Model 1 holds true where:
B Blue (dominant)
b Pink (recessive).
If this were so, then the parents of Cage 2, the blue male and female, would be BB or Bb. This leads to a few possible cases:
Cases 1: Blue Male (BB) x Blue Female (BB) All the Offspring should be BB – Blue. All offspring were blue thus this is consistent with the data.
Case 2: Blue Male (Bb) x Blue Female (Bb) In this case 25% of the offspring should be BB, 50% should be Bb and 25% should be bb, thus ¾ of the offspring should be blue and ¼ should be pink. No pink offspring were observed, so this is not consistent with the data.
Case 3: Blue Male (BB) x Blue Female (Bb) Half the offspring should be BB and the other half should be Bb. Therefore all the offspring should be blue, which is consistent with the data.
Since two cases were consistent either both parents were homozygous dominant (blue) and or one is heterozygous.
Model 2: Pink is dominant
P Pink (dominant)
p Blue (recessive)
Parents: Blue Male (1) x Pink Female (1) gave offspring: Cage 4: 14 Blue Males; 11 Blue Females.
If this were so, using the data of Cage 4, the blue male would be pp. The Pink female could be PP or Pp. This leads to 2 possible cases:
Case 1: Blue Male (pp) x Pink Female (PP) All the offspring should be Pp – pink. No pink offspring were observed so this is not consistent with the data.
Cases 2: Blue Male (pp) x Pink Female (Pp) Half the offspring should be Pp – pink. No pink offspring were observed, so this is not consistent with the data.
Therefore, Model 1, which holds consistent with the data, is correct.
Phenotype: Body Shape (Short or Long)
Model 1: Long is dominant (Supported Hypothesis)
Allele Contribution to Phenotype
XL Long (dominant)
Xl Short (recessive)
OR
Model 2: Short is dominant
Allele Contribution to Phenotype
XS Short (dominant)
Xs Long (recessive)
In both Models the trait is sex-linked, because in Cage 3 it was observed that a Long Male and a Long Female crossed gave offspring to only short males, long males and long females which indicates it is sex-linked since only the males are short because they only inherit one allele if the gene is one the x chromosome unlike the females who inherit two alleles.
List of Crosses for Body Shape Phenotype
Cage 1: 10 Short Males; 8 Short Females; 4 Long Males; 6 Long Females.
Parents: Short Male (1) x Short Female (1) gave offspring: Cage 2: 18 Short Males; 11 Short Females.
Parents: Long Male (1) x Long Female (1) gave offspring: Cage 3: 8 Short Males; 9 Long Males; 8 Long Females.
Parents: Short Male (1) x Long Female (1) gave offspring: Cage 4: 16 Long Males; 15 Long Females.
Parents: Long Male (4) x Long Female (4) gave offspring: Cage 5: 5 Short Males; 8 Long Males; 13 Long Females.
Parents: Short Male (3) x Short Female (2) gave offspring: Cage 6: 17 Short Males; 14 Short Females.
Parents: Short Male (6) x Long Female (5) gave offspring: Cage 7: 18 Long Males; 17 Long Females.
Parents: Long Male (7) x Short Female (6) gave offspring: Cage 8: 12 Short Males; 16 Long Females.
Parents: Short Male (6) x Long Female (8) gave offspring: Cage 9: 7 Short Males; 9 Short Females; 13 Long Males; 8 Long Females.
Parents: Short Male (3) x Short Female (9) gave offspring: Cage 10: 17 Short Males; 18 Short Females.
Using the data provided Model 2 proves inconsistent with the data because if the long phenotype were in fact recessive than when a long male is crossed with a long female all the offspring must be long.
Parents: Long Male (1) x Long Female (1) gave offspring: Cage 3: 8 Short Males; 9 Long Males; 8 Long Females.
In Cage 3 the offspring of a long male and long female are both short and long therefore long must be dominant. Since the trait is sex-linked and the males were the only short ones, the female (parent) must be heterozygous where only the males are capable of being short since they only inherit one X chromosome from the mother.
Model 1 holds true where:
Allele Contribution to Phenotype
XL Long (dominant)
Xl Short (recessive)
If this were so, then the parents of Cage 7, Short Male and Long Female, The Short Male would be XlY and the Long Female would be XLXL.
Short Male (XlY) x Long Female (XLXL) All the Offspring would be Long where the Males should be XLY – Long and the females should be XLXl - Long. All offspring in the observed data were long thus this is consistent with the data.
Model 2: Short is dominant
Allele Contribution to Phenotype
XS Short (dominant)
Xs Long (recessive)
Parents: Short Male (1) x Long Female (1) gave offspring: Cage 4: 16 Long Males; 15 Long Females.
If this were so, then the parents of Cage 2 the Short Male would be XSY and the Long Female would be XsXs. This leads to 1 possible case:
Short Male (XSY) x Long Female (XsXs) Half the Offspring should be short females (XSXs) - Short. No short offspring were observed, so this is not consistent with the data.
Therefore, Model 1, which holds consistent with the data, is correct.
Phenotype: Eye Color (Red or Green)
Model 1: Red is dominant (Supported Hypothesis)
Allele Contribution to Phenotype
XR Red (dominant)
Xr Green (recessive)
OR
Model 2: Green is dominant
Allele Contribution to Phenotype
XG Green (dominant)
Xg Red (recessive)
In both Models the trait is sex-linked, because in Cage 2 and 8 it was observed that a Red Male crossed with a Green Female gave offspring only red females and green males which indicates it is sex-linked. Since only the males are green they only inherit one allele if the gene is on the x chromosome from the mother and the mother is in fact green so the males will be as well.
List of Crosses for Eye Color Phenotype
Cage 1: 5 Red Males; 4 Red Females; 4 Green Males; 7 Green Females.
Parents: Red Male (1) x Green Female (1) gave offspring: Cage 2: 16 Green Males; 14 Red Females.
Parents: Green Male (1) x Red Female (1) gave offspring: Cage 3: 9 Red Males; 10 Red Females; 10 Green Males; 6 Green Females.
Parents: Red Male (3) x Red Female (3) gave offspring: Cage 4: 3 Red Males; 18 Red Females; 3 Green Males.
Parents: Green Male (3) x Green Female (3) gave offspring: Cage 5: 17 Green Males; 22 Green Females.
Parents: Green Male (2) x Red Female (2) gave offspring: Cage 6: 10 Red Males; 1 Red Female; 8 Green Males; 6 Green Females.
Parents: Red Male (6) x Green Female (6) gave offspring: Cage 7: 4 Red Males; 17 Red Females; 15 Green Males.
Parents: Red Male (6) x Green Female (6) gave offspring: Cage 8: 12 Red Females; 8 Green Males.
Parents: Green Male (6) x Red Female (6) gave offspring: Cage 9: 6 Red Males; 9 Red Females; 7 Green Males; 7 Green Females.
Parents: Green Male (9) x Green Female (9) gave offspring: Cage 10: 12 Green Males; 13 Green Females.
Using the data provided Model 2 proves inconsistent with the data because if the red phenotype were in fact recessive than when a red male is crossed with a red female all the offspring must be red.
Parents: Red Male (3) x Red Female (3) gave offspring: Cage 4: 3 Red Males; 18 Red Females; 3 Green Males.
In Cage 4 the offspring of a red male and red female are both red and green therefore red must be dominant. Since the trait is sex-linked and the males were the only green ones, the female (parent) must be heterozygous where only the males are capable of being green since they only inherit one X chromosome from the mother.
Model 1 holds true where:
Allele Contribution to Phenotype
XR Red (dominant)
Xr Green (recessive)
If this were so, then the parents of Cage 2, Red Male and Green Female, the Red Male would be XRY and the Green Female would be XrXr.
Red Male (XRY) x Green Female (XrXr) All the male offspring should be XrY - green and all the females should be XRXr - red. All the male offspring were green and all the female offspring were red in the observed data thus this is consistent with the data.
Model 2: Green is dominant
Allele Contribution to Phenotype
XG Green (dominant)
Xg Red (recessive)
Parents: Red Male (6) x Green Female (6) gave offspring: Cage 8: 12 Red Females; 8 Green Males.
If this were so, then the parents of Cage 8 the Red Male would be XgY and the Green Female would be XGXG or XGXg. This leads to 2 possible cases:
Case 1:Red Male (XgY) x Green Female (XGXG)
All the offspring should be green; the males should be XGY and the females should be XGXg. Red offspring were observed in the data, so this is not consistent with the data.
Case 2: Red Male (XgY) x Green Female (XGXg)
The offspring should be both green and red; the males should be XGY or XgY and the females should be XGXg or XgXg. No red female or green male offspring were observed in the data, so this is not consistent with the data.
Therefore, Model 1, which holds consistent with the data, is correct.
Conclusion
The purpose of this lab was to deduce the gene interactions and the mode of inheritance of three traits in virtual insects. Knowledge of genetics are utilized to design and interpret crosses and to figure out how alleles of different genes interact (dominant/recessive) and are transmitted (autosomal/sex-linked). The Virtual Genetics Lab is a computer stimulation of genetics of imaginary insects. The computer randomly picks a character with more than one phenotype. It then randomly chooses which form of the character will be dominant and which will be recessive and whether the gene involved is autosomal or sex-linked. As in a real genetics lab, the insects are kept in cages; Cage 1 contains the Field population, individuals collected in the wild. One can select any two insects (male and female) and cross them; the computer automatically puts their offspring in a new cage.
Evaluation
When the crosses were made one must recognize how to analyze if a trait is dominant or recessive, for instance if a red male was crossed with a green female and all the offspring were red, the conclusion would be that red must be dominant and green must be recessive because (RR x rr = Rr), therefore both parents were homozygous but one had the dominant and the other had the recessive trait thus all the offspring were heterozygous and red. This experiment consisted of several models but it was very simple to determine whether they were consistent or not by using punnett squares. This lab does not have a margin of error because one must create models to interpret the crosses and using punnett squares and the knowledge of genetics, if a model was not consistent there was always an alternative model which would be consistent.
