TS Inter 1st Year Botany Study Material Chapter 10 Biomolecules

Telangana TSBIE TS Inter 1st Year Botany Study Material 10th Lesson Biomolecules Textbook Questions and Answers.

TS Inter 1st Year Botany Study Material 10th Lesson Biomolecules

Very Short Answer Type Questions

Question 1.
Medicines are either man made (i.e., synthetic) or obtained from living organisms like plants, bacteria, animals etc. and hence the latter are called natural products. Sometimes natural products are chemically altered by man to reduce toxicity or side effects. Write against each of the following whether they were initially obtained as a natural product or as a synthetic chemical. [ Mar. ’20]
a. Penicillin b. Sulfonamide c. Vitamin C d. Growth Hormone
Answer:
a. Penicillin-Natural product
b. Sulfonamide-Synthetic chemical
c. Vitaminc-Natural product
d. Growth Hormone-Natural product

Question 2.
Select an appropriate chemical bond among ester bond, glycosidic bond, peptide bond and hydrogen bond and write against each of the following.
a. Polysaccharide
b. Protein
c. Fat d. Water
Answer:
a. Polysaccharide – Glycosidic bond
b. Protein – Peptide bond
c. Fat – Ester bond
d. Water – Hydrogen bond

Question 3.
Give one example for each of aminoacids, sugars, nucleotides and fatty acids. [Mar. ’13]
Answer:

1. Amino acid – Eg: Glycine
2. Sugars – Eg : Glucose
3. Nucleotide – Eg: Adenylic acid
4. Fatty acids – Eg : Palmitic acid

Question 4.
Explain the Zwitterionic form of an amino acid. [Mar. ’14]
Answer:
1) is a zwitterionic form a neutral form due to equal positive and negative charges.
2) Amino acid contains both acidic (carboxylic acid) and basic (amino group) centres and hence shows both positive and negative charge.

Question 5.
What constituents of DNA are linked by glylosidic bond?
Answer:

1. Nitrogen base and pentose (deoxy ribose) sugar, linked by glylosidic bond.
2. This bond is formed by dehydration.

Question 6.
Glycine and Alanine are different with respect to one substituent on the α – carbon. What are the other common substituent groups?
Answer:

1. – H and – CH₃ are substituent groups respectively in Glycine and Alanine – at α – carbon.
2. Both as them contain – H, – COOH and – NH₂ substituent groups in common.

Question 7.
Starch, Cellulose, Glycogen, Chitin are polysaccharides found among the following. Choose the one appropriate and write against each. [Mar. – 2018 m Mar. 17 – A.P & T.S ; Mar. ’15 – T.S]
a. Cotton fibre ………..
b. Exoskeleton of cockroach ……………
c. Liver ………………
d. Peeled potato …………..
Answer:
a. Cotton fibre – Cellulose
b. Exoskeleton of cockroach – Chitin
c. Liver – Glycogen
d. Peeled potato – Starch

Short Answer Questions

Question 1.
Explain briefly the metabolic basis for ‘living’.
Answer:

1. Metabolic pathways can lead to a more complex structure from a simpler structure. For example, formation of sucrose from water and CO₂ in mesophyll. They are called biosynthetic or anabolic pathways.
2. Some metabolic pathways may lead to a simpler structure from a complex structure. For example, glucose becomes lactic acid in our skeletal muscle. They are called as degradative or catabolic pathways.
3. Anabolic pathways, consume energy. For instance, assembly of a protein from amino acids requires energy input.
4. Catabolic pathways lead to the release of energy. For instance, the glycolytic pathway leading to the formation of lactic acid from glucose and releases energy. It consists of 10 metabolic steps.
5. Living organisms have learned to trap the energy liberated during degradation and store it in the form of chemical bonds. This stored bond is utilized as and when biosynthetic, osmotic and mechanical work is performed.
6. The most important form of energy currency in living systems is the bond energy in a chemical called Adenosine Triphosphate (ATP).

Question 2.
Is rubber a primary metabolite or a secondary metabolite? Write four sentences about rubber.
Answer:

1. Rubber is a secondary metabolite.
2. Metabolic products that do not have identifiable functions in the host organisms are called secondary metabolites.
3. Rubber is produced from latex of Hevea and Ficus elastica.
4. Latex is produced in special type of tissues called laticiferous tissues.
5. Tyres of vehicles are made from volcanization rubber.

Question 3.
Schematically represent primary, secondary and tertiary structures of a hypothetical polymer using protein as an example.
Answer:

Primary structure	Secondary structure	Tertiary structure
1. Proteins are made of amino acids which have carboxyl (- COOH) and amino (- NH2). The -COOH end of an amino acid is joined to – NH2 end of the other amino acid. Many amino are joined by peptide bonds which held them together in a particular sequence and constitute the primary structure of proteins. This structure does not make a protein functional.	1. Afunctional protein has 3 – dimentional configu­ration. It has one or more polypeptide chains. The sequence of amino acids determines where the chain will bend through and the formation of H-bonds peptide chains assume may be in the form of twisted helix or pleated sheet.	1. When individual peptide chains of secondary structure of protein are further coiled and folded into sphere like shapes with the H-bonds between NH2 and COOH groups. Various other kinds of bonds cross linking on chain to another. They form tertiary structure.
2. It is linear sequence of amino acids.	2. Have α helices and β- sheets held in place of amino acids.	2. Final folding and twisting of poly peptide.

Question 4.
Nucleic acid exhibits secondary structure, justify with example. [Mar. ’15 – T.S.]
Answer:

1. Nucleic acid exhibits a wide variety of secondary structures.
2. For example, one of the secondary structures exhibited by DNA is a famous Watson-Crick Model.
3. According to this model, DNA exists as a double helix. The two strands of polynucleotides are anti parallel i.e., run in the opposite direction.
4. The backbone is formed by the sugar – phosphate sugar chain.
5. The nitrogen bases are projected more or less perpendicular to the back bone but face inside. Adenine (A) and Guanine (G) of one strand pairs with Thymine (T) and Cytosine (C) respectively, on the other strand. Each step is represented by a pair.
6. Coiling occurs at an angle of 360°. At each step turn is 36°. One full turn of the helical strand involves 10 base pairs.
7. The length of each turn is 34A.
8. The distance between two steps is 3.4A.
9. This form of DNA with above features is called B – DNA.

Question 5.
Comment on the statement “living state is a non-equilibrium steady-state to be able to perform work.”
Answer:

1. A living organism consists of tens and thousands of chemical compounds called metabolites or biomolecules.
2. Biomolecules are present at concentrations characteristic of each of them. For examples, the blood concentration of glucose in a normal healthy individual to 4.5 to 5.0 mm, while that of hormones would be nanograms / ml.
3. All living organisms exist in a steady state characterized by concentrations of each of these biomolecules, that are in a metabolic flux.
4. Any chemical or physical process moves spontaneously to equilibrium. The steady state is a non-equilibrium state. As per the physics, the systems at equilibrium cannot perform work.
5. Living organisms work continuously, they can not afford to reach equilibrium. Hence the living state is a non-equilibrium steady – state to be able to perform work.
6. Living process is a constant effort to present falling into equilibrium, which is acheived by energy input.
7. Metabolism provides a mechanism for the production of energy. Hence the living state and metabolism are synonymous.
8. Thus, without metabolism there can not be a living state.

Question 6.
Dynamic state of body constituents is a more realistic concept than the fixed concentrations of body constituents at any point of time – Elaborate.
Answer:

1. Living organisms like simple bacterial cell, a protozoan, a plant or an animal contain thousands of organic compounds, the biomolecules.
2. The biomolecules are present in certain concentration, and ore expressed as mols/cell or mols/litre etc.,
3. All the biomolecules have a turn over. It means that they are constantly being changed into some other biomolecules and are also made from some other biomolecules. This breaking and making is through chemical reactions that are called metabolism.
4. Each of the metabolic reactions results in the transformation of biomolecules. For example, removal of CO₂ from amino acids making an amino acid into an amine.
5. Majority of these metabolic reactions do not occur in isolation, but are always linked to some other reactions. It means, metabolites are converted into each other in a series of linked reactions called metabolic pathways.
6. The metabolic pathways are either linear or circular, they may criss – cross each other. Flow of metabolites through the pathway has a definite rate and direction.
7. The metabolite flow is called the dynamic state of body constituents. Interlinked metabolic traffic is very smooth and without a single reported mishap for healthy conditions.
8. Every chemical reaction in a metabolic pathway is a catalysed reaction. There is no uncatalysed metabolic conversion in living systems.
9. Proteins with catalytic power are named as enzymes. They hasten the rate of a given metabolic conversion.

Long Answer Type Questions

Question 1.
What are secondary metabolites? Enlist them indicating their usefulness to man.
Answer:
Metabolic products that do not have identifiable functions in the host organism are called secondary metabolites. Secondary metabolites are alkaloides, flavonoides, rubber, essential oils, antibiotics, coloured pigments, scents, gums, spices etc. Many of these secondary metabolites are useful to man.

1. Pigments – Eg : Carotenoids, Anthocyanins etc.
2. Alkaloids – Eg : Morphine, Codeine etc.
3. Terpenoids – Eg: Monoterpenes, Diterpenes etc.
4. Essential oils – Eg : Lemon, grass oil etc.
5. Toxins – Eg : Abrin, Ricin
6. Drugs – Eg : Vinblastin, Curcumin etc.
7. Polymeric substances – Eg : Rubber, gums, cellulose etc.

Question 2.
What are the processes used to analyse elemental composition, organic constituents and inorganic constituents of living tissue? What are the inferences on the most abundant constituents of living tissue? Support the inferences with appropriate data.
Answer:
To analyse elemental composition, organic constituents and inorganic constituents of living tissue one has to perform chemical analysis.

Analyse Organic Compounds:

1. Take any living tissue (a vegetable or a piece of liver etc) and grind it in trichloro acetic acid (Cl₃C COOH) using a motor and a pestle.
2. We obtain a thick slurry.
3. If we were to strain this slurry through a cheese cloth or cotton, we would obtain two fractions. One is called the filtrate or acid soluble pool and the other is retentate or acid insoluble fraction.
4. To identify a particular compound, one has to use various separation technique and separate an organic compound from the rest.
5. Analytical techniques when applied to the compound gives us the molecular formula and probable structure of the compound.
6. All the carbon compounds that we get from living tissues can be called biomolecules.

Analyse inorganic compounds:

1. One weighs a small amount of a living tissue (say a leaf or liver and this is called wet weight) and dry it.
2. All the water evaporates.
3. The remaining material gives dry weight.
4. Now if the tissue is fully burnt, all the carbon compounds are oxidised to gaseous form and are removed.
5. The remaining is called ash.
6. This ash contains inorganic elements like calcium, magnesium etc.
7. Inorganic elements are also in acid solution fraction.

Elemental analysis:
Elemental analysis gives elemental composition of living tissues in the form of hydrogen, oxygen chlorine, carbon etc.

Analysis of compounds gives an idea of the kind of organic and inorganic compounds.

Question 3.
Nucleic acids exhibit secondary structure. Describe through Watson – Crick Model.
Answer:

1. Nucleic acid exhibits a wide variety of secondary structures.
2. For example, one of the secondary structures exhibited by DNA is a famous Watson-Crick Model.
3. According to this model, DNA exists as a double helix. The two strands of polynucleotides are anti parallel i.e., run in the opposite direction.
4. The backbone is formed by the sugar – phosphate sugar chain.
5. The nitrogen bases are projected more or less perpendicular to the back bone but face inside. Adenine (A) and Guanine (G) of one strand pairs with Thymine (T) and Cytosine (C) respectively, on the other strand. Each step is represented by a pair.
6. Coiling occurs at an angle of 360°. At each step turn is 36°. One full turn of the helical strand involves 10 base pairs.
7. The length of each turn is 34A.
8. The distance between two steps is 3.4A.
9. This form of DNA with above features is called B – DNA.

Question 4.
What is the difference between a nucleotide and nucleoside ? Give two examples of each with their structure.
Answer:

Nucleotide	Nucleoside
1. Nucleotide is made up of nitrogen base sugar and phosphoric acid.	1. Nucleoside is made up of nitrogen base and sugar.
2. Nucleotide of RNA is called ribonucleotide and nucleotide of DNA is called deoxyribo nucleotide.	2. Nucleoside with ribose sugar is called riboside of ribo-nucleoside. Nucleoside with deoxyribose sugar is called deoxyribonucleoside.
3. Example : Adenylic acid, guanylic acid, cytidyolic acid, thymidylic acid, uridylic acid, AMP	3. Example : Adeniosine, guanosine, cytidine, thymidine and uridine

Question 5.
Describe various forms of lipid using a few examples.
Answer:

• Lipids are water insoluble.
• Lipids could be simple fatty acids.
• A fatty acid has a carboxyl group attached to an R-group.
  The R – group could be methyl (-CH₃) or ethyl (- C₂H₅) or higher number of – CH₂ groups (1 carbon to 19 carbons). For example, palmitic acid has 16 carbons including carboxyl carbon.
• Arachidonic acid has 20 carbons including carboxyl carbon.
• Fatty acids could be saturated (without double bond) or unsaturated (with one or more C = C double bonds)
• Simple lipid is glycerol which is trihydroxy propane.
  
• Many lipids have both glycerol and fatty acids. Here the fatty acids are found esterified with glycerol. They are then called monoglycerides, diglycerides and triglycerides.
  
• These are also called fats and oils based on melting point.
• Oils have lower melting point (Eg. Gingely oil) and hence remain as oil in winters.
• Some lipids have phosphorous and a phosphorylated organic compound in them. These are called phospholipids. They are found in cell membrane. One example is Lecithin.
• Some tissues especially the neural tissues have lipids with more complex structures.
• If a phosphate group is also found esterified to the sugar, they are called nucleotides. Example of nucleotides are adenylic acid, thymidylic acid, guanylic acid, uridylic acid and cytidylic acid.

Intext Question Answers

Question 1.
What are macro molecules? Give examples.
Answer:
Macro molecules are large sized biomolecules that have high molecular weight, lower solubility and complex molecular structure. It occurs in collaidal state. Macro molecules are formed by polymerisation of large number of micro molecules. They belong to four classes of organic compounds – carbohydrates, lipids, proteins and nucleic acids.

Question 2.
Illustrate a glycosidic, peptide, and a phospho-diester bond.
Answer:
Glycosidic bond :
In a polysaccharide, the individual monosaccharides are linked by means of glycosidic bond. This bond is formed by dehydration. This bond is formed between two carbon atoms of two adjacent monosaccharides.

Peptide bond :
In a polypeptide or a protein amino acids are linked by peptide bonds. These bonds are formed by the reaction between carbocyl group (- COOH) of one amino acid with the amino group (-NH₂) of the next amino acid, with the elimination of water.

Phospho-diester bond :
In a nucleic acid a phosphate moiecty links the 3′ – carbon of one sugar of one nucleotide to the 5′ carbon of sugar of the succeeding nucleotide. The bond between the phosphate and hydroxyl group of sugar is an ester bond. As there is one such ester bond on either side, it is called phospho-diester bond.
5′ carbon end

Question 3.
What is meant by tertiary structure of proteins?
Answer:
When a long protein chain of secondary structure is folded upon itself like a hollow woolen ball, it give rise to tertiary structure.

Tertiary structure is necessary for many biological activities of protein.

This gives us a 3 – dimension view of a protein.

Question 4.
Find and write down structures of 10 interesting small molecular weight biomolecules. Find if there is any industry which manufactures the compounds by isolation. Find out who are the buyers.
Answer:

Question 5.
Proteins have primary structure. If you are given a method to know which amino acid is at either of the two termini (ends) of a protein, can you connect this information to purify or homogeneity of a protein?
Answer:
The primary sturcture of protein is based on the number type and order of amino acid present in the chain. A protein has a linear structure in which the left end of line represents the first and the right end represents the last amino acid. The number of amino acid in between the two ends determine the purity or homogeneity of proteins.

Question 6.
Find out and make a list of proteins used as therapeutic agents. Find other applications of proteins (Eg : Cosmetics etc.)
Answer:
Proteins used as therapeutic agents are thrombin, fibrinogen, enkephalins, antigens, antibodies, streptokinase, protein tyrosine kinase, diastase, renin, insulin, oxytocin, vasopressin, etc.

Other applications :
Proteins are also used in cosmetics, dairy industries, textile industries, research techniques etc.

Question 7.
Explain the composition of triglyceride.
Answer:
The components of triglyceride are single molecule of glycerol and 3 fatty acids. In glycerol 3 carbon atoms are present along with 30 n groups. Fatty acids consists of long chain hydrocarbon with a carboxylic group at one end. Both of them form ester bond. This bond is saturated when single bonded carbons are present and unsaturated when double bonded carbon atoms are present.

Question 8.
Can you describe what happens when milk is converted into curd or yoghurt, based on your Understanding of proteins?
Answer:
Milk is converted into curd or yughurt due to denaturation of proteins. The configuration of protein is lost. In denaturation disruption of bonds that maintains secondary and tertiary structure leads to the conversion of globular proteins into fibrous proteins. This involves a change in physical, chemical and biological properties of protein molecules.

Question 9.
Can you attempt building models of biomolecules using commercially availalble atomic models (Ball and Stick models)?
Answer:
Yes, models of biomolecules can be prepared using commercially available atomic models. Ball and stick models and space filling models are 3D models which serve to display the structure of chemical products and substances or biomolecules.

With ball and stick models, the centers of the atoms are connected by straight lines which represents covalent bonds. Double and triple bonds are often represented by springs.

The bond angles and bond lengths reflects the actual relationship. While the space occupied by the atoms is either not represented at all or only denoted essentially by the relative sizes of the sphere.

Question 10.
Attempt titrating an amino acid against a weak base and discover the number of dissociating (ionizable) functional groups in the amino acid.
Answer:
The existence of different ionic forms of amino acids can be easily understood by the titration curves. The number of dissociating functional group is one in case of neutral and basic amino acids and two in case of acidic amino acids.

Question 11.
Draw the structure of the amino acid, alanine.
Answer:

Question 12.
What are gums made of? Is Fevicol different?
Answer:
Gums are secondary metabolites. It is made up of compounds present in plant, fungi and microbial cells. Yes, Fevicol is different from gum. It is synthetic resin made by polymerisation manufactured by esterification of organic compounds.

Question 13.
Find out a qualitative test for proteins, fats and oils, amino acids and test any fruit juice, saliva, sweat and urine for them.
Answer:
Biuret test for protein :
The biuret test is a chemical test used for determining the presence of peptide bonds, in a positive test, a copper II ion (Cu²⁺ ion) is reduced to copper I (Cu⁺) which forms a complex with the nitrogen and carbon of peptide bonds in an alkaline solution. A violet colour indicates the presence of protein.

Ninhydrin test for amino acid :
Ninhydrin is a chemical used to detect ammonia or primary and secondary amines. When reacting with these free amines, a deep blue or purple colour known as Ruhemann’s purple is evolved. Most of the amino acids are hydrolyzed and reacted with ninhydrin except proline (a secondary amine).

Solubility test for fats and oils :
A positive solubility test for fats is that the fat dissolves in lighter fluid and not in water. In this test, 5 drops of fat or oi! are added in two test tubes containing 10 drops of lighter fluid and 10 drops of cold water respectively.

Fruit juice :
Fruit juice contains sugar. So it cannot be tested by the above mentioned test.

Saliva :
Saliva contains proteins, mineral salts, amylase etc. So.it can be tested for proteins and amino acids.

Sweat :
Sweat contains NaCl salts.

Urine :
Urine contains proteins. So it can be tested for proteins.

Question 14.
Find out how much cellulose is made by all the plants in the biosphere and compare it with how much of paper is manufactured by man and hence what is the consumption of plant material by man annually. What a loss of vegetation!
Answer:
About 100 billion tonnes of cellulose is prepared per year by the plants of the world. The increase in industrialization increased the use of paper. Due to this reason vegetation is being lost to a great extent.

Question 15.
All life forms exhibit “Unity in diversity” – Give reasons.
Answer:
There is a wide diversity of all living organisms but their chemical composition and metabolic reactions appear to be similar. The most abundant chemical in all life forms in water. Living organisms contain more carbon, hydrogen, and oxygen than animate matter.