Chemistry-class-12-html_c-12biomolecules

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **MCQ (Single correct option)** </Primary>

<hr>

<main>

**Answer**(b)

Glycogen is a branched chain polymer of $\alpha D$ glucose units in which chain is formed by $C 1-C 4$ glycosidic linkage whereas branching occurs by the formation of $C 1-C 6$ glycosidic linkage. Structure of glycogen can be shown below similar to the structure amylopectin.

**Structure of amylopectine**

Glycogen is also known as animal starch present in liver, muscles and brain.

</div>
<div  style='float: right; width:45%; text-align: left;font-size:21px'>

</div>

</main>

<hr>

<footer> <font size = "4"> <span style="color:blue">MCQ (Single correct option)</span> $\rarr$ MCQ (More than one correct questions) $\rarr$ Short answer type questions $\rarr$ Matching the columns $\rarr$ Assertion and Reason $\rarr$ Long answer type questions  </font></footer>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **MCQ (Single correct option)** </Primary>

<hr>

<main>

**Answer**(c)

Sucrose (cane sugar) is a disaccharide. One molecule of sucrose on hydrolysis gives one molecule of glucose and one molecule of fructose.

**Note** Sucrose is a dextro-rotatory sugar on hydrolysis produces a laevorotatory mixture so, known as invert sugar. Sucrose is a non-reducing sugar while maltose and lactose are reducing sugar.

</div>

</main>

<hr>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **MCQ (Single correct option)** </Primary>

<hr>

<main>

**Answer**(c)

Secondary structures of protein denotes the shape in which a long polypeptide chain exists. The secondary structure exist in two type of structure $\alpha$ - helix and $\beta$ - pleated structure.

In $\alpha$ - helix structure, a polypeptide chain forms all possible hydrogen bonds by twisting into a right handed screw with $-NH$ group of each amino acid rest hydrogen bonded to $C C=O$ of adjacent amino acid, which form a helix.

</div>

</main>

<hr>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **MCQ (Single correct option)** </Primary>

<hr>

<main>

**Ques.** In disaccharides, if the reducing groups of monosaccharides, i.e., aldehydic or ketonic groups are bonded, these are non-reducing sugars. Which of the following disaccharide is a non-reducing sugar?
 
(a)
<img src="https://cdn.mathpix.com/cropped/2024_01_24_76a1a59bab5d20ee6cc6g-328.jpg?height=281&width=706&top_left_y=968&top_left_x=487" height="130px">
(b)
<img src="https://cdn.mathpix.com/cropped/2024_01_24_76a1a59bab5d20ee6cc6g-328.jpg?height=309&width=754&top_left_y=1254&top_left_x=458" height="130px">

</div>

</main>

<hr>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **MCQ (Single correct option)** </Primary>

<hr>

<main>

**Answer**(b)

</div>
<div  style='float: right; width:45%; text-align: left;font-size:21px'>

This structure represents sucrose in which $\alpha-D$ glucose and $\beta$-D- fructose is attached to each other by $C _1-C _2$ glycosidic linkage.

Since, reducing groups of glucose and fructose are involved in glycosidic bond formation, this is considered as non-reducing sugar.

</div>

</main>

<hr>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **MCQ (Single correct option)** </Primary>

<hr>

<main>

**Answer**(a)

Nucleoside Species formed by the attachment of a base to $1^{\prime}$ position of sugar is known as nucleoside. The sugar carbon are numbered as $1^{\prime}, 2^{\prime}, 3^{\prime}$,...to distinguish them from bases.

Structure of nucleoside

</div>

</main>

<hr>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **MCQ (Single correct option)** </Primary>

<hr>

<main>

**Ques.** There cyclic structures of monosaccharides are given below which of these are anomers.

(i)
<img src="https://cdn.mathpix.com/cropped/2024_01_24_76a1a59bab5d20ee6cc6g-332.jpg?height=349&width=256&top_left_y=475&top_left_x=331" height="200px">
(ii)
<img src="https://cdn.mathpix.com/cropped/2024_01_24_76a1a59bab5d20ee6cc6g-332.jpg?height=349&width=257&top_left_y=470&top_left_x=677" height="200px">
(iii)
<img src="https://cdn.mathpix.com/cropped/2024_01_24_76a1a59bab5d20ee6cc6g-332.jpg?height=394&width=261&top_left_y=428&top_left_x=1027" height="200px">

</div>

</main>

<hr>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **MCQ (Single correct option)** </Primary>

<hr>

<main>

**Ques.** Optical rotations of some compounds alongwith their structures are given below which of them have $D$ configuration.

(a) I, II, III

(b) II, III

(d) III

</div>

</div>
<div  style='float: right; width:59%; text-align: left;font-size:21px'>

</div>

</main>

<hr>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **MCQ (Single correct option)** </Primary>

<hr>

<main>

**Answer**(a)

$D$ and $L$ configuration are relative configuration decided by relating structure of given saccharide with $D$ or $L$ glyceraldehyde.

When $OH$ an lowest asymmetric carbon is written at right hand side, it is represented as $D$ configuration and when $OH$ is written on left hand side, it is represented as $L$ configuration.

</div>
<div  style='float: right; width:45%; text-align: left;font-size:21px'>

</div>

</main>

<hr>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **MCQ (Single correct option)** </Primary>

<hr>

<main>

**Ques.** Structure of disaccharide formed by glucose and fructose is given below. Identify anomeric carbon atoms in monosaccharide units.

(a) ' $a$ ' carbon of glucose and ' $a$ ' carbon of fructose

(b) ' $a$ ' carbon of glucose and ' $e$ ' carbon of fructose

(d) ' $f$ ' carbon of glucose and ' $f$ ' carbon of fructose

</div>

</main>

<hr>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **MCQ (Single correct option)** </Primary>

<hr>

<main>

(ii)

</div>
<div  style='float: right; width:45%; text-align: left;font-size:21px'>

(iii)

</div>

</main>

<hr>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **MCQ (Single correct option)** </Primary>

<hr>

<main>

In this way, numbering for the disaccharides can be done as

</div>
<div  style='float: right; width:49%; text-align: left;font-size:21px'>

(ii)
<img src="https://cdn.mathpix.com/cropped/2024_01_24_76a1a59bab5d20ee6cc6g-335.jpg?height=640&width=454&top_left_y=317&top_left_x=288" height="280px">

</div>

</main>

<hr>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **MCQ (More than one correct questions)** </Primary>

<hr>

<main>

**Answer**$(b, d)$

Sucrose on hydrolysis produces equimolar mixture of $\alpha-D(+)$ glucose and $B-D(-)$. fructose. Since in sucrose $C-1$ of glucose and $C-2$ of fructose are linked with each other So, they are non-reducing in nature.

</div>
<div  style='float: right; width:45%; text-align: left;font-size:21px'>

</div>

</main>

<hr>

<footer> <font size = "4"> MCQ (Single correct option) $\rarr$ <span style="color:blue">MCQ (More than one correct questions)</span> $\rarr$ Short answer type questions $\rarr$ Matching the columns $\rarr$ Assertion and Reason $\rarr$ Long answer type questions  </font></footer>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **MCQ (More than one correct questions)** </Primary>

<hr>

<main>

**Answer**$(b, d)$

Amylopectin and glycogen have almost similar structure in which glucose are linked linearly to each other by $C _1-C _4$ glycosidic linkage and branched at $C _1-C _6$ glycosidic linkage.

**Structure of amylopectin**

Glycogen are carbohydrates stored in animal body. The structure to similar to amylopectin and is rather more highly branched.

</div>
<div  style='float: right; width:45%; text-align: left;font-size:21px'>

</div>

</main>

<hr>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **MCQ (More than one correct questions)** </Primary>

<hr>

<main>

**Ques.** Amino acids are classified as acidic, basic or neutral depending upon the relative number of amino and carboxyl groups in their molecule. Which of the following are acidic?

(a) $(CH _3) _2    CH -\underset{NH _2}{\underset{|}{CH}}-COOH$

(b) $HOOC-CH _2-CH _2\underset{NH _2}{\underset{|}{CH}}-COOH$

(d) $HOOC-CH _2-\underset{NH _2}{\underset{|}{CH}} H-COOH$

</div>

</main>

<hr>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **MCQ (More than one correct questions)** </Primary>

<hr>

<main>

**Explanation**

This problem is based on concept of nature of amino acid, that either it is acidic, basic or neutral.

Depending upon the number of acidic $COOH$ group, and basic $-NH _2$ group amino acid, proteins can be classified as

(i) If number of $COOH$ groups = number of $NH _2$ groups, amino acid is neutral.

(ii) If number of $COOH$ groups $>$ number of $NH _2$ groups, amino acid is acidic.

(iii) If number of $COOH$ group < number of $NH _2$ group, amino acid is basic.

</div>

</main>

<hr>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **MCQ (More than one correct questions)** </Primary>

<hr>

<main>

**Answer**$(a, b)$

Purines consist of six membered and five membered nitrogen containing ring fused together.

Guanine and adenine are purine bases whose structures are

While thymine and uracil are pyrimidene bases.

Hence (a) and (b) are correct choices.

</div>
<div  style='float: right; width:45%; text-align: left;font-size:21px'>

</div>

</main>

<hr>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **Short answer type questions** </Primary>

<hr>

<main>

**Answer**

Sugar present in milk is known as lactose sugar. Two units of monosaccharides $\beta$-D-galactose and $\beta$-D-glucose are linked together.

Hence, are known as disaccharides.

</div>
<div  style='float: right; width:45%; text-align: left;font-size:21px'>

</div>

</main>

<hr>

<footer> <font size = "4">  MCQ (Single correct option) $\rarr$ MCQ (More than one correct questions) $\rarr$ <span style="color:blue">Short answer type questions</span> $\rarr$ Matching the columns $\rarr$ Assertion and Reason $\rarr$ Long answer type questions  </font></footer>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **Short answer type questions** </Primary>

<hr>

<main>

**Answer**

Monosaccharides contain carbonyl group.

Hence, are classified as aldose or ketose.

When aldehyde group is present, the monosaccharides are known as aldose.

When ketone group is present, the monosaccharides are known as ketose. Fructose has molecular formula $C _6 H    _{12} O _6$ containing 6 carbon and keto group and is classified as ketohexose.

</div>
<div  style='float: right; width:45%; text-align: left;font-size:21px'>

</div>

</main>

<hr>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **Short answer type questions** </Primary>

<hr>

<main>

**Ques.** The letters ' $D$ ' or ' $L$ ' before the name of a stereoisomer of a compound indicate the correlation of configuration of that particular stereoisomer. This refers to their relation with one of the isomers of glyceraldehyde. Predict whether the following compound has ' $D$ ' or ' $L$ ' configuration

</div>

</main>

<hr>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **Short answer type questions** </Primary>

<hr>

<main>

**Explanation**

This problem is based on relative configuration i.e., $D$ and $L$ configuration. This can be done by relating structure of monosaccharides with structure of glyceraldehyde.

If $OH$ is present at right side of second last carbon of monosaccharide is considered as D configuration.

If $OH$ is present at left side of second last carbon of monosaccharide is considered as Lconfiguration.

</div>

</main>

<hr>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **Short answer type questions** </Primary>

<hr>

<main>

**Answer**

<img src="https://cdn.mathpix.com/cropped/2024_01_24_76a1a59bab5d20ee6cc6g-341.jpg?height=278&width=293&top_left_y=1619&top_left_x=469" height="200px">
$\beta$-D-ribose
5
<img src="https://cdn.mathpix.com/cropped/2024_01_24_76a1a59bab5d20ee6cc6g-341.jpg?height=241&width=278&top_left_y=1652&top_left_x=874" height="200px">
$\beta$-D-2-deoxyribose

In case of cyclic structure of saccharide, if $-OH$ group present at second last carbon is present at bottom side, then it is considered as $D$ configuration (as shown above)

</div>

</main>

<hr>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **Short answer type questions** </Primary>

<hr>

<main>

(iii) Phosphoric acid $(H _3 PO _4)$.

Nucleosides are joined together by phosphodiester linkage between $5^{\prime}$ and $3^{\prime}$ carbon atoms of pentose sugar and a dinucleotide with phosphoric acid $(CH _3 PO _4)$ is formed

</div>
<div  style='float: right; width:45%; text-align: left;font-size:21px'>

</div>

</main>

<hr>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **Short answer type questions** </Primary>

<hr>

<main>

**Answer**

Monosaccharides units present in starch, cellulose and glucose can be determined by knowing the product of their hydrolysis.

(i) Starch is a polysaccharide of $\alpha$-glucose in which two types of linkage are observed i.e., $C _1-C _6$ and $C _1-C _4$ glycosidic linkage.

(ii) Cellulose is a straight chain polysaccharide of $\beta$-D glucose in which glucose are linked together by $C _1-C _4$ glycosidic linkage.

(iii) Glucose is a monosaccharide.

</div>

</main>

<hr>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **Short answer type questions** </Primary>

<hr>

<main>

**Answer**

At the surface of enzyme, active sites are present. These active sites of enzymes hold the substrate molecule in a suitable position, so that it can be attacked by the reagent effectively. This reduces the magnitude of activation energy.

Enzymes contains cavities of characteristics shape and possessing active groups known as active centre on the surface. The molecules of the reactant (substrate) having complementary shape, fit into these cavities. On account of these active groups, an activated complex is formed which then decomposes to yield the products.

</div>

</main>

<hr>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **Short answer type questions** </Primary>

<hr>

<main>

**Answer**

All naturally occurring $\alpha$-amino acids (except glycine) are optically active due to the presence of chiral carbon atom. These have either $D$ - or $L$-configuration. $D$-form means that, the amino $(-NH _2)$ group is present towards the right hand side. L-form shows the presence of $-NH _2$ group on the left hand side.

<img src="https://cdn.mathpix.com/cropped/2024_01_24_76a1a59bab5d20ee6cc6g-346.jpg?height=217&width=227&top_left_y=995&top_left_x=521" height="150px">
$D$-alanine
<img src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/0c47c323-70d6-417f-94e2-c2fb4ab69e00/public" height="150px">
$L$-alanine

</div>

</main>

<hr>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **Short answer type questions** </Primary>

<hr>

<main>

**Answer**

**Denaturation of proteins** Protein present in egg white has an unique three dimensional structure. When it is subjected to physical change like change in temperature. i.e., on boiling, coagulation of egg white occurs due to denaturation of protein.

During denaturation hydrogen bonds are disturbed due to this globules unfold and helix gets uncoiled and protein looses its biological activity.

</div>

</main>

<hr>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **Matching the columns** </Primary>

<hr>

<main>

**Ques.** Match the vitamins given in Column I with the deficiency disease they cause given in Column II.

|  | Column I <br> (Vitamins) | |Column II <br> (Diseases) |  
| :--- | :--- | :--- | :--- |
| A. | Vitamin A | 1. | Pernicious anaemia |
| B. | Vitamin $B _1$ | 2. | Increased blood clotting time |
| C. | Vitamin $B    _{12}$ | 3. | Xerophthalmia |
| D. | Vitamin C | 4. | Rickets |
| E. | Vitamin D | 5. | Muscular weakness |
| F. | Vitamin E | 6. | Night blindness |
| G. | Vitamin K | 7. | Beri-beri |
|  |  | 8. | Bleeding gums |
|  |  | 9. | Osteomalacia |

</font>

</div>

</main>

<hr>

<footer> <font size = "4"> MCQ (Single correct option) $\rarr$ MCQ (More than one correct questions) $\rarr$ Short answer type questions $\rarr$ <span style="color:blue">Matching the columns</span> $\rarr$ Assertion and Reason $\rarr$ Long answer type questions  </font></footer>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **Matching the columns** </Primary>

<hr>

<main>

**Answer**

A. $\rightarrow(3,6)$

B. $\rightarrow(7)$

C. $\rightarrow$ (1)

D. $\rightarrow$ (8)

E. $\rightarrow(4,9)$

F. $\rightarrow$ (5)

G. $\rightarrow(2)$

</font>

</div>
<div  style='float: right; width:45%; text-align: left;font-size:21px'>

|Column I (Vitamins)|Column II (Diseases)|
|-|-|
|A.  Vitamin A | Xerophthalmia |
| | Night blindness |
|B. Vitamin $B_1$ | Beri beri |
|C. Vitamin $B_{12}$ | Pernicious anaemia |
|D. Vitamin C | Bleeding gums |
|E. Vitamin D | Rickets |
| | Osteomalacia |
|F. Vitamin E | Muscular weakness |
|G. Vitamin K | Incresed blood clotting time |

</div>

</main>

<hr>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **Matching the columns** </Primary>

<hr>

<main>

**Ques.** Match the following enzymes given in Column I with the reactions they catalyse given in Column II.

|  | Column I <br> (Enzymes) | | Column II <br> (Reactions) |  
| :--- | :--- | :--- | :--- |
| A. | Invertase | 1. | Decomposition of urea into $NH _3$ and $CO _2$. |
| B. | Maltase | 2. | Conversion of glucose into ethyl alcohol. |
| C. | Pepsin | 3. | Hydrolysis of maltose into glucose. |
| D. | Urease | 4. | Hydrolysis of cane sugar. |
| E. | Zymase | 5. | Hydrolysis of proteins into peptides. |

</div>

</main>

<hr>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **Matching the columns** </Primary>

<hr>

<main>

**Answer**

A. $\rarr(4)$

B. $\rarr(3) \quad$

C. $\rarr(5)$

D. $\rarr(1) \quad$

E. $\rarr(2)$

</div>
<div  style='float: right; width:65%; text-align: left;font-size:21px'>

</div>

</main>

<hr>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **Assertion and Reason** </Primary>

<hr>

<main>

#### **Assertion** and **Reason**

In the following questions a statement of Assertion (A) followed by a statement of Reason ( $R$ ) is given. Choose the correct answer out of the following choices.
 
(a) Assertion and reason both are correct statements and reason explains the assertion.

(b) Both assertion and reason are wrong statements.

(d) Assertion is wrong statement and reason is correct statement.
 
(e) Assertion and reason both are correct statements but reason does not explain assertion.

</div>

</main>

<hr>
<footer> <font size = "4"> MCQ (Single correct option)  $\rarr$ MCQ (More than one correct questions) $\rarr$ Short answer type questions $\rarr$ Matching the columns $\rarr$ <span style="color:blue">Assertion and Reason</span> $\rarr$ Long answer type questions  </font></footer>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **Assertion and Reason** </Primary>

<hr>

<main>

**Answer**(d)

Assertion is wrong statement and reason is correct statement. $\alpha$-glycosidic linkage is present in maltose
 
<img src="https://cdn.mathpix.com/cropped/2024_01_24_76a1a59bab5d20ee6cc6g-348.jpg?height=352&width=684&top_left_y=1550&top_left_x=466" height="170px">

Because maltose is composed of two glucose unit in which $C-1$ of one glucose unit is linked to $C-4$ of another glucose unit.

</div>

</main>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **Assertion and Reason** </Primary>

<hr>

<main>

**Answer**(e)

Assertion and reason both are correct and reason does not explain assertion. All naturally occurring $\alpha$-amino except glycine are optically active.

Glycine is optically inactive because glycine does not have all four different substituent as shown below.

</div>

</main>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **Assertion and Reason** </Primary>

<hr>

<main>

**Answer**(b)

Both assertion and reason are wrong statements. Deoxyribose $C _5 H    _{10} O _4$ is a carbohydrate because it follow $C _5(H _2 O) _2$ formula and exist as polyhydroxy carbonyl compound whose cyclic structure is as shown below

<img src="https://cdn.mathpix.com/cropped/2024_01_24_76a1a59bab5d20ee6cc6g-349.jpg?height=246&width=359&top_left_y=927&top_left_x=636" height="170px">
$\beta$-o-2 deoxyribose

</div>

</main>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **Long answer type questions** </Primary>

<hr>

**Answer**

Chemical reactions of D-glucose which can't be explained by its open chain structure are

(i) Glucose does not give Schiff's test and does not produce hydrogensulphite addition product with $NaHSO _3$, despite having aldehyde group

(ii) The pentaacetate of glucose does not react with hydroxylamine.

In actual, glucose exist in two different crystalline form $\alpha$ form and $\beta$ form. It was proposed that one of the $OH$ groups may add to the $-CHO$ group and form cyclic hemiacetal structure. Glucose forms a 6 membered pyranose structure.

Cyclic structure exist in equilibrium with open structure and can be represented as

</div>

</main>

<hr>

<footer> <font size = "4"> MCQ (Single correct option)  $\rarr$ MCQ (More than one correct questions) $\rarr$ Short answer type questions $\rarr$ Matching the columns $\rarr$  Assertion and Reason  $\rarr$ <span style="color:blue">Long answer type questions</span>  </font></footer>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **Long answer type questions** </Primary>

<hr>

**Answer**

Evidences on the basis of which glucose was assigned the following structure are as follows

(i) Glucose on reaction with $HI$ produces $n$ hexane which indicates presence of six carbon atom linked in a having straight chain.

$
C _6 H    _{12} O _6 \xrightarrow{HI} n \text{ hexane }
$

(ii) Glucose on reaction with acetic anhydride produces glucose penta acetate which indicates presence of five $OH$ groups.

$
C _6 H    _{12} O _6 \xrightarrow{Ac _2 O} \text{ Glucose pentaacetate }
$

</div>

</main>

<hr>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **Long answer type questions** </Primary>

<hr>

**Answer**

Carbohydrates that are used as storage molecules in plants and animals are as follows

(i) Plant contains mainly starch, cellulose, sucrose etc.

(ii) Animal contain glycogen in their body. So, glycogen is also known as animal starch. Glycogen is present in liver, muscles and brain when body needs glucose, enzyme breaks glycogen down to glucose.

(iii) Cellulose is present in wood, and fibre of clothes.

</div>

</main>

<hr>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **Long answer type questions** </Primary>

<hr>

**Answer**

Primary structure of proteins Proteins consist of one or more polypeptide chains. Each polypeptide is a protein contains amino acids joined with each other in a specific sequence. Secondary structure of proteins it refers to the shape in which a long polypeptide chain can exist.

| $\alpha$ helix structure                                                                                                                                                                                                                              | $\beta $ pleated sheet structure                                                                                                                                                                                                                                                                           |
|-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|
| A structure of twisting of all a polypeptide chain formed by possible H-bonds into a right handed screw (helix ) with the -NH froup of each amino acid,and residuw hydrogen binded to the -CO- of an adjacent tum of the helix Hence, called $\alpha$-helix| All peptude chains are tretched out to nearly maximum extension and then laid side by side which are held together by intermolecular hydrogen bonds. This structure resembles the hydrogen bonds. This structure resembles the pleated folds of the drapery . Hence, called $\beta$-pleated sheet structure |

</div>

</main>

<hr>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **Long answer type questions** </Primary>

<hr>

$\alpha$-helix structure of proteins

</div>
<div  style='float: right; width:45%; text-align: left;font-size:21px'>

$\beta$-pleated sheet structure of proteins

</div>

</main>

<hr>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **Long answer type questions** </Primary>

<hr>

**Answer**

On complete hydrolysis of DNA, following fragments are formed a pentose sugar ( $\beta$-D-2-deoxyribose) phosphoric acid $(H _3 PO _4)$ and bases (nitrogen containing heterocyclic compounds).

**Structures**

</div>
<div  style='float: right; width:70%; text-align: left;font-size:21px'>

(i) Sugar

<img src="https://cdn.mathpix.com/cropped/2024_01_24_76a1a59bab5d20ee6cc6g-352.jpg?height=269&width=344&top_left_y=659&top_left_x=304" height="200px">
$\beta$-D-2-deoxyribose 
<img src="http://13.201.94.86:82//subject-images/chemistry/p104.png" height="200px">

</div>

</main>

<hr>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **BIOMOLECULES**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **Long answer type questions** </Primary>

<hr>

A unit formed by the attachment of a base to 1'-position of sugar is called nucleoside. When nucleoside links to phosphoric acid at 5'-position of sugar moiety, a nucleotide is formed. Nucleotides are joined together by phosphodiester linkage between $5^{\prime}$ and $3^{\prime}$ carbon atoms of the pentose sugar.

In DNA two nucleic acid chains are wound about each other and held together by hydrogen bonds between pairs of bases.

The two strands are complementary to each other because hydrogen bonds are formed between specific pair of base adenine form hydrogen bonds with thymine whereas cytosine form hydrogen bonds with guanine.

</div>
<div  style='float: right; width:34%; text-align: left;font-size:21px'>

</div>

</main>

<hr>