Unformatted text preview: Biology 1M03 Tutorial 4 Data Collection & PTC Tasting Student Package Objectives !"Understand the nature of dominant and recessive inheritance of a trait ! Design and execute an experimental test of a hypothesis ! Apply Chi square statistical test to population data ! Report findings in a scientific format ! Read the Introduction in this tutorial manual (pages 1‐4) ! Complete the ELM pre‐tutorial quiz Preparation Introduction Phenylthiocarbamide, or PTC, is a chemical compound that is notable for its duality in taste; it is either bitter and unpleasant or completely tasteless, depending on one’s genetic makeup. The ability to taste PTC is a dominant genetic trait, meaning an individual needs to possess only one copy of the “taster” allele to exhibit the “taster” phenotype. Recall that the phenotype describes the physiological traits of the organism, while the genotype is the organisms’ genetic makeup. Approximately 75% of people can taste PTC, though this ranges by ethnicity, from less than 60% in certain racial groups to nearly 100% in others. A single gene, TAS2R38, on chromosome 7 has been implicated in the ability to taste PTC (Kim et al 2003). The ability to taste PTC is associated with the ability to taste other bitter foods and substances, and evolutionary biologists suggest PTC tasting had an adaptive value for early humans, encouraging the avoidance of many bitter substances, which are often noxious or toxic. PTC tasters reportedly tend to avoid foods containing bitter compounds, including cruciferous vegetables such as cabbage, broccoli, brussel sprouts, turnips (Online Mendelian Inheritance in Man, 2008), and some reports suggest that PTC tasters are less likely to smoke, or drink coffee or wine (Kim et al, 2004). While some studies have shown that homozygous tasters (those with two copies of the taster gene) experience a more intense bitter sensation than heterozygous tasters; other studies suggest a second gene on chromosome 16 may mediate taste sensitivity. In either case, two things appear to be true about PTC tasting: 1) The ability to taste PTC is inherited as a dominant allele, and 2) some tasters experience a more intense bitterness, a phenomenon sometimes referred to as super‐tasting. In this tutorial, we will simply classify individuals as either “tasters” or “non‐tasters”. In this tutorial, you will measure some trait in your classmates that you predict may be associated with the ability to taste PTC. Once you measure the prevalence of your trait, you want to test whether the trait is distributed as one would expect if there were no significant relationship between your trait and the ability to taste PTC. To do this, we will conduct a chi square test, which is used to determine if the observed frequency of some event differs from the expected frequency for that event. Chi square testing is particularly useful when there are more than two possible outcomes, because the formula allows us to combine the differences between the observed and expected for any number of events: 1 The Chi square Test Chi square tests allow us to measure whether the observed outcomes for a particular set of events are statistically different from the expected outcomes. The example of the roll of a die provides an excellent illustration of the usefulness of the chi square test. Suppose you have a normal, six‐sided die and you would like to test the hypothesis that eachof the six sides may turn up with equal probability (i.e. 1/6). Thus, your null hypothesis is that there is no underlying factor causing any one side turn up more than 1/6 of the time. You decide to experiment, and you roll your die 30 times. Your record your results in a table : Die shows # times occured 6 4 4 7 4 5 You can use the chi‐square equation to determine whether your hypothesis that each side has a 1/6 probability of turning up is statistically significant. Your observed values are recorded in the table above, and your expected value for each of the six “outcomes” is 5, since 1/6 x 30 =5. !2 = ! (O‐ E)2 = ( 6 ‐ 5 )2 + ( 4 ‐ 5 )2 + ( 4 ‐ 5 )2 + ( 7 ‐ 5 )2 + ( 4 ‐ 5 )2 + ( 5 ‐ 5 )2 E 5 5 5 5 5 5 2 = (1) +(‐1)2 +(‐1)2 + (2)2 + (‐1)2 + (0)2 5 = 8/5 =1.6 So, your chi square statistic is 1.6. In order to reject the null hypothesis, the chi‐square statistic must exceed the critical value found on the chi square distribution chart corresponding to the degrees of freedom (Df) in our experiment and the desired level of statistical significance. The degrees of freedom in a chi square test is equal to the number of possible outcomes minus one. So here Df =6‐1 = 5. The desired level of statistical significance in this case is 0.05, which means there that, if we were to obtain a chi square value larger than the critical value, there is 95% certainty that the observed difference is real, and not due to chance. Statistical significance Df 0.5 0.10 0.05 0.02 0.01 0.001 1 0.455 2.706 3.841 5.412 6.635 10.827 2 1.386 4.605 5.991 7.824 9.210 13.815 3 2.366 6.251 7.815 9.837 11.345 16.268 4 3.357 7.779 9.488 11.668 13.277 18.465 5 4.351 9.236 11.070 13.388 15.086 20.517 2 Here we see the critical value is 11.070. Since 1.6 < 11.070, we cannot reject the null hypothesis; we conclude that our data do not provide any evidence that the dice are unfair. However, if the chi square statistic exceeds the critical value, we reject the null hypothesis. Now imagine you’d like to test whether some characteristic, or “outcome”, is randomly distributed or whether it is associated with the ability to taste PTC. You have a sample of 30 individuals, and first you test their ability to taste PTC. The results are as follows: Initial Sample: n=30 Taster type Non‐tasters Tasters n 9 21 Percentage 9/30 = 30% 21/30 = 70% You suspect that the ability to taste PTC is sex‐linked, so you re‐assess the original sample group recording the sex of the tasters and the non‐tasters. Of the original group of 30, 12 are male (6 tasters and 6 non‐tasters) and 18 are female (15 tasters and 3 non‐tasters). How can you determine whether PTC tasting ability is distributed within the two sex groups in a manner that is expected based on your previous data? We can tally the number of non‐tasters and tasters and within the two groups. This will give us the data to perform a two‐way chi‐square test to determine if the observed data differs from the expected. In other words, our null hypothesis is that the distribution of PTC tasting within the two groups (males and females) will be the same as the distribution within your original sample. We can begin by entering the observed values for the number of individuals in each category in the chart below. Note that the Totals should sum to 30 for both the total number of tasters, and the sum of males and females. OBS EXP Female OBS EXP Total Observed (by tasting ability) 9 21 30 3 15 18 Non‐tasters 7 Tasters 5 Total Observed (by gender) 12 Male To get the expected values, we must multiply the number of individuals in each sub‐population by the expected proportions derived from our original sample. 1. Males Expected tasters = (nmales) (expected proportion of tasters) = (12)(0.70) = 8.4 Expected non‐tasters = (nmales)(expected proportion of non‐tasters) = (12)(0.30) = 3.6 2. Females Expected tasters = (nfemales)(expected proportion of tasters) = (18)(0.70) = 12.6 Expected non‐tasters = (nfemales)(expected proportion of non‐tasters) = (18)(0.30) = 5.4 3. Fill in your expected values into the above table. 3 Now we are ready to conduct a two‐way chi square test. Recall the formula: ! = ! (O‐ E)2 E 2 !2 = ! (O‐ E)2 = (7 ‐ 3.6)2 + (5 ‐ 8.4)2 + ( 3– 5.4)2 + (15 – 12.6)2 E 3.6 8.4 5.4 12.6 = 11.56 + 11.56 + 5.76 + 5.76 3.6 8.4 5.4 12.6 = 3.211 + 1.376 + 1.067 + 0.457 = 6.111 Using the chi‐square distribution chart, we find the critical value that corresponds to one degree or freedom with a statistical significance, or p value, of 0.05: (Note: in a two‐way chi square test we have one degree of freedom because we multiply the number of alternatives minus one (n‐1), for each dimension. In this example we have (2‐1)(2‐1) = (1)(1) = 1 Statistical significance Df 0.5 0.01 0.05 0.02 0.01 0.001 1 0.455 2.706 3.841 5.412 6.635 10.827 2 1.386 4.605 5.991 7.824 9.210 13.815 3 2.366 6.251 7.815 9.837 11.345 16.268 4 3.357 7.779 9.488 11.668 13.277 18.465 5 4.351 9.236 11.070 13.388 15.086 20.517 Our chi‐square statistic, 6.111, is higher than 3.841. To report these results we can write: !2 (1, n = 30) = 6.111, p = 0.05 Degrees of freedom Level of statistical significance Sample size Chi square statistic Therefore, our results are significant and we can reject the null hypothesis. We can conclude that the ability to taste PTC is related to gender. 4 In‐ tutorial Exercise This tutorial will provide you with the opportunity to collect data on your classmate’s ability to taste PTC, then generate a hypothesis about how PTC tasting might be related to other preferences or traits such as smoking, or tolerating spicy foods. Step 1: Whole Class Data Obtain a strip of PTC paper from your TA. Give it a taste! Record your ability to taste the chemical on the board by making a tick mark under the category that applies to you: Non‐taster, Taster, or Super‐taster). When everyone has recorded their own result, your TA will tally the results for your class. Record the class results in the following table: Step 2: Data Collection and Hypothesis Testing Your TA will assign you a group. Within your group, come up with a prediction about a variable you suspect may co‐vary with the ability to taste PTC. Tolerance or preference for another strong taste, such as coffee or spicy foods might be a possible variable to investigate, but you are free to choose any thing you think might be worth analyzing. Be sure your variable has only two possible values, usually “yes” or “no”. In the box below, state the variable your group will assess, and state your null hypothesis. (HINT refer to the section on the null hypothesis in the Introduction) Variable: Null Hypothesis: Survey the class to determine the observed values for your chosen variable. In the data table on the next page: ‐Label the two rows at lefts with a description of your variable’s value (i.e. smoker/non smoker). ‐Fill in the observed values in the upper left of each divided cell. ‐Fill in the total observed values for each column and row. n % Whole Class: Sample size n= Non‐tasters Tasters 5 Data Table Variable OBS EXP Total Observed (by tasting ability) OBS EXP Non‐tasters Tasters Total Observed (by variable x) Now you need to calculate the expected values for your sample. Use the space below to calculate the Expected values, as demonstrated in the above example. Fill these values into your data table Step 3: Data Analysis Calculate the chi‐square statistic for your data in the space below. Does this value exceed the value found in the chi‐square distribution chart? "2 = !#(O‐ E)2 = E Step 4: Reporting your Data For the post‐tutorial, INDIVIDUAL assignment, you will report your group’s results using the handout provided on the next page. 6 Name: Tutorial # Group members: Question: Hypothesis: Null hypothesis: Methods: Explain your Methods in the space below. Be concise (2‐3 sentences) but be sure that another scientist could replicate your experiment based on the information you provide. You must also describe what statistical analysis you conducted. Tutorial 4: PTC Data Analysis Post‐tutorial Assignment: Report of Experimental Findings Student #:______ Results: Use space below to show your results. Present your results in a table and be sure to include the sample size, variables measured, and the observed and expected values. N= PTC taster? Taster Non‐taster Variable Observed Expected Data Analysis Use the space below to show the results of your data analysis. You do not need to include all of your calculations, but be sure to use proper format for reporting your chi‐square statistic. !2 ( ) = , p = Conclusion: State whether results are significant and whether they support the initial hypothesis Hand in your report to the 1M03 drop boxes by 1pm on the day following your tutorial. 7 References Kim UK, Jorgenson E, Coon H, Leppert M, Risch N, Drayna D (2003). Positional cloning of the human quantitative trait locus underlying taste sensitivity to phenylthiocarbamide. Science 299:1221–1225. Kim UK, Breslin PA, Reed D, Drayna D. (2004). Genetics of human taste perception. Journal of Dental Research 83(6):448– 53. Online Mendelian Inheritance in Man. 171200: Thiourea tasting. [Accessed August 5, 2008] http://www.ncbi.nlm.nih.gov/entrez/dispomim.cgi?id=171200 Wooding S, Kim UK, Bamshad MJ, Larsen J, Jorde LB, Drayna D. (2000). Natural selection and molecular evolution in PTC, a bitter‐taste receptor gene. American Journal of Human Genetics 74(4):637–46. 8 ...
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 Spring '11
 JonathanStone,JamesQuinn
 Biology, PTC, CHI SQUARE, null hypothesis, ptc tasting

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