BIOMOLECULES

CARBOHYDRATES:
These are natural substances which are obtained from plants or animals, they are poly hydroxy
aldehydes or ketones which are made up of C H O with a general formula Cx(H2O)y.
Carbohydrates are the source of energy for the living beings, plants store carbohydrates in the form of
food.
Classification:
Reducing sugars Non-reducing sugars
Carbohydrates that reduce Tollen’s and Carbohydrates that don’t reduce Tollen’s and
Fehling’s sol Fehling’s sol
Ex: Glucose, Fructose and Galactose Ex: Sucrose, Lactose

On the basis of Hydrolysis:

Mono Saccharides Di Saccharides Oligo Saccharides Poly Saccharides
Doesn’t give any Give two mono Give 2-8 mono They are long chains
carbohydrates upon saccharides on Saccharides on of mono saccharides
hydrolysis Hydrolysis hydrolysis linked by glycosidic
bonds
Ex: Glucose, Ex: Sucrose, Ex: Raffinose, Ex: Starch,
Fructose, Galactose Maltose, Lactose Sucrose, Maltose Glycogen, Cellulose

On the basis of functional group and No. of hydroxyl groups:

OH Aldoses Ketoses
Groups
3 Aldotriose Ketotriose
5 Aldopentose Ketopentose
6 Aldohexose ketohexose
Structural elucidation of Glucose:
- Glucose on reaction with Red P/HI shows that glucose has six carbons in straight chain.
- Upon reaction with HCN to give Cyanohydrin explains the presence of Carbonyl group
- Glucose gets oxidised with mild oxidising agent like Bromine water to give Gluconic acid which means
presence of aldehyde group
- On reduction and acetylation with acetic anhydride gives Penta acetyl derivative that confirms the presence
of 5 -OH groups.
-On oxidation with Nitric acid it is converted into saccharic acid which shows the presence of Primary –OH
group on one the terminal carbons.
Osazone test:
When carbohydrate reacts with phenyl hydrazine it forms hydrazone due to the presence of aldehyde or keto
group and presence of CH-OH group in C2 is also converted to hydrazone by undergoing oxidation to ketone
and then condensation.
This test is given by Glucose, Fructose and Mannose as they have same molecular structure.

All three molecules react with three equivalents of phenyl hydrazine to produce yellow needle shaped
crystals of Osazone.
Chemical Reactions of Glucose:

Oxidative cleavage of Carbohydrates:

Carbohydrates upon reacting Per-Iodic acid undergo oxidative cleavage to give a mixture of carbonyl
compounds and carboxylic acids.
- Primary alcohol groups –CH2OH converted to HCHO
- Secondary alcohol groups >CH-OH converted to HCOOH
- Aldehyde group –CHO converted to HCOOH
- Keto group >C=O converted to CO2
Ex:
Oxidative cleavage of D-Glucose takes 5 moles of HIO4 to give 5 moles HCOOH and 1 mole HCHO

Ex:
Oxidative cleavage of D-Fructose takes 5 moles of HIO4 to give 3 moles HCOOH and 1 mole HCHO and CO2

FISCHER structure of Carbohydrates:

D and L Configurations
- D and L notations are given to carbohydrates with respect to glyceraldehyde
 D-Aldoses and their derivatives:

HAWORTH structure of Carbohydrates:

The letters D & L represent stereo isomers of Carbohydrate, D – isomer has –OH group on the right for the
lowest asymmetric carbon and L-isomer has –OH group on to the left side for the lowest asymmetric carbon.
 Cyclic structure of glucose is supported by the following facts and open chain is not supporting for the
same
1) Though it has aldehyde group it doesn’t give Schiff’s test and doesn’t react with NaHSO3
2) Glucose doesn’t react with hydroxylamine which shows that it doesn’t have free aldehyde group
3) Glucose is found to exist in two crystalline forms namely α & β glucopyranose structures
as shown above. These isomers are called Anomers as they differ by the configuration of –OH group in C 1
atom.
-Pyranose name is given as it resembles the structure of Pyran and for fructose the Furanose is given as it
resembles the structure of Furan.
Mutarotation:
This is the change in the optical rotation due to the change in the equilibrium between the two anomers of
Glucose in the aqueous solution. The equilibrium mixture will have 64% β-D glucose and 36% α-D glucose.

Epimers:
The difference in the configuration of any of the carbon atoms of the two isomers are called epimers

- D (+) Glucose and D (+) galactose differ in their configuration at C4 carbon, so these are called C4 epimers
- D (+) Glucose and D (+) mannose differ in their configuration at C2 carbon, so these are called C2 epimers
HAWORTH PROJECTIONS OF HEXOSES:
 Lobry deBruyn Van Ekenstein Rearrangement Reaction:
Glucose gives a positive test for Tollen’s and Fehling’s due to the presence of –CHO group, but Fructose also
gives a positive test for the same due to its conversion into a mixture of Mannose and Glucose. This
conversion is due to isomerisation in the presence of Base. This is one kind of tautomerism between the
aldehyde and to enol and to its keto form and vice versa.

Aldehyde Ketone

Killiani Fischer Synthesis:

This is the method to convert and aldopentose to aldohexose.

Aldopentose C2 epimers γ- Lactone mix. Aldohexose (C3 epimers)

Ruff degradation:
This is the method to convert and aldohexose to aldopentose.
 Wohl Degradation:
This is the method to convert and aldohexose to aldopentose.
Also used in the conversion of hexose to pentose.

PREVIOUS YEAR QUESTIONS:

1) Which of the given statement is incorrect about glucose?
(JEE 2020)
a. Glucose exists in two crystalline forms α and β b. Glucose gives Schiff's test.
c. Penta acetate of glucose does not form oxime. d. Glucose forms oxime with
hydroxylamine.
ANS: (b)
SOL: Glucose exists in two crystalline forms α and β which are anomers of each other
Glucose does not react with Schiff’s reagent because after the internal cyclisation, it forms either α-
anomer or β-anomer. In these forms, free aldehydic group is not present.
Glucose forms open chain structure in aqueous solution which contains aldehyde at chain end. This
aldehydic group reacts with NH4OH to form oxime. On the other hand, glucose penta acetate being a
cyclic structure even in aqueous form does not have terminal carbonyl group. Therefore it will not react
with NH4OH.
2) The correct statement about gluconic acid is:
(JEE 2020)
a. It is prepared by oxidation of glucose with HNO3
b. It is obtained by partial oxidation of glucose
c. It is a dicarboxylic acid
d. It forms hemiacetal or acetal
ANS: (b)
3) Glucose and Galactose are having identical configuration in all the positions except position.
(JEE 2019)
(1) C – 2 (2) C – 5 (3) C – 3 (4) C – 4
ANS: (4)
SOL: Galactose and glucose are C-4 epimers.
 DISACCHARIDES:
Haworth Structures of Disaccharides:
Sucrose:
C12H22O11 (Sucrose) + H2O C6H12O6 (Glucose) + C6H12O6 (Fructose)

The link between the two carbohydrates is called Glycosidic linkage which is obtained by the condensation of
any two carbohydrates.
-The two monosaccharides are linked by C1 of Glucose and C2 of fructose
-Sucrose is dextro rotatory but after hydrolysis it gives dextro Glucose and Leavo Fructose, but the rotation of
fructose is -92.4o which is more than glucose which is +52.5O. So, the mixture becomes Leavo rotatory, hence
sucrose is called Invert Sugar
Lactose:
C12H22O11 (Lactose) + H2O C6H2O6 (Glucose) + C6H12O6 (Galactose)
-The two monosaccharides are linked by C1 of Galactose and C4 of Glucose.

Maltose:
C12H22O11 (Maltose) + H2O C6H2O6 (Glucose) + C6H12O6 (Glucose)
-The two monosaccharides are linked by C1 of Glucose and C4 of another Glucose
Polysaccharides:
Starch: Starch is the main storage polysaccharide of plants and important dietary source of Humans. It
consists of mainly two components one is amylose and the other amylopectin
Amylosis water soluble and Amylopectin is insoluble in water, Amylose is a linear chain with C 1-C4
glycosidic linkages and amylopectin is cross linked chain with C1 –C4 and C1-C6 glycosidic linkages.

Cellulose:
Cellulose occurs extensively in plants and it is an important constituent of cell wall of plants. It is a straight
chain polysaccharide having β-D glucose monomeric units having C1-C4 glycosidic linkages.

Glycogen:
 Glycogen is stored in animal body and present in liver, muscles and brain and when the body needs glucose it
breaks up to give the same. It is also found in Yeast and Fungi.
Monomer is made up of α- D- Glucose where it forms C1-C4 & C1- C6 cross links with Amylopectin chain.
Dextrin and Inulin are also other examples of Polysaccharides.

CARBOHYDRATE TYPE REDUCING/ MONOMERIC UNITS LINKAGES

NON-REDUCING
Glucose Monosaccharide Reducing Glucose NA
Fructose Monosaccharide Reducing Fructose NA
Sucrose Disaccharide Non-Reducing α-D-Glucose & C1 & C2 resp.
β-D-Fructose
Lactose Disaccharide Reducing β-D-Glucose & C1 & C4 resp.
β-D-Galactose
Maltose Disaccharide Reducing α-D-Glucose & C1 & C4 resp.
α-D-Glucose
Starch Polysaccharide Non-Reducing Amylose & Amylose C1 & C4
Amylopectin Amylopectin C1 & C6
Glycogen Polysaccharide Non-Reducing Amylopectin C 1 & C 4, C 1 & C 6
Cellulose Polysaccharide Non-Reducing β-D-Glucose C1 & C4 resp.

Detection Test for Carbohydrates:

TEST REAGENT RESULT
Molisch’s Test Con.H2SO4 Purple Color
Any carbohydrate
Fehling’s Test Alkaline Cu(OH)2 Yellow or Red Cu2O ppt
(Only reducing Sugars)
Benedict’s Test Alkaline Cu(OH)2 Yellow or Red Cu2O ppt
(Only reducing Sugars)
Barfoed’s Test AcOH + Cu(OAc)2 Red CuO ppt
(Monosaccharides)
Seliwanoff’s Test Con.H2SO4 + Red Colour
(Ketoses) Resorcinol
Iodine Test I2 solution Blue-Black Color
(Starch)
Osazone Test Phenyl Hydrazine Yellow Osazone crystals
(Aldoses & Ketoses)
------------------------------------------------------------------------------------------------------------------------------------------
1)

(JEE 2020)
 Which of the following options is correct?
A B C
a. Lactose Glucose Albumin
b. Lactose Glucose Alanine
c. Lactose Fructose Alanine
d. Glucose Sucrose Albumin

2) Which of the given statements is INCORRECT about glycogen?

(JEE 2019)
(1) It is present in some yeast and fungi. (2) It is a straight chain polymer similar to amylose.
(3) It is present in animal cells. (4) Only  C1 & C4 -linkages are present in the molecule.
ANS: (2)
3) Amylopectin is composed of: (JEE 2019)

ANS: (4)
4) Products formed by hydrolysis of maltose are: (JEE
2020)
a. α-D-Glucose, α -D-Glucose b. α -D-Glucose, β-D-Glucose
c. α -D-Galactose, β -D-Glucose d. β -D-Galactose, α -D-Glucose
ANS: (a)
5) Maltose on treatment with dil.HCl gives: (JEE
2019)
1) D-Galactose 2) D-Glucose D-Fructose 3) D-Glucose 4) D-Fructose
ANS: (3)
6) Which of the following statements is not true about sucrose?
(JEE 2019)
(1) The glycosidic linkage is present between C 1 of α-glucose and C1 of β-fructose
(2) On hydrolysis, it produces glucose and fructose
(3) It is a non-reducing sugar (4) It is also named as invert sugar
ANS: (1)\
7) Among the following, the incorrect statement is : (JEE 2018)
(1) Cellulose and amylose has 1,4-glycosidic linkage. (2) Lactose contains -D-galactose and -D-glucose.
(3) Maltose and lactose has 1,4-glycosidic linkage. (4) Sucrose and amylose has 1,2-glycosidic linkage.
ANS: (4)
SOL: In amylose 1,4-glycosidic linkage is present.

AMINO ACIDS
These are the building blocks of proteins i.e., they are the monomer of proteins which are containing Amine
and Carboxylic acid groups.
-Amino acids are colourless, crystalline and water soluble which forms a dipolar ion or Zwitter ion as the
proton from –COOH group migrates to -NH2.
α-Amino Acids:
Carboxylic acids in which one α-hydrogen atoms of alkyl group is substituted by amino (–NH2)
group are called α-Amino acids. The general formula is

- Zwitter ion upon electrolysis in acidic medium migrates toward Cathode and electrolysis in basic medium
migrates towards Anode. When amino acids have both +ve and –ve charge they don’t migrate towards any
electrode and the pH at this point is called isoelectric point.
What is the isoelectric point of Histidine?
The neutral form exists under more basic conditions when the extra +ve has been neutralised. For example,
for histidine, the neutral form is dominant between pH 6.00 and 9.17, pI is halfway between these two values,
i.e., pI = 1/2 (pKa2 + pKa3), so pI = 7.59.

Classification:
On the basis of number of functional groups:
Acidic Amino acids where no. of –COOH groups are more than –NH2 groups.
Ex: Aspartic acid, Glutamic acid etc.
Basic Amino acids where no. of –NH2 groups are more than –COOH groups.
Ex: Lysine, Arginine etc.

Neutral Amino acids where no. of –COOH groups = no. of –NH2 groups.
Ex: Glycine, Alanine etc.,

On the basis of diet:

There are a total of 20 amino acids existing in which 10 are essential and 10 are non-essential amino acids
Essential amino acids are to be consumed in the diet as they are not synthesised in the body
Ex: Valine, Leucine, Arginine etc.
Non-essential amino acids are synthesized in the body.
Ex: Glycine, Alanine, Aspartic acid etc.
Ninhydrin test:
Test for amino acids

1) The correct sequence of amino acids present in the tripeptide given below is:
(JEE 2019)

(1) Leu - Ser - Thr (2) Thr - Ser- Leu (3) Thr - Ser - Val (4) Val - Ser – Thr
ANS: (4)

2) The peptide that gives positive ceric ammonium nitrate test and carbylamine test is:
(JEE 2019)
(1) Ser - Lys (2) Lys – Asp (3) Gln - Asp (4) Asp - GlU
ANS: (1)
Lysine has −NH2 group hence gives positive carbyl amine test and serine has −OH group hence gives positive
cerric ammonium nitrate test.

SOL: Ceric ammonium nitrate test is given by alcohol. Only serine(ser) contain –OH group.

3) The increasing order of pKa of the following amino acids in aqueous solutions is:
(JEE 2019)
Gly Asp Lys Arg
(1) Asp < Gly < Arg < Lys (2) Arg < Lys < Gly < Asp (3) Gly < Asp < Arg < Lys (4) Asp < Gly <
Lys < Arg
ANS: (4)
SOL: Order of acidic strength: Asp> Gly> Lys > Arg and hence opp is the pKa order.

4) The correct structure of ‘P’ of the following reaction is: (JEE

2019)
ANS: (1)
SOL: Asn-Ser is a dipeptide having the following structure:

5) Which of the following tests cannot be used for testing amino acids:
(JEE 2019)
(1) Biuret test (2) Xanthoproteic test (3) Barfoed test (4) Ninhydrin test
ANS: (3)

6) The correct match between item I and item II is: (JEE

2019)

ANS: (1)

7) The correct match between item I and item II is: (JEE

2019)

ANS: (1)
SOL:
8) Among the following compounds most basic amino acid is:
(JEE 2019)
(1) Lysine (2) Asparagine (3) Serine (4) Histidine
ANS: (4)

9) The correct structure of Histidine in the strongly acidic solution is (pH=2)

(JEE 2019)

ANS: (1)

SOL:

PROTEINS
Proteins are the polymers of amino acids, which are natural substance obtained by the condensation
polymerisation of a α – amino acids. Proteins are essential for the growth of living beings.
- A polypeptide with more than 100 amino acid residues having molecular mass higher than 10,000 u is
called protein
- Insulin is a protein which contains 51 amino acids only
Synthesis of Proteins:
Amino acids are the monomers of proteins and amino acids undergo polymerisation to give proteins
-The amino acids undergoing polymerisation can be of same type or different type due to which specific
proteins are obtained.

→ →
NH2-CH(R)-COOH + NH2-CH(R)-COOH – H 2 O nNH2-CH(R)-CO-NH-CH(R)-COOH ( n−1 ) H 2 O[-
NH-CH(R)-CO-NH-CH(R)-CO-]n
‘n’ molecules of amino acids undergo condensation with the loss of water to give amide link or peptide
bond, there will be (n-1) peptide bonds for n molecules of amino acids.
-So, a dipeptide has one peptide bond and two amino acids
Classification:
Fibrous Proteins Globular Proteins
They have long thread like structure They have spherical shape with a
molecular globule
They are insoluble in water They are soluble in water
They are held together by H- bonds They have three-dimensional spherical
shape as the polypeptides coil around
Ex: Keratin in Hair, nails, wool and silk Ex: Egg albumin, Insulin
Myosin in muscles

Proteins are also classified on the basis of structure:

-Primary structure which has single strand held together by peptide bonds.

Secondary structure which is helical i.e., obtained by the H- bonds

Secondary structure of proteins can be of two types which is
α-Helix & β-Pleated structure

Ex: α- Helix- Keratin, Myosin β- Pleated- Fibroin, Silk

α- Helix
Polypeptide chain form all possible hydrogen bonds by twisting into the right-handed screw. In this -NH
group of each amino acid, residue forms a hydrogen bond with C=O group of adjacent turn of helix.
β pleated
When polypeptide chains are arranged side by side, these chains are held together by large number of
intermolecular hydrogen bonds (between C = O and NH groups of different chains) gives β pleated sheet
structure

-Tertiary structure which is obtained on further folding of these helical chains.

The types of bonds between polypeptides chains in tertiary structure are:
Disulphide bonds, Ionic Bonds & Hydrogen Bonds

-Quaternary structure is complex structure having mixed random folds of helical chains.

Denaturation of Proteins:
The process of converting quaternary, tertiary, secondary structured proteins into primary proteins by
heating or addition of suitable enzymes.
Ex: Boiling of egg, curdling of milk by the enzyme lactobacillus.
Biuret Test:

Detection Test for Proteins:

TEST REAGENT RESULT
Biuret Test Alkaline Cu2+ solution Pink-Purple

Millon’s Test Con.HNO3/ Hg Reddish Brown ppt

Ninhydrin 2,2 Dihydroxyindane 1, 3- Deep Blue or Purple

Test Dione color

Xanthoprotein Con.HNO3 Intense Yellow ppt

 Test
NUCLEIC ACIDS
Nucleic Acids are Biopolymers essential to all forms of life, they are made up of Nucleotides which are
the monomers, and these monomers are made of three components Ribose sugar group, Phosphate group and
Nitrogenous Base group.

Nitrogenous bases are of two types Purine Bases & Pyrimidine Bases, Purine bases are 9 membered
heterocyclic rings and Pyrimidine bases are 6 membered heterocyclic rings.
Purine and pyrimidine bases are as shown below:

Nucleotides are the combination of Nucleosides and phosphate link, so nucleoside is a unit comprises of sugar
link and Nitrogenous base link.

Nucleic acids are of two types Deoxyribonucleic acid (D.N.A) and Ribonucleic acid (R.N.A)
In the polymerisation of Nucleotides the Sugar link is attached to the phospate link of another nucleotide
through condensation, In DNA the heterocyclic bases are held together by Hydrogen bonds

Poly Nucleotide:
- The nucleic acid chain is written as

- Sequence of nucleotides in the chain of nucleic acid is called its primary structure.
- Nucleic acids also contain secondary structure.
- DNA has a double strand helix structure.
- Two nucleic acid chains are wound about each other and held together by hydrogen bonds between pairs of
bases
- Adenine (A) forms two hydrogens with thymine (T)
- Cytosine (C ) forms three hydrogen bonds with guanine (G)
DNA linking is shown below:

RNA linking is shown below:

 Differences between DNA & RNA:
DNA RNA
DNA is double stranded RNA is single stranded
It has Deoxy ribose sugar It has Ribose sugar
Bases are Adenine, Guanine, Cytosine & Bases are Adenine, Guanine, Cytosine &
Thymine Uracil
Adenine Pairs with Thymine Adenine pairs with Uracil
No of Purines are same as pyrimidine Bases No of Purines are not same as pyrimidine
bases
DNA is genetic material in all living beings RNA is genetic material in some viruses
DNA is specific for a given organism Three types of RNA are present in an
organism- m-RNA, r-RNA & t-RNA
Length of DNA is quite large consisting of Length of RNA is short consisting of only
millions of nucleotides few thousands of nucleotides
Complementary Nucleotides are present It forms hair pin loop like structure which is
throughout the length of DNA helical which can have few complementary
nucleotides
Found in nucleus, chloroplast and Found in nucleus, cytoplasm and ribosome
mitochondria

There are three major classes of RNA, each with specific functions in protein synthesis
(i) mRNA
- Messenger RNA is produced by DNA; the process is called transcription.
- Messenger RNA encodes the amino acid sequence of a protein in their nucleotide base sequence.
- A triplet of nitrogenous bases specifying an amino acid in mRNA is called codon.
(ii) tRNA
- tRNA is also known as soluble RNA (sRNA) as it is soluble in 1 molar solution of sodium chloride.
- tRNA identifies amino acids in the cytoplasm and transports them to the ribosome.
- Molecules of tRNA are single-stranded and relatively very small.
- Anticodon is a three-base sequence in a tRNA molecule that forms complementary base pairs with
a codon of mRNA.
- All transfer RNA possesses the sequence CCA at their three ends; the amino acid is attached to the
terminal as residue.
(iii) rRNA
- Ribosomal RNA is found in ribosomes of cell and is also called insoluble RNA.
- The main function of rRNA is to attract provide large surface for spreading of mRNA over ribosome’s
during translocation process of protein synthesis.

DNA finger printing:

-It is used in identifying fire accident victims by comparing with the DNA of relation
-In forensic laboratories for identification of criminals
-To identify new species and compare with the existing ones.
-To determine the paternity of the individuals.
 Detection test for Nucleic Acids:
DISCHE Diphenyl Blue color if DNA
TEST Amine Green color if RNA

LIPIDS:
Lipids are the Biomolecules which are usually long hydrocarbons of fatty acids attached to Phosphate, sugar,
or acid groups.
They are highly soluble in non-polar solvents and insoluble in water.
Lipids are the store house of energy in the form of Oils, Fats or wax present in the plants and animals.

Types of Lipids:
- GlycoLipids = (Lipid + Sugar)
-PhosphoLipids = (Phosphate group + Fatty acid)
- Wax = (Long chain ester of alcohol & acid)
- Fats & Oils = (Glycerol + Long Chain Fatty acid)

PREVIOUS YEAR QUESTIONS:

1) Among the following compound which one is found in RNA? (JEE 2019)

ANS: (3)
SOL: Uracil is found in RNA
2) Which of the following statements is not true about RNA? (JEE
2019)
(1) It usually does not replicate (2) It is present in the nucleus of the cell
(3) It controls the synthesis of protein (4) It has always double stranded -helix
Structure

ANS: (4)

3) Which one of the following bases is not present in DNA? (JEE

2018)
(1) Cytosine (2) Thymine (3) Quinoline (4) Adenine
ANS: (3)
SOL: Adenine, Thymine, Cytosine, Guanine are bases present in DNA.
Quinoline an aromatic compound is NOT present in DNA.
 VITAMINS
A vitamin is an organic compound and an essential nutrient in the diet in limited quantities. Vitamins are key
to defence mechanism in the body deficiency of which we are succumbed to diseases
Vitamins are classified in to two groups
Fat Soluble Vitamins: These are absorbed by dietary Fat and are stored in adipose tissue and liver
Ex: Vitamins A, D, E & K
Water soluble vitamins: These are dissolved easily in water and readily excreted through kidneys
Ex: Complexes of Vitamin B & C

DEFICIENCY
VITAMIN SOURCE DISEASES
Fat Soluble: Milk, Fish liver Oil, Night blindness/
Vitamin-A (Retinol) Carrots & Butter Xerophthalmia
Vitamin-D (Calciferol) Exposure to sunlight, Fish Rickets/Softness of Bones
& egg yolk
Vitamin-E (Tocopherol) Vegetable oils Fragility/Haemolysis of
RBC
Vitamin-K (Mendione) Green leafy vegetables Failure of blood clotting
Water Soluble: Citrus fruits, amla & Scurvy/ Lower body
Vitamin-C (Ascorbic Acid) green leafy vegetables resistance to infection
Vitamin-B Complex Yeast, milk, cereals, green Beri-Beri / heart Failure
 Vitamin-B1 (Thiamine) vegetables
Vitamin-B2 (Riboflavin) Milk, egg white, liver, Anaemia/Inflammation in
kidney mouth
Vitamin-B6 (Pyridoxine) Yeast, milk, egg yolk, Skin disorders
cereals
Vitamin-B5 (Nicotinic Acid) Broccoli, mushrooms, Diarrhoea, Dermatitis&
Avocados & Salmon Dementia
Vitamin-B12 (Cobalamin) Meat, fish, egg & curd Pernicious Anaemia

1) Match (JEE 2020)

ANS: (b)