size. In 2 different experiments where the deviations are the same, a smaller Chi-square value will be found in the experiment with the largest sample size.

The larger the number of classes in the cross, the larger the Chi-square value that will be obtained with the Chi-square formula.

To illustrate the importance of the 3 factors cited above and the use of Critical Values of the Chi-square Distribution table, determine the Chi-square value for each of the following hypothetical tobacco seedling experiments, experiment A and experiment B.

To find your expected values (e), you need to know the expected ratio for the particular cross under investigation. In this example the ratio is 3 green :1 albino. So, 75% of the total count (160 seedlings) should be green and 25% albino. For the green seedlings, your expected number is 75% of 160 (.75 x 160). For the albino seedlings, the expected number is 25% of 160 (.25 x 160).

X² = S (o - e)2/e =.21 + .63 = .84

X² = S (o - e)2/e =

Below is the Critical Values of the Chi-square Distribution Table, which has a column showing the number of classes of data in the experiment minus one (C - 1). The expression (C - 1) is used to determine the number of degrees of freedom (df) for a particular experiment.

The minimum probability (P) often used in biology for rejecting the null hypothesis is 5% (0.05). If the probability realized from a Chi-square value for an experiment is equal to or less than 5% (P < 0.05), the null hypothesis would be rejected. This can also be stated as follows: If the probability (P) is equal to or less than 5% (P < 0.05), then the deviations from the expected count are considered significant. If a Chi-square value is large enough to give a P value of 1% (0.01), then the deviations from the expected are considered highly significant. From a statistical viewpoint, if the results from a hybrid cross have a probability of occurring only 5% of the time; the results would not be considered due to random chance alone. These results would constitute a significant deviation from what is expected.
 

 

EXPERIMENT A: The number of degrees of freedom (df) for this experiment is 1 (C-1 = 1). Thus, you would look across row 1 to determine where the Chi-square value of 0.84 falls in the abbreviated table of Chi-square values on page 103. Looking across row 1, you will see that the value 0.84 falls between 0.30 and 0.50. This means that 30% to 50% of the time similar results would occur in a monohybrid cross due to random chance alone. Since the deviations yielded a probability greater than 5%, the deviation from the expected results would be considered insignificant. The null hypothesis would not be rejected in this experiment.

EXPERIMENT B: The number of degrees of freedom is the same in this experiment (df = 1) and you would again look across row 1 to determine where the Chi-square value falls.

• What is the Chi-square value for Experiment B?
• Where does this value fall in the table?
• Are the observed results significantly different from the expected? Explain.
• Would the null hypothesis be rejected in this experiment?

Complete a Chi-square Test on the Results from Your Monohybrid Cross with Complete Dominance. Use the Chi-square test to determine if the results observed in your cross were significantly different from the expected results.

X² = S (o - e)2/e =

• What is the Chi-square value for your experiment?
• Are your observed results significantly different from the expected? Explain.

B. Monohybrid cross of Tomato with incomplete dominance

In this experiment half of the student pairs will be given a flat containing young tomato plants demonstrating incomplete dominance. The seeds were parented by a pair of heterozygous tomato plants (Yy).

In this cross the homozygous dominant plants (YY) will be yellow, the heterozygotes (Yy) will be yellow-green, and the homozygous recessive plants (yy) will be green. Since it is sometimes difficult to differentiate between the plants, it is best to count the green plants first, then the yellow-green, and finally the yellow.

Based on your knowledge of monohybrid crosses with incomplete dominance, you would expect a phenotypic ratio of 1 green plant : 2 yellow-green plants : 1 yellow plant. The Punnett Square shows the results of this cross. The only 2 possible gametes from each heterozygous parent in this cross are Y and y.
 

	Y	y
Y	YY	yY
y	Yy	yy

Classify and count the seedlings in your flat and record the results in the table below. Note that there are 3 classes of data in this monohybrid cross with incomplete dominance.

Complete a Chi-square Test on the Results from Your Monohybrid Cross with Incomplete Dominance:

Use the Chi-square test to determine if the results observed in your cross were significantly different from the expected results. With 3 classes of data, how many degrees of freedom (df) will there be?

X² = S (o - e)2/e =

• What is the Chi-square value for your experiment?
• Are your observed results significantly different from the expected? Explain.

C. Dihybrid Cross of Corn with Complete Dominance

Each pair of students will be given an ear of corn (F2 generation) that is the result of a cross between 2 hybrid plants differing in 2 traits (kernel color and kernel texture). The ear of corn represents the F2 generation of a cross between these plants. The F1 offspring were hybrids for the 2 traits. One pair of alleles controls color (purple or yellow) and the other pair control texture (smooth or wrinkled).

Based on your knowledge of dihybrid crosses and Mendelian genetics, answer the following questions before you proceed with the remainder of the exercise.

• What is the genotype of individuals in the F1 generation?
• What is the phenotype of individuals in the F1 generation?
• In the space below, complete a Punnett Square to help you determine the genotypic and phenotypic ratios of the F2 generation?
• List the possible F2 phenotypes in the space below along with the expected phenotypic ratios.
• List the possible genotypes in the F2 generation.

Classify and count each of the kernels on 8 randomly selected rows of kernels on your corncob. Make sure you do not count the same row twice. Record the counts in the table below.

• Which of the alleles for color is dominant?
• Which of the alleles for texture is dominant?
• How do you know that the color and texture alleles are located on different chromosomes?
• What is meant by principle of independent assortment?

Use the Chi-square test to determine if the results observed in your cross were significantly different from the expected results. With 4 classes of data, how many degrees of freedom (df) will there be?

X² = S (o - e)2/e =

• What is the Chi-square value for your experiment?
• Are your observed results significantly different from the expected? Explain.