GENERAL BIOLOGY LAB 1 (BSC1010L) Lab #8: Mendelian Genetics

GENERAL BIOLOGY LAB 1 (BSC1010L) Lab #8: Mendelian Genetics OBJECTIVES: Understand Mendel s laws of segregation and independent assortment. Differentiate between an organism s genotype and phenotype. Recognize different patterns of inheritance. Perform monohybrid and dihybrid crosses. Use pedigree analysis to identify inheritance patterns. INTRODUCTION: Through his studies of the inheritance patterns of the garden pea, Pisum sativum, Gregor Mendel changed our understanding of heredity. Mendel studied characters/traits that differed between plants and designed cross-fertilization experiments to understand how these characters transmit to the next generation. The results of Mendel s work refuted the prevailing hypothesis of blending inheritance and provided a new framework for understanding genetics. Ultimately, Mendel postulated two laws to explain heredity: (1) the law of segregation and (2) the law of independent assortment. Monohybrid crosses and the law of segregation The law of segregation, also termed the first law, states that during gamete formation the alternate forms of a gene (i.e. alleles) on a pair of chromosomes segregate randomly so that each allele in the pair is received by a different gamete. For example, if you were to examine the gene responsible for petal color, you may discover that the gene can be expressed as either yellow or white flowers. In this scenario, the gene is petal color, while the alleles are yellow and white. Depending on which allele is expressed, petal color will vary. Examine Figure 1 below making sure that you can follow the path of each allele from parent to offspring. Figure 1. Schematic of Mendel s law of segregation 1

In diploid organisms, all alleles exist in pairs; identical alleles within a pair are homozygous, while different alleles are heterozygous. Allele forms are represented by a single letter that explains whether a particular trait is dominant or recessive. Dominant alleles are assigned an uppercase letter (E), while recessive alleles are lowercase (e).in general, a dominant trait is expressed when at least one of the alleles present in the resulting allelic pair is dominant (EE or Ee). In contrast, for a recessive trait to be expressed, both alleles within the pair must be recessive (ee). For example, when considering ear lobe shape, two forms (attached and unattached) are apparent (Fig. 2). This trait is regulated by a single gene where unattached ear lobes are dominant (E) while attached ear lobes (e) are recessive. Figure 2. (a) Unattached (EE or Ee) vs. (b) attached earlobes (ee) An organism s genotype (EE, Ee, ee) is the combination of alleles present whereas the phenotype is the physical expression of the genotype. In the earlobe shape example above, an individual can have a genotype of EE, Ee or ee. People with EE or Ee genotypes have the unattached earlobe phenotype (Fig 2a), while those with an ee genotype express the attached earlobe form (Fig 2b). Note that dominant traits can be either homozygous (EE) or heterozygous (Ee) while recessive traits are always homozygous (ee). Question: Given that the allele for brown eyes (B) is dominant and the allele for blue eyes (b) is recessive, which of the following genotypes would result in individuals with brown eyes? Which genotype(s) is/are homozygous and which is/are heterozygous? BB: Bb: bb: 2

In today s lab you will use the concepts of Mendelian Genetics to solve problems regarding inheritance. TASK 1 Patterns of Inheritance I: Simple Dominance Simple dominance is the term used to describe a common outcome of allelic combinations, where one allele, if present, will dominate over the other and will be expressed. Information about alleles present in a parental population can be used to determine the probability of different genotypic and phenotypic ratios for a variety of traits in the offspring. In instances when only 1 or 2 traits are being considered the Punnett square (Fig. 3) approach is used to predict the possible outcomes of the parental cross. When only one trait is being considered the cross is monohybrid while a dihybrid cross involves 2 traits. General instructions on how to perform a cross using the Punnett square approach: 1. Write down the genotypes of the parents 2. Note the gametes that each parent can contribute 3. Draw a Punnett Square 4. Across the top write the gametes that one parent contributes and along the side write the gametes contributed by the other parent 5. Perform the cross 6. Determine the genotypic and phenotypic ratios Figure 3. Example of a Monohybrid cross In the example above (Fig. 3), the genotypic ratio is 1:2:1 (1: CC, 2: Cc, 1: cc) while the phenotypic ratio is 3:1. Since C = curly hair and c = straight hair, ¾ of the possible offspring will have curly hair while only ¼ will have straight hair. 3

Procedure: 1. You will now simulate a cross between two heterozygous individuals, Tt and Tt. Each group should obtain two coins from your TA. You will flip the coins simultaneously to represent the potential outcomes of a cross between two Tt individuals. A head represents the dominant tall allele (T) while a tail symbolizes the recessive dwarf allele (t). Before you begin flipping the coins, perform the Tt x Tt cross in the Punnett square below to estimate the expected genotypic and phenotypic ratios. Parent 1 Parent 2 Based on this cross, what do you anticipate the genotypic and phenotypic ratios to be? Write your hypotheses (H o and H a ) in Table 1. Table 1: Expected Genotypic Ratio Expected Phenotypic Ratio 2. Begin flipping the two coins simultaneously for a total of 64 times. Record your results in Table 2. 4

Table 2: Response Number TT Tt tt Questions: a. What ratio of allele combinations did you observe? b. What genotypes and phenotypes result from these crosses? c. What are the genotypic and phenotypic ratios? d. How did your results compare to your expectations? Do your results support or reject your null hypothesis? 5

e. Do you think your results would have been closer if you flipped the coins 6400 times instead of just 64? Why or why not? 3. Albinism, a recessively inherited trait, results in organisms that lack pigment in the skin, hair or eyes. A female with normal pigmentation, but who had an albino mother, mates with an albino male. They have one child. Using the information you have learned so far complete Table 3. Table 3: Genotype of child s mother Genotype of child s father Possible gametes of mother Possible gametes of father Possible genotype and phenotype of the offspring Genotypic ratio of children Phenotypic ratio of children TASK 2 - Patterns of inheritance II: Incomplete vs. Complete Dominance & Codominance Inheritance of traits can occur in multiple forms. So far you have considered complete dominance, where a homozygous dominant or a heterozygous individual expresses the dominant phenotype, while an individual that is homozygous recessive expresses the recessive phenotype. However, in certain cases a cross between two different allele forms results in a phenotypic expression that combines the two allelic traits. This type of inheritance is known as incomplete dominance. For example, if an offspring resulting from a cross between a red (RR) and a white (rr) snapdragon plant receives the dominant allele for red flower color (R) from one parent and the allele for white flower color (r) from the other, the resulting genotype will be Rr. The heterozygous form (Rr) of the plant will bear pink flowers since neither allele is completely dominant over the other (Fig. 4). 6

Figure 4. Pink snapdragons are an example of incomplete dominance 1. Determine the possible phenotypes of the F1 offspring when two pink snapdragons are crossed. Show your work in the space provided below. Parent 2 Parent 1 2. What would be the resulting genotypes of a cross between a pink and a white snapdragon? Show your work in the space provided below. Parent 2 Parent 1 7

Expression of both alleles of a particular gene is known as codominance. When alleles are inherited codominantly, both phenotypes are expressed at the same time in the heterozygous condition in contrast to incomplete dominance where the heterozygote is an intermediate between the two homozygotes (Fig. 5). Figure 5. Different types of inheritance The ABO blood type system is an excellent example of codominance. Humans have four blood types, namely, A, B, AB and O. All individuals carry two alleles, one from each parent. In this system, both alleles inherited determine one s blood type, where a person with Type AB blood possesses phenotypic traits of both A and B blood types (Table 4). Blood type (phenotype) Genotype Type A I A I A or I A i, Type B I B I B or I B i, Type AB I A I B Type O i i Table 4. Relationship between blood type and genotype For example, an individual with Type B blood can have two possible genotypes, I B I B or I B i, where I (dominant) and i (recessive) represent an allele from each parent (Table 4). The different blood types are characterized by the presence of a particular sugar molecule attached to the proteins on the surface of red blood cells (Fig 4). In Type A blood, the attached sugar molecule is galactosamine, while in Type B blood it is galactose. In contrast, individuals with Type O blood, have no sugars present on the surface of their red blood cells. These protein-sugar complexes are antigens that act as recognition markers for the immune system. The immune system is tolerant to its own antigens but produces antibodies against antigens that differ from 8

its own. The antibodies formed bind to the antigens causing agglutination (clumping) and lysis of the foreign red blood cells. Therefore, an individual with Type A blood could not receive a blood transfusion from a Type B blood donor because the antigens on the donor s red blood cells will trigger an immune response from the recipient s antibodies. Thus, the Type A recipient will produce antibodies against the donor s Type B antigens. Figure 6. Human ABO blood types When the wrong blood type is given to a patient, agglutination of the blood occurs and this can ultimately lead to death. Table 5 provides a quick overview of which blood types can donate to which, and which types can receive from which. A plus indicates that mixing of the donor and recipient blood types results in agglutination whereas a blank cell means that no agglutination occurs when the blood types are mixed. Recipient (Antibodies) Donor A B AB O (Antigens) A + + B + + AB + + + O Questions: Table 5. Effects of mixing different blood types 1. Given the information above, why do you think that Type O is the universal donor and AB is the universal recipient? 9

2. There was a mix of up children in the maternity ward of a hospital. The children in question and their blood types are listed below. Child 1: type A (genotype I A I A or I A i) Child 2: type B (genotype I B I B or I B i) Child 3: type AB (genotype I A I B ) Child 4: type O (genotype ii) Which child or children could belong to a couple having AB and O blood types? 3. Based on the previous question, is it possible to prove paternity based on blood types? Explain. 4. A woman with Type O blood has a child with the same blood type. Can the child s father have Type AB blood? Why or why not? Another trait involved in blood typing is the Rh factor. The Rh factor works along the principle of simple dominance instead of codominance. An individual who is Rh positive possesses the Rh antigen on his/her blood cells while someone who is Rh negative lacks Rh antigens. Generally, the Rh status of an individual is always included with the blood type. For example, a person that is AB+ has Type AB blood and is Rh positive. This information is 10

particularly important during pregnancy since Rh incompatibility can develop in women that are Rh- and who have an Rh+ baby. Mixing of maternal and fetal blood through the placenta can cause the mother to develop antibodies against the Rh antigens from the baby. This condition is usually not harmful to the first child but may cause mild to severe symptoms (depending on the amount of Rh antibodies created) during subsequent pregnancies since the mother s Rh antibodies attack the Rh antigens of the developing fetus. Determining blood type Procedure: 1. Prepare your station by obtaining the following supplies: a. 4 small plastic blood typing trays b. Toothpicks for mixing c. Four bottles of blood (on your table) representing four different individuals d. One bottle representing A antibodies e. One bottle representing B antibodies f. One bottle representing Rh antibodies 2. You will use each blood typing tray to determine the blood type of a particular individual. Note that each tray contains 3 wells, labeled A, B and Rh. 3. Add 3 drops of blood from individual 1 to every well in a blood typing tray. 4. Add 3 drops from the bottle labeled A antibodies to the well labeled A. 5. Add 3 drops of B antibodies to the well labeled B. 6. Add 3 drops of Rh factor solution to the well labeled Rh. 7. Mix each well with a toothpick. Note: Use a different toothpick for each well and tray. 8. After 1 min, examine the tray for the presence of crystals. Presence of crystals, indicates agglutination and a positive test* for a particular blood type and Rh factor. 9. Repeat steps 3-8 for the remaining individuals. *Important note: In this particular experiment, agglutination indicates a positive test for a particular blood type. For example, if agglutination occurs in well A, then the individual has blood type A. However, when working with real blood (i.e. for the purposes of transfusions), agglutination would be a negative result. With regular blood, agglutination would indicate that the antibodies of one s blood detected a foreign substance (an antigen), causing an immune response and cell lysis. 10. Record your results in Table 6. Note which wells agglutination occurs in each tray. Based on your results, determine the blood type of the four individuals examined. 11

Table 6: Agglutination (Yes or No) Individual Well A Well B Rh factor (+/-) Blood type TASK 3 - Patterns of Inheritance III: Lethal Inheritance Lethal inheritance, the last pattern to be examined, occurs when an offspring dies after acquiring a particular trait. An example of such a scenario would be albinism in plants. 1. Why is albinism fatal in plants? 2. Using the tray of plants in the front of the lab, count the number of green and albino plants you observe. Using this information, determine the ratio of green to albino plants. Green : Albino 3. Based on the above ratio, predict the genotypes of the parental generation. 12

Dihybrid crosses and the law of independent assortment Mendel s second law or the law of independent assortment states that alleles from different genes assort independently from one another during meiosis if they are located on separate chromosomes (Fig. 7). For instance, alleles for the attached earlobe gene will assort independently from those that determine height as long as they are located on different chromosomes. This seems easy enough to grasp but is often the bane of biology students. Think of it like this - if you are flipping a coin and recording the results of each toss (heads or tails), and your friend is doing the same thing, do your tosses have any effect on the results of his? No, because your two actions are not linked. The chance of getting any two events to happen at the same time is simply the product of the chance of each event happening at all. For example, the chance of getting a heads in a coin flip is 50%. However, the chance that you and your friend get heads on the same toss is.5 x.5 =.25 or 25%. Figure 7. Schematic of Mendel s law of independent assortment TASK 4 - Drosophila Virtual Genetics Lab Procedure: 1. Click to open the Drosophila Genetics Lab Program on the desktop. 2. Each group will perform one of the Single Gene Inheritance experiments (1, 2 or 3) per group. 3. Follow the instructions listed in the pdf under "Procedure" and then complete the "Results" and "Questions" sections. 4. Once completed, perform one of the Double Gene Inheritance experiments (4 or 5) and repeat step 3. 5. Lastly, perform experiment 15 on Lethal Inheritance and repeat step 3. 13

TASK 5 - Testing Mendel s laws of inheritance in Brassica rapa Throughout the semester you have been growing the rapid-cycling Brassica rapa plants. As you noted in previous labs, your plants have different phenotypes, i.e., some have purple stems and green leaves while others have green stems and yellow-green leaves. In fast plants, the presence of the anthocyanin pigment produces a purple color (P) that is apparent on the stems and leaf tips of the developing plant. In the absence of this pigment, the stems and leaves are bright green (p) in color. The second trait, yellow-green leaf color (y), being observed by some of you is the result of a recessive gene, that when present in the homozygous condition, produces leaves, stems and seed pods that are yellow-green in color. In contrast, plants that are homozygous dominant (YY) or heterozygous (Yy) for this trait produce leaves and stems that are green in color. The seeds you planted during the last lab period are the F2 generation that resulted from a cross of your F1 plants (see Fig. 8 and 9). The objective of this task is to investigate the pattern of gene inheritance (i.e. dominant vs. recessive) as well as Mendel s laws of segregation and independent assortment. You will also learn how to use a statistical test to examine how well the inheritance patterns observed in class agree with expected statistical ratios. Note: Before beginning your analyses, you will need to determine whether you are investigating single-gene (Fig. 8) or double gene (Fig. 9) inheritance patterns. This semester two separate types of F1 plants were grown and since both inheritance patterns can produce plants that bear purple stems and green leaves, you will need to check for the presence of trichomes (hairs) on the stem. Since the number of hairs is controlled by more than one gene, we will not consider this trait in the current exercise. Therefore if your plant s stem is hairy, then your investigation will be of monohybrid inheritance patterns since only the anthocyanin gene will be considered (see Fig 8). Figure 8. Possible F2 Genotypes and phenotypes for the Anthocyanin gene 14

Figure 9. Possible F2 Genotypes and phenotypes for the Anthocyanin and YellowGreen Leaf genes 1. Based on your records, what were the parental phenotypes (F1) of your group s plants? i. What are the possible genotypes that produced these phenotypes? 2. Examine the seedlings in your group and record the following information in Table 7. i. phenotypes of the plants in your group ii. genotype for each phenotype observed (see Fig. 8 and 9) iii. quantity of each genotype 15

Table 7: Group # # seedlings Phenotypes present (can be > 1) Genotype Quantity of Genotype 1 2 3 4 5 6 3. Combine your results with the data from the rest of the class in Table 8. Table 8: Phenotype Purple stem, green leaf Quantity Purple stem, yellow-green leaf Non-purple stem, green leaf Non-purple stem, yellow-green leaf Total number of plants 16

4. Depending on which inheritance pattern you are investigating, calculate the expected phenotypic ratios of the F2 population resulting from the cross of F1 individuals, either Pp x Pp or Ppyy x PpYy. Parent 1 Parent 1 Parent 2 Parent 2 5. Based on your expected ratios formulate null and alternate hypotheses (H o and H a ) regarding the phenotypic pattern you should observe in your F2 plants. 17

6. Chi-Square analyses: You will use this statistical test to examine how the predicted phenotypic ratios of Brassica rapa compare to your observed data. A test that is often applied to determine how well observed ratios fit expected statistical ratios is the chi-square (χ 2 ) or goodness of fit test. This test calculates the deviations of observed numbers from expected numbers into a single numerical value called χ 2. The difference between the number observed and the number expected for a particular phenotype is squared and then divided by the number expected. This is repeated for each phenotype class. The χ 2 value consists of the summation of these values for all classes. The formula for χ 2 (Suzuki, et al., 1986) is: χ 2 = (observed - expected) 2 expected Degrees of Freedom (df) 1 2 3 4 5 6 7 8 9 Possibility of Chance Occurrence in Percentage (5% or Less Considered Significant) 90% 80% 70% 50% 30% 20% 10% 5% (sig.) 1% 0.016 0.211 0.584 1.064 1.610 2.204 2.833 3.490 4.168 0.064 0.446 1.005 1.649 2.343 3.070 3.822 4.594 5.380 0.148 0.713 1.424 2.195 3.000 3.828 4.671 5.527 6.393 0.455 1.386 2.366 3.357 4.351 5.348 6.346 7.344 8.343 1.074 2.408 3.665 4.878 6.064 7.231 8.383 9.524 10.656 1.642 3.219 4.642 5.989 7.289 8.558 9.083 11.030 12.242 2.706 4.605 6.251 7.779 9.236 10.645 12.017 13.362 14.684 3.841 5.991 7.815 9.488 11.070 12.592 14.067 15.507 16.919 6.635 9.210 11.341 13.277 15.086 16.812 18.475 20.090 21.666 Table 9. χ 2 Values and Probabilities Associated with each χ 2 value is a probability that indicates the chance that, in repeated experiments, deviations from the expected would be as large as or even larger than the ones observed in this experiment. The calculated χ 2 value can be used to obtain probabilities, or p values, from a chi square table (Table 9). These probabilities indicate the likelihood that the observed deviations are due to random chance alone. The most widely accepted p value is 0.05, which means that in cases where the p < 0.05, the chance of obtaining the observed results by chance alone is 5% or less. In general, low χ 2 values indicate a high probability that the observed deviations could be due to random chance alone, while high χ 2 values indicate a low probability that the observed deviations can be explained by chance events and the deviations are likely the result of the experimental treatment. If the χ 2 value results in p < 0.05, there is less than a 5% chance that the observed results occurred by chance alone and the observed results are significantly different from the expected (i.e. you reject the null hypothesis). If p 0.05, then 5% of the time or more, any deviation from the expected results is due to chance only (i.e. you fail to reject the null hypothesis). The χ 2 test does not prove that a hypothesis is correct but evaluates whether or not the data and the hypothesis have a good fit. In Table 9, which lists probabilities and χ 2 values, note the Degrees of Freedom column. In any experiment, the degree of freedom (df) is a measure of the number of categories that are independent of each other. The df is one less than the number of different phenotypes 18

possible. In the current experiment we have four possible phenotypes so there are 4-1 = 3 df. Example: If, in an F 2 population of 100 plants, results are 60 wild type: 40 mutant (expected ratio would be 75 wild type : 25 mutant), then: χ 2 = (60-75) 2 + (40 25) 2 75 25 = 3 + 9 = 12.0 Because we have two possible phenotypes (mutant and wild), we have 2-1 df. Looking in the χ 2 table for χ 2 = 12 with 1 degree of freedom, probability is < 0.01; therefore, these results are not supportive of a 3:1 ratio since the probability is less than 5% (0.05). To read the χ 2 table, locate the df specific to your experiment in the first column. Now look across the row to find the probability column that best matches your calculated χ 2 value. The row matching 1 degree of freedom is marked by an orange square in Table 9. You will notice that our calculated χ 2 value is too high and the closest number to that value is 6.635 in the last column (circled in green). This means that the probability of obtaining the expected 3:1 ratio simply by chance is less than 1% therefore the null hypothesis must be rejected. 7. Now calculate the χ 2 based on your class data. Show all of your calculations in the space provided below. 8. Note the following information: Degrees of freedom: χ 2 : Probability of chance occurrence (from Table 9): 19

9. Based on your results, what can you conclude about your H o? 10. How do your observed ratios compare to those expected? Explain. 11. What do your class results suggest about the inheritance patterns of Brassica rapa? TASK 6 - Analyzing Pedigrees A pedigree is a map of relatives that is used to determine the inheritance pattern of a particular disease or trait. This map usually includes the gender of each family member, how each is related (through lines connecting individuals) and also provides information about genetic traits. Certain symbols are used to indicate these variables (see Fig. 10). 20

Figure 10. Symbols used in pedigree analysis Questions: 1. What would you look for in a pedigree to determine if a trait was dominant? 2. What would you look for in a pedigree to determine if a trait was recessive? 21

3. The following is a pedigree for albinism. Note: the half-filled circles and squares in this pedigree represent carriers but in other pedigrees carriers may not be clearly denoted as in this case. Can you determine if this trait is dominant or recessive? Explain. 4. Below is a pedigree where multiple individuals are afflicted with Brachydactyly or shortening of the digits. Can you determine if this trait is dominant or recessive? Explain. LOOK AHEAD: Before coming to lab next week, make sure to read the Molecular Biology task sheet as well as Chapter 7 (pgs. 71-74) in your lab manual. REFERENCES: http://www.fastplants.org/ 22