Genetics of Organisms (Drosophila) Lab #5

1. Genetics of Organisms (Drosophila)Lab #5 Objectives: • Learn the life cycle for Drosophila melanogaster (egg-larval-pupal-adult) • Distinguish between male & female flies • Study & examine mutational characteristics of: Eye color # Bristles Wing size (long vs. short) Antennae size (vestigial vs. normal

2. Why use Drosophila??? • Simple food requirements • Occupies little space • Hardy (survives well) • Completes life cycle around 12 days (temp) • Produces large numbers of offspring • Easily immobilized (chilling on ice) to examine • Several hereditary variations • Small # of GIANT chromosomes (four pairs) • Chromosomes easily located in the salivary gland cells

3. Key Factors of this Lab • The flies live for about a month • Females will store up sperm in her receptacles & can continue to fertilize eggs • Virgins must be separated from males early on (distinguishing males & females) • Temperature is the most important factor that determines the length of life cycle

4. Distinguishing between Male & Females Males -Smaller • Dark, blunt abdomen • Sex comb on forelimb Females • Larger • Lighter pointed abdomen (ovipositer) • Transverse stripes on abdomen • No sex comb

5. Three Genetic Crosses • Examine THREE genetic crosses: -Monohybrid: ss (sepia) x SS (normal-red) -Dihybrid: Normal eyes, vestigial wings x sepia, normal wings SSvv x ssVV -Sex-Linked: White eyed female x Normal red-eyed male XrXr x XRy

6. Table 7.1: F1 Generation Data Parental cross was w/white eyed-females x Red-Eyed Males

7. F2 Generation: Sex-Linked trait Red-Eyed heterozygous female crossed with white-eyed male XRXr x XrY (these came from F1 generation) Table 7.2 F2 Generation Data Assume that the following Data was obtained and determine if it is significant by calculating a Chi-Square value. Do this on the top of Pg 85. Remember to calculate the expected (theoretical) values from the cross that produced the results above. The cross is written below.

8. Answer for First Cross X2 = (78-72)2 + (69-71)2 + (62-71)2 + (73-71)2 71 71 71 71 X2 = 0.69 + 0.06 + 1.14 + 0.06 X2 = 1.95 Degrees of freedom: (4 categories) = 3 Critical Value: see Table7.3 (pg.85) = 7.82 The chi-squared value is less than 7.82 Our null hypothesis is accepted: That there is not a significant difference between the observed value and accepted value. We must then accept that our results follows the expected (theoretical value)

9. Other Crosses to do • Pgs 85-86: Green & Albino seedlings • Table 7.3 # Expected Green Calculate Albino Calculate • Pg.88: Long & short winged flies **complete and calculate Chi-Square values to determine if the results are significant and acceptable (statistically valid) • Pg 89 Table 7.8

10. 2nd Genetic Cross (pg. 83) • Red-Eyed: Dominant S • Sepia eyes: Recessive s (3) Parental cross: SS X ss (show a punnett square in box) (3) F1 Cross: Ss X Ss (show a punnett square in box) **Fill in this info. For #4 Table F1: All Ss F2: 3:1 (red:sepia) #5 There should be no deviations in the F2 results. Sometimes there could be fewer red-eyed individuals than expected and more individuals with sepia eyes.

11. Data for Second Cross (Pg. 84) #6. The mutation will be Autosomal The mutation will be recessive The cross is monohybrid #7 **Answer Questions “a” and “b”

12. Page 85 • Answer Questions #1-3 based on the information written in the beginning of the lab.