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**Virtual Fly Lab AP Biology**

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Key Concepts Physical characteristics are transmitted from generation to generation according to some general patterns. When we have sufficient information about the parents, we can predict the occurrence of traits in offspring. And by analyzing the offspring, we can discern the mode of transmission – monohybrid or dihybrid, sex linked or autosomal.

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Key Concepts To do this laboratory, you are expected to understand basic Mendelian genetics and chi-square analysis. If any of the following concepts are unclear to you, review the genetics section in your textbook before continuing: Genotype, phenotype Parental generation, F1, F2 Monohybrid cross, dihybrid cross Sex-linked traits Crossing over, linked genes Punnett square

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**How to Tell the Males from the Females**

Abdomen is blunt and darker Dark bristles (sex combs) found on the inner surface of the forelimbs Females Abdomen is long, pointed, and lightly striped

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**Life Cycle of Drosophila**

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Phenotypes Flies with characteristics representative of a normal fruit fly (lacks mutations) is said to be a wild-type fly. Flies with differences from the wild-type are said to be mutants. Presentation_2.html

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**Overview of Experiment**

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Analysis of Results In the virtual laboratory you will select and breed flies and analyze the results of the breeding through the F2 generation. One way to discover patterns of inheritance is by working backward. In other words, you determine the genotype of the original parental generation by careful analysis of the F1 and F2 generations.

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Case 1 Based on the data obtained, is this cross: monohybrid or dihybrid? Is the cross: sex-linked or autosomal? monohybrid autosomal

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**Case 2 Sex-linked X+Xw x X+Y**

Based on the data obtained, which is the most likely mode of interitance? From the data presented, determine the genotype of the parental (before the F1 generation; not shown here) generation. + = wild type (red eyes) w = white eyes Sex-linked X+Xw x X+Y

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**Chi-Square Analysis of Data**

From Case 1, the F1 generation: +se x +se (+ = red; se = sepia ) From this conclusion, you could write the following hypothesis: "If the parents are heterozygous for eye color, there will be a 3:1 ratio of red eyes to sepia eyes in the offspring." Do your results support this hypothesis?

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**Chi-Square Analysis of Data**

The actual results of an experiment are unlikely to match the expected results precisely. But how great a variance is significant? Chi-square (ϰ2 ) tests the validity of a null hypothesis, which states that there is no statistically significant difference between the observed results of your experiment and the expected results. When there is little difference between the observed results and the expected results, you obtain a very low chi-square value; your hypothesis is supported.

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**How to Calculate the Expected Value**

In a cross between two heterozygous individuals, the offspring would be expected to show a 3 : 1 ratio. For example, in Case 1, three-fourths of the individuals would have red (wild-type) eyes, and one-fourth would have sepia eyes. If there are 44 offspring, how many are expected to have red eyes? We expect three-fourths to have red eyes. If there are 44 offspring, how many are expected to have sepia eyes? Now you are ready to calculate chi-square.

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**Calculating Chi-Square**

ϰ2 = (Observed – Expected)2 Expected ∑ Phenotype Observed No. (o) Expected No. (e) (o-e) (o-e)2 (o-e)2 / e Red eyes 31 33 -2 4 0.12 Sepia eyes 13 11 2 0.36 ϰ2 = 0.48

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**Using the Chi-Square Critical Values Table**

The chi-square critical values table provides two values that you need to calculate chi-square: Degrees of freedom This number is one less than the total number of classes of offspring in a cross. In a monohybrid cross, such as our Case 1, there are two classes of offspring (red eyes and sepia eyes). Therefore, there is just one degree of freedom.

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**Using the Chi-Square Critical Values Table**

The chi-square critical values table provides two values that you need to calculate chi-square: Probability The probability value (p) is the probability that a deviation as great as or greater than each chi-square value would occur simply by chance. Many biologists agree that deviations having a chance probability greater than 0.05 (5%) are not statistically significant. Therefore, when you calculate chi-square you should consult the table for the p value in the 0.05 row.

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**Degrees of Freedom (df)**

Critical Values Table Degrees of Freedom (df) Probability (p) 1 2 3 4 5 0.05 3.84 5.99 7.82 9.49 11.1 0.01 6.64 9.21 11.3 13.2 15.1 0.001 10.8 13.8 16.3 18.5 20.5 Determine the degrees of freedom. This is the number of categories (red eyes or sepia eyes) minus one. For the data in Case 1, the number of degrees of freedom is ___ 1 The probability (p) value for 1 degree of freedom in the 0.05 row ____ 3.85

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**Degrees of Freedom (df)**

Critical Values Table Degrees of Freedom (df) Probability (p) 1 2 3 4 5 0.05 3.84 5.99 7.82 9.49 11.1 0.01 6.64 9.21 11.3 13.2 15.1 0.001 10.8 13.8 16.3 18.5 20.5 Since our ϰ2 value of 0.48 is less than 3.84, we can accept our null hypothesis. The results support the hypothesis that the parents are heterozygous for eye color.

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