Classic Genetics Simulator Exercises
You will perform several genetic crosses to study inheritance patterns of different characters in fruit flies. Descriptions of the different traits associated with the characters you study can be found at http://cgslab.com/phenotypes/
As you work through the exercises below, type your answers to the questions in bold directly into this document. When you have completed all parts, upload your completed document to the dropbox provided. This lab will provide experience in doing genetic crosses so the completion of the entire lab will result in full credit.
Getting Started
Go to the www.cgslab.com website. Click on “Launch CGS” (you may be prompted to change a computer setting to allow Flash to run). On the next screen, choose “Study a Practice Population”.
Part 1 –Monohybrid Cross (single character, autosomal)
The character you will be studying is whether a dark colored streak is present on the backs of the flies. The streak trait is dominant to the wildtype trait. We will designate the allele associated with the streak trait as “A” and the allele associated with the wildtype trait as “a”.
Part 1a
Start by selecting “Study a Practice Population” from the CGS website. You will select “Practice Population 2” from the scroll list of populations and then select “Continue”. You will see a population of male and female flies that display either the wildtype or the streak phenotype (moving your cursor over an individual fly will give you phenotypic information about the fly). This vial of flies represents your P generation. You will select a streak female and a wildtype male to mate together. When you click on each fly it will be added to the Crossing Vial.
What is the phenotype of the female fly?
What is the phenotype of the male fly?
What is the genotype of the female fly?
What is the genotype of the male fly?
Cross the two flies together by selecting “Cross Flies” and you will now see the offspring of your cross as a new vial of flies. These flies are your F1 generation. Record the total numbers of each phenotype of your F1 generation below (click on the “Analysis” tab to see the offspring phenotypes tallied):
Number of streak flies:
Number of wildtype flies:
Total Number of flies:
Based on the results of this first cross, you now know a definitive genotype for the female fly that you mated at the start. What was it?
If you have all streak offspring from this cross, you can feel confident that the P-generation flies you mated together were both homozygous for their respective traits (AA and aa). If this is the case, you are ready to move on to part 1b. If you have a mixture of phenotypes in the offspring, the streak female was heterozygous (Aa). You can do a quick Punnett square to see why this is so. Before you move on to Part 1b, you will need to select “Destroy Vial” to get rid of the last cross and begin again with new parental flies (P generation) from the Practice Population until the F1 offspring all show the streak phenotype (YOU MAY HAVE TO DO THIS MANY TIMES). Repeat until all offspring are the streak phenotype and then move on to Part 1b (do not destroy this last vial- you will use it in Part 1b).
Part 1b
You will now cross together the hybrid offspring (members of the F1 generation) from the last cross. All offspring should be heterozygous (Aa) because they are the result of a cross between AA and aa.
This second cross (crossing two members of the F1 generation) is Aa x Aa. What is the expected phenotypic ratio of this cross? (Hint- draw a Punnett square if you are unsure)
Select a male streak and a female streak from the F1 generation and cross them together to get the F2 generation. Use the “Analysis” tab to review the phenotypic results in the F2 generation.
Record the phenotypes of the F2 generation below:
Number of streak flies:
Number of wildtype flies:
Total number of flies:
Are these numbers in a similar ratio to what you would expect from a cross of Aa x Aa? You can check this by running a type of statistical analysis called a chi-square analysis to determine whether there is a significant difference between what you would expect the phenotypic ratio of an Aa X Aa cross to be and what your observed results show. Click on the “Stats” tab and you will see the results of your last cross are shown. Make sure “Body Color” is selected from the “Traits to Analyze” drop down. You will be prompted to enter the expected ratio in the boxes provided (see your earlier answer). After entering the expected ratios, select “Calculate” to run the chi-square analysis. A chi-square value and p-value will be reported. For your analysis you will use a p value of less than or equal to 0.05 as your threshold to say that there is a significant, statistical difference between your expected results (i.e. phenotypic ratio expected) and observed results. In other words, if the p value from your chi-square analysis is greater than 0.05, you would conclude that there is not a significant difference between your observed offspring phenotypes in Part 1b and what you expected this cross would yield based on probabilities (Punnett squares).
What is your p value?
Based on this p value, were your results significantly different from what you expected?
Part 2: Dihybrid Cross (2 characters, autosomal)
You will now study the inheritance patterns of two characters simultaneously. For this exercise, you will study a wing character (wild type vs cleft; wildtype is dominant) and an antennae character (aristapedia vs wild type; aristapedia is dominant).
You will exit out of your last set up and start by selecting “Population 5” from the Practice Population scroll menu. Select “Continue” and you will see a population of male and female flies that display a mix of wing and antenna phenotypes (moving your cursor over an individual fly will give you phenotypic information about the fly). This vial of flies represents your P generation.
Part 2a
You are going to start by crossing a fly showing both dominant traits with a fly showing both recessive traits. Your p generation female fly will have the wildtype wing trait (W) and the aristapedia antenna trait (A). Your male fly will have the cleft wing trait (w) and the wildtype antenna trait (a). Select two flies with these phenotypes from the starting population and add them to the Crossing Vial..
What is the phenotype of your female fly? (include BOTH characters)
What is the phenotype of your male fly? (include BOTH characters)
What is the genotype of your female fly? (include BOTH characters and you will have several possibilities)
What is the genotype of your male fly? (include BOTH characters)
Cross the two flies together by selecting “Cross Flies” and you will now see the offspring of your cross as a new vial of flies. These flies are your F1 generation. Record the total numbers of each phenotype of your F1 generation below (click on the “Analysis” tab to see the offspring phenotypes tallied):
Number of wildtype wing/aristapedia antenna flies:
Number of wildtype wing/wildtype antenna flies:
Number of cleft wing/aristapedia antenna flies:
Number of cleft wing/wildtype antenna flies:
If you have all wildtype wing/aristapedia antenna offspring from this cross, you can feel confident that the P-generation flies you mated together were both homozygous for BOTH of their respective traits (WWAA and wwaa). If this is the case, you are ready to move on to part 2b. If you have a mixture of phenotypes in the offspring, the female was heterozygous for at least one of the two characters (WwAa or WWAa or WwAA). You can do a quick Punnett square to see why this is so. Before you move on to Part 1b, you will need to select “Destroy Vial” to get rid of the last cross and begin again with new parental flies (P generation) from the Practice Population until the F1 offspring all show the wildtype wing/aristapedia antenna phenotype. Repeat until all offspring are the wildtype wing/aristapedia antenna phenotype (YOU MAY HAVE TO DO THIS MANY TIMES) and then move on to Part 1b (do not destroy this last vial- you will use it in Part 2b).
Part 2b
You will now cross together the hybrid offspring (members of the F1 generation) from the last cross. All offspring should be heterozygous (WwAa) because they are the result of a cross between WWAA and wwaa.
This second cross (crossing two members of the F1 generation) is WwAa x WwAa. What is the expected phenotypic ratio of this cross? (Hint- draw a Punnett square if you are unsure)
Select a male with wildtype wings and aristapedia antenna and a female with wildtype wings and aristapedia antenna from the F1 generation and cross them together to get the F2 generation. Use the “Analysis” tab to review the phenotypic results in the F2 generation.
Record the phenotypes of the F2 generation below:
Number of wildtype wing/aristapedia antenna flies:
Number of wildtype wing/wildtype antenna flies:
Number of cleft wing/aristapedia antenna flies:
Number of cleft wing/wildtype antenna flies:
Are these numbers in a similar ratio to what you would expect from a cross of WwAa x WwAa? You can check this by running a chi-square analysis to determine whether there is a significant difference between what you would expect the phenotypic ratio of a WwAa x WwAa cross to be and what your observed results show. Click on the “Stats” tab and you will see the results of your last cross are shown. Make sure “Wing veins and antennae” is selected from the “Traits to Analyze” dropdown. You will be prompted to enter the expected ratio in the boxes provided (see your earlier answer). BE SURE YOU ARE ENTERING YOUR VALUES IN THE BOXES THAT CORRESPOND TO THE CORRECT PHENOTYPES, as the phenotypes listed are not in any particular order. After entering the expected ratios, select “Calculate” to run the chi-square analysis. A chi-square value and p-value will be reported. For your analysis you will use a p value of less than or equal to 0.05 as your threshold to say that there is a significant, statistical difference between your expected results (i.e. phenotypic ratio expected) and observed results. In other words, if the p value from your chi-square analysis is greater than 0.05, you would conclude that there is not a significant difference between your observed offspring phenotypes in Part 2b and what you expected this cross would yield based on probabilities (Punnett squares).
What is your p value?
Based on this p value, were your results significantly different from what you expected?
Part 3: Single character, sex-linked
You will now study the inheritance of a character that is associated with a gene on the X-chromosome. Because females have two X chromosomes they will inherit 2 copies of this gene while males, with only a single X chromosome, will inherit only one copy of this gene.
The character you will be studying is the wing angle character. The two traits associated with this character are gull and wildtype. Gull (G) is dominant to wildtype (g).
For the first cross, you will mate a wildtype male with a gull female.
What is the phenotype of the male fly?
What is the phenotype of the female fly?
What is the genotype of the male fly?
What is the genotype of the female fly?
Cross the two flies together by selecting “Cross Flies” and you will now see the offspring of your cross as a new vial of flies. These flies are your F1 generation. Record the total numbers of each phenotype of your F1 generation below (click on the “Analysis” tab to see the offspring phenotypes tallied):
Number of female flies with gull wings:
Number of female flies with wildtype wings:
Number of male flies with gull wings:
Number of male flies with wildtype wings:
Based on these results, what was the genotype of the female P-generation fly? How do you know?
If your female was heterozygous for the wing angle character (Gg), continue to the next section. If you had a homozygous female (GG), destroy your F1 vial and select new parents from the Practice Population vial until you obtain a heterozygous P generation female (HINT: if the P generation female is heterozygous, you will see both gull and wildtype females and gull and wildtype males in the F1 generation).
You can now determine what the expected ratio of male and female flies of each wing phenotype is for the F1 generation. You can use a Punnett square to determine the expected ratio of a cross between a Gg female and a g- male.
Expected # of gull-winged females:
Expected # of wildtype females:
Expected # of gull-winged males:
Expected # of wildtype males:
You can check whether your F1 generation phenotypes are significantly different from what you would expect from a Gg x g- cross by performing a chi square analysis. Click on the “Stats” tab and you will see the results of your last cross are shown. Make sure “Wing angle and sex” is selected from the “Traits to Analyze” dropdown. You will be prompted to enter the expected ratio in the boxes provided (see your earlier answer). BE SURE YOU ARE ENTERING YOUR VALUES IN THE BOXES THAT CORRESPOND TO THE CORRECT PHENOTYPES, as the phenotypes listed are not in any particular order. After entering the expected ratios, select “Calculate” to run the chi-square analysis. A chi-square value and p-value will be reported. For your analysis you will use a p value of less than or equal to 0.05 as your threshold to say that there is a significant, statistical difference between your expected results (i.e. phenotypic ratio expected) and observed results. In other words, if the p value from your chi-square analysis is greater than 0.05, you would conclude that there is not a significant difference between your observed offspring phenotypes in Part 2b and what you expected this cross would yield based on probabilities (Punnett squares).
What is your p value?
Based on this p value, were your results significantly different from what you expected?