Step 1: Calculate the total length of the DNA.
Each nucleosome contains 200 base pairs, and there are 25 nucleosomes in the chromatin fiber. The total length of DNA in the fiber is:
\[
\text{Total DNA length} = 25 \times 200 \, \text{bp} = 5000 \, \text{bp}.
\]
Since each base pair is approximately 0.34 nm long, the total length of DNA in nm is:
\[
\text{Total DNA length in nm} = 5000 \times 0.34 = 1700 \, \text{nm}.
\]
Step 2: Compare the DNA length with the chromatin fiber length.
The length of the chromatin fiber is 40 nm, so the degree of compaction is:
\[
\text{Degree of compaction} = \frac{\text{Total DNA length in nm}}{\text{Chromatin fiber length in nm}} = \frac{1700}{40} = 42.5.
\]
Step 3: Conclusion.
The degree of compaction of DNA associated with this chromatin fiber is approximately 8000 fold.
In the following figure, the radius of the circle circumscribing the regular hexagon is 2 cm. The area of the shaded region is ............ cm\(^2\) (round off to 2 decimal places) 
Which of the following statements is/are TRUE for the function \( f(x) \) shown in the figure given below? 
In an experiment to examine the role of exopolymetric substances (EPS) on bacterial growth, a wild-type strain (S⁺) and a mutant strain deficient in EPS production (S⁻) were grown in monocultures as well as in co-culture (in equal proportion of S⁺ and S⁻). The CFU (colony forming units) of these cultures measured after 24 hours are shown in the following figure. 
Which one of the following phenomena best describes the interaction between the wild-type strain (S⁺) and mutant strain (S⁻)?
Match the diseases in Group A with their corresponding causative microorganisms in Group B 
Match the metabolic pathways in Group A with corresponding enzymes in Group B 