Thursday, December 5, 2019
Biology Simulator Assignment free essay sample
In vial one, a wild type hyper female Drosophila and wild type mellow male Drosophila were crossed respectively to determine the dominant behavioural phenotype. The first reciprocal cross was then carried out in vial two to define whether or not the behavioural phenotype is x-linked recessive. According to the results obtained from vial one, when the wild type hyper female and wild type mellow male was crossed, the resulting offspringÃ¢â¬â¢s phenotypes were all wild type hyper. This cross shows that the hyper behavioral phenotype can be considered dominant in comparison to mellow. In the reciprocal cross, the behavioural phenotypes were isolated from one another while the wing veins were kept constant (in this case both wild type). When the wild type mellow female Drosophila and wild type hyper male Drosophila were crossed, all female offspring obtained a similar phenotype to that of the male parental while all the male offspring had phenotypical combinations similar to that of the female parent. We will write a custom essay sample on Biology Simulator Assignment or any similar topic specifically for you Do Not WasteYour Time HIRE WRITER Only 13.90 / page This set of results shows that the mellow behavioral phenotype is a recessive x-linked gene carried by the female because the resulting male offspring showed the same characteristics to that of the female parent (received X Ã¢â¬â chromosome from female parent). On an additional note, there was a higher frequency of females in comparison to men; 121 and 105 respectively. In addition, the behavioral gene is sex -linked also because of the different resulting phenotypical ratio in comparison to the cross carried out in vial one. Another cross was carried out in vial three to determine the dominant wing vein phenotype. A wild type hyper female and veinless hyper male were crossed and as a result, all male offspring consisted phenotypical characteristics similar to that of the female parent while the female offspring had the same characteristics to that of the male parent. The results from this cross are somewhat similar to that of the reciprocal cross carried out in vial one. All male offspring had features exactly alike to that of the female parent, thus showing that wild type gene is x-linked recessive as all male offspring were wild type and hyper. It is also important to note that in this cross, a higher population of males existed than females (112 hyper wild type and 104 veinless hyper respectively). The second reciprocal cross was carried out in vial four directly after the above cross. This cross displayed a concise picture of which behavioral and wing vein phenotypes were dominant. Unlike the cross the done above, this time, a veinless hyper female and wild type hyper male were crossed and the resulting male and female offspring were all veinless hyper. Thus, proving veinless and hyper behaviour are the respective dominant phenotypes among the female and male Drosophila. Another cross was carried out to derive the first filial generation in which a veinless hyper female was crossed with a wild type mellow male and as a result, the offspring were all veinless hyper (a total of 214 offspring) which can be seen on the cross Ã¢â¬â record sheet under vial seven. A second filial generation was derived from a veinless hyper female and male Drosophila were crossed to distinguish a gene linkage relationship among the two phenotypes. Due to the genes being sex-linked, specifically x-linked recessive, the typical 9:3:3:1 ratio cannot be expected. However, a chi-squared test was carried out to determine how Ã¢â¬ËÃ¢â¬â¢offÃ¢â¬â¢ our values were from the expected critical value of 7. 815 (obtained at a p-value of 0. 05) since such a ratio was not expected. A chi value of 31. 1 was achieved which is greater than the critical value. Thus, the null hypothesis is rejected because the genes are linked which means that they are located very closely together on corresponding chromosomes. However, hypothetically, if a 9:3:3:1 ratio had been expected and if the chi value had been less than the critical value, than the null hypothesis would have been accepted as this indicates that the genes are not linked and are independently assorted (less cross overs /variation).