Fruit Fly Exercise 1- Level 1

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Name StarGenetics Fruit Fly Exercise 1- Level 1 Description of StarGenetics In this exercise you will use StarGenetics, a software tool that simulates mating experiments, to analyze the nature and mode of inheritance of specific genetic traits. Getting started with StarGenetics To get to StarGenetics, please navigate to: http://web.mit.edu/star/genetics/. Click on the Start button to launch the application. Click Trust when a prompt appears asking if you trust the certificate. Click on File New in the drop-down menu in the upper left hand corner. Click on the Fruit Fly Exercise 1 Level 1file. You have begun a research project in a genetics laboratory that studies the fruit fly Drosophila melanogaster. Your research advisor decides to give you an interesting mutant fly called grounded that has just been discovered in one of the fly vials in the lab. In addition, you are also given two true-breeding wild type flies (male and female). All 3 flies can be found within the Strains box. 1 First, describe the sex and phenotype of the mutant fly. Describe the phenotype as it compares to wild type. The sex and phenotype of each fly should be visually obvious. You can also obtain this information by clicking on the fly and looking at the Properties window. 2 You want to determine the genotype of grounded. You are unsure whether its phenotype is caused by a dominant or recessive allele. You are also not sure if grounded is true-breeding or not. To determine its genotype, cross grounded with one of the wild type flies. Set up a cross between grounded and a wild type fly by dragging each to the Mating Site and by clicking on the Mate button. Each resulting offspring can be viewed within the Individual tab or a summary of the results is available on the Summary tab. Number of F1 flies that look like the wild type parent: Number of F1 flies that look like the grounded mutant parent: Total number of progeny generated: % of progeny that is wild type: % of progeny that is mutant: Ver. 3 L. Alemán, S. Bumgarner, & B. Lau 1

3 Based on the results that you obtained in question 2, does the mutant allele seem to be dominant or recessive to the wild type allele? Why? 4 Based on the results that you obtained in question 2, is the grounded parent fly a truebreeding fly? What is the likely genotype of this grounded fly? Please explain. You can use the Punnett Square tool to help determine genotypes for a given trait. In the Punnett Square tool, click on the different genotype options to see the resulting genotypic ratios. 5 Now you have determined some facts about the grounded allele and the trait that it causes. Given what you know, do you expect the mutant F1 flies to be homozygous or heterozygous for the allele that causes the grounded trait? According to your reasoning, if you mated two mutant F1 flies, what percentage of flies would you expect to be wild type versus mutant in the F2 progeny? Draw a Punnett square of this cross to justify your answer. 6 Now mate a mutant F1 female fly with a mutant F1 male fly. What percentage of flies are wild type and what percentage are mutant in these 50 F2 progeny? You can collect individual flies that you have generated (for use in future experiments) by dragging them to the Strains box or by clicking the Add to strains button within the Strains window. To start a new mating experiment, click on the Save experiment button. Number of F2 flies that look like the wild type parent: Number of F2 flies that look like the grounded parent: % of progeny that is wild type: % of progeny that is mutant: Ver. 3 L. Alemán, S. Bumgarner, & B. Lau 2

7 From the cross in question 6, you notice that the percentage of wild type and mutant progeny generated is not quite what you had predicted in question 5. Continue to mate the mutant F1 female fly with the mutant F1 male fly to produce 1000 progeny flies. What percentage of flies are wild type and what percentage are mutant in the 1000 F2 progeny? To add additional flies to one cross click on the Mate button again after the initial mating between the two F1 mutant flies. Number of F2 flies that look like the wild type parent: Number of F2 flies that look like the grounded parent: % of progeny that is wild type: % of progeny that is mutant: 8 Record the percentages of wild type and mutant flies that you obtained for both the 50 and the 1000 progeny sample sizes on the white board! 9 Do you notice a difference between the percentages of wild type and mutant progeny generated when you observe 50 F2 flies versus 1000 F2 flies? Which of these two sample sizes, the smaller set of 50 flies or the larger set of 1000 flies, do you think offers a more accurate view of the inheritance pattern of the grounded trait? Explain you answer. 10 You are surprised by the percentages of wild type and mutant flies that you observe in question 6. Based on the Punnett square that you drew in question 5, do you expect to find a mutant F2 fly that, when crossed with a wild type fly, produces only mutant progeny? See if you can find this fly by performing a few crosses. Are you surprised by your results? Explain your answer. Hint: You do not need to cross all of the mutant F2 flies to answer this question, just try some. 11 You are still very puzzled by your results. Then, you remember that you noticed an usually large number of underdeveloped dead embryos in your vial from the matings between the two F1 mutant flies (in question 6). Aha! State a new hypothesis to explain your data. Explain your reasons for proposing this hypothesis. Ver. 3 L. Alemán, S. Bumgarner, & B. Lau 3

12 Record your new hypothesis on the whiteboard to share with your classmates! 13 As you can see by comparing your data with that of your classmates, a lot of variability can be observed in the results from genetic crosses. You might have noticed that your results differ to some extent from what you expected, given the hypothesis that you suggested in question 10. How can you tell if the observed data diverge TOO MUCH from what you expected? That is, when must you conclude that your hypothesis couldn t be true, given your data? There is a statistical test that allows you to determine if any differences that you see in your genetic data are likely due to chance or, alternatively, if they indicate that your hypothesis is incorrect. This statistical analysis is called the Chi Square Test. It takes into account the size of your progeny sample and the amount of variation (for instance, the number of different phenotypes) that is being assessed in any particular cross. The formula for the Chi Square Test is: This formula is easier to understand and apply when expressed as shown in the chart below. a) Based on the Punnett square that you drew in question 5, you probably predicted that 75% of the F2 progeny flies would be mutant and 25% would be wild type. This prediction is where the numbers in the Expected column come from in the chart below. Fill in the numbers that you actually observed in question 7 in the Observed column below. Then perform the calculations indicated to complete the chart. Observed Expected (0 E) (0 E) 2 (0 E) 2 / E Mutant 750 Wild type 250 Total 1000 1000 - - χ2 = You can use this calculated Chi Square value ( χ2 ) to determine the probability (also called the p value) that any deviation that you saw between the Expected and Observed data is merely due to chance. To do this, you ll also need to determine the Degrees of freedom in your data. This is a measure of the amount of variation that existed in your experiment. There were only two different phenotypic classes (wild type and mutant) in your cross. You can calculate Degrees of freedom using the following formula: Degrees of freedom = Number of classes - 1 Enter the results of your calculations below: Ver. 3 L. Alemán, S. Bumgarner, & B. Lau 4

Chi Square (χ2) value = Degrees of freedom = Now you can use the Chi Square Table (included at the end of this exercise) to determine the p value from this test. probability (p) value = A p value of 0.05 or less indicates that your hypothesis was incorrect and that you should reject it. This p value of 0.05 means that any difference that you observed between your expected data and your observed data would only occur JUST BY CHANCE 5% of the time. Based on the p value you obtained, should you reject your original hypothesis (from question 5)? a) Now repeat the Chi Square Test using the predictions from your new hypothesis (from question 10b) and the data generated in question 7. Could your new hypothesis explain the observed data? Fill in the chi square table below and explain your reasoning. Mutant Wild type Observed Expected (0 E) (0 E) 2 (0 E) 2 / E Total 1000 1000 - - χ2 = Chi Square (χ2) value = Degrees of freedom = probability (p) value = Ver. 3 L. Alemán, S. Bumgarner, & B. Lau 5

Reference Chi Square Table Ver. 3 L. Alemán, S. Bumgarner, & B. Lau 6