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Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Zoology Guide Pdf Chapter 5 Digestion and Absorption Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Zoology Guide Chapter 5 Digestion and Absorption

11th Bio Zoology Guide Digestion and Absorption Text Book Back Questions and Answers

Part I

I. Choose The Best Options

Question 1.
Choose the incorrect sentence from the following:
a. Bile juice emulsifies the fat
b. Chyme is a digestive acidic food in stomach
c. Pancreatic juice converts lipid into fatty acid and glycerol
d. Enterokinase stimulates the secretion of pancreatic juice
Answer:
d. Enterokinase stimulates the secretion of pancreatic juice

Question 2.
What is chyme?
a. The process of conversion of fat into small droplets.
b. The process of conversion of micelles substances of glycerol into fatty droplet.
c. The process of preparation of incompletely digested acidic food through gastric juice.
d. The process of preparation of completely digested liquid food in midgut.
Answer:
d. The process of preparation of completely digested liquid food in midgut.

Question 3.
Which of the following hormones stimulate the production of pancreatic juice and bicarbonate?
a. Angiotensin and epinephrine
b. Gastrin ¿md insulin
c. Cholecystokinin and secretin
d. Insulin and glucagon
Answer:
c. Cholecystokinin and secretin

Question 4.
The sphincter of Oddi guards
a. Hepatopanci’eatic duct
b. Common bile duct
c. Pancreatic duct
d. Cystic duct
Answer:
a. Hepatopanci’eatic duct

Question 5.
In small intestine, active absorption occurs in case of
a. Glucose
b. Amino acids
c. Na⁺
d. All the above
Answer:
d. All the above

Question 6.
Which one is incorrectly matched?
a. Pepsin – stomach
b. Renin – liver
c. Trypsin – intestine
d. Ptyalin – mouth
Answer:
b. Renin – liver

Question 7.
Absorption of glycerol, fatty acid and monoglycerides takes place by
a. Lymph vessels within villi
b. Walls of stomach
c. Colon
d. Capillaries within villi
Answer:
a. Lymph vessels within villi

Question 8.
First step in digestion of fat is
a. Emulsification
b. Enzyme action
c. Absorption by lacteals
d. Storage in adipose tissue
Answer:
a. Emulsification

Question 9.
Enterokinase takes part in the conversion of
a. Pepsinogen into pepsin
b. Trypsinogen into trypsin
c. Protein into polypetide
d. Caseinogen into casein
Answer:
b. Trypsinogen into trypsin

Question 10.
Which of the following combinations are not matched?

Answer:
a. Bilirubin and biliverdin – (i) Intestinal juice

Question 11.
Match column I with column II and choose the correct option

a. (P-iv) (Q-iii) (R-i) (S-ii)
b. (P- iii) (Q-ii) (R-i) (S-iv)
c. (P-iv) (Q-iii) (R-ii) (S-i)
d. (P-ii) (Q-iv) (R-iii) (S-i)
Answer:
a. (P-iv) (Q-iii) (R-i) (S-ii)

Question 12.
Match column I with column II and choose the correct option

a. (P-iv) (Q- ii) (R- i) (S-iii)
b. (P- ii) (Q- iv) (R- i) (S-iii)
c. (P-i) (Q-iii) (R-ii) (S-iv)
d. (P-iii) (Q-i) (R-ii) (S-iv)
Answer:
b. (P- ii) (Q- iv) (R- i) (S-iii)

Question 13.
Match column I with column II and choose the correct option

a. (P-iv) (Q-ii) (R-i) (S- iii)
b. (P- iii) (Q- iv) (R- ii) (S- i)
c. (P- iv) (Q- iii) (R-ii) (S- i)
d. (P- iii) (Q- ii) (R- iv) (S- i)
Answer:
c. (P- iv) (Q- iii) (R-ii) (S- i)

Question 14.
Which of the following is not the function of the liver?
a. Production of insulin
b. Detoxification
c. Storage of glucogen
d. Production of bile
Answer:
a. Production of insulin

Question 15.
Assertion (A): Large intestine also shows the presence of a villi-like small intestine.
Reason (B): Absorption of water takes place in the large intestine
a. Both A and B are true and B is the correct explanation of A
b. Both A and B are true but B is not the correct explanation of A
c. A is true but B is false
d. A is false but B is true
Answer:
a. Both A and B are true and B is the correct explanation of A

Question 16.
Which of the following is not true regarding intestinal villi?
a. They possess microvilli
b. They increase the surface area
c. They are supplied with capillaries and the lacteal vessels
d. They only participate in the digestion of fats
Answer:
d. They only participate in the digestion of fats

Question 17.
Which of the following combinations are not matched?
a. Vitamin D – Rickets
b. Thiamine – Beriberi
c. Vitamin K – Sterlity
d. Miacin – Pellagea.
Answer:
c. Vitamin K – Sterility

Question 18.
Why are villi present in the intestine and not in the stomach?
Answer:
In the small intestine, digestion gets completed and the absorption of digested food materials like glucose, amino acids, fatty acids, and glycerol takes place. The food materials are to be retained in the intestine by increasing the surface area. Hence villi are present in the intestine. The stomach is the temporary storing organ of food. In the stomach, HCl, pepsin, renin, and lipase are secreted. These are concerned with digestion. Hence villi are not present in the stomach.

Question 19.
Bile juice contains no digestive enzymes. yet it is important for digestion. Why?
Answer:

• The pile contains bile pigments (Bilirubin and biliverdin)
• The pile pigments are broken down products of heamoglobin of dead RBC’s
• Bile salts, cholesterol, and phospho lipids.
• Bile has no enzyme.
• Bile helps in the emulsification of fats.
• Bile salts reduce the surface tension of fat droplets and break them into small globules.
• Bile also activates lipase to digest lipids.

Question 20.
List the chemical changes that starch molecule undergoes from the time it reaches the small intestine.
Answer:

Question 21.
How do proteins differ from fats in their energy value and their role in the body?
Answer:

Question 22.
Digestive secretions are secreted only when needed discuss.
Answer:

• The saliva is secreted by the salivary gland in the mouth Saliva
• The saliva contains water.
• Electrolytes – Na⁺, K⁺ , Cl^–, HCo₃^–
• Salivary amylase (ptyalin)
• Mucus (a glycoprotein)
• Polysaccharides, starch is hydrolyzed by the salivary amylase enzyme into disaccharides (maltose)

Stomach

• The gastric j uice contains HCI and proenzymes
• Proenzyme pepsinogen on exposure to HCI gets converted into active enzyme pepsin.

• The HCI provides an acidic medium (pH=1.8) which is optimum for pepsin, kills bacteria and other harmful organisms and avoids putrification.
• Proteolytic enzyme found in gastric juice of Infants is rennin helps in the digestion of milk protein caseinogen to casein in the presence of calcium.

Small Intestine

Pancreatic juice:
Enzymes: Trypsinogen, Chymotrypsinogen, Carboxypeptidases, Pancreatic, Amalyse, Pancreatic Lipase, and Nucleases.

Trypsinogen is activated by an enzyme enterokinase, secreted by the intestinal mucosa into active trypsin, which in turn activates the enzyme chymotrypsinogen in the pancreatic juice.

Bile Juice:
The bile contains bile pigment (Bilirubin, and biliverdin) as the breakdown product of heamoglobin of dead RBCs, Bile salts, Cholesterol, and phospholipids. But has no enzymes. Bile helps in the emulsification of fats. Bile salts reduce the surface tension of fat droplets and break them into small globules, bile also activates lipases to digest lipids.

Pancreatic juice action:
Trypsin hydrolyses protein into polypeptides and peptones. While chymotrypsin hydrolyses peptide bonds associated with specific amino acids.

Succus enterius:
The secretions of the Brunner’s gland along with the secretions of the intestinal glands constitute the intestinal juice or succus entericus.
Enzymes: Maltase, lactase, sucrase (invertase), dipeptidases, lipases, nucleosidases.

Bicarbonate ions from the Pancreas provide an alkaline medium (pH=7.8) for the enzymatic action.
All macromolecules → Micromolecules
Carbohydrate → Monosaccharides
Protein → Aminoacid
Lipids → Fatty acids and Glycerol

Question 23.
Label the given diagram.
Answer:

A – Right hepatic duct of the liver
B – Common bile duct
C – Pancreatic duct (duct of wirsung)
D – Hepatopancreatic duct
E – Cystic duct

Part II

11th Bio Zoology Guide Digestion and Absorption Additional Important Questions and Answers

I. Choose The Best Option.

Question 1.
……………………… litres of digestive juice is poured into the alimentary canal and are reabsorbed each day.
a) 6 – 7 lit
b) 0 – 7 lit
c) 5-7 lit
d) 3 – 7 lit
Answer:
b) 0 – 7 lit

Question 2.
The Hard chewing surface of the teeth is made of ………………………… and helps in mastication of food.
a) Enamel
b) Crown
c) Denton
d) Plague
Answer:
a) Enamel

Question 3.
Which is the correct statement?
a) Tongue is a freely movable muscular organ attached at the anterior end by the frenulum to the floor of the buccal cavity.
b) Tongue is a freely movable muscular organ not attached at the posterior end by the frenulum to the floor of the buccal cavity.
c) Tongue is a freely movable muscular organ attached at the posterior end by the frenulum to the floor of the buccal cavity.
d) Tongue is a freely movable muscular organ attached at the anterior, posterior end by the frenulum to the floor of the buccal cavity.
Answer:
c) Tongue is a freely movable muscular organ attached at the posterior end by the frenulum to the floor of the buccal cavity.

Question 4.
Which is the correct sequence?
a) Gullet → Glottis → Epiglottis
b) Epiglottis → Glottis → Gullet
c) Glottis → Gullet → Epiglottis
d) Gullet → Epiglottis → Glottis
Answer:
a) Gullet → Glottis → Epiglottis

Question 5.
Which is a false statement?
a) Stomach divided into three region
b) Cardiac, fundic, pyloric regions
c) Pyloric region found between duodenum and jejunum
d) Cardiac region has a sphincter
Answer:
c) Pyloric region found between duodenum and jejunum

Question 6.
Find out the incorrect pair.
a) Starch – Amylase
b) Protein – Pepsin
c) Casein – Trypsin
d) Lipid – Lipase
Answer:
c) Casein – Trypsin

Question 7.
Find out the correct pair
a) Duodenum – 25 m
b) Jejunum – 2.4 m
c) Ileum – 3.7 m
d) Oesophagus – 10 m
Answer:
b) Jejunum – 2.4 m

Question 8.
Which is a wrong statement
a) Brunner’s gland doesn’t secrete mucus and enzymes
b) Brunner’s gland secretes mucus and enzymes
c) It is found in duodenum
d) It is theopenningofcaecum
Answer:
a) Brunner’s gland doesn’t secrete mucus and enzymes

Question 9.
Where is crypts of Leiberkuhn seen?
a) Small Intestine
b) Oesophages
c) Stomach
d) Rectum
Answer:
a) Small Intestine

Question 10.
The anal column may get enlarged and causes
a) Haemoralds
b) Haemorhoids
c) Elaemorods
d) Elaemorals
Answer:
b) Haemorhoids

Question 11.
Find the correct statement
a) Serosa – The outer layer formed of connective tissue
b) Serosa – Connective tissue, epithelial tissue
c) Serosa – Connective tissue, striated cells
d) Serosa – Connective tissue, thin squanmous epithelium
Answer:
d) Serosa – Connective tissue, thin squanmous epithelium

Question 12.
Match the following
1. Parotid gland – i) Pepsin
2. Sub maxillary gland – ii) Stenson’s duct
3. Sublingual gland – iii) Wharton’s duct
4. Stomach – iv) Duct of Rivinis
a) (1-ii) (2-iii) (3-iv) (4-i)
b) (1-i) (2-ii) (3-iii) (4-iv)
c) (1-ii) (2-iii) (3-i) (4-iv)
d) (1-iii) (2-ii) (3-iv) (4-i)
Answer:
a) (1-ii) (2-iii) (3-iv) (4-i)

Question 13.
Where is castle intrinsic factor secreted?
a) Intestine
b) Digestive passage
c) Stomach
d) Large intestine
Answer:
c) Stomach

Question 14.
The hepatic lobules are covered by …………………….. a thin connective tissue sheath.
a) Glisson’s capsule
b) Cardiac membrane
c) Renal membrane
d) Cystic membrane
Answer:
a) Glisson’s capsule

Question 15.
Find the correct statement. The differentiation of Liver
a) 4-5 week
b) 3-4 week
c) 4-7 week
d) 12-3 week
Answer:
b) 3-4 week

Question 16.
Find the wrong statement
a) Saliva – Ptyalin
b) Digestive tract – Mucous membrane
c) Stomach – Pepsin
d) Small intestine – Glucokinase
Answer:
d) Small intestine – Glucokinase

Question 17.
What is pH of food at the time of absorption?
a) 7.3
b) 7.5
c) 7.8
d) 7.7
Answer:
c) 7.8

Question 18.
True or false
a) Carbohydrate – Glucose
b) Protein – Aminoacid
c) Fat – Fatty acid
d) Bile – Pepsin

Answer:-
a) True
b) True
c) False
d) False

Question 19.
Find x-part the diagram

a) Common bile duct
b) Pancreatic duct
c) Jejunum
d) Gall bladder
Answer:
b) Pancreatic duct

Question 20.
Find the correct pair.
a) Carbohydrate -50% -400-500 gm
b) Fat-15%65-75 gm
c) Carbohydrate -51 % 400 -450 gm
d) Fat-15% -70-75 gm
Answer:
a) Carbohydrate -50% -400-500 gm

Question 21.
Name the vitamin synthesized by bacteria of the large intestine
a) D
b) K
c) C
d) E
Answer:
b) K

Question 22.
Find the correct statement.
a) Unused protein – stored in the liver
b) Unused protein – stored in the muscle
c) Unused protein – excretes as nitrogen
d) Unused protein – excretes through faeces
Answer:
c) Unused protein – excretes as nitrogen

Question 23.
What is the nature of food in the stomach?
a) Chyme
b) Fermented
c) Solid
d) Semisolid
answer:
a) Chyme

Question 24.
Which is the longest part of the digestive system?
a) Large intestine
b) Small intestine
c) Oesophages
d) Stomach
Answer:
b) Small intestine

Question 25.
Where is pyloric muscle present?
a) Junction between oesophagus and stomach
b) Junction between the large intestine and small intestine
c) Junction between small intestine and stomach
d) Junction between large intestine and rectum
Answer:
c) Junction between small intestine and stomach

Question 26.
How much protein is needed for a day?
a) 1 gm per kg
b) 2 gm per kg
c) 1.5gmperkg
d) 2.5gmperkg
Answer:
a) 1 gm per kg

(2 marks)

II. Very Short Questions

Question 1.
What are the uses of food?
Answer:
The food we eat provides energy and organic substances for growth and the replacement of worn-out and damaged tissues. It regulates and coordinates the various activities that take place in the body.

Question 2.
What are the special features that help in absorbing digested food?
Answer:

• There is an increase in the small intestine surface area.
• The villi are present in the inner walls of the intestine.
• The villi is the absorbtive unit
• The microvilli present in the villi increase the absorptive surface.

Question 3.
Why do we need a digestive system?
Answer:
The food that we eat is macromolecules, and inabsorbable. These are to be broken down into smaller macro-molecules in absorbable forms. This is done by the digestive system.

Question 4.
How is fat and other nutrients of bile helped in digestion?
Answer:
It helps in emulsifying fat. The bile salt decreases the surface tension of fat molecules and converting it to chilo micron.

Question 5.
What is the function of the digestive system?
Answer:
The function of the digestive system is to bring the nutrients, water, and electrolytes from the external environment into every cell in the body through the circulatory system.

Question 6.
What happens when there is no secretion of HCI in the stomach?
Answer:

• The HCI in the stomach coverts the inactivated pepsinogen into active pepsin.
• The activated pepsin acts on protein and converts them into proteases and peptones
• HCI provides an acidic medium which is optimum for pepsin action.

Question 7.
List out the processes starting from the ingestion of protein and storning in the muscle cells and converting them in to the parts of cytoplasm?
Answer:
Stomach:
The gastric juice contains pepsin. This is the first enzyme that works on protein.

• Rennin is present in the gastric juice of infants
• It helps in the digestion of caesinogen and converts into casein.

Pancreas:

• Trypsin hydrolyses proteins in to polypeptides and peptones.
• Chymotrypsin hydrolyses peptide bonds associated with specific amino acids.

Succus Entricus
The peptidases present in the intestinal juice convert the di and polypeptides to amino acids.

The end product of digestion the amino acids that are absorbed by the villi and reach the blood.

Question 8.
What is diphyodont dentition?
Answer:
Human beings and many mammals form two sets of teeth during their lifetime, a set of 20 temporary milk teeth which gets replaced by a set of 32 permanent teeth. This type of dentition is called diphyodont dentition.

Question 9.
Why the food prepared in the house is better than the food which is prepared by causing preservative and artificial enhancers?
Answer:
The food prepared by using artificial enhancers and preservatives creates so many diseases.
Diseases

• Heart problems
• Hypertension
• Sterility
• Stomach disorders
• Attainment of early puberty in girl children.

Question 10.
What is known as the dental formula of human beings?
Answer:
The arrangement of teeth in each half of the upper and lower jaw in the order of I, C, P, and M can be represented by the dental formula. The dental formula of man is 2123 / 2123.

Question 11.
What are the steps to be taken to care for our alimentary tract?
Answer:

• We have to take healthy foods.
• We have to take plenty of water.
• We have to regulate our stress.
• We have to take probiotics daily.
• We have to do exercise daily.

Question 12.
What are the functions of soluble and insoluble fibres?
Answer:
The food contain two types of fibres.
Soluble fibre: It soaks up toxins and waste in the digestive system.
Insoluble fibre: Roughage. It moves bulk through the intestine to help with regular bowel movements.
This upper surface of the tongue has small projections called Papillae.

Question 13.
What is the function of the tongue?
Answer:
The tongue helps in the intake of food, chew and mix food with saliva, swallow food, and also speak. The upper surface of the tongue has small projections called papillae with taste buds.

Question 14.
What are Oesophages?
Answer:
Oesophages connect the buccal cavity and stomach.

Question 15.
What is gastro oesophagus reflux disorder?
Answer:
If the cardiac sphincter does not contract properly during the churning action of the stomach the gastric juice with acid may flow back into the oesophagus and cause heart bum, resulting in GERD (Gastro Oesophagus Reflex Disorder).

Question 16.
How larger food molecules are converted into small molecules?
Answer:

Question 17.
What are gastric rugae?
Answer:
The inner wall of the stomach has many folds called gastric rugae which unfold to accommodate a large meal.

Question 18.
What is meant by colities?
Answer:

• The bacterial infection may cause inflammation of the innerlining of colon called colitis.
• The most common symptoms of colitis are rectal bleeding abdominal cramps and diarrhoea.

Question 19.
What is indigestion?
Answer:

• It is a digestive disorder in which the food is not properly digested leading to a feeling of fullness of stomach.
• It may be due to in adequate enzyme secretion anxiety food poisoning overeating and spicy food.

Question 20.
Give notes on vomiting?
Answer:
It is reverse peristalsis. Harmful substances are ejected through the mouth. This action is controlled by the vomit centre located in the medulla oblongate a feeling of nausea precedes vomiting.

Question 21.
What is meant by digestion? What is the different processes of digestion?
Answer:
The break down of the macromolecules of food in to the micro molecules of food is known as digestion.
Stages

1. Ingestion
2. Digestion
3. Absorption
4. Assimilation
5. Elimination of undigested substances digestion

Question 22.
What is Frenulum?
answer:
The tongue is attached at the posterior end to the floor of the buccal cavity by the structure frenulum and the tongue is free in the front.

Question 23.
What is meant by GERD – GASTERO oesophagus reflex disorder?
Answer:

• There are two sphincter muscles namely cardiac sphincter and pyloric sphincter present in the stomach.
• If the sphineter does not contract properly during the churning action of the stomach of the gastric juice with acid may flow back in to the oesophagus and cause heart bum resulting in GERD.

Question 24.
How is piles or haemorrhoides formed?
Answer:

• The anal mucosa is folded into several vertical folds contains arteries and veins called anal columns.
• if these anal columns get enlarged and causes piles or haemorrhoides.

Question 25.
Name the enzyme which converts the inactivated enzymes into the active enzyme.
Answer:
1. Enterokinase:
It converts the inactivated Trypsinogen in to Trypsin.

2. Trypsin:
The inactive chymotrypsinogen is converted into chymotrypsin

Question 26.
What are the food components needed for a person for healthy living?
Answer:

• Carbohydrates
• Proteins
• Lipids
• Vitamin
• Minerals
• Fibre
• Water

Question 27.
Define Thecodont?
Answer:
Each tooth is embedded in a socket in the jaw bone; this type of attachment is called thecodont.

Question 28.
What is meant by assimilation?
answer:
All the body tissues utilize the absorbed substances for their activities and incorporate in to their protoplasm this process is called assimilation.

Question 29.
Define Plaque.
Answer:
Minerals salts like Calcium and Magnesium are deposited on the teeth and form a hard layer of tartar or calculus called plaque.

Question 30.
What is Papillae?
Answer:
This upper surface of the tongue has small projections called Papillae.

Question 31.
What are the parts of Stomach?
Answer:

• A cardiac portion
• A fundic portion
• A pyloric portion

Question 32.
What is the portion of small intestine?
Answer:
Duod enum – 25Cm
Jejunum-2.4m
Ileum-3.5m

Question 33.
What is Gastric rugae?
Answer:
The inner wall of the stomach has much folds called gastric rugae which unfolds to accommodate a large meal.

Question 34.
What are the parts of large Intestine?
Answer:

• Caecum
• Colon
• Rectum

Question 35.
What are the regions of colon.
Answer:
The colon is divided into four region.

1. An ascending region
2. A Transverse region
3. A Descending region
4. A Sigmoid region

Question 36.
What are the layers found in the alimentary canal?
Answer:

• Serosa
• Muscularis
• Sub – mucosa
• Mucosa

Question 37.
What are the elements found in Saliva?
Answer:

• Water
• Electrolytes (Na⁺, K⁺, Cl^–, HCO₃^–₎
• Salivary Amylase (Ptyalin)
• Anti bacterial agent Lysozyme
• Lubricating agent mucus (glycoprotein).

Question 38.
What are the components present in bile?
Answer:

• Bilirubin
• Biliverdin
• Bile Salts
• Cholesterol
• Phospholipids

Question 39.
Name the gastric juices found in the stomach.
Answer:

• Hydrochloric acid (PH 1.8)
• Proenzyme – Pepsinogen
• Pepsin Rennin

Question 40.
What is the function of Pyloric Sphincter?
Answer:

• The opening of the stomach into the duodenum is guarded by the Pyloric Sphincter.
• It periodically allows partially digested food to enter the duodenum and also prevents regurgitation of food.

Question 41.
What is the Calorific value of carbohydrates?
Answer:

• The caloric value of Carbohydrates is 4.1 calories /gram.
• The physiological fuel value is 4 Kcal / gram.

Question 42.
A person is suffering from a digestion problem. What may be the reason?
Answer:
This person may be suffering from constipation.
Constipation:
The faeces are retained within the rectum, because of irregular bowel movement due to poor intake of fibre in the diet and lack of physical activities.

Question 43.
What is oral hydration therapy?
Answer:
If there is more loss of water due to diarrhea dehydration may occur. Treatment is known as oral hydration therapy.
This involves drinking plenty of fluids sipping small amounts of water at a time interval to rehydrate the body.

Question 44.
Define Obesity.
Answer:
It is caused due to the storage of excess of body fat in adipose tissue.
It may induce hypertension, atherosclerotic heart disease and diabetes.

Question 45.
What is BMI Calculation?
Answer:
BMI is calculated as body weight in Kg, divided by the square of height in meter.
\(\mathrm{BMI}=\frac{\text { Body Weight in } \mathrm{Kg}}{\text { (Body Height) }^{2} \text { in meter }}\)
For example :
A person Weight = 50 Kg
Height = 1.6m
\(=50 / 1.6^{2}\)
BMI = 19.5

(3 marks)

III. Short Questions

Question 1.
Define Gingivitis?
Answer:
The plaque formed on teeth is not removed regularly, it would spread down the tooth into the narrow gap between the gums and enamel and cause inflammation, called gingivitis.
Symptoms;
It leads to redness and bleeding of gums and leads to bad smell.

Question 2.
What is Heterodont?
Answer:
The permanent teeth are of four different types (heterodont).
Incisors – Chisel like cutting teeth
Caniues – Dogger shaped tearing teeth
Premolar -Grinding Molar – Grinding and Crushing
\(\frac{2123}{2123} \times 2=\frac{16}{16}\)
Upper Jaw – 16 teeth
Lower Jaw – 16 teeth

Question 3.
What are the signifance of Liver?
Answer:
1. Destroy aging and defective bloodcells.
2. Stored glucose in the form of glycogen or disperses glucose into the blood stream with the help of pancreatic hormones.
3. Stores fat soluble vitamins and iron.
4. Detoxifies toxic substances.
5. Involves in the synthesis of non – essential aminoacids and urea.

Question 4.
Explain about the protein deficiency disease.
Answer:
Protein Energy Malnutrition (PEM):

• Marasmus
• Kwarshiorkor Marasmus:
• Children are suffering from diarrhoea Body becomes lean and weak
• Reduced fat and muscle tissue with thin and folded skin.

Kwashiorkor:

• Dry skin
• Potbelly
• Edema in the legs and face
• Stunted growth
• Changes in hair colour
• Weakness and irritability

Question 5.
Name the digestive secretions.
Answer:

• Salivary glands
• Bile juice
• Pancreatic juice Gastric juice Small Intestinal juice.

Question 6.
What are the types of Salivary glands and their ducts?
Answer:

Question 7.
What are the cells of gastric gland and their Secretions?
Answer:

Question 8.
Draw and label the layers of the alimentary canal.
Answer:

A – Microvilli
B – Circular muscle
C – Mucous
D – Muscular layer

Question 9.
Explain the protein deficiency diseases.
Answer:
Growing children require more amount of protein for their growth and development. Protein deficient diet during the early stage of children may lead to protein-energy malnutrition such as Marasmus and Kwashiorkor. Symptoms are dry skin, pot-belly, oedema in the legs and face, stunted growth, changes in hair colour, weakness, and irritability.

Marasmus is an acute form of protein malnutrition. This condition is due to a diet with inadequate carbohydrates and protein. Such children are suffering from diarrhea, the body becomes lean and weak (emaciated) with reduced-fat and muscle tissue with thin and folded skin.

Question 10.
What are the ill effects of adulteration of food?
Answer:

• Food adulteration causes harmful effects in the form of head ache palpitations allergies, cancers.
• It reduces the food quality common adulteration are addict onto citric acid to lemon juice.
• Papaya seeds to pepper melamine to milk.

Question 11.
A person has diet control in particular time, he takes large amount of rice, curd, buttermilk and onion why? and write about it?
Answer:
Yes the person is suffering from Jaundice.
Jaundice:

• It is the condition in which liver is affected and the defective liver fails to break down haemoglobulin and to remove bile pigments from the blood.
• Deposition of these pigments changes the colour of eyes and skin yellow.
• Jaundice is caused due to hepatitis Viral Infection

Question 12.
What is the effects of crystallized cholesterol?
Answer:
The effects of crystallized cholesterol is Gall Stones.

Gall Stones:

• Any alteration in the composition of the bile can cause the formation of stones in the gall bladder.
• The stones are mostly formed of crystallized cholesterol in the bile.
• The gall stone causes obstruction in the cystic duct, hepatic duct and also hepatopancreatic duct, causing pain, Jaundice and pancreatitis.

Question 13.
What is indigestion?
Answer:
It is a digestive disorder in which the food is not properly digested leading to a feeling of fullness of stomach. It may be due to inadequate enzyme secretion, anxiety, food poisoning, overeating and spicy food.

Question 14.
Writes notes on Peptic Ulcer.
Answer:

• It refers to an eroded area of the tissue lining (Mucosa) in the stomach or duodenum.
• A duodenal ulcer occurs in people in the age group of 25-45 years.
• Gastric ulcer is more common in person above the age of 5üyears.
• Ulcer mostly due to infections caused by the bacterium Helicobacter pylon.
• It may be due to uncontrolled usage of aspirin or certain anti-inflammatory drugs.
• It is caused due to smoking, alcohol, caffeine, and psychological stress.

Question 15.
What is a hiatus hernia or diaphragmatic hernia?
Answer:
It is a structural abnormality in which the superior part of the stomach protrudes slightly above the diaphragm. The exact cause of hiatus hernias is not known. In some people, injury or other damage may weaken muscle tissue, by applying too much pressure (repeatedly) on the muscles around the stomach while coughing, vomiting, and straining during bowel movement and lifting heavy objects.

Question 16.
Give notes on the stomach.
Answer:

• Stomach functions as the temporary storage organ for food.
• It consists of three parts cardiac fundic and pyloric stomach.
• The oesophagus opens into a cardiac stomach and guarded by cardiac sphincter.
• The pyloric stomach opens into duodenum and is guarded by the pyloric sphincter.

• It allows partially digested food to enter the duodenum and prevents regurgitation of food.
• The inner walls of stomach has many folds which unfolds to accommodate a large meal.

Question 17.
Give short notes on intestinal villi?
Answer:

• The ileal mucosa has numerous vascular projections called villi which are involved in the process of absorption.
• The cells lining the villi produce numerous microscopic projections called microvilli giving a brush border appearance and increase the surface area enormously.
• Along with villi the clear mucosa contain mucous secreting goblet cell and peyer patches which produce lymphocytes.
• The wall of the small intestine bears crypts between the base of villi called crypts of leiberkuhn.

Question 18.
Write a paragraph on peptic ulcers.
Answer:
It refers to an eroded area of the tissue lining (mucosa) in the stomach or duodenum. A duodenal ulcer occurs in people in the age group of 25 – 45 years. Gastric ulcer is more common in persons above the age of 50 years. An ulcer is mostly due to infections caused by the bacterium Helicobacter pylori. It may also be caused due to uncontrolled usage of aspirin or certain anti-inflammatory drugs. An ulcer may also be caused due to smoking, alcohol, caffeine, and psychological stress.

Question 19.
Give an account of the Pancreas.
Answer:

• The Pancreatic is the second-largest gland in the digestive system which is a yellow coloured compound organ.
• It consists of exocrine and endocrine cells.
• It is situated between the limbs of the ‘U’ shaped duodenum.
• The exocrine portion secretes trypsin, pancreatic lipase, amylase.
• The islets of Langerhans cells of the pancreas secrete insulin and glucogen hormone.

Question 20.
Name the alimentary canal parts and the absorptive substance.
Answer:

(5 marks)

IV. Essay Questions

Question 1.
Describe the structure of the large intestine with a diagram.
Answer:

1. The Caecum:

• It is a small blind pouch-like structure that opens into the colon and it possesses a narrow finger-like tubular projection called vermi form appendix.
• Caecumand vermiform appendix is large in herbivorous animals and act as an important site for cellulose digestion with the help of symbiotic bacteria.

2. The Colon:
The colon is divided in to four regions an ascending transverse a descending part and a sigmoid Colon. The Colon is lined by dilations called haustra.

3. Sigmoid Colon:

• ‘S’ shaped sigmoid colon opensinto the rectum.
• The anus is guarded by two anal sphincter muscles. The anal mucosa is folded in to several vertical folds and contains arteries and veins called anal column.
• Anal colomn may get enlarged and causes piles.

Question 2.
Describe the structure of liver with a diagram.
Answer:

• The liver is the largest gland in our body.
• It is situated below the diaphragm.
• The liver consists of two major left and right lobes and two minor lobes.
• Each lobe has many hepatic lobules called a functional unit of liver and is covered by a thin connective sheath called Glissons capsule.
• Liver cells secrete and is stored in gall bladder. The duct of gall bladder and the hepatic duct form the common bile duct.
• The bile duct and the pancreatic duct joined to gether formed a common duct and opens into duodenum and is guarded by a sphincter of oddi.
• Liver has high power of regeneration and liver cells are replaced by new ones every 3-4 weeks.

Question 3.
Describe the process of digestion in the mouth.
Answer:

• The smell the sight and taste as well as the mechanical stimulation of food in the mouth trigger a reflex action that results in the secretion of saliva.
• The mechanical digestion starts in the mouth by grinding and chewing of good.
• The saliva contains water electrolytes like Na, K, C1, HCO3 salivary amylase or ptyalin antibacterial agent lysozyme and a lubrication agent mucus.
• The saliva moistening lubricating and adhering the masticated food into a bolus.
• The ptyalin in the saliva hydrolyzes 30% of the poly saccharide into disaccharides.
• The bolus is passed into the pharynx and then into the oesophagus by swallowing or deglutition.
• The bolus reaches the stomach by successive waves of muscular contraction called peristalsis.

Question 4.
Describe the process of digestion in the stomach
Answer:

• The secretion of gastric juice begins when the food is in the mouth.
• The gastric juice contains HCI pepsinogen renin etc.
• The HCI changes the pepsinogen into pepsin.

• Pepsin acts on protein and converts into proteoses and peptones.

• The HCI provides an acidic medium which is optimum for pepsin kills bacteria and other harmful organisms and avoids putrefaction.
• The mucous and bicarbonates protect the stomach from acidic HCl.
• The rennin converts the milk protein caesinogen to casein in the presence of calcium ions.

Question 5.
Describe the process of digestion in the small intestine.
Answer:

The bile pancreatic juice and intestinal juice the secretions released into the small intestine.

Bile:

• The bile contains bile pigments bilirubin and biliverdin as the break down products of haemoglobin of  dead RBCs bile salts.
• Bile helps in emulsification of fats Bile salts reduce the surface tension of the fat droplets and break them into small globules.
• Bile also activates lipases to digest lipids.

Pancreas:

• The pancreatic juice contains enzymes such as trypsinogen, Chymotrypsinogen.
• Trypsinogen is activated by an enzyme enterokinase into active trypsin.
• Trypsin activate the chymotrypsinogen into chymotrypsin.
• Trypsin hydrolyses protein into polypeptides and peptones.

Chymotrypsin hydrolyses peptide bonds associated with specific aminoacids.
The amylase converts glycogen and starch into maltose.
Lipase acts on tri glycerides and hydrolyes them into free fatty acid and mono glycerides.

Succtts entricus:

Question 6.
What is meant by absorption? How is digested food absorbed in the digestive system?
Answer:
Absorption is a process by which the end product of digestion passes through the intestinal mucosa in to the blood and lymph.
Process of absorption:

1. Simple Diffusion:
Small amounts of glucose amino acids and chloride ions are absorbed by simple diffusion.

2. Facilitated Transport:
Fructose arc absorbed with the help of the carrier ions like Na.

3. Active Transport:
Aminoacids, Glucose and Sodium are absorbed by active transport.

4. Passive Transport:
Fatty acid are absorbed by the Lacteals of Villi.

Question 7.
What is the caloric value of carbohydrates, proteins and fats?
Answer:
We obtain 50% of energy from carbohydrates 35 % from fats and 15 % from proteins.
We require about 400 – 500 gm of carbohydrates. 60 – 70 gm of fat 65 to 75 gm of proteins per day.

Carbohydrate:
The caloric valve of Carbohydrate is 4.1 calories gram and its physiological fuel value is 4 Kcal per gram.

Lipid:
Fat hasa caloric valve of 9.45 KCal and a physiological fuel value of 9 KCal per gram.

Protein:
The caloric and physiological fuel value of one gram of protein are 5.65 Kcal and 4 KCal respectively.

Question 8.
What is meant by Hiatus hernia or Oesophagus Hernia

• It is a structural abnormality in which superior part of the stomach protrudes slightly above the diaphragm.
• Applying toomuch pressure on the muscles around the stomach while coughing, vomiting and straining during bowel movement and lifting heavy object it may weaken the muscle tissues of stomach.
• In some people injury or the damage may weaken muscle tissue.
• Heart burn is common in this disease.
• The stomach contents travel back into the oesophagus or even into oral cavity and causes pain in the centre of the chest due to the eroding nature of acidity.

Question 9.
Obesity – Explain.
Answer:

• It is caused due to the storage of excess of body fat in adipose tissue.
• Obesity may be genetic or due to excess intake of food endocrine and metabolic disorders.
• Degree of obesity is assessed by body mass index (BMI).
• A Normal BMI range for adult is 19 – 25 above 25 is obese.
• BMI is calculated as body weight in Kg divided by the square of body height in metres.
• For example a 50 Kg person with a height of 160 Cms would have a BMI of 19.5.
• That is BMI \(=50 / 1.6^{2}\) = 19.5

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Zoology Guide Pdf Chapter 4 Organ and Organ Systems in Animals Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Zoology Guide Chapter 4 Organ and Organ Systems in Animals

11th Bio Zoology Guide Organ and Organ Systems in Animals Text Book Back Questions and Answers

Part I

I. Choose The Best Options

Question 1.
The clitellum is a distinct part in the body of earthworm Lampito mauritii, it is found in?
a. Segments 13-14
b. Segments 14-17
c. Segments 12 -13
d. Segments 14-16
Answer:
b. Segments 14-17

Question 2.
Sexually, earthworms are
a. Sexes are separate
b. Hermaphroditic but not self – fertilizing
c. Hermaphroditic and self – fertilizing
d. Parthenogenic
Answer:
b. Hermaphroditic but not self – fertilizing

Question 3.
To sustain themselves, earthworms must guide their way through the soil using their powerful muscles. They gather nutrients by ingesting organic matter and soil, absorbing what they need into their bodies. Say whether the statement is True or False: The two ends of the earthworm can equally ingest soil.
Answer:
a. True
b. False

Question 4.
The head region of Cockroach ……………….. pairs of …………….. and …………….. shaped eyes occur.
a. One pair, sessile compound and kidney shaped
b. Two pairs, stalked compound and round shaped
c. Many pairs, sessile simple and kidney shaped
d. Many pairs, stalked compound and kidney shaped
Answer:
a. One pair, sessile compound and kidney shaped

Question 5.
The location and numbers of malpighian tubules in Periplaneta.
a. At the junction of midgut and hindgut, about 150.
b. At the junction of foregut and midgut, about 150.
c. Surrounding gizzard, eight.
d. At the junction of colon and rectum, eight.
Answer:
a. At the junction of midgut and hindgut, about 150.

Question 6.
The type of vision in Cockroach is
a. Three dimensional
b. Two dimensional
c. Mosaic
d. Cockroachdonot have vision
Answer:
c. Mosaic

Question 7.
How many abdominal segments are present in male and female Cockroaches?
a. 10,10
b. 9,10
c. 8,10
d. 9,9
Answer:
d. 9,9

Question 8.
Which of the following have an open circulatory system?
a. Frog
b. Earthworm
c. Pigeon
d. Cockroach
Answer:
d. Cockroach

Question 9.
Buccopharyngeal respiration in frog
a. is increased when nostrils are closed
b. Stops when there is pulmonary respiration
c. is increased when it is catching fly
d. stops when mouth is opened.
Answer:
b. Stops when there is pulmonary respiration

Question 10.
Kidney of frog is
a. Archinephros
b. Pronephros
c. Mesonephros
d. Metanephros
Answer:
c. Mesonephros

Question 11.
Presence of gills in the tadpole of frog indicates that
a. fishes were amphibious in the past
b. fishes involved from frog -like ancestors
c. frogs will have gills in future
d. frogs evolved from gilled ancestor
Answer:
d. frogs evolved from a gilled ancestor

Question 12.
Choose the wrong statement among the following:
a. In earthworm, a pair of male genital pore is present.
b. Setae help in the locomotion of earthworms.
c. Muscular layer in the body wall of an earthworm is made up of circular muscles and longitudinal muscles
d. Typhlosole is part of the intestine of earthworms.
Answer:
d. Typhlosole is part of the intestine of earthworm.

Question 13.
Which of the following are the sense organs of Cockroach?
a. Antennae, compound eyes, maxillary palps, anal cerci
b. Antennae, compound eye, maxillary palps and tegmina
c. Antennae, ommatidia, maxillary palps, sternum and anal style.
d. Antennae, eyes, maxillary palps, and tarsus of walking legs and coxa
Answer:
a. Antennae, compound eyes, maxillary palps, anal cerci

(2 marks)

II. Very Short Questions

Question 14.
What characteristics are used to identify the earthworms?
Answer:
In gardens, earthworms can be traced by their fecal deposits known as worm castings on the soil surface.
The earthworms can be identified using the following characteristics:

• Long and cylindrical narrow body.
• Bilateral symmetry
• It is light brown in colour with a purple tinge at the anterior end.
• The division of the body into many segments or metameres.
• The dorsal surface of the body is marked by a dark mid-dorsal line.
• In mature worms, segments 14-17 may be found swollen with a glandular thickening of the skin called the clitellum.

Question 15.
What are earthworm casts?
Answer:
The undigested particles along with soil are passed out through the anus as worm castings or vermicasts.

Question 16.
How do earthworms breathe?
Answer:

• There are no lungs and gills.
• They respire through the body surface.
• The surface blood vessel helps in gaseous exchange.

Question 17.
Why do you call cockroaches a pest?
Answer:
Cockroaches destroy food and contaminate it with their offensive odour. They are carriers of a number of bacterial diseases. The cockroach allergen can cause asthma in sensitive people.

Question 18.
Comment on the functions of alary muscles?
Answer:
The triangular muscles that are present on both side of the heart are responsible for blood circulation.

Question 19.
Name the visual units of the compound eyes of cockroach.
Answer:

• The photoreceptors of the cockroach consist of a pair of compound eye on the dorsal surface of the head.
• Each eye is formed of about 2000 simple eyes called the ommatidia.

Question 20.
How does the male frog attract the female for mating?
Answer:

• The male frog has a pair of vocal sacs and a copulatory or nuptial pad on the ventral side of the first digit of each forelimb.
• Vocal sacs produce croaking sound to attract females and nuptial pad is also helpful in mating.

Question 21.
Write the types of respiration seen in frogs.
Answer:
Frog respires on land and in the water by two different methods. In water, the skin acts as an aquatic respiratory organ (cutaneous respiration). Dissolved oxygen in the water gets exchanged through the skin by diffusion. On land, the buccal cavity, skin, and lungs act as the respiratory organs. In buccal respiration on land, the mouth remains permanently closed while the nostrils remain open.

The floor of the buccal cavity is alternately raised and lowered, so air is drawn into and expelled out of the buccal cavity repeatedly through the open nostrils. Respiration by the lungs is called pulmonary respiration. The lungs are a pair of elongated, pink-colored sac-like structures present in the upper part of the trunk region (thorax). Air enters through the nostrils into the buccal cavity and then to the lungs. During aestivation and hibernation, gaseous exchange takes place through the skin.

Question 22.
Differentiate between peristomium and prostomium in earthworms.
Answer:

Question 23
Give the location of clitellum and spermathecal openings in Lampito mauritii.
Answer:

• Clitellum -14 -17 segment.
• Spermatheca – There are three pairs of spermathecae lying in segments 6/ 7,7/ 8,8/9

Question 24.
Differentiate between tergum and a sternum.
Answer:

Question 25.
Head of cockroach is called hypognathous. Why?
Answer:
The mouth parts of cockroach are directed downwards. The head is small, triangular lies at a right angle to the longitudinal body axis. Hence it is called hypognathous.

Question 26.
What are the components of blood in frogs?
Answer:
1. Plasma-60%
2. Red blood cells, white blood cells platelets 40 %

Question 27.
Draw a neat labeled diagram of the digestives system of frog.
Answer:

Question 28.
Explain the reproductive system of frog
1. Male reproductive organ:

• There are pair of testes. Each test is attached with kidney and dorsal body wall with the peritoneal membrane mesorchium.
• The vasa efferentia arises from each test is opened into the bladder canal and it communicates with the urinogenital duct that comes out of kidneys and opens into the cloaca.

2. Female reproductive system:

• It consists of paired ovaries attached to the kidney and dorsal body wall by folds of peritoneum called mesovarium.
• Each oviduct opens into the body cavity at the anterior end by a funnel like opening called ostia and posteriority the oviducts dilated to form ovisac before they open into cloaca.

• Fertilization is external.
• The eggs hatch into tadpoles.
• Tadpole develops three pairs of gills.
• The tadpole grows into an air-breathing carnivorous adult frog through a process metamorphosis.

Part II

11th Bio Zoology Guide Organ and Organ Systems in Animals Additional Important Questions and Answers

I. Choose The Best Options

Question 1.
We can locate the earth worm’s living area through ……………………….
a. Small hole in the ground
b. Worm costings
c. Debris
d. Excreta of cattle.
Answer:
b. Worm costings

Question 2.
The region between 14 – 17 segments in earthworm is called as …………………
a. Pygidium
b. Prostomium
c. Clitellum
d. Peristomium
Answer:
c. Clitellum

Question 3.
What is the shape of the setae?
a. ‘S’ shaped
b. C shaped
c. Round shaped
d. Curved shaped
Answer:
a. ‘S’ shaped

Question 4.
Find the correct pair.
a. First segment -Clitellum
b. Last segment – Peristomium
c. 14-17 -Pygidium
d. Vascular fold -Typhlosole
Answer:
d. Vascular fold -Typhlosole

Question 5.
In which segment the lateral hearts are situated?
a. 7-13
b. 6-13
c. 6-15
d. 7-15
Answer:
b. 6-13

Question 6.
Find out the correct statement
a. In male cockroach the reproductive sac lie anteriorily.
b. In female cockroaches chitinous plates gonapophyses are present around the female genital aperature.
c. In male cockroach the sternum of 10th segment have pair of anal cerci.
d. In the 12th segment anal styes are seen.
Answer:
b. In female cockroaches chitinous plates gonapophyses are present around the female genital aperature.

Question 7.
What is the length of lampito mauritii.
a. 80-210 mm
b. 85-350 mm
c. 80 – 220 mm
d. 80 – 200 mm
Answer:
a. 80-210 mm

Question 8.
The pale brown the purplish tinge colour of earthworm is due to the pigment
a. Haemolymph
b. Porphyrin
c. Choloroquin
d. Flaemoglobin
Answer:
b. Porphyrin

Question 9.
In which segments the spermatheca are situated
a. 6-7 segments, 7-8 segments 8-9 segments
b. 6-7 segments, 8-9 segments 9-10 segments
c. 8-9 segments, 9-10 segments 10-11 segments
d. 7-8 segments, 8-9 segments 9-10 segments
Answer:
a. 6-7 segments, 7-8 segments 8-9 segments

Question 10.
Find out the wrong statement?
a. The wall of earthworm is thin and moist.
b. There are cuticle and epithelial layer.
c. The body cavity which lies between the digestive system and body wall does not act as a fluid-filled structure.
d. The coelomic fluid is alkaline and milky in nature.
Answer:
c. The body cavity which lies between the digestive system and body wall does not act as a fluid-filled structure.

Question 11.
Whether the following statement is correct or wrong. Justify.
a. In earthworm the digestive tract runs from the mouth to anus.
b. In earthworm the mouth is seen in the first segment.
c. In the second segment lies buccal cavity.
d. In the 3 – 4th segment lies the pharynx.
a. True, False, False, True
b. True, True, True, True
c. False, False, False, True
d. True, False, True, False
Answer:
b. True, True, True, True

Question 12.
The large complex molecules which consist of organic-rich soil eaten by earthworm with the help of digestive enzymes is converted into the simple absorptive unit is
a. Intestinal digestion
b. Digestion
c. Rectal digestion
d. Enzymatic digestion
Answer:
b. Digestion

Question 13.
Find the odd one out.
The earthworm receptors are
a. Photoreceptors
b. Vision receptors
c. Taste receptors
d. Gustatory receptors
Answer:
b. Vision receptors

Question 14.
Which of the following is true with an excretory system of earthworm.
a. Nephridia
b. Nephron minute tubules
c. Nephridia – minute coiled tubules
d. Nephron – coiled tubule
Answer:
c. Nephridia – minute coiled tubules

Question 15.
Apart from nephridia, there is specialised cell present in the intestinal walls.
a. Chlorogogen
b. Chloricgen
c. Chlorajan
d. Chlorojin
Answer:
a. Chlorogogen

Question 16.
What is the other name for the seminal funnel?
a. Ciliary
b. Ciliary rosettes
c. Ciliary flagella
d. Ciliary antennae
Answer:
b. Ciliary rosettes

Question 17.
How many days are taken by the earthworm to complete its life cycle?
a. 70 days
b. 65 days
c. 69 days
d. 60 days
Answer:
d. 60 days

Question 18.
What is the fluid manure of earthworm consist of?
a. Vermicomposting
b. Vermiculture
c. Vermiwash
d. Earthworm manure
Answer:
c. Vermiwash

Question 19.
Find out the unrelated one
a. Vermi compose
b. Vermin
c. Vermiculture
d. Vermi wash
Answer:
b. Vermin

Question 20.
Match and find out correct
I. Coxa – a. Thick
II. Trochanter – b. Long
III. Femur – c. Small
IV. Tibia – d. Large
a. I – a, II – b, III – c, IV – d
b. I-d, II-c, III-b, IV-a
c. I-a, II-c , III-d ,IV-b
d. I – b, II – a, III – c, IV – d
Answer:
b. I-d, II-c, III-b, IV-a

Question 21.
Find out the wrong pair of cockroaches.
a. Tarsus – Podomeres
b. Genital opening Sclerites – Parametabolus
c. Sclerites of the dorsal side – Tergites
d. Sclerites of the ventral side – Sternites
Answer:
b. Genital opening Sclerites – Parametabolus

Question 22.
Find out the wrong pair of cockroaches.
a. Spiracles – Stigmata
b. Ostia – Colourless coelomic fluid
c. Ostia – Digestive system cockroach
d. Supra oesophageal ganglion – Brain
Answer:
c. Ostia – Digestive system cockroach

Question 23.
Which is the ancient organism of insect?
a. Cockroach
b. Cricket
c. Grasshopper
d. Scorpion
Answer:
a. Cockroach

Question 24.
Cockroach belongs to …………………………………………… period about 320 million years ago.
a. Devonian
b. Carboniferous
c. Missisipian
d. Pensylvanian.
Answer:
b. Carboniferous

Question 25.
One of the fastest moving land insect is the cockroach. What is it’s speed?
a. 6.4 km/hr
b. 5.0 km/hr
c. 5.4 km/hr
d. 6.5 km/hr
Answer:
c. 5.4 km/hr

Question 26.
Which makes the wings of insect?
a. Chitin
b. Pecten
c. Cellulose
d. Hemicellulose
Answer:
a. Chitin

(2 marks)

II. Very Short Questions

Question 1.
Name the earthworms of India.
Answer:

• Lampito mauritii (Megascolex mauritii)
• Perioynx excavatus
• Metaphire posthuma (Pheretima Posthuma)

Question 2.
What are the regions of clitellum?
Answer:

• Preclitellar region (1st – 13th segments)
• Clitellar region (14th – 17th segment)
• Post – Clitellar region (after 17th segment)

Question 3.
Name the structure that helps in locomotion where is it seen?
Answer:
Locomotion is effected through setae. In all the segments of the body except the first last and clitellum there is a ring of chitinous body setae.

Question 4.
What is the composition of the coelom of earthworms?
Answer:
The coelomic fluid is milky and alkaline. It consists of granulocytes or eleocytes amoebocytes, mucocytes and leucocytes.

Question 5.
Classify earthworms based on their ecological strategies.
Answer:

• Earthworms are classified as epigeics, anecics and endogeics based on their ecological strategies.
• Epigeics are the surface dwellers e.g., Perionyx excavaus and Eudrilus eugeniae.
• Anecics are found in the upper layers of the soil e.g., Lampiro mauritii, Lumbricus terrestris.
• Endogeics are found in deeper layers of the soil e.g., Octochaetona thursoni.

Question 6.
What are the mouthparts of the cockroach?
Answer:

• Labrum (i) pair of mandibles Labrum (ii) pair of maxillae
• Labium and hypopharynx or tongue.

Question 7.
Give notes on sclerites?
Answer:
In each segment, exoskeleton has hardened plates called sclerites, which are joined together by a delicate and elastic articular membrane or arthrodial membrane.

Question 8.
When is cockroach evolved?
Answer:
The cockroaches are ancient among all groups of insects dating back to the carboniferous period about 320 million years ago.

Question 9.
Name the five segments of the leg of the cockroach?
Answer:

1. Coxa -Large
2. Trochanter-Small
3. Femur -Long and broad
4. Tibia – Long and thick
5. Tarsus -has five movable joints

Question 10.
Where are hepatic caeca seen in cockroaches?
Answer:
At the junctional region of the gizzard are eight finger-like tubular blind processes called hepatic caecae.

Question 11.
Trace the air paths of respiration.
Answer:
Spiracle trachea tracheoles Tissues.

Question 12.
Write a note on coelom of earthworm.
Answer;
A spacious body cavity called the coelom is seen between the alimentary canal and the body wall. The coelom contains the coelomic fluid and serves as a hydrostatic skeleton, in which the coelomocytes are known to play a major role in regeneration. immunity and wound healing. The coelomic fluid of the earthworm is milky and alkaline, which consists of granulocytes or cicocytes. amoebocytes, mucocytes and leucocytes.

Question 13.
What are the structures that is not present in frog?
Answer:
In frog there is no external ear neck and tail.

Question 14.
Give notes on chyme?
Answer:
Digestion of food takes place by the action of hydrochloric acid and gastric juices secreted from the walls of the stomach. This partially digested food is called as chyme.

Question 15.
What are the regions of nervous system?
Answer:
Central nervous system peripheral nervous system autonomous nervous system.

Question 16.
A cockroach produces nutritionally dense milk to feed their young ones. It may be considered as a superfood of the future. How?
Answer:

1. It contains crystalline milk.
2. It is synthesised by diploptera punctata.

Question 17.
What are the economic importance of frog?
Answer:

• Frogs feed on insects and helps in reducing insect pest population.
• Frogs are used in traditional medicine for controlling blood pressure and for antiaging properties.

Question 18.
What are the types of cockroach?
Answer:

• American cockroach
• Brown – banded cockroach
• German cockroach
• Oriental cockroach
• Viviparous cockroach

Question 19.
Name the cells that helps in excretion of cockroach?
Answer:

• Fat bodies
• Nephrocytes
• Cuticle
• Urecose glands

Question 20.
Define uricotelic organism.
Answer:
The nitrogenous wastes are eliminated through uric acid. (Eg.) Hence cockroach excretes uric acid as a waste it is said to be uricotelic.

Question 21.
What is typhiosole?
Answer:

• The dorsal wall of the intestine of earthworm is folded into the cavity as the typhiosole.
• This fold contains blood vessels and increases the absorptive area of the intestine.

Question 22.
What are the glands seen in male reproductive system ?
Answer:

• Mushroom-shaped gland
• conglobate gland.

Question 23.
What is clitellum?
Answer:
In mature worms 14 – 17 segments may be found swollen with a glandular thickening of the skin called the clitellum. This helps in the formation of cocoon.

Question 24.
Where is spermathecal openings seen in the earthworm?
Answer:
They are lying inter segmentally between the grooves of the segments 6/7,7/S and 8/9.

Question 25.
Where is genital openings seen in the earthworm?
Answer:

• The female genital aperture lies on the ventral side in the 14th segment.
• A pair of male genital apertures are situated latero-ventrally in the 18th segment.

Question 26.
Name the body muscles of earthworm.
Answer:

• Cuticle
• Epidermis
• Coelomic epithelium

Question 27.
Name the cells t at makes the epidermis?
Answer:

• Supportive cells
• Glandular cells
• Basal cells
• Sensory cells

Question 28.
What is the functions of coelomocytes of earthworm?
Answer:
Uses of coelomocytes

1. Regeneration
2. Immunity
3. Woundhealing

Question 29.
Name the cells of coelom of earthworm
Answer:

• Granulocytes or eleocytes
• Amoebocytes
• Mucocytes
• Leucocytes

Question 30.
What are lateral hearts ?
Answer:
In the anterior part of body of earthworm the dorsal vessel is connected with the ventral vessel by eight pairs of commissural vessels or lateral hearts lying in the 6th – 13th segment.

Question 31.
Give notes on nephrostome.
Answer:
The mega nephridium of earthworm has an internal funnel like opening called the nephrostome which is fully ciliated.

Question 32.
What is chloragogen cell?
Answer:
Besides nephridia special cells on the coelomic wall of the intestine called chloragogen cells are present. They excrete nitrogenous wastes in the blood.

Question 33.
Whatisprotandrous?
Answer:

• The two sex organs of earthworm mature at different times and hence self fertilisation is prevented.
• The sperm develops earlier than the production of ova. This process is known as protandrous.
  It transmits the diseases like cholera, dysentery and tuberculosis hence it is known as vectors.

Question 38.
Whatishypognathous?
Answer:
The mouth parts of cockroach are directed downwards so its is hypognathous.

Question 39.
What is compound eyes?
Answer:
The head of cockroach bears a pair of large sessile and reniform compound eyes. Each eye is formed of about 2000 simple eyes called the ommatidia and the vision caused by the ommatidia is mosaic vision.

Question 40.
Name the segments of the legs of cockroach?
Answer:
There are five segments in the legs of cockroach.

1. Coxa
2. Trochanter
3. Femur
4. Tibia
5. Tarsus

Question 41.
What is podomeres?
Answer:
The last segment of the leg tarsus has five movable joints called podomeres or tarsomeres.

Question 42.
Give notes on wings of cockroach?
Answer:
Cockroach has two pairs of wing. The first pair of wings protects the hind wings when rest is called elytra or tegmina. The second pair of wings used in flight.

Question 43.
Name the plates of the abdomen of cockroach?
Answer:
There are 10 segments in the abdomen. The sclerites of the dorsal side are called tergites.
The sclerites on the ventral side are called sternites and the sclerites on the lateral sides are called pleurites.

Question 44.
What are the sensory receptors seen in cockroach?
Answer:

• Antenna
• Compound eyes
• Labrum
• Mandibles
• Labialpalps
• Analcerci

Question 45.
Name the fat bodies of cockroach?
Answer:
Nephrocytes, Cuticle, Urecose glands.

Question 46.
What are the glands seen in the male cockroach?
Answer:

• Mushroom shaped gland
• Conglobate gland.

Question 47.
What is meant by paurometabolus?
Answer:

• In cockroach the embryonic development occurs in the ootheca for 5-13 weeks.
• The development of cockroach is gradual through nymphal stages. Hence it is called as paurometabolus.

Question 48.
What are pokilotherms?
Answer:
The organisms which change its temperature according to the temperature of the environment is known as pokilotherms.

Question 49.
What is nictitating membrane?
Answer:
The third eyelid of frog is nictitating membrane. It protects the eye.

Question 50.
What is cloaca?
Answer:
As the digestive excretory reproductive system opens commonly through a aperture this is called as cloaca.

Question 51.
What is spiracle?
Answer:
In cockroach the trachea open through 10 pairs of small holes called spiracles.

Question 52.
What is meant by chordotonal receptor?
Answer:
Chordotonal receptor is found on the anal cerci which are receptive to vibrations in air and land.

Question 53.
How can a earthwom senses its burrow?
Answer:
In the prostomium of earthworm there are thermal and chemical receptors with the help of this they can find it’s habitat.

Question 54.
Compare the respiration of human with the respiration of cockroach?
Answer:
In the respiratory system of cockroach there are spiracles and trachea. Each spiracles can open and close. During inspiration spiracles open. This oxygen enters into the haemocoel through spiracles and exchange of gases taking place.

Question 55.
List the very special features of cockroach.
Answer:

• Cockroach can survive being submerged under water upto 45 minutes.
• They hold their breath often to help regulate loss of water.

Question 56.
Cockroach can live without ahead? How?
Answer:
A cockroach can live for a week without its head. There is no connection between head and respiration. There is no nostrils and lungs.
The abdomen has 10 pairs of spiracles. These spiracles are communicated with the tracheoles and haemolymph and exchange of gases taking place.

Question 57.
Why is mosaic vision with less resolution in cockroaches ?
Answer:
The unit of compound eye is ommatidium. There are hundreds of ommatidia. Each ommatidium forms a image. Each image formed in all the ommatidia forms a vision. This image is a mosaic vision.

Question 58.
List the charcteristic features of order Aneura?
Answer:
Frogs and Toads have elongated hindlimbs. This helps in jumping, Frogs can live in water and on trees. Parental care is seen in few species.

Question 59.
Differentiate the compound eyes from simple eye.
Answer:

Question 60.
Why three chambered heart of frog is not as efficient as the four chambered heart of birds and mammals?
Answer:

• The heart of birds and mammals have four chambers. The oxygenated and deoxygenated blood is carried by separate blood vessels and transports to body parts and the purifying organ.
• The frog has three chambered heart. The oxygenated and deoxygenated blood mixeshere. This mixed blood is reaching all the parts.

Question 61.
What is meant by cloaca a Common digestive and excretary opening ?
Answer:

• In the elasmobranhs amphibians, reptiles egg laying mammals the faeces and urine pass through this opening.
• This passage is also a genital passage for the deposition of sperm. This is called cloacal aperature.

Question 62.
Give notes on setae of earthworm?
Answer:

• Earthworm have setae which are small hair like bristles. They are not composed of the same material as human hair.
• They will be helpful in feeding mating and locomotion.

Question 63.
Give notes on intestinal caeca of earthworm?
Answer:
In 26th segment of metaphire posthuma a pair of cone shaped bulging is seen. It is known as intestinal caecum. This secretes amylolytic enzymes. This helps in starch digestion.

( 3 marks)

III. Short Questions

Question 1.
Based on their ecological strategies classify the earthworms?
Answer:

Question 2.
Where is longest earthworm seen?
Answer:

• Micro chaetus rappi is an African giant earthworm can reach a length of 6.7 meter (22 feet)
• Drawida nilamburansis is a species of earthworm in Kerala reaches a maximum length upto 1 meter (3 feet).

Question 3.
What is the significance of coelomic fluid of earthworm?
Answer:

• In the coelomic fluid coelomocytes are present.
• It helps in regeneration.
• It helps in immunity and healing of wounds.

Question 4.
What are the sensory receptors seen in the earthworm?
Answer:

• Photo receptors – Found on the dorsal surface of the body.
• Gustatory – Sense of taste are found in the buccal cavity.
• Tactile receptors Sense of touch
• Chemo receptors Seen in the prostomium and the bodywall.
• Thermo receptors

Question 5.
Draw the diagram of nephridia of earthworm and name the parts.
Answer:

Question 6.
What are the other name for sclerites ?
Answer:
On the basis of their location they gets their name.

• Dorsal sclerites – Tergites
• Ventral sclerites – Stemites
• Lateral sclerites – Plurites

Question 7.
The cockroach can survive with out the head. Whether the statement is correct or wrong if it is so give reason?
Answer:
This statement is correct.
Reason:

• A cockroach can live for a week without its head.
• Due to their open circulatory system they breath through little holes on each of their body segment hence they are not dependent on the mouth or head to breath.

Question 8.
What are the parts of the nervous system ?
Answer:
Supraoesophagial nerve ganglion or brain sub- oesophagial ganglion – circum oesophageal connectives double ventral nerve cord.

Question 9.
What are the significance of nervous system.
Answer:

• Brain or supra oesophageal ganglion or brain.
• It acts as a sensory and an endocrine centre.
• Sub – oesophageal ganglion
• It acts as a motor centre controls the movements of the mouthparts legs and wings.

Question 10.
Give notes on ommatidia?
Answer:

• The photo receptors of the cockroach consists of a pair of compound eyes at the dorsal surface of the head.
• Each eye is formed of about 2000 simple eyes called ommatidia.

Question 11.
Give notes on ‘Mosaic vision’?
Answer:

• The cockroach perceives the vision through each ommatidia. This vision is mosaic vision.
• Though there is sensitivity but the vision is not a clear one.

Question 12.
Why is sexual dimorphism exhibited clearly during breeding season in frog?
Answer:

• During breeding the sexual dimorphism is seen clearly.
• The male frog has a pair of vocal sac and a ; nuptial pad on the ventral side of the first digit i of each fore limb.
• Vocal sacs assist in amplifying the croaking sound of frog.

Question 13.
How will you classify the earthworm based on their living in relation to ecological strata?
Answer:

1. Epigeics – Surface living (Eg.) Eudrilus eugeniae
2. Anecics-Found in upper layers of the soil. (Eg.) Lampito mauritii.
3. Endogeics – Found in deeper layers of the soil. (Eg.) Octochaetona thurstoni.

Question 14.
Give an account of respiratory system of earthworm?
Answer:

• Earthworm has no special respiratory organ like lungs or gills.
• Respiration takes place through the body wall.
• The outer surface of the skin is richly supplied with blood capillaries which helps in the diffusion of gases.
• Oxygen diffuses through the skin into the blood.
• Carbondi-oxide from the blood diffuses out.
• The skin is kept moist by mucous and coelomic j fluid and facilitates exchange of gases.

Question 15.
Give an account of nervous system of earthworm?
Answer:

• The brain composed of bilobed mass of supra- pharyngeal ganglia. On the third segment j supra-pharyngeal nerve ganglion and on the 4th segment sub-pharyngeal nerve ganglion is seen.
• The brain and the sub-pharyngeal ganglia are connected by a pair of cirum-pharyngeal connectives.
• The double ventral nerve cord runs backward from the sub-pharyngeal ganglion.

Question 16.
What is the excretory organ of earthworm? What are its type?
Answer:
The nephridia is the excretory organ of earthworm
They are three types.

1. Pharyngeal or tufted nephridia seen in 5-9 segments
2. Micro nephridia or Integumentary nephridia seen 14 – 19th -segment.
3. Mega nephridia or septal nephridia seen from 19th – last segment.

Question 17.
Give notes on vermiwash.
Answer:

• Vermi wash is a liquid manure or plant tonic obtained from earthworm.
• It is used as a foliar spray and helps to induce plant growth.
• It is a collection of excretory products, mucus secretion micro nutrients from the soil organic molecules.

Question 18.
What is wormery or wormbin?
Answer:
Earth worm can be used for recycling of waste food leaf litter and biomass to prepare a good fertilizer in container is known as wormery or wormbin. It makes superior compost.

Question 19.
Give the systematic classification of earthworm?
Answer:

• Phylum – Annelida
• Class – Oligocheata
• Order – Haplotaxida
• Genus – Lampito
• Species – Mauriitii

Question 20.
In which part of the cockroach’s body the sensory receptors are seen?
Answer:

Question 21.
Give the systematic classification of frog?
Answer:

• Phylum – Chordata
• Class – Amphibia
• Order – Aneura
• Genus – Rana
• Species – Hexatacdyla

Question 22.
Give the systematic classification of cockroach?
Answer:

• Phylum – Arthropoda
• Class – Insecta
• Order – Orthroptera
• Genus – Periplanata
• Species – Americana

Question 23.
Give an account of exo skeleton of cockroach?
Answer:

• The entire body is covered by a hard chitinous exoskeleton.
• In each segment exoskeleton has hardened plates called sclerites which are joined together by a delicate and elastic articular membrane.
• The sclerites of the dorsal side are called tergites. Ventral side are called sternites lateral side are called pleurites.

Question 24.
Give an account of mouth parts of cockroach?
Answer:

• The appendages form the mouth parts which are of biting and chewing type.
• These are mandibulate or orthopterus.

Mouth parts

1. Labrum – Upperlip
2. A pair of mandibles
3. Pair of maxillae.
4. Labium-Lower lip
5. Tongue – Hypopharynx.

Question 25.
Name the digestive glands of cockroach.
Answer:

• Salivary glands.
• Hepatic caeca or entericcaeca

Question 26.
Draw the male frog with vocal sac and nuptial pad and marks the parts.
Answer:

Question 27.
Give an account of buccal cavity of frog.
Answer:

• The wide mouth opens into the buccal cavity.
• On the floor of the buccal cavity lies a large muscular sticky tongue.
• The tongue is attached in front and free behind.
• The free edge of the tongue is forked.
• A row of small and maxillary teeth is found on the inner region of the upper jaw.
• Vomerine teeth are present one on each side of the internal nosteils.

Question 28.
Give an account of blood of frog ?
Answer:
60% of frog’s blood is plasma and 40% is red blood cells. The blood cells composed of red blood cells, white blood cells and platelets.
White blood cells

1. Neutrophil
2. Basophil
3. Eosinophils
4. Lymphocytes
5. Monocytes

 ( 5 marks)

IV. Essay Questions

Question 1.
Describe the external features of the earthworm?
Answer:

• Earthworm has along and cylindrical body.
• It is 80-210 mm in length. It is light brown in colour.
• The body is encircled by a large number of grooves which divides it into a number of compartments called segments ormetameres.
• The mouth is found in the centre of the first segment of the body called the peristomium.
• Overhanging the mouth is a small flab called prostomium.

• The 14 – 17th segments become swollen called clitellum.
• There are pair of female genital opening in the 14th segment and pair of male genital opening in the 18th segment.
• In the segments 6/7, 7/8, 8/9 lies the spermatheca.
• In all the segments of the body except the first last and clitellum there is ring of chitinous body setae. They all involved in locomotion.
• The last segments bear anus.

Question 2.
Give an account of digestive system of earthworm?
Answer:
1. The alimentary canal runs as a straight tube throughout the length of the body from the mouth to anus.
2. The mouth opens into the buccal cavity which occupies the 1st and 2nd segments.
3.  The thick muscular pharynx lies in the 3rd and 4th segment and is surrounded by the pharyngeal glands.
4. A small narrow oesophagus lies in the 5th segment and 6th segment contains muscular gizzardes. Which helps in grinding the soil and decaying leaves.
5. The intestine starts from the 7th segment and ’ continues upto the last segment.
6. The dorsal wall of the intestine is folded into the vascular cavity called typhlosole.

Question 3.
Describe the structure of circulatory system of earthworm?
Answer:

• In earthworm closed type of blood vascular system is seen which contains blood vessels, capillaries and lateral hearts.
• Two median longitudinal vessels run above and below the alimentary canal as dorsal and ventral vessels of the earthworm.
• There are paired valves in the dorsal vessels which prevent the backward flow of the blood.
• From 6 to 13 segments with the 8 pairs of commissural vessels which connects the dorsal and the ventral vessel called lateral hearts.
• The blood is pumped from the dorsal vessel to the ventral vessel.

The blood glands present in the anterior segments of the earthworm produce blood cells and haemoglobin and gives red colour to the blood.

Question 4.
Describe the structure of reproductive system of earthworm ?
Answer:

Earth worm is a hermaphrodite organism the male and female reproductive organs are found in the same individual.

Male reproductive system:

• Two pairs of testes are present in the 10th and 11th segments. The testes give rise to the germ cellor spermatogonia.
• Two pairs of seminal funnels called ciliary rosettes are situated in the same segments as the testes. Three pairs of spermathecalies in the 7,8,9 segments.
• The vas deferens arise from the ciliary rosettes run upto the 18th segment and open exterior through the male genital aperture which contains two pairs of penial setae.
• A pair of prostate glands lie in the 18th and 19th segments.
• The secretion of prostate cement the spermatozoa into a bundles of spermatophores.

Female reproductive system:

• A pair of ovaries lying in the 13th segment. Ovarian funnels are present beneath the ovaries continue as a oviduct and opens in the 14th segment as a female genital opening. There are three pairs of spermathecae lie in the 7th, 8th and 9th segment.
• They receive spermatozoa during copulation.
• The two earthworm mate juxta position opposite gonadal openings and exchanging sperms mature egg cells in the nutritive fluid are deposited in the cocoons produced by the glands cells of the citellum which also collects the partner’s sperm from the spermathecae.
• Fertilization and developments occurs in the cocoon.
• After 3 weeks baby earthworm are released.

Question 5.
Give an account of locomotion of earthworm?
Answer:

• The earthworm normally crawls with the help of their body muscles setae and buccal chamber. The outer circular and inner longitudinal muscle layers lies below the epidermis of the body wall.
• The contraction of circular muscles make the body long and narrow while the longitudinal muscles make the body short and broad and hence due to the contraction of longitudinal muscle the earthworm moves.
• The alternate waves of extensions and contractions are aided by the leverage afforded by the buccal chamber and setae.

Question 6.
Describe the morphological features of cockroach?
Answer:

• Cockroach is a bilaterally symmetrical segmented animal which is divisible into head thorax and abdomen.
• The entire body is covered with chitinous exo-skeleton.
• Each segment consists of sclerites. The head is small and triangular and the mouth parts are directed downwards hence known as hypognathous. The head bears a pair of compound eye. Each compound eye is composed of unit of ommatidia.
• The mouth parts are of mandibulatetype. It consists of labrum pair of mandibles a pair of maxillae labium and tongue.
• The thorax consists of prothorax mesothorax and metathorax. Each thoracic segment bears a pair of walking legs.
• Due to the presence of 3 pairs of leg they are called as hexapoda. Each leg consists of five segments they are coxa, trochanter femur tibia and tarsus.
• It has two pairs of wings. It is called as tegmina or elytra. The wings arise from the mesothorax protects the hind wings when at rest. The second pair of wings arise from metathoraxa and used inflight.

Question 7.
Draw the diagram of mouth parts of cockroach ?
Answer:

Question 8.
Describe the structure of digestive system of cockroach with a diagram ?
Answer:

The alimentary canal is divided into three regions namely foregut midgut and hindgut.

Foregut:

• It includes pre-oral cavity mouth pharynx oesophagus and the posterior region contains crop.
• The food is stored in the crop. The crop is followed by gizzard which have chitinous teeth helps in the grinding of the food particles.

Midgut:

• At the junctional region of the gizzard are eight finger like tubular blind processes called the hepatic caecae or enteric caecae.
• At the junction of midgut and hind gut lies 100 – 150 yellow coloured malphigian tubules. It excretes the nitrogenous wastes from the haemolymph.

Hindgut:

• The hind gut is broader than the midgut.
• It consists of ileum colon andrectum. The rectum opens out through anus.
  Digestive glands
  Salivary glands
  Hepatic caeca

Question 9.
Describe the structure of circulatory system of cockroach ?
Answer:

Cockroach has an open type of circulatory system.
The coelom is filled with haemolymph. Heart is an elongated tube with muscular wall lying mid dorsally beneath the thorax.
The heart consists of 13 chambers with ostia on either side.
The blood from the sinuses enter the heart through the ostia and is pumped anteriorilly to sinuses again. In each segment there is triangular muscle called alary muscles are seen. It is responsible for blood circulation.
There is a pulsatile vesicle lies at the base of each antenna also pumps blood.

Question 10.
Give an account of excretory system of cockroach?
Answer:

• The malphighian tubules are the main excretory organs of cockroach which help in eliminating the nitrogenous wastes from the body in the form of uric acid. Excretion is uricotelic.
• In addition fat body, nephrocytes cuticle and urecose glands are also excretory in function. The malpighian tubules are attached at the junction of midgut and hindgut. There are about 100-150 in number present in 6 – 9 bundles.
• Each tubule is lined by glandular and ciliated cells and the waste is excreted out through the hindgut.
• The glandular cells of malpighian tubules absorb water salts and nitrogenous wastes. The cells of the tubules reabsorb water and inorganic salts.
• By the contraction of the tubules nitrogenous waste is pushed in to the ileum. The remaining waste with solid uric acid is exceeded along with the faecal matter.

Question 11.
Describe the structure of reproductive system of male cockroach?
Answer:

• The male reproductive system consists of a pair of testes vasa deferentia an ejaculatory duct utricular gland phallic gland and the external genitalia.
• A pair of 3 lobed testes lies on the 4th and 6th abdominal segments. The vas deferens opens into the male gonopore which lies ventral to anus.
• The mushroom shaped gland is a large reproductive gland which opens into the anterior part of the ejaculatory duet.
• The sperms are stored in the seminal vesicles as bundles of spermatophores. Surrounding the male genital opening are few chitinous structures called phallomeres or gonopophyses which help in copulation.

Question 12.
Describe the structure of female reproductive system of cockroach?
Answer:

• The female reproductive system consists of a pair of ovaries vagina genital pouch collaterial glands speromthecae and the external genitalia.
• A pair of ovaries lie in the segmetn of 2nd and 6th abdominal segment. Each ovary is formed of eight ovarian tubules. Oviducts of each ovary unite into a common oviduct known as vagina which opens into the genital chamber.
• A pair of spermathecae is present in the 6th segment and opens in to the genital pouch.
• During copulation the ova descend to the genital chamber and fertilised by the sperm. The collateral gland secreta a hard case called ootheeca around the egg.
• The ootheca is dropped to a crack or crevice of high relative humidity near a food source. The nymphs are released from this ootheca and they grows by moulting about 13 times to reach the adult form.

Question 13.
Describe the morphological features of frog?
Answer:

The body of frog is streamlined to help in swimming. Body is divided into head and trunk.

Head:

• The head is triangular and has an apex which forms the snout.
• The mouth is at the anterior end on the head contains pair of external nostrils, pair of eyes with unmovable upper eyelid movable lower eye lid which protects the eye.
• The nictitating membrane protects the eye when the frog is underwater.
• A pair of ear drum lies behind the eyes. There is no external ear neck and tail.

Trunk:

• It bears a pair of fore limbs and a pair of hind limbs. The hind limbs are longer than the forelimbs. At the posterior end between the hind limbs is the cloacal aperture.
• Fore limbs help to bear the weight of the body. It consists of upper arm fore arm and a hand. The hind limbs consist of thigh shank and foot.
• Foot bears five long webbed toes and one small spot called the sixth toe cloacal aperture.
• Fore limbs help to bear the weight of the body. It consists of upper arm fore arm anda hand.
• The hind limbs consist of thigh shank and foot. Foot bears five long webbed toes and one small spot called the sixth toe.

Question 14.
Describe about the digestive system of frog?
Answer:
The digestive system extends from mouth to the cloaca.
Digestive tract organs:

• The alimentary canal consists of the buccal cavity pharynx oesophagus duodenum ileum and the rectum which opens outside by the cloacal aperture.
• The mouth opens into the buccal cavity.

Tongue:

• On the floor of the buccal cavity lies a large muscular sticky tongue. The tongue is attached in front and free behind.
• The free edge of tongue is forked.

Teeth:

• A row of small and pointed maxillary teeth is found on the inner region of the upper jaw. Vomerine teeth are also present as two groups. One on each side of the internal nostrils.
• The lower jaw is devoid of teeth. The mouth opens into the buccal cavity and then to esophagus through pharynx. Oesophagus opens into stomach
• Stomach opens in to intestine then to rectum and finally to the cloaca. Liver, Pancresa are the structures of digestive system.

Question 15.
Describe the structure of the heart of frog ?
Answer:

• The heart consists of three chamber. Two auricle and one ventricle. Heart is covered by pericardium. On the dorsal side of the heart is a triangular chamber called sinus venosus.
• Truncus arteriosus is a thick walled structure which is obliquely placed on the ventral surface of the heart.
• It divides into right and left aortic trunk. Each divides into carotid systemic and pulmocutaneous arteries.
• The systemic trunk of each side is joined posteriorly to form the dorsal aorta. They supply blood to the posterior part of the body. The pulmo-cutaneous trunk supplies blood to lungs and skin.
• The sinus venosus receives the deoxygenated blood from the pre and post venacava and delivers the blood to the right auricle.
• The left auricle receives oxygenated blood through pulmonary vein.

Question 16.
Describe the structure of the respiratory system of frog?
Answer:

• Frog respires on land and in the water by two different methods. In water the dissolved oxygen in the water gets exchanged through the skin by diffusion.
• On land the buccal cavity skin and lungs act as the respiratory organ.

Buccal respiration:

• Mouth remains permanently closed while the nostrils remain open. The floor of the buccal cavity is raised and lowered so air drawn into and expelled out of the buccal cavity repeatedly through the open nostrils.
• A pair of elongated pink coloured sac like lungs are present in the upper part of the trunk.
• Air enters through the nostrils into the buccal cavity and then to the lungs.

Question 17.
Describe the structure of the nervous system
Answer:

• The nervous system is divided into the central nervous system, peripheral nervous system and the autonomous nervous system.
• Peripheral nervous system consists of 10 pairs of cranial nerves and 10 pairs of spinal nerves. The autonomous nervous system is divided into sympathetic and parasympathetic nervous system.
• CNS consists of brain and spinal cord. Brain is covered with pia mater and dura mater. The brain is divided into forebrain mid brain and hind brain
• Fore brain Consists of a pair of olfactory lobes and cerebral hemisphere and diencephalon. The olfactory lobes contain a small cavity called olfactory ventricle.
• The midbrain includes two large optic lobes and has cavities called optic ventricles.
• Hind brain consists of cerebellum and medulla oblongata. The medulla oblongata passes out through the foramen magnum and continues as spinal cord.

Question 18.
Draw the diagram of buccal cavity of frog and name the parts?
Answer:

Question 19.
Describe the locomotion of earthworm?
Answer:

• The earthworm normally crawl with the help of their body muscles setae and buccal chamber.
• The outer circular and inner longitudinal muscle layers lies below the epidermis of the body wall.
• The contraction of circular muscles make the body long and narrow while the longitudinal muscle make the body short and broad and hence due to the contraction of longitudinal muscle the earthworm moves.
• The alternate waves of extensions and contractions are aided by the leverage afforded by the buccal chamber and setae.

Question 20.
Tabulate the morphological differences between lampito mauritii and metaphire posthuma.
Answer:

Question 21.
Tabulate anatomical differences between lampitomaritii and metaphire posthuma
Answer:

Question 22.
Differentiate the male cockroach from female cockroach.
Answer:

23. Draw the life cycle of lampito mauritii.
Answer:

Question 24.
Differentiate the frog from toad.
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Zoology Guide Pdf Chapter 3 Tissue Level of Organisation Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Zoology Guide Chapter 3 Tissue Level of Organisation

11th Bio Zoology Guide Tissue Level of Organisation Text Book Back Questions and Answers

Part I

I. Choose The Best Options.

Question 1.
The main function of the cuboidal epithelium is
a. Protection
b. Secretion
c. Absorption
d. Both (b) and (c)
Answer:
d. Both (b) and (c)

Question 2.
The ciliated epithelium lines the
a. Skin
b. Digestive tract
c. Gall bladder
d. Trachea
Answer:
d. Trachea

Question 3.
What type of fibres are found in connective tissue matrix?
a. Collagen
b. Areolar
c. Cartilage
d. Tubular
Answer:
a. Collagen

Question 4.
Prevention of substances from leaking across the tissue is provided by
a. Tight junction
b. Adhering junction
c. Gap junction
d. Elastic junction
Answer:
a. Tight junction

Question 5.
Non-shivering thermogenesis in neonates produces heat through
a. White fat
b. Brown fat
c. Yellow fat
d. Colourless fat
Answer:
b. Brown fat

Question 6.
Some epithelia are pseudostratified. What does this mean?
Answer:
Pseudostratified epithelial cells are columnar but unequal in size. Although the epithelium is single-layered yet it appears to be multilayered due to the fact that nuclei lie at different levels in different cells.

Question 7.
Differentiate white adipose tissue from brown adipose tissue
Answer:

Question 8.
Why blood is considered as a typical connective tissue?
Answer:
Blood is considered a typical connective tissue because it is the fluid connective tissue containing plasma, RBCs, WBCs, and platelets. It functions as the transport medium for the cardiovascular system carrying nutrients, nitrogenous wastes, and respiratory gases throughout the body.

Question 9.
Differentiate between elastic fibres and elastic connective tissue.
Answer:

Question 10.
Name any four important functions of epithelial tissue and provide at least one example of a tissue that exemplifies each function.
Answer:

Question 11.
Write the classification of connective tissue and their functions.
Answer:

Question 12.
What is an epithelium? Enumerate the characteristic features of different epithelia. Epithelial tissue is a sheet of cells that covers the body surface or lines the body cavity.
Answer:

Part II

11th Bio Zoology Guide Tissue Level of Organisation Additional Important Questions and Answers

I. Choose The Best Options

Question 1.
What are the types of epithelium
a. Simple squamous epithelium
b. Simple cuboidal epithelium
c. Simple columnar epithelium
d. Stratified epithelium
Answer:
a. Simple squamous epithelium

Question 2.
Which one of the following is not the functions of the epithelium.
a. Protection
b. Absorption
c. Reproduction
d. Excretion
Answer:
c. Reproduction

Question 3.
Find out the epithelium with irregular boundaries
a. Ciliated epithelium
b. Squamous epithelium
c. Columnar epithelium
d. Pseudostratified epithelium
Answer:
b. Squamous epithelium

Question 4.
Name the epithelium which helps in protection, absorption and secretion.
a. Pseudostratified epithelium
b. Compound epithelium
c. Cuboidal epithelium
d. Columnar epithelium
Answer:
a. Pseudostratified epithelium

Question 5.
Name the tissue which have numerous mitochondria ?
a. Brown adipose tissue
b. White adipose tissue
c. Dense connective tissue
d. Loose connective tissue
Answer:
a. Brown adipose tissue

Question 6.
Match and find the correct answers
I. Ciliated epithelium – a. Outer skin
II. Ciliated epithelium – b. Heart
III. Squamous epithelium – c. Gall bladder
IV. Compound epithelium – d. Ureter
a. I – c, II – b, III – d, IV – a
b. I – b, II – c, III – d, IV – a
c. I – a, II – b, III – c, IV – d
d. I – d, II – c, III – b, IV – a
Answer:
d. I – d, II – c, III – b, IV – a

Question 7.
Find out the wrong pair.
a. Exocrine glands – Saliva
b. Endocrine glands – Hormones
c. Ants – Adipocytes
d. Blood – Fluid connective tissue
Answer:
c. Ants – Adipocytes

Question 8.
Name the tissues present in osteocytes
a. Connective tissue
b. Bone tissue
c. Blood
d. Adipose
Answer:
b. Bone tissue

Question 9.
What are myofibrils?
a. Minute fibrils of muscle fibres
b. Fibers of epithelial tissues
c. The end of nerve tissue
d. In cardiac muscles
Answer:
a. Minute fibrils of muscle fibres

Question 10.
Match and find the correct answer
I. Simple squamous epithelium – a. Respiratory tract
11. Simple Cuboidal epithelium – b. Intestine
III. Simple columnar epithelium – c. Kidney
IV. Ciliated epithelium – d. Alveoli
a. I – a, II – b, III – c, IV – d
b. I – d, II – c, III – b, IV – a
c. I – c, II – d, III – a, IV – b
d. I – a, II – c, III – b, IV – d
Answer:
b. I – d, II – c, III – b, IV – a

(2 marks)

II. Very Short Questions

Question 1.
Define tissues.
Answer:
Group of cells that are similar in structure and perform common or related functions are called tissues.

Question 2.
Define organ system?
Answer:
If two or more organs perform common physical and chemical functions they are called “organ systems”.

Question 3.
What are the four types of tissues?
Answer:
Simple epithelium:

• It consists of a simple layer.
• It helps in protection, absorption, filtration, excretion, secretion, and sensory reception.

Compound epithelium:

• It is multilayered.
• It provides protection against chemical and mechanical stresses.

Question 4.
What is epithelial tissue? What are its types?
Answer:
It is a sheet of cells that covers the body surface or lines the body cavity.
Types:

1. Simple epithelium
2. Compound epithelium

Question 5.
What are the functions of epithelial tissues?
Answer:

• Outer covering
• Protection
• Absorption
• Excretion
• Secretion

Question 6.
What is unicellular glandular epithelium?
Answer:
It consists of isolated glandular cells.
(Eg.) Goblet cells of the alimentary canal.

Question 7.
Based on the secretion how are exocrine glands classify?
Answer:

1. Merocrine
2. Holocrine
3. Apocrine

Question 8.
Where are connective tissues originated from?
Answer:
Connective tissues originated from Mesoderm.
Types of connective tissue: Bones and blood, Cartilage.

Question 9.
What are the functions of connective tissue?
Answer:

• Binding
• Support
• Protection
• Insulation
• Transportation

Question 10.
What are the fibres present in the connective tissues?
Answer:

• Collagen
• Elastic
• Reticular

Question 11.
What is meant by myofibrils?
Answer:
Each muscle is made of many long cylindrical fibers arranged in parallel arrays known as myofibrils.

Question 12.
What are involuntary muscles?
Answer:
Smooth muscles are involuntary as their functions cannot be directly controlled. (Eg.) Blood vessels, Stomach intestine

Question 13.
What is the unit of nervous system and name the tissues which made the nervous system?
Answer:
The unit of nervous system is neuron. Cells:

• Excitable cells
• Neuroglial cells.

Question 14.
What is the function of compound stratified epithelium and where is it seen?
Answer:
Uses:
Protection, secretion and absorption.

Site of occurrence:
Ciliated epithelium — Respiratory tract. Nonciliated epithelium – Epididymis urethra of male.

Question 15.
What is meant by tissue fluid? What is its composition?
Answer:
The areolar connective tissue acts as a reservoir of water and salts for the surrounding body tissue. Hence it is called tissue fluid.
Composition:

• Fibroblasts
• Macrophages
• Mast cells

Question 16.
What is Ehler’s Danlos syndrome?
Answer:
Defect in the synthesis of collagen in the joints heart values organ walls and arterial walls.

Question 17.
What is stickler syndrome?
Answer:
It is a defect which affects collagen and results in facial abnormalities.

Question 18.
What is Rhabdo Myo sarcoma?
Answer:
It is a life-threatening soft tissue tumour of head neck and urinogenital tract.

Question 19.
What is Rheumatoid arthritis?
Answer:
The immune cell attack and inflame the membranes around the joints.

Question 20.
What is Sjogren’s syndrome?
Answer:
It is a disease in which progressive inability to secrete saliva and tears.

Question 21.
What is Palmaris muscle?
Answer:
It is a long narrow muscle run from the elbow to the wrist and is important for hanging and climbing in primates.

Question 22.
What is Parkinson’s disease?
Answer:
It is a degenerative disorder of the nervous system that affects movement often including tremors.

Question 23.
What is Alzheimer’s disease?
Answer:
It is a chronic neurodegenerative disease which includes the symptoms of difficulty in remembering recent events.

Question 24.
What is Biopsy?
Answer:
It is an examination of tissue or liquid removed from a living body to discover presence cause or extent of a disease.

Question 25.
What is an autopsy?
Answer:
It is a dissection of a dead body (Post – mortem) examination to discover the cause of death or the extent of disease.

Question 26.
What is Forensic science?
Answer:
It is the field of science that effectively uses histological techniques to trace out crimes.

Question 27.
The multicellular epithelium helps protect and prevent friction. What is special about the unicellular epithelium
Answer:
The unicellular epithelium is made up of a single layer of cells. These are seen in the organs which do the functions of absorption secretion and filtration.

Question 28.
What is acinus?
Answer:
It is the unit of secretion.

Question 29.
What are adipocytes?
Answer:
This is composed of big adipose tissue in the centre and the cytoplasm is seen as a thin covering layer.

Question 30.
What is the substrate of bone tissue?
Answer:
The mineral hydroxyapatite is a substrate of bone tissue.

(3 marks)

III. Short Questions 

Question 1.
Explain the types of simple epithelium.
Answer:
Simple epithelium is a simple layered sheet of cells that covers the body surface or lines the body cavity.
Types:
1. Squamous epithelium:
It is made of flattened cells with irregular boundaries. It is found in glomeruli, air sacs of lungs, the lining of heart, blood vessels.

2. Cuboidal epithelium:
It is made of cube-like cells. It is found in kidney tubules, ducts, and glands. It is important for secretion and absorption.

3. Columnar epithelium:
It is made of column-like cells. It lines the digestive tract. It is important for secretion and absorption.

4. Ciliated epithelium:
It has cilia at the free end. It is found in bronchi, uterine tubes. It is helpful in propelling materials.

5. Glandular epithelium:
Cuboidal or columnar epithelium specialized for secretion is called the glandular epithelium. E.g., goblet cells and salivary gland.

Question 2.
Compare dense regular connective tissue with dense irregular connective tissues?
Answer:

Question 3.
Classify the muscles and their location?
Answer:
Muscles are of three types.

1. Skeletal muscles
2. Smooth muscles
3. Cardiac muscles.

1. Skeletal muscles:
These muscle fibres are bundled together in a parallel fashion.
Location: They are closely attached to skeletal
bones.

2. Smooth muscles:
These fibres are tapered at both ends and do not show striations.
Location: Blood vessels stomach intestine.

3. Cardiac Muscles:
In this muscles, cell junction fuse the plasma membranes of cardiac muscle cells and make them stick together.
Location: Heart.

Question 4.
Explain compound epithelium.
Answer:

• The compound epithelium is made up of multilayered cells.
• These protect organs against chemical and mechanical stresses.
• These cover the dry surface of the skin, the moist surface of the buccal cavity, pharynx, inner lining of ducts of salivary glands, and pancreatic ducts.

(5 marks)

IV. Essay Questions

Question 1.
What is glandular epithelium? Describe its types?
Answer:
Some of the epithelial cells get specialized for secretion they are called the glandular epithelium.
I. Based on cellular structure
They are classified as Unicellular (Eg.) Goblet cells of the alimentary canal. Multicellular (Eg.) Salivary gland

II. Based on mode of pouring
Exocrine glands – The products are released through ducts. (Eg.) Mucus secreting glands, Saliva secreting glands.
Endocrine glands – They do not have ducts. Their secretions directly secreted into the fluid bathing the glands.
(Eg.) Pituitary gland.

a) Exocrine glands based on cellular nature

•  Unicellular
• Multicellular

b) Exocrine based on the structure

•  Simple
• Compound glands

c) Based on their secretary units

• tubular
• alveolar
• tubulo alveolar

d) Based on their mode of secretion

• Merocrine
• Holocrine
• Apocrine.

Question 2.
What is meant by compound gland? What are its uses and its types?
Answer:
It is made up of more than one layer of cells.

Uses:
It helps in providing protection against chemical and mechanical stresses.

Location:
Buccal cavity, Pharynx salivary glands, Pancreatic ducts

Types:

• Stratified squamous epithelium. It is present in the dry epidermis of the skin,
• Keratinized type-Mouth and vagina.
• Non – Keratinized type
• Stratified cuboidal epithelium – Sweat glands, Mammary gland
• Columnar epithelium-Pharynx urethra
• Transitional epithelium – Ureters, urinary bladder.

Question 3.
Explain the types of muscle.
Answer:
Each muscle is made of long, cylindrical fibres. They are composed of fine fibrils called myofibrils. Muscle fibres contract and relax. Skeletal muscle is attached to skeletal bones. It is striped or striated. It is a voluntary muscle. The smooth muscle fibres are fusiform and do not have striations. It is an involuntary muscle. Cardiac muscle tissue is present in the heart. It is striated and branched and involuntary.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Chemistry Guide Pdf Chapter 9 Solutions Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

11th Chemistry Guide Solutions Text Book Back Questions and Answers

Textbook Evaluation:

I. Choose the best answer:

Question 1.
The molality of a solution containing 1.8 g of glucose dissolved in 250 g of water is
a) 0.2 M
b) 0.01 M
c) 0.02 M
d) 0.04 M
Answer:
d) 0.04 M

Question 2.
Which of the following concentration terms is / are independent of temperature
a) molality
b) molarity
c) mole fraction
d) a and b
Answer:
d) a and b

Question 3.
Stomach acid, a dilute solution of HCl can be neutralized by reaction with aluminium hydroxide Al(OH)₃ + 3HCl (aq) -> AlCl₃ + 3H₂O. How many milliliters of 0.1 M Al(OH)₃ solution is needed to neutralize 21 ml of 0.1 M HCl?
a) 14 mL
b) 7 mL
c) 21 mL
d) none of these
Answer:
b) 7 mL

Question 4.
The partial pressure of nitrogen in air is 0.76 atm and its Henry’s law constant is 7.6 × 10⁴ atm at 300 K. What is the mole fraction of nitrogen gas in the solution obtained when air is bubbled through water at 300 K ?
a) 1 × 10^-4
b) 1 × 10⁴
c) 2 × 10^-5
d) 1 × 10^-5
Answer:
d) 1 × 10^-5

Question 5.
The Henry’s law constant for the solubility of Nitrogen gas in water at 350 K is 8 × 10⁴ atm. The mole fraction of nitrogen in air is 0.5. The number of moles of Nitrogen from air dissolved in 10 moles of water at 350 K and 4 atm pressure is
a) 4 × 10^-4
b) 4 × 10⁴
c) 2 × 10^-2
d) 2.5 × 10^-4
Answer:
d) 2.5 × 10^-4

Question 6.
Which one of the following is incorrect for an ideal solution?
a) ∆H_mix = 0
b) ∆U_mix =0
c) ∆P = P_observed – P_{calculated by Raoults law} = 0
d) ∆G_mix = 0
Answer:
d) ∆G_mix = 0

Question 7.
Which one of the following gases has the lowest value of Henry’s law constant?
a) N₂
b) He
c) CO₂
d) H₂
Answer:
c) CO₂

Question 8.
P₁ and P₂ are the vapour pressures of pure liquid components, 1 and 2 respectively of an ideal binary solution If x₁ represents the mole fraction of component 1, the total pressure of the solution formed by 1 and 2 will be
a) P₁ + x₁ (P₂ – P₁)
b) P₂ – x₁ (P₂ + P₁)
c) P₁ – x₂(P₁ – P₂)
d) P₁ + x₂(P₁ – P₂)
Answer:
c) P₁ – x₂(P₁ – P₂)

Question 9.
Osomotic pressure (π) of a solution is given by the relation
a) π = nRT
b) πV = nRT
c) πRT = n
d) none of these
Answer:
b) πV = nRT

Question 10.
Which one of the following binary liquid mixtures exhibits positive deviation from Raoults law?
a) acetone + chloroform
b) water + nitric acid
c) HCl + water
d) ethanol + water
Answer:
d) ethanol + water

Question 11.
The Henry’s law constants for two gases A and B are x and y respectively. The ratio of mole fractions of A to B 0.2. The ratio of mole fraction of B and A dissolved in water will be
a) \(\frac{2 x}{y}\)

b) \(\frac{y}{0.2 x}\)

c) \(\frac{0.2 x}{y}\)

d) \(\frac{5 x}{y}\)
Answer:
d) \(\frac{5 x}{y}\)

Question 12.
At 100°C the vapour pressure of a solution containing 6.5g a solute in 100g water is 732 mm. If K_b = 0.52, the boiling point of this solution will be
a) 102°C
b) 100°C
c) 101°C
d) 100.52°C
Answer:
c) 101°C

Question 13.
According to Raoults law, the relative lowering of vapour pressure for a solution is equal to
a) mole fraction of solvent
b) mole fraction of solute
c) number of moles of solute
d) number of moles of solvent
Answer:
b) mole fraction of solute

Question 14.
At same temperature, which pair of the following solutions are isotonic?
a) 0.2 M BaCl₂ and 0.2 M urea
b) 0.1 M glucose and 0.2 M urea
c) 0.1 M NaCl and 0.1 M K₂SO₄
d) 0.1 M Ba(NO₃)₂ and 0.1 M Na₂SO₄
Answer:
d) 0.1 M Ba(NO₃)₂ and 0.1 M Na₂SO₄

Question 15.
The empirical formula of a non – electrolyte (X) is CH₂O. A solution containing six grams of X exerts the same osmotic pressure as that of 0.025 M glucose solution at the same temperature. The molecular formula of X is
a) C₂H₄O₂
b) C₈H₁₆O₈
c) C₄H₈O₄
d) CH₂O
Answer:
b) C₈H₁₆O₈

Question 16.
The K_H for the solution of oxygen dissolved in water is 4 × 10⁴ atm at a given temperature. If the partial pressure of oxygen in air is 0.4 atm, the mole fraction of oxygen in solution is
a) 4.6 × 10³
b) 1.6 × 10⁴
c) 1 × 10^-5
d) 1 × 10⁵
Answer:
c) 1 × 10^-5

Question 17.
Normality of 1.25 M sulphuric acid is
a) 1.25 N
b) 3.75 N
c) 2.5 N
d) 2.25 N
Answer:
c) 2.5 N

Question 18.
Two liquids X and Y on mixing gives a warm solution. The solution is
a) ideal
b) non-ideal and shows positive deviation from Raoults law
c) ideal and shows negative deviation from Raoults Law
d) non-ideal and shows negative deviation from Raoults Law
Answer:
d) non-ideal and shows negative deviation from Raoults Law

Question 19.
The relative lowering of vapour pressure of a sugar solution in water is 2.5 × 10^-3. The mole fraction of water in that solution is
a) 0.0035
b) 0.35
c) 0.0035/18
d) 0.9965
Answer:
d) 0.9965

Question 20.
The mass of a non – volatile solute (molar mass 80 g mol^-1) which should be dissolved in 92g of toluene to reduce its vapour pressure to 90%
a) 10 g
b) 20 g
c) 9.2 g
d) 8 g
Answer:
d) 8 g

Question 21.
For a solution, the plot of osmotic pressure (π) versus the concentration (c in mol L^-1) gives a straight line with slope 310 R where ‘R’ is the gas constant. The temperature at which osmotic pressure measured is
a) 310 × 0.082 K
b) 310° C
c) 37°C
d) \(\frac{310}{0.082}\) K
Answer:
c) 37°C

Question 22.
200 ml of an aqueous solution of a protein contains 1.26 g of protein. At 300 K, the osmotic pressure of this solution is found to be 2.52 × 10^-3 bar. The molar mass of protein will be (R = 0.083 L bar mol^-1 K^-1}
a) 62.22 kg mol^-1
b) 12444 g mol^-1
c) 300 g mol^-1
d) None of these
Answer:
a) 62.22 kg mol^-1

Question 23.
The Van’t Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is
a) 0
b) 1
c) 2
d) 3
Answer:
d) 3

Question 24.
Which is the molality of a 10% w/w aqueous sodium hydroxide solution?
a) 2.778
b) 2.5
c) 10
d) 0.4
Answer:
b) 2.5

Question 25.
The correct equation for the degree of an associating solute, ‘n’ molecules of which undergoes association in solution, is
a) α = \(\frac{\mathrm{n}(\mathrm{i}-1)}{\mathrm{n}-1}\)

b) α² = \(\frac{n(1-i)}{(n-1)}\)

c) α = \(\frac{\mathrm{n}(\mathrm{i}-1)}{1-\mathrm{n}}\)

d) α = \(\frac{\mathrm{n}(1-\mathrm{i})}{\mathrm{n}(1-\mathrm{i})}\)
Answer:
c) α = \(\frac{\mathrm{n}(\mathrm{i}-1)}{1-\mathrm{n}}\)

Question 26.
Which of the following aqueous solutions has the highest boiling point?
a) 0.1 M KNO₃
b) 0.1 M Na₃PO₄
c) 0.1 M BaCl₂
d) 0.1 M K₂SO₄
Answer:
b) 0.1 M Na₃PO₄

Question 27.
The freezing point depression constant for water is 1.86° K Kg mol^-1. If 5 g Na₂SO₄ is dissolved in 45 g water, the depression in freezing point is 3.64°C. The Vant Hoff factor for Na₂SO₄ is
a) 2.57
b) 2.63
c) 3.64
d) 5.50
Answer:
a) 2.57

Question 28.
Equimolal aqueous solutions of NaCl and KCl are prepared,. If the freezing point of NaCl is -2°C, the freezing point of KCl solution is expected to be
a) -2°C
b) -4°C
c) -1°C
d) 0°C
Answer:
a) -2°C

Question 29.
Phenol dimerises in benzene having van’t Hoff factor 0.54. What is the degree of association?
a) 0.46
b) 92
c) 46
d) 0.92
Answer:
d) 0.92

Question 30.
Assertion:
An ideal solution obeys Raoults Law.
Reason:
In an ideal solution, solvent – solvent as well as solute – solute interactions are similar to solute-solvent interactions.
a) both assertion and reason are true and reason is the correct explanation of assertion
b) both assertion and reason are true but reason is not the correct explanation of assertion
c) assertion is true but reason is false
d) both assertion and reason are false
Answer:
a) both assertion and reason are true and reason is the correct explanation of assertion

II. Write brief answer to the following questions:

Question 31.
Define:
(i) Molality
(ii) Normality
Answer:
(i)Molality :
Molality (m) is defined as the number of moles of the solute dissolved in one kilogram (Kg) of the solvent. The units of molality are moles per kilogram, i.e., mole kg^-1. The molality is preferred over molarity if volume of the solution is either expanding or contracting with temperature.
molality (m) = \(\frac{\text { Number of mole of solute }}{\text { mess of solvent in } \mathrm{kg}}\)

ii) Normality:
Normality (N) of a solution is defined as the number of gram equivalents of the solute present in one liter of the solution. Normality is used in acid-based redox titrations.
Normality (N) = \(\frac{\text { Number of gram equivalents of solute }}{\text { Volume of solution in litre }}\)

Question 32.
a) What is a vapour pressure of liquid?
Answer:
“The pressure exerted by the vapors above the liquid surface which is in equilibrium with the liquid at a given temperature is called vapor pressure”.

b) What is a relative lowering of vapour pressure?
Answer:
The relative lowering of vapour pressure is defined as the ratio of lowering of vapour pressure to the vapour pressure
of pure solvent (P₀) RLVP = \(\frac{p^{0}-P}{P^{0}}\)

Question 33.
State and explain Henry’s law.
Answer:
“The partial pressure of the gas in vapor phase (vapour pressure of the solute) is directly proportional to the mole fraction (x) of the gaseous solute in the solution at low concentrations”. This statement is known as Henry’s law.
Henry’s law can be expressed as,
P_solute α x_{solute in solution}
P_solute = K_H x_{solute solution}
Here, P_solute represents the partial pressure of the gas in vapour state which is commonly called as vapour pressure. X_solute in solution represents the mole fraction of solute in the solution. K_H is a empirical constant with the dimensions of pressure.

Question 34.
State Raoult law and obtain expression for lowering of vapour pressure when nonvolatile solute is dissolved In solvent.
Answer:
In an ideal solution, the vapour pressure of the solution is decreased when a non-volatile solute is dissolved in a solvent. The magnitude of decrease in the vapour pressure of the solution depends on the amount of solute added.
Let us consider the solution with the following features.
Mole fraction of the solvent = x_A
Mole fraction of the solute = x_B
Vapour pressure of the pure solvent = P°_A
Vapour pressure of solution = P
As the solute is nonvolatile, the vapour pressure of the solution is only due to the solvent. Therefore, the vapour pressure of the solution (P) will be equal to the vapour pressure of the solvent (P_A) over the solution.
i.e., P = P_A
According to Raoult’s law, the vapour pressure of solvent over the solution is equal to the product or its vapour pressure in a pure state and its mole fraction.
P_A = P°_A x_A or
P = P°_A x_A

Question 35.
What is molal depression constant? Does it depend on nature of the solute?
Answer:
If m = 1 then ∆T_f = K_f
“Then K_f is equal to the depression in freezing point for 1 molal solution”. No, it does not depends on nature of the solute.

Question 36.
What is osmosis?
Answer:
“The phenomenon of the flow of solvent through a semipermeable membrane from pure solvent to the solution is called osmosis”. Osmosis can also be defined as “the excess pressure which must be applied to a solution to prevent the passage of solvent into it through the semipermeable membrane”. Osmotic pressure is the pressure applied to the solution to prevent osmosis.

Question 37.
Define the term ‘isotonic solution’.
Answer:
Two solutions having same osmotic pressure at a given temperature are called isotonic solutions.

Question 38.
You are provided with a solid. ‘A’ and three solutions of A dissolved in water – one saturated, one unsaturated, and one supersaturated. How would you determine which solution is which?
Answer:
(A) Unsaturated solution:
It can dissolve salt an additional to it.
(B) Saturated solution:
Further solubility of salt does not takes place but solubility can takes place on heating.
(c) Supersaturated solution:
Solubility of salt do not takes place on even an further heating.

Question 39.
Explain the effect of pressure on the solubility.
Answer:
Generally, the change in pressure does not have any significant effect in the solubility of solids and liquids as they are not compressible. However, the solubility of gases generally increases with increase of pressure.

Consider a saturated solution of a gaseous solute dissolved in a liquid solvent in a closed container. In such a system, the following equilibrium exists.
Gas                      ⇌     Gas
(in gaseous state) (in solution)

According to Le-Chatelier principle, the increase in pressure will shift the equilibrium in the direction which will reduce the pressure. Therefore, more number of gaseous molecules dissolves in the solvent and the solubility increases.

Question 40.
A sample of 12 M Concentrated hydrochloric acid has a density 1.2 M gL^-1 calculate the molality.
Solution:
Given:
Molarity = 12 M HCl
density of solution = 1.2 g L^-1
In 12 M HCl solution, there are 12 moles of HCl in 1 litre of the solution.
Molality = \(\frac{\text { no of moles of solute }}{\text { mass of solvent (in } \mathrm{kg} \text { ) }}\)
Calculate mass of water(solvent)
mass of 1 litre HCl solution = density × volume
= 1.2 × gmL^-1 × 1000 mL = 1200 g
mass of HCl = no. of moles of HCl × molar mass of HCl
= 12 mol × 36.5 g mol^-1
= 438 g.
mass of water = mass of HCl solution – mass of HCl
mass of water = 1200 – 438 = 762 g
molality(m) = \(\frac{12}{0.762}\) = 15.75 m

Question 41.
A 0.25 M glucose solution, at 370.28 K has approximately the pressure as blood does what is the osmotic pressure of blood?
Solution:
C = 0.25 M
T = 370.28 K
(π)_glucose = CRT
(π) = 0.25 mol L^-1 × 0.082L atm K^-1mol^-1 × 370.28K
= 7.59 atm

Question 42.
Calculate the molality of a solution containing 7.5 g glycine(NH₂-CH₂-COOH) dissolved in 500g of water.
Solution:

Question 43.
Which solution has the lower freezing point? 10 g of methanol (CH₃OH) in 100g g of water (or) 20 g of ethanol (C₂H₅OH) in 200 g of water.
Solution:
∆T_f = K_f i.e
∆T_f α m
m_CH3-OH = \(\frac{\left(\frac{10}{32}\right)}{0.1}\)
= 3.125 m

m_C2H5-OH = \(\frac{\left(\frac{20}{46}\right)}{0.2}\)
= 2.174 m

∴ Depression in freezing point is more in methanol solution and it will have lower freezing point.

Question 44.
How many moles of solute particles are present in one liter of 10^-4 M potassium sulphate?
Answer:
In 10^-4 M K₂SO₄ solution, there are 10^-4 moles of potassium sulphate.
K₂SO₄ molecule contains 3 ions (2K+ and 1 SO₄^2-)
1 mole of K₂SO₄ molecule contains 3 × 6.023 × 10²³ ions
10^-4 mole of K₂SO₄ contains 3 × 6.023 × 10²³ × 10^-4 ions
= 18. 069 × 10¹⁹

Question 45.
Henry’s law constant for solubility of methane in benzene is 4.2 × 10^-5 mm Hg at a particular constant temperature. At this temperature calculate the solubility of methane at
i) 750 mm Hg
ii) 840 mm Hg.
Solution:
(K_H)_benzene = 4.2 × 10^-5 mm
Solubility of methane =?
P = 750 mm Hg P = 840 mm Hg
According to Henrys Law,
P = K_H X_{in solution}
750 mm Hg = 4.2 × 10^-5 mm Hg. X_{in solution}
⇒ X_{in solution} = \(\frac{750}{4.2 \times 10^{-5}}\)

i. e solubility = 178. 5 × 10⁵
similarly at P = 840 mm Hg
solubility = \(\frac{840}{4.2 \times 10^{-5}}\) = 200 × 10^-5

Question 46.
The observed depression in freezing point of water for a particular solution is 0.093°C calculate the concentration of the solution in molality. Given that molal depression constant for water is 1.86 K Kg mol^-1.
Solution:
∆T_f = 0.093°C = 0.093 K, m = ?
K_f = 1.86 K Kg mol^-1
∆T_f = K_f.m
∴ m = \(\frac{\Delta \mathrm{T}_{\mathrm{f}}}{\mathrm{K}_{\mathrm{f}}}=\frac{0.093 \mathrm{~K}}{1.86 \mathrm{~K} \mathrm{Kg} \mathrm{mol}^{-1}}\)
= 0.05 mol Kg^-1 = 0.05 m

Question 47.
The vapour pressure of pure benzene (C₆H₆) at a given temperature is 640 mm Hg. 2.2 g of non – volatile solute is added to 40 g of benzene. The vapour pressure of the solution is 600 mm Hg. Calculate the molar mass of the solute?
Solution:
P°C₆H₆ = 640 mm Hg
W₂ = 2.2 g (non volatile solute)
W₁ = 40 g (benzene)
P_solution = 600 mm Hg
M₂ =?

11th Chemistry Guide Solutions Additional Questions and Answers

I. Choose the best answer:

Question 1.
6.02 × 10²⁰ molecules of urea ate present in 200 ml of its solution, The concentration of urea solution is (N₀ = 6.02 × 10²³ mol^-1)
a) 0.001 M
b) 0.01M
c) 0.02 M
d) 0.10 M
Answer:
c) 0.02 M

Question 2.
Calculate the molarity and normality of a solution containing 0.5 g of NaOH dissolved in 500 ml solution
a) 0.0025 M, 0.025 N
b) 0.025 M, 0.025 N
c) 0.25 M, 0.25 N
d) 0.025M, 0.0025 N
Answer:
b) 0.025 M, 0.025 N

Question 3.
5 ml of N HCl, 20 ml of N/2 H₂SO₄ and 30 ml of N/3 HNO₃ are mixed together and volume made to one liter. The normality of the resulting solution is
a) \(\frac{\mathrm{N}}{40}\)

b) \(\frac{\mathrm{N}}{10}\)

c) \(\frac{\mathrm{N}}{20}\)

d) \(\frac{\mathrm{N}}{5}\)
Answer:
a) \(\frac{\mathrm{N}}{40}\)

Question 4.
At 25°C, the density of 15 M H₂SO₄ is 1.8 g cm^-3. Thus, mass percentage of H₂SO₄ in aqueous solution is
a) 2%
b) 81.6%
c) 18%
d) 1.8%
Answer:
b) 81.6%

Question 5.
Mole fraction of C₃H₅(OH)₃ in a solution of 36 g of water and 46 g of glycerine is :
a) 0.46
b) 0.36
c) 0.20
d) 0.40
Answer:
c) 0.20

Question 6.
The molality of a urea solution in which 0.0100 g of urea, [(NH₂)₂CO] is added to 0.3000 dm³ of water at STP is
a) 0.555 m
b) 5.55 × 10^-4
c) 33.3 m
d) 3.33 × 10^-2 m
Answer:
b) 5.55 × 10^-4

Question 7.
15 grams of methyl alcohol is dissolved in 35 grams of water. What is the mass percentage of methyl alcohol in solution?
a) 30%
b) 50%
c) 70%
d) 75%
Answer:
a) 30%

Question 8.
A 3.5 molal aqueous solution of methyl alcohol (CH₃OH) is supplied. What is the mole fraction of methyl alcohol in the solution?
a) 0.100
b) 0.059
c) 0.086
d) 0.050
Answer:
b) 0.059

Question 9.
In which mode of expression of concentration of a solution remains independent of temperature?
a) Molarity
b) Normality
c) Formality
d) Molality
Answer:
d) Molality

Question 10.
Calculate the molarity of pure water (d = 1 g/L)
a) 555 M
b) 5.55 M
c) 55. 5 M
d) None
Answer:
c) 55. 5 M

Question 11.
Calculate the quantity of sodium carbonate (anhydrous) required to prepare 250 ml solution
a) 2.65 grams
b) 4.95 grams
c) 6.25 grams
d) None of these
Answer:
a) 2.65 grams

Question 12.
Find the molality of H₂SO₄ solution whose specific gravity is 1.98 g ml^-1 and 95 % by volume H₂SO₄
a) 7.412
b) 8.412
c) 9.412
d) 10.412
Answer:
c) 9.412

Question 13.
Calculate molality of 1 liter solution of 93 % H₂SO₄ by volume. The density of solution is 1.84 g ml^-1
a) 9.42
b) 10.42
c) 11.42
d) 12.42
Answer:
b) 10.42

Question 14.
Calculate the molality and mole fraction of the solute in aqueous solution containing 3.0 g of urea per 250 gm of water (Mol.wt. of urea = 60).
a) 0.2 m, 0.00357
b) 0.4 m, 0.00357
c) 0.5 m, 0.00357
d) 0.7 m, 0.00357
Answer:
a) 0.2 m, 0.00357

Question 15.
Calculate normality of the mixture obtained by mixing 100 ml of 0.1 N HCl and 50 ml of 0.25 N NaOH solution.
a) 0.0467 N
b) 0.0367 N
c) 0.0267 N
d) 0.0167 N
Answer:
d) 0.0167 N

Question 16.
300 ml 0.1 M HCl and 200 ml of 0.03 M H₂SO₄ are mixed. Calculate the normality of the resulting mixture
a) 0.084 N
b) 0.84 N
c) 2.04 N
d) 2.84 N
Answer:
a) 0.084 N

Question 17.
What weight of oxalic acid (H₂C₂O₄.2H₂O) is required to prepare, 1000mL of N/10 solution?
a) 9.0 g
b) 12.6 g
c) 6.3 g
d) 4.5 g
Answer:
c) 6.3 g

Question 18.
Which of the following units is useful in relating concentration of solution with its vapour pressure?
a) Mole fraction
b) Parts per million
c) Mass percentage
d) Molality
Answer:
a) Mole fraction

Question 19.
The pressure under which liquid and vapour can co-exist at equilibrium is called the
a) Limiting vapour pressure
b) Real vapour pressure
c) Normal vapour pressure
d) Saturated vapour pressure
Answer:
b) Real vapour pressure

Question 20.
CO(g) is dissolved in H₂O at 30°C and 0.020 atm. Henry’s law constant for this system is 6.20 × 10⁴ atm. Thus, mole fraction of CO(g) is
a) 1.72 × 10^-7
b) 3.22 × 10^-7
c) 0.99
d) 0.01
Answer:
b) 3.22 × 10^-7

Question 21.
H₂S gas is used in qualitative analysis of inorganic cations. Its solubility in water at STP is 0.195 mol kg^-1. Thus, Henry’s law constant ( in atm raolaT1) for H₂S is
a) 2.628 × 10^-4
b) 5.128
c) 0.185
d) 3.826 × 10³
Answer:
b) 5.128

Question 22.
Which of the following is correct for a solution showing positive deviations from Raoult’s law?
a) ∆V = +ve, ∆H = + ve
b) ∆V = -ve, ∆H = – ve
c) ∆V = + ve, ∆H = -ve
d) ∆V = – ve, ∆H = +ve
Answer:
a) ∆V = +ve, ∆H = + ve

Question 23.
If liquids A and B form an ideal solution
a) The entropy of mixing is zero
b) The Gibbs free energy is zero
c) The Gibbs free energy as well as the entropy of mixing are each zero
d) The enthalpy of mixing is zero
Answer:
d) The enthalpy of mixing is zero

Question 24.
Water and ethanol form non – ideal solution with positive deviation from Raoult’s law. This solution, will have vapour pressure
a) equal to vapour pressure of pure water
b) less than vapour pressure of pure water
c) more than vapour pressure of pure water
d) less than vapour pressure of pure ethanol
Answer:
c) more than vapour pressure of pure water

Question 25.
Which of the following is less than zero for ideal solutions?
a) ∆H_mix
b) ∆V
c) ∆G_mix
d) ∆S_mix
Answer:
c) ∆G_mix

Question 26.
Which of the following shows negative deviation from Raoult’s law?
a) CHCl₃ and CH₃COCH₃
b) CHCl₃ and C₂H₅OH
c) C₆H₅CH₃ and C₆H₆
d) C₆H₆ and CCl₄
Answer:
a) CHCl₃ and CH₃COCH₃

Question 27.
Given at 350 K, P°_A = 300 torr and P°_B = 800 torr, the composition of the mixture having a normal boiling point of 350 K is :
a) X_A = 0.08
b) X_A = 0.06
c) X_A = 0.04
d) X_A = 0.02
Answer:
a) X_A = 0.08

Question 28.
In mixture A and B, components show – ve deviation as :
a) ∆V_mix is + ve
b) A – B interaction is weaker than A – A and B – B interaction
c) ∆H_mix is + ve
d) A – B interaction is stronger than A – A and B – B interaction
Answer:
d) A – B interaction is stronger than A – A and B – B interaction

Question 29.
If liquid A and B form ideal solution, then:
a) ∆V_mix is = 0
b) ∆V_mix = 0
c) ∆G_mix =0, ∆S_mix = 0
d) ∆S_mix = 0
Answer:
b) ∆V_mix = 0

Question 30.
Which liquid pair shows a positive deviation from Raoult’s law ?
a) Acetone – chloroform
b) Benzene – methanol
c) Water – nitric acid
d) Water – hydrochloric acid
Answer:
b) Benzene – methanol

Question 31.
For A and B to form an ideal solution which of the following conditions should be satisfied ?
a) ∆H_mixing =0
b) ∆V_mixing =0
c) ∆S_mixing =0
d) All three conditions mentioned above
Answer:
d) All three conditions mentioned above

Question 32.
Two liquids are mixed together to form a mixture which boils at same temperature, and their boiling point is higher than the boiling point of either of them so they shows.
a) no deviation from Raoult’s law
b) positive, deviation from Raoult’s law
c) negative-deviation from Raoult’s law
d) positive or negative deviation from Raoult’s law depending upon the composition
Answer:
c) negative-deviation from Raoult’s law

Question 33.
Molal elevation constant of liquid is:
a) the elevation in b.p. which would be produced by dissolving one mole of solute in 1oo g of solvent
b) the elevation of b.p. which would be produced by dissolving 1 mole solute in 10 g of solvent
c) elevation in b.p. which would be produced by dissolving 1 mole of solute in 1000g of solvent
d) none of the above
Answer:
c) elevation in b.p. which would be produced by dissolving 1 mole of solute in 1000g of solvent

Question 34.
The vapour pressure of pure liquid solvent is 0.50 atm. When a non – volatile solute B is added to the solvent, its vapour pressure drops to 0.30 atm. Thus, mole fraction of the component B is
a) 0.6
b) 0.25
c) 0.45
d) 0.75
Answer:
a) 0.6

Question 35.
The mass of a non – volatile solute (molecular mass = 40) which should be dissolved in 114 g octane to reduce its vapour pressure to 80 % will be
a) 20 g
b) 30 g
c) 10 g
d) 40 g
Answer:
c) 10 g

Question 36.
The vapour pressure of pure liquid solvent A is 0.80 atm. When a non – volatile substance B is added to the solvent, its vapour pressure drops to 0.60 atm. Mole fraction of the component B in the solution is:
a) 0.50
b) 0.25
c) 0.75
d) 0.40
Answer:
b) 0.25

Question 37.
18 g of glucose (C₆H₁₂O₆) is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at 100°C is :
a) 752.40 torr
b) 759.00 torr
c) 7.60 torr
d) 76.00 torr
Answer:
a) 752.40 torr

Question 38.
Calculate the vapour pressure of a solution at 100°C containing 3 g of cane sugar in 33 g of water, (at wt. C = 12, H = 1, O = 16)
a) 760 mm
b) 756.90 mm
c) 758.30 mm
d) None of these
Answer:
b) 756.90 mm

Question 39.
Lowering of vapour pressure due to a solute in 1 molal aqueous solution at 100°C is
a) 13.44 mm Hg
b) 14.12 mm Hg
c) 31.2 mm Hg
d) 35.2 mm Hg
Answer:
a) 13.44 mm Hg

Question 40.
The vapour pressure of a dilute aqueous solution of glucose is 750 mm Hg at 373 K. The mole fraction of the solute is
a) \(\frac{1}{76}\)

b) \(\frac{1}{7.6}\)

c) \(\frac{1}{38}\)

d) \(\frac{1}{10}\)
Answer:
a) \(\frac{1}{76}\)

Question 41.
When 3 g of a nonvolatile solute is dissolved in 50 g of water, the relative lowering of vapour pressure observed is 0.018 Nm^-2. Molecular weight of the substance is
a) 60
b) 30
c) 40
d) 120
Answer:
a) 60

Question 42.
Elevation in boiling point of a molar (1M) glucose solution (d = 1.2 gmL^-1) is
a) 1.34 K_b
b) 0.98 K_b
c) 2.40 K_b
d) K_b
Answer:
b) 0.98 K_b

Question 43.
Given, H₂O (l) ⇌ H₂O (g) at 373 K, ∆H° = 8.31 kcal mol^-1. Thus, boiling point of 0.1 molal sucrose solution is
a) 373. 52 K
b) 373.052 K
c) 373.06 K
d) 374.52 K
Answer:
c) 373.06 K

Question 44.
A solution of 0.450 g of urea (mol. Wt. 60) in 22.5 g of water showed 0.170°C of elevation in boiling point. Calculate the molal elevation constant of water.
a) 0.17°C
b) 0.45°C
c) 0.51°C
d) 0.30°C
Answer:
c) 0.51°C

Question 45.
At higher altitudes, water boils at temperature < 100°C because
a) temperature of higher altitudes is low
b) atmospheric pressure is low
c) the proportion of heavy water increases
d) atmospheric pressure becomes more
Answer:
b) atmospheric pressure is low

Question 46.
Which aqueous solution exhibits highest boiling point?
a) 0.015 M glucose
b) 0.01 M KNO₃
c) 0.015 M urea
d) 0.01 M Na₂SO₄
Answer:
d) 0.01 M Na₂SO₄

Question 47.
A solution of urea in water has boiling point of 100.15°C. Calculate the freezing point of the same solution if K_f and K_b for water are 1.87 K kg mol^-1 and 0.52 K kg mol^-1 respectively
a) – 0.54°C
b) – 0.44°C
c) – 0.64°C
d) – 0.34°C
Answer:
a) – 0.54°C

Question 48.
Which will have largest ∆T_b?
a) 180 g glucose in 1 kg water
b) 342 g sucrose in 1,000 g water
c) 18 g glucose in 100 g water
d) 65 g urea in 1kg water
Answer:
d) 65 g urea in 1kg water

Question 49.
An aqueous solution of glucose boils at 100.01°C. The molal elevation constant for water is 0.5 K mol^-1 kg. The number of molecules of glucose in the solution containing 100 g of water is
a) 6.023 × 10²³
b) 12.046 × 10²²
c) 12.046 × 10²⁰
d) 12.046 × 10²³
Answer:
c) 12.046 × 10²⁰

Question 50.
The latent heat of vaporization of water is 9700 cal/mole and if the b.p. is 100°C, ebullioscopic constant of water is
a) 0.513°C
b) 1.026°C
c.) 10.26°C
d) 1.832°C
Answer:
a) 0.513°C

Question 51.
If for a sucrose solution elevation in boiling point is 0.1 °C then what will be the boiling point of NaCl solution for same molal concentration
a) 0.1°
b) 0.2°C
c) 0.08°C
d) 0.01°C
Answer:
b) 0.2°C

Question 52.
The molal boiling point constant for water is 0.513°C kg mol^-1. When 0.1 mole of sugar is dissolved in 200 ml of water, the solution boils under a pressure of one atmosphere at
a) 100.513°C
b) 100.0513°C
c) 100.256°C
d) 101.025°C
Answer:
c) 100.256°C

Question 53.
The boiling point of 0.1 m K₄[Fe(CN)₆] is expected to be (K_b for water = 0.52 K kg mol’1)
a) 100.52°C
b) 100.10°C
c) 100.26°C
d) 102.6°C
Answer:
c) 100.26°C

Question 54.
The value of K_f for the water is 1.86K Kg mole^-1, calculated from glucose solution. The value of K_f for water calculated for NaCl solution will be :
a) = 1.86
b) < 1.86
c) > 1.86
d) zero
Answer:
a) = 1.86

Question 55.
The amount of urea to be dissolved in 500 cc of water (K_f = 1.86) to produce a depression of 0.186°C in the freezing point is :
a) 9 g
b) 6 g
c) 3 g
d) 0.3 g
Answer:
c) 3 g

Question 56.
Freezing point of an aqueous solution is – 0.186°C. Elevation of boiling point of the same solution is if K_b = 0.512 K molality^-1 and K_f= 1.86 K molality^-1
a) 0.186°C
b) 0.0512°C
c) 0.092°C
d) 0.237°C
Answer:
b) 0.0512°C

Question 57.
What should be the freezing point of aqueous solution containing 17 g of C₂H₅OH in 1000 g of water (K_f for water = 1.86 deg kg mol^-1)?
a) – 0.69°C
b) 0.34°C
c) 0.0°C
d) – 0.34°C
Answer:
a) – 0.69°C

Question 58.
The freezing point of equimolal aqueous solution will be highest for:
a) C₆H₅NH₃Cl
b) Ca(NO₃)₂
c) La(NO₃)₂
d) C₆H₁₂O₆
Answer:
d) C₆H₁₂O₆

Question 59.
Cryoscopic constant of a liquid
a) is the decrease in freezing point when 1 g of solute is dissolved per kg of the solvent
b) is the decrease in the freezing point when 1 mole of solute is dissolved per kg of the solvent
c) is the elevation for 1 molar solution
d) is a factor used for calculation of depression in freezing point
Answer:
b) is the decrease in the freezing point when 1 mole of solute is dissolved per kg of the solvent

Question 60.
Which of the following solution will have highest freezing point?
a) 2 M NaCl solution
b) 1.5 M AlCl₃ solution
c) 1 M Al₂(SO₄)₃ solution
d) 3 M Urea solution
Answer:
d) 3 M Urea solution

Question 61.
0.48 g of a substance is dissolved in 10.6 g of C₆H₆. The freezing point of benzene is lowered by 1.8°C. what will be the mol.wt. of the substance (K_f for benzene = 5)
a) 250.2
b) 90.8
c) 125.79
d) 102.5
Answer:
c) 125.79

Question 62.
Which of the following aqueous molal solution have highest freezing point?
a) Urea
b) Barium chloride
c) Potassium bromide
d) Aluminium sulphate
Answer:
a) Urea

Question 63.
What weight of NaCl is added to one liter of water so that ∆T_f/K_f = 1?
a) 5.85 g
b) 0.585 g
c) 0.0585 g
d) 0.0855 g
Answer:
c) 0.0585 g

Question 64.
A solution of glucose (C₆H₁₂O₆) is isotonic With 4 g of urea (NH₂ – CO – NH₂) per liter of solution. The concentration of glucose is :
a) 4 g/L
b) 8 g/L
c) 12 g/L
d) 14 g/L
Answer:
c) 12 g/L

Question 65.
A 5% solution of cane sugar (molar mass = 342) is isotonic with 1% of a solution of unknown solute. The molar mass of unknown solute in g/mol is
a) 136.2
b) 171.2
c) 68.4
d) 34.2
Answer:
c) 68.4

Question 66.
The weight of urea dissolved in 100 ml solution which produce an osmotic pressure of 20.4 atm, will be
a) 5 g
b) 4 g
c) 3 g
d) 6 g
Answer:
a) 5 g

Question 67.
In the phenomenon of osmosis, the membrane allow passage of _________.
a) Solute only
b) Solvent only
c) Both solute and solvent
d) None of these
Answer:
b) Solvent only

Question 68.
A 5.8% (wt./vol.) NaCl solution will exert an osmotic pressure closest to which one of the following:
a) 5.8% (wt./vol.) sucrose solution
b) 5.8% (wt./vol.) glucose solution
c) 2 molal sucrose solution
d) 1 molal glucose solution
Answer:
c) 2 molal sucrose solution

Question 69.
Osmotic pressure of a sugar solution at 24°C is 2.5 atmospheres. Determine the concentration of the solution in gram mole per liter.
a) 0.0821 moles/liter
b) 1.082 moles/liter
c) 0.1025 moles/liter
d) 0.0827moles/liter
Answer:
c) 0.1025 moles/liter

Question 70.
What is the freezing point of a solution that contains 10.0g of glucose C₆H₁₂O6 in 100 g of H₂O? K_f = 1.86° C/m.
a) – 0.186°C
b) + 0.186°C
c) – 0.10°C
d) – 1.03°C
Answer:
d) – 1.03°C

Question 71.
The order of osmotic pressure of equimolar solutions of BaCl2, NaCl and glucose will be:
a) BaCl₂ > NaCl > glucose
b) NaCl > BaCl₂ > glucose
c) glucose > BaCl₂ > NaCl
d) glucose > NaCl > BaCl₂
Answer:
a) BaCl₂ > NaCl > glucose

Question 72.
The wt. of urea dissolved in 100 ml solution which produce an osmotic pressure of 20.4 atm, will be
a) 5 g
b) 4 g
c) 3 g
d) 6 g
Answer:
a) 5 g

Question 73.
A compound MX₂ has observed and normal molar masses 65.6 and 164 respectively. Calculate the apparent degree of ionization of MX₂:
a) 75%
b) 85%
c) 65%
d) 25%
Answer:
a) 75%

Question 74.
The freezing point of 0.2 molal K₂SO₄ is – 1.1°C. Calculate van’t Hoff facor and percentage degree of dissociation of K₂SO₄. K_f for water is 1.86°
a) 97.5
b) 90.75
c) 105.5
d) 85.75
Answer:
a) 97.5

Question 75.
For 0.1M solution, the colligative property will follow the order
a) NaCl > Na₂SO₄ > Na₃PO₄
b) NaCl > Na₂SO₄ ≈ Na₃PO₄
c) NaCl < Na₂SO₄ < Na₃PO₄
d) NaCl < Na₂SO₄ = Na₃PO₄
Answer:
c) NaCl < Na₂SO₄ < Na₃PO₄

Question 76.
P_H of a 0.1M monobasic acid is found to be 2. Hence its osmotic pressure at a given temp. T K is
a) 0.1 RT
b) 0.11 RT
c) 1.1 RT
d) 0.01 RT
Answer:
b) 0.11 RT

Question 77.
Which has the highest boiling point?
a) 0.1 m Na₂SO₄
b) 0.1 m Al(NO₃)₃
c) 0.1 m MgCl₂
d) 0.1 m C₆H₁₂O₆ (glucose)
Answer:
b) 0.1 m Al(NO₃)₃

Question 78.
Aluminium phosphate is 100% ionized in 0.01 molal aqueous solution. Hence ∆T_b/ K_b is:
a) 0.01
b) 0.015
c) 0.0175
d) 0.02
Answer:
d) 0.02

Question 79.
1.0 molal aqueous solution of an electrolyte X₃Y₂ is 25% ionized. The boiling point of the solution is (K_b for H₂O = 0.52 K kg/mol)
a) 373.5 K
b) 374.04 K
c) 377.12 K
d) 373.25 K
Answer:
b) 374.04 K

Question 80.
The freezing point of 0,05 m solutions of a non – electrolyte in water is
a) -1.86 °C
b) -0.93°C
c)-0.093°C
d) 0.93°C
Answer:
c)-0.093°C

Question 81.
For an ideal solution containing a non – volatile solute, which of the following expression is correctly represented?
a) ∆T_b = K_b × m
b) ∆T_b = K_b × M
c) ∆T_b = K_b × 2m
d) ∆T_b = K_b × 2M
Where m is the molality of the solution and K_b is molal elevation constant.
Answer:
a) ∆T_b = K_b × m

Question 82.
If 5.85 g of NaCl are dissolved in 90 g of water, the mole fraction of NaCl is
a) 0.1
b) 0.2
c) 0.3
d) 0.0196
Answer:
d) 0.0196

Question 83.
What will be the molarity of a solution containing 5g of sodium hydroxide in 250 ml solution?
a) 0.5
b) 1.0
c) 2.0
d) 0.1
Answer:
a) 0.5

Question 84.
If 5.85 g of NaCl (molecular weight 58.5) is dissolved in water and the solution is made up to 0.5 liter, the molarity of the solution will be
a) 0.2
b) 0.4
c) 1.0
d) 0.1
Answer:
a) 0.2

Question 85.
To prepare a solution of concentration of 0.03 g/ml of AgNO₃, what amount of AgNO₃ should be added in 60ml of solution
a) 1.8
b) 0.8
c) 0.18
d) None of these
Answer:
a) 1.8

Question 86.
How many g of dibasic acid (mol.wt. 200) should be present in 100ml of its aqueous solution to give decinormal strength?
a) 1 g
b) 2 g
c) 10 g
d) 20 g
Answer:
a) 1 g

Question 87.
The molarity of a solution of Na₂CO₃ having 10.6 g/500 ml of solution is
a) 0.2 M
b) 2 M
c) 20 M
d) 0.02 M
Answer:
a) 0.2 M

Question 88.
Molecular weight of glucose is 180, A solution of glucose which contains 18 g per liter is
a) 2 molal
b) 1 molal
c) 0.1 molal
d) 18 molal
Answer:
c) 0.1 molal

Question 89.
0.5 M of H₂SO₄ is diluted from lliter to 10 liters, normality of resulting solution is
a) 1 N
b) 0.1 N
c) 10 N
d) 11 N
Answer:
b) 0.1 N

Question 90.
An aqueous solution of glucose is 10% in strength. The volume in which 1 g mole of it is dissolved will be
a) 18 liters
b) 9 liters
c) 0.9 liters
d) 1.8 liters
Answer:
d) 1.8 liters

Question 91.
When 1.80 g glucose dissolved in 90 g of H₂O, the mole fraction of glucose is
a) 0.00399
b) 0.00199
c) 0.0199
d) 0.998
Answer:
b) 0.00199

Question 92.
A 5 molar solution of H₂SO₄ is diluted from 1 liter to 10 liters. What is the normality of the solution?
a) 0.25 N
b) 1 N
c) 2N
d) 7 N
Answer:
b) 1 N

Question 93.
Normality of 2 M sulphuric acid is
a) 2 N
b) 4 N
c) N/2
d) N/4
Answer:
b) 4 N

Question 94.
What is the molarity of H₂SO₄ solution, that has a density 1.84 g/cc at 35°C and Contains solute 98% by weight
a) 4.18 M
b) 8.14 M
c) 18.4 M
d) 18 M
Answer:
c) 18.4 M

Question 95.
Which of the following is a colligative property?
a) Osmotic pressure
b) Boiling point
c) Vapour pressure
d) Freezing point
Answer:
a) Osmotic pressure

Question 96.
The vapour pressure of benzene at a certain temperature is 640 mm of Eg. A non – volatile and non – electrolyte solid weighing 2.175 g is added to 39.08 g of benzene. The vapour pressure of the solution is 600 mm of Hg. What is the molecular weight of solid substance?
a) 49.50
b) 59.6
c) 69.5
d) 79.8
Answer:
c) 69.5

Question 97.
The average osmotic pressure of human blood is 7.8 bar at 37°C. What is the concentration of an aqueous NaCl solution that could be used in the Mood stream?
a) 0.16 mol/L
b) 0.32 mol/L
c) 0.60 mol/L
d) 0.45 mol/L
Answer:
b) 0.32 mol/L

Question 98.
The osmotic pressure in atmospheres of 10% solution of cane sugar at 69°C is
a) 724
b) 824
c) 8.21
d) 7.21
Answer:
c) 8.21

Question 99.
The molal boiling point constant for water is 0.513°C kg mol^-1. When 0.1 mole of sugar is dissolved in 200 ml of water, the solution boils under a pressure of one atmosphere at
a) 100.513°C
b) 100.0513°C
c) 100.256°C
d) 101.025°C
Answer:
c) 100.256°C

Question 100.
The freezing point of a solution prepared from 1.25 g of a non – electrolyte and 20 g of water is 271.9 K. If molar depression constant is 1.8 K mole^-1 then molar mass of the solute will be
a) 105.7
b) 106.7
c) 115.3
d) 93.9
Answer:
a) 105.7

Question 101.
Osmotic pressure of 0.1 M solution of NaCl and Na₂SO₄ will be
a) same
b) osmotic pressure of NaCl solution will be more than Na₂SO₄ solution
c) osmotic pressure of Na₂SO₄ solution will be more than NaCl
d) osmotic pressure of NaSO₄ will be less than that of NaCl solution
Answer:
c) osmotic pressure of Na₂SO₄ solution will be more than NaCl

Question 102.
At 25 °C the highest osmotic pressure is exhibited by 0.1 M solution of
a) CaCl₂
b) KCl
c) Glucose
d) Urea
Answer:
a) CaCl₂

Question 103.
Azeotropic mixture of HCl and water has
a) 84% HCl
b) 22.2% HCl
c) 63 % HCl
d) 20.2 HCl
Answer:
d) 20.2 HCl

Question 104.
The boiling point of water (100°C) becomes 100.25°C, if 3 grams of a nonvolatile solute is dissolved in 200 ml of water. The molecular weight of solute is (K_b for water is 0.6 K Kg mol^-1)
a) 12.2 g mol^-1
b) 15.4 g mol
c) 17.3 g mol^-1
d) 20.4 g mol
Answer:
c) 17.3 g mol^-1

II. Very short question and answer(2 Marks):

Question 1.
Define solution.
Answer:
A solution is homogeneous mixture of two or more substances, consisting of atom, ions or molecules. The constituent of the homogeneous mixture present in a lower amount is called the solute, and the one present in larger amount is called the solvent. For example, when a small amount of NaCl dissolved in water.

Question 2.
What is saturated solution?
Answer:
A saturated solution is one that contains the maximum amount of a solute that can dissolve in a solvent at a specific temperature.
For example, the solubility of NaCl in 100 g of water at 20°C is 36 g but at other temperatures, or in other solvents, is different.

Question 3.
What is unsaturated solution?
Answer:
An unsaturated solution is the one that contains less amount of solute than its capacity to dissolve.

Question 4.
What is Supersaturated solution?
Answer:
A super saturated solution is one that contains more dissolved solute than the saturated solution. It is generally not stable and eventually the dissolved solute will separate as crystals.

Question 5.
What is mass percentage?
Answer:
The mass percentage of a component in a solution is the mass of the component present in 100 g of the solution.
Mass percentage of component = \(\frac{\text { Mass of the component in the solution } \times 100}{\text { Total mass of the solution }}\)

Question 6.
What is parts per million (ppm)?
Answer:
If the amount of solute in solution is very much less, then the concentration is expressed as parts per million (ppm).
Parts per million (ppm) = \(\frac{\text { Mass of the solute }(\mathrm{mg})}{\text { Mass of the solvent }} \times 10^{6}\)

Question 7.
What is Molarity?
Answer:
Molarity (symbol M) is defined as the number of moles of solute present in a liter of solution. The units of molarity are moles per liter (mol L^-1) or moles per cubic decimeter (mol dm^-3)
Molarity(m) = \(\frac{\text { Number of ntoles of solute }}{\text { Volume of solution in liter }}\)

Question 8.
What is Non – ideal solution?
Answer:
The solutions which do not obey Raoult’s law over the entire range of concentration, are called non – ideal solutions. For a non – ideal solution, there is a change in the volume and enthalpy upon mixing, i.e.
∆H_mixing ≠ 0 & ∆V_mixing ≠ 0.

Question 9.
State Raoult’s law.
Answer:
According to Raoult’s law, the vapor pressure of solvent over the solution is equal to the product of its vapor pressure in pure state and its mole fraction.
P_A = P°_A X_A or
P = P°_AX_A

Question 10.
Define boiling point.
Answer:
The boiling point of a liquid is the temperature at which its vapour pressure becomes equal to the atmospheric pressure (1 atm).

Question 11.
Define freezing point.
Answer:
Freezing point is defined as “the temperature at which the solid and the liquid states of the substance have the same vapour pressure”.

Question 12.
What is Osmotic pressure?
Answer:
Osmotic pressure can be defined as “the pressure that must be applied to the solution to stop the influx of the solvent (to stop osmosis) through the semi permeable membrane”.

Question 13.
State Dalton’s law.
Answer:
According to Dalton’s law of partial pressure the total pressure in a closed vessel will be equal to the sum of the partial pressures of the individual components.

Question 14.
What is elevation of boiling point? Give it.
Answer:
The temperature difference between the solution and pure solvent is called elevation of boiling point,
∆T_b = T – T°
Unit of ∆T_b is K Kg mole^-1.

III. Short Question and answers(3 Marks):

Question 1.
What is mole fraction?
Answer:
Mole fraction, x, of solute is defined as the ratio of the number of moles of the components divided by the total number of the moles of all the component present in the solution.
Mole fraction of the solute x_solvent = \(\frac{\text { Moles of the component }}{\text { total number of moles of all the component in solution }}\)

Question 2.
What is volume percentage?
Answer:
It is defined as the volume of the component present in 100mL of the solution. If V_a is a volume of solute and V_b is volume of solvent, then
Vollume percentage of solute = \(\frac{V_{a} \times 100}{V_{a}+V_{b}}\)

Question 3.
What are the advantages of using standard solution?
Answer:

1. The error due to weighing the solute can be minimized by using concentrated stock solution that requires large quantity of solute.
2. We can prepare working standards of different concentrations by diluting the stock solution, which is more efficient since consistency is maintained.
3. Some of the concentrated solutions are more stable and are less likely to support microbial growth than working standards used in the experiments.

Question 4.
Limitations of Henry’s law.
Answer:
Henry’s law is applicable at moderate temperature and pressure only. Only the less soluble gases obeys Henry’s law. The gases reacting with the solvent do not obey Henry’s law.
Example:
Ammonia or HCl reacts with water and hence does not obey this law.
NH₃ + H₂O ⇌ NH₄⁺ + OH^–

Question 5.
What is Van’t Hoff factor?
Answer:
It is defined as the ratio of the actual molar mass to the abnormal (calculated) molar mass of the solute. Here, the abnormal molar mass is the molar mass calculated using the experimentally determined colligative property.
i = \(\frac{\text { Normal (actual) molar mass }}{\text { obseved (abnormal) molar mass }}\)

Question 6.
What is ideal solution?
Answer:
An ideal solution is a solution in which each component i.e. the solute as well as the solvent obeys the Raoult’s law over the entire range of concentration.
For an ideal solution,

1. There is no change in the volume on mixing the two components (solute & solvents) (∆V_mixing =0).
2. There is no exchange of heat when the solute is dissolved in solvent (∆H_mixing =0).
3. escaping tendency of the solute and the solvent present in it should be same as in pure liquids.
   Example:
   benzene & toluene; h-hexane & n-heptane; ethyl bromide & ethyl iodide; chlorobenzene & bromobenzene.

Question 7.
What are colligative properties?
Answer:
“ The properties of the solutions which depend only on the number of solute particles but not on the nature of the solute are called colligative properties”. The following four colligative properties are very important.

1. Relative lowering of vapor pressure (∆P)
2. Elevation of boiling point (∆T_b)
3. Depression of freezing point (∆T_b)
4. Osmotic pressure (π)

Question 8.
What are the significances of Osmotic pressure?
Answer:
Unlike elevation of boiling point (for 1 molal solution the elevation in boiling point is 0.512°C for water) and the depression in freezing point (for 1 molal solution the depression in freezing point is 1.86°C for water), the magnitude of osmotic pressure is large.

The osmotic pressure can be measured at room temperature enables to determine the molecular mass of bio molecules which are unstable at higher temperatures. Even for a very dilute solution the osmotic pressure is large.

Question 9.
0.24 g of a gas dissolves in 1L of water at 1.5 atm pressure. Calculate the amount of dissolved gas when the pressure is raised to 6.0 atm at constant temperature?
Solution:
P_solute = K_H X_{solute in solution}
At pressure 1.5 atm,
P₁ = K_HX₁ …………..(1)
At pressure 6.0 atm,
p₂ = K_HX₂ …………….(2)
Dividing equation (1) by (2)
From equation = \(\frac{P_{1}}{P_{2}}=\frac{X_{1}}{X_{2}}\)

There fore
\(\frac{0.24 \times 6.0}{1.5}\) = 0.96 g/L

Question 10.
Why the carbonated drinks are stored in a pressurized container?
Answer:
The carbonated beverages contain carbon dioxide dissolved in them. To dissolve the carbon dioxide in these drinks, the CO₂ gas is bubbled through them under high pressure. These containers are sealed to maintain the pressure. When we open these containers at atmospheric pressure, the pressure of the CO₂ drops to the atmospheric level and hence bubbles of CO₂ rapidly escape from the solution and show effervescence. The burst of bubbles is even more noticeable, if the soda bottle in warm condition.

Question 11.
An aqueous solution of 2% non volatile solute exerts a pressure of 1.004 bar at the boiling point of the solvent. What is the molar mass of the solute when P°_A is 1.013 bar?
Solution:
\(\frac{\Delta P}{P_{A}^{0}}=\frac{W_{B} \times M_{B}}{M_{B} \times W_{A}}\)

In a 2 % solution weight of the solute is 2 g and solvent is 98g
∆P = P_solution – P°_A = 1.013 – 1.004 bar
= 0.009 bar
M_B = \(\frac{\mathrm{P}_{\mathrm{A}}^{0} \times \mathrm{W}_{\mathrm{B}} \times \mathrm{M}_{\mathrm{A}}}{\Delta \mathrm{P} \times \mathrm{W}_{\mathrm{A}}}\)

M_B = 2 × 18 × 1.013/(98 × 0.009) = 41.3 g mol^-1

Question 12.
Ethylene glycol (C₂H₆O₂) can be at used as an antifreeze in the radiator of a car. Calculate the temperature when ice will begin to separate from a mixture with 20 mass percent of glycol in water used in the car radiator. K_f for water = 1.86K Kg mol^-1 and molar mass of ethylene glycol is 62 g mol^-1.
Solution:
Weight of solute (W₂) = 20 mass percent of solution means 20 g of ethylene glycol
Weight of solvent (water) W₁ = 100 – 20 = 80 g
∆T_f = K_f m
= \(\frac{\mathrm{K}_{\mathrm{f}} \times \mathrm{W}_{2} \times 1000}{\mathrm{M}_{2} \times \mathrm{W}_{1}}=\frac{1.86 \times 20 \times 1000}{62 \times 80}\)
= 7.5 K
The temperature at which the ice will begin to separate is the freezing of water after the addition of solute i.e 7.5 K lower than the normal freezing point of water (273 – 7.5 K) = 265.5 K

Question 13.
At 400 K 1.5 g of an unknown substance is dissolved in solvent and the solution is made to 1.5 L. Its osmotic pressure is found to be 0.3 bar. Calculated the molar mass of the unknown substance.
Solution :

Question 14.
The depression in freezing point is 0.24 K obtained by dissolving 1 g NaCl in 200g water. Calculate van’t – Hoff factor. The molar depression constant in 1.86 K Kg mol^-1
Solution :
Molar mass of solute = \(\frac{1000 \times K_{f} \times \text { mass of } N a C l}{\Delta T_{f} \times \text { mass of solvent }}\)

= \(\frac{1000 \times 1.86 \times 1}{0.24 \times 200}\)
= 38.75 g mol^-1

Theoretical molar mass of NaCl is = 58.5 g mol^-1

i = \(\frac{\text { Theoretical molar mass }}{\text { Experimental molar mass }}\)
= \(\frac{58.5}{38.75}\) = 1.50

IV. Long question and answers(5 Marks):

Question 1.
Explain the factors including the solubility of solute?
Answer:
Factors influencing the solute:
The solubility of a solute generally depends on the nature of the solute and the solvent in which it is dissolved. It also depends on the temperature and pressure of the solution.

Nature of solute and solvent:
Sodium-chloride, an ionic compound, dissolves readily iff a polar solvent such as water, but it does not dissolve in non polar-organic solvents such as benzene or toluene. Many organic compounds dissolve -readily in organic solvents and do not dissolve in water. Different gases dissolve in water to different extents: for example, ammonia is more soluble than oxygen in water.

Effect of temperature:

Solid solute in liquid solvent:
Generally, the solubility of a solid solute in a liquid solvent increases with increase in temperature. When the temperature is increased, the average kinetic energy of the molecules of the solute and the solvent increases. The increase in kinetic energy facilitates the solvent molecules to break the intermolecular attractive forces that keep the solute molecules together and hence the solubility increases.

When a solid is added to a solvent, it begins to dissolve. i.e. the solute leaves from the solid state (dissolution). After some time, some of the dissolved solute returns back to the solid state (recrystallisation). If there is excess of solid present, the rate of both these processes becomes equal at a particular stage. At this stage an equilibrium is established between the solid solute molecules and dissolved solute molecules.

Solute(solid) ⇌ Solute(dissolved)

According to Le-Chatelier principle, if the dissolution process is endothermic, the increase in temperature will shift the equilibrium towards left i.e solubility increases, for an exothermic reaction, the increase in temperature decreases the solubility. The solubilities, of ammonium nitrate, calcium Chloride, ceric sulphate nano-hydrate, and sodium chloride in water at different temperatures are given in the following graph.

Plot of solubility versus temperature for selective compounds

The following conclusions are drawn from the above graph:

1. The solubility of sodium chloride does not vary appreciably as the maximum solubility is achieved at normal temperature. In fact, there is only 10 % increase in solubility between 0° to 100 °C.
2. the dissolution process of ammonium nitrate is endothermic, the solubility increases steeply with increase in temperature.
3. In the case of ceric sulphate, the dissolution is exothermic and the solubility decreases with increase in temperature.
4. Even though the dissolution of calcium chloride is exothermic, the solubility increases moderately with increase in temperature. Here, the entropy factor also plays a significant role in deciding the position of the equilibrium.

Gaseous solute in liquid solvent:
In the case of gaseous solute in liquid solvent, the solubility decreases with increase in temperature. When a gaseous solute dissolves in a liquid solvent, its molecules interact with solvent molecules with weak intermolecular forces. When the temperature increases, the average kinetic energy of the molecules present in the solution also increases.

The increase in kinetic energy breaks the weak intermolecular forces between the gaseous solute and liquid solvent which results in the release of the dissolved gas molecules to the gaseous state. Moreover, the dissolution of most of the gases in liquid solvents is an exothermic process, and in such processes, the increase in temperature decreases the dissolution of gaseous molecules.

Question 2.
Explain vapour pressure of liquid in liquids binary solution?
Answer:
Now, let us consider a binary liquid solution formed by dissolving a liquid solute ‘A’ in a pure solvent B in a closed vessel. Both the components A and B present in the solution would evaporate and an equilibrium will be established between the liquid and vapour phases of the components A and B.

The French chemist Raoult, proposed a quantitative relationship between the partial pressures and the mole fractions of two components A & B, which is known as Raoult’s Law. This law states that “in the case of a solution of volatile liquids, the partial vapour pressure of each component (A & B) of the solution is directly proportional to its mole fraction”.
According to Raoult’s law,
P_A ∝ x_A
P_A = k x_A
when x_A = 1, k = P°_A
where p°A is the vapour pressure of pure component A’ at the same temperature. Therefore,
P_A = P°_A x_A
Similarly, for component ‘B’
P_B =P°_B x_B
x_A and x_B are the mole fraction of the components A and B respectively.

According to Dalton’s law of partial pressure the total pressure in a closed vessel will be equal to the sum of the partial pressures of the individual components.
Hence,
P_total = P_A + P_B
Substituting the values of P_A and P_B from equations in the above equation,
P_total = X_A P°_A + X_BP°_B
We know that X_A + X_B = 1 or X_A = 1 – X_B
Therefore,
P_total = (1 – X_B)P°_A + X_BP°_B
P_total = P°_A + X_B(P°_B – P°_A)
The above equation is of the straight¬line equation form y = mx + c. The plot of P_totalversus x_B will give a straight line with (P°_B – P_A) as slope and P°_A as the y intercept.

Let us consider the liquid solution containing toluene (solute) in benzene (solvent).

The variation of vapour pressure of pure benzene and toluene with its mole fraction is given in the graph.

Solution of benzene in toluene obeying Raoult’s law

The vapour pressures of pure toluene and pure benzene are 22.3 and 74.7 mmHg, respectively. The above graph shows, the partial vapour pressure of the pure components increases linearly with the increase in the mole fraction of the respective components. The total pressure at any composition of the solute and solvent is given by the following straight line (represented as red line) equation.
P_Solution = P°_toluene + X_benzene(P_benzene – P_toluene)

Question 3.
Explain Raoult’s law for the binary solution of Non-volatile solutes in liquids?
Answer:
When a nonvolatile salute js dissolved in a pure solvent, the vapour pressure of the pure solvent will decrease. In such solutions, the vapour pressure of the -Solution will depend only on the solvent ‘molecules as the solute is nonvolatile.

For example, when sodium chloride is added to the water, the vapor pressure of the salt solution is lowered. The vapour pressure of the solution is determined by the number of molecules of the solvent present in the surface at any time and is proportional to the mole fraction of the solvent.

Rate of vapourization reduced by presence of nonvolatile solute.

P_solution ∝ X_A
Where X_A is the mole fraction of the solvent
P_solution = kX_A
When X_A = 1, k = P°_solvent
(P°_solvent is the partial pressure of pure solvent)
P_Solution = P°_Solvent
\(\frac{P_{\text {solution }}}{P_{\text {solvent}}^{a}}\) = X_A

1 – \(\frac{p_{\text {Solution }}}{p_{\text {Solvent }}^{0}}\) = 1 – X_A

\(\frac{p_{\text {Solvent }}^{o}-p_{\text {Solurion }}}{P_{\text {Solvent }}^{0}}\) = X_B
Where X_B is the fraction of the solute
(∴ x_A + x_B = 1, X_B = 1 – X_A)

The above expression gives the relative lowering of vapour pressure. Based on this expression, Raoult’s Law can also be stated as “the relative lowering of vapour pressure of an ideal solution containing the nonvolatile solute is equal to the mole fraction of the solute at a given temperature”.

Question 4.
What is ideal solution? Write special features and characters of ideal solution.
Answer:
An ideal solution is a solution in which each component i.e. the solute as well as the solvent obeys the Raoult’s law over the entire range of concentration. Practically no solution is ideal over the entire range of concentration. However, when the concentration of solute is very low, the dilute solution behaves ideally.

If the two components present in the solution (A and B) are identical in size, structure, and having almost similar intermolecular attractive forces between them (i.e. between A-A, B-B, and B-A) and then the solution tends to behave like an ideal solution.

For an ideal solution:
(i) there is no change in the volume on mixing the two components, (solute & solvents). (∆V_mixing = 0)
(ii) there is no exchange of heat when the solute is dissolved in solvent (∆H_mixing = 0).
(iii) escaping tendency of the solute arid the solvent present in it should be same as in pure liquids.
Example:
benzene & toulene; n-hexane & n-heptane; ethyl bromide & ethyl iodide; chlorobenzene & bromobenzene.

Question 5.
Explain Non – ideal solution with strong positive deviation.
Answer:
The solutions which do not Raoult’s law over the entire range, of concentration, are called non-ideal solutions. For a non-ideal solution, there is a change in the volume and enthalpy upon mixing, i.e.
∆H_mixing ≠ 0 &, ∆V_mixing ≠ 0. The deviation of the non-ideal solutions from the Raoult’s law can either be positive or negative.

Non-ideal solutions – positive deviation from Rauolt’s Law:
The nature of the deviation from the Rauolt’s law can be explained in terms ©f the intermolecular interactions between solute and solvent (B). Consider a case in which the intermolecular attractive forces between A and B are weaker than those between the molecules of A (A-A) and molecules of B (B-B).

The molecules present in such a solution have a greater tendency to escape from the solution when compared to the ideal solution formed by A and B, in which the intermolecular attractive forces (A-A, B-B, A-B) are almost similar. Consequently, the vapour pressure of such non-ideal solution increases and it is greater than the sum of the vapour pressure of A and B as predicted by the Raoult’s law. This type of deviation is called positive deviation.

Here, P_A > p°_A X_A and p_B > P°_B X_B
Hence P_total > p°_A X_A + P_B X_B

Let us understand the positive deviation by considering a solution of ethyl alcohol and water. In this solution the hydrogen bonding interaction between ethanol and water is weaker than those hydrogen bonding interactions amongst themselves (ethyl alcohol-ethyl alcohol and water-water interactions).

This results in the increased evaporation of both components from the aqueous solution of ethanol. Consequently, the vapour pressure of the solution is greater than the vapour pressure predicted by Raoults law. Here, the mixing process is endothermic i.e. ∆H_mixing > 0 and there will be slight increase in volume(∆V_mixing > 0).

Example:
Ethyl alcohol cyclohexane, Benzene & acetone, Carbon tetrachloride & chloroform, Acetone & ethyl alcohol, Ethyl alcohol & water.

Question 6.
Explain Non – ideal solution with strong negative deviation.
Answer:
Let us consider a case where the attractive forces between solute (A) and solvent t (B) are stronger – thtSSar intermolecular attractive forces between the individual components (A – A & B – B). Here, the escaping tendency of A and B will be lower when compared with an ideal solution formed by A and B. Hence, the vapour pressure of such solutions will be lower than the sum of the vapour pressure of A and B. This type of deviation is called negative deviation. For the negative deviation,
P_A < P°_A X_A and P_B < P°_B X_B.

Let us consider a solution of phenol and aniline. Both phenol and aniline form hydrogen bonding interactions amongst themselves. However, when mixed with aniline, the phenol molecule forms hydrogen bonding interactions with aniline, which are stronger than the hydrogen bonds formed amongst themselves.

Formation of new hydrogen bonds considerably reduce the escaping tendency of phenol and aniline from the solution. As a result, the vapour pressure of the solution is less and there is a slight decrease in volume (∆V_mixing < 0) on mixing. During this process evolution of heat takes place i.e. ∆H_mixing < 0 (exothermic)

Example:
Acetone + chloroform, Chloroform + diethyl ether, Acetone + aniline, Chloroform + Benzene.

Question 7.
Explain the factors responsible for deviation from Raoult’s law.
Answer:
Factors responsible for deviation from Raoult’s law:
The deviation of solution from ideal behavior is attributed to the following factors.

i) Solute-solvent interactions:
For an ideal solution, the interaction between the solvent molecules (A-A), the solute molecules (B-B) and between the solvent & solute molecules (A-B) are expected to be similar. If these interactions are dissimilar, then there will be a deviation from ideal behavior.

ii) Dissociation of solute:
When a solute present in a solution dissociates to give its constituent ions, the resultant ions interact strongly with the
solvent and cause deviation from Raoult’s law.
For example, a solution of potassium chloride in water deviates from ideal behavior because the solute dissociates to give K and Cl ion which form strong ion-dipole interaction with water molecules.
KCl(s) + H₂O (l) → K⁺(aq)+ Cl^–(aq)

iii) Association of solute:
Association of solute molecules can also cause deviation from ideal behaviour. For example, in solution, acetic acid exists as a dimer by forming intermolecular hydrogen bonds, and hence deviates from Raoult’s law.

iv) Temperature:
An increase in temperature of the solution increases the average kinetic energy of the molecules present in the solution which causes decrease in the attractive force between them. As result, the solution deviates from ideal behaviour.

v) Pressure:
At high pressure the molecules tend to stay close to each other and therefore there will be an increase in their intermolecular attraction. Thus, a solution deviates from Raoult’s law at high pressure.

vi) Concentration:
If a solution is sufficiently dilute there is no pronounced solvent-solute interaction because the number of solute molecules are very low compared to the solvent. When the concentration is increased by adding solute, the solvent-solute interaction becomes significant. This causes deviation from the Raoult’s law.

Question 8.
How would you determine the molar mass of solute from Tb?
Answer:
The elevation of boiling point is directly proportional to the concentration of the solute particles.
∆T_b ∝ m
m is the concentration of solution expressed in molality.
∆T_b = K_bm
Where
K_b = molal boiling point elevation constant or Ebullioscopic constant.
∆T_b = \(\frac{K_{b} \times W_{B} \times 1000}{M_{B} \times W_{A}}\)

M_b = \(\frac{\mathrm{K}_{\mathrm{b}}}{\Delta \mathrm{Tb}} \times \frac{\mathrm{W}_{\mathrm{B}} \times 1000}{\mathrm{~W}_{\mathrm{A}}}\)

Question 9.
How would you determine the molar mass of solute from T?
Answer:
1f the solution is prepared by dissolving
W_B g of solute in W_B g of solvent, then the molality is,

m = \(\frac{\text { Number of moles of solute } \times 1000}{\text { weight of solvent in grams }}\)

Number of moles of solute = \(\frac{W_{B}}{M_{B}}\)
Where, M_B = molar mass of the solute
Therefore,

m = \(\frac{\mathrm{W}_{\mathrm{B}} \times 1000}{\mathrm{M}_{\mathrm{B}} \times \mathrm{W}_{\mathrm{A}}}\)

∆T_f = \(\frac{K_{f} \times W_{B} \times 1000}{M_{B} \times W_{A}}\)

Molar mass can be calculated using

M_B = \(\frac{\mathrm{Kb} \times \mathrm{W}_{\mathrm{B}} \times 1000}{\Delta \mathrm{T}_{\mathrm{b}} \times \mathrm{W}_{\mathrm{A}}}\)

Question 10.
How would you determine the molar mass of solute form A?
Answer:
According to van’t Hoff equation π = cRT
c = \(\frac{n}{V}\)
Here, n = number of moles of solute dissolved in ‘V’ liter of the solution.
Therefore,
π = \(\frac{n}{V}\)RT
πV = nRT
If the solution is prepared by dissolving W_Bg of nonvolatile solute in W_A g of solvent,
then the number of moles ‘n’ is,
n = \(\frac{\mathrm{W}_{\mathrm{B}}}{\mathrm{M}_{\mathrm{B}}}\)
since, M_B = molar mass of the solute
Substituting the ‘n’ value, we get,
π = \(\frac{\mathrm{W}_{\mathrm{B}}}{\mathrm{V}} \frac{\mathrm{RT}}{\mathrm{M}_{\mathrm{B}}}\)

M_B = \(\frac{W_{B}}{V} \frac{R T}{\pi}\)
From the above equation molar mass of the solute can be calculated.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Zoology Guide Pdf Chapter 2 Kingdom Animalia Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Zoology Guide Chapter 2 Kingdom Animalia

11th Bio Zoology Guide Kingdom Animalia Text Book Back Questions and Answers

Part I

I. Choose The Best Options

Question 1.
The symmetry exhibited in cnidarians is
a. Radial
b. Bilateral
c. Pentamerous radial
d. Asymmetrical
Answer:
a. Radial

Question 2.
Sea anemone belongs to phylum
a. Protozoa
b. Porifera
c. Coelenterata
d. Echinodermata
Answer:
c. Coelenterata

Question 3.
The excretory cells that are found in platyhelminthes are
a. Protonephridia
b. Flame cells
c. Solenocytes
d. All of these
Answer:
b. Flame cells

Question 4.
In which of the following organisms, self fertilization is seen.
a. Fish
b. Round worm
c. Earthworm
d. Liver fluke
Answer:
d. Liver fluke

Question 5.
Nephridia of Earthworms are performing the same functions as
a. Gills of prawn
b. Flame cells of Planaria
c. The trachea of insects
d. Nematoblasts of Hydra
Answer:
b. Flame cells of Planaria

Question 6.
Which of the following animals has a true coelom?
a. Ascaris
b. Pheretima
c. Sycon
d. Taenia solium
Answer:
b. Pheretima

Question 7.
Metameric segmentation is the main feature of
a. Annelida
b. Echinodermata
c. Arthropoda
d. Coelenterata
Answer:
a. Annelida

Question 8.
In Pheretima locomotion occurs with the help of
a. circular muscles
b. longitudinal muscles and setae
c. circular, longitudinal muscles and setae
d. parapodia
Answer:
c. circular, longitudinal muscles and setae

Question 9.
Which of the following have the highest number of species in nature?
a. Insects
b. Birds
c. Angiosperms
d. Fungi
Answer:
a. Insects

Question 10.
Which of the following is a crustacean?
a. Prawn
b. Snail
c. Sea anemone
d. Hydra
Answer:
a. Prawn

Question 11.
The respiratory pigment in cockroach is
a. Haemoglobin
b. Haemocyanin
c. Heamoerythrin
d. None of the above
Answer:
d. None of the above

Question 12.
Exoskeleton of which phylum consists of chitinous cuticle?
a. Annelida
b. Porifera
c. Arthropoda
d. Echinodermata
Answer:
c. Arthropoda

Question 13.
Lateral line sense organs occur in
a. Salamander
b. Frog
c. Water snake
d. Fish
Answer:
d. Fish

Question 14.
The limbless amphibian is
a. Icthyophis
b. Hyla
c. Rana
d. Salamander
Answer:
a. Icthyophis

Question 15.
Four chambered heart is present in
a. Lizard
b. Snake
c. Scorpion
d. Crocodile
Answer:
d. Crocodile

Question 16.
Which of the following is not correctly paired?
a. Humans – Ureotelic
b. Birds – Uricotelic
c. Lizards – Uricotelic
d. Whale – Ammonotelic
Answer:
d. Whale – Ammonotelic

Question 17.
Which of the following is an egg laying mammal?
a. Delphinus
b. Macropus
c. Ornithorhynchus
d. Equus
Answer:
c. Omithorhynchus

Question 18.
Pneumatic bones are seen in
a. Mammalia
b. Aves
c. Reptilia
d. Sponges
Answer:
b. Aves

Question 19.
Match the following columns and select the correct option.

a. p – (ii), q – (i), r – (iii), s – (iv)
b. p – (iii), q – (iv), r – (ii), s – (i)
c. p – (ii), q – (iv), r – (i), s – (iii)
d. p – (i), q – (ii), r – (iii), s – (iv)
Answer:
b. p – (iii), q – (iv), r – (ii), s – (i)

Question 20.
In which of the following phyla, the adult shows radial symmetry but the larva shows bilateral symmetry?
a. Mollusca
b. Echinodermata
c. Arthropoda
d. Annelida
Answer:
b. Echinodermata

Question 21.
Which of the following is correctly matched?
a. Physalia – Portuguese man of war
b. Pennatula – Sea fan
c. Adamsia – Sea pen
d. Gorgonia-Sea anemone
Answer:
a. Physalia – Portuguese man of war

Question 22.
Why are spongin and spicules important to a sponge?
Answer:
Spongin and spicules provide support and support the soft body parts of the sponges. The spicules give the sponges rigidity and form to the sponges.

Question 23.
What are the four characteristics common to most animals?
Answer:

1. Cellular structure
2. The nature of coelom ;
3. Notochord
4. Segmentation or absence of segmentation.

Question 24.
List the features that all vertebrates show at some point in their development.
Answer:
All vertebrates possess notochord during the embryonic stay. li is repLaced by vertebra) column. All vertebrates possess pained appendages such as fins or lunits. Skin is covered by a protective skeleton comprising of scales. fiathcrs hairs, claws, nails, etc. Respiration is aerobic through gills, skin. buccopharyngeal cavity’ and lungs. All vertebrates have a muscular heart with two, three, or four chambers and kidneys for excretion and osmoregulation.

Question 25.
Compare closed and opened circulatory system
Answer:

Question 26.
Compare Schizocoelom with enterocoelom
Answer:

Question 27.
Identify the structure that the archenteron becomes in a developing animal.
Answer:
The archenteron becomes the cavity of the digestive tract.

Question 28.
Observe the animal below and answer the following questions
a. Identify the animal
b. What type of symmetry does this animal exhibit?
c. Is this animal Cephalized?
d. How many germ layers does this animal have?
e. How many openings does this animal’s digestive system have?
f. Does this animal have neurons?
Answer:
a) Sea anemone
b) Bilateral symmetry
c) It is not a cephalized animal
d) Diploblastic animal
e) One
f) Yes.

Question 29.
Choose the term that does not belong in the following group and explain why it does not belong?
Answer:

• The notochord, cephalization, dorsal nerve cord, and radial symmetry.
• Notochord, cephalization, and dorsal nerve cord are the characteristic features of chordates.
• The radial symmetry is not a characteristic feature of chordate.
• It is the feature of cnidarian and adult echinoderms. Hence it does not belong to the group.

Question 30.
Why flatworms are called acoelomates?
Answer:
The body cavity is formed from mesoderm but in flatworms, there is nobody cavity their body is solid with a perivisceral cavity.

Question 31.
What are flame cells?
Answer:
Flame cells are the specialized excretory cells in flatworms. They help in excretion and osmoregulation.

Question 32.
Concept Mapping – Use the following terms to create a concept map that shows the major characteristic features of the phylum Nematoda: Roundworms, pseudocoelomates, digestive tract, cuticle, parasite, sexual dimorphism
Answer:

Question 33.
In which phyla is the larva trochophore found?
Answer:
Trochopore larva is seen in the Phylum – Annelida.

Question 34.
Which of the chordate characteristics do tunicates retain as adults?
Answer:
Ventral and tabular heart. Respiration is through gill slits.

Question 35.
List the characteristic features that distinguish cartilaginous fishes from living jawless fishes.
Answer:

Question 36.
List three features that characterise bony fishes.
Answer:

1. These fishes have a bony endoskeleton.
2. The skin is covered by ganoid, cycloid or ctenoid scales.
3. Gills are covered by an operculum.
4. They are ammonotelic.
5. They have mesonephric kidneys.
6. External fertilization is seen.

Question 37.
List the functions of air bladder in fishes.
Answer:

• Air bladder may be connected to the gut or not.
• They help in gaseous exchange.
• In ray-finned fishes, they help in buoyancy.

Question 38.
Write the characteristics that contribute to the success of reptiles on land.
Answer:

• The characteristics that contribute to the success of reptiles on land are as follows:
• The presence of dry and cornified skin with epidermal scales or scutes prevents the loss of water.
• The presence of metanephric kidney.
• They are uricotelic (they excrete uric acid to prevent the loss of water).

Question 39.
List the unique features of a bird’s endoskeleton.
Answer:

• The endoskeleton is fully ossified.
• The long bones are hollow with air cavities. So that they can easily fly with lesser weight.

Question 40.
Could the number of eggs or young ones produced by an oviparous and viviparous female be equal? Why?
Answer:
No. The number of eggs or young ones produced by an oviparous and viviparous female cannot be equal. When the oviparous animals lay eggs in the external environment or in the medium, the chance of survival and successful development into the adults are not certain. But in the case of viviparous animals, young ones are nurtured by the adult animals. Hence, oviparous animals lay more eggs if they are fertilized in the medium or in water.

Part II

11th Bio Zoology Guide Kingdom Animalia Additional Important Questions and Answers

I. Choose The Best Options

Question 1.
…………………. is the first group of animals to exhibit tissue-level organization.
a. Cnidaria
b. Porifera
c. Mollusca
d. Echinodermata
Answer:
a. Cnidaria

Question 2.
Name the organs formed from ectoderm.
a. Heart
b. Hair
c. Muscle
d. Intestine
Answer:
b. Hair

Question 3.
The mesoglea seen in between the ectoderm and endoderm is present in …………………. phylum.
a. Platyhelminthes
b. Arthropoda
c. Annelida
d. Coelenterates
Answer:
d. Coelenterates

Question 4.
Inporiferans through ………………. pores water enters into the body and goes out through.
a. Osculum Ostia
b. Ostia Osculum
c. Mouth Ostia
d. Mouth Osculum
Answer:
b. Ostia Osculum

Question 5.
Choose the correct option.
a. Segmentation – Annelida
b. Archenteron – Heart Formation
c. Ostia – Sea anemone
d. Polyp Medusa – Phylum Ptenopora
Answer:
a. Segmentation – Annelida

Question 6.
Find out the correct and wrong statement and find out the correct sequence
I. In the phylum cnidaria on the tentacles nematocysts are present.
II. In evolutionary history the annelid is the first segmented animal.
III. The roundworms are diploblastic bilateral animals.
IV. The arthropods excrete through flame cells.
a. I – False, II – False, III – True, IV – True
b. I – True, II – False, III – False, IV – True
c. I – True, II – True, III – False, IV – Flase
d. I – False, II – True, III – True, IV – False
Answer:
c. I – True, II – True, III – False, IV – False

Question 7.
The organism that shows the regeneration character
a. Planaria
b. Liver fluke
c. Tapeworm
d. Leech
Answer:
a. Planaria

Question 8.
What is the excretory organ of roundworm?
a. Flame cells
b. Rennet glands
c. Green glands
d. Malphigeal tubules
Answer:
b. Rennet glands

Question 9.
The coelom of phylum Arthropoda is
a. Pseudo coelom
b. Eucoelom
c. Schizo coelom
d. Enter coelom
Answer:
c. Schizo coelom

Question 10.
Name the respiratory organ of Mollusca.
a. Ctenidia
b. Gills
c. Book lungs
d. Trachea
Answer:
a. Ctenidia

Question 11.
Name the excretory organ of Cephalo Chordata.
a. Mesonephridia
b. Metanephridia
c. Protonephridia
d. Flame cells
Answer:
d. Flame cells

Question 12.
Name the organism which has both features of chordate and non-chordates?
a. Balanoglossus
b. Ascidia
c. Amphioxces
d. Salpa
Answer:
a. Balanoglossus

Question 13.
…………………….. are called as tunicates
a. Urochordates
b. Cephalo chordates
c. Vertebrata
d. Hemi Chordata
Answer:
a. Urochordates

Question 14.
The eggs of birds are ………………..
a. Megalecithal
b. Mesolecithal
c. Telolocithal
d. Alecithal
Answer:
a. Megalecithal

Question 15.
Find the correct answer by matching.
A. Sponges – I. Mesoglea
B. Open circulation – II. Asymmetrical
C. Diploblastic animal – III. Echinodermata
D. Snails – IV. Coanocytes
a. A – IV, B -I, C – II, D – III
b. A -I, B – II, C – III, D – IV
c. A – IV, B – III, C -I, D – II
d. A – IV, B – II, C – III, D -I
Answer:
C. Diploblastic animal – III. Echinodermata

Question 16.
Find out the correct pair.
a. Planula – Planeria
b. Regeneration – Annelida
c. Trochopore larva – Cnidaria
d. Veliger larva – Mollusca
Answer:
d. Veliger larva – Mollusca

Question 17.
Which one of the following is not correctly paired?
a) Ctenophora – Veliger
b) Annelida – Trochophore
c) Cnidaria – Planula
d) Porifera – Parenchymula
Answer:
a) Ctenophora – Veliger

Question 18.
Find out the wrong statement.
a) In most animals the coelom lies between the body wall and the alimentary canal
b) In acoelomate organisms the free movement of the interval organ is restricted.
c) The cavity formed from the mesoderm is pseudo coelom.
d) If in a body cavity scattered pouches are seen then the coelom is pseudo coelom.
Answer:
c) The cavity formed from the mesoderm is pseudo coelom.

(2 marks)

II. Very Short Questions

Question 1.
What are pinococytes?
Answer:
In sponges, the outer surface is formed of plate-like cells that maintain the size and structure of the sponges are called pinococytes.

Question 2.
What are choanocytes?
Answer:
The inner layer of sponges is formed of flagellated collar cells called coanocytes. They maintain water flow through the sponges thus facilitating respiratory and digestive functions.

Question 3.
Define tissue.
Answer:
Cells that perform similar functions are aggregated to form tissues.

Question 4.
Define organ? Which was the first animal to have organ system?
Answer:

• Different kinds of tissues aggregate to form an organ to perform a specific function.
• In phylum Platyhelminthes, the organ level of organisation is first formed.

Question 5.
Differentiate between a complete digestive system from an incomplete digestive system.
Answer:

Question 6.
What are diploblastic animals?
Answer:

• Animals in which the cells are arranged in two embryonic layers the ectoderm and endoderm are diploblastic animals.
• The ectoderm gives rise to the epidermis.
• The endoderm gives rise to the tissue lining the gut.

Question 7.
What is radial symmetry?
Answer:
When any plane passing through the central axis of the body divides an organism into two identical parts, it is called radial symmetry, e.g. Cnidarian.

Question 8.
What is protostomia?
Answer:
In Eumetazoans, the embryonic blastopore that develops into the mouth are known as protostomia.

Question 9.
What are deutrostomia ?
Answer:
Eumetazoans in which the anus is formed from or near the blastopore and the mouth is formed away from the blastopore are deuterostomes.

Question 10.
List the excretory organs of phylum Arthropoda?
Answer:

• Malphigean tubules
• Green glands
• Coxal glands

Question 11.
Differentiate the respiratory pigment haemoglobin from haemocyanin.
Answer:

Question 12.
What are the advantages of bilaterally symmetrical animals?
Answer:
The bilaterally symmetrical animals can seek food, locate mates, escape from predators and move more efficiently. These animals have dorsal-ventral sides and anterior, posterior ends, right and left sides. They exhibit cephalization with sense organs and brain at the anterior end of the animal.

Question 13.
What is cleidoic egg?
Answer:
If the female organisms lay cleidoic eggs or shelled egg then it is known as cleidoic eggs.

Question 14.
What are the extraembryonic membranes present in reptiles?
Answer:

1. Amnion
2. Allantois
3. Chorion
4. Yolk sac

Question 15.
Name the muscles that help pigeons to fly. Write the kingdom, phyllum, and class for pigeon.
Answer:
a. Pectoralis major b. Pectoralis minor
(i) kingdom – Animalia
(ii) phylum – Chordata
(iii) class – Aves

( 3 marks)

III. Short Questions

Question 1.
What are the structures formed from ectoderm endoderm and mesoderm?
Answer:

1. Edoderm – Skin, Hair, Nerves, Nail, Teeth
2. Mesoderm – Muscles, Bones, Heart
3. Endoderm – Intestine, Lungs, Liver.

Question 2.
Name the parts A, B, and C in the diagram?
Answer:

A) Ectoderm
B) Pseudo coelom
c) Mesodorm

Question 3.
Differentiate parazoa from eumetazoa?
Answer:

Question 4.
Match


a) I – b, II – d, III – a, IV – c
b) I – a, II – b, III – d, IV – c
c) I – b, II – a, III – d, IV – c
Answer:
a) I – b, II – d, III – a, IV – c

Question 5.
Distinguish between hibernation and aestivation.
Answer:
Hibernation:

• The dormancy period for animals during winter is called hibernation.
• It is known as winter sleep.

Aestivation:

• The dormancy period for animals during summer is called Aestivation.
• It is known as summer sleep.

Question 6.
In the given diagram Balanoglossus mark A, B, and C.

Answer:
A) Proboscis
B) Collarette
c) Genital wings

Question 7.
Give any five characteristic features of Urochordata?
Answer:

1. They are exclusively marine.
2. They are mostly sessile some pelagic or free-swimming.
3. The body is covered by a tunic.
4. The coelom is absent.
5. The notochord is present only in the tail region of the larval stage.
6. The circulatory system is open type.

Question 8.
Look at the picture given below and answer questions.
a) What is the name of this fish?
b) What is the name of the larva of this fish?
c) What is the shape of the mouth?

Answer:
a) iprey
b) Ammocete
c) Circular

Question 9.
What are the characteristic features of amphibia?
Answer:

• Amphibians live both in aquatic as well as terrestrial habitats.
• They are poikilothermic.
• They have two pairs of limbs.
• They may have a tail or may not be present.
• Their skin is smooth or rough.
• The heart is three-chambered.
• They excrete urea as an excretory product.
• The kidneys are mesonephric.
• They are oviparous and development is indirect.

(5 Marks)

IV. Essay Questions

Question 1.
Explain various patterns of organisation in animals.
Answer:
Animals exhibit different patterns of organisation:
The cellular level of organisation:

• Cells are loosely arranged without the formation of tissues.
• There is a division of labour among the cells, e.g., sponges.

Tissue level of organisation:

• Cells which perform a similar function are grouped into tissues.
• The tissues perform a common function, e.g., cnidarians.

Organ level of organisation:
Different kinds of tissues aggregate to form an organ to perform a specific function e.g., flatworms and other hyper phyla.

Organ system level of organisation:

• The tissues are organised to form organs and organ systems.
• All the organ system function in a coordinated manner.

Question 2.
What is coelom? Describe its types?
Answer:
Body cavity lined with mesoderm is meant as a coelom. This lies between the body wall and the alimentary canal.

1.Pseudo coelom:
The body cavity is not lined by the mesodermal epithelium and the mesoderm is formed as scattered pouches between the ectoderm and endoderm. (Eg.) Roundworm

2.Eucoelom:
The coelom is a fluid-filled cavity that develops within the mesoderm and is lined by mesodermal epithelium called the peritoneum.

3.Schizocoelomates:
In these animals, the body cavity is formed by splitting mesoderm. (Eg.) Annelids.

4.Entero coelomate:
The body cavity is formed from the mesodermal pouches of the archenteron. (Eg.) Echinodermata

Question 3.
Compare Platyhelminthes with Aschelminthes?
Answer:

Question 4.
Classify animals based on coelom.
Answer:
The cavity between the body wall and the gut wall is called coelom. If the animals do not have a coelom, they are called acoelomates. e.g., flatworms. In some animals, the body cavity is not fully lined by the mesodermal epithelium. The mesoderm is formed as scattered pouches between the ectoderm and endoderm. Such a body cavity is called a pseudocoel. The animals which have pseudocoel e.g. roundworms.

If the coelom develops within the mesoderm and is lined by mesodermal epithelium it is called eucoelom. The animals which have true coelom are called eucoelomates. If the body cavity is formed by splitting mesoderm, the animals are called schizocoelomates e.g., Annelids, arthropods, and mollusks. If the body cavity is formed from the mesodermal ‘ pouches of archenteron, the animals are called enterocoelomate animals, e.g., echinoderms, hemichordates, and chordates.

Question 5.
What are the characteristic features of Hemichordata?

• They possess the characters of invertebrates and chordates.
• This phylum consists of soft worm-like organisms.
• They are triploblastic coelomate animals.
• They are bilaterally symmetrical.
• Their circulatory system is simple and open type.
• They are ciliary feeders.
• Respiration is through paired gill silts opening into the pharynx.
• Excretion is through the glomerulus.
• The nervous system is primitive sexes are separate.
• In its development, there is a free-swimming to maria larva.

Question 6.
What are the general characters of the phylum Vertebrata?
Answer:

1. They possess notochord during the embryonic stage only.
2. The notochord is replaced by a cartilaginous or bony vertebral column in the adult.
3. They possess paired appendages such as fins or limbs.
4. Skin is covered by skeleton consists of scales, feathers, hairs, claws, nails.
5. Respiration is through the gills skin buccopharyngeal cavity and lungs.
6. The heart is with two or three or four chambers.
7. Kidneys are for excretion and osmoregulation.

Question 7.
Write the general characters of the phylum cnidaria.
Answer:

• The cnidaria is aquatic, radially symmetrical, and diploblastic.
• The tentacles have stinging cells called cnidocytes or cnidoblasts or nematocysts.
• They exhibit a tissue level of organisation.
• They have a central gastrovascular cavity called coelenteron.
• Digestion is both extracellular and intracellular.
• Alternation of generation is seen in cnidarians which have polyp and medusa forms.
• Development is indirect with planula larva e.g. Physalia.

Question 8.
What are the general characters of mammals?
Answer:

• The body is covered by hairs.
• They are found in a variety of habitats.
• The presence of the mammary gland is the most unique feature of mammals.
• They have two pairs of limbs.
• The skin consists of sweat glands and sebaceous glands.
• Exo skeleton includes horns spines, scales claws, etc.
• Teeth are thecodont heterodont and diphyodont.
• The heart is four-chambered and posses a left systematic arch.
• Mammals have a large brain when compared to other animals.
• Their kidneys are metanephric and are ureotelic.
• All are homeothermic.

Question 9.
Give three distinct features of all chordates that are seen at some stage of their life cycle? What is the fate of two characters out of three in the matured adults?
Answer:

1. Presence of notochord below the nerve chord and above the alimentary canal.
2. The presence of the nerve cord lies above the notochord and below the dorsal body wall.
3. Presence of pharyngeal gill slits in all chordates at some stage of their life cycle.

Features saw in the matured adult animals

Question 10.
Compare the chordates with non-chordates?
Answer:

Question 11.
What are the parts of ABCD in the model diagram of Chordata?

Answer:
A) Dorsal Nerve Cord
B) Notochord
C) Mouth
D) Pharyngeal gill clefts
E) Muscle segment

Question 12.
Look into the given diagram and answer the question.
a) What is the name of the organism.
b) What is the respiratory organ of this animal.
c) What type of metamorphosis is seen?
d) Whether this organism contain nerve card?
e) What is the outer covering of it’s body?

Answer:
a) Ascidian
b) Gill clefts
c) Retrogressive metamorphosis
d) The larva consists of a nerve cord.
e) Tunic

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Zoology Guide Pdf Chapter 1 The Living World Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Zoology Guide Chapter 1 The Living World

11th Bio Zoology Guide Living World Text Book Back Questions and Answers

Part I

Question 1.
A living organism is differentiated from a non-living structure based on
a. Reproduction
b. Growth
c. Metabolism
d. All the above
Answer:
d. All the above

Question 2.
A group of organisms having similar traits of a rank is
a. Species
b. Taxon
c. Genus
d. Family
Answer:
b. Taxon

Question 3.
Every unit of classification regardless of its rank is
a. Taxon
b. Variety
c. Species
d. Strain
Answer:
a. Taxon

Question 4.
Which of the following is not present in the same rank?
a. Primata
b. Orthoptera
c. Diptera
d. Insecta
Answer:
a. Primata

Question 5.
What taxonomic aid gives comprehensive information about a taxon?
a. Taxonomic Key
b. Herbarium
c. Flora
d. Monograph
Answer:
a. Taxonomic Key

Question 6.
Who coined the term biodiversity?
a. Walter Rosen
b. AG Tansley
c. Aristotle
d. AP de Candole
Answer:
a. Walter Rosen

Question 7.
Cladogram considers the following characters
a. Physiological and Biochemical
b. Evolutionary and Phylogenetic
c. Taxonomic and systematic
d. None of the above
Answer:
b. Evolutionary and Phylogenetic

Question 8.
The molecular taxonomic tool consists of
a. DNA and RNA
b. Mitochondria and Endoplasmic reticulum
c. Cell wall and Membrane proteins
d. All the above
Answer:
a. DNA and RNA

Question 10.
Differentiate between probiotics and pathogenic bacteria
Answer:

Question 11.
Why mule is sterile?
Answer:
Mule gets one set of chromosomes (32) from the male parent, horse and one set of chromosomes (31) from the female parent, donkey. These two sets of chromosomes do not match with each other and cannot produce gametes by meiosis. Hence mule is sterile in nature.

Question 12.
What is the role of Charles Darwin in relation to the concept of species?
Answer:
Charles Darwin’s book on Origin of Species explains the evolutionary connections of species by the process of natural selection.

Question 13.
Why elephants and other wild animals are entering the human living areas?

• For the construction of houses, dams, and factories forests are destroyed. The area surface of forests is also getting reduced.
• As the bull elephant is hunted for their tusks the cow elephant during breeding season enters in to the dwelling area of people.

Question 14.
What is the difference between a Zoo and a wildlife sanctuary?

Question 15.
Can we use recent molecular tools to identify and classify organisms?
Answer:
The recent molecular taxonomical tools can be used to identify and classify the organism. The following molecular techniques and approaches are used in molecular tools.

1. DNA barcoding – Short genetic marker in an organism’s DNA to identify whether it belongs to a particular species.
2. DNA hybridization – Measures the degree of genetic similarity between pools of DNA sequences.
3. DNA fingerprinting – to identify an individual from a sample of DNA by looking at unique patterns in their DNA.
4. Restriction Fragment Length Polymorphism (RFLP) Analysis – the difference in homologous DNA sequences can be detected by the presence of fragments of different lengths after digestion of DNA samples.
5. Polymerase chain reaction (PCR) sequencing- to amplify a specific gene or portion of the gene.

Question 16.
Explain the role of Latin and Greek names in Biology.
Answer:
Aristotle (384 to 322 BC) was the first to classify all animals in his Historia Animalium in Latin. He classified the living organisms into plants and animals. Animals were classified as walking (terrestrial), flying (birds), and swimming (aquatic) based on their locomotion.

He classified the animals with red blood cells as Enaima and those without red blood cells as Anima. Though his method of classification had limitations, his contribution to biology was remarkable. Theophrastus did his research on the classification of plants. He was known as the Father of Botany.

Part II 

11th Bio Zoology Guide The Living World Additional Important Questions and Answers

Question 1.
Biodiversity is
a. A species live in a particular ecosystem.
b. Presence of a large number of species in a particular ecosystem.
c. A species live in a different ecosystem.
d. Many species live in more than one ecosystem.
Answer:
b. Presence of a large number of species in a particular ecosystem.

Question 2.
Aristotle has classified organisms based on the following category of locomotion.
a. Walking & bore dwellers
b. Flying & arboreal
c. Swimmers & aquatic
d. All the above.
Answer:
c. Swimmers & aquatic
d. All the above.

Question 3.
Who is “Father of Botany”?
a. Theophrastus
b. John Ray
c. Carolus Linnaeus
d. Aristotle
Answer:
a. Theophrastus

Question 4.
Whose researchers confirm that species is a fundamental unit of classification.
a. John Ray
b. R.H. Whittaker
c. CarlWoese
d. Cavalier-Smith
Answer:
a. John Ray

Question 5.
Find the correct pair.
1. Domestic Cat – Felis silvestris
2. Wildcat – Felis margarita
3. Wildcat – Felis Domestica
4. Tiger – Panthera tigers
Answer:
4. Tiger – Panthera tigers

Question 6.
Who has developed binomial nomenclature.
a. Carolous Linnaeus
b. Augustin
c. Aristotle
d. Ernst Haeckel
Answer:
a. Carolous Linnaeus

Question 7.
Find the unrelated pair.
a. Carl Woese – Trinominal hypothesis
b. Cavalier-Smith – Seven kingdom system
c. Male Lion and female Tiger results in – Hinny
d. Male Tiger and female Lion results in – Tigon
Answer:
c. Male Lion and female Tiger results in – Hinny

Question 8.
The three domains classification is based on the difference in the gene.
a. 60s rRNA
b. 70s rRNA
c. l6s rRNA
d. m RNA
Answer:
c. l6s rRNA

Question 9.
The prokaryotes that produce methane gas belongs to …………………… kingdom.
a. Monera
b. Eukarya
c. Bacteria
d. Archaea
Answer:
d. Archaea
Question 10.
Find out the correct sequence by matching.
A. Augustin Pyramus de Candole – Father of Botany
B. Aristotle – Father of Modern Taxonomy
C. Carolous Linnaeus – Father of Taxonomy
D. Theophrastus – Introduces Taxonomy
Answer:
D. Theophrastus – Introduces Taxonomy

Question 11.
Crosses between animals – Match.
A. Male Horse + Female Donkey – Tigon
B. Male Donkey + Female Horse – Tiger
C. Male Lion + Female Tiger – Mule
D. Male Tiger + Female Lion – Hinny
a) A-II, B -1, C – IV, D – III
b) A-IV, B -1, C – II, D – III
c) A-I, B-II, C-III, D-IV
d) A-IV, B-I, C-II, D-III
Answer:
A-II, B -I, C – IV, D – III

Question 12.
Three domain classification was proposed by:
a. Cavalier-Smith
b. R.H. Whittaker
c. Carolus Linnaeus
d. Carl Woese
Answer:
d. Carlwoese

Question 13.
Find out the wrong pair
a. Peacock – Pavocristatus
b. Tiger – Pantheratigeris
c. Man -Homosapiens
d. Domestic crow – Salcopopsindica
Answer:
d. Domestic crow – Salcopopsindica

Question 14.
Find the correct match.
1. John ray -a. Five kingdom concept
2. Linnaeus -b. Cladogram
3. Ernest Haeckel -c. Binomial nomenclature
4. R.H. Whittaker – d. Methodus Plantarum
a. 1 -d,2-c,3-b,4-a
b. 1-a,2-b,3-c,4-d
c. 1 – c, 2 – a, 3 – b, 4 – d
d. 1 – d, 2 – c, 3 – a, 4 – b
Answer:
a. 1 -d,2-c,3-b,4-a

Question 15.
Where are the 80s and 70s ribosomes seen in Eukaryotic cells?
a. Cytoplasm – Chloroplast
b. Mitochondrial – Golgi apparatus
c. Chloroplast – Endo plasm reticulum
d. Nucleus – Lysosomes
Answer:
a. Cytoplasm – Chloroplast

( 2 marks)

II. Very Short Questions

Question 1.
Classification of organisms is necessary.
Answer:
Classification of organisms is necessary to recognize, identify them, and differentiate closely related species.

Question 2.
What are the unique characteristic features of living organisms?
Answer:

• Cellular organization
• Nutrition
• Respiration
• Metabolism
• Movement
• Reproduction
• Excretion
• Homeostasis

Question 3.
The mating between different species produces sterile offsprings.
Answer:
The maternal and paternal chromosomes of the offsprings produced by the mating between different species are not identical and hence gametes are not produced by meiotic division.

Question 4.
What are the scientific stages of taxonomy?
Answer:

• Characterization
• Identification
• Nomenclature
• Classification

Question 5.
Why are molecular tools used now to study taxonomy?
Answer:
Molecular tools are accurate and authentic. Hence they are used to study taxonomy.

Question 6.
What is phylogenetic or cladistics classification?
Answer:
It is a classification based on evolution and genetic relationship.

Question 7.
What is the phylogenetic tree?
Answer:
It’s a method of representing evolutionary relationships with the help of a tree diagram known as a cladogram.

Question 8.
What is a cladogram?
Answer:
Arranging organisms on the basis of their similar or derived characters produced a phylogenetic tree or cladogram.

Question 9.
What are the three domains of life indicate?
Answer:
This system emphasizes the separation of prokaryotes into two domains.

Question 10.
How Archaea differ from bacteria?
Answer:
If differs in cell wall composition and in membrane composition and rRNA type.

Question 11.
What is the seven taxonomic hierarchy?
Answer:

1. Kingdom
2. Phyla
3. Class
4. Order
5. Family
6. Genus
7. Species

Question 12.
Define species?
Answer:
It is a group of animals having similar morphological features and is reproductively isolated to produce fertile offspring.

Question 13.
Define ‘Family’?
Answer:
It is a taxonomic category which includes a group of related genera with less similarity as compared to genus and species.

Question 14.
Define order?
Answer:
Order is an assemblage of one or more related families which show few common features. (Eg) Family Candiae and Felidae are placed in the order Carnivora.

Question 15.
Define class.
Answer:
Class includes one or more related orders with some common characters.

Question 16.
Define Phylum.
Answer:
The group of classes with similar distinctive characteristics constitute phylum.

Question 17.
Define animal kingdom.
Answer:
All living animals belonging to various phyla are included in the kingdom.

Question 18.
What are the features that we have to keep in mind in naming them scientifically?
Answer:

• Morphology
• Genetic information
• Habitat
• Feeding pattern
• Adaptations
• Evolutions

Question 19.
On whose guidelines naming animals in a scientific way is done?
Answer:
The naming of the organism is based on the guidelines of the international code of Zoological nomenclature.

Question 20.
What are taxonomical keys?
Answer:
Keys are based on a comparative analysis of the similarities and dissimilarities of organisms. There are separate keys for different taxonomic categories.

Question 21.
What is a museum?
Answer:
Biological museums have a collection of preserved plants and animals for study and ready reference.

Question 22.
Define Zoological parks.
Answer:
These are places where wild animals are kept in protected environments under human care.
It enables us to study their food habits and behaviour.

Question 23.
What are marine parks?
Answer:
Marine organisms are maintained in protected environments.

Question 24.
What are printed taxonomical tools?
Answer:

• Identification cards
• Description
• Field guides
• Manuals

Question 25.
What is the phylogenetic tree?
Answer:
It is the inferred evolutionary relationships upon similarities and differences in their physical or genetic characters.

Question 26.
Define phylogeny.
Answer:
Relationships among various biological species based upon similarities and differences in their physical or genetic characteristics.

Question 27.
What are shared characters?
Answer:
A shared character is one that two lineages have in common.

Question 28.
What are derived characters?
Answer:
Derived character is one that evolved in the lineage leading up to a clade.

Question 29.
Vandaloor Zoological park.
Answer:

• It is situated in the South-Western Part of Chennai.
• It spreads over an area of 1500 acres.
• It is one of the largest zoological parks in India.
• The Zoo houses 2553 species of both flora and fauna.

( 3 marks)

III. Short Questions

Question 1.
Define ecosystem.
Answer:
The ecosystem is defined as a community of living organisms (plants and animals), non-living things (minerals, climate, soil, sunlight, and water), and their interrelationships, e.g. Forest and grassland.

Question 2.
On which criteria the systematic classification is done?
Answer:

• Evolutionary history.
• Environmental adaptations.
• Environmental relationship.
• The interrelationship between species.

Question 3.
Give an account of Aristotle’s classification?
Answer:

•  In his book ‘History of Animals,’ he classifies plants and animals into two categories.
• Based on locomotion walking, flying, swimming,
• He classifies the organisms on the basis of blood.
• He classifies the animals into two as ‘Enaima’ with blood and those without blood as’ Anaima’

Question 4.
Who has developed the five kingdom classification?
Answer:

1. R.H. Whittaker proposed the five-kingdom classification.
2. It is based on cell structure.
3. Mode of nutrition.
4. Mode of reproduction.
5. Phylogenetic relationships.

The kingdoms are

• Monera
• Protista
• Fungi
• Plantae
• Animalia

Question 5.
What are the special features of frogs that are identified in Western Gauts?
Answer:

• This frog has shiny purple skin.
• There is a light blue ring around the eyes.
• It has a pointy big nose.
• It’s Zoological name Nasikabatrachus Bhupathi.

Question 6.
What is trinomial nomenclature
Answer:
Giving three names to the species is meant as trinomial nomenclature.
When members of any species have large variations then a trinomial system is used.
The species is classified into subspecies and this is an extension of binominal nomenclature system which has an addition of subspecies. Followed by Genus name species subspecies name is also added.

Question 7.
What are the limitations of Aristotle’s classification?
Answer:
Many organisms were not fitting into his classification. Frogs have lungs and they are amphibians while their larva, the tadpole is aquatic and respires through gills. It is difficult to classify frogs according to his method. All flying organisms such as birds, bats, flying insects were grouped together. Ostrich, emu and a penguin are flightless birds and hence they cannot be classified by his method.

 (5 marks)

V. Essay Questions

Question 1.
List the defects of Aristotle’s classification.
Answer:

• Aristotle’s classification system had limitations and many organisms were not fitting into his classification.
• The tadpoles of frog are born in water and have gills but when they metamorphosed into adult frogs they have lungs and can live both in water and on land. There is no answer to this question.
• Based on locomotion birds bats and flying insects were grouped either just by observing one single characteristic feature the flying ability.
• On the contrary to the above-said example, the ostrich emu and penguin are all birds but cannot fly. He did not classify them as birds.

Question 2.
What is special about the Domain Archaea?
Answer:

• This domain includes single-celled organisms the prokaryotes.
• They have the ability to grow in extreme conditions like volcano vents hot springs and polar ice caps hence are called extremophiles.
• They are capable of synthesizing their food without sunlight and oxygen by utilizing hydrogen sulphide and other chemicals from the volcanic vents.
• Some of them produced methane.
• Few live in salty environments and called Halophiles.
• Some thrive in acidic environments and are called thormoacidophiles.

Question 3.
What is special about the domain bacteria?
Answer:

1. Bacterias are prokaryotic.
2. They do not have a definite nucleus and do not have histones.
3. They have circular DNA.
4. They do not possess membrane-bound organelles except for 70s ribosomes.
5. Their cell wall contains peptidoglycans.
6. Many are decomposers. Some are photo-synthesizers and few cause diseases.
7. There are beneficial probiotic bacteria. (Eg.) Cyanobacteria produce oxygen.

Question 4.
What is the significance of cladistic classification?
Answer:
Cladistic classification takes into account ancestral characters (traits commons for the entire group) and derived characters (traits whose structure and function differ from the ancestral characters). The accumulation of derived characters resulted in the formation of new subspecies.

Question 5.
What are the basic roles to be followed in naming the animals?
Answer:

• The scientific name should be italicized in printed form and if handwritten it should be underlined separately.
• The generic name’s first alphabet should be in uppercase.
• The specific name should be in lower case.
• The scientific names of any two organisms are not similar.
• The name of the scientist who first publishes the scientific name may be written after the species name along with the year of publication.
• (Eg.) Lion – Felis Leo Linn . 1758 (or) Felis Leo L. 1758

Question 6.
What are the rules to be followed in the nomenclature of organisms?
Answer:
The scientific name should be italicized in printed form and the generic name and specific name should be underlined separately if it is handwritten.

• The first alphabet of the generic name should be of uppercase.
• The specific name (species) should be in lower case letters.
• The name or abbreviated name of the scientist who first published the scientific name may be written after the specific (species) name along with the year of publication, e.g. Felis Leo Linn., 1958.
• If the specific (species) name is framed after any person’s name, the name of the species shall end with i, ii, or ae. e.g. Ground – dwelling lizard Cyrtodactylus varadgirii.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Botany Guide Pdf Chapter 11 Transport in Plants Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 11 Transport in Plants

11th Bio Botany Guide Transport in Plants Text Book Back Questions and Answers

Part – I

Question 1.
In a fully turgid cell:
(a) DPD = 10 atm; OP = 5 atm; TP = 10 atm
(b) DPD = 0 atm; OP =10 atm; TP = 10 atm
(c) DPD = 0 atm; OP = 5 atm; TP = 10 atm
(d) DPD = 20 atm; OP = 20 atm; TP = 10 atm
Answer:
(b) DPD = 0 atm; OP =10 atm; TP = 10 atm

Question 2.
Which among the following is correct?
i) apoplast is fastest and operate in nonliving part
ii) Transmembrane route includes vacuole
in) Symplast interconnect the nearby cell through plasma desmata
iv) Symplast and the transmembrane route is in the living part of the cell
a) i and ii
b) ii and iii
c) iii and iv
d) i, ii, iii, iv
Answer:
d) i, ii, iii, iv

Question 3.
What type of transpiration is possible in the xerophyte Opuntia?
(a) Stomatal
(b) Lenticular
(c) Cuticular
(d) All the above
Answer:
(b) Lenticular

Question 4.
Stomata of a plant open due to
a) Influx of K⁺
b) Effrilx of K⁺
c) Influx of Cl^–
d) Influx of OH^–
Answer:
a) Influx of K⁺

Question 5.
Munch hypothesis is based on:
(a) translocation of food due to TP gradient and imbibition force
(b) ranslocation of food due to TP
(c) translocation of food due to imbibition force
(d) None of the above
Answer:
(b) ranslocation of food due to TP

Question 6.
If the concentration of salt in the soil is too high and the plants may wilt even if the field is thoroughy irrigated. Explain
Answer:
High salt concentration results in high be osmotic potential of the soil solution, so the plant has to use more energy to absorb water. Under extreme salinity conditions, plants may be unable to absorb water and will wilt even if the surrounding soil is thoroughly irrigated. This is also referred to as the osmotic or water deficit effect of salinity.

Question 7.
How phosphorylase enzyme open the stomata in starch sugar interconversion theory?
Answer:

• The discovery of enzyme phosphorylase in guard cells by Hanes (1940) greatly supports the starch-sugar interconversion theory.
• The enzyme phosphorylase hydrolyses starch into sugar and high PH followed and the opening takes place during the night.

Day:

1.
2. Photosynthesis occur
3. pH – increased
4. Movement of water from
5. subsidiary cells to guard cells
6. Guard cells become turgid
7. Opening of stomata

Night:

1.

2. No Photosynthesis
3. pH – lowered
4. Movement of water from guard cells
5. Guard cells become flaccid
6. Closure of stomata

Question 8.
List out the non-photosynthetic parts of a plant that need a supply of sucrose?
Answer:
The non-photosynthetic parts of a plant that need a supply of sucrose:

1. Roots
2. Tubers
3. Developing fruits and
4. Immature leaves.

Question 9.
What are the parameters which control water potential?
Answer:
1. Slatyer and Taylor (1960) introduced the concept of water potential.
Definition – water potential is the potential energy of water in a system – compared to pure water when temperature and pressure are kept constant.

2. It is also a measure of how freely water molecules can move in a particular environment or system. Water potential is denoted by the Greek symbol  Ψ (psi) and measured in Pascal (Pa). At standard temperature, the water potential of pure water is zero

3. Addition of solute to pure water decreases the kinetic energy thereby decreasing the water potential, from zero to negative.

4. So, Comparatively a solution always has low water potential than pure water. In a group of cells with different water potential, a water potential gradient is generated.

5. Water will move from higher water potential to lower water potential.
When potential ( Ψ) can be determined by. Solute concentration or Solute potential ( Ψ_s) Pressure potential ( Ψ_p)
By correlating two factors, water potential is written as (Ψw=Ψ_s)+Ψ_p

a) Solute potential (Ψ_s) or Osmotic potential

• Denotes the effect of dissolved solute on water potential.
• In pure water, the addition of solute reduces its free energy and lowers the water potential value from zero to negative.
• Thus the value of solute potential is always negative. In a solution at standard atmospheric pressure, water potential is always equal to solute potential (Ψw = Ψ_s ).

b) Pressure Potential (Ψ_p)

• Pressure potential is a mechanical force working against the effect of solute potential.
• Increased pressure potential will increase water potential and water enters cells and cells become turgid.
• This positive hydrostatic pressure within the cell is called Turgor, pressure likewise, withdrawal of water from the cell decreases the water potential and the cell becomes flaccid.

Question 10.
An artificial cell made of selectively permeable membrane immersed in a beaker (in the figure) Read the values and answer the following questions?

a) Draw an arrow to indicate the direction of water movement
b) Is the solution outside the cell isotonic, hypotonic or hypertonic?
c) Is the cell isotonic, hypotonic, or hypertonic?
d) Will the cell become more flaccid, more turgid or stay in original size?
e) With reference to artifical cell state, the process is endomosis or exomosis? Give reasons
Answer:
a)

(b) Outside solution in hypotonic.
(c) The cell is hypertonic.
(d) The cell becomes more turgid.
(e) The process is endo – osmosis because the solvent (water) moves inside the cell.

Reason: Endomosis is defined as the osmotic entry of solvent into a cell when it is placed in pure water/Hypotonic solution. The solution in the beaker outside the cell is pure water. ( Ψw = 0), and water enters into the artificial cell which is placed inside the beaker of pure water, (i.e) from hypotonic to hypertonic solution.

Part II 

11th Bio Botany Guide Transport in Plants Additional Important Questions and Answers

I – Choose The Correct Answers

Question 1.
In plants, cell to cell transport is aided by:
(a) diffusion alone
(b) osmosis alone
(c) imbibition alone
(d) all the three above
Answer:
(d) all the three above

Question 2.
The smell from a lightened incense stick or mosquito coil or open perfume bottle in a closed room is due to
a) Osmosis
b) Facilitated diffusion
c) Simple diffusion
d) imbibition
Answer:
c. Simple diffusion

Question 3.
Which of the following statements are correct?
(i) Cell membranes allow water and non-polar molecules to permeate by simple diffusion.
(ii) Polar molecules like amino acids can also diffuse through the membrane.
(iii) Smaller molecules diffuse faster than larger molecules.
(iv) Larger molecules diffuse faster than smaller molecules.

(a) (i) and (iv) only
(b) (i) and (iii) only
(c) (i) and (ii) only
(d) (ii) and (iv) only
Answer:
(b) (i) and (iii) only

Question 4.
Solute potential is also known as
a) Water potential
b) Pressure potential
c) Osmotic potential
d) Maic potential
Answer:
c. Osmotic potential

Question 5.
The swelling of dry seeds is due to a phenomenon called:
(a) osmosis
(b) transpiration
(c) imbibition
(d) none of the above
Answer:
(c) imbibition

Question 6.
Cell A has an osmotic potential of -20 bars and a pressure potential of +6 bars. What will be its water potential?
a) -14 bars
b) +14 bars
c) -20 bars
d) +20 bars
Answer:
a. -14 bars

Question 7.
The OP and TP of two pairs of cells A – B, and X-Y are under
a) Cell A: OP=-I0atm, TP=4atm
b) Cell B : OP = l0atm, TP = 6atm
c) Cell X: Op =-l0atm, TP = 4atm
d) CeIlY: OP = -Katm, TP = 4atm
The net movement of water shall be from
a) A toB and X to Y
b) A to B and Y toX
c) B to A and X to Y
d) B to A and Y to X
Answer:
d. B to A and Y to X

Question 8.
Water potential is influenced by which of the two factors among the given four
I) Concentration
II) Pressure
III) Temperature
IV) gravity
a) I & II
b) II & III
c) III & IV
d) I & IV
Answer:
a) I & II

Question 9.
………………………. is equal to TP and is positive except plasmolysed cell and in xylem vessel where it is negative
a) Water potential
b) Pressure potential
c) Solute potential
d) Hydrostatic potential
Answer:
b. Pressure potential

Question 10.
Diffusion Pressure Deficit (DPD) was termed by Meyer in:
(a) 1928
(b) 1828
(c) 1936
(d) 1938
Answer:
(d) 1938

Question 11.
Imbibants present in plants are generally
a) Hydrothermic
b) Hydrostatic
c) Hydrophilic
d) Hydrophobic
Answer:
c. Hydrophilic

Question 12.
Kramer (1949) recognised two distinct mechanisms, which independently operate in the absorption of water in plants are:
(a) osmosis and diffusion
(b) imbibition and diffusion
(c) diffusion and absorption
(d) active absorption and passive absorption
Answer:
(d) active absorption and passive absorption

Question 13.
Root pressure is totally apsent in Gymnosperms because
a) Trachea absent
b) Tracheids absent
c) Trees are tall
d) Trees are comparatively short
Answer:
a. Trachea absent

Question 14.
When respiratory inhibitors like KCN, chloroform are applied:
(a) there is a decrease in the rate of respiration and an increase in the rate of absorption of water.
(b) there is an increase in the rate of respiration and a decrease in the rate of absorption of water.
(c) there is a decrease in the rate of respiration and also a decrease in the rate of absorption of water.
(d) there is an increase in the rate of respiration and also in the rate of absorption of water.
Answer:
(c) there is a decrease in the rate of respiration and also a decrease in the rate of absorption of water.

Question 15.
Find the DPD in a flaccid cell if its OP is 10
a) 20
b) 30
c) 10
d) 40
Answer:
c.10

Question 16.
Pulsation theory was proposed by:
(a) Strasburger
(b) Godsey
(c) J.C. Bose
(d) C.V. Raman
Answer:
(c) J.C. Bose

Question 17.
When a cell is kept in 0.5m solution of sucrose it’s volume does not alter. If the same cell is placed in 0.5M solution of sodium chloride, the volume of the cell
a) Increase
b) Decrease
c) cell will be pIasrnoysed
d) Will does not show any change
Answer:
d. Will does not show any change

Question 18.
Indicate the correct statements:
(i) Root pressure is absent in gymnosperms.
(ii) Root pressure is totally absent in angiosperms.
(iii) There is a relationship between the ascent of sap and root pressure.
(iv) There is no relationship between the ascent of sap and root pressure.

(a) (i) and (ii)
(b) (ii) and (iii)
(c) (ii) and (iv)
(d) (i) and (iv)
Answer:
(d) (i) and (iv)

Match The Following & Find Out The Correct Order

Question 19.
I) Water potential – A) Turgor pressure
II) Solute potential – B) Osmotic potential + Pressure potential
III) Matric potential – C) Osmotic potential
IV) Pressure potential – D) Imbibition pressure

Answer:
b) B C D A

Question 20.
I) Leaves – A) Antitransport
II) Seed – B) Transpiration
III) Roots – C) Negative osmotic potential
IV) Aspirin – D) Imbibition
V) Plasmolyced cell – E. Absorption

Answer:
a) C B D E A

Question 21.
I) Transport of substance from a region of lower concentration to a region of higher concentration is with the expenditure of energy – A. Antiport
II) The movement of two types of molecules across the membrane in opposite direction – B. Symport The movement of a molecule across III) a membrane independent of other molecules – C. Active port
IV) The movement of two types of molecules across the membrane in the same direction – D. Uniport

Answer:
c) C D A B

Question 22.
I) Passive transport – A) Uphill transport
II) Active transport – B) Short distance transport
HI) Cell to cell transport – C) Long-distance transport
IV) Ascent of sap – D) Downhill transport

Answer:
b) D A B C

Question 23.
The length and breadth of stomata is:
(a) about 10 – 30μ and 2 – 10μ respectively
(b) about 10 – 14μ and 3 – 10μ respectively
(c) about 10 – 40μ and 3 – 10μ respectively
(d) about 5 – 30μ and 5 – 10μ respectively
Answer:
(c) about 10 – 40μ and 3 – 10μ respectively

Question 24.
A membrane that permits the solvent and not the solute to pass through it is termed is
a) Permeable,
b) impermeable
c) semipermeable
d) differentially permeable
Answer:
c. Semi permeable

Question 25.
Who did observe that stomata open in light and close in the night:
(a) Unger
(b) Sachs
(c) Boehm
(d) Von Mohl
Answer:
(d) Von Mohl

Question 26.
The phosphorylase enzyme in guard cells supports the starch-sugar interconversion theory. The above reaction is:
(a) oxidation reaction
(b) hydrolyses reaction
(c) reduction reaction
(d) none of the above
Answer:
(b) hydrolyses reaction

Question 27.
If a cell kept in a solution of unknown concentration gets deplasmolysed the solution is
a) hypotonic
b) hypertonic
c) isotonic
d) detonic
Answer:
a. hypotonic

Question 28.
A cell placed in a strong salt solution will shrink because
a) the cytoplasm will decompose
b) mineral salts will break the cell wall
c) salt will leave the cell
d) water will leave by exosmosis
Answer:
d. water will leave by exosmosis

Question 29.
Phenyl Mercuric Acetate (PMA), when applied as a foliar spray to plants:
(a) induces partial stomatal closure for two weeks.
(b) induces partial stomatal opening for two weeks.
(c) induces partial stomatal closure for four weeks.
(d) induces stomatal closure permanently
Answer:
(a) induces partial stomatal closure for two weeks.

Question 30.
The osmotic pressure of cell sap is maximum in
a) Hydrophytes
b) Halophytes
c) Xerophytes
d) Mesophytes
Answer:
b. Halophytes

Question 31.
Say true or false and on that basis choose the right answer.
I) In facilitated diffusion, molecules move across the cell membrane with the help of special proteins, with the expenditure of energy
II) Porin is a larger transport protein, facilitates smaller molecules to pass through.
III) Aquaporins are recognized to transport urea, CO₂, NH3 metalloid & ROS
IV) The carrier proteins structure does not get modified due to its association with the molecules

Answer:
d. False True True False

Question 32.
I) Hypertonic is a strong solution (low solvent/high solute/ low Ψ )
II) Hypotonic is a weak solution (high solvent/low or zero solutes/ high Ψ)
III) Hypertonic is the weak solution (high solvent/low or zero solutes/high Ψ)
IV) Hypotonic is a strong solution (low solvent / high solute/low Ψ)

Answer:
b. True True False False

Question 33.
From sieve elements sucrose is translocated into sink organs such as root, tubers etc and this process is termed as:
(a) Xylem unloading
(b) Xylem uploading
(c) Phloem unloading
(d) Phloem uploading
Answer:
(c) Phloem unloading

Question 34.
The value of pure water is zero in which three aspects of the given options
I) Osmotic pressure
II) Osmotic potential
III) Water potential
IV) Pressure potential
a) I, II, & III
b) II, III & IV
c) I, Ill & IV
d) I, II & IV
Answer:
a. I, II & III

Question 35.
Gases such as oxygen and carbon dioxide cross the cell membrane by
a) Passive diffusion through the lipid bilayer
b) Primary active transport
c) Specific gas transport proteins
d) Secondary active transport
Answer:

Question 36.
Hydathodes are generally present in plants that grow in:
(a) dry places
(b) moist and shady places
(c) sunny places
(d) deserts
Answer:
(b) moist and shady places

Question 37.
Why sugars are transported in the form of su-crose in phloem?
a) It is inactive and highly soluble
b) It is active
c) It yields high ATP
d) It is lighter in weight.
Answer:
a. It is inactive and highly soluble

Question 38.
Unloading of pholem at sink includes
a) Passive transport
b) diffusio
c) Osmosis
d) Active transport
Answer:
d. Active transport

Question 39.
The liquid coming out of the hydathode of grasses is:
(a) pure water
(b) not pure water
(c) a solution containing a number of dissolved substances
(d) saltwater
Answer:
(c) a solution containing a number of dissolved substances

Question 40.
In a flaccid cell
a) DPD = OP
b) DPD = TP
c) TP = OP
d) OP = O
Answer:
a. DPD = OP

Question 41.
The pathway of water movement involving living part of a cell is
a) Apoplast pathway
b) symplast pathway
c) Transmembrane pathway
d) Lateral conduction
Answer:
b. Symplast pathway

Question 42.
The ascent of sap is
a) Upward movement of water in plants
b) downward movement of water in plants
c) upward and downward movement of water plants
d) None of the above
Answer:
a. upward movement of the water plants

Question 43.
High tensile strength of water is due to
a) Adhesion only
b) cohesion only
c) Both (a) and (b)
d) None of these
Answer:
c. Both (a) and (b)

Question 44.
Maximum transpiration occur in
a) Mesophytes
b) Xerophytes
c) Hydrophytes
d) Epiphytes
Answer:
a. Mesophytes

Question 45.
Supply ends in transport of solutes are
a) green leaves
b) root and stem
c) xylem and phloem
d) Hormones and enzymes
Answer:
c. Xylem and phloem

Question 46.
For guttation in plants, the process responsible is
a) Root pressure
b) Atmospheric pressure
c) Imbibition
d) None of these
Answer:
a. Root pressure

Question 47.
Which of the following theories for Ascent of sap was proposed by famous Indian scientist. J.C. Bose.
a) Transpiration pull theory
b) Pulsation theory
c) Root pressure theory
d) Atmospheric pressure theory
Answer:
b. Pulsation theory

Question 48.
Which of the following plant material is an efficient water imbibant?
a) Lignin
b) Pectin
c) Cellulose
d) Agar
Answer:
d. Agar

Question 49.
Which of the following helps in the Ascent of sap?
a) Root pressure
b) Transpiration
c) Capillarity
d) All the above
Answer:
d. All the above

Question 50.
In a girdled plant which of the following dies first?
a) Shoot
b) root
c) Both die simultaneously
d) None – the plant survives
Answer:
b. root

Question 51.
Assertion:-A Imbibition is also diffusion
Reason -R The movement of water in the above process is along a concentration gradient.
a) Both A and Rare true and R is correct explanation of A
b) Both A and R are true but R is not the correct explanation of A
c) A true but R false
d) Both A and Rare false
Answer:
a) Both A and R are True and R is correct explanation of A

Question 52.
Assertion: – A In rooted plant, the transport of water and minerals in xylem is essentially multi-directional
Reason – R Organic compound and nuitrient undergoes undirectional transport only
Answer:
d) Both A and R are false

Question 53.
Assertion: – A The adsorption of water by solid particles of an adsorbant with out forming a solution is known as imbibition
Reason: – R The liquid which is imbided is known as imbibate
Answer:
b) Both A and R are true but R is not the correct explanation of A

Question 54.
Assertion: – A In phloem loading, food is transported to the sink
Reason – R Food is transported from source to sink ‘
Answer:
d) Both Assertion ‘A’ and Reason ‘R’ are false

Question 55.
Assertion – A: Xylem a principal water conducting ’
Reason -R: It has been recognised by girdling or ringing experiments
Answer:
a) Both A and R are True R is the correct explanation of A

Question 56.
Assertion: – A In phloem, sugar are translocated in non reducing form
Reason – R Non reducing sugars are most reactive sugars
Answer:
c) Assertion is true but Reason is false

Question 57.
Assertion: AIn ringing experiment a narrow continuous band of tissues external to the phloem is removed
Reason: R Ringing experiment proves that phloem is involved in water transport ’
Answer:
d) Both A and R are false

II. Two Mark Questions

Question 1.
What is the need for the transport of materials in plants?
Answer:
Water absorbed from roots must travel up to leaves by xylem for food preparation by photosynthesis. Likewise, food prepared from leaves has to travel to all parts of the plant including roots.

Question 2.
What is osmosis
Answer:
It is a special type of diffusion almost same like simple diffusion but has a selectively permeable membrane is here, through which osmosis occur.
(OR)
It is the movement of water molecules from a place of its higher concentration, to the place of its lower concentration through a semipermeable membrane.

Question 3.
Define the term diffusion.
Answer:
The net movement of molecules from a region of their higher concentration to a region of their lower concentration along a concentration gradient until an equilibrium is attained.

Question 4.
The touch me plant closes its leaves at the touch – Explain.
Answer:

• In the ‘Touch me not’ plant the touching act as stimulus, and it closes the leaves.
• When we touch the plant, at that time the stem releases some chemicals, which force water to move out of the cell leading to the loss of Turgor pressure and the leaves droop down However after sometime they become normal.

Question 5.
What is meant by Porin?
Answer:
Porin is a large transporter protein found in the outer membrane of plastids, mitochondria and bacteria which facilitates smaller molecules to pass through the membrane.

Question 6.
Define water potential
Answer: