Materials Included In Kit

Additional Materials Required

Prelab Preparation

1. Fill two 400-mL beakers with 300 mL of distilled water in each beaker.
2. Add five drops of blue food coloring to one beaker and five drops of yellow food coloring to the other beaker. Mix well with the stirring rod.
3. Place in an accessible spot for students to use during Part D of the lab.

Safety Precautions

Food coloring will stain skin and clothes. Wear chemical splash goggles and chemical-resistant gloves. Instruct students to follow normal laboratory safety guidelines. Wash hands thoroughly with soap and water before leaving the laboratory.

Disposal

Please consult your current Flinn Scientific Catalog/Reference Manual for general guidelines and specific procedures, and review all federal, state and local regulations that may apply, before proceeding. The colored water may be rinsed down the drain with excess water according to Flinn Suggested Disposal Method #26b. The bingo chips and reaction plates used in this activity should be stored for future use.

Lab Hints

• Enough materials are provided in this kit for 30 students working in pairs, or for 15 groups of students. This laboratory activity can reasonably be completed in one 50-minute class period. The prelaboratory assignment may be completed before coming to lab, and the Post-Lab Questions may be completed the day after the lab.
• Demonstrate the lab procedure briefly before beginning this lab. Also emphasize the importance of placing the correct number of each color chips in the bag for each determination.

Teacher Tips

• Discuss the results to enhance students’ understanding of the trend. Did any offspring’s score vary by more than 2 from their parents? If so how many groups received this data?
• If classroom has access to a digital camera, allow students to take pictures of their colored reaction plate phylogeny tree.

Answers to Prelab Questions

1. A heterozygous individual has the genotype WwXXYyZz for hair color. In this hypothetical example, how many genes code for hair color?

   According to the above example, four genes code for hair color.

2. Using the genotype above, how many dominant alleles does this person have? How many recessive alleles?

   This individual has five dominant alleles and three recessive alleles.

3. Jon and Katie are having a baby. If Jon’s genotype for height is AabbCc and Katie’s is AaBbcc, what is the maximum number of dominant alleles the baby could have?

   The maximum number of dominant alleles would be four.

Sample Data

Answers to Questions

1. Why is it necessary to return the bingo chips to the male and female bags after each draw?

   This is necessary because each child has an equal probability of inheriting each allele. The first child receiving seven dominant genes has no effect on what the second child will receive.

2. Look at the phylogenetic tree in the data table. How do the eye colors of the offspring in Generation III compare to that of their parents in Generation II?

   Overall the offspring of Generation II (Generation III) were similar to their parents. All of the offspring in Generation III had a score within two of at least one of their parents.

3. In humans, tall parents tend to have tall children and short parents tend to have short children. However, the average adult height in humans varies greatly within a population. Why is this so?

   Height is a trait that follows polygenic inheritance. Traits that are inherited by more than one gene often form a bell-shaped curve with few individuals on the extreme ends and most falling in the middle region.

4. If two heterozygous individuals AaBbCc are crossed with each other, what is the probability they will produce heterozygous offspring AaBbCc? Explain. Hint: Figure out the probability of inheriting each gene and multiply those three probabilities together.

   The odds of this genotype being produced is ⅛. When each allele segregates independently the odds of obtaining Aa are ½, the odds of obtaining Bb are ½ and the odds of obtaining Cc is ½. Therefore when those fractions are multiplied together the odds of obtaining AaBbCc together is ⅛.

5. Mendel’s Law of Independent Assortment states that allele pairs separate independently during the formation of gametes. How is it possible for an offspring to have darker- or lighter-shaded eyes than either of his or her parents?

   Inheritance of eye color depends on the total number of dominant alleles carried by each parent—that is, if a particular trait follows polygenic inheritance and is carried by five genes. The father carries 5 dominant alleles and the mother carries 7 dominant alleles. It is possible the offspring could inherit 10 dominant alleles but it is much more likely the offspring will carry closer to 5, 6 or 7.

References

Bonner, Sheveeta & Brooks, Susan. “The Science of Skin Color.” Science Scope. Nov. 2007; Vol. 31 #3.

Campbell, N.A. Biology: Seventh Edition; Pearson Education Inc; San Francisco, CA; 2005; pp. 259 & 263.

Introduction

Have you ever been told you look like your parents? In this simple and fun activity, perform a simulation of the inheritance of eye color and create a visual representation of the simulation to enhance understanding of the genetics of eye color.

Concepts

• Genetics
• Independent assortment
• Monogenic vs. polygenic

Background

A trait is an observable or detectable variation in genetic makeup. Character is easily determined by an observable heritable feature. Some traits are monogenic such as the ability to taste the chemical phenylthiocarbamide (PTC)—the trait is controlled by one gene. Monogenic inheritance is determined by either the presence or absence of the gene. There is no partial inheritance, which means either the trait is expressed or it is not expressed.

Not all traits are monogenic (in fact, most traits are not). Instead, several traits follow polygenic inheritance. Polygenic inheritance is defined as an additive effect of two or more genes on a single observable trait. Each gene may be either expressed or not expressed, leading to a wide variation of that trait within a population. Characteristics such as eye color and height follow polygenic inheritance.

There is evidence that eye color in humans is controlled by at least three genes, probably more. Consider three genes with a dominant allele for each gene (A, B or C), where each contributes one “unit” of darkness to the phenotype and being incomplete dominance with respect to the other allele (a, b or c). Eye color is a result of the number of dominant alleles inherited. Basically, the more dominant alleles a person has the darker their eyes will be. An AABBCC person would have very dark eyes while an aabbcc person would be very light blue. An AbBbCc person’s eyes would be an intermediate shade. Because the alleles have a cumulative effect, the genotypes AaBbCc and AABbcc would also display the same eye color. All three genotypes make the same genetic contribution of three “units” of darkness. Therefore it does not matter on which gene the dark alleles are carried—eye is instead determined by the total number of dark alleles present.

A simplified model of polygenic inheritance is displayed in Figure 1. If two heterozygotes (AaBbCc) mate the odds are much more likely they will produce offspring with three dominant alleles (AaBbCc) than completely recessive (aabbcc) or completely dominant (AABBCC).

Experiment Overview

The purpose of this lab activity is to create a mathematical and visual example of how traits such as eye color are passed from generation to generation. This laboratory activity will be conducted assuming five genes code for eye color. The actual number of genes is not exactly known but is around five.

Materials

Prelab Questions

1. A heterozygous individual has the following genotype, WwXXYyZz, for hair color. In this hypothetical example, how many genes code for hair color?
2. Using the genotype above, how many dominant alleles does the person have? How many recessive alleles?
3. Jon and Katie are having a baby. If Jon’s genotype for height is AabbCc and Katie’s is AaBbcc, what is the maximum number of dominant alleles the baby could have?

Safety Precautions

Food coloring will dye skin and clothes. Wear chemical splash goggles and chemical-resistant gloves. Wash hands thoroughly with soap and water before leaving the laboratory. Please follow all laboratory safety guidelines.

Procedure

Part A. Determining Parental Pigmentation in Generation I

1. Obtain two brown paper bags. Label one bag male and the other bag female.
2. Add five blue chips to each bag and five yellow chips to each bag, for a total of 10 chips in each bag. Shake the bag to mix the chips.
3. Without looking inside the bags draw five chips from each bag. Combine the 10 chips drawn from the bags and count the number of blue chips. This number represents the male parent’s gene combination. Record this number inside the (Generation I-1) square on the Eye Color and Polygenic Inheritance Worksheet.
4. Place the 10 chips that were drawn in step 3 back into their original bags. Both the male and female bag should each contain five blue and five yellow chips again.
5. Without looking inside the bags, draw five chips from each bag. Combine the 10 chips drawn from the bag, and count the number of blue chips. This number represents the female parent’s gene combination. Record this number inside the (Generation I 2) circle on the Eye Color and Polygenic Inheritance Worksheet.
6. Before moving on to Part B, remove the original bingo chips from each bag and replace with the genetic combination obtained for each parent in steps 3 and 5. For example, if seven blue chips and three yellow chips were drawn in step 3, the male bag should now contain seven blue chips and three yellow chips for a total of 10 chips.

7. Determine the genetic makeup for the first offspring (Generation II-2). Without looking, one group member should draw five chips from the male bag and five chips from the female bag.
8. Record the total number of blue chips obtained in step 7 in the (Generation II-2) square of the worksheet. Replace the chips into their proper bags and repeat step 7 to find the number of blue chips for the remaining offspring in Generation II (persons 3, 5 and 7).
9. The spouses of Generation II offspring were chosen at random. Fill in the phylogenetic tree as follows:
   {10917_Procedure_Table_1}

10. Begin by determining the genetic makeup for offspring of Generations II-1 and II-2 which are (Generations III-1 and III-2). Place the appropriate number of blue chips in the male and female bags based on the numbers in circle (Generation II-1) and square (Generation II-2). Once the blue chips have been added to each bag, add yellow ships until each bag has 10 chips total.
11. Draw five chips from each bag to determine the genetic makeup (number of blue chips) offspring using the same procedure as in Part B, step 7.
12. Repeat step 10 to determine the offspring of (Generations II-3 and II-4), (Generations II-5 and II-6), and (Generations II-7 and II-8). Be sure to place the appropriate number of blue and yellow chips into each bag corresponding to the parent’s scores each time. Each bag should have a total of 10 chips.

13. Obtain 10–15 mL of both blue water and yellow water in separate graduated cylinders or small beakers.
14. Obtain a 96-well reaction plate and place it on a white piece of paper.
15. A phylogenic tree will be constructed using the colored water and reaction plates. See the diagram below to reference where each individual will be placed in the reaction plate. Exact placement is not crucial—overall the reaction plate should represent the phylogenetic tree with the correct parents above the correct offspring.
    {10917_Procedure_Figure_2}
16. Place a 96-well reaction plate on a piece of white paper.
17. Look at the numbers recorded for the first generation. If I-1 has a score of 6 (representing 6 blue chips), use a thin-stem pipet to place 6 drops of blue water and four drops of yellow water in the appropriate well for I-1.
18. Repeat step 17 for all individuals in Generations I, II and III.
19. Observe the difference in color between an individual with a high score versus a low score. This phylogenetic tree is used to observe the general inheritability of eye color.

Student Worksheet PDF

10917_Student1.pdf