Biomolecules

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 $14+27=41$

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.