Lab Instructions_ Genetics and Evolution Act I Mission Memo

Step 1: Anticipate your analysis. Determine what you should observe if the concentrations of cadmium in water samples exceed the recommended limit. This step will help us identify the evidence needed to build an argument in Step 3. Step 2: Calculate the probability that the concentration of cadmium in water exceeds the recommended limit of 100 mg L -1 . Determine whether the concentration of cadmium in water likely exceeds the recommended limit. This step gives us the evidence needed to build an argument in Step 3, when we will conclude if an elevated concentration of cadmium in water could explain the widespread death of sunstalks. Step 3: Weigh the evidence and conclude if an elevated concentration of cadmium in water could explain the widespread death of sunstalks. Construct an argument to answer the question “Could an elevated concentration of cadmium have caused the widespread death of sunstalks in the Sarcannian forest?” Your argument should draw on your calculations in Steps 1 and 2. Step 1: Anticipate your analysis. To construct a sound argument, one must anticipate the evidence needed to support a claim. In this assignment, you can choose between two claims: Potential Claim 1: Yes, the concentration of cadmium exceeds the limit for healthy life. Potential Claim 2: No, the concentration of cadmium does not exceed the limit for healthy life. Directions : For question 1, evaluate each of the three figures in the question below and determine which one would support the claim “Yes, the concentration of cadmium exceeds the limit for healthy life.” 1. The answer choices below show the normal probability distribution that best represents the frequency distribution of the concentration of cadmium (mg L -1 ) in 20 samples taken from three different locations. In each figure, the y-axis represents the probability of observing a concentration of cadmium less than the corresponding concentration of cadmium on the x-axis. The x-axis represents the concentration of cadmium (mg L -1 ), with higher values indicating more cadmium

should follow the formatting guidelines listed below. Save your plot as an JPG, PDF, or .xls/.xlsx. Formatting Instructions ● Chart type: Column ● Y-axes title: “Frequency”; Font size = 18 ● Y-axis numbers: Font size = 14 ● X-axis title: “Concentration of cadmium (mg L -1 )”; Font size = 18 ● X-axis numbers: Font size = 14 ● Bins = Use the following 12 bins: 80 to 120, 120 to 160, 160 to 200, 200 to 240, 240 to 280, 280 to 320, 320 to 360, 360 to 400, 400 to 440, 440 to 480, 480 to 520, 520+ 3. Based on the frequency distribution, does a normal probability distribution reasonably model the concentration of cadmium in water? 4. Explain your answer to the previous question. Be sure to discuss the assumptions of a normal probability distribution and why, based on the frequency distribution, these assumptions seem appropriate (or not) for modeling the concentration of cadmium in water. Now that we have determined whether the data can be reasonably modeled by a normal probability distribution, we need to decide how to estimate the probability of observing a certain concentration of cadmium in the Sarcannian forest groundwater. Your approach to estimating this probability depends on whether the data can be reasonably modeled by a normal probability distribution. As a quick refresher, let’s walk through and practice using two approaches to estimating this probability.

8. Summarize the evidence that supports your claim, including how you determined whether the concentration of cadmium in water exceeds the recommended limit (or not), based on probability. Use quantitative evidence when possible.

Step 1: Calculate the probability that the concentration of other heavy metals in the water exceeds the recommended limit of 100 mg L -1 . Directions: Use the mean and standard deviations for each heavy metal listed at the beginning of this appendix to answer questions 9 - 12. Assume the data presented at the beginning of the appendix are reasonably modeled by a normal probability distribution. Use Excel for calculations, modeling, and graphing. Round all calculated values to the nearest one's place. For example, if you calculate the value as 3.8218%, round to 4%. 9. Calculate the probability that a water sample from the Sarcannian Forest soil has a concentration of chromium greater than 100 mg L -1 (> 100 mg L -1 ). Express your answer as a percentage. Probability that a water sample has a concentration of chromium greater than 100 mg L -1 expressed as a percentage = 10. Calculate the probability that a water sample from the Sarcannian Forest soil has a concentration of copper greater than 100 mg L -1 (> 100 mg L -1 ). Express your answer as a percentage. Probability that a water sample has a concentration of copper greater than 100 mg L -1 expressed as a percentage = 11. Calculate the probability that a water sample from the Sarcannian Forest soil has a concentration of selenium greater than 100 mg L -1 (> 100 mg L -1 ). Express your answer as a percentage. Probability that a water sample has a concentration of selenium greater than 100 mg L -1 expressed as a percentage =

produce sets of offspring, referred to as families. If all offspring are raised in the same environment, variation in phenotypes within and among families likely depends on the alleles inherited from parents. In this example, three families are shown, each resulting from the mating of parents with identical phenotypes (the shading of the icons represents the phenotype). The offspring in each family have the same phenotype as their parents. The phenotypes of the parents and offspring enable one to infer the effect of the genotype on the phenotype. Since you left the Sanctuary, GUS and IRMA have directed a team of agricultural robots to prepare our breeding experiment. This team produced offspring from three sets of parents: (1) a metal-tolerant parent mated to a metal-tolerant parent (T x T); (2) a metal- tolerant parent mated to a metal-intolerant parent (T x I); and (3) a metal-intolerant parent mated to a metal-intolerant parent (I x I). If metal tolerance depends on a gene, each type of offspring should have a distinct genotype, which produces a distinct phenotype. I will need you to help me design an experiment to quantify the effect of heavy metals on the phenotype of each genotype in Appendix 4 of this Mission Memo. But first, we need to predict the phenotypes that we expect to observe from our breeding treatments. And to do that, we must understand how offspring inherit their genotype and phenotype from their parents. We will follow three steps to predict the phenotypes that we expect to observe. Step 1: Determine the relationship between alleles for a gene. Determine the nature of the relationship between two alleles of the same gene (dominant/recessive) and explain why. This step will help us to predict the phenotypes of offspring in Steps 2 and 3. Step 2: Practice predicting the probability of observing genotypes and phenotypes in offspring. Predict the genotypic and phenotypic outcomes of a cross between two parents, including estimating probabilities of observing certain genotypes and phenotypes. This step will help us to predict the phenotypes of sunstalk offspring from the breeding experiment in Step 3. Step 3: Predict the probability of observing metal tolerant offspring resulting from breeding of sunstalks. Predict the genotypes and phenotypes of sunstalk offspring. This step will ultimately prepare us to analyze the results of the breeding experiment that you will design in Appendix 4 of this Mission Memo.

To explore the relationship between genotype and phenotype, let’s consider an example in humans: cystic fibrosis. This genetic disorder causes a thick mucus to accumulate in a person's lungs, causing infections and other illnesses. Using our understanding of genes, we can predict the probability that a child will be born with cystic fibrosis. Cystic fibrosis is caused by an allele of the gene that codes for a protein called CFTR. This gene has more than 1200 alleles, but we simplify our analysis by focusing on two of these alleles. We’ll use the abbreviation C 1 to represent the allele that codes for a functional copy of the CFTR protein. When lung cells use the C 1 allele to produce the CFTR protein, the person has no risk of cystic fibrosis. Another allele, which we will abbreviate as C 2 , codes for a dysfunctional copy of the CFTR protein. When lung cells use the C 2 allele to produce the CFTR protein, the protein folds abnormally and fails to function as expected; consequently, a person with the C 2 allele may develop cystic fibrosis (Figure 5). Figure 5. This figure provides a generic illustration of the difference between two alleles, C 1 and C 2 , for the CFTR gene . The top row shows the amino acid sequence and subsequent protein structure produced by the C 1 allele of the CFTR gene , while the

genotype do not express the functional version of the CFTR protein, which means they do have CF. 16. Why is the C 1 allele dominant to the C 2 allele? Your answer should minimally consider how the function or dysfunction of the CFTR protein affects the phenotype. The dominant C 1 allele codes for a functional protein, while the recessive C 2 allele codes for a non functional protein. However, a person will only have cystic fibrosis if neither copy of the CFTR gene codes for a functional protein. In the case of the genotype C 1 C 1 , or C 1 C 2 , cells can produce some functional CFTR proteins, resulting in healthy lungs. This example underscores the primary cause of dominance: the dominant allele codes for a functional version of a protein, and a cell requires only one copy of this allele to produce enough copies of this protein to generate a phenotype. Step 2: Practice predicting the probability of observing genotypes and phenotypes in offspring. So how can we predict the probability that a person will develop cystic fibrosis? Imagine that a male human and a female human decide to mate. Both parents are heterozygous for the allele associated with cystic fibrosis; in other words, both parents have the genotype C 1 C 2 . The male’s sperm and the female’s egg have one copy of each gene. In this case, each cell has either the C 1 or C 2 allele. Given that both parents have one copy of the C 1 and one copy of the C 2 allele, each parent is equally likely to contribute the C 1 allele or the C 2 allele. However, we cannot know which allele resides inside the egg and which allele resides inside the sperm. Thus, we must consider all possible genotypes resulting from the mating of these parents. 17. If a biological parent has the C 1 C 2 genotype, what genotypes of sperm or egg could be produced?

Recall that a parent contributes only one of its two alleles to an offspring. For a parent that is heterozygous for the CFTR gene (C 1 C 2 ), only two possibilities exist for the genotype of the sperm or egg: 1) either a C 1 allele or a C 2 allele. Figure 7 (below) illustrates this phenomenon. Figure 7. This figure illustrates the possible genotypes of gametes produced by a biological mother and a biological father if both have the C 1 C 2 genotype for the CFTR gene , which is located on chromosome 7 in humans. Half of the gametes produced by each individual have the C 1 allele, while the other half have the C 2 allele. Now that you’ve considered the genotypes of sperm or egg that could be produced by the parents, you’re ready to predict the genotypes and phenotypes of their offspring. In sexually reproducing species, we must consider all possible combinations of eggs and sperm when predicting the genotypes of offspring. Give it a try. 18. If a male and a female both have the C 1 C 2 genotype, what are the possible genotypes of their offspring?

19. If a male and a female both have the C 1 C 2 genotype, what are the possible phenotypes of their offspring? If both parents have the C 1 C 2 genotype, their sperm or eggs could contain either the C 1 or C 2 allele. The sperm and egg each contain one copy of each gene from the male and female, respectively. After a sperm fertilizes an egg, the fertilized egg will contain two sets of each chromosome and thus two copies of each gene (half from each parent). Since we cannot know which sperm will fertilize which egg, we must consider all possible combinations of sperm and egg to predict the genotype and phenotype of an offspring. Let's expand on Figure 7. Figure 8. This figure illustrates the possible genotypic and phenotypic combinations that could be produced from all possible pairings of the gametes of the biological parents.

As Figure 8 shows, four combinations of egg and sperm may occur. The possible genotypes resulting from these combinations are as follows: Pairing 1: Genotype = C 1 C 1 ; Phenotype = no cystic fibrosis Pairing 2: Genotype = C 2 C 1 ; Phenotype = no cystic fibrosis Pairing 3: Genotype = C 1 C 2 ; Phenotype = no cystic fibrosis Pairing 4: Genotype = C 2 C 2 ; Phenotype = cystic fibrosis This process may seem a bit complicated at first. Thankfully, your fellow scientists on Earth have devised a quick method to determine every possible combination of alleles. This technique uses a set of boxes called a Punnett square. A Punnett square enables one to determine the possible combinations of alleles produced when two parents breed. In the case of a single gene, such as the CFTR gene, we can use a Punnett square to determine every combination of alleles. The first row and the first column on the outer portion of the Punnett square represents the alleles of each parent. Each cell in the Punnett square represents a potential genotype of a single offspring. For a gene in a diploid organism (2 copies of each gene) such a human or a sunstalk, four combinations of alleles can occur, each represented by one of the boxes (Figure 9).

Figure 9. This figure shows an example of a Punnett square and breaks down what each row, column, and cell represents. A Punnett square enables one to determine the possible combinations of alleles produced when two parents reproduce. The first row and the first column on the outer portion of the Punnett square represents the alleles of each biological parent. Each cell in the Punnett square represents a potential genotype of a single offspring. For a gene in a diploid organism (2 copies of each gene) such a human, four combinations of alleles can occur, each represented by one of the boxes. Let’s use the Punnett square to check our answers to questions 18 and 19. 20. On a sheet of paper, draw the Punnett square illustrating the combinations of alleles that could occur when both a male and a female have the C 1 C 2 genotype, and upload it to Canvas. Compare the results of your analysis to your answers to questions 18 - 19. Figure 10 (below) shows all combinations of alleles that could occur when both a male and a female have the C 1 C 2 genotype.

Figure 10. In this Punnett square, the biological mother contributes eggs that contain either the C 1 or C 2 allele of the CFTR gene , while the biological father contributes sperm that contain either the C 1 or C 2 allele of the CFTR gene . Each of the four boxes in the Punnett square represents the genotype of an offspring. For a gene in a diploid organism (2 copies of each gene), four combinations of alleles can occur, with each combination being represented by one of the boxes. Here are the results of the Punnett square in Figure 10 (from left to right, top to bottom): Upper left cell: Genotype = C 1 C 1 ; Phenotype = no cystic fibrosis Upper right cell: Genotype = C 2 C 1 ; Phenotype = no cystic fibrosis Lower left cell: Genotype = C 1 C 2 ; Phenotype = no cystic fibrosis Lower right cell: Genotype = C 2 C 2 ; Phenotype = cystic fibrosis The results in Figure 8 and Figure 10 are the same, but a Punnett square ensures that we

consider all possible combinations of alleles. Scientists on Earth express such results as a probability. The probability of observing a specific genotype, such as C 2 C 2 , depends on the relative frequency of observing a given genotype relative to all possible genotypes. Let’s consider the Punnett square in Figure 10 as an example. One of the four possible genotypes of the offspring is C 2 C 2 , therefore the relative frequency of observing the genotype C 2 C 2 is 1/4, or 0.25. In other words, the probability of obtaining an offspring with the C 2 C 2 genotype equals 25%. Help me determine the probability of observing other genotypes or phenotypes in this example. As a note, the order in which alleles are written in a genotype does not matter. For example, the genotype C 2 C 1 and the genotype C 1 C 2 are the same genotype. Directions: For questions 21 - 24, round all calculated values to the nearest one’s place. For example, if you calculate the value as 3.8218%, round to 4%. 21. What is the probability of observing an offspring with the C 1 C 1 genotype? Express your answer as a percentage. 22. What is the probability of observing an offspring with a heterozygous genotype (C 1 C 2 or C 2 C 1 )? Express your answer as a percentage. 23. What is the probability of observing an offspring that does not have cystic fibrosis? Express your answer as a percentage. 24. What is the probability of observing an offspring that has cystic fibrosis? Express your answer as a percentage. For more complex scenarios, we might write out the steps involved in calculating the probability of observing each genotype or phenotype in an offspring. For example, we

could list the steps in calculating the probabilities for the cross in Figure 10. Genotypic probabilities - Probability of C 1 C 1 = (1 combinations of alleles as C 1 C 1 ) / (4 potential combinations of alleles) = 1/4 = 0.25 - Probability of C 1 C 2 = (2 combinations of alleles as C 1 C 2 ) / (4 potential combinations of alleles) = 2/4 = 0.50 - Probability of C 2 C 2 = (1 combinations of alleles as C 2 C 2 ) / (4 potential combinations of alleles) = 1/4 = 0.25 Phenotypic probabilities - Probability of an offspring that does not have cystic fibrosis = (3 combinations of alleles that have at least one C 1 allele) / (4 potential combinations of alleles) = 3/4 = 0.75 - Probability of an offspring that has cystic fibrosis = (1 combinations of alleles as C 2 C 2 ) / (4 potential combinations of alleles) = 1/4 = 0.25 Now that we understand how genes pass from parents to offspring, we’re ready to predict the genotypes and phenotypes of offspring from experimental breeding of sunstalks. Step 3: Predict the probability of observing metal-tolerant offspring resulting from the experimental breeding of sunstalks. If the metal tolerance of sunstalks has a genetic basis, at least one gene must contribute to the degree of metal tolerance observed in a sunstalk. As with many species on Earth, sunstalks have diploid cells, each containing two copies of each gene. We’ll define these hypothetical alleles as A 1 and A 2 . Assume that the A 1 allele codes for metal in tolerance, while the A 2 allele codes for metal tolerance. We will conduct a breeding experiment to confirm the presence of a gene that affects metal tolerance. The results of this experiment will also tell us whether the allele that confers metal tolerance is dominant, recessive, or incompletely dominant to the allele that confers metal intolerance.

In the experiment, we will breed three sets of parents: 1. Metal tolerant phenotype (A 2 A 2 ) x metal tolerant phenotype (A 2 A 2 ) 2. Metal tolerant phenotype (A 2 A 2 ) x metal intolerant phenotype (A 1 A 1 ) 3. Metal intolerant phenotype (A 1 A 1 ) x metal intolerant phenotype (A 1 A 1 ) Before conducting this experiment, we should consider each model of genetics and predict the phenotypes that we should observe if this model is correct. For each set of parents, we must predict the probability of observing the metal tolerance phenotype in the offspring of sunstalks, given that the allele for metal tolerance (A 2 ) is dominant, recessive, or incompletely dominant to the allele for metal intolerance (A 1 ). Once we have completed the breeding experiment, we will determine whether the data support any of these models. Model 1: A dominant allele for metal tolerance (A 2 ) exists at a single locus. In the first model, a single gene exists in which an allele that confers metal tolerance (A 2 ) is dominant to an allele that confers metal intolerance (A 1 ). For each set of parents, use this model to predict the probability of observing the metal tolerance phenotype in the offspring of each set of parents. Directions : For questions 25 - 27, assume that the allele for metal tolerance (A 2 ) is dominant to the allele for metal intolerance (A 1 ). Round all calculated values to the nearest one’s place. For example, if you calculate the value as 3.8218%, round to 4%. For each question, estimate the probability of observing the metal tolerance phenotype in the offspring produced by each set of parents: 1. Metal tolerant phenotype (A 2 A 2 ) x metal tolerant phenotype (A 2 A 2 ) 2. Metal tolerant phenotype (A 2 A 2 ) x metal intolerant phenotype (A 1 A 1 ) 3. Metal intolerant phenotype (A 1 A 1 ) x metal intolerant phenotype (A 1 A 1 ) 25. For the first set of parents (A 2 A 2 x A 2 A 2 ), what is the probability of observing an offspring that has the metal tolerance phenotype? Express your answer as a percentage.

26. For the second set of parents (A 2 A 2 x A 1 A 1 ), what is the probability of observing an offspring that has the metal tolerance phenotype? Express your answer as a percentage. 27. For the third set of parents (A 1 A 1 x A 1 A 1 ), what is the probability of observing an offspring that has the metal tolerance phenotype? Express your answer as a percentage. Model 2: A recessive allele for metal tolerance (A 2 ) exists at a single locus. In the second model, a single gene exists in which an allele that confers metal tolerance (A 2 ) is recessive to an allele that confers metal intolerance (A 1 ). For each set of parents, use this model to predict the probability of observing the metal tolerance phenotype in the offspring of each set of parents. Directions : For questions 28 - 30, assume that the allele for metal tolerance (A 2 ) is recessive to the allele for metal intolerance (A 1 ). Round all calculated values to the nearest one’s place. For example, if you calculate the value as 3.8218%, round to 4%. For each question, estimate the probability of observing the metal tolerance phenotype in the offspring produced by each set of parents: 1. Metal tolerant phenotype (A 2 A 2 ) x metal tolerant phenotype (A 2 A 2 ) 2. Metal tolerant phenotype (A 2 A 2 ) x metal intolerant phenotype (A 1 A 1 ) 3. Metal intolerant phenotype (A 1 A 1 ) x metal intolerant phenotype (A 1 A 1 ) 28. For the first set of parents (A 2 A 2 x A 2 A 2 ), what is the probability of observing an offspring that has the metal tolerance phenotype? Express your answer as a percentage.

29. For the second set of parents (A 2 A 2 x A 1 A 1 ), what is the probability of observing an offspring that has the metal tolerance phenotype? Express your answer as a percentage. 30. For the third set of parents (A 1 A 1 x A 1 A 1 ), what is the probability of observing an offspring that has the metal tolerance phenotype? Express your answer as a percentage. Model 3: An incompletely dominant allele for metal tolerance (A 2 ) exists at a single locus. In the third model, a single gene exists in which an allele that confers metal tolerance (A 2 ) is incompletely dominant to an allele that confers metal intolerance (A 1 ). What do we mean by incomplete dominance? When an allele is incompletely dominant to another allele, the heterozygous genotype (i.e., A 1 A 2 ) has a phenotype that lies in between the phenotypes of the homozygous genotypes (i.e., A 1 A 1 and A 2 A 2 ). Put another way, the phenotype of the heterozygous genotype appears to be a blend of the phenotypes of the homozygous genotypes. Let’s consider an example of incomplete dominance that occurs on Earth. Snapdragons are a type of flowering plant in which flower color depends on two alleles that are incompletely dominant to one another. We will label these alleles for the gene that codes for flower color as R 1 and R 2 . The R 1 allele codes for a red pigment, while the R 2 allele codes for a white pigment. When a plant is homozygous for R 1 (i.e., R 1 R 1 ), it produces red flowers. When a plant is homozygous for R 2 (i.e., R 1 R 1 ), it produces white flowers. However, when a plant is heterozygous (i.e., R 1 R 2 ), it produces pink flowers—a color that is intermediate to red and white (Figure 11). Figure 11 shows a Punnett square of a breeding between a snapdragon with red flowers and a snapdragon with white flowers. Even in the case of incomplete dominance, we can use a Punnett square to predict the probability that an offspring will inherit a particular genotype or phenotype. For example, in Figure 11, the chance that an offspring will inherit the heterozygous genotype and produce pink flowers is 4/4, or 100%.

Figure 11. This Punnett square illustrates the possible genotypic and phenotypic outcomes of snapdragon offspring for the trait flower color. In this example, assume flower color is controlled by a single gene that has two alleles, R 1 and R 2 . The R 1 allele codes for a red pigment, while the R 2 allele codes for a white pigment. These alleles are incompletely dominant to one another. This means that when a plant is heterozygous (i.e., R 1 R 2 ), it produces pink flowers—a color that is intermediate to red and white. Directions : For questions 31 – 33, assume that the allele for metal tolerance (A 2 ) is incompletely dominant to the allele for metal intolerance (A 1 ). Round all calculated values to the nearest one’s place. For example, if you calculate the value as 3.8218%, round to 4%. For each question, estimate the probability of observing the metal tolerance phenotype in the offspring produced by each set of parents: 1. Metal tolerant phenotype (A 2 A 2 ) x metal tolerant phenotype (A 2 A 2 ) 2. Metal tolerant phenotype (A 2 A 2 ) x metal intolerant phenotype (A 1 A 1 )

3. Metal intolerant phenotype (A 1 A 1 ) x metal intolerant phenotype (A 1 A 1 ) 31. For the first set of parents (A 2 A 2 x A 2 A 2 ), what is the probability of observing an offspring that has the metal tolerance phenotype (both partial or full)? Express your answer as a percentage. 32. For the second set of parents (A 2 A 2 x A 1 A 1 ), what is the probability of observing an offspring that has the metal tolerance phenotype (both partial or full)? Express your answer as a percentage. 33. For the third set of parents (A 1 A 1 x A 1 A 1 ), what is the probability of observing an offspring that has the metal tolerance phenotype (both partial or full)? Express your answer as a percentage. Appendix 4 How should we design a breeding experiment to determine the genetic basis of metal tolerance? Thanks to your knowledge of genetics, we now know the genotypes and phenotypes of offspring produced by three sets of parents: (1) a homozygous metal-tolerant parent (A 2 A 2 ) mated to a homozygous metal-tolerant parent (T x T); (2) a homozygous metal-

tolerant parent mated to a homozygous metal-intolerant parent (A 1 A 1 ) (T x I); and (3) a homozygous metal-intolerant parent mated to a homozygous metal-intolerant parent (I x I). In this experiment, we’ll be testing three models, described below. ● Model 1: The metal tolerance allele (A 2 ) is dominant to the metal intolerance allele (A 1 ) and there is a single gene for metal tolerance located at a single locus. ● Model 2: The metal tolerance allele (A 2 ) is recessive to the metal intolerance allele (A 1 ) and there is a single gene for metal tolerance located at a single locus. ● Model 3: The metal tolerance allele (A 2 ) and the metal intolerance allele (A 1 ) are incompletely dominant to one another and there is a single gene for metal tolerance located at a single locus. A well-designed experiment should enable us to quantify the effect of heavy metals on the health of sunstalks. By exposing the genotypes of sunstalk offspring to different concentrations of heavy metals (the independent variable) and observing the health of these offspring (the dependent variable), we can document the degree of metal tolerance associated with each genotype. From these data, we can infer whether a gene affects metal tolerance and, if so, the relationship between the alleles for that gene. To draw accurate conclusions from our experiment, we must accomplish three goals: 1. Design treatments that control variables affecting the health of sunstalks. 2. Assign sunstalk offspring from each set of parents randomly to each treatment. 3. Include a sufficient number of sunstalks to replicate observations of metal tolerance. Therefore, we will follow three steps to design a breeding experiment to determine whether metal tolerance depends on a dominant, recessive, or incompletely dominant allele. Step 1: Identify appropriate treatments for controlling variables. A well-designed experiment controls the values of variables to create treatments. The value of a certain variable should differ among treatments, while the values of other variables remain constant among treatments. By identifying the appropriate values of each variable, we can ensure that our experiment isolates the effect of a single independent variable (genotype) on the dependent variable (metal tolerance).

Step 2: Randomize subjects among treatments and blocks. A well-designed experiment will randomize subjects among treatments distributed over space and time. By randomly distributing sunstalks among treatments, we decouple the variables that we are manipulating in the experiment from any variables associated with the identity of a sunstalk. By randomly distributing sunstalks in each treatment among blocks, we decouple our treatments from any variables associated with position of a sunstalk in space. Step 3: Determine the sample size in our experiment. A well-designed experiment will replicate observations in each treatment. By identifying the number of independent observations in each treatment—also known as the sample size—we can decide whether to generalize our conclusions about the genetic basis of metal tolerance in sunstalks. Step 1: Identify appropriate treatments for controlling variables. By raising sunstalks in controlled environments, we can set the concentration of heavy metals in soil while keeping the values of other variables constant. Before we consider the treatments for our experiment, let's review how to design treatments for any experiment. An experiment should have at least two treatments, so we can quantify the effect of an independent variable on a dependent variable. The value of the independent variable should differ between the treatments. The values of all other variables should remain constant between treatments. For example, imagine that you want to determine whether the concentration of nitrogen in soil affects the growth of a plant. You should design at least two treatments in which the concentration of nitrogen differed between treatments, while other variables had the same values in both treatments. Table 1 describes two treatments that accomplish this goal. Notice that the values of all but one variable remain the same between treatments. The concentration of nitrogen in soil is the only variable whose value differs between treatments. The soil in treatment 1 contains 15 ppm of nitrogen, whereas the soil in treatment 2 contains 40 ppm of nitrogen.

Variable Treatment 1: Low Nitrogen Treatment 2: High Nitrogen concentration of nitrogen in soil (ppm) 15 ppm 40 ppm concentration of water in soil (%) 40% 40% temperature of air (°C) 25°C 25°C light intensity (lux) 7500 lux 7500 lux If these treatments are maintained during the experiment, any difference in the growth of plants observed between treatments would likely have resulted from the difference in the concentration of nitrogen in soil. Therefore, if the plants in the treatment defined by 40 ppm of nitrogen grow more than the plants in the treatment defined by 15 ppm of nitrogen, we could infer that additional nitrogen caused these plants to grow more. Designing treatments to quantify the effect of heavy metals on the health of sunstalks Now, let's determine the treatments needed to determine whether the concentration of heavy metals affects the health of sunstalks. Consider the four experimental designs shown in Figure 12. Each design includes two environmental treatments that control the values of five variables.

Figure 12. Four experimental designs that differ in the values of five variables controlled in two treatments. The first column lists the names of variables that could affect the growth of a plant. The second and third columns list the values of these variables in the first and second treatment, respectively. Directions: Use the experimental designs shown in Figure 12 above to answer questions 34 - 35. 34. Which experimental design would enable one to infer the effect of heavy metals on the health of sunstalks? A. Design A B. Design B C. Design C D. Design D

35. Justify your response to the previous question. For the experimental design that you selected, explain why this design would enable you to conclude whether the concentration of heavy metals affects the health of a sunstalk. For each experimental design that you rejected, explain why this design would not enable you to conclude whether the concentration of heavy metals affects the health of a sunstalk. Step 2: Randomize subjects among treatments and blocks. Now that you have chosen the environmental treatments for our experiment, we need to assign the offspring to these treatments. Recall that these offspring resulted from breeding three sets of parents: (1) offspring of a homozygous metal-tolerant parent mated to a homozygous metal-tolerant parent (T x T); (2) offspring of a homozygous metal- tolerant parent mated to a homozygous metal-intolerant parent (T x I); and (3) offspring of a homozygous metal-intolerant parent mated to a homozygous metal-intolerant parent (I x I). Because each female sunstalk produces hundreds of seeds, we can have more than enough offspring for our experiment. GUS and IRMA will direct a team of agricultural robots to select 144 offspring from each set of parents, for a total of 432 sunstalks. Each of these sunstalks will experience one of the two environmental treatments that you selected. Although all offspring have been handled similarly, we cannot assume they have experienced identical conditions prior to the experiment. Therefore, we should randomly assign offspring of the three sets of parents to the two treatments. We will use stratified randomization to distribute the 432 offspring between our environmental treatments. Stratified randomization is the process of dividing subjects according to a categorical variable (e.g., genotypes of parents) and randomly assigning subjects in each category to each treatment. This process will ensure that we assign the same number of sunstalks from each set of parents to each environmental treatment. Before we begin, let's review an example of stratified randomization. Example: Stratified Randomization

Imagine that you want to determine whether the concentration of nitrogen in soil affects the growth of a plant. You designed two treatments that differ only in the concentration of nitrogen in the soil. All other variables—including water, temperature, and light—have the same values in both treatments. The subjects of this experiment are 30 plants of the same species. These plants have one of three genotypes, either A, B, or C. The sample of 30 plants comprises 10 plants of each genotype. How should we distribute the 30 plants between the treatments? Because the genotype of a plant could affect its growth, we should assign half of the plants of each genotype to each treatment. Otherwise, the genotypes of the plants could bias our observations about the effect of nitrogen on growth. Microsoft Excel enables one to randomly divide subjects within categories. Let's learn how to use the random function and the sort command in Excel to perform stratified randomization. The random function has the following syntax: =rand() When entering this syntax into a cell, the function returns a number between 0 and 1. This number will have an infinite number of decimal places, but Excel will display these numbers as rounded to a certain decimal place. In this example, we will use random numbers rounded to the nearest hundredth of a decimal place. Figure 13 contains a spreadsheet from Microsoft Excel. Rows 3 to 32 show data for 30 plants, each identified by a unique number (1-30) in column A and a genotype of offspring (A-C) in column B. In column C, the random function has been used to generate a random number for each plant.

Figure 13. This spreadsheet lists the identification numbers and genotypes of 30 plants in columns A and B, respectively. For each plant, the random function was used to generate a random number between 0 and 1 in column C. By associating a random number with each plant, we can use the sorting command on Excel to perform stratified randomization. First, we highlight all three columns of data

(cells A3-C32). Then, we select "Custom Sort" from the command menu. Finally, we sort first by genotype (column B) and then by random number (smallest to largest) within each genotype. Refer to the “ Using the Random Function to Assign Subjects to Experimental Treatments ” Excel tutorial for more information about how to use Microsoft Excel to perform stratified randomization. Figure 14 shows the result of sorting the 30 plants by their random numbers. Columns E, F and G contain the same data as columns A, B, and C, except that rows 3 through 32 have been sorted according to the random numbers in column G. To ensure an equal distribution of each genotype between treatments, the sorting by random numbers occurred within each genotype. After sorting, the five plants with the lowest random numbers were assigned to treatment 1, and the five plants with the highest random numbers were assigned to treatment 2. These assignments were recorded in column H.

Figure 14. This spreadsheet lists the identification numbers and genotypes of 30 plants in columns A and B, respectively. For each plant, the random function was used to generate a random number between 0 and 1 in column C. Columns E, F and G contain the same data as columns A, B, and C, except that rows 3 through 32 have been sorted according to the random numbers in column G. After sorting, the five plants with the lowest random numbers were assigned to treatment 1, and the five plants with the highest random numbers were assigned to treatment 2. These assignments were recorded in column H.

Using stratified randomization to assign sunstalks of each genotype to the treatments Now, we are ready to use stratified randomization to distribute the 432 offspring in our experiment between the two environmental treatments. Remember that we want to randomly assign an equal number of sunstalks from each set of parents to each environmental treatment. Given that we have 144 sunstalks from each set of parents, we should end up with 72 offspring from each set of parents in each environmental treatment. Directions: For question 36, download the Excel file, “Data: Stratified Randomization,” which contains a list of identification numbers and parental genotypes for 432 offspring. Use Excel to generate random numbers, sort sunstalks, and assign treatments. Generate random numbers in column C of the spreadsheet. Record the treatment assigned to each sunstalk (1 or 2) in column D. 36. Use stratified randomization to assign the 144 offspring of each set of parents to the two environmental treatments. When you complete this task, save the Excel file and upload this new version to Canvas. Randomly assign sunstalks to experimental blocks During the experiment, the sunstalks will be located in a large underground laboratory, designed for growing flora under controlled conditions. Think of this space as an artificial greenhouse, in which light comes from a source other than your sun. Although many variables will be controlled in this laboratory, most variables are impossible to control perfectly throughout a large space. Therefore, we will design experimental blocks to prevent any bias from uncontrolled variables associated with each sunstalk's location in the laboratory. Consider the four experimental designs in Figure 15, each of which contains six blocks of sunstalks. A well-designed experiment would contain blocks that randomize location of subjects to minimize any bias caused by uncontrolled variables.

Figure 15. Each of these four experimental designs contains six blocks. Each block is outlined by a dashed box. Each block consists of six sunstalks represented by circles. Gray circles represent sunstalks grown in contaminated soil (100 ppm of heavy metals), and white circles represent sunstalks grown in clean soil (0 ppm of heavy metals). The label in each circle represents the genotype of the sunstalk: the offspring of two metal- tolerant parents (T x T), the offspring of a metal-tolerant parent and a metal-intolerant parent (T x I), or the offspring of two metal-intolerant parents (I x I). Directions: Use Figure 15 to answer questions 37 - 38. 37. Which experimental design randomly distributes the genotypes and environmental treatments over space to minimize any bias caused by uncontrolled variables associated with the location of a sunstalk? A. Design A B. Design B C. Design C D. Design D

38. Justify your response to the previous question. For the experimental design that you selected, explain why this design minimizes any bias caused by uncontrolled variables associated with the location of a sunstalk. For each design that you rejected, explain why this design does not minimize any bias caused by uncontrolled variables associated with the location of a sunstalk. Step 3: Determine the sample size in our experiment. Now that we have designed the treatments and randomly distributed the offspring between these treatments, let's ensure that our experimental design includes sufficient replication to accurately quantify the effect of heavy metals on the health of sunstalks. In this experiment, we will observe the health of each sunstalk when grown in one of two environmental treatments: metal-contaminated soil (300 ppm of heavy metals) or clean soil (0 ppm of heavy metals). Therefore, the concentration of heavy metals in soil is an independent variable, and the health of a sunstalk is a dependent variable. The health of each sunstalk will be quantified as the percentage of healthy leaf tissue. When grown in clean soil, all genotypes should have a very high percentage of healthy leaf tissue. When grown in metal-contaminated soil, the percentage of healthy tissue should vary among the genotypes. The exact pattern of variation among genotypes depends on whether the allele that confers metal tolerance (A 2 ) is dominant, recessive, or incompletely dominant to the allele that confers metal intolerance (A 1 ). Therefore, the genotype of a sunstalk is also an independent variable in our experiment. In addition to controlling variables and randomizing subjects, we must replicate observations to accurately quantify the effects of the genotype and concentration of heavy metals in soil on the health of a sunstalk. Replication is a process in which multiple subjects are given the same treatment independently. The amount of replication in an experiment equals the number of independent observations of the dependent variable, also called the sample size. To see how we define replication in practice, let's consider an example. Imagine that you wanted to quantify how the concentration of nitrogen in soil affects the growth of a plant. You designed two treatments that differ only in the concentration of nitrogen in soil. You obtained a set of 30 plants, representing 3 genotypes: A, B and C.

Figure 16 contains a spreadsheet from Microsoft Excel in which rows 3 to 32 show data for 30 plants, representing three genotypes. Each plant is identified by a unique number (1-30) in column A and its a genotype (A-C) in column B. In column C, the plants have been randomly distributed between two treatments (1 or 2). Figure 16. This spreadsheet lists the identification numbers and genotypes of 30 plants in columns A and B, respectively. Each plant

has been randomly assigned to a treatment in column C. Let's evaluate the amount of replication in this experiment to determine the sample. To do so, we must ask a fundamental question. What is the subject of the experiment? In other words, what are we studying? Recall that we wanted to quantify how the concentration of nitrogen in soil affects the growth of plants. Therefore, plants are the subjects of this experiment. And we have 30 plants in our experiment. Based on this information, we might be tempted to conclude that we have a sample size of 30 subjects. However, that conclusion ignores the fact that these 30 plants were divided between two treatments. When quantifying the growth of plants in each treatment, the sample size is actually 15 plants. Now, let's consider how the genotypes of these plants affect the sample size. If we do not want to conclude anything about the effect of genotype on the growth of a plant, we do not need to account for the genotype when calculating the sample size. But what if we were interested in knowing whether the three genotypes respond differently to the concentration of nitrogen in soil? In that case, our sample size is much fewer than 15 plants. Instead, we count the number of subjects of each genotype in each treatment, which means we would have a sample size of only 5 plants per genotype per treatment. That's probably too few observations to accurately detect a pattern. From this example, you can appreciate how difficult it is to replicate observations in an experiment. Generally, one should aim for a sample size of at least 30 subjects per group, with the group being defined by the combination of effects that we want to quantify. For the example described above, we would need to independently expose 30 plants of each genotype to each treatment. That's a lot of plants! And if we wanted to consider the effect of a third variable, such as light, we would need to double that number of plants to ensure 30 plants of each genotype in each combination of nitrogen and light. Based on this example, you should be able to determine the sample size in the experiment that we designed to test our hypotheses about the effect of heavy metals on the health of sunstalks representing each of the three genotypes. Directions: Use the information provided above to answer questions 39 – 40. 39. What is the sample size in our experiment?

A. The sample size is 432 sunstalks, because we have 432 sunstalks in the experiment. B. The sample size is 206 sunstalks, because we have 206 sunstalks in each environmental treatment. C. The sample size is 144 sunstalks, because we have 144 sunstalks of each genotype in the experiment. D. The sample size is 72 sunstalks, because we have 72 sunstalks of each genotype in each environmental treatment. E. The sample size is 1 sunstalk, because we have 1 sunstalk of each genotype in each environmental treatment in each experimental block. 40. Justify your response to the previous question. For the sample size that you selected, explain why our experimental design offers this number of independent observations per group. For each sample size that you rejected, explain why our experimental design would not provide this number of independent observations per group.