Biomolecules
A known artificial sweetener $\mathbf{X}$ is composed of 4-chloro-4-deoxy- $\alpha$-D-galactose and 1,6-dichloro-1,6-dideoxy- $\beta$-D-fructose joined by a glycosidic linkage. Structure of D-galactose is given below :
The correct structure of $\mathbf{X}$ is
The correct statement(s) regarding sugars is(are) :
On treatment with HNO3, gluconic acid is oxidized to saccharic acid, whereas glucose is not oxidized to saccharic acid.
Fructose gives a positive Fehling’s test because it isomerises to glucose and another aldohexose in the presence of Fehling’s reagent.
Invert sugar is an equimolar mixture of D-glucose and D-fructose formed after hydrolysis of the corresponding disaccharide.
Specific rotation of invert sugar is −40°.
A linear octasaccharide (molar mass $=1024 \mathrm{~g} \mathrm{~mol}^{-1}$ ) on complete hydrolysis produces three monosaccharides: ribose, 2-deoxyribose and glucose. The amount of 2-deoxyribose formed is $58.26 \%(\mathrm{w} / \mathrm{w})$ of the total amount of the monosaccharides produced in the hydrolyzed products. The number of ribose unit(s) present in one molecule of octasaccharide is $\qquad$ .
Use: Molar mass $\left(\right.$ in g $\left.\mathrm{mol}^{-1}\right)$ : ribose $=150,2$-deoxyribose $=134$, glucose $=180$;
Atomic mass (in amu): $\mathrm{H}=1, \mathrm{O}=16$
Explanation:
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.
For a double strand DNA, one strand is given below:
The amount of energy required to split the double strand DNA into two single strands is _______ kcal $\operatorname{mol}^{-1}$.
[Given: Average energy per H-bond for A-T base pair $=1.0 ~\mathrm{kcal}~ \mathrm{mol}^{-1}$, G-C base pair $=1.5 ~\mathrm{kcal}$ $\mathrm{mol}^{-1}$, and A-U base pair $=1.25 ~\mathrm{kcal} ~\mathrm{mol}^{-1}$. Ignore electrostatic repulsion between the phosphate groups.]
Explanation:
$\mathrm{A}=\mathrm{T} \quad \Rightarrow 2 \mathrm{H}$-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 compound(s), which on reaction with HNO3 will give the product having degree of rotation, [$\alpha$]D = $-$52.7$^\circ$ is (are)
If the absolute values of the net charge of the peptide at $\mathrm{pH}$ $=2$, $\mathrm{pH}=6$, and $\mathrm{pH}=11$ are $\left|Z_1\right|,\left|Z_2\right|$, and $\left|Z_3\right|$, respectively, then what is $\left|Z_1\right|+\left|Z_2\right|+\left|Z_3\right|$?
Explanation:
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
The correct structure(s) of $\beta $-$L$-glucopyranose is (are) :
For "invert sugar", the correct statement(s) is(are)
(Given : specific rotations of (+)-sucrose, (+)-maltose, L-($-$)-glucose and L-(+)-fructose in aqueous solution are +66$^\circ$, +140$^\circ$, $-$52$^\circ$ and +92$^\circ$, respectively.)
The structure of D-(+)-glucose is
The structure of L-($-$)-glucose is
Explanation:
(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).
Explanation:
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.
A tetrapeptide has $-$COOH group on alanine. This produces glycine (Gly), valine (Val), phenyl alanine (Phe) and alanine (Ala), on complete hydrolysis. For this tetrapeptide, the number of possible sequences (primary structures) with $-$NH2 group attached to a chiral center is _______.
Explanation:
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
The substituents R1 and R2 for nine peptides are listed in the table given below. How many of these peptides are positively charged at pH = 7.0 ?
| Peptide | ${R_1}$ | ${R_2}$ |
|---|---|---|
| I | H | H |
| II | H | $C{H_3}$ |
| III | $C{H_2}COOH$ | H |
| IV | $C{H_2}CON{H_2}$ | ${(C{H_2})_4}N{H_2}$ |
| V | $C{H_2}CON{H_2}$ | $C{H_2}CON{H_2}$ |
| VI | ${(C{H_2})_4}N{H_2}$ | ${(C{H_2})_4}N{H_2}$ |
| VII | $C{H_2}COOH$ | $C{H_2}CON{H_2}$ |
| VIII | $C{H_2}OH$ | ${(C{H_2})_4}N{H_2}$ |
| IX | ${(C{H_2})_4}N{H_2}$ | $C{H_3}$ |
Explanation:
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.
A decapeptide (mol. wt. 796) on complete hydrolysis gives glycine (mol. wt. 75), alanine and phenylalanine. Glycine contributes 47.0% to the total weight of the hydrolysed products. The number of glycine units present in the decapeptide is _________.
Explanation:
(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.
The following carbohydrate is
The total number of basic groups in the following form of lysine is
Explanation:
There are two basic groups in lysine
The correct statement about the following disaccharide is :
The correct statement(s) about the following sugar X and Y is(are)
Cellulose upon acetylation with excess acetic anhydride/H$_2$SO$_4$ (catalytic) gives cellulose triacetate whose structure is :
Statement-1 : Glucose gives a reddish-brown precipitate with Fehling's solution.
Statement-2 : Reaction of glucose with Fehling's solution gives $\mathrm{CuO}$ and gluconic acid.
Which of the following disaccharide will not reduce Tollen's reagent?
Explanation:
(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.