7 (4pts). You join a summer research program in a lab that is trying to understa

Question

7 (4pts). You join a summer research program in a lab that is trying to understand cells response to cold. You grow half the cells at room temperature (23°C) and the other half at 15°C. After two hours incubation, you isolate the DNA from cells using a gentle procedure that leaves nucleosomes and some higher order chromatin structures intact. You then treat the DNA for a short time with a low concentration of M-nuclease, an enzyme that degrades protein-free stretches of DNA. After removing proteins, you separate the resulting DNA on the basis of length using gel electrophoresis. Finally, you visualize the DNA fragments from a region near the Brrl gene or the Brr2 gene shown on the gel below. Remember that smaller DNA molecules move faster through a gel and are found near the bottom of the gel. Darker spots contain more DNA than fainter spots. The lanes are as follows: 1. "marker" containing known DNA fragments of indicated lengths. 2. Cells grown at 23°C, visualized DNA near Brrl gene. 3. Cells grown at 15°C, visualized DNA near Brrl gene. 4.Cells grown at 23°C, visualized DNA near Br2 gene. 5. Cells grown at 15°C, visualized DNA near Br2 gene. A. There is a fragment of DNA at 150 nucleotides in each lane 2-5. What would this smallest fragment of DNA correspond to and why is it 150 nucleotides in length? B. Lane 2 has a ladder of spots with longer lengths. Why are the spots evenly spaced? What do they correspond to? C. Notice the faint spots and extensive smearing in lane 3, suggesting the DNA could be cut almost anywhere near the Brrl gene after growth of the cells at 15°C. This was not observed in the other lanes. What probably happened to the DNA to change the pattern between lanes 2 and 3? D. Describe the relative expression levels of Brrl and Brr2 at 15°C and 23°C. Where is gene expression high and low?

Explanation / Answer

Question 7.

A. In the gel electrophoresis image, the lowest spot (150bp) represents DNA length which is similar to that of the segment of DNA found in a nucleosome. Semi digestion with M-nuclease causes breaks in the DNA backbone present within the linker DNA or other DNA segments which are not bound tightly to histone proteins. So, this bands probably comprise the DNA that are tightly bound to a single histone octamer and it has generated by cutting the linker DNA outside a single nucleosome.

B. The bands with longer lengths probably corresponds to stretches of DNA associated with increasing numbers of nucleosomes .In the experimental data, adjacent bands differ in size by roughly 200bp (length of DNA found in a nucleosome along with adjacent linker DNA). This interpretation suggests that the M-nuclease did not digest all the DNA completely. If all non-nucleosomal DNA were digested completely, the samples would contain only the 150-base-pair fragments. Here the gel image shows many bands.

C. From the data given that M-nuclease has the ability to cut anywhere near Brr1 gene after growth in 15’c, which means the DNA is no longer protected from digestion by binding to the histone proteins. It can be suggested that the wrapping of DNA within the nucleosomes has been loosened. This change in the nucleosomes must be specific to the Brr1 gene throughout the genome.

D. As the chromatin has loosened near Brr1, it suggests that Brr1 gene expression is increased when cells are grown in different temperatures, whereas Brr2 gene expression is likely to be the same under the two conditions. Probably, the Brr1 gene posses regulators for a protein that is required for cells to grow at 23’ but not in 15’.

Question 8:

A. Protein B and C are transcription activators as there binding to promoter regions leads to mRNA production. (experiment 3 and 4)

B. Protein A is repressors as it's binding to promoter site don't transcribe mRNA (Experiment 2)

C. Protein A clearly binds stronger to site one than protein B because binding of protein A (experiment 2) represses transcription,, binding of protein B alone activates transcription (exp 3) but when both of them are added in the in vitro transcription system (exp 5),, the transcription represses which clearly signifies that A binds stronger than B.

D. In exp 8,, A is bound to site 1 as it have stronger binding affinity than b and its clearly protein c which is bound to site 2 driving the mRNA production by transcription no matter what is bound to the upstream site 1 to it.

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