To determine the total number of hydrogen bonds in a DNA-double helix strand based on the given sequence, we must consider how base pairing occurs. In DNA, the bases Guanine (G) and Cytosine (C) pair together with 3 hydrogen bonds, while Adenine (A) and Thymine (T) pair with 2 hydrogen bonds.
The provided sequence is:
5′-G-G-C-A-A-A-T-C-G-G-C-T-A-3′
Let's count the hydrogen bonds for each base pair:
Adding all these together gives the total number of hydrogen bonds: 6 + 3 + 6 + 2 + 3 + 6 + 3 + 2 + 2 = 33.
Thus, the total number of hydrogen bonds is 33, which falls within the specified range (33,33).
To determine the total number of hydrogen bonds in a DNA double-helix strand, we need to understand the base pairing rules of DNA:
Given the DNA sequence:
5'-G-G-C-A-A-A-T-C-G-G-C-T-A-3'
Let's pair each base with its complementary base on the opposite strand and count the hydrogen bonds:
| Base Pair | Hydrogen Bonds |
| G - C | 3 |
| G - C | 3 |
| C - G | 3 |
| A - T | 2 |
| A - T | 2 |
| A - T | 2 |
| T - A | 2 |
| C - G | 3 |
| G - C | 3 |
| G - C | 3 |
| C - G | 3 |
| T - A | 2 |
| A - T | 2 |
Now, sum the total number of hydrogen bonds:
Total hydrogen bonds = 3 + 3 + 3 + 2 + 2 + 2 + 2 + 3 + 3 + 3 + 3 + 2 + 2 = 33
The total number of hydrogen bonds is 33, which is within the expected range of 33 to 33.
Fat soluble vitamins are :
A. Vitamin B\( _1 \)
B. Vitamin C
C. Vitamin E
D. Vitamin B\( _{12} \)
E. Vitamin K
Choose the correct answer from the options given below :
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
Two blocks of masses \( m \) and \( M \), \( (M > m) \), are placed on a frictionless table as shown in figure. A massless spring with spring constant \( k \) is attached with the lower block. If the system is slightly displaced and released then \( \mu = \) coefficient of friction between the two blocks.
(A) The time period of small oscillation of the two blocks is \( T = 2\pi \sqrt{\dfrac{(m + M)}{k}} \)
(B) The acceleration of the blocks is \( a = \dfrac{kx}{M + m} \)
(\( x = \) displacement of the blocks from the mean position)
(C) The magnitude of the frictional force on the upper block is \( \dfrac{m\mu |x|}{M + m} \)
(D) The maximum amplitude of the upper block, if it does not slip, is \( \dfrac{\mu (M + m) g}{k} \)
(E) Maximum frictional force can be \( \mu (M + m) g \)
Choose the correct answer from the options given below:
Let \( f : \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \[ (\sin x \cos y)(f(2x + 2y) - f(2x - 2y)) = (\cos x \sin y)(f(2x + 2y) + f(2x - 2y)), \] for all \( x, y \in \mathbb{R}. \)
If \( f'(0) = \frac{1}{2} \), then the value of \( 24f''\left( \frac{5\pi}{3} \right) \) is: