Arrange the following in the descending order of their number in restriciton site of E.coRI.
I. Number of hydrogen bonds broken .
II. Number of covalent bonds broken .
III. Total number of hydrogen bonds .
IV Number of hydrogen bonds left intact after treating with enzyme .
V. Number of unpaired nucleotides left at each end of the fragment .

Options

A. III,V , IV ,I,II
B. III,I,IV,II
C. III,I,IV,V,II
D. I,III,V,IV,II

Answer

To arrange the given items in descending order based on their numbers in the restriction site of EcoRI, we will analyze each item step by step.

### Step-by-Step Solution:

1. **Identify the Restriction Site of EcoRI**:
The restriction site for EcoRI is:
**GAATTC** (which is palindromic, meaning it reads the same forwards and backwards).

2. **Calculate the Total Number of Hydrogen Bonds**:
- Between G and C: 3 hydrogen bonds.
- Between A and T: 2 hydrogen bonds.
- Total for GAATTC:
- G-C: 3
- A-T: 2 (for two A-T pairs)
- Total = 3 (G-C) + 2 (A-T) + 2 (A-T) + 3 (C-G) = 14 hydrogen bonds.
- **Total Number of Hydrogen Bonds = 14**.

3. **Determine the Number of Covalent Bonds Broken**:
- EcoRI cuts between G and A, breaking the phosphodiester bonds.
- There are 2 covalent bonds broken (one on each strand).
- **Number of Covalent Bonds Broken = 2**.

4. **Calculate the Number of Hydrogen Bonds Broken**:
- When EcoRI cuts, it breaks the hydrogen bonds between A and T pairs.
- For the two A-T pairs, 2 hydrogen bonds per pair are broken.
- Total broken = 2 (A-T) + 2 (A-T) + 3 (G-C) + 3 (C-G) = 8 hydrogen bonds broken.
- **Total Number of Hydrogen Bonds Broken = 8**.

5. **Determine the Number of Hydrogen Bonds Left Intact**:
- After the enzyme acts, the remaining hydrogen bonds are those that were not broken.
- Initially, there were 14 hydrogen bonds, and 8 were broken.
- Remaining = 14 - 8 = 6 hydrogen bonds.
- **Number of Hydrogen Bonds Left Intact = 6**.

6. **Calculate the Number of Unpaired Nucleotides Left at Each End**:
- After the cut, EcoRI leaves unpaired nucleotides at the ends.
- For the sticky ends, there are 2 unpaired nucleotides (one at each end).
- Therefore, total unpaired nucleotides = 4 (2 at each end).
- **Number of Unpaired Nucleotides Left = 4**.

### Summary of Values:
- I. Total Number of Hydrogen Bonds = **14**
- II. Number of Covalent Bonds Broken = **2**
- III. Total Number of Hydrogen Bonds Broken = **8**
- IV. Number of Hydrogen Bonds Left Intact = **6**
- V. Number of Unpaired Nucleotides Left = **4**

### Arranging in Descending Order:
1. Total Number of Hydrogen Bonds (14)
2. Total Number of Hydrogen Bonds Broken (8)
3. Number of Hydrogen Bonds Left Intact (6)
4. Number of Unpaired Nucleotides Left (4)
5. Number of Covalent Bonds Broken (2)

### Final Arrangement:
- I. 14
- III. 8
- IV. 6
- V. 4
- II. 2

### Final Answer:
**Descending Order: I (14), III (8), IV (6), V (4), II (2)**