AP Lab #7Genetics of Organisms Virtual Fly Lab
Instructions Instructions: 1. Find the Fly Lab Manual in your box of materials use the access code found in the manual to register for the lab at www.biologylabsonline.com 2. Read through the introduction and objectives for the lab. 3. Then click “START LAB”
Your Choice Monohybrid cross Create a monohybrid cross between a wild-type female and male fly with sepia-colored eyes. Sepia is a nonlethal recessive allele located on Chromosome III. Dihybrid Cross Design a dihybrid cross by selecting and crossing an ebony body female fly with a male fly that has the vestigial mutation for wing size. Sex-linked cross Design and perform the following crosses to examine the inheritance of sex-linked alleles in Drosophila. Cross a female fly with a tan body with a wild-type male.
Now What? Based on what you know about the principles of Mendelian genetics, predict the phenotypic ratio that you would expect to see for the F1 offspring of this cross and describe the phenotype of each fly. (In other words, use a punnett square) Select 10,000 flies as the number of offspring for this mating. Click on the Mate button. Scroll down to see the wild type offspring. Record this on your data sheet. These offspring are the F1 generation. Note: The actual number of F1 offspring created by FlyLab does not exactly equal the 10,000 offspring that you selected. This difference represents the experimental error introduced by the FlyLab. Are the phenotypes of the F1 offspring what you would have predicted for this cross? Why or why not?
Next Step To set-up the F2 generation, click “Select” under the F1 female and Click “Select” under the F1 male. And mate them. Record F2 offspring –use the scroll buttons to view all the phenotypes of the F2 offspring. Are phenotypes of the F2 offspring what you would have predicted for this cross? Why or why not? From pg. 85 of AP Biology Lab Manual: “Statistics can be used to determine if differences among groups are significant or simply the result of predictable error. The statistical test most frequently used to determine whether data obtained experimentally provide a good fit or approximation to the expected or theoretical data is the chi square test.”
kahy-skwair Determine the number of degrees of freedom (df) based on the number of different phenotypes minus 1. The null hypothesis states that there is no statistically significant difference between the observed and expected data. If our cross resulted in a phenotype ratio that was expected, we would seek to accept our null hypothesis. If our actual data do not fit the expected ratios, then we would reject the null hypothesis. The minimum probability for rejecting a null hypothesis is 0.05, so use this row in the chi-square table provided in your manual. These results are said to be significant at a probability of p=0.05, which means that only 5% of the time would you expect to see similar data if the null hypothesis was correct. Fill in the chi-square chart on your data sheet with the data for the F2 cross.
Chi What? Lets do a sample analysis.
Suppose you preformed a simple monohybrid cross between two individuals that were heterozygous for the trait of interest. Results of a monohybrid cross between two heterozygotes for the 'a' gene. Our expected numbers due to our prediction through a punnett square 25 AA 50 Aa 25 aa 10 AA 75 Aa 15 aa The phenotypic ratio 85 of the A type and 15 of the a-type (homozygous recessive). In a monohybrid cross between two heterozygotes, however, we would have predicted a 3:1 ratio of phenotypes. In other words, we would have expected to get 75 A-type and 25 a-type. Are our results different?
Calculate the chi square statistic x2 by completing the following steps: 1. For each observed number in the table subtract the corresponding expected number (O — E). 2. Square the difference [ (O —E)2 ]. 3. Divide the squares obtained for each cell in the table by the expected number for that cell [ (O - E)2 / E ]. 4. Sum all the values for (O - E)2 / E. This is the chi square statistic.
We now have our chi square statistic (x2 = 5.33), our predetermined alpha level of significalnce (0.05), and our degrees of freedom (df =1). Entering the Chi square distribution table (pg. 86) with 1 degree of freedom and reading along the row we find our value of x2 5.33) lies between 3.841 and 5.412. The corresponding probability is 0.05<P<0.02. This is smaller than the conventionally accepted significance level of 0.05 or 5%, so the null hypothesis that the two distributions are the same is rejected. In other words, when the computed x2 statistic exceeds the critical value in the table for a 0.05 probability level, then we can reject the null hypothesis of equal distributions. Since our x2 statistic (5.33) exceeded the critical value for 0.05 probability level (3.841) we can reject the null hypothesis that the observed values of our cross are the same as the theoretical distribution of a 3:1 ratio.