The naturally occurring amino acid that contains only one basic functional group in its chemical structure is
The products formed in the above reaction are
L-isomer of tetrose $\mathrm{X}\left(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{4}\right)$ gives positive Schiff's test and has two chiral carbons. On acetylation, '$\mathrm{X}$' yields triacetate. '$\mathrm{X}$' also undergoes following reactions
'$\mathrm{X}$' is
The one that does not stabilize 2$^\circ$ and 3$^\circ$ structures of proteins is
Match List I with List II
| LIST I Natural amino acid |
LIST II One letter code |
||
|---|---|---|---|
| A. | Glutamic acid | I. | Q |
| B. | Glutamine | II. | W |
| C. | Tyrosine | III. | E |
| D. | Tryptophan | IV. | Y |
Choose the correct answer from the options given below:
Sulphur (S) containing amino acids from the following are:
(a) isoleucine (b) cysteine (c) lysine (d) methionine (e) glutamic acid
Match List I with List II
| LIST I Natural Amino Acid |
LIST II One Letter Code |
||
|---|---|---|---|
| A. | Arginine | I. | D |
| B. | Aspartic acid | II. | N |
| C. | Asparagine | III. | A |
| D. | Alanine | IV. | R |
Choose the correct answer from the options given below:
Match List I with List II
| LIST I Enzymatic reaction |
LIST II Enzyme |
||
|---|---|---|---|
| A. | Sucrose $\to$ Glocuse and Fructose | I. | Zymase |
| B. | Glucose $\to$ ethyl alcohol and CO$_2$ | II. | Pepsin |
| C. | Starch $\to$ Maltose | III. | Invertase |
| D. | Proteins $\to$ Amino acids | IV. | Diastase |
Choose the correct answer from the options given below:
All structures given below are of vitamin C. Most stable of them is :
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : $\alpha$-halocarboxylic acid on reaction with dil. $\mathrm{NH_3}$ gives good yield of $\alpha$-amino carboxylic acid whereas the yield of amines is very low when prepared from alkyl halides.
Reason (R) : Amino acids exist in zwitter ion form in aqueous medium.
In the light of the above statements, choose the correct answer from the options given below :
The correct representation in six membered pyranose form for the following sugar [X] is
Match List I with List II
| List I | List II | ||
|---|---|---|---|
| Test | Functional group / Class of Compound | ||
| A. | Molisch's Test | I. | Peptide |
| B. | Biuret Test | II. | Carbohydrate |
| C. | Carbylamine Test | III. | Primary amine |
| D. | Schiff's Test | IV. | Aldehyde |
Choose the correct answer from the options given below :
A protein '$\mathrm{X}$' with molecular weight of $70,000 \mathrm{~u}$, on hydrolysis gives amino acids. One of these amino acid is
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : Ketoses give Seliwanoff's test faster than Aldoses.
Reason (R) : Ketoses undergo $\beta$-elimination followed by formation of furfural.
In the light of the above statements, choose the correct answer from the options given below :
(F, L, D, Y, I, Q, P are one letter codes for amino acids)
Number of cyclic tripeptides formed with 2 amino acids A and B is :
Match items of Row I with those of Row II.
Row I :
Row II :
(i) $\alpha$-$\mathrm{D}$-$(-)$-$\mathrm{Fructofuranose}$
(ii) $\beta$-D-$(-)$-Fructofuranose
(iii) $\alpha$-D-$(-)$ Glucopyranose
(iv) $\beta$-D-$(-)$-Glucopyranose
Correct match is
In an oligopeptide named Alanylglycylphenyl alanyl isoleucine, the number of $\mathrm{sp}^{2}$ hybridised carbons is __________.
Explanation:
The given Oligopeptide has the following structure
It has 10 sp2 hybridised C atoms given with star in the above structure.
Number of compounds from the following which will not produce orange red precipitate with Benedict solution is ___________.
Glucose, maltose, sucrose, ribose, 2-deoxyribose, amylose, lactose
Explanation:
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.
Testosterone, which is a steroidal hormone, has the following structure.
The total number of asymmetric carbon atom/s in testosterone is ____________.
Explanation:
Explanation:
The peptide has seven amino acid units therefore it has six peptide bonds.
Total number of tripeptides possible by mixing of valine and proline is ___________
Explanation:
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 a base present in RNA with the following structure. % of N in uracil is ___________
Given:
Molar mass N = 14 g mol$^{-1}$
O = 16 g mol$^{-1}$
C = 12 g mol$^{-1}$
H = 1 g mol$^{-1}$
Explanation:
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 \% $
From the following, the correct statements about polysaccharides are
I. Starch is a polymer of $\alpha-\mathrm{D}(+)$-glucose.
II. Amylose component of starch is not soluble in water.
III. Amylose is a branched chain polymer of $\alpha-\mathrm{D}(+)$-glucose.
IV. Cellulose is a straight chain polymer of $\beta-\mathrm{D}(+)$-glucose units.
I and IV only
II and III only
II and IV only
I and III only
The deficiency of which vitamin causes pernicious anaemia?
Vitamin $B_2$
Vitamin $\mathrm{B}_6$
Vitamin E
Vitamin $B_{12}$
A vitamin $X$ is soluble in fat and its source is egg yolk. Deficiency of $X$ causes the disease
scurvy
convulsions
xerophthalmia
rickets
$ \text { The Haworth projection shown below represents } $
β-D-(-)fructofuranose
β-D-(+)glucopyranose
$\alpha$-D-(-)fructofuranose
α-D-(+)glucopyranose
The functional group present in asparagine, a non-essential amino acid, are
Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: Amylose is insoluble in water.
Reason R: Amylose is a long linear molecule with more than 200 glucose units.
In the light of the above statements, choose the correct answer from the options given below.
The formulas of A and B for the following reaction sequence
are
For the below given cyclic hemiacetal (X), the correct pyranose structure is :
Match List - I with Match List - II.
| List - I | List - II | ||
|---|---|---|---|
| (A) | Glucose + HI | (I) | Gluconic acid |
| (B) | Glucose + Br$_2$ water | (II) | Glucose pentacetate |
| (C) | Glucose + acetic anhydride | (III) | Saccharic acid |
| (D) | Glucose + HNO$_3$ | (IV) | Hexane |
Choose the correct answer from the options given below:
A sugar 'X' dehydrates very slowly under acidic condition to give furfural which on further reaction with resorcinol gives the coloured product after sometime. Sugar 'X' is
Animal starch is the other name of
Which one of the following is a reducing sugar?
Glycosidic linkage between C1 of $\alpha$-glucose and C2 of $\beta$-fructose is found in
During the denaturation of proteins, which of these structures will remain intact?
The sugar produced after complete hydrolysis of DNA is
The structure of protein that is unaffected by heating is
Sugar moiety in DNA and RNA molecules respectively are
When sugar 'X' is boiled with dilute H2SO4 in alcoholic solution, two isomers 'A' and 'B' are formed. 'A' on oxidation with HNO3 yields saccharic acid where as 'B' is laevorotatory. The compound 'X' is :
Stability of $\alpha$-Helix structure of proteins depends upon
Given below are two statements.
Statement I : Maltose has two $\alpha$-D-glucose units linked at C1 and C4 and is a reducing sugar.
Statement II : Maltose has two monosaccharides : $\alpha$-D-glucose and $\beta$-D-glucose linked at C1 and C6 and it is a non-reducing sugar.
In the light of the above statements, choose the correct answer from the options given below :
L-isomer of a compound 'A' (C4H8O4) gives a positive test with [Ag(NH3)2]+. Treatment of 'A' with acetic anhydride yields triacetate derivative. Compound 'A' produces an optically active compound (B) and an optically inactive compound (C) on treatment with bromine water and HNO3 respectively. Compound (A) is :
