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AP Biology Lab Seven: Genetics of Organisms.Hypothesis In the sex linked cross of Drosophila Melanogaster, a phenotypic ratio of 1:1 will be obtained.

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Introduction

Transfer-Encoding: chunked AP Biology Lab 7 :Genetics of Organisms Introduction Drosophila Melanogaster, the fruit fly, is a great organism for genetic use because it has simple food requirements, occupies little space, is hardy, completes its life cycle in 12 days, makes a large number of offspring, can be knocked out easily, and it has many types of hereditary variations that can be seen with low power microscopes. Drosophila has a small number of chromosomes, four pairs. They are easily located in the large salivary glands. The Drosophila can be obtained from many places. Research of Drosophilae has led to a lot of knowledge about many of its genes. Many factors combine to affect the length of the Drosophila life cycle. Temperature affects the life cycle the most. At room temperature the average life cycle of the Drosophila is about 12 days. Eggs of the Drosophila are small, oval shaped, and have two filaments at one end. They are usually laid on the surface of the culture medium, and with practice, can be seen with the naked eye. After one day the eggs hatch into the larva. The larval stage of the Drosophila eats all the time. Larvae tunnel into the culture medium when they eat. ...read more.

Middle

Describe the observed mutations? In the F1 generation the males had white eyes and the females had red eyes. In the F2 generation the males and females could have had either red or white eyes. 2. Write a hypothesis which describes the mode of inheritance of the trait you studied. This is your null hypothesis ( as described in the Statistical Analysis Section). For a sex linked cross there will always be a one to one ratio of the phenotypes. In the F1 generation there will be a one to one ratio of red eyed females to the number of white eyed males. In the F2 generation there will be a one to one ratio of red eyed females to white eyed females. There will also be a one to one ratio of red eyed males to white eyed males. 3. Refer to a textbook and review Punnett squares. In the space below construct two Punnett squares to predict the expected results of both the parental and F1 crosses from your null hypothesis. Parental cross Y Xr Xr YXr Xr Xr Xr YXr Xr Xr F1 cross Y Xr Y XR XR YXR XR Xr YXR XRXR XR YXR XR Xr YXr XR Xr Xr YXr XrXr YXr X RXr Xr YXr XrXr YXr XRXr 4. ...read more.

Conclusion

There will be 333 long winged flies. 666 intermediate winged flies. There will be 33 short wing flies. c. Complete the table Table 7.8 Observed Phenotypes Expected (e) (o-e) (o-e)2 (o-e)2/ e LL 333 -103 10609 31.86 Ll 666 -156 24336 36.54 ll 333 -73 5329 16.00 84.4 1. How many degrees of freedom are there? There are 2 degrees of freedom. 2. Chi-Square= 84.4 3. Referring to the critical values chart, what is the probability value for these data? Less than .001. 4. According to the probability value can you except or reject the null hypothesis? I can reject the null hypothesis because the Chi-square answer is greater than the critical value from the table. Error Analysis Results from this lab could have been affected by many things. The constant knocking out of flies could have caused some of the larvae to not hatch therefore affecting our numbers. Also, incorrectly identifying the characteristics of the flies could have also greatly affected the results received. Improper calculation of numbers could have also caused inaccurate results. Finally, some flies could have gotten stuck in the medium and could have been identified. Conclusion From the results of the experiment I can conclude that I received results that were close to a 1:1 ratio. The Chi- Square worked from my data was accepted at a possibility greater than .05. The null hypothesis in this case can be accepted. ...read more.

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