Carbohydrates Questions in English

(B) Disaccharides are carbohydrates that yield two molecules of monosaccharides upon hydrolysis.
This process can be catalyzed by dilute acids or specific enzymes (e.g.,sucrase,maltase).
Therefore,statement $P$ is correct.
Upon hydrolysis,the glycosidic bond linking the two monosaccharide units is broken,releasing the constituent monosaccharides.
Therefore,statement $Q$ is also correct.

(A) Statement $P$ is true because the general formula for carbohydrates is indeed represented as $C_n(H_2O)_n$ or $C_n(H_2O)_m$,where $n$ and $m$ are integers.
Statement $Q$ is true because for monosaccharides (simple sugars),the ratio of hydrogen to oxygen atoms is $2:1$,which is the same as in water. Thus,for monosaccharides,the formula simplifies to $C_n(H_2O)_n$,meaning $n = m$.
Since both statements are correct and the equality of $n$ and $m$ explains why the general formula can be written as $C_n(H_2O)_n$,Statement $Q$ is the correct explanation for Statement $P$.

(B) Statement $X$ is correct because starch is the primary storage polysaccharide in plants.
Statement $Y$ is incorrect because glucose is not the stored food in animals; instead,animals store food in the form of glycogen.
Therefore,Statement $X$ is correct and Statement $Y$ is incorrect.

(B) The structural formula shows a linear chain of glucose units linked by $\beta-1,4$-glycosidic bonds.
In this structure,every alternate glucose unit is inverted (rotated by $180^{\circ}$),which is a characteristic feature of cellulose.
Starch and glycogen consist of $\alpha$-glucose units with $\alpha-1,4$ and $\alpha-1,6$ linkages,whereas cellulose is a polymer of $\beta$-glucose.
Therefore,the given structure represents cellulose.

(A) The provided figure represents a chain of glucose units linked together.
By observing the linkage between the carbon atoms of adjacent glucose rings,we can see that the bond connects the $C-1$ position of one glucose unit to the $C-4$ position of the next glucose unit.
The orientation of the $-OH$ group at the $C-1$ position is in the $\beta$ configuration (pointing upwards relative to the ring plane).
Therefore,the linkage is a $\beta\ 1-4$ glycosidic bond,which is characteristic of cellulose.

(C) The given structural formula shows the condensation reaction between two glucose molecules to form a disaccharide.
Specifically, it depicts the formation of maltose from two $\alpha$-$D$-glucose units.
The bond labeled '$A$' is an $\alpha$-$1,4$-glycosidic bond.
Lactose is formed from glucose and galactose, while sucrose is formed from glucose and fructose.
Starch is a polysaccharide, not a disaccharide.

(B) The provided image shows the condensation reaction between two monosaccharide units (glucose molecules) to form a disaccharide.
In this reaction,a water molecule $(H_2O)$ is released,and the two sugar units are linked together by an oxygen atom.
This specific linkage between two monosaccharide units is known as a glycosidic bond.
Therefore,$A$ indicates a glycosidic bond.

(C) The reaction shows the condensation of two glucose molecules to form a disaccharide (maltose).
In the first glucose molecule,the hydroxyl group $(-OH)$ is at the $C-1$ position,and in the second glucose molecule,the hydroxyl group is at the $C-4$ position.
The bond formed between the $C-1$ of the first glucose and the $C-4$ of the second glucose is known as an $\alpha\ 1-4$ glycosidic bond.
Therefore,$A$ indicates an $\alpha\ 1-4$ glycosidic linkage.

(B) The reaction shows the condensation of two glucose molecules to form a disaccharide.
Specifically,it depicts the formation of a glycosidic bond between two $\alpha$-$D$-glucose units.
The condensation of two glucose molecules results in the formation of maltose and the release of one water molecule $(H_2O)$.
Therefore,the product formed is maltose.

(D) The given structural formula represents the formation of lactose.
Lactose is a disaccharide formed by the condensation reaction between one molecule of glucose and one molecule of galactose.
In the reaction,a water molecule $(H_2O)$ is released,and a glycosidic bond is formed between the two hexose sugar units.
Therefore,both glucose and galactose are involved in the structure.

(D) The correct matching is as follows:
$(1)$ $C_n(H_2O)_m$ represents the general formula for Carbohydrates $(c)$.
$(2)$ $C_n(H_2O)_{n-1}$ represents the general formula for Disaccharides $(d)$, where $n$ is the number of carbon atoms.
$(3)$ $C_n(H_2O)_n$ represents the general formula for Monosaccharides $(a)$, such as glucose $(C_6H_{12}O_6)$.
$(4)$ $n(C_5H_{10}O_5)$ represents the general formula for Polysaccharides $(b)$, which are polymers of monosaccharide units.
Therefore, the correct sequence is $(1-c, 2-d, 3-a, 4-b)$.

(B) The $-C-O-C-$ linkage is formed when two monosaccharides are joined together by a dehydration reaction (removal of a water molecule).
This specific type of ether linkage between two sugar molecules is known as a glycosidic bond.
Peptide bonds are found in proteins $(-CO-NH-)$,while ester bonds are typically found in lipids and nucleic acids.

(B) The formation of $Sucrose$ from $Glucose$ and $Fructose$ is a condensation reaction.
In this process,a molecule of water $(H_2O)$ is removed from the two monosaccharides to form a glycosidic bond.
This removal of a water molecule is known as dehydration or a dehydration synthesis reaction.
Therefore,'$X$' represents dehydration.

(B) The provided structure represents a disaccharide where two glucose units are linked together.
In this structure,the glycosidic bond is formed between the carbon-$1$ of the first glucose molecule and the carbon-$4$ of the second glucose molecule.
This type of linkage is known as a $1, 4$-glycosidic bond.
Therefore,the correct option is $B$.

(B) The correct matching is as follows:
$(1)$ Starch is the primary storage polysaccharide in plants, stored as starch granules, so $(1-d)$.
$(2)$ Glycogen is the storage polysaccharide in animals, primarily stored in the liver and muscles, so $(2-c)$.
$(3)$ Ribose is a pentose sugar that is a structural component of nucleotides like $ATP$, $RNA$, etc., so $(3-b)$.
$(4)$ Glucose is the most common monosaccharide used as a primary fuel source in cellular respiration, so $(4-a)$.
Therefore, the correct sequence is $(1-d), (2-c), (3-b), (4-a)$.

(A) Carbohydrates are the primary source of energy for the human body. In our daily activities,we consume carbohydrates (such as starch,glucose,and sucrose) in the largest quantities compared to other macromolecules like proteins,lipids,or nucleic acids to meet our metabolic energy requirements.

(A) The structure shown is a pentose sugar.
Specifically,it is a ribose sugar because it has a hydroxyl group $(-OH)$ attached to the $2'$ carbon atom.
In deoxyribose sugar,the $2'$ carbon atom has only a hydrogen atom $(-H)$ instead of a hydroxyl group $(-OH)$.
Therefore,the given structure is ribose sugar.

(C) The provided structure is a five-membered ring (furanose ring) containing an oxygen atom.
By examining the hydroxyl $(-OH)$ groups attached to the carbon atoms,we can identify the sugar.
Specifically,there are $-OH$ groups at both the $C-2$ and $C-3$ positions of the pentose ring.
This specific arrangement of hydroxyl groups is characteristic of ribose sugar.
Deoxyribose would lack the $-OH$ group at the $C-2$ position.
Glucose and galactose are hexose sugars (six-membered rings),which do not match this structure.
Therefore,the correct answer is Ribose sugar.

(C) $ATP$ (Adenosine Triphosphate) consists of an adenosine molecule (adenine + ribose sugar) linked to three phosphate groups.
The first phosphate group is attached to the $5'$ carbon of the ribose sugar via a phosphoester bond.
The subsequent two phosphate groups are linked to each other by high-energy phosphoanhydride bonds.
Therefore,the bond between the adenosine (specifically the ribose sugar) and the first phosphate group is a phosphoester bond.

(B) $Lactose$, also known as milk sugar, is a disaccharide composed of two monosaccharide units: $Glucose$ and $Galactose$.
Upon hydrolysis in the presence of the enzyme $Lactase$, the glycosidic bond between these two units is broken, resulting in the production of one molecule of $Glucose$ and one molecule of $Galactose$.
$Lactose + H_2O \xrightarrow{Lactase} Glucose + Galactose$.

(C) Lactose is a disaccharide found in milk.
It consists of two monosaccharide units: $D$-glucose and $D$-galactose.
Upon hydrolysis,the glycosidic bond between these two units is broken,yielding one molecule of glucose and one molecule of galactose.
The reaction is: $\text{Lactose} + H_2O \xrightarrow{\text{Lactase}} \text{Glucose} + \text{Galactose}$.

(B) Maltose is a disaccharide composed of two glucose units linked by an $\alpha-1,4$-glycosidic bond.
Upon hydrolysis,catalyzed by the enzyme maltase or by dilute acids,maltose breaks down into its constituent monosaccharides.
The chemical reaction is: $\text{C}_{12}\text{H}_{22}\text{O}_{11} + \text{H}_2\text{O} \xrightarrow{\text{Maltase}} 2\text{C}_6\text{H}_{12}\text{O}_6$ (Glucose + Glucose).
Therefore,the hydrolysis of maltose yields two molecules of glucose.

(D) High-energy compounds are those that release a large amount of free energy upon hydrolysis,typically due to the presence of phosphoanhydride bonds.
$ATP$ (Adenosine triphosphate) and $GTP$ (Guanosine triphosphate) contain two high-energy phosphoanhydride bonds.
Creatine phosphate is a well-known high-energy storage molecule in muscle cells.
$AMP$ (Adenosine monophosphate) contains only one phosphate group attached to the ribose sugar via a phosphoester bond,which is a low-energy bond. Therefore,$AMP$ is not considered a high-energy compound.

(A) The given structure represents a pentose sugar,specifically ribose.
In a ribose molecule,the carbon at the $2'$ position (marked as '$X$') is bonded to a hydrogen atom $(H)$ and a hydroxyl group $(OH)$.
Looking at the provided structure,the $X$ position corresponds to the $2'$ carbon of the ribose ring.
Therefore,the group attached at this position is a hydrogen atom above and a hydroxyl group below,which matches option $A$.

(B) The given structural formula represents a pentose sugar in its cyclic form.
By observing the carbon atoms,we can see that at the $2'$ position,there is a hydrogen atom $(H)$ instead of a hydroxyl group $(-OH)$.
Ribose has an $-OH$ group at the $2'$ position,whereas deoxyribose lacks an oxygen atom at this position,hence the name 'deoxy'.
Therefore,the structure shown is $2-$deoxyribose.

(B) The given structure is a pentose sugar (a $5$-carbon sugar) in its cyclic form.
By observing the $2^{nd}$ carbon atom $(C_2)$,we can see an $-OH$ group attached to it.
In $Deoxyribose$ sugar,the $2^{nd}$ carbon atom has only a hydrogen atom $(-H)$ attached,lacking the oxygen atom.
Since the $2^{nd}$ carbon in this figure has an $-OH$ group,it is a $Ribose$ sugar.
Therefore,the correct option is $B$.

(A) The correct answer is $A$.
Chitin is a structural polysaccharide that forms the exoskeleton of arthropods.
It is a complex carbohydrate (a nitrogen-containing polysaccharide) in which $N$-acetyl glucosamine monomers are linked together by $(1, 4)$ $\beta$-glycosidic linkages.
This chitinous exoskeleton provides structural strength,protection,and elasticity to the bodies of arthropods.

(D) Starch is the major storage carbohydrate in plants.
For a substance to be an ideal storage material,it should be chemically stable (non-reactive) and should not affect the water potential of the cell (osmotically inactive).
$(2)$ Chemically non-reactive: Starch is a large,complex polysaccharide that does not interfere with the metabolic activities of the cell.
$(4)$ Osmotically inactive: Being a large,insoluble molecule,it does not dissolve in the cell sap,thus it does not affect the osmotic pressure of the cell.
Therefore,properties $(2)$ and $(4)$ make starch an ideal storage material.
Thus,the correct option is $(D)$.

(A) glycosidic bond is a type of covalent bond that joins a carbohydrate (sugar) molecule to another group,which may or may not be another carbohydrate.
Starch is a polysaccharide composed of glucose units linked together by $\alpha-1,4$ and $\alpha-1,6$ glycosidic bonds.
Proteins are composed of amino acids linked by peptide bonds.
Lipids (specifically triglycerides) are composed of fatty acids and glycerol linked by ester bonds.
Therefore,the glycosidic bond is characteristic of carbohydrates like starch.

(D) Fructose is a monosaccharide with the chemical formula $C_6H_{12}O_6$.
It contains a ketone functional group,making it a ketose.
Since it contains $6$ carbon atoms,it is classified as a hexose.
Therefore,fructose is a keto hexose sugar.

(D) Carbohydrates are classified into monosaccharides,disaccharides,and polysaccharides based on their complexity.
Monosaccharides are the simplest form of carbohydrates,which cannot be further hydrolyzed into smaller sugar units.
Glucose,Galactose,and Ribose are all examples of monosaccharides.
Therefore,all the given options are correct.

(C) ketohexose is a monosaccharide containing six carbon atoms and a ketone functional group.
Glucose and Galactose are aldohexoses,meaning they contain six carbon atoms and an aldehyde functional group.
Ribose is an aldopentose,containing five carbon atoms and an aldehyde functional group.
Fructose is a ketohexose,as it contains six carbon atoms and a ketone group at the $C-2$ position.
Therefore,the correct option is $C$.

(D) Starch is a polysaccharide that consists of two components: amylose and amylopectin.
Amylose is a helical molecule that can trap iodine molecules $(I_2)$ within its structure.
When starch reacts with iodine,it forms a characteristic blue-colored complex.
This reaction is commonly used as a diagnostic test for the presence of starch in biological samples.

(D) Sucrose,Lactose,and Maltose are disaccharides,which are composed of two monosaccharide units linked together.
Cellulose is a polysaccharide,which is a complex carbohydrate composed of a long chain of many monosaccharide units (specifically glucose).
Therefore,Cellulose is the odd one out.

(A) Cotton fibres are primarily composed of cellulose.
Cellulose is a polysaccharide consisting of a linear chain of several hundred to many thousands of $\beta(1 \rightarrow 4)$ linked $D$-glucose units.
It is the most abundant organic compound in the plant kingdom and provides structural support to the cell walls of plants.

(A) Chitin is a complex carbohydrate,specifically a structural polysaccharide.
It is a polymer of $N$-acetylglucosamine units.
It is found in the cell walls of fungi and the exoskeletons of arthropods like insects and crustaceans.
Therefore,it is classified as a polysaccharide.

(B) Ribose is a monosaccharide with $5$ carbon atoms,which classifies it as a pentose sugar.
It contains an aldehyde functional group at the $C-1$ position,which classifies it as an aldose.
Therefore,ribose is an aldo pentose sugar.

(C) The biological importance of carbohydrates includes:
$1$. Structural components: Cellulose is a major carbohydrate found in the cell wall of plants.
$2$. Food storage: Starch in plants and glycogen in animals serve as primary energy storage molecules.
$5$. Nucleic acid structure: Ribose and deoxyribose sugars are essential components of $RNA$ and $DNA$ respectively.
Statements $3$ (protein synthesis) and $4$ (enzymes) are primarily functions of proteins,not carbohydrates.
Therefore,the correct statements are $1, 2,$ and $5$.

(D) Polysaccharide like $Glycogen$ is a polymer of glucose.
In a polysaccharide chain, the end is called the reducing end and the end is called the non-reducing end.
$Glycogen$ is a branched polymer where each branch has one non-reducing end, while the main chain has one reducing end.
$DNA$, $RNA$, and $Proteins$ do not follow this specific nomenclature of reducing and non-reducing ends in the same context as carbohydrates.

(A) In a polysaccharide,individual monosaccharide units are linked together by a glycosidic bond.
This bond is formed by a dehydration reaction (removal of a water molecule) between two carbon atoms of adjacent monosaccharides.
Peptide bonds link amino acids in proteins,ester bonds are found in lipids,and phosphodiester bonds link nucleotides in nucleic acids.

(C) Glycogen is a polysaccharide that serves as a form of energy storage in animals and fungi.
It is a branched polymer composed of glucose units.
Glucose is a type of sugar that belongs to the category of aldohexoses because it contains an aldehyde group and six carbon atoms.
Therefore,the repeating unit of glycogen is glucose,which is an aldohexose.

(A) Glucose $(C_6H_{12}O_6)$ is a simple sugar classified as an aldohexose because it contains an aldehyde group $(-CHO)$ and six carbon atoms.
Glucose exists in both open-chain and cyclic (ring) forms in aqueous solutions.
It is a fundamental monosaccharide and serves as the primary energy source for living organisms.
Therefore,the statement that it is an aldohexose is correct.

(B) The correct pair is $Polysaccharide - glycogen$.
$1$. $Alanine$ is an amino acid,not a nitrogen base.
$2$. $Glycogen$ is a polymer of glucose and is classified as a polysaccharide.
$3$. $Trypsin$ is an enzyme (protein),not an amino acid.
$4$. $Insulin$ is a hormone (protein),not an enzyme.

(C) Fructose is a simple monosaccharide found in many foods and is one of the three dietary monosaccharides,along with glucose and galactose,that are absorbed directly into the blood during digestion.
It has the chemical formula $C_6H_{12}O_6$.
Since it contains $6$ carbon atoms,it is classified as a hexose sugar.

(D) Monosaccharides are classified based on the number of carbon atoms they contain.
Trioses are monosaccharides that contain $3$ carbon atoms.
$A$. Ribose is a pentose sugar ($5$ carbons).
$B$. Ribulose is a pentose sugar ($5$ carbons).
$C$. Glucose is a hexose sugar ($6$ carbons).
$D$. Glyceraldehyde is a triose sugar ($3$ carbons).
Therefore,the correct answer is $D$.

(D) Maltose is a disaccharide composed of two glucose units linked by an $\alpha-1,4$-glycosidic bond.
Upon hydrolysis,the glycosidic bond is broken by the addition of a water molecule.
This reaction results in the formation of two molecules of glucose.
Therefore,the hydrolysis of maltose yields $Glucose + Glucose$.

(B) Cellulose is a polysaccharide,which is a homopolymer of glucose units. Therefore,statement $A$ is incorrect because it is a homopolymer,not a heteropolymer.
Statement $R$ is also incorrect because cellulose is a carbohydrate (sugar),not a protein,and it is composed of glucose monomers,not amino acids.

(C) Cellulose is a polysaccharide composed of a linear chain of several hundred to many thousands of $\beta(1 \rightarrow 4)$ linked $D$-glucose units.
These glucose units are joined together by $\beta-1,4$-glycosidic bonds.
Therefore,the bond found in cellulose is a glycosidic bond.

(D) monosaccharide is the simplest form of carbohydrate that cannot be further hydrolyzed into smaller sugar units.
$Sucrose$ is a disaccharide composed of glucose and fructose.
$Starch$ and $Glycogen$ are polysaccharides (polymers of glucose).
$Glyceraldehyde$ is a triose sugar, which is a type of monosaccharide.

(D) Glycogen is a polysaccharide,which is a complex carbohydrate.
It serves as the primary storage form of glucose in animals and fungi.
Structurally,it consists of a branched chain of glucose units linked by $\alpha-1,4$ and $\alpha-1,6$ glycosidic bonds.
Therefore,it is a polymer of glucose.