Bio Exam Essay.

BioSci 315

Take Home Essay Questions - Set 2 Name:_Laayla M. Tariq

Download this MS-Word document. Fill in your name at the top and type your answers below. Save the questions and your answers as an MS Word document with the .doc file extension (not .docx). This assignment is due in the D2L Dropbox on Sunday, 04-11-10 at 11:59pm.

A. For 5 points, correctly answer only one of the questions below (use as much space as needed):

Pts
1. Progesterone is known to activate transcription of the ovalbumin gene in chick oviduct.
2. You isolate Chromatin from untreated and progesterone-treated oviduct cells
3. Digest it four different times with deoxyribonuclease I (DNase I).
4. After you purify the DNA from the digested chromatin samples, you denature the DNA and “spot” equal amounts from each extract on filters.
5. The filters are exposed to denatured, radioactive ovalbumiun gene DNA (the probe), allowing it to hybridize with any ovalbumin gene DNA (the target) remaining after digesting the chromatin.
6. After making an autoradiograph of the filters, the following results are obtained:


2.0 a. State the hypothesis being tested in this experiment.


Progesterone is a hormone that activates transcription and therefore experimentally digesting progesterone treated oviduct cells with deoxyribonuclease I (DNase I) will result in denaturing of the DNA overtime, compared to the DNA from digests of chromatin from untreated tissue.






3.0 b. Explain these results.

The autoradiograph is an “X-ray photograph made by bringing an object containing radioactive material into close contact with the emulsion on a film or plate” (YourDictionary). When the chromatin is digested with DNase I, it becomes radioactively active. The autoradiograph results reveal that chromatin from progesterone treated oviduct cells digested with DNase I will contain DNA that will denature or structurally breakdown in order for transcription to take place since progesterone is known to activate transcription. We can see this because as time goes on, from 2 minutes to 20 minutes, the equal amount of “spots” from each extract is no longer equal because the treated cells are denaturing overtime and the spots are becoming lighter and smaller. DNase I is therefore, acting on the progesterone treated chromatin, which it cannot do with non-treated cells because chickens do not carry high levels progesterone, besides the low natural amount their bodies carry. The low initial amount of progesterone, which is a natural sex hormone, is not relevant to the fact that sample B was additionally injected with “more progesterone.” The denaturing is independent of the initial amount and is dependent on the overall level of progesterone in the samples. Therefore, progesterone does activate transcription of the ovalbumin gene in chick oviduct due to the results shown in autoradiograph.


"Autoradiograph." Dictionary Definitions. LoveToKnow, n.d. Web. 2 April 2010. .


B. For 5 points, correctly answer one of the questions below (use as much space as needed):

Pts
1. Consider the 10nm fiber and solenoid extracts from rat nuclei shown below:



You have purified the proteins from each of these chromatin extracts. After running polyacrylamide gel electrophoresis of the purified proteins and staining the gels, you see the following:



5 Explain these results.

The purification of the proteins from both of the chromatin extracts (of rat nuclei) being run by polyacrylamide gel electrophoresis of which the gels are then stained, provide very similar results when compared to each other. The proteins purified from the 10 nm fiber chromatin contains H1, 2H3, 2H2A, 2H2B, 2H4. The proteins purified from solenoid chromatin contain all that but H1. First of all, we are told that the image on the right is of 30 nm solenoid chromatin, obviously longer than the 10 nm fiber. With the way gel electrophoresis works, the migrating proteins are divided into individual sections related to their gel mediated mobilities (Garfin 201). These denatured proteins are coated with a negative charge and end up having an overall negative charge with similar charge to mass ratios. Therefore, the rate of these proteins migrating depends solely on the size. The smaller the protein, the longer they will be able to migrate to the top, being able to travel through the many holes of the gel. The size of the pores of the gel is hard to measure, yet it is defined by the size limits of the proteins (Garfin 201). Therefore, these results make sense, with proteins purified from solenoid chromatin not being able to pass through the H1 gel hole due to the bigger size, compared to the proteins purified from 10 nm fiber chromatin, which were able to migrate through. This could be supported by the fact that the proteins from the solenoid chromatin had higher levels of salt in them. As salt decreases, protein solubility increases, and therefore, the proteins would be able to dissolve and migrate through (Young 1). On the other hand, as salt increases, such as in 30 nm solenoid chromatin sample, solubility decreases, which justifies the fact they weren’t able to migrate to H1.


Garfin, David. "Gel Electrophoresis of Proteins." Essential Cell Biology: Cell Structure, A Physical Approach 1. (2003): 197-268. Web. 11 Apr 2010. .

Derek R. Young. Introduction to Biochemical Engineering Term Project. 1994. .