Biomolecules

Two nucleic acid chains are wound about each other and held together by H bonds between pair of bases.

Adenine from two hydrogen bonds with thymine and Guanine form three hydrogen bond with cytosine.

$5^{\prime} \mathrm{G}-\mathrm{G}-\mathrm{C}-\mathrm{A}-\mathrm{A}-\mathrm{A}-\mathrm{T}-\mathrm{C}-\mathrm{G}-\mathrm{G}-\mathrm{C}-\mathrm{T}-\mathrm{A}-3^{\prime}$

In given DNA strand total seven guanine and cytosine bases which form total 21 H -bonds and six adenine and thymine base which will form total 12 H -bonds with other DNA strand.

Total no. of H bonds $=7 \times 3+6 \times 2=33$

Ans. 33

Arginine, Phenylalanine, Histidine, Valine are essential amino acids.

The structure of tyrosine is :

Number of C-atoms = 9

To understand the number of tripeptides formed by three different amino acids using each amino acid once, consider the amino acids as distinct objects we need to arrange. In this case, we have 3 amino acids (let’s name them A, B, and C for simplicity). The question asks how many unique sequences (tripeptides) we can form if we use each amino acid exactly once.

Arranging 3 distinct items in a sequence is a fundamental combinatorial problem. The number of ways to arrange n distinct objects is given by the factorial of n, denoted as n!. The factorial function (n!) means multiplying all whole numbers from n down to 1. Therefore, for 3 amino acids, the number of unique tripeptides we can form is calculated by 3! (3 factorial).

3! = 3 × 2 × 1 = 6

This means there are 6 possible unique tripeptides that can be formed using these three different amino acids exactly once, which are: ABC, ACB, BAC, BCA, CAB, CBA.

Vitamins A, D, E, K and B$_{12}$ are stored in liver and adipose tissue.

A. RNA is regarded as the reserve of genetic information. (False)

B. DNA molecule self-duplicates during cell division. (True)

C. DNA synthesizes proteins in the cell. (False)

D. The message for the synthesis of particular proteins is present in DNA. (True)

E. Identical DNA strands are transferred to daughter cells. (True)

The given Oligopeptide has the following structure

It has 10 sp² hybridised C atoms given with star in the above structure.

Benedict's solution is a reagent that tests for reducing sugars. A reducing sugar is one that has a free aldehyde or ketone group, which can reduce the copper(II) ions in Benedict's reagent to copper(I) oxide, which forms a brick-red or orange-red precipitate.

Given the compounds :

1. Glucose - is a reducing sugar, as it has a free aldehyde group.

2. Maltose - is a reducing sugar. It has a free anomeric carbon that can form an aldehyde group.

3. Sucrose - is not a reducing sugar. Its glycosidic bond locks the anomeric carbons, preventing them from forming an aldehyde or ketone.

4. Ribose - is a reducing sugar, as it has an aldehyde group.

5. 2-deoxyribose - is a reducing sugar. It has a free anomeric carbon that can form an aldehyde group.

6. Amylose - is a polysaccharide and does not typically behave as a reducing sugar in Benedict's test. Even though one end of the amylose chain can technically open to reveal an aldehyde group (the so-called reducing end), the reaction would likely be much less dramatic than with true reducing sugars, and might not occur at all under typical test conditions.

7. Lactose - is a reducing sugar, as it has a free anomeric carbon that can form an aldehyde group.

Hence, out of the seven compounds listed, sucrose and amylose will not likely produce an orange-red precipitate with Benedict's solution. Thus, the number of compounds that will not produce an orange-red precipitate with Benedict's solution is 2.

The peptide has seven amino acid units therefore it has six peptide bonds.

To calculate the total number of tripeptides possible by mixing valine and proline, we need to use the fundamental principle of counting.

Since a tripeptide is made up of three amino acids, and we have two choices for each position (either valine or proline), we can use the multiplication principle to find the total number of possible tripeptides.

That is, the total number of tripeptides possible by mixing valine and proline is:

2 x 2 x 2 = 8

Therefore, there are a total of 8 possible tripeptides that can be formed by mixing valine and proline.

Uracil is $\mathrm{C}_{4} \mathrm{H}_{4} \mathrm{~N}_{2} \mathrm{O}_{2}$

Mol. Wt of $\mathrm{C}_{4} \mathrm{H}_{4} \mathrm{~N}_{2} \mathrm{O}_{2}$ = 112

$\%$ by mass of $N=\frac{14 \times 2}{112} \times 100$

$ =25 \% $

In tetrapeptide

No. of amino acids = 4

No. of peptide bonds = 3

Hence, (1)

$\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6 \stackrel{\text { Zymase }}{\longrightarrow} \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} \stackrel{\mathrm{NaOl}}{\longrightarrow} \underset{\text { (B) }}{\mathrm{HCOONa}}+\mathrm{CHI}_3 $

Uracil is the base which only present is RNA.


Structure of nucleotides number of 0-9.

Since dipeptides and free amino acids do not give biuret test. Hence glycine and glycylalanine do not give this test.

The given pentapeptide is

$\text { ALA - GLY - LEU - ALA - VAL }$

It has 4 peptide linkages.

Combination of n amino acids gives a polypeptide with (n – 1) peptide linkages.

Similarly combination of four amino acids gives a tetrapeptide with three peptide linkages.

Total negative charge produced = 4.

Total number of chiral carbon in sucrose = 9

The amino acids present in the given tripeptide I$l$e–Arg–Pro are isoleucine, arginine and proline.

Number of chiral carbons present in the given tripeptide is 4.

The structure of threonine is
No. of chiral centres present in it = 2

Structure of Tripeptide Asp – Glu – Lys

The problem involves determining the composition of an octasaccharide that, upon complete hydrolysis, yields three types of monosaccharides: ribose, 2-deoxyribose, and glucose.

First, consider the balanced chemical equation for the hydrolysis of the octasaccharide:

$ \text{Octasaccharide} + 7 \text{H}_2\text{O} \rightarrow \text{Ribose} + \text{2-deoxyribose} + \text{Glucose} $

The initial molar mass of the octasaccharide is $1024 \, \text{g/mol}$, and it requires 7 water molecules (each with a molar mass of $18 \, \text{g/mol}$, thus totaling $126 \, \text{g/mol}$) to undergo hydrolysis. Therefore, the total mass on the reactant side is:

$ 1024 + 126 = 1150 \, \text{g} $

According to the given data, the 2-deoxyribose formed constitutes $58.26\%$ of the total mass of the monosaccharides. To find the mass of 2-deoxyribose, calculate:

$ 1150 \times \frac{58.26}{100} = 669.99 \, \text{g} \approx 670 \, \text{g} $

The molar mass of 2-deoxyribose is $134 \, \text{g/mol}$, so the number of units produced is:

$ \frac{670}{134} = 5 \, \text{units} $

Assuming there is one unit of glucose (molar mass $180 \, \text{g/mol}$), the remaining units in the octasaccharide must be ribose. Given five units of 2-deoxyribose and one unit of glucose, the potential number of ribose units can be calculated as the difference to reach a total of eight saccharide units, ensuring:

$ 5 \text{ (2-deoxyribose)} + 1 \text{ (glucose)} + x \text{ (ribose)} = 8 $

Solving this gives:

$ x = 8 - 5 - 1 = 2 $

To verify the setup, the total mass of the hydrolysis products equals the mass at the reactant side:

$ 670 \, \text{(2-deoxyribose)} + 180 \, \text{(glucose)} + (2 \times 150 \, \text{(ribose)}) = 1150 \, \text{g} $

Thus, the octasaccharide contains 2 ribose units. Therefore, the number of ribose units present in one molecule of the octasaccharide is 2.

$\mathrm{A}=\mathrm{T} \quad \Rightarrow 2 \mathrm{H}$-bond

$\mathrm{G} \equiv \mathrm{C} \quad \Rightarrow 3 \mathrm{H}$-bond

Number of $\mathrm{A}=\mathrm{T}$ pair $=7$

Number of $\mathrm{G} \equiv \mathrm{C}$ pair $=6$

Number of $\mathrm{H}$-bond involve in $\mathrm{A}=\mathrm{T}=7 \times 2=14$

Number of $\mathrm{H}$-bond involve in $\mathrm{G} \equiv \mathrm{C}=6 \times 3=18$

Energy required for $\mathrm{A}=\mathrm{T}=14 \times 1=14$

Energy required for $\mathrm{G} \equiv \mathrm{C}=18 \times 1.5=27$

Total energy required $14+27=41$

At pH = 2,

There are two $-$NH₂ group, and + 1 charge on each group because all amino groups exist in the form of $-$NH$_3^ \oplus $.

Therefore, |Z₁| = 2.

At pH = 6,

NH₂ of lysine (+ 1) (pH = 9.47) and COOH ($-$1) of glutamic (pH = 3.08) acid, so because of dipolar ion exists, therefore |Z₂| = 0.

At pH = 11,

COOH of glutamic acid has ($-$1), COOH of lysine ($-$1) and OH of phenol ($-$1).

Therefore, |Z₃| = |$-$3| = 3 (All COOH and OH exist in the form of $-$COO^$-$ and $-$O^$-$).

$ \therefore $ |Z₁| + |Z₂| + |Z₃| = 2 + 0 + 3 = 5

Chemical reaction and product formed after hydrolysis of given peptide can be represented as


(A) is glycine which is only naturally occurring amino acid. While (B), (C) and (D) are not the naturally occurring amino acids. Hence, correct integer is (1).

The structure of melamine is :

Each nitrogen has one lone pair of electrons. Number (no.) of nitrogen in a molecule $=6$

Total no. of lone pairs in melamine

$ \begin{aligned} & =\text { No. of nitrogen } \times \text { lone pair } \\\\ & =6 \times 1=6 \end{aligned} $

Hence, total number of lone pair on nitrogen is 6.

The possible combinations with C-terminal as alanine and N-terminal with chiral carbon (i.e. excluding glycine) are four.

Val$-$Phe$-$Gly$-$Ala

Val$-$Gly$-$Phe$-$Ala

Phe$-$Val$-$Gly$-$Ala

Phe$-$Gly$-$Val$-$Ala

For basic amino acids with pH > 7, peptides will exist as cations. For example, when the substituents are basic, that is R₁ = CH₂CONH₂ and R₂ = (CH₂)₄NH₂ or when R₁ = (CH₂)₄NH₂ and R₂ = (CH₂)₄NH₂ or when R₁ = CH₂OH and R₂ = (CH₂)₄NH₂ or when R₁ = (CH₂)₄NH₂ and R₂ = CH₃.

(i) A decapeptide has nine peptide bonds which hydrolyzes to give ten amino acids. Each peptide bond hydrolyses, to form one molecule of water. Hence, nine molecules of water are required to hydrolysis nine peptide bonds.

$\text{Decapeptide} \xrightarrow{\text{hydrolyse} +9\, \text{H}_2\text{O}} \text{Amino acids}$

(ii) On hydrolysis a molecule of water (equivalent to 18 g) is added across each amino acid.

Mass of hydrolysed decapeptide = Mass of decapeptide + 9 \times \text{mass of each water molecule}

$= 796 \, \text{g mol}^{-1} + 9 \times 18 \, \text{g mol}^{-1} $

$= (796 + 162) \, \text{g mol}^{-1} $

$= 958 \, \text{g mol}^{-1} $

Mass of glycine in hydrolysed decapeptide

$= \frac{47}{100} \times 958 \, \text{g mol}^{-1} $

$= 450.26 \, \text{g mol}^{-1} $

Mass of each glycine = 75 $\, \text{g mol}^{-1}$

Number of glycine units

$= \frac{\text{Mass of hydrolysed decapeptide}}{\text{Mass of each glycine}} $

$n = \frac{450.26 \, \text{g mol}^{-1}}{75 \, \text{g mol}^{-1}} $

$n = 6.00$

Hence, there are six molecules of water in decapeptide.

There are two basic groups in lysine

(A) The structure of disaccharide is shown below.

The given disaccharide consists of two rings, first ring is acetal whereas second is hemiacetal. Due to presence of hemiacetal ring, the Tollen’s reagent will be able to react with this disaccharide. This is a reducing sugar.

(B) The structure of disaccharide is shown below.

The given disaccharide is made up of two rings, both are acetal rings. Thus, this disaccharide is unable to react with Tollen’s reagent. This is a non-reducing sugar.

Final Answer :

(A) The disaccharide will reduce Tollen’s reagent.

(B) The disaccharide will not reduce Tollen’s reagent.

Hints :

The Tollen’s reagent is silver ammoniacal solution and is basic in nature. The Tollen’s reagent is used for distinguishing reducing and non-reducing sugars. The hemiacetal reacts with Tollen’s reagent, but acetal does not react with Tollen’s reagent.