however, you need to purchase the Fly Lab manual from the bookstore in order to obtain a login name and password.

Transcription

1 Laboratory 3- The Virtual Fly The Virtual Fly Laboratory uses a computer program that simulates crosses for 29 common morphological variants (bristles, eye colour, body colour, wing size, etc.) based on the actual dominant/recessive inheritance patterns and linkage relationships known for these traits in Drosophila melanogaster. For information on Drosophila melanogaster that you will need for this lab, go to Print it off and bring to the lab. The principle advantages of a computer simulation for our purposes is that crosses and the resulting progeny can be produced in only a few seconds (instead of several days), and that large number of progeny can be produced in each cross, which increases the statistical reliability of experimental results. Further advantages are that computer simulations are not subject to escaped flies, attempts to mate flies of identical sex, misidentification of variants, and do not require preparations of malodorous media or any other joys that afflict experiments with living organisms. The Virtual Fly Website can be found at: however, you need to purchase the Fly Lab manual from the bookstore in order to obtain a login name and password. For each cross, follow the instructions on the computer to select the phenotypes of the parents. Record the phenotypes and results in the tables provided. 1. Click on the box: Design a Cross Between Two Flies Select the appropriate variant(s) for your cross. All other variations are automatically set as homozygous wild-type for both parents. Also note that the program does not use the standard notation for Drosophila variants; you are expected to interpret the dominance or recessiveness of particular variants from the crosses. Use the correct notation (lower and upper case, and + ) to write the genotypes. To help you with this, appendices will be available in the lab or go to : and refer to Appendix B and C. 2. Click on the box: Mate Designed Flies You will be given the F 1 results of the cross: record the phenotypes of the F 1 males and females. 3. To produce the second generation, click on the circles below the diagrams of the F 1 offspring, then click on the box: Mate Selected Flies You will be given the results of the F 2 generation: record them in the table provided. If any calculations are required, they should be done before performing another cross. If you repeat a cross, the computer will give you a new set of data. Use the BACK icon to return to the beginning of the program to perform another cross.

2 Virtual Fly Part 1 Due at end of lab period Fall 06 Name Student ID Lab Slot 1 - MONOHYBRID CROSS - AUTOSOMAL

3 PURPLE EYE x WILD-TYPE EYE CROSS A PARENTS PURPLE EYE WILD-TYPE EYE WILD-TYPE EYE PURPLE EYE F 1 F 2 # MALES # FEMALES # MALES # FEMALES WILD-TYPE EYE PURPLE EYE Table is worth 1.0 mark Observed F 2 Phenotype Ratio (round to one decimal place): 0.5 Cross A Wild-type: Purple Cross B Wild-type: Purple QUESTIONS Which of the alleles of this gene for eye colour is dominant? Are the results of a cross involving an autosomal trait the same for reciprocal crosses? 3. In either Cross A or Cross B, is there any difference between results for males and results for females -.25 in the F 1?.25 in the F 2? Does it make any difference to the F 2 generation if the mutant allele is carried by the male or female parent? /2.75

4 5. Refer to Appendices B and C for the correct way to write genotypes and using the correct symbols for the mutants involved complete the cross diagram to determine the EXPECTED F 2 RATIOS PARENT P 1 GENOTYPE P 1 GAMETES F 1 F 1 GENOTYPE F 1 GAMETES CROSS A PURPLE EYE WILD EYE WILD EYE PURPLE EYE 2 marks total for the table PUNNETT SQUARES using the F 1 gametes & % CROSS A % & 1 mark total for the whole table EXPECTED F 2 RATIOS (don t forget to write the phenotype with the number ratio) 0.5 Cross A: 0.5 Cross B: 2 - MONOHYBRID CROSS - SEX-LINKED /4

5 WHITE EYE x WILD-TYPE EYE PARENTS CROSS A WHITE EYE WILD-TYPE EYE WILD- TYPE EYE WHITE EYE F 1 F 2 # MALES # FEMALES # MALES # FEMALES WILD-TYPE EYE WHITE EYE 1.0 mark total for the table Observed F 2 Phenotype Ratio (round to one decimal place): 0.5 Cross A: wild-type male: wild-type female: white female: white male Cross B: wild-type male: wild-type female: white female: white male QUESTIONS Which of the alleles of this gene for eye colour is dominant? Are the results of a cross involving a sex-linked trait the same for reciprocal crosses? Is there any difference between results for males and results for females - in the F 1? (Compare Cross A results with Cross B) in the F 2? (Compare the # of males with the # of females in Cross A) Does it make any difference to the F 2 generation if the mutant allele is carried by the male or female parent? 5. Look for the mutants purple eye and white eye on the gene map..25 Which chromosome is the white gene on.25 Which chromosome is the purple gene on /3.25

6 6. Refer to Appendices B and C for the correct way to write genotypes and using the correct symbols for the mutants involved complete the cross diagram to determine the EXPECTED F 2 RATIOS CROSS A PARENT P 1 GENOTYPE WHITE EYE WILD EYE WILD EYE WHITE EYE P 1 GAMETES F 1 F 1 GENOTYPE F 1 GAMETES This table is worth 2 marks total PUNNETT SQUARES using the F 1 gametes & % CROSS A % & 1 mark total for this table EXPECTED F 2 RATIOS (don t forget to write the phenotype with the number ratio and include sex) 0.5 Cross A: 0.5 Cross B: /4

7 3 - DIHYBRID CROSS - AUTOSOMAL INDEPENDENT BODY x WING PARENTS CROSS A BODY WING WING BODY F 1 Phenotypes F 2 Phenotypes # MALES # FEMALES # MALES # FEMALES WILD TYPE & 1.0 mark for this table in total Observed F 2 Phenotype Ratio (round to one decimal place and exclude the sex): 0.5 Cross A wild-type: ebony: vestigial: ebony & vestigial Cross B wild-type: ebony: vestigial: ebony & vestigial /1.5

8 4 - DIHYBRID CROSS -test cross BODY x WING PARENTS CROSS A WILD WING BODY CROSS A- F 1 WING BODY F 1 Phenotypes F 2 Phenotypes # MALES # FEMALES F 1 : # MALES F 1 # FEMALES WILD-TYPE PURPLE & Table is worth 1.0 mark in total NOTE: To do Cross B, you need to start over and redo Cross A BUT instead of mating the F 1 male with a F 1 female as in Cross A, you redesign the F 1 female to be vestigial and ebony (2 recessive traits). In the Table for Cross B, when recording data for Cross B, you don t go to the F 2. The # of males and # of females you record is for the F 1 of THAT cross. Phenotype Ratios excluding sex: 0.5 F 2 Cross A wild-type: ebony: vestigial: ebony & vestigial F 1 Cross B wild-type: ebony: vestigial: ebony & vestigial 1.0 What does the results of this test cross show you?

9 QUESTIONS / Look at the F 1 results of crosses 3 and 4. For each of the three genes involved state whether the mutant allele of that gene is dominant or recessive The mutant allele of the Ebony gene is The mutant allele of the Vestigial gene is 2. Can you tell from the F 1 results if two autosomal loci are independent? 0.5 Cross 3: Cross 4: 3. Is there a difference in the phenotypes from the males and females of the F 2 generation? 0.5 Cross 3: Cross 4: For an autosomal trait (e.g. vestigial wing) does it make any difference in F 2 results which parent carries the mutant allele? Can you tell from the F 2 results if two autosomal loci are independent(ie on separate chromosomes)? 1 How? 6. Find the locus for each of these genes on the gene map 0.75 Ebony 0.75 Purple 0.75 Vestigial Gene Chromosome # Position Notation Did the results obtained in the above crosses correspond to the independence shown by the gene map? Did the dominant/recessive alleles correspond with the notation given on the gene map? /6

10 9. Use the information from questions 1-3 and both Appendix B and C to determine the correct genotypes and complete the cross diagrams to find the EXPECTED RATIO for each cross. 3 - DIHYBRID CROSS - AUTOSOMAL INDEPENDENT PARENT BODY WING P GENOTYPE P GAMETES F 1 GENOTYPE [% & & ] F 1 F 1 GAMETES 3 marks for this table PUNNETT SQUARE & % 2 marks for the table 1 Expected F 2 Phenotype Ratio excluding sex: /6