All Exams >
Chemistry >
Topicwise Question Bank for IIT JAM/CSIR/GATE Chemistry >
All Questions
All questions of Natural Products Chemistry for Chemistry Exam
The most basic amino acid among the following is:- a)Tyrosine
- b)Methionine
- c)Arginine
- d)Glutamine
Correct answer is option 'C'. Can you explain this answer?
The most basic amino acid among the following is:
a)
Tyrosine
b)
Methionine
c)
Arginine
d)
Glutamine
|
Kunal Goyal answered |
Basic Amino Acid: Arginine
Amino acids are the building blocks of proteins. There are 20 different types of amino acids commonly found in proteins. Out of these, some amino acids are basic, some are acidic, and some are neutral.
Basic amino acids are those that have a basic or alkaline side chain in their chemical structure. These amino acids have a positive charge at neutral pH (pH 7) due to the presence of an amino group (-NH2) in their structure.
Among the given options, the most basic amino acid is Arginine.
Explanation:
Arginine has a guanidine group (-NH-C(NH2)-NH2) in its side chain, which makes it a basic amino acid. This group has a pKa value of 12.5, which means that at neutral pH, the guanidine group will be mostly protonated (+NH3) and have a positive charge.
Methionine has a sulfur-containing side chain, but it is not basic in nature. Tyrosine has a hydroxyl group (-OH) in its side chain, which is not basic either. Glutamine has a neutral side chain and is not basic.
Hence, among the given options, Arginine is the most basic amino acid.
Amino acids are the building blocks of proteins. There are 20 different types of amino acids commonly found in proteins. Out of these, some amino acids are basic, some are acidic, and some are neutral.
Basic amino acids are those that have a basic or alkaline side chain in their chemical structure. These amino acids have a positive charge at neutral pH (pH 7) due to the presence of an amino group (-NH2) in their structure.
Among the given options, the most basic amino acid is Arginine.
Explanation:
Arginine has a guanidine group (-NH-C(NH2)-NH2) in its side chain, which makes it a basic amino acid. This group has a pKa value of 12.5, which means that at neutral pH, the guanidine group will be mostly protonated (+NH3) and have a positive charge.
Methionine has a sulfur-containing side chain, but it is not basic in nature. Tyrosine has a hydroxyl group (-OH) in its side chain, which is not basic either. Glutamine has a neutral side chain and is not basic.
Hence, among the given options, Arginine is the most basic amino acid.
How many of the following are reducing sugar:Sucrose, Lactose, Maltose, Glucose, Galactose, Fructose, Glycogen, Ribose.
Correct answer is '7'. Can you explain this answer?
How many of the following are reducing sugar:
Sucrose, Lactose, Maltose, Glucose, Galactose, Fructose, Glycogen, Ribose.
|
Pooja Choudhury answered |
The aldehyde group of aldoses is very susceptible to oxidation, whereas ketoses are less so, but can easily be oxidized if, like fructose, they contain an α-hydroxyl and can tautomerize to an aldose. Most monosaccharides are reducing sugars. This includes all of the common ones galactose, glucose, fructose, ribose, xylose, and mannose. Some disaccharides, such as lactose and maltose are reducing sugars since they have at least one anomeric carbon free, allowing that part of the sugar to linearize and yield an aldose.
Which of the following pairs give posit ive Tollen’s test:- a)Glucose, Sucrose
- b)Glucose, Fructose
- c)Hexanal, acetophenone
- d)Fructose, Sucrose
Correct answer is option 'B'. Can you explain this answer?
Which of the following pairs give posit ive Tollen’s test:
a)
Glucose, Sucrose
b)
Glucose, Fructose
c)
Hexanal, acetophenone
d)
Fructose, Sucrose
|
|
Ameya Rane answered |
It seems that the sentence is incomplete as it does not provide the options or pairs to choose from. Can you please provide the pairs so I can assist you further?
D-Glucose exist in x different forms. The value of x(stereoisomer) is:
Correct answer is '3'. Can you explain this answer?
D-Glucose exist in x different forms. The value of x(stereoisomer) is:
|
Pranavi Mishra answered |
**D-Glucose Stereoisomers**
D-Glucose is a monosaccharide, a type of carbohydrate, and it exists in different forms known as stereoisomers. Stereoisomers are compounds that have the same molecular formula and connectivity but differ in the spatial arrangement of their atoms. In the case of D-Glucose, the arrangement of its hydroxyl groups around the asymmetric carbon atoms gives rise to different stereoisomers.
**Definition of Stereoisomers**
Stereoisomers can be categorized into two main types: enantiomers and diastereomers.
1. **Enantiomers**: Enantiomers are a pair of stereoisomers that are non-superimposable mirror images of each other. They have the same connectivity but differ in their three-dimensional orientation. Enantiomers are optically active, meaning they rotate plane-polarized light in opposite directions. In the case of D-Glucose, there are two enantiomers: D-Glucose and L-Glucose.
2. **Diastereomers**: Diastereomers are stereoisomers that are not mirror images of each other. They have different connectivity or spatial arrangement at one or more stereocenters. Diastereomers are not optically active. In the case of D-Glucose, there are three diastereomers: α-D-Glucose, β-D-Glucose, and open-chain D-Glucose.
**Explanation of the Answer: 3**
The question asks for the number of different forms (stereoisomers) in which D-Glucose exists. The correct answer is '3'. Here's the breakdown of the three stereoisomers:
1. **Enantiomers (1 stereoisomer)**:
- D-Glucose (Dextrose): It is the naturally occurring form of glucose found in living organisms.
- L-Glucose: It is the mirror image of D-Glucose and is not commonly found in nature.
2. **Diastereomers (2 stereoisomers)**:
- α-D-Glucose: It is a cyclic form of glucose where the hydroxyl group on the anomeric carbon is in the axial position.
- β-D-Glucose: It is another cyclic form of glucose where the hydroxyl group on the anomeric carbon is in the equatorial position.
Therefore, considering both enantiomers and diastereomers, D-Glucose exists in a total of 3 different forms or stereoisomers.
D-Glucose is a monosaccharide, a type of carbohydrate, and it exists in different forms known as stereoisomers. Stereoisomers are compounds that have the same molecular formula and connectivity but differ in the spatial arrangement of their atoms. In the case of D-Glucose, the arrangement of its hydroxyl groups around the asymmetric carbon atoms gives rise to different stereoisomers.
**Definition of Stereoisomers**
Stereoisomers can be categorized into two main types: enantiomers and diastereomers.
1. **Enantiomers**: Enantiomers are a pair of stereoisomers that are non-superimposable mirror images of each other. They have the same connectivity but differ in their three-dimensional orientation. Enantiomers are optically active, meaning they rotate plane-polarized light in opposite directions. In the case of D-Glucose, there are two enantiomers: D-Glucose and L-Glucose.
2. **Diastereomers**: Diastereomers are stereoisomers that are not mirror images of each other. They have different connectivity or spatial arrangement at one or more stereocenters. Diastereomers are not optically active. In the case of D-Glucose, there are three diastereomers: α-D-Glucose, β-D-Glucose, and open-chain D-Glucose.
**Explanation of the Answer: 3**
The question asks for the number of different forms (stereoisomers) in which D-Glucose exists. The correct answer is '3'. Here's the breakdown of the three stereoisomers:
1. **Enantiomers (1 stereoisomer)**:
- D-Glucose (Dextrose): It is the naturally occurring form of glucose found in living organisms.
- L-Glucose: It is the mirror image of D-Glucose and is not commonly found in nature.
2. **Diastereomers (2 stereoisomers)**:
- α-D-Glucose: It is a cyclic form of glucose where the hydroxyl group on the anomeric carbon is in the axial position.
- β-D-Glucose: It is another cyclic form of glucose where the hydroxyl group on the anomeric carbon is in the equatorial position.
Therefore, considering both enantiomers and diastereomers, D-Glucose exists in a total of 3 different forms or stereoisomers.
The pH value of a solution in which a polar amino acid does not migrate under the influence of electric field is called:- a)Isoelectric po int
- b)Isoelectronic point
- c)Neutralizat ion point
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
The pH value of a solution in which a polar amino acid does not migrate under the influence of electric field is called:
a)
Isoelectric po int
b)
Isoelectronic point
c)
Neutralizat ion point
d)
None of these
|
Ishaan Sengupta answered |
The pH value of a solution in which a polar amino acid does not migrate under the influence of an electric field is called the isoelectric point.
Explanation:
Polar Amino Acids:
Amino acids are organic compounds that contain both an amino group (-NH2) and a carboxyl group (-COOH). They are the building blocks of proteins. Some amino acids have polar side chains, which means they have a partial positive or negative charge.
Electric Field and Migration:
When an electric field is applied to a solution containing charged particles, such as polar amino acids, the charged particles will migrate towards the oppositely charged electrode. The migration of charged particles in an electric field is called electrophoresis.
Isoelectric Point:
The isoelectric point (pI) is the pH value at which a polar amino acid does not migrate under the influence of an electric field. At this pH, the net charge on the amino acid is zero, and it remains stationary in the electric field.
Charge of Amino Acids:
The charge on an amino acid depends on the pH of the solution. At low pH values (acidic conditions), the amino acid is protonated and carries a positive charge. At high pH values (basic conditions), the amino acid is deprotonated and carries a negative charge. At the isoelectric point, the amino acid is in its zwitterionic form, which means it has both a positive and a negative charge that cancel each other out, resulting in a net charge of zero.
pH and Isoelectric Point:
The pH at which the isoelectric point occurs depends on the specific amino acid. Each amino acid has a unique isoelectric point based on the properties of its side chain. For example, the isoelectric point of lysine is around pH 9.8, while the isoelectric point of glutamic acid is around pH 3.2.
Conclusion:
The pH value at which a polar amino acid does not migrate under the influence of an electric field is called the isoelectric point. At this pH, the amino acid is in its zwitterionic form and carries no net charge.
The complementary strand of DNA for the following single stranded DNA sequence: 5' - A - T - C- A - T - G - C-3' is:- a)5' - A - T - C- A - T - G - C-3'
- b)5' - T - A - G - T - A - C- G -3'
- c)5' - G - C- A - T - G - A - T -3'
- d)5' - C- G - T - A - C- T - A -3'
Correct answer is option 'C'. Can you explain this answer?
The complementary strand of DNA for the following single stranded DNA sequence: 5' - A - T - C- A - T - G - C-3' is:
a)
5' - A - T - C- A - T - G - C-3'
b)
5' - T - A - G - T - A - C- G -3'
c)
5' - G - C- A - T - G - A - T -3'
d)
5' - C- G - T - A - C- T - A -3'
|
Vedika Singh answered |
Correct Answer :- C
Explanation : In DNA, Adenine (A) pairs with Thymine (T) and Guanine (G) pairs with Cytosine (C).
Hence, on replication of DNA in M phase, complementary nucleotides for each base pair will be added as follows:
A - T
T - A
G - C
C - G
Since replication occurs in 5'-3' direction, so the complementary strand will be formed in 3'-5' direction.
The pKa values for the three ionizable groups X, Y and Z or Glutamic acid are 4.3, 9.7 and 2.2 respectively:
The isoelectric point for the amino acid is: - a)7.00
- b)3.25
- c)4.95
- d)5.95
Correct answer is option 'B'. Can you explain this answer?
The pKa values for the three ionizable groups X, Y and Z or Glutamic acid are 4.3, 9.7 and 2.2 respectively:
The isoelectric point for the amino acid is:
a)
7.00
b)
3.25
c)
4.95
d)
5.95
|
Sandeep Sahoo answered |
Isoelectric point is calculated by the average of Pka values of left right side group.
Two forms of D-glucopyranose, are called:- a)Enantiomers
- b)Diastereomers
- c)Epimers
- d)Anomers
Correct answer is option 'D'. Can you explain this answer?
Two forms of D-glucopyranose, are called:
a)
Enantiomers
b)
Diastereomers
c)
Epimers
d)
Anomers
|
|
Kalpana Pandey answered |
Because, the configuration is changed only at anomeric carbon and rest are same...
Which of the following pairs is/are correctly matched:- a)Galactose-Lactose
- b)Fructose-Glucose
- c)Fructose-Galactose
- d)Ribose-Deoyribose
Correct answer is option 'B,C,D'. Can you explain this answer?
Which of the following pairs is/are correctly matched:
a)
Galactose-Lactose
b)
Fructose-Glucose
c)
Fructose-Galactose
d)
Ribose-Deoyribose
|
Aditya Deshmukh answered |
(b) Cassini–Huygens is an unmanned spacecraft sent to the planet Saturn. Therefore, option 1 is wrong, this eliminates option (a), (c) and (d). Now we are left with final answer B only 2 and 3. MESSENGER is a robotic NASA spacecraft orbiting the planet Mercury. Voyager 1 (September 1977) and voyager 2 (Aug 1977) were launched to study the outer Solar System
A disaccharide does not give a posit ive test for Tollen’s reagent. Upon acidic hydrolysis, it gives an equimolar mixture of two different monosaccharides, both the which can be oxidized by bromine water. This disaccharide is:- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
A disaccharide does not give a posit ive test for Tollen’s reagent. Upon acidic hydrolysis, it gives an equimolar mixture of two different monosaccharides, both the which can be oxidized by bromine water. This disaccharide is:
a)
b)
c)
d)
|
|
Subha Som answered |
D is ketal which doesn't break in aq sol
D-glucose and D-galactose are:
- a)C2-epimers
- b)C3-epimers
- c)C4-epimers
- d)Diastereomers
Correct answer is option 'C'. Can you explain this answer?
D-glucose and D-galactose are:
a)
C2-epimers
b)
C3-epimers
c)
C4-epimers
d)
Diastereomers
|
|
Vedika Singh answered |
Epimers are carbohydrates which vary in one position for the placement of the -OH group. The best examples are for D-glucose and D-galactose. Both monosaccharides are D-sugars, meaning that the -OH group on carbon-5 of these hexoses is located on the right in Fischer Projection.
Among the following, the incorrect statement(s) is/are:- a)Guanine is a Purine nucleobase
- b)Glycine and proline are achiral amino acids
- c)DNA contains glycosidic bonds and pentose sugars
- d)Sucrose is a non-reducing sugar.
Correct answer is option 'B'. Can you explain this answer?
Among the following, the incorrect statement(s) is/are:
a)
Guanine is a Purine nucleobase
b)
Glycine and proline are achiral amino acids
c)
DNA contains glycosidic bonds and pentose sugars
d)
Sucrose is a non-reducing sugar.
|
Priya Saha answered |
Incorrect Statements
Glycine and proline are achiral amino acids.
Explanation
Achiral molecules are those that do not have a mirror image that is different from the original molecule. Glycine and proline are not achiral amino acids.
Glycine is the simplest amino acid and has a hydrogen atom as its side chain (R-group). It is not chiral because it has two identical hydrogen atoms as its R-group.
Proline, on the other hand, has a unique structure due to the presence of a cyclic side chain that forms a five-membered ring. This ring structure causes proline to be a chiral molecule.
Therefore, statement B is incorrect.
Correct Statements
A) Guanine is a Purine nucleobase
B) DNA contains glycosidic bonds and pentose sugars
C) Sucrose is a non-reducing sugar.
Conclusion
In summary, glycine and proline are not achiral amino acids, making statement B incorrect. The other statements are correct. Guanine is a purine nucleobase, DNA contains glycosidic bonds and pentose sugars, and sucrose is a non-reducing sugar.
Glycine and proline are achiral amino acids.
Explanation
Achiral molecules are those that do not have a mirror image that is different from the original molecule. Glycine and proline are not achiral amino acids.
Glycine is the simplest amino acid and has a hydrogen atom as its side chain (R-group). It is not chiral because it has two identical hydrogen atoms as its R-group.
Proline, on the other hand, has a unique structure due to the presence of a cyclic side chain that forms a five-membered ring. This ring structure causes proline to be a chiral molecule.
Therefore, statement B is incorrect.
Correct Statements
A) Guanine is a Purine nucleobase
B) DNA contains glycosidic bonds and pentose sugars
C) Sucrose is a non-reducing sugar.
Conclusion
In summary, glycine and proline are not achiral amino acids, making statement B incorrect. The other statements are correct. Guanine is a purine nucleobase, DNA contains glycosidic bonds and pentose sugars, and sucrose is a non-reducing sugar.
In a double stranded DNA, if the sequence 5’AGATCC3’ appears on one strand of DNA, what sequence appears on the complementary strand?- a)5’AGATCC3’
- b)5’CCTAGA3’
- c)5’GGATCT3’
- d)5’TCTAGG3’
Correct answer is option 'D'. Can you explain this answer?
In a double stranded DNA, if the sequence 5’AGATCC3’ appears on one strand of DNA, what sequence appears on the complementary strand?
a)
5’AGATCC3’
b)
5’CCTAGA3’
c)
5’GGATCT3’
d)
5’TCTAGG3’
|
Harshitha Sharma answered |
The complementary sequence to 5'-ATCGGCTA-3' is 3'-TAGCCGAT-5'. This is because in DNA, adenine (A) always pairs with thymine (T), and cytosine (C) always pairs with guanine (G). Therefore, the complementary strand will have T instead of A, G instead of C, C instead of G, and A instead of T in the corresponding positions.
The decreasing order of isoelectric point for the following a-amino acids is:
(I) Lysine (II) Alanine (III) Glutamic acid- a)II > I > III
- b)I > II > III
- c)III > I > II
- d)I > III > II
Correct answer is option 'B'. Can you explain this answer?
The decreasing order of isoelectric point for the following a-amino acids is:
(I) Lysine (II) Alanine (III) Glutamic acid
(I) Lysine (II) Alanine (III) Glutamic acid
a)
II > I > III
b)
I > II > III
c)
III > I > II
d)
I > III > II
|
|
Soumya Sharma answered |
I. Lysine: pI = 9.87
II. Alanine: pI = 6.02
III. Glutamic acid: pl = 3.22
Ribose and 2-deoxyribose can be different iated by:- a)Fehling’s reagent
- b)Tollen’s reagent
- c)Osazone formation
- d)Barfoed’s reagent
Correct answer is option 'C'. Can you explain this answer?
Ribose and 2-deoxyribose can be different iated by:
a)
Fehling’s reagent
b)
Tollen’s reagent
c)
Osazone formation
d)
Barfoed’s reagent
|
|
Ameya Rane answered |
B)Tollen's reagent
c)Benedict's reagent
d)Seliwanoff's reagent
c)Benedict's reagent
d)Seliwanoff's reagent
The following carbohydrate is: 
- a)A ketohexose
- b)An aldohexose
- c)A β-pyranose
- d)An α-pyranose
Correct answer is option 'B,C'. Can you explain this answer?
The following carbohydrate is:
a)
A ketohexose
b)
An aldohexose
c)
A β-pyranose
d)
An α-pyranose
|
Suraj Srj answered |
Aldohexose - four asymmetric centers, an aldehyde group, six carbon atoms and five alcoholic group in Fischer projection.
Pyran stands for six membered ring (one hetroatom included).
Alpha- pyranose if OH group at C-1 is axial downward.
Beta- pyranose if OH group at C-1 is equatorial upward.
Therefore, above given cyclic hemiacetal structure is an aldohexose and a beta- pyranose.
The correct statement about the following disaccharide is:
- a)Ring I is pyranose with α-glycosidic link
- b)Ring I is furanose with α-glycosidic link
- c)Ring II is furanose with α-glycosidic link
- d)Ring II is pyranose with β-glycosidic link
Correct answer is option 'A'. Can you explain this answer?
The correct statement about the following disaccharide is:
a)
Ring I is pyranose with α-glycosidic link
b)
Ring I is furanose with α-glycosidic link
c)
Ring II is furanose with α-glycosidic link
d)
Ring II is pyranose with β-glycosidic link
|
Sonu Choudhary answered |
As we know that ,the pyranose is 6 membered ring and furanose is 5 membered ring therfore otion A is correct ....
Which one is not a constituent of nuclei acid:- a)Uracil
- b)Guanidine
- c)Phosphoric acid
- d)Ribose sugar
Correct answer is option 'B'. Can you explain this answer?
Which one is not a constituent of nuclei acid:
a)
Uracil
b)
Guanidine
c)
Phosphoric acid
d)
Ribose sugar
|
Anirban Khanna answered |
A constituent of nucleic acid are uracil, phosphorid acid, ribose sugar and deoxyribose sugar. So, guanidine is not a constituent of nucleic acid.
Biuret test is not given by:- a)Proteins
- b)Urea
- c)Polypeptide
- d)Carbohydrates
Correct answer is option 'D'. Can you explain this answer?
Biuret test is not given by:
a)
Proteins
b)
Urea
c)
Polypeptide
d)
Carbohydrates
|
|
Garima Chavan answered |
Biuret Test and its purpose:
The Biuret test is a chemical test used to detect the presence of proteins or polypeptides in a given sample. It is based on the reaction between copper ions and peptide bonds in proteins, resulting in a color change. The test is named after the compound biuret, which contains the necessary functional groups for this reaction to occur.
Understanding the Biuret Test:
The test involves adding a few drops of a Biuret reagent, typically a solution of copper sulfate and sodium hydroxide, to the sample being tested. If proteins or polypeptides are present, a violet or pink color will develop. This color change occurs due to the formation of a coordination complex between copper ions and the peptide bonds in the proteins.
Explanation of the given options:
a) Proteins: Proteins are composed of amino acids, which are linked together by peptide bonds. Therefore, proteins are capable of giving a positive Biuret test.
b) Urea: Urea is a compound that contains a carbonyl group and an amine group, both of which are involved in the formation of peptide bonds. Thus, urea can also give a positive Biuret test.
c) Polypeptide: Polypeptides are chains of amino acids connected by peptide bonds, just like proteins. Hence, polypeptides can also produce a positive Biuret test.
d) Carbohydrates: Carbohydrates, on the other hand, do not contain peptide bonds. They are composed of simple sugars (monosaccharides) or their derivatives, such as disaccharides and polysaccharides. As a result, carbohydrates do not react with the Biuret reagent and do not give a positive test.
Therefore, the correct answer is option 'D' - Carbohydrates. Unlike proteins, urea, and polypeptides, carbohydrates lack the peptide bonds necessary for the formation of the coordination complex with copper ions in the Biuret test. Hence, they do not give a positive result in this test.
The Biuret test is a chemical test used to detect the presence of proteins or polypeptides in a given sample. It is based on the reaction between copper ions and peptide bonds in proteins, resulting in a color change. The test is named after the compound biuret, which contains the necessary functional groups for this reaction to occur.
Understanding the Biuret Test:
The test involves adding a few drops of a Biuret reagent, typically a solution of copper sulfate and sodium hydroxide, to the sample being tested. If proteins or polypeptides are present, a violet or pink color will develop. This color change occurs due to the formation of a coordination complex between copper ions and the peptide bonds in the proteins.
Explanation of the given options:
a) Proteins: Proteins are composed of amino acids, which are linked together by peptide bonds. Therefore, proteins are capable of giving a positive Biuret test.
b) Urea: Urea is a compound that contains a carbonyl group and an amine group, both of which are involved in the formation of peptide bonds. Thus, urea can also give a positive Biuret test.
c) Polypeptide: Polypeptides are chains of amino acids connected by peptide bonds, just like proteins. Hence, polypeptides can also produce a positive Biuret test.
d) Carbohydrates: Carbohydrates, on the other hand, do not contain peptide bonds. They are composed of simple sugars (monosaccharides) or their derivatives, such as disaccharides and polysaccharides. As a result, carbohydrates do not react with the Biuret reagent and do not give a positive test.
Therefore, the correct answer is option 'D' - Carbohydrates. Unlike proteins, urea, and polypeptides, carbohydrates lack the peptide bonds necessary for the formation of the coordination complex with copper ions in the Biuret test. Hence, they do not give a positive result in this test.
Which one among the following is a Sesquiterpene:- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Which one among the following is a Sesquiterpene:
a)
b)
c)
d)
|
|
Madhu Bala answered |
Total number of carbon atoms= 15
Therefore the number of isopreane units =3
it is sesquiterpene.
Therefore the number of isopreane units =3
it is sesquiterpene.
Which one of the following compounds will form an osazone derivative:- a)CH3CH2CHOHCH2OH
- b)CH3COCH2CH2OH
- c)CH3CH2COCH2OH
- d)CH3CH2COCH2OCH3
Correct answer is option 'C'. Can you explain this answer?
Which one of the following compounds will form an osazone derivative:
a)
CH3CH2CHOHCH2OH
b)
CH3COCH2CH2OH
c)
CH3CH2COCH2OH
d)
CH3CH2COCH2OCH3
|
|
Sarthak Chavan answered |
When sugars are reacted with excess of phenylhydrazine osazone formation takes place.Osazone will form in that sugar only where keto hydroxy group is present.
Thus CH3CH2COCH2OH will form an osazone derivative; as it has keto hydroxy group.
Which of the following pairs give the same osazone?- a)D-glucose; D-mannose
- b)D-galactose; D-tallose
- c)D-gulose; D-idose
- d)D-altrose; D-glucose
Correct answer is option 'A,B,C'. Can you explain this answer?
Which of the following pairs give the same osazone?
a)
D-glucose; D-mannose
b)
D-galactose; D-tallose
c)
D-gulose; D-idose
d)
D-altrose; D-glucose
|
|
Kalpana Pandey answered |
As the given pairs are epimer of each other and therefore they give the same osazone
Which of the following pairs form the same osazone?- a)Galactose
- b)Glucose
- c)Fructose
- d)Arabinose
Correct answer is option 'B,C'. Can you explain this answer?
Which of the following pairs form the same osazone?
a)
Galactose
b)
Glucose
c)
Fructose
d)
Arabinose
|
|
Jay Nambiar answered |
When sugars react with an excess of phenylhydrazine, a class of compounds known as osazone is formed.
Glucose and fructose differ in the arrangement of atoms around the C−1 and C−2 carbon. When both of them react with three molecules of phenylhydrazine, the formation of osazone crystals makes the difference between the carbon atoms in their chain and hence they form the same osazone.
Which of the following regarding steroids is/are correct:- a)Cortisone
- b)Progesterone
- c)Glycolipid
- d)Ergosterol
Correct answer is option 'A,B,D'. Can you explain this answer?
Which of the following regarding steroids is/are correct:
a)
Cortisone
b)
Progesterone
c)
Glycolipid
d)
Ergosterol
|
Mrinalini Sen answered |
But if the inflammation is more severe or widespread, systemic steroids may be required. Corticosteroids can be used for many medical conditions that cause inflammation. Injected corticosteroids, often called cortisone shots, are often used for arthritis, tendonitis, and bursitis.
Tollens test will be negative for:- a)Glucose
- b)Mannose
- c)Sucrose
- d)Galactose
Correct answer is option 'C'. Can you explain this answer?
Tollens test will be negative for:
a)
Glucose
b)
Mannose
c)
Sucrose
d)
Galactose
|
|
Niharika Kulkarni answered |
Tollens test and its principle:
Tollens test is a chemical test used to detect the presence of aldehydes in a given sample. It is also known as the silver mirror test. The test is based on the principle that aldehydes are easily oxidized to carboxylic acids and can be reduced to form silver mirror-like precipitates in the presence of Tollens reagent.
Negative Tollens test for Sucrose:
Sucrose is a disaccharide composed of glucose and fructose monomers. It is a non-reducing sugar, meaning that it does not have an aldehyde or ketone functional group that could be oxidized to form a silver mirror-like precipitate. Therefore, Tollens test will not give a positive result for sucrose.
Positive Tollens test for Glucose, Mannose and Galactose:
Glucose, Mannose, and Galactose are monosaccharides that have an aldehyde functional group capable of being oxidized by Tollens reagent. When these sugars are treated with Tollens reagent, they will undergo oxidation and reduce the Tollens reagent to form a silver mirror-like precipitate. Hence, Tollens test will give a positive result for glucose, mannose, and galactose.
Conclusion:
In conclusion, Tollens test is a chemical test used to detect the presence of aldehydes in a given sample. Sucrose is a non-reducing sugar, and it does not have an aldehyde functional group, so Tollens test will not give a positive result for sucrose. On the other hand, glucose, mannose, and galactose are reducing sugars, and they have an aldehyde functional group, so Tollens test will give a positive result for these sugars.
Tollens test is a chemical test used to detect the presence of aldehydes in a given sample. It is also known as the silver mirror test. The test is based on the principle that aldehydes are easily oxidized to carboxylic acids and can be reduced to form silver mirror-like precipitates in the presence of Tollens reagent.
Negative Tollens test for Sucrose:
Sucrose is a disaccharide composed of glucose and fructose monomers. It is a non-reducing sugar, meaning that it does not have an aldehyde or ketone functional group that could be oxidized to form a silver mirror-like precipitate. Therefore, Tollens test will not give a positive result for sucrose.
Positive Tollens test for Glucose, Mannose and Galactose:
Glucose, Mannose, and Galactose are monosaccharides that have an aldehyde functional group capable of being oxidized by Tollens reagent. When these sugars are treated with Tollens reagent, they will undergo oxidation and reduce the Tollens reagent to form a silver mirror-like precipitate. Hence, Tollens test will give a positive result for glucose, mannose, and galactose.
Conclusion:
In conclusion, Tollens test is a chemical test used to detect the presence of aldehydes in a given sample. Sucrose is a non-reducing sugar, and it does not have an aldehyde functional group, so Tollens test will not give a positive result for sucrose. On the other hand, glucose, mannose, and galactose are reducing sugars, and they have an aldehyde functional group, so Tollens test will give a positive result for these sugars.
Synthesis of each molecule of glucose in photosynthesis involves:- a)18 molecules of ATP
- b)10 molecules of ATP
- c)8 molecules of ATP
- d)6 molecules of ATP
Correct answer is option 'A'. Can you explain this answer?
Synthesis of each molecule of glucose in photosynthesis involves:
a)
18 molecules of ATP
b)
10 molecules of ATP
c)
8 molecules of ATP
d)
6 molecules of ATP
|
|
Saikat Ghoshal answered |
Overview:
The synthesis of each molecule of glucose in photosynthesis involves the utilization of ATP (adenosine triphosphate) as an energy source. ATP is a high-energy molecule that provides the necessary energy for various cellular processes, including glucose synthesis. In the process of photosynthesis, plants convert sunlight, water, and carbon dioxide into glucose and oxygen. The synthesis of glucose occurs through a series of complex reactions known as the Calvin cycle.
Explanation:
The Calvin cycle is the set of reactions that occur in the stroma of chloroplasts during photosynthesis. It involves a series of enzyme-catalyzed reactions that utilize ATP and NADPH (nicotinamide adenine dinucleotide phosphate) generated during the light-dependent reactions. The primary purpose of the Calvin cycle is the fixation of carbon dioxide and the synthesis of carbohydrates, particularly glucose.
Steps involved in glucose synthesis:
1. Carbon fixation:
- In the first step of the Calvin cycle, carbon dioxide molecules are incorporated into an organic molecule called ribulose-1,5-bisphosphate (RuBP).
- This reaction is catalyzed by the enzyme ribulose bisphosphate carboxylase/oxygenase (Rubisco).
- This process results in the formation of two molecules of 3-phosphoglycerate (3-PGA).
2. Reduction:
- The second step involves the conversion of 3-PGA into glyceraldehyde-3-phosphate (G3P).
- This conversion requires ATP and NADPH as energy and reducing power sources, respectively.
- Each molecule of 3-PGA is phosphorylated by ATP and reduced by NADPH to form G3P.
3. Regeneration:
- In the final step, G3P molecules are used to regenerate RuBP, which is necessary for the continuation of the Calvin cycle.
- Several ATP molecules are utilized in this step to convert G3P into RuBP.
ATP utilization in glucose synthesis:
- During the Calvin cycle, a total of 18 molecules of ATP are utilized in the synthesis of each molecule of glucose.
- ATP is involved in various steps of the cycle, including the phosphorylation of 3-PGA to form G3P and the regeneration of RuBP.
- The energy provided by ATP is essential for driving the endergonic reactions of glucose synthesis.
Therefore, the correct answer is option 'A' - 18 molecules of ATP.
The synthesis of each molecule of glucose in photosynthesis involves the utilization of ATP (adenosine triphosphate) as an energy source. ATP is a high-energy molecule that provides the necessary energy for various cellular processes, including glucose synthesis. In the process of photosynthesis, plants convert sunlight, water, and carbon dioxide into glucose and oxygen. The synthesis of glucose occurs through a series of complex reactions known as the Calvin cycle.
Explanation:
The Calvin cycle is the set of reactions that occur in the stroma of chloroplasts during photosynthesis. It involves a series of enzyme-catalyzed reactions that utilize ATP and NADPH (nicotinamide adenine dinucleotide phosphate) generated during the light-dependent reactions. The primary purpose of the Calvin cycle is the fixation of carbon dioxide and the synthesis of carbohydrates, particularly glucose.
Steps involved in glucose synthesis:
1. Carbon fixation:
- In the first step of the Calvin cycle, carbon dioxide molecules are incorporated into an organic molecule called ribulose-1,5-bisphosphate (RuBP).
- This reaction is catalyzed by the enzyme ribulose bisphosphate carboxylase/oxygenase (Rubisco).
- This process results in the formation of two molecules of 3-phosphoglycerate (3-PGA).
2. Reduction:
- The second step involves the conversion of 3-PGA into glyceraldehyde-3-phosphate (G3P).
- This conversion requires ATP and NADPH as energy and reducing power sources, respectively.
- Each molecule of 3-PGA is phosphorylated by ATP and reduced by NADPH to form G3P.
3. Regeneration:
- In the final step, G3P molecules are used to regenerate RuBP, which is necessary for the continuation of the Calvin cycle.
- Several ATP molecules are utilized in this step to convert G3P into RuBP.
ATP utilization in glucose synthesis:
- During the Calvin cycle, a total of 18 molecules of ATP are utilized in the synthesis of each molecule of glucose.
- ATP is involved in various steps of the cycle, including the phosphorylation of 3-PGA to form G3P and the regeneration of RuBP.
- The energy provided by ATP is essential for driving the endergonic reactions of glucose synthesis.
Therefore, the correct answer is option 'A' - 18 molecules of ATP.
One sequence of amino acids respects for long distances in silk protein. Complete hydrolysis of one mole of a fragment with this sequence gives 2 mol alanine, 3 mol glycine, and 1 mol serine. Partial hydrolysis yields Ala-Gly-Ala, Gly-Ala-Gly, Gly-Ser-Gly, and Ser-Gly-Ala peptides. What is the amino acid repeat:- a)Gly-Gly-Ser-Ala-Gly-Ala
- b)Gly-Gly-Gly-Ala-Ala-Ser
- c)Ser-Ala-Ala-Gly-Gly-Gly
- d)Gly-Ser-Gly-Ala-Gly-Ala
Correct answer is option 'B'. Can you explain this answer?
One sequence of amino acids respects for long distances in silk protein. Complete hydrolysis of one mole of a fragment with this sequence gives 2 mol alanine, 3 mol glycine, and 1 mol serine. Partial hydrolysis yields Ala-Gly-Ala, Gly-Ala-Gly, Gly-Ser-Gly, and Ser-Gly-Ala peptides. What is the amino acid repeat:
a)
Gly-Gly-Ser-Ala-Gly-Ala
b)
Gly-Gly-Gly-Ala-Ala-Ser
c)
Ser-Ala-Ala-Gly-Gly-Gly
d)
Gly-Ser-Gly-Ala-Gly-Ala
|
|
Dipanjan Sharma answered |
Explanation:
- The given information states that one mole of a fragment with a specific sequence of amino acids gives 2 mol alanine, 3 mol glycine, and 1 mol serine upon complete hydrolysis.
- Partial hydrolysis yields four peptides: Ala-Gly-Ala, Gly-Ala-Gly, Gly-Ser-Gly, and Ser-Gly-Ala.
- It can be observed that Gly occurs the most frequently in the partial hydrolysis peptides, and it also occurs the most frequently in the complete hydrolysis product.
- Therefore, the amino acid repeat should start with Gly.
- The sequence of partial hydrolysis peptides can be used to determine the order of the amino acids in the repeat.
- The partial hydrolysis peptides are Gly-Ala-Gly, Ala-Gly-Ala, Gly-Ser-Gly, and Ser-Gly-Ala.
- The repeat should be a sequence that includes all of these peptides, in the order they are given.
- Gly occurs twice in a row in the first partial hydrolysis peptide, so the repeat should have two consecutive Gly.
- Ala occurs before and after Gly in the first and second partial hydrolysis peptides, respectively, so the repeat should have two consecutive Ala.
- Ser occurs before and after Gly in the third and fourth partial hydrolysis peptides, respectively, so the repeat should have Ser in between two Gly.
- Therefore, the amino acid repeat is Gly-Gly-Ala-Ala-Gly-Gly.
- Option B is the correct answer.
From a carboxymethyl-cellulose column at pH 6.0, Arginine, valine and Glutamic acid will elute in the order:- a)Arginine, valine, Glutamic acid
- b)Arginine, Glutamic acid, valine
- c)Glutamic acid, Arginine, valine
- d)Glutamic acid, valine, Arginine
Correct answer is option 'D'. Can you explain this answer?
From a carboxymethyl-cellulose column at pH 6.0, Arginine, valine and Glutamic acid will elute in the order:
a)
Arginine, valine, Glutamic acid
b)
Arginine, Glutamic acid, valine
c)
Glutamic acid, Arginine, valine
d)
Glutamic acid, valine, Arginine
|
|
Aryan Choudhary answered |
Elution order of Arginine, Valine, and Glutamic acid from carboxymethyl-cellulose column at pH 6.0
Carboxymethyl-cellulose (CMC) column is a type of ion exchange chromatography in which anion exchange resins interact with positively charged molecules. The elution order of amino acids from CMC column depends on their pKa values and the pH of the buffer used. At pH 6.0, the carboxyl groups on CMC resin are partially ionized, and the amino acids are in their zwitterionic form.
The pKa values of Arginine, Valine, and Glutamic acid are as follows:
- Arginine: pKa1=2.17, pKa2=9.04, pKa3=12.48
- Valine: pKa1=2.32, pKa2=9.62
- Glutamic acid: pKa1=2.19, pKa2=4.25, pKa3=9.67
Based on the pKa values and the pH of the buffer, the following elution order is expected:
- Glutamic acid (pI=3.22) will elute first because it has the lowest pI value among the three amino acids and will be negatively charged in the column.
- Valine (pI=6.01) will elute next because it is neutral at pH 6.0 and does not interact strongly with the CMC resin.
- Arginine (pI=10.76) will elute last because it has the highest pI value and will be positively charged in the column, interacting strongly with the negatively charged CMC resin.
Therefore, the correct elution order of Arginine, Valine, and Glutamic acid from CMC column at pH 6.0 is:
- Glutamic acid, Valine, Arginine
Carboxymethyl-cellulose (CMC) column is a type of ion exchange chromatography in which anion exchange resins interact with positively charged molecules. The elution order of amino acids from CMC column depends on their pKa values and the pH of the buffer used. At pH 6.0, the carboxyl groups on CMC resin are partially ionized, and the amino acids are in their zwitterionic form.
The pKa values of Arginine, Valine, and Glutamic acid are as follows:
- Arginine: pKa1=2.17, pKa2=9.04, pKa3=12.48
- Valine: pKa1=2.32, pKa2=9.62
- Glutamic acid: pKa1=2.19, pKa2=4.25, pKa3=9.67
Based on the pKa values and the pH of the buffer, the following elution order is expected:
- Glutamic acid (pI=3.22) will elute first because it has the lowest pI value among the three amino acids and will be negatively charged in the column.
- Valine (pI=6.01) will elute next because it is neutral at pH 6.0 and does not interact strongly with the CMC resin.
- Arginine (pI=10.76) will elute last because it has the highest pI value and will be positively charged in the column, interacting strongly with the negatively charged CMC resin.
Therefore, the correct elution order of Arginine, Valine, and Glutamic acid from CMC column at pH 6.0 is:
- Glutamic acid, Valine, Arginine
Which pair(s) of amino acids cannot have hydrophobic interactions:- a)Leucine and alanine
- b)Arginine and Glutamic acid
- c)Glycine and asparagine
- d)Glutamic acid and serine
Correct answer is option 'B,C,D'. Can you explain this answer?
Which pair(s) of amino acids cannot have hydrophobic interactions:
a)
Leucine and alanine
b)
Arginine and Glutamic acid
c)
Glycine and asparagine
d)
Glutamic acid and serine
|
|
Shilpa Datta answered |
Explanation:
Hydrophobic interactions refer to the tendency of nonpolar molecules to cluster together in an aqueous solution. Amino acids can be classified as hydrophobic or hydrophilic based on their polarity.
A) Leucine and alanine:
- Leucine is a hydrophobic amino acid with a nonpolar side chain.
- Alanine is also a hydrophobic amino acid with a nonpolar side chain.
- Therefore, these two amino acids can have hydrophobic interactions.
B) Arginine and glutamic acid:
- Arginine is a hydrophilic amino acid with a positively charged side chain.
- Glutamic acid is a hydrophilic amino acid with a negatively charged side chain.
- Since both amino acids are hydrophilic, they cannot have hydrophobic interactions.
C) Glycine and asparagine:
- Glycine is a hydrophilic amino acid with a small, nonpolar side chain.
- Asparagine is a hydrophilic amino acid with a polar side chain.
- Since both amino acids are hydrophilic, they cannot have hydrophobic interactions.
D) Glutamic acid and serine:
- Glutamic acid is a hydrophilic amino acid with a negatively charged side chain.
- Serine is a hydrophilic amino acid with a polar side chain.
- Since both amino acids are hydrophilic, they cannot have hydrophobic interactions.
Therefore, the correct answer is options B, C, and D.
Hydrophobic interactions refer to the tendency of nonpolar molecules to cluster together in an aqueous solution. Amino acids can be classified as hydrophobic or hydrophilic based on their polarity.
A) Leucine and alanine:
- Leucine is a hydrophobic amino acid with a nonpolar side chain.
- Alanine is also a hydrophobic amino acid with a nonpolar side chain.
- Therefore, these two amino acids can have hydrophobic interactions.
B) Arginine and glutamic acid:
- Arginine is a hydrophilic amino acid with a positively charged side chain.
- Glutamic acid is a hydrophilic amino acid with a negatively charged side chain.
- Since both amino acids are hydrophilic, they cannot have hydrophobic interactions.
C) Glycine and asparagine:
- Glycine is a hydrophilic amino acid with a small, nonpolar side chain.
- Asparagine is a hydrophilic amino acid with a polar side chain.
- Since both amino acids are hydrophilic, they cannot have hydrophobic interactions.
D) Glutamic acid and serine:
- Glutamic acid is a hydrophilic amino acid with a negatively charged side chain.
- Serine is a hydrophilic amino acid with a polar side chain.
- Since both amino acids are hydrophilic, they cannot have hydrophobic interactions.
Therefore, the correct answer is options B, C, and D.
There are 20 naturally occurring amino acids. The maximum number of tripeptides that can be obtained is:- a)6470
- b)7465
- c)8000
- d)5360
Correct answer is option 'C'. Can you explain this answer?
There are 20 naturally occurring amino acids. The maximum number of tripeptides that can be obtained is:
a)
6470
b)
7465
c)
8000
d)
5360
|
|
Swara Reddy answered |
Naturally occurring amino acids are 20. Hence, number of possible tripeptides
= 203 = 8000
Ident ify the correct set of stereochemical relat ionships amongst the following monosaccharides I-IV:
- a)I and II are anomers; III and IV are epimers
- b)I and III are epimers; II and IV are anomers
- c)I and II are epimers; III and IV are anomers
- d)I and III are anomers; I and II are epimers
Correct answer is option 'D'. Can you explain this answer?
Ident ify the correct set of stereochemical relat ionships amongst the following monosaccharides I-IV:
a)
I and II are anomers; III and IV are epimers
b)
I and III are epimers; II and IV are anomers
c)
I and II are epimers; III and IV are anomers
d)
I and III are anomers; I and II are epimers
|
|
Subha Som answered |
I, III are C2 epimer II,IV are C1 ie anomer
How many different tripeptides can be obtained from alanine, glycine and phenylalanine, each tripeptide containing all the three amino acids?
Correct answer is '6'. Can you explain this answer?
How many different tripeptides can be obtained from alanine, glycine and phenylalanine, each tripeptide containing all the three amino acids?
|
|
Arnab Pillai answered |
Introduction:
In this question, we need to determine the number of different tripeptides that can be obtained from the amino acids alanine, glycine, and phenylalanine. Each tripeptide must contain all three amino acids.
Explanation:
To solve this problem, we need to consider the different arrangements of the three amino acids in a tripeptide. Since each tripeptide must contain all three amino acids, we can use the concept of permutations to find the number of different tripeptides.
Permutations:
Permutations are arrangements of objects where the order matters. In this case, the order of the amino acids in the tripeptide matters.
Using the formula for permutations:
The number of permutations of n objects taken r at a time is given by the formula:
P(n, r) = n! / (n - r)!
Applying the formula:
In this case, we have 3 amino acids (alanine, glycine, and phenylalanine) that need to be arranged in tripeptides. Since each tripeptide contains exactly 3 amino acids, we can use the formula for permutations with n = 3 and r = 3.
P(3, 3) = 3! / (3 - 3)!
= 3! / 0!
The factorial of 3 is 3! = 3 * 2 * 1 = 6.
The factorial of 0 is defined as 1.
Therefore, the calculation simplifies to:
P(3, 3) = 6 / 1 = 6
Conclusion:
Hence, there are 6 different tripeptides that can be obtained from the amino acids alanine, glycine, and phenylalanine, where each tripeptide contains all three amino acids.
In this question, we need to determine the number of different tripeptides that can be obtained from the amino acids alanine, glycine, and phenylalanine. Each tripeptide must contain all three amino acids.
Explanation:
To solve this problem, we need to consider the different arrangements of the three amino acids in a tripeptide. Since each tripeptide must contain all three amino acids, we can use the concept of permutations to find the number of different tripeptides.
Permutations:
Permutations are arrangements of objects where the order matters. In this case, the order of the amino acids in the tripeptide matters.
Using the formula for permutations:
The number of permutations of n objects taken r at a time is given by the formula:
P(n, r) = n! / (n - r)!
Applying the formula:
In this case, we have 3 amino acids (alanine, glycine, and phenylalanine) that need to be arranged in tripeptides. Since each tripeptide contains exactly 3 amino acids, we can use the formula for permutations with n = 3 and r = 3.
P(3, 3) = 3! / (3 - 3)!
= 3! / 0!
The factorial of 3 is 3! = 3 * 2 * 1 = 6.
The factorial of 0 is defined as 1.
Therefore, the calculation simplifies to:
P(3, 3) = 6 / 1 = 6
Conclusion:
Hence, there are 6 different tripeptides that can be obtained from the amino acids alanine, glycine, and phenylalanine, where each tripeptide contains all three amino acids.
The number of amino acids and number of peptides bonds in a linear tetrapeptide (made of different amino acids) are respectively:- a)4 and 4
- b)5 and 5
- c)5 and 4
- d)4 and 3
Correct answer is option 'D'. Can you explain this answer?
The number of amino acids and number of peptides bonds in a linear tetrapeptide (made of different amino acids) are respectively:
a)
4 and 4
b)
5 and 5
c)
5 and 4
d)
4 and 3
|
Nikhil Thakur answered |
@----@-----@-----@ consider @ as a amino acid and dotted lines as a peptide bond we clearly see that four amino acid have 3 bond .
Among the D-glucopyranose forms given below, which is more stable:- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Among the D-glucopyranose forms given below, which is more stable:
a)
b)
c)
d)
|
Saurabh Chavan answered |
There should be 0 or minimum number of OH or CH2OH group on axial position hence then gives more stability.
In fructose the possible optical iso mer are:
Correct answer is '8'. Can you explain this answer?
In fructose the possible optical iso mer are:
|
|
Pragati Sharma answered |
Fructose is a monosaccharide or simple sugar that is found in many fruits, vegetables and honey. It has the chemical formula C6H12O6 and is an important source of energy for the body. When fructose is dissolved in water, it can exist in different forms or isomers, depending on the spatial arrangement of its atoms.
Possible Optical Isomers of Fructose
Fructose has four chiral carbon atoms, which means that it can have up to 2^4 or 16 possible stereoisomers. However, not all of these isomers are optically active or have the ability to rotate plane-polarized light. Only those isomers that are non-superimposable mirror images of each other are optically active and are called enantiomers or optical isomers.
Fructose exists in two main forms: alpha fructose and beta fructose. These two forms differ in the orientation of their hydroxyl groups around the anomeric carbon atom, which is the carbon atom that is bonded to both an oxygen atom and another carbon atom. Alpha fructose has its hydroxyl group pointing downwards, while beta fructose has its hydroxyl group pointing upwards.
Each of these forms can exist in two possible configurations, depending on the orientation of the hydroxyl groups around the other three chiral carbon atoms. These configurations are designated as either D or L, depending on the orientation of the hydroxyl group on the last chiral carbon atom. Therefore, fructose can exist in four possible isomers: alpha-D-fructose, alpha-L-fructose, beta-D-fructose, and beta-L-fructose.
Each of these four isomers can exist in two possible enantiomeric forms, depending on whether the molecule is a right-handed or left-handed form. Therefore, fructose can exist in eight possible optical isomers: alpha-D-fructose, alpha-L-fructose, beta-D-fructose, beta-L-fructose, (alpha-D)-fructose, (alpha-L)-fructose, (beta-D)-fructose, and (beta-L)-fructose.
Conclusion
In summary, fructose has eight possible optical isomers, which are determined by the orientation of its hydroxyl groups around its four chiral carbon atoms. These isomers are important for understanding the biological activity and metabolism of fructose, as well as for the production of pure fructose for use in food and pharmaceutical applications.
Possible Optical Isomers of Fructose
Fructose has four chiral carbon atoms, which means that it can have up to 2^4 or 16 possible stereoisomers. However, not all of these isomers are optically active or have the ability to rotate plane-polarized light. Only those isomers that are non-superimposable mirror images of each other are optically active and are called enantiomers or optical isomers.
Fructose exists in two main forms: alpha fructose and beta fructose. These two forms differ in the orientation of their hydroxyl groups around the anomeric carbon atom, which is the carbon atom that is bonded to both an oxygen atom and another carbon atom. Alpha fructose has its hydroxyl group pointing downwards, while beta fructose has its hydroxyl group pointing upwards.
Each of these forms can exist in two possible configurations, depending on the orientation of the hydroxyl groups around the other three chiral carbon atoms. These configurations are designated as either D or L, depending on the orientation of the hydroxyl group on the last chiral carbon atom. Therefore, fructose can exist in four possible isomers: alpha-D-fructose, alpha-L-fructose, beta-D-fructose, and beta-L-fructose.
Each of these four isomers can exist in two possible enantiomeric forms, depending on whether the molecule is a right-handed or left-handed form. Therefore, fructose can exist in eight possible optical isomers: alpha-D-fructose, alpha-L-fructose, beta-D-fructose, beta-L-fructose, (alpha-D)-fructose, (alpha-L)-fructose, (beta-D)-fructose, and (beta-L)-fructose.
Conclusion
In summary, fructose has eight possible optical isomers, which are determined by the orientation of its hydroxyl groups around its four chiral carbon atoms. These isomers are important for understanding the biological activity and metabolism of fructose, as well as for the production of pure fructose for use in food and pharmaceutical applications.
Glucose and mannose are:- a)keto-hexoses
- b)Anomers
- c)Epimers
- d)Disaccharides
Correct answer is option 'C'. Can you explain this answer?
Glucose and mannose are:
a)
keto-hexoses
b)
Anomers
c)
Epimers
d)
Disaccharides
|
Avinash Mehta answered |
Epimers are compounds having the same chemical formula but differ in the spatial arrangement around a single carbon atom. D-Galactose, on the other hand, differs from D-Glucose ONLY at the 4th Carbon atom. Hence, it is a C-4 epimer of D-Glucose. D-Mannose and D-Galactose.
From the below list, identify the sugars which on reduction give the same glucitol:- a)D-glucose: D-allose
- b)D-glucose: D-gulose
- c)D-glucose: D-galactose
- d)D-glucose: D-mannose
Correct answer is option 'B'. Can you explain this answer?
From the below list, identify the sugars which on reduction give the same glucitol:
a)
D-glucose: D-allose
b)
D-glucose: D-gulose
c)
D-glucose: D-galactose
d)
D-glucose: D-mannose
|
|
Akash Kulkarni answered |
Explanation:
When a sugar is reduced with a reducing agent such as sodium borohydride, it can form a sugar alcohol or polyol known as a glucitol. The formation of glucitol depends on the stereochemistry of the reducing sugar.
Reducing Sugars:
Reducing sugars are those sugars that have a free aldehyde or ketone group that can be oxidized to form a carboxylic acid. The reducing sugars in the given options are:
a)D-glucose: D-allose
b)D-glucose: D-gulose
c)D-glucose: D-galactose
d)D-glucose: D-mannose
Glucitol:
Glucitol is a sugar alcohol derived from glucose. It is also known as sorbitol. The chemical formula of glucitol is C6H14O6.
Answer:
The correct answer is option B, i.e., D-glucose and D-gulose.
Both D-glucose and D-gulose are epimers of each other, differing only in the stereochemistry at the C-2 carbon. When reduced with a reducing agent, both D-glucose and D-gulose form the same glucitol, i.e., D-sorbitol. This is because the stereochemistry at the C-2 carbon does not affect the formation of the glucitol.
On the other hand, D-allose, D-galactose, and D-mannose are not epimers of D-glucose and D-gulose at the C-2 carbon. Therefore, they will not form the same glucitol upon reduction.
When a sugar is reduced with a reducing agent such as sodium borohydride, it can form a sugar alcohol or polyol known as a glucitol. The formation of glucitol depends on the stereochemistry of the reducing sugar.
Reducing Sugars:
Reducing sugars are those sugars that have a free aldehyde or ketone group that can be oxidized to form a carboxylic acid. The reducing sugars in the given options are:
a)D-glucose: D-allose
b)D-glucose: D-gulose
c)D-glucose: D-galactose
d)D-glucose: D-mannose
Glucitol:
Glucitol is a sugar alcohol derived from glucose. It is also known as sorbitol. The chemical formula of glucitol is C6H14O6.
Answer:
The correct answer is option B, i.e., D-glucose and D-gulose.
Both D-glucose and D-gulose are epimers of each other, differing only in the stereochemistry at the C-2 carbon. When reduced with a reducing agent, both D-glucose and D-gulose form the same glucitol, i.e., D-sorbitol. This is because the stereochemistry at the C-2 carbon does not affect the formation of the glucitol.
On the other hand, D-allose, D-galactose, and D-mannose are not epimers of D-glucose and D-gulose at the C-2 carbon. Therefore, they will not form the same glucitol upon reduction.
Most abundant RNA in the cell- a)rRNA
- b)mRNA
- c)tRNA
- d)tRNA threonine
Correct answer is option 'A'. Can you explain this answer?
Most abundant RNA in the cell
a)
rRNA
b)
mRNA
c)
tRNA
d)
tRNA threonine
|
|
Pooja Choudhury answered |
Ans: a
Explanation: Ribosomal ribonucleic acid (rRNA) is the most abundant type of RNA. It constitutes the material inside the ribosome and directly helps in the translation of mRNA into proteins.
Which gives red colour with Fehling’s solut ion:- a)Glucose
- b)Cellulose
- c)Benzaldehyde
- d)Cane sugar
Correct answer is option 'A'. Can you explain this answer?
Which gives red colour with Fehling’s solut ion:
a)
Glucose
b)
Cellulose
c)
Benzaldehyde
d)
Cane sugar
|
|
Sandeep Sahoo answered |
All monosacharides are reducing sugar. Glucose is a monosacharide and reducing sugar. Reducing sugars reacts with feeling's solutions, benedict's solutions and tollens reangets. So glucose gives red colour with fehling solutions.
Which of the following regarding to the examples of Keto sugar is/are correct:- a)Fructose
- b)Ribulose
- c)Ribose sugar
- d)Dihydroxy acetose
Correct answer is option 'A,B,D'. Can you explain this answer?
Which of the following regarding to the examples of Keto sugar is/are correct:
a)
Fructose
b)
Ribulose
c)
Ribose sugar
d)
Dihydroxy acetose
|
|
Biswa Nath answered |
Keto sugar means which contains >C=O groups.
Fructose, Ribulose, Dihydroxy acetose all of them contains >C=O group.
But, in Ribose sugar there have no such groups.
Fructose, Ribulose, Dihydroxy acetose all of them contains >C=O group.
But, in Ribose sugar there have no such groups.
Which one of the following is an essential amino acid:- a)Methionine
- b)Tyrosine
- c)Proline
- d)Glycine
Correct answer is option 'A'. Can you explain this answer?
Which one of the following is an essential amino acid:
a)
Methionine
b)
Tyrosine
c)
Proline
d)
Glycine
|
|
Niharika Kulkarni answered |
Essential Amino Acid: Methionine
Definition: Essential amino acids are amino acids that cannot be synthesized by the body and must be obtained from the diet.
Methionine: Methionine is an essential amino acid that plays an important role in the synthesis of proteins and other important molecules in the body. It is classified as a sulfur-containing amino acid due to the presence of a sulfur atom in its chemical structure.
Sources: Methionine can be found in a variety of foods, including meat, fish, dairy products, and eggs. It can also be found in some plant-based sources, such as nuts and seeds.
Functions: Methionine is important for a number of functions in the body, including:
- Protein synthesis: Methionine is necessary for the formation of proteins in the body.
- Methylation: Methionine is a precursor to S-adenosylmethionine (SAMe), which plays a role in methylation reactions in the body. Methylation is important for a variety of processes, including DNA synthesis and repair, neurotransmitter synthesis, and detoxification.
- Antioxidant defense: Methionine is a precursor to glutathione, an important antioxidant in the body that helps to protect against oxidative stress.
Conclusion: Methionine is an essential amino acid that plays an important role in protein synthesis, methylation, and antioxidant defense. It can be found in a variety of foods, and is particularly abundant in meat, fish, dairy products, and eggs.
Definition: Essential amino acids are amino acids that cannot be synthesized by the body and must be obtained from the diet.
Methionine: Methionine is an essential amino acid that plays an important role in the synthesis of proteins and other important molecules in the body. It is classified as a sulfur-containing amino acid due to the presence of a sulfur atom in its chemical structure.
Sources: Methionine can be found in a variety of foods, including meat, fish, dairy products, and eggs. It can also be found in some plant-based sources, such as nuts and seeds.
Functions: Methionine is important for a number of functions in the body, including:
- Protein synthesis: Methionine is necessary for the formation of proteins in the body.
- Methylation: Methionine is a precursor to S-adenosylmethionine (SAMe), which plays a role in methylation reactions in the body. Methylation is important for a variety of processes, including DNA synthesis and repair, neurotransmitter synthesis, and detoxification.
- Antioxidant defense: Methionine is a precursor to glutathione, an important antioxidant in the body that helps to protect against oxidative stress.
Conclusion: Methionine is an essential amino acid that plays an important role in protein synthesis, methylation, and antioxidant defense. It can be found in a variety of foods, and is particularly abundant in meat, fish, dairy products, and eggs.
A nannopeptide contains ………….. pepetide linkages:- a)10
- b)8
- c)9
- d)18
Correct answer is option 'B'. Can you explain this answer?
A nannopeptide contains ………….. pepetide linkages:
a)
10
b)
8
c)
9
d)
18
|
|
Milan Majumdar answered |
Ans.
Option (b)
The peptide bond is formed between two amino acids by the elimination of a water molecule. A dipeptide contains one peptide linkage. A tripeptide contains two peptide linkages. Similarly, a nanopeptide contains 8 peptide linkages.
Chapter doubts & questions for Natural Products Chemistry - Topicwise Question Bank for IIT JAM/CSIR/GATE Chemistry 2025 is part of Chemistry exam preparation. The chapters have been prepared according to the Chemistry exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Chemistry 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
Chapter doubts & questions of Natural Products Chemistry - Topicwise Question Bank for IIT JAM/CSIR/GATE Chemistry in English & Hindi are available as part of Chemistry exam.
Download more important topics, notes, lectures and mock test series for Chemistry Exam by signing up for free.
Signup to see your scores go up within 7 days!
Study with 1000+ FREE Docs, Videos & Tests
10M+ students study on EduRev
|
© EduRev
|
Education Revolution
|
|
Signup on EduRev and stay on top of your study goals
10M+ students crushing their study goals daily