Match List - I with List - II.
List - I (Saccharides) List - II (Glycosidic linkages found)
(A) Sucrose (I) \( \alpha 1 - 4 \)
(B) Maltose (II) \( \alpha 1 - 4 \) and \( \alpha 1 - 6 \)
(C) Lactose (III) \( \alpha 1 - \beta 2 \)
(D) Amylopectin (IV) \( \beta 1 - 4 \)
Choose the correct answer from the options given below:
Match List - I with List - II.
List - I (Saccharides) List - II (Glycosidic linkages found)
(A) Sucrose (I) \( \alpha 1 - 4 \)
(B) Maltose (II) \( \alpha 1 - 4 \) and \( \alpha 1 - 6 \)
(C) Lactose (III) \( \alpha 1 - \beta 2 \)
(D) Amylopectin (IV) \( \beta 1 - 4 \)
Choose the correct answer from the options given below:
Show Hint
- (A)-(II), (B)-(IV), (C)-(III), (D)-(I)
- (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
- (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
- (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
The Correct Option is B
Solution and Explanation
- Sucrose consists of glucose and fructose linked by an \( \alpha 1-4 \) glycosidic bond, so it matches with (I).
- Maltose has two glucose units connected by an \( \alpha 1-4 \) bond, and it also has the possibility of forming a linkage with \( \alpha 1-6 \), which matches with (II).
- Lactose is a disaccharide formed from one molecule of glucose and one of galactose, with an \( \alpha 1-\beta 2 \) glycosidic bond, so it matches with (III).
- Amylopectin is a highly branched polymer of glucose with \( \beta 1-4 \) linkages in its chains, so it matches with (IV).
Thus, the correct matching is (A)-(I), (B)-(II), (C)-(III), (D)-(IV).
Top JEE Main Chemistry Questions
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- At equilibrium the concentrations of and in a sealed vessel at . What will be for the reaction
- Find out the major products from the following reactions
Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
- The correct order of atomic radii is:
Top JEE Main Hybridisation Questions
- Which of the following has a square pyramidal shape?
- How many of the following compounds have \(sp^3\) hybridized central atoms?
\(H_2O, NH_3, SiO_2, SO_2, CO, CH_4\) and \(BF_3\) - Which of the following have a trigonal bipyramidal shape?
\(PF_5, PBr_5, [PtCl_4], SF_6, BF_3, BrF_5, PCl_5, [Fe(CO)_5]\) - Identify the following species in which \(d^2sp^3\) hybridization is shown by the central atom.
- Given below are two statements: Statement I: $\text{PF}_5$ and $\text{BrF}_5$ both exhibit $\text{sp}^3\text{d}$ hybridisation.Statement II: Both $\text{SF}_6$ and $[\text{Co}(\text{NH}_3)_6]^{3+}$ exhibit $\text{sp}^3\text{d}^2$ hybridisation. In the light of the above statements,choose the correct answer from the options given below:
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.