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Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 14 Human Nutrition Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Biology Solutions Chapter 14 Human Nutrition

1. Choose correct option

Question A.
Acinar cells are present in ……………..
a. liver
b. pancreas
c. gastric glands
d. intestinal glands
Answer:
b. pancreas

Question B.
Which type of teeth are maximum in number in human buccal cavity?
a. Incisors
b. Canines
c. Premolars
d. Molars
Answer:
d. Molars

Question C.
Select odd one out on the basis of digestive functions of tongue.
a. Taste
b. Swallowing
c. Talking
d. Mixing of saliva in food
Answer:
c. Talking

Question D.
Complete the analogy:
Ptyalin: Amylase : : Pepsin : …………….. .
a. Lipase
b. Galactose
c. Proenzyme
d. Protease
Answer:
d. Protease

2. Answer the following questions

Question A.
For the school athletic meet, Shriya was advised to consume either Glucon-D or fruit juice but no sugarcane juice. Why it must be so?
Answer:
Sugarcane juice contain disaccharides. Disaccharides take time to digest i.e. breaking into monosaccharides, Glucon — D and fruit juices contain monosaccharide. Therefore, for instant supply of energy during athletic meet Glucon – D or fruit juices are preferred and not sugarcane.

Question B.
Alcoholic people may suffer from liver disorder. Do you agree? Explain your answer.
Answer:

1. Liver disorder in alcoholic people may occur after years of heavy drinking.
2. Most of the alcohol in the body is broken down in the liver by an enzyme called alcohol dehydrogenase, which transforms ethanol into a toxic compound called acetaldehyde (CH₃CHO).
3. ver consumption of alcohol leads to cirrhosis (distorted or scarred liver) and eventually to liver failure.
   Therefore, alcoholic people may suffer from liver disorder.

Question C.
Digestive action of pepsin comes to a stop when food reaches small intestine. Justify.
Answer:
Pepsin acts in acidic medium thus it is active in stomach. There is alkaline condition in the small intestine. pH of small intestine is very high for pepsin to work. Therefore, pepsin gets denatured in the small intestine.

Question D.
Small intestine is very long and coiled. Even if we jump and run, why it does not get twisted? What can happen if it gets twisted?
Answer:

1. Mesentery is a tissue that is located in the abdomen. It attaches the small intestine to the wall of the abdomen and keeps it in place and therefore it does not get twisted while running and jumping.
2. If small intestine gets twisted, the affected spot may block the food, liquid passing through it. It may sometimes cut off the blood flow if the twist is very severe. If this happens the surrounding tissue may die and can cause serious problems.

3. Write down the explanation

Question A.
Digestive enzymes are secreted at appropriate time in our body. How does it happen?
Answer:

1. The digestive enzymes and juices are produced in sequential manner and at a proper time.
2. These secretions are under neurohormonal control.
3. Sight, smell and even thought of food trigger saliva secretion.
4. Tenth cranial nerve stimulates secretion of gastric juice in stomach.
5. Even the hormone gastrin brings about the same effect.

B. Explain the structure of tooth. Explain why human dentition is considered as thecodont, diphydont and heterodont.
Answer:

1. Structure of tooth:
   • A tooth consists of the portion that projects above the gum called crown and the root that is made up of two or three projections which are embedded in gum.
   • A short neck connects the crown with the root.
   • The crown is covered by the hardest substance of the body called enamel which is made up of calcium phosphate and calcium carbonate.
   • Basic shape of tooth is derived from dentin which is a calcified connective tissue.
   • The dentin encloses the pulp cavity. It is filled with connective tissue pulp. It contains blood vessels and nerves.
   • Pulp cavity has extension in the root of the tooth called root canal.
   • The dentin of the root of tooth is covered by cementurn which is a bone like substance that attaches the root to the surrounding socket in the gum.
2. Human dentition is described as thecodont, diphyodont and heterodont.
3. It is called the codont type because each tooth is fixed in a separate socket present in the jaw bones by gomphosis type of joint.
4. It is called diphyodont type because we get only two sets of teeth, milk teeth and permanent teeth.
5. It is called heterodont type because humans have four different type of teeth like incisors, canines, premolars and molars.
   

Question C.
Explain heterocrine nature of pancreas with the help of histological structure.
Answer:
Pancreas:

1. Pancreas is a leaf shaped heterocrine gland present in the gap formed by bend of duodenum under the stomach.
2. Exocrine part of pancreas is made up of acini, the acinar cells secrete alkaline pancreatic juice that contains various digestive enzymes.
3. Pancreatic juice is collected and carried to duodenum by pancreatic duct.
4. The common bile duct joins pancreatic duct to form hepato-pancreatic duct. It opens into duodenum.
5. Opening of hepato-pancreatic duct is guarded by sphincter of Oddi.
6. Endocrine part of pancreas is made up of islets of Langerhans situated between the acini.
7. It contains three types of cells a-cells which secrete glucagon, P-cells which secretes insulin and 5 cells secrete somatostatin hormone.
8. Glucagon and insulin together control the blood-sugar level.
9. Somatostatin hormone inhibits glucagon and insulin secretion.

4. Write short note on

Question A.
Position and function of salivary glands.
Answer:
Salivary Glands:

• There are three pairs of salivary glands which open in buccal cavity.
• Parotid glands are present in front of the ear.
• The submandibular glands are present below the lower jaw.
• The glands present below the tongue are called sublingual.
• Salivary glands are made up of two types of cells.
• Serous cells secrete a fluid containing digestive enzyme called salivary amylase.
• Mucous cells produce mucus that lubricates food and helps swallowing.

Question B.
Jaundice
Answer:

1. Jaundice is a disorder characterized by yellowness of conjunctiva of eyes and skin and whitish stool.
2. It is a sign of abnormal bilirubin metabolism and excretion.
3. Jaundice develops if excessive break down of red blood cells takes place along with increased bilirubin level than the liver can handle or there is obstruction in the flow of bile from liver to duodenum.
4. Bilirubin produced from breakdown of haemoglobin is either water soluble or fat soluble.
5. Fat soluble bilirubin is toxic to brain cells.
6. There is no specific treatment to jaundice.
7. Supportive care, proper rest are the treatments given to the patient.
   [Note: Treatment ofjaundice will depend on the underlying cause of it. For example, hepatitis-induced jaundice would require treatment which includes antiviral or steroid medications ]

Question 5.
Observe the diagram. This is histological structure of stomach. Identify and comment on significance of the layer marked by arrow.

Answer:
The layer marked in the diagram represents glandular epithelium of mucosa.
Significance of the glandular epiihelium of mucosa:
Goblet cells of the epithelial layer of a mucous membrane secrete mucus which lubricates the lumen of the alimentary canal. This helps in movement of food through the gastrointestinal tract.

Question 6.
Find out pH maxima for salivary amylase, trypsin, nucleotidase and pepsin and place on the given pH scale

Answer:
Salivary amylase = 6.8
Trypsin = 8
Nucleotidase = 7.5
Pepsin = 2

Question 7.
Write the name of a protein deficiency disorder and write symptoms of it.
Answer:

1. Kwashiorkor is a protein deficiency disorder.
2. This protein deficiency disorder is found generally in children between one to three years of age.
3. Children suffering from Kwashiorkor are underweight and show stunted growth, poor brain development, loss of appetite, anaemia, protruding belly, slender legs, bulging eye, oedema of lower legs and face, change in skin and hair colour.

Question 8.
Observe the diagram given below label the A, B, C, D, E and write the function of A, C in detail.

Answer:
A- Bile duct, B- Stomach, C- Common hepatic duct, D- Pancreas, E- Gall Riadder

Functions: Bile duct: It carries hile from the gall bladder and empties it into the tipper part of the small intestine. Common hepatic duct: It drains bile from the liver. It helps in transportation of waste from liver and helps in digestion by releasing bile.
[Note: Labels (A) and (O) have been modified for the better understanding of the students]

Practical / Project : Here are the events in the process of digestion. Fill in the blanks and complete the flow chart.

Answer:

11th Biology Digest Chapter 14 Human Nutrition Intext Questions and Answers

Can you recall? (Textbook Page No. 161)

Question 1.
What is nutrition?
Answer:

1. Nutrition is the sum of the processes by which an organism consumes and utilizes food substances,
2. WHO defines nutrition as the intake of food, considered in relation to the body’s dietary needs.
3. The term nutrition includes the process like ingestion, digestion, absorption, assimilation and egestion.

Question 2.
Enlist life processes that provide us energy to perform different activities.
Answer:
The life processes which are essential and provide us energy are nutrition and respiration.

Think about it (Textbook Page No. 161)

Question 1.
Our diet includes all necessary nutrients. Still we need to digest it. Why is it so?
Answer:

1. Digestion is a very important process of converting complex, noil-diffusible and non-absorbable food substances into simple, diffusible and assimilable substances.
2. Our diet includes all necessary nutrients, which are in the form of complex substances like carbohydrates, proteins, fats and vitamins.
3. These complex substances are converted into simple, diffusible and assimilable substances through the process of digestion.
   Hence, there is a need for digestion of food.

Human Digestive System (Textbook Page No. 161)

Question 1.
Label the diagram
Answer:

Do you know? (Textbook Page No. 162)

Question 1.
Who controls the deglutition?
Answer:
The process of swallowing is called deglutition. Medulla oblongata controls the deglutition.

Question 2.
Is deglutition voluntary or involuntary?
Answer:

• Deglutition consists of three phases: oral phase, pharyngeal phase and oesophagal phase.
• The oral phase is voluntary whereas the pharyngeal and oesophagal phases are involuntary.
  [Source: Goya!, R. K., & Mashimo, H. (2006,.). Physio!o’ of oral, pharyngeal, and esophageal motility. GI Motility online.]

Use your brain power (Textbook Page No. 165)

Question 1.
Draw a neat labelled diagram of human alimentary canal and associated glands in situ.
Answer:

Question 2.
Write a note on human dentition.
Answer:

1. Human dentition is described as thecodont, diphyodont and heterodont.
2. It is called thecodont type because each tooth is fixed in a separate socket present in the jaw bones by gomphosis type of joint.
3. It is called diphyodont type because we get only two sets of teeth, milk teeth and permanent teeth.
4. It is called heterodont type because humans have four different type of teeth like incisors, canines, premolars and molars.

Question 3.
Muscularis layer in stomach is thicker than that in intestine. Why is it so?
Answer:
Muscularis layer in stomach is thicker than that of intestine because food is churned and gastric juices are mixed in the stomach whereas in intestine only absorption takes place.

Question 4.
Liver is a vital organ. Justify.
Answer:

1. Kupffer cells of liver destroy toxic substances, dead and worn-out blood cells and microorganisms.
2. Bile juice secreted by liver emulsifies fats and makes food alkaline.’
3. Liver stores excess of glucose in the form of glycogen.
4. Deamination of excess amino acids to ammonia and its further conversion to urea takes place in liver.
5. Synthesis of vitamins A, D, K and BI2 takes place in liver.
6. It also produces blood proteins like prothrombin and fibrinogen.
7. During early development, it acts as haemopoietic organ.
   Therefore, liver is a vital organ.

Internet my friend: (Textbook Page No. 171)

Question 1.
Collect the different videos of functioning of digestive system,
Answer:
[Note: Students can scan the adjacent Q.R code to get conceptual clarity with the aid of a relevant video.]

Find out (Textbook Page No. 162)

Question 1.
What will be the dental formula of a three years old child?
Answer:
The dental formula of a three-year-old child will be: I \(\frac{2}{2}\), C \(\frac{1}{1}\), M \(\frac{2}{2}\) = \(\frac{2,1,2}{2,1,2}\)
i. e. 5 × 2 = 10 teeth in each jaw = 20 teeth.
As a child has 20 teeth by the age of three.

Question 2.
What is dental caries and dental plaque? How can one avoid it?
Answer:

• Dental caries are tooth decay or cavities caused by acids secreted by bacteria. Dental caries may be yellow or black in color.
• Dental plaques also known as tooth plaque is a soft, sticky film which forms on the teeth regularly. It is colourless to pale yellow in colour.
• Tooth decay and dental plaque can be prevented by brushing teeth twice a day with a fluoride containing tooth paste.
• Rinsing mouth thoroughly with a mouth wash and use of dental floss or interdental cleaners to clean teeth daily can help to avoid dental caries and dental plaque.

Internet my friend (Textbook Page No. 162)

Question 1.
Find out the role of orthodontist and dental technician.
Answer:
a. Orthodontics is a specialization in dental profession. Orthodontist straightens the crooked teeth, locates problem in patients’ teeth and their overall oral development. They might use X-rays, plaster molds or dental appliances like retainers and space maintainers to correct the problems,

b. Dental technicians are the ones which improves patients’ appearance, ability to chew and speech. They make dentures, crowns, bridges and dental braces.

Question 2.
What is a root canal treatment?
Answer:

• Root canal treatment is also known as endodontic treatment.
• It is a dental treatment of removing infection from inside of a tooth.
• Root canal is hollow section of tooth which contains the nerve tissue, blood vessels and other cells, this is also known as pulps.
• Crown and root are a part of tooth. Crown is present above the gum while root is embedded in the gum.
  e. Pulp which is present inside the root canal nourishes the tooth and provides moisture to the surrounding material.
• The nerves present inside the pulp sense hot cold temperatures as pain.
• First step of a root canal treatment is removal of dead pulp tissues by making a hole on the surface of tooth.
• In second step, the dentist cleans and decontaminates the area and fills the hollow area with adhesive cement in order to seal the canal completely.
• The tooth is dead after the therapy and the patient no longer feel any pain but the tooth becomes more fragile than ever.
• The last step of root canal is adding a crown or filling. Until the crown or filling is complete, patient is not supposed to chew or bite using that tooth. After the crown or filling patient can use that tooth as before.

Find out (Textbook Page No. 163)

Question 1.
You must have heard about appendicitis. It is inflammation of appendix. Find more information about this disorder.
Answer:

1.  Appendicitis is a condition where there is inflammation of appendix.
2. Appendix is a vestigial organ. It is a linger shaped pouch that projects from colon on the lower right side of the abdomen.
3. Appendicitis pain is very severe. It initially starts from the navel and then moves.
4. It occurs in the people of age group between 10 to 30.
5. Surgical removal is the standard treatment for appendicitis.
6. Symptoms: Nausea and vomiting, loss of appetite, low grade fever, constipation, abdominal bloating, severe pain in the right side of the abdomen.
7. Appendicitis is caused when there is blockage in the lining of the appendix that results in infection. The bacteria multiply rapidly and causes inflammation and it is then filled with pus.
8. If not treated properly appendix can rupture which can lead to further complications.
   [Students can use above answer for reference and find more information about appendicitis.]

Question 2.
What is heartburn? Why do we take antacids to control it?
Answer:
Heart burn is a problem created when stomach contents (acid) are forced back up to oesophagus. It causes a burning pain in lower chest.

Antacids are bases and help to treat heartburn by neutralizing the stomach acid. The key ingredients of antacids are calcium carbonate, magnesium hydroxide, aluminium hydroxide or sodium bicarbonate.

Activity (Textbook Page No. 163)

Make a model of human digestive system in a group.
Answer:
[Students are expected to perform this activity on their own.]

Always Remember (Textbook Page No. 166)

Question 1.
Food remains for a very short time in mouth but action of salivary amylase continues for further IS to 30 minutes till gastric juice mixes with food in the stomach. Why do you think it stops after the food gets mixed with gastric juice?
Answer:

1. The gastric juices are mixed with food in the stomach.
2. The pH of the stomach is 1.0-2.0 which is very acidic. Such high level of acidity leads to denaturation of salivary amylase’s protein structure.
3. On the other hand, pH 6.8 is required for salivary amylase to carry out the activity which is not found in stomach. Thus, activity of salivary amylase is stopped when food is mixed with gastric juice.

Internet my friend (Textbook Page No. 167)

Question 1.
How are bile pigments formed?
Answer:

1. When old and worn out red blood cells are destroyed by macrophages in liver, the globin portion of hemoglobin is split off and heme is converted to biliverdin.
2. Most of this biliverdin is converted to bilirubin, which gives bile its major pigmentation.
   [Source http://www.biologydiscussion.com/human-physiology/digestive-system/bile-pigments/bile-pigments-origin-and-formation-digestive-juice-human-biology/81803]

Think about it (Textbook Page No. 167)

Question 1.
How can I keep my pancreas healthy? Can a person live without pancreas?
Answer:

1. Pancreas can be kept healthy by:
   • Eating proper balanced and low-fat diet, with plenty of whole grains, fruits and vegetables.
   • Regular exercise and maintaining a healthy weight.
   • Limiting alcohol consumption and avoid smoking.
   • Adequate intake of water.
   • Regular checkups.
2. The pancreas is a gland that secretes digestive enzymes and insulin which is needed for a person to survive.
3. Without pancreas the person will develop diabetes and will have to take insulin for the rest of the life.
4. Without pancreas the body’s ability to absorb nutrients also decreases.
   Hence, though a person can survive without pancreas he may have to remain dependent on the medicines for survival.

Do it yourself? (Textbook Page No. 167)

Question 1.
You have studied the representation of enzymatic actions in the form of reactions.
Write the reactions of pancreatic enzymes.
Answer:

Do it yourself (Textbook Page No. 168)

Question 1.
Observe the following reactions and explain in words.

Answer:

1. Maltase acts on maltose to form glucose.
2. Sucrase acts on sucrose to form glucose and fructose.
3. Lactase acts on lactose to form glucose and galactose.
4. Dipeptidase acts on dipeptides to form amino acids.
5. Emulsified fats are converted into fatty acids and glycerol by lipase.

Use your brain power (Textbook Page No. 168)

Question 1.
Make a flow chart for digestion of carbohydrate.
Answer:

Question 2.
What is a proenzyme? Enlist various proenzymes involved in process of digestion and state their function.
Answer:
Proenzymes are synthesized in cells as an inactive precursor that undergo some modification before becoming catalytically active.
The various proenzymes involved in process of digestion are as follows:

• Pepsinogen: Pepsinogen when converted into its active form pepsin acts on proteins to form peptones and proteoses.
• Trypsinogen: Trypsinogen when converted to it active form trypsin converts proteins, proteoses and peptones to polypeptides.
• Chymotrypsinogen: Chymotrypsinogen when converted to active form chymotrypsin it converts polypeptides to dipeptides.

Question 3.
Differentiate between Chyme and Chyle.
Answer:

No.	Chyme	Chyle
a.	Chyme is a semi-fluid acidic mass of partially digested food.	Chyle is an alkaline slurry which contains various nutrients ready for absorption.
b.	Chyme leaves stomach and enters the small intestine.	Chyle leaves small intestine and enters large intestine.

Question 4.
Digestion of fats take place only after the food reaches small intestine. Give reason.
Answer:
Digestion of fats takes place in small intestine because the presence of fats in small intestine stimulates the release of pancreatic lipase from pancreas and bile from liver. Pancreatic lipases hydrolyze fat molecules into fatty acids and monoglycerides and bile brings about emulsification of fats. Therefore, digestion of fats occur when food reaches small intestine.

Observe and Discuss (Textbook Page No. 169)

Question 1.
Action of digestive juice in your group.
Answer:

Digestive juices	Action
Saliva	Saliva contains salivary amylase which breaks down starch into maltose.
Gastric juice	HC1 breaks converts inactive pepsinogen into its active form pepsin. Pepsin then breakdown proteins into peptones and proteoses.
Pancreatic juice	Pancreatic amylase acts on glycogen and starch and converts those into disaccharides. Enterokinase converts trypsinogen into trypsin (active form). Trypsin converts proteins, proteoses, peptones to polypeptides. Chymotrypsin converts polypeptides to dipeptides. Nucleases digest nucleic acids to pentose sugar.
Intestinal enzymes	Maltase converts maltose to glucose. Sucrase converts sucrose to glucose and fructose. Lactase converts lactose to glucose and galactose. Dipeptidases converts dipeptides to amino acids. Lipase converts emulsified fats into fatty acids and monoglycerides.
Bile juice	It brings about emulsification of fats.

Can you recall? (Textbook Page no. 170)

Question 1.
What is balanced diet?
Answer:
Balanced diet is a diet which contains proper amount of carbohydrates, fats, vitamins, proteins and minerals to maintain a good health.

Question 2.
Explain the terms undernourished, over-nourished and malnourished in details.
Answer:

• Undernourished: When supply of nutrients is less than the minimum amount of nutrients or food required for good health is called undernourished.
• Over-nourished: The intake of nutrients is excessive. In over-nourished the amount of nutrients exceeds the amount required for normal growth.
• Malnourished: Malnourished is a condition where a person’s diet does not contain right amount of nutrients.

Do you know? (Textbook Page No. 170)

Question 1.
What is gross calorific value?
Answer:
The amount of heat liberated by complete combustion of lg food in a bomb calorimeter is termed as gross calorific (gross energy) value.

Question 2.
What is physiological value?
Answer:
The actual energy produced by 1 g food is its physiological value.

Question 3.
Name the following
Energy content of food in animals is expressed in terms of?
Answer:
Heat Energy

Question 4.
Complete the following table representing Gross calorific value and physiological value of food component.

Food Component	Gross calorific value (Kcal/g)	Physiological value (Kcal/g)
Fats	(A)	9.0
(B)	5.65	4.0
Carbohydrates	(C)	(D)

Answer:

Food Component	Gross calorific value (Kcal/g)	Physiological value (Kcal/g)
Fats	9.45	9.0
Proteins	5.65	4.0
Carbohydrates	4.1	4.0

Find out (Textbook Page No. 171)

Question 1.
Find out the status of nialnutrition among children in Maharashtra and efforts taken by the government to overcome the situation. Search for various NGOs working in this field.
Answer:
93,783 children have been diagnosed with severe acute malnutrition and 5.7 lakh with moderate acute malnutrition in Maharashtra.
Steps taken by government to overcome malnutrition:

1. Promotion of infant and young child feeding practices.
2. Management of malnutrition at community and facility level by trained service providers.
3. Treatment of children with severe acute malnutrition at special units called the Nutrition Rehabilitation Centres (NRCs), set up at public health facilities.
4. A special program to combat micronutrient deficiencies of Vitamin A, Iron and Folic acid.
5. The initiatives like Mother and Child protection card, village health and nutrition days, are taken by the government for addressing the nutrition concerns in children, pregnant women and lactating mothers.

Various NCOs working in this field:

1. Akshay Patra
2. Fight Hunger Foundation,
3. Feeding India,
4. No Hungry child
   [Source: http://pib.nic.in/newsite/PrintRelease.aspx?relid=l 13725; https://yourstory.com/2016/10/world- food-day-ngosj
   [Note: Students can use above answer as reference and find more information from the internet.]

Question 2.
Are jaundice and hepatitis same disorders?
Answer:
Jaundice and Hepatitis are two different disorders.

Jaundice: Jaundice occurs when the rate of bilirubin production exceeds the rate of its elimination. It causes yellowing of skin and eyes.

Hepatitis: It is a disease where there is inflammation of liver. It may be caused because of infection, over alcohol consumption, immune system disorder etc.

Do you know (Textbook Page No. 171)

Question 1.
Alcoholism causes different disorders of liver like steatosis (fatty liver), alcoholic hepatitis, fibrosis and cirrhosis. Collect more information on these disorders and try to increase awareness against alcoholism in society. Collect information about NGOs working against alcoholism.
Answer:
Steatosis (fatty liver): Steatosis is accumulation of fat in the liver. Treatment can help but it cannot be cured. Major risk factors are obesity and Diabetes type II, it is also associated with excessive alcohol consumption. Fatigue, weight loss and abdominal pain are some symptoms. It is a benign condition but in very smaller number of patients it can lead to liver failure. Treatment involves diet and exercise to reduce obesity.

Alcoholic hepatitis: Alcoholic Hepatitis is liver inflammation caused by excessive consumption of alcohol. It occurs in people who drink heavily for many years. Symptoms like yellowing of skin and eye, accumulation of fluid in stomach which leads to increase in stomach size. Treatments like completely stopping of alcohol consumption, hydration and nutrition care are carried out. Administration of steroid drugs reduces liver inflammation.

Fibrosis: There is significant scarring of liver tissue in this condition. Fibrosis itself does not cause any symptoms. Diagnosis includes doctor’s evaluation, blood tests and imaging tests, liver biopsy. Treatments include stopping the consumption of alcohol. There are no such effective drugs for curing of fibrosis.

Cirrhosis: It is a chronic liver damage caused due to various reasons which leads to irreversible scarring of liver and liver failure. Causes of cirrhosis are chronic alcohol abuse and hepatitis. Patients may experience fatigue, weakness and weight loss. In later stages, patients may develop jaundice, abdominal swelling and gastrointestinal bleeding. In advanced stage, a liver transplant is required.

NGOs working against alcoholism:

1. Muktangan Rehabilitation Centre
2. Anmol Jeevan Foundation
3. Sankalp Rehabilitation Trust
4. Kripa Foundation
5. Harmony Foundation
6. Hands for you Rehab Centre

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 1 Reproduction in Lower and Higher Plants Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Biology Solutions Chapter 1 Reproduction in Lower and Higher Plants

1. Multiple Choice Questions

Question 1.
Insect pollinated flowers usually possess ………………
(a) sticky pollen with a rough surface
(b) large quantities of pollens
(c) dry pollens with a smooth surface
(d) light colored pollens
Answer:
(a) sticky pollen with a rough surface

Question 2.
In ovule, meiosis occurs in ………………
(a) Integument
(b) Nucellus
(c) Megaspore
(d) Megaspore mother cell
Answer:
(d) Megaspore mother cell

Question 3.
The ploidy level is NOT the same in ………………
(a) Integuments and nucellus
(b) Root tip and shoot tip
(c) Secondary nucleus and endosperm
(d) Antipodals and synergids
Answer:
(c) Secondary nucleus and endosperm

Question 4.
Which of the following types require pollination but result is genetically similar to autogamy?
(a) Geitonogamy
(b) Xenogamy
(c) Apogamy
(d) Cleistogamy
Answer:
(a) Geitonogamy

Question 5.
If diploid chromosome number in a flowering plant is 12, then which one of the following will have 6 chromosomes?
(a) Endosperm
(b) Leaf cells
(c) Cotyledons
(d) Synergids
Answer:
(d) Synergids

Question 6.
In angiosperms, endosperm is formed by/ due to ………………
(a) free nuclear divisions of megaspore
(b) polar nuclei
(c) polar nuclei and male gamete
(d) synergids and male gametes
Answer:
(c) polar nuclei and male gamete

Question 7.
Point out the odd one.
(a) Nucellus
(b) Embryo sac
(c) Micropyle
(d) Pollen grain
Answer:
(d) Pollen grain

2. Very Short Answer Questions

Question 1.
The part of gynoecium that determines the compatible nature of pollen grain.
Answer:
Stigmatic surface.

Question 2.
How many haploid cells are present in a mature embryo sac?
Answer:
6 cells, 2 synergids, 1 egg cell, 3 antipodals.

Question 3.
Even though each pollen grain has 2 male gametes why at least 20 pollen grains are required to fertilize 20 ovules in a particular carpel?
Answer:
Angiosperms have phenomenon of double fertilization in which both the male gametes are utilized, one for fusion with egg cell to form zygote and other for fusion with secondary nucleus to form endosperm.

Question 4.
Megasporogenesis
Answer:
It is the process of formation of haploid megaspores from diploid megaspore mother cell.

Question 5.
What is hydrophily?
Answer:
Transfer of pollen grains in pollination process through agency of water is known as hydrophily.

Question 6.
The layer which supplies nourishment to the developing pollen grains.
Answer:
Tapetum

Question 7.
Parthenocarpy
Answer:
The condition in which fruit is developed without the process of fertilization is called parthenocarpy.

Question 8.
Are pollination and fertilization necessary in apomixis?
Answer:
Apomixis is formation of embryos without formation of gametes hence there is no need of pollination and fertilization.

Question 9.
The part of pistil which develops into fruit and seed.
Answer:
Ovary develops into fruit and ovules into seed.

Question 10.
What is the function of filiform apparatus ?
Answer:
Filiform apparatus guides the pollen tube towards egg cell.

3. Short Answer Questions

Question 1.
How polyembryony can be commercially exploited?
Answer:

1. Polyembryony is the development of more than one embryo inside the seed.
2. When such polyembryonic seed germinate we get multiple seedlings from it.
3. This condition increases the chances of survival of new plants.
4. Nucellar embryos are genetically identical to parent plants hence we get uniform plants.
5. In horticulture we can utilize these as rootstock for grafting, hence they have significant role in fruit breeding programmes e.g. Citrus, Mango.

Question 2.
Pollination and seeds formation are very crucial for the fruit formation, Justify.
Answer:

1. After fertilization, ovary is transformed into fruit, where ovary wall becomes fruit wall, i.e pericarp.
2. Mature ovules are transformed into seeds after fertilization.
3. Fertilization is a process where male gametes unites with female gamete to form zygote which develops into embryo.
4. In pollination process pollen grains carrying non-motile male gamete are transferred on stigma.
5. Seeds have embryo which germinate into new plant hence the goal of reproduction to create offspring for next generation is achieved. Hence these are the crucial events for fruit formation.

Question 3.
Incompatibility is a natural barrier in the fusion of gametes. How will you explain this statement?
Answer:

1. Self incompatibility or self-sterility is a genetic mechanism that prevents germination of pollen on stigma of same flower. This favours cross pollination. E.g. Tobacco.
2. In pollen-pistil interaction, when pollen grain is deposited on stigma, pistil has the ability to recognize and allow germination of right type of pollen.
3. Special type of proteins on stigmatic surface determine compatibility or incompatibility.
4. A physiological mechanism operates to ensure successful germination of compatible pollen.
5. Compatible pollen absorbs water and nutrients from stigmatic surface that are absent in pollen and then pollen tube emerges which grow-s through style.

Question 4.
Describe three devices by which cross pollination is encouraged in Angiosperms by avoiding self-pollination?
Answer:

1. Unisexuality, dichogamy, prepotency, heteromorphy and herkogamy are the outbreeding devices.
2. Unisexuality : The plants bear either male or female flowers. Due to unisexual nature, self-pollination is avoided. Plants are either dioecious, e.g. Papaya or monoecious, e.g. maize.
3. Heteromorphy : In same plants different types of flowers are produced. In these flowers, stigmas and anthers are situated at different levels. There is heterostyly and heteroanthy. This prevents self-pollination e.g. Primrose.
4. Herkogamy : In bisexual flowers we may come across mechanical device to prevent self-pollination. Natural physical barrier avoids contact of pollens with stigma. E.g. Calotropis where pollinia are situated below the stigma.

4. Long Answer Questions

Question 1.
Describe the process of double fertilization.
Answer:
Double fertilization:
(1) Out of the two male gametes produced by the male gametophyte in angiosperms, one unites with the female gamete and the other with the secondary nucleus. Since both the male gametes take part in fertilization and fertilization occurs twice, it is called double fertilization.

(2) During double fertilization, the pollen tube on reaching the ovule enters the embryo sac through micropyle and bursts in one of the synergids. Owing to this, the two male gametes contained in the pollen tube, are set free.

(3) Out of the two male gametes, one unites with the egg or female gamete and the other unites with the secondary nucleus of the embryo sac, forming a triploid or triple fusion nucleus, called the primary endosperm nucleus. The process involving the fusion of one of the male gametes with the egg nucleus, resulting in the formation of a diploid zygote is called syngamy.

(4) The reproductive process in which non-motile male nuclei are carried to the egg cell through a pollen tube is called siphonogamy.

(5) After fertilization, zygote develops into an embryo. Certain changes take place in the ovule leading to the development of a seed.

Question 2.
Explain the stages involved in the maturation of microspore into male gametophyte.
OR
Describe the development of male gametophyte before pollination in angiosperms.
OR
Sketch and label male gametophyte in angiosperm.
Answer:

1. Microspore or pollen grain is first cell of male gametophyte.
2. The protoplast of pollen grain divides mitotically to form two unequal cells – a small thin walled generative cell and a large naked vegetative or tube cell.
3. The generative cell possesses thin cytoplasm and a nucleus. It separates and floats in the cytoplasm of vegetative cell.
4. The vegetative, possesses thick cytoplasm, irregular shaped nucleus and the reserved food.
5. In majority of the angiosperms, the pollen grains are liberated at two-celled stage after the dehiscence of the anther.
6. The generative cell of the pollen grain divides by mitosis to form two male non-motile gametes.

Question 3.
Explain the development of dicot embryo.
Answer:
Development of embryo (dicot) in angio- sperm:

The oospore undergoes a transverse division to form a large basal cell towards the micropyle and a small apical or terminal cell towards the chalaza of the embryo sac. This two celled structure is called proembryo. The basal cell or suspensor initial undergoes repeated transverse divisions to form a multicellular structure called suspensor. The suspensor pushes the embryo towards the endosperm to draw its nutrition.

1. The development of embryo from a zygote is called embryogenesis.
2. The fusion of male gamete and an egg cell during fertilization results in the formation of a diploid zygote. The zygote develops a wall around it and is converted into oospore.
3. The apical cell or embryonal initial of the proembryo undergoes a transverse division followed by two vertical divisions at right angles to form an octant stage.
4. From octant, the lower four cells form hypocotyl and radicle while four cells of upper side form plumule with two cotyledons.
5. The lowermost cell of suspensor is hypophysis and by its further division forms part of radicle and root cap.
6. The cells from upper side of octant divide repeatedly to form heart shaped which elongated further to form two lateral cotyledons.
7. Enlargement of hypocotyl and cotyledon results into curved embryo which appears horse shoe shaped.

Question 4.
Draw a diagram of the L.S of anatropous ovule and list the components of embryo sac and mention their fate after fertilization.
Answer:
Components of Embryo sac.

1. Mature embryo sac is 7-celled and 8 nucleate.
2. Egg apparatus at micropylar end – with 2 synergids and egg cell.
3. Central cell with secondary nucleus formed by 2 polar nuclei
4. Antipodal cells at chalazal end – 3 cells.
5. Pollen tube enters the synergids, Synergids guide the growth of pollen tube towards egg.
6. Male gamete fuses with female gamete, i.e. syngamy to form zygote which develops into embryo.
7. One male gamete fuses with secondarynucleus to form primary endosperm nucleus (PEN) which forms endosperm, nutritive tissue for embryo.

5. Fill in the Blanks

Question 1.
The ……………… collects the pollen grains.
Answer:
biotic agents

Question 2.
The male whorl, called the ……………… produces ………………
Answer:
androecium, pollen grains

Question 3.
The pollen grains represent the ………………
Answer:
male

Question 4.
The ……………… contains the egg or ovum.
Answer:
embryo sac

Question 5.
…………….. takes place when one male gamete and the egg fuse together. The fertilized egg grows vs into seed from which the new plants can grow.
Answer:
Fertilization

Question 6.
The ……………… is the base of the flower to which other floral parts are attached.
Answer:
thalamus

Question 7.
……………… is the transfer of pollen grains from anther of the flower to the stigma of the same or a different flower.
Answer:
Pollination

Question 8.
Once the pollen reaches the stigma, pollen tube traverses down the ……………… to the ovary where fertilization occurs.
Answer:
style

Question 9.
The ……………… are coloured to attract the insects that carry the pollen. Some flowers also produce ……………… or ……………… that attracts insects.
Answer:
petals, fragrance, nectar

Question 10.
The whorl ……………… is green that protects the flower until it opens.
Answer:
Calyx.

6. Label the Parts of seed.

Answer:

7. Match the following

Column I (Structure Before seed formation)	Column II (Structure After seed formation)
A. Funiculus	i. Hilum
B. Scar of Ovule	ii. Tegmen
C. Zygote	iii. Testa
D. Inner Integument	iv. Stalk of Seed
	v. Embryo

(a) A-v, B-i, C-ii, D-iv
(b) A-iii, B-iv, C-i, D-v
(c) A-iv, B-i, C-v D-ii
(d) A-iv, B-v C-iii, D-ii
Answer:
(c) A-iv, B-i, C-v D-ii

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 15 Excretion and Osmoregulation Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Biology Solutions Chapter 15 Excretion and Osmoregulation

1. Choose correct option

Question (A).
Which one of the following organisms would spend maximum energy in production of nitrogenous waste?
a. Polar bear
b. Flamingo
c. Frog
d. Shark
Answer:
b. Flamingo

Question (B).
In human beings, uric acid is formed due to metabolism of __________.
a. amino acids
b. fatty acids
c. creatinine
d. nucleic acids
Answer:
d. nucleic acids

Question (C).
Visceral layer : Podocytes :: PCT : _______
a. Cilliated cells
b. Squamous cells
c. Columnar cells
d. Cells with brush border
Answer:
d. Cells with brush border

Question (D).
Deproteinised plasma is found in __________.
a. Bowman’s capsule
b. Descending limb
c. Glomerular capillaries
d. Ascending limb
Answer:
a. Bowman’s capsule, b. Descending limb, d. Ascending limb

Question (E).
Specific gravity of urine would _______ if level of ADH increases.
a. remain unaffected
b. increases
c. decreases
d. stabilise
Answer:
b. increases

Question (F).
What is micturition?
a. Urination
b. Urine formation
c. Uremia
d. Urolithiasis
Answer:
a. Urination

Question (G).
Which one of the following organisms excrete waste through nephridia?
a. Cockroach
b. Earthworm
c. Crab
d. Liver Fluke
Answer:
c. Crab

Question (H).
Person suffering from kidney stone is advised not to have tomatoes as it has _______.
a. seeds
b. lycopene
c. oxalic acid
d. sour taste
Answer:
c. oxalic acid

Question (I).
Tubular secretion does not take place in ________.
a. DCT
b. PCT
c. collecting duct
d. Henle’s loop
Answer:
b. PCT

Question (J).
The minor calyx ____________.
a. collects urine
b. connects pelvis to ureter
c. is present in the cortex
d. receives column of Bertini
Answer:
a. collects urine

Question (K).
Which one of the followings is not a part of human kidney?
a. Malpighian body
b. Malpighian tubule
c. Glomerulus
d. Loop of Henle
Answer:
b. Malpighian tubule

Question (L).
The yellow colour of the urine is due to presence of ___________
a. uric acid
b. cholesterol
c. urochrome
d. urea
Answer:
c. urochrome

Question (M).
Hypotonic filtrate is formed in _______
a. PCT
b. DCT
c. LoH
d. CT
Answer:
a. PCT

Question (N).
In reptiles, uric acid is stored in _____
a. cloaca
b. fat bodies
c. liver
d. anus
Answer:
a. cloaca

Question (O).
The part of nephron which absorbs glucose and amino acid is______
a. collecting tubule
b. proximal tubule
c. Henle’s loop
d. DCT
Answer:
b. proximal tubule

Question (P).
Bowman’s capsule is located in kidney in the ________
a. cortex
b. medulla
c. pelvis
d. pyramids
Answer:
a. cortex

Question (Q).
The snakes living in desert are mainly__________
a. aminotelic
b. ureotelic
c. ammonotelic
d. uricotelic
Answer:
d. uricotelic

Question (R).
Urea is a product of breakdown of ___________
a. fatty acids
b. amino acids
c. glucose
d. fats
Answer:
b. amino acids

Question (S).
Volume of the urine is regulated by__________
a. aldosterone
b. ADH
c. both a and b
d. none
Answer:

Question 2.
Answer the following questions

Question (A).
Doctors say Mr. Shaikh is suffering from urolithiasis. How it could be explained in simple words?
Answer:
Urolithiasis is the condition of having calculi in the urinary tract (which also includes the kidneys), which may pass into urinary bladder.

Question (B).
Anitaji needs to micturate several times and feels very thirsty. This is an indication of change in permeability of certain part of nephron. Which is this part?
Answer:

1. Need to micturate several times (polyuria) and feeling very thirsty (polydipsia) is a symptom of diabetes insipidus (imbalance of fluids in the body).
2. ADH prevents diuresis and due to absence of ADH, large amount of dilute urine is excreted.
3. ADH stimulates reabsorption of water from last part of DCT and entire collecting duct by increasing the permeability of cells.
4. If the permeability of these cells changes, it will result in increase in urine volume (frequent micturition) and increase in the osmolarity of blood. An imbalance in volume and osmolarity of body fluids increases thirst.

[Note: Water is reabsorbed by osmosis in PCT, DCT and descending limb of loop of Henle)

Question (C).
Effective filtration pressure was calculated to be 20 mm Hg; where glomerular hydrostatic pressure was 70 mm of Hg. Which other pressure is affecting the filtration process? How much is it?
Answer:
The other pressure affecting the filtration process is osmotic pressure of blood and filtrate hydrostatic pressure. Commonly effective filtration pressure (EFP) is represented as;
EFP = Glomerular Hydrostatic pressure in glomerulus – (Osmotic pressure of blood + Filtrate Hydrostatic pressure)
If EFP = 20 mmHg and Glomerular Hydrostatic pressure = 70 mmHg
20 = 70 – (Osmotic pressure of blood + Filtrate hydrostatic pressure)
∴ (Osmotic pressure of blood + Filtrate hydrostatic pressure) = 70-20
Then (Osmotic pressure of blood + Filtrate Hydrostatic pressure) = 50 mmHg .

[Note: Given values are insufficient to calculate the exact osmotic pressure of blood and filtrate hydrostatic pressure. The sum of the two values can be calculated to be 50 mmHg ]

Question (D).
Name any one guanotelic organism.
Answer:
Spiders, scorpions and penguins are guanotelic organisms as they excrete guanine.

Question (E).
Why are kidneys called ‘retroperitoneal’?
Answer:
Kidneys are located in abdomen. Kidneys are not surrounded by peritoneum instead they are located posterior to it. Thus, kidneys are called retroperitoneal.

Question (F).
State role of liver in urea production.
Answer:

1. Ammonia formed during the breakdown of amino acids is converted into urea in the liver of ureotelic animals.
2. This conversion takes place by the help of the ornithine / urea cycle.
3. 3 ATP molecules are used to produce one molecule of urea using the ornithine/ urea cycle. Since, the liver contains carrier molecules and enzymes necessary for urea cycle, it plays a major role in urea production.

Question (G).
Why do we get bad breath after eating garlic or raw onion?
Answer:

1. Raw onion and garlic contain volatile sulphur-containing compounds.
2. Sulphur-containing compounds have a distinctive odour which remain in the mouth after consumption of onion and garlic.
3. Also, volatile compounds (like certain sulphur containing compounds) in foodstuffs are generally excreted through the lungs and may result in bad breath.

3. Answer the following questions

Question (A).
John has two options as treatment for his renal problem : Dialysis or kidney transplants. Which option should he choose? Why?
Answer:

1. If John has two options of dialysis and kidney transplant, readily available he must opt for kidney transplant.
2. A kidney transplant, if successful, can improve the quality of life of a patient and reduce the risk of death.
3. The patient would not have to endure frequent dialysis procedures. Repeated visits for dialysis takes time and may not allow the patient to perform normal activities or go to office regularly.
4. Dialysis is regarded as a holding measure until kidney transplant can be performed or a supportive measure in those for whom a transplant would be inappropriate. However, dialysis cannot replace all the functions of a normal kidney such as production of hormones like erythropoietin, calcitriol and renin. Hence, if John has an option of kidney transplant, he must opt for it.

Question (B).
Amphibian tadpole can afford to be ammonotelic. Justify.
Answer:

1. Tadpole (larval stage of life cycle of amphibian) is aquatic. They are ammonotelic as they excrete nitrogenous waste in the form of ammonia.
2. Ammonia is very toxic and requires large amount of water for its elimination.
3. It is readily soluble in water and diffuses across the body surface and into the surrounding water.
4. Also, the water lost during excretion can be made up through the surrounding water in ammonotelic organisms.

Hence, amphibian tadpole can afford to be ammonotelic.

Question (C).
Birds are uricotelic in nature. Give reason.
Answer:

1. Birds are capable of converting ammonia into uric acid by ‘inosinic acid pathway’ in their liver.
2. Uric acid is least toxic and hence, it can be retained in the body for some time.
3. It is least soluble water hence, negligible amount of water is required for its elimination.
4. This mode of excretion can also help reduce body weight (an adaptation for flight) and those animals which
   need to conserve more water follow uricotelism.

Hence, in order to conserve water as an adaptation for flight, birds are uricotelic in nature.

Question 4.
Write the explanation in your word

Question (A).
Nitya has been admitted to hospital after heavy blood loss. Till proper treatment could be given; how did Nitya’s body must have tackled the situation?
Answer:

1. Heavy blood loss is called haemorrhage. In case of haemorrhage or severe dehydration, the osmoreceptors stimulate Antidiuretic hormone (ADH) secretion.
2. ADH is important in regulating water balance through the kidneys. For detailed mechanism of reabsorption by ADH:

Regulating water reabsorption through ADH:
Hypothalamus in the midbrain has special receptors called osmoreceptors which can detect change in osmolarity (measure of total number of dissolved particles per liter of solution) of blood.

If osmolarity of blood increases due to water loss from the body (after eating namkeen or due to sweating), osmoreceptors trigger release of Antidiuretic hormone (ADH) from neurohypophysis (posterior pituitary). ADH stimulates reabsorption of water from last part of DCT and entire collecting duct by increasing the permeability of cells.

This leads to reduction in urine volume and decrease in osmolarity of blood.

Once the osmolarity of blood comes to normal, activity of osmoreceptor cells decreases leading to decrease in ADH secretion. This is called negative feedback.
In case of hemorrhage or severe dehydration too, osmoreceptors stimulate ADH secretion. ADH is important in regulating water balance through kidneys.

In absence of ADH, diuresis (dilution of urine) takes place and person tends to excrete large amount of dilute urine. This condition called as diabetes insipidus.

[Note: Hypothalamus is a part of forebrain]

Another regulatory mechanism that must have been activated is RAAS. For detailed mechanism of electrolyte reabsorption:

Electrolyte reabsorption through RAAS:
Another regulatory mechanism is RAAS (Renin Angiotensin Aldosterone System) by Juxta Glomerular Apparatus (JGA).

Whenever blood supply (due to change in blood pressure or blood volume) to afferent arteriole decreases (e.g. low BP/dehydration), JGA cells release Renin. Renin converts angiotensinogen secreted by hepatocytes in liver to Angiotensin I. ‘Angiotensin-converting enzyme’ further modifies Angiotensin I to Angiotensin II, the active form of hormone. It stimulates adrenal cortex to release another hormone called aldosterone that stimulates DCT and collecting ducts to reabsorb more Na⁺ and water, thereby increasing blood volume and pressure.

Question 5.
Complete the diagram / chart with correct labels / information. Write the conceptual details regarding it

Question (A).

Answer:

The composition of urine depends upon food and fluid consumed by an individual. There are two ways in which it the composition is regulated. They are as follows:

i. Regulating water reabsorption through ADH
ii. Electrolyte reabsorption though RAAS
iii. Atrial Natriuretic Peptide

i. Regulating water reabsorption through ADH:
Hypothalamus in the midbrain has special receptors called osmoreceptors which can detect change in osmolarity (measure of total number of dissolved particles per liter of solution) of blood.

If osmolarity of blood increases due to water loss from the body (after eating namkeen or due to sweating), osmoreceptors trigger release of Antidiuretic hormone (ADH) from neurohypophysis (posterior pituitary). ADH stimulates reabsorption of water from last part of DCT and entire collecting duct by increasing the permeability of cells.
This leads to reduction in urine volume and decrease in osmolarity of blood.

Once the osmolarity of blood comes to normal, activity of osmoreceptor cells decreases leading to decrease in ADH secretion. This is called negative feedback.
In case of hemorrhage or severe dehydration too, osmoreceptors stimulate ADH secretion. ADH is important in regulating water balance through kidneys.

In absence of ADH, diuresis (dilution of urine) takes place and person tends to excrete large amount of dilute urine. This condition called as diabetes insipidus.

[Note: Hypothalamus is a part of forebrain]

ii. Electrolyte reabsorption through RAAS:
Another regulatory mechanism is RAAS (Renin Angiotensin Aldosterone System) by Juxta Glomerular Apparatus (JGA).

Whenever blood supply (due to change in blood pressure or blood volume) to afferent arteriole decreases (e.g. low BP/dehydration), JGA cells release Renin. Renin converts angiotensinogen secreted by hepatocytes in liver to Angiotensin I. ‘Angiotensin converting enzyme’ further modifies Angiotensin I to Angiotensin II, the active form of hormone. It stimulates adrenal cortex to release another hormone called aldosterone that stimulates DCT and collecting ducts to reabsorb more Na⁺ and water, thereby increasing blood volume and pressure.

iii. Atrial natriuretic peptide (ANP): A large increase in blood volume and pressure stimulates atrial wall to produce atrial natriuretic peptide (ANP). ANP inhibits Na⁺ and Cl^– reabsorption from collecting ducts inhibits release of renin, reduces aldosterone and ADH release too. This leads to a condition called Natriuresis (increased excretion of Na⁺ in urine) and diuresis.

Question (B).

Answer:

1. Nephrons are structural and functional units of kidney.
2. Each nephron consists of a 4 – 6 cm long, thin-walled tube called the renal tubule and a bunch of capillaries known as the glomerulus.
3. The wall of the renal tubule is made up of a single layer of epithelial cells.
4. Its proximal end is wide, blind, cup-like and is called as Bowman’s capsule, whereas the distal end is open.
5. The nephron is divisible into Ilowman’s capsule, neck, proximal convoluted tubule (PCT), Loop of Henle (LoH), distal convoluted tubule (DCT) and collecting tubule (CT).
6. The glomerulus is present in the cup-like cavity of Bowman’s capsule and both are collectively known as renal corpuscle or Malpighian body.

Question (C)

Answer:
Nephron is the structural and functional unit of kidney.
Structure of nephron:
A nephron (uriniferous tubule) is a thin walled, coiled duct, lined by a single layer of epithelial cells. Each nephron is divided into two main parts:

i. Malpighian body
ii. Renal tubule

i. Malpighian body: Each Malpighian body is about 200pm in diameter and consists of a Bowman’s capsule and glomerulus.

a. Glomerulus:
Glomerulus is a bunch of fine blood capillaries located in the cavity of Bowman’s capsule.
A small terminal branch of the renal artery, called as afferent arteriole enters the cup cavity (Bowman capsule) and undergoes extensive fine branching to form network of several capillaries. This bunch is called as glomerulus.
The capillary wall is fenestrated (perforated).

All capillaries reunite and form an efferent arteriole that leaves the cup cavity.
The diameter of the afferent arteriole is greater than the efferent arteriole. This creates a high hydrostatic pressure essential for ultrafiltration, in the glomerulus.

b. Bowman’s capsule:
It is a cup-like structure having double walls composed of squamous epithelium.
The outer wall is called as parietal wall and the inner wall is called as visceral wall.
The parietal wall is thin consisting of simple squamous epithelium.
There is a space called as capsular space / urinary space in between two walls.
Visceral wall consists of special type of squamous cells called podocytes having a foot-like pedicel. These podocytes are in close contact with the walls of capillaries of glomerulus.
There are small slits called as filtration slits in between adjacent podocytes.

ii. Renal tubule:

a. Neck:
The Bowman’s capsule continues into the neck. The wall of neck is made up of ciliated epithelium. The lumen of the neck is called the urinary pole. The neck leads to proximal convoluted tubule.

b. Proximal Convoluted Tubule :
This is highly coiled part of nephron which is lined by cuboidal cells with brush border (microvilli) and surrounded by peritubular capillaries. Selective reabsorption occurs in PCT. Due to convolutions (coiling), filtrate flows slowly and remains in the PCT for longer duration, ensuring that maximum amount of useful molecules are reabsorbed.

c. Loop of Henle :
This is ‘U’ shaped tube consisting of descending and ascending limb.
The descending limb is thin walled and permeable to water and lined with simple squamous epithelium.
The ascending limb is thick walled and impermeable to water and is lined with simple cuboidal epithelium.
The LoH is surrounded by capillaries called vasa recta.
Its function is to operate counter current system – a mechanism for osmoregulation.
The ascending limb of Henle’s loop leads to DCT.

d. Distal convoluted tubule:
This is another coiled part of the nephron.
Its wall consists of simple cuboidal epithelium.
DCT performs tubular secretion / augmentation / active secretion in which, wastes are taken up from surrounding capillaries and secreted into passing urine.
DCT helps in water reabsorption and regulation of pH of body fluids.

e. Collecting tubule:
This is a short, straight part of the DCT which reabsorbs water and secretes protons.
The collecting tubule opens into the collecting duct.

Question (D).

Answer:

The composition of urine depends upon food and fluid consumed by an individual. There are two ways in which it the composition is regulated. They are as follows:

i. Regulating water reabsorption through ADH
ii. Electrolyte reabsorption though RAAS
iii. Atrial Natriuretic Peptide

i. Regulating water reabsorption through ADH:
Hypothalamus in the midbrain has special receptors called osmoreceptors which can detect change in osmolarity (measure of total number of dissolved particles per liter of solution) of blood.

If osmolarity of blood increases due to water loss from the body (after eating namkeen or due to sweating), osmoreceptors trigger release of Antidiuretic hormone (ADH) from neurohypophysis (posterior pituitary). ADH stimulates reabsorption of water from last part of DCT and entire collecting duct by increasing the permeability of cells.
This leads to reduction in urine volume and decrease in osmolarity of blood.

Once the osmolarity of blood comes to normal, activity of osmoreceptor cells decreases leading to decrease in ADH secretion. This is called negative feedback.
In case of hemorrhage or severe dehydration too, osmoreceptors stimulate ADH secretion. ADH is important in regulating water balance through kidneys.

In absence of ADH, diuresis (dilution of urine) takes place and person tends to excrete large amount of dilute urine. This condition called as diabetes insipidus.

[Note: Hypothalamus is a part of forebrain]

ii. Electrolyte reabsorption through RAAS:
Another regulatory mechanism is RAAS (Renin Angiotensin Aldosterone System) by Juxta Glomerular Apparatus (JGA).
Whenever blood supply (due to change in blood pressure or blood volume) to afferent arteriole decreases (e.g. low BP/dehydration), JGA cells release Renin. Renin converts angiotensinogen secreted by hepatocytes in liver to Angiotensin I. ‘Angiotensin-converting enzyme’ further modifies Angiotensin I to Angiotensin II, the active form of hormone. It stimulates adrenal cortex to release another hormone called aldosterone that stimulates DCT and collecting ducts to reabsorb more Na⁺ and water, thereby increasing blood volume and pressure.

iii. Atrial natriuretic peptide (ANP): A large increase in blood volume and pressure stimulates atrial wall to produce atrial natriuretic peptide (ANP). ANP inhibits Na⁺ and Cl^– reabsorption from collecting ducts inhibits release of renin, reduces aldosterone and ADH release too. This leads to a condition called Natriuresis (increased excretion of Na⁺ in urine) and diuresis.

Question (E).

Answer:

1. When renal function of a person falls below 5 – 7 %, accumulation of harmful substances in blood begins. In such a condition the person has to go for artificial means of filtration of blood i.e. haemodialysis.
2. In haemodialysis, a dialysis machine is used to filter blood. The blood is filtered outside the body using a dialysis unit.
3. In this procedure, the patients’ blood is removed; generally from the radial artery and passed through a cellophane tube that acts as a semipermeable membrane.
4. The tube is immersed in a fluid called dialysate which is isosmotic to normal blood plasma. Hence, only excess salts if present in plasma pass through the cellophane tube into the dialysate.
5. Waste substances being absent in the dialysate, move from blood into the dialyzing fluid.
6. Filtered blood is returned to vein.
7. In this process it is essential that anticoagulant like heparin is added to the blood while it passing through the tube and before resending it into the circulation, adequate amount of anti-heparin is mixed.
8. Also, the blood has to move slowly through the tube and hence the process is slow.

Question 6.
Prove that mammalian urine contains urea.
Answer:

1. Urea is a nitrogenous waste formed by breakdown of protein (deamination of amino acids).
2. During this process, amino groups are removed from the amino acids present in the proteins and converted to highly toxic ammonia. The ammonia is finally converted to area through ornithine cycle. Thus, the urea formed is passed to kidneys and excreted out of the body through urine.
3. Reabsorption of urea (proximal tubule, collecting ducts) and active secretion of urea (Henle loop) leads to a urea circulation (urea recycling) between the lumen of the nephron and renal medulla, which is an important element of the renal urine concentration.
4. About 54 g of urea is filtered per day in the glomerular capsule, of which approximately 30 g is excreted in the urine and 24 g is reabsorbed into blood (assuming GFR is 180 litres/day).
5. Urinalysis can help detect the amount of urea in urine (Urine urea nitrogen test, urease test, etc.).

Practical / Project :

Visit to a nearby hospital or pathological laboratory and collect detailed information about different blood and urine tests.
Answer:
Testing the urine is known as urinalysis. It generally has three parts:

1. Visual examination: Check sample colour and clearness.
2. Dipstick examination: Checks for abnormal amounts of glucose, protein, etc.
3. Microscopic examination: Check for presence of RBCs. WBCs, bacteria, crystals, etc.
4. Apart from routine urine examination, specific tests may also be done. They are as follows:
   • BUN (Blood Urea Nitrogen) Test: It measures the amount of nitrogen in blood and evaluates kidney function.
   • Urease Test/ Urea Nitrogen Test: It is done to check the amount of urea in urine sample.
   • Urine albumin to creatine ratio (UACR) test: Estimates the amount of albumin in urine.

[Students are expected to collect more information and perform the given activity on their own]

12th Biology Digest Chapter 15 Excretion and Osmoregulation Intext Questions and Answers

Can you recall? (Textbook Page No. 174)

Question 1.
Why are various waste products produced in the body of an organism?
Answer:
Metabolism produces a variety of by-products, some of which need to be eliminated. Such by-produçts are called metabolic waste products.

Question 2.
How are these waste eliminated?
Answer:
Depending on the type of waste product, they are eliminated through various organs of the body:

The various excretory products produced by the human body are as follows:

1. Fluids such as water; gaseous wastes like CO₂ nitrogenous wastes like ammonia, urea and uric acid, creatinine; minerals; salts of sodium, potassium. calcium, etc. if present in body in excess are excreted through urine, faeces and sweat.
2. Pigments formed due to breakdown of haemoglobin like bilirubin (excreted through faeces) and urochrome (eliminated through urine).
3. The pigments present in consumed foodstuffs like beet root or excess of vitamins, hormones and drugs.
4. Volatile substances present in spices (eliminated through lungs).

Have you ever observed? (Textbook Page No. 174)

Question 1.
When does urine appear deeply coloured?
Answer:
Urine can appear deeply coloured due to various reasons:

• Severe dehydration resulting in production of concentrated urine.
• Consumption of foodstuff like beet root, which contain coloured pigments.
• Some medications can also cause the urine to appear deeply coloured.

Think about it. (Textbook Page No. 174)

Question 1.
Do organisms differ in type of metabolic wastes they produce?
Answer:
Yes, organisms differ in the type of metabolic wastes they produce. Some organisms excrete ammonia while some excrete urea or uric acid as metabolic wastes.

Question 2.
Do environment or evolution have any effect on type of waste produced by an organism?
Answer:

• The theory of evolution proposes that life started in an aquatic environment.
• Aquatic organisms are generally ammonotelic. It is believed that the urea cycle evolved to adapt to a changing environment when terrestrial life forms evolved.
• Arid conditions probably led to the evolution of the uric acid pathway as a means of conserving water.
• However, the correlation between evolution and type of waste production is uncertain.

Question 3.
How do thermoregulation and food habits affect saste production?
Answer:

1. To generate heat. endotherms convert the food that they eat into energy through a process called metabolism. Hence, they consume more tì.od in order to meet their energy requirements.
2. Also, carnivorous diet contains more proteins than herbivores.
3. Consumption of high protein or more food containing proteins can result in production of large amount of nitrogenous waste
4. These animals would also require more energy to eliminate the high levels oF nitrogenous wastes which build up when animal protein is digested.

Use your brain power. (Textbookpage No. 175)

Question 1.
Why ammonia is highly toxic?
Answer:

1. Ammonia is basic in nature and its retention in the body would disturb the pH of the body.
2. An increase in pH would disturb all enzyme catalysed reactions in the body and also make the plasma membrane unstable.

Hence, ammonia is highly toxic to the body.

Find out. (Textbook page No. 175)

You will study about a type of arthritis called gouty arthritis caused due to accumulation of uric acid in joints. Where does uric acid come from in case of ureotelic human beings?
Answer:

1. Uric acid produced as a waste product during the normal breakdown of nucleic acids (purines) and certain naturally occurring substances found in foods such as mushrooms. Mackerel, dried beans. etc.
2. This uric acid is generally excreted out along with urine.
3. If uric acid is produced in excess or not excreted, it accumulates in joints causing gouty arthritis.

Think about it. (Textbook Page No.175)

Endotherms consume more food in order to meet energy requirements. Also, carnivorous diet contains more proteins than herbivorous. Does it affect excretion of nitrogenous waste?
Answer:

1. To generate heat, endotherms convert the food that they eat into energy through a process called metabolism. Hence, they consume more food in order to meet their energy requirements.
2. Also, carnivorous diet contains more proteins than herbivores.
3. Consumption of high protein or more food containing proteins can result in production of large amount of nitrogenous waste.
4. These animals would also require more energy to eliminate the high levels of nitrogenous wastes which build up when animal protein is digested.

Observe and Discuss (Textbook Page No. 176)

Question 1.
These are blood reports of patients undergoing investigations for kidney function. What is creatinine? What is your observation and opinion about the findings? Why is it used as an index of kidney function?


Answer:
i. Creatinine:

• Plasma creatinine is produced from catabolism of creatinine phosphate during skeletal muscle contraction.
• It provides a ready source of high energy phosphate.

ii. Observations and Opinion:
Report A indicates a value of creatinine which is higher than the normal range. This would indicate impaired kidney function.
Report B indicates high fasting blood sugar and detection of sugar in the blood is known as glucosuria. High fasting blood sugar (>126 mg/dL) is usually indicative of diabetes.

iii. Creatinine used as an index of kidney function:

• Normally blood creatinine levels remain steady because the rate of production matches his excretion in urine.
• Hence, plasma creatinine is used as an index of kidney function and its level above normal is an indication of poor renal function.

Think about it. (Textbook Page No. 176)

Question 1.
During summer, we tend to produce less urine, is it so?
Answer:

1. Generally, excess water containing wastes is lost from the body in the time of urine. sweat and faeces.
2. During summer when the surrounding temperature is high. we also lose a considerable amount of water in the form of sweat.

Thus, the kidneys retain water for maintaining the concentration of body fluids and reduce the amount of water lost through urine.

Question 2.
Marine birds like Albatross spend their life on the sea. That means the water the, drink is salty. how do they manage osnioregulation then?
Answer:

1. Marine birds like Albatross have special glands called salt glands near their nostrils.
2. These glands are capable of secreting salts by active transport and help to manage osmotic balance,

[Note: The salt glands in Albatross are located in or on the skull in the area of eyes.]

Question 3.
like ectothermic and endothermic animals, do organisms differ in the way they maintain salt balance?
Answer:
Yes, organisms differ in the way they maintain salt balance.

1. Animals can be either isosmotic to the surrounding (osmoconformers or control the internal environment independent of external environment (osmoregulators).
2. Marine organisms are mostly osmoconfòrmers because their body fluids and external environment are isosmotic in nature while fresh water forms and terrestrial organisms are osmoregulators,
3. Generally, most organisms can tolerate only a narrow range of salt concentrations. Such organism are known as stenohaline organisms.
4. Organisms which are capable of handling a wide change in salinity are called euryhaline organisms.e.g. Barnacles, clams etc.

Find out. (Textbook Page No 176)

Question 1.
How do freshwater fishes and marine fishes carry out osmoregulation?
Answer:
Osmoregulation is the process of maintaining an internal concentration of salt and water in the body of fishes.

i. Freshwater fishes:
The salt concentration inside the body of freshwater fishes is higher than their surrounding water. Due to this, water enters the body due to osmosis.
If the flow of water into the body is not regulated. fishes would swell and get bigger.
To compensate this, the kidneys produce a large amount of urine,
Excretion of large amounts of urine regulates the level of water in the body hut results in the loss of salts.
Thus, in order to maintain a sufficient salt level, special cells in the gills (chloride cells) take tip ions from
the water, which are then directly transported into the blood.

ii. Marine fishes:
Since the salt content in blood of marine fishes is much lower than that of seawater, they constantly tend to lose water and build up salt.
To replace the water loss, they continually need to drink seawater.
Since their small kidney can only excrete a relatively small amount of urine, salt is additionally excreted through gills, where chloride cells work in reverse as in freshwater fishes.

Make a table. (Textbook Page No. 178)

Question 1.
The details of modes of excretion of nitrogenous wastes.
Answer:
The three main modes of excretion in animals are as follows:

i. Ammonotelism
ii. Ureotelism
iii. Uricotelism

i. Ammonotelism:

1. Elimination of nitrogenous wastes in the form of ammonia is called as ammonotelism.
2. Ammonia is basic in nature and hence it can disturb the pH of the body, if not eliminated immediately.
3. Any change in pH would disturb all enzyme catalyzed reactions in the body and would also make the plasma membrane unstable.
4. Ammonia is readily soluble in water and needs large quantity of water to dilute and reduce its toxicity.
5. This is however an energy saving mechanism of excretion and hence all animals that have plenty of water available for dilution of ammonia, excrete nitrogenous wastes in the form of ammonia.
6. Animals that follow this mode of excretion are known as ammonotelic animals.
7. 1 gm ammonia needs about 300 – 500 ml of water for elimination.
8. Ammonotelic animals excrete ammonia through general body surface (skin), gills and kidneys.
   e.g. Ammonotelism is found in aquatic invertebrates, bony fishes, and aquatic / larval amphibians. Animals without excretory system (Protozoa) are also ammonotelic.

ii. Ureotelism:

1. Elimination of nitrogenous wastes in the form of urea is called as ureotelism.
2. Urea is comparatively less toxic and less water-soluble than ammonia. Hence, it can be concentrated to some extent in body.
3. The body requires less water for elimination.
4. Since it is less toxic and less water soluble, ureotelism is suitable for animals that need to conserve water to some extent. Hence, ureotelism is common in terrestrial animals, as they have to conserve water.
5. It takes about 50 ml H₂O for removal of 1 gm NH₂ in form of urea.
6. Ureotelic animals generally convert ammonia to urea in the liver by operating ornithine / urea cycle in which 3 ATP molecules are used to produce one molecule of urea.
   e.g. Mammals, cartilaginous fishes (sharks and rays), many aquatic reptiles, most of the adult amphibians, etc. are ureotelic.

iii. Uricotelism:

1. Elimination of nitrogenous wastes in the form of uric acid is called as uricotelism.
2. Uric acid is least toxic and hence, it can be retained in the body for some time in concentrated form.
3. It is least soluble in water. Hence there is minimum (about 5 – 10 ml for 1 gm) or no need of water for its elimination.
4. Those animals which need to conserve more water follow uricotelism. However, these animals need to spend more energy.
5. Ammonia is converted into uric acid by ‘inosinic acid pathway’ in the liver of birds, e.g. Birds, some insects, many reptiles, land snails, are uricotelic.

No.	Ureotelism	Uricotelism
i.	It is the elimination of nitrogenous waste in the form of urea.	It is the elimination of nitrogenous waste in the form of uric acid.
ii.	Excretion of urea requires less (moderate ) amount of water.	Excretion of uric acid requires negligible amount of water.
iii.	Removal of 1 gm of urea requires 50 ml of water.	Removal of 1 gm of uric acid requires 5 – 10 ml of
iv.	rea is less toxic.	Uric acid is least toxic.
e.g.	It is generally seen in terrestrial animals. Mammals, cartilaginous fishes (sharks and rays), many aquatic reptiles, most adult amphibians, etc.	It is seen in birds, some insects, many reptiles, land snails, etc.

No.	Ammonotelism	Uricotelism
i.	It is the elimination of nitrogenous waste in the form of ammonia.	It is the elimination of nitrogenous waste in the form of uric acid.
ii.	Excretion of ammonia requires plenty of water.	Excretion of uric acid requires negligible amount of water.
iii.	Removal of 1 gm of ammonia requires 300 – 500 ml of water.	Removal of 1 gm of uric acid requires 10ml of water.
iv.	Ammonia is very toxic.	Uric acid is less toxic.
e.g.	It is found in aquatic invertebrates, bony fishes and aquatic/ larval amphibians, etc.	It is seen in birds, some insects, many reptiles, land snails, etc.

[Students can Refer these and make a chart on their own.]

Use your brain power. (Textbook Page No. 178)

Question 1.
Creatinine is considered as index of kidney function. Give reason.
Answer:

1. Plasma creatinine is produced from catabolism of creatinine phosphate during skeletal muscle contraction.
2. It provides a ready source of high energy phosphate.
3. Normally blood creatinine levels remain steady because the rate of production matches its excretion in urine.
4. Hence, plasma creatinine is used as an index of kidney function and its level above normal is an indication of poor renal function.

[Note: Plasma creatinine is a waste product produced by muscles from the breakdown of a compound called ‘creatine phosphate ’.]

Make a table. (Textbook Page No. 178)

Question 1.
The excretory organs found in various animal phyla.
Answer:

Sr. No.	Animal Phyla	Excretory organs
i.	Porifera	Lack excretory organ instead rely on water transport system/ Canal system
ii.	Coelenterata	Lack specialised excretory organs. Excretion takes place through simple diffusion or through the mouth.
iii.	Ctenophora	Lack specialised excretory organs
iv.	Platyhelminthes	Protonephridia or Flame cells
v.	Aschelminthes	Excretory tube and pore
vi.	Annelida	Nephridia
vii.	Arthropoda	Malpighian tubules
viii.	Mollusca	Organ of Bojanus
ix.	Echinodermata	Lack specialized excretory organs, waste materials directly diffuse into water or are excreted through tube feet
x.	Hemichordata	Proboscis gland
xi.	Chordata	Kidney

Observe and complete. (Textbook Page No. 178)

Question 1.
Label the diagram and complete following paragraphs.

i. Kidney: A pair of ____ shaped kidneys are present on either side of the ____ from 12^th thoracic to 3^rd lumbar vertebra. Kidneys are present behind ___. Hence are called retroperitoneal. Dimensions of
each kidney are 10 × ____ × ____ cms. Average weight is ___ g in males and 135 g in ____. Outer surface is ___ and inner is concave. Notch on the inner concave surface is called ___. Renal artery enters and renal vein as well as ureter leave the kidney through hilus. Each kidney has almost 1 million functional units called ___.

ii. Ureters: A pair of ureters arise from ___ of each kidney. Each ureter is a long muscular tube 25 – 30 cm in length. Ureters open into ___ by separate openings, which are not guarded by valves. They pass obliquely through the wall of urinary bladder. This helps in prevention of ___ of urine due to compression of ureters while bladder is filled.

iii. Urinary bladder: It is a median ___ sac. A hollow muscular organ, the bladder is situated in pelvic cavity posterior to pubic symphysis. At the base of the ___ there is a small inverted triangular area called trigone. At the apex of this triangle is opening of urethra. At the two points of the base of the triangle are openings of ureters. Urinary bladder is covered externally by peritoneum. Inner to peritoneum is muscular layer. It is formed by detrusor muscles which consist of three layers of smooth muscles. Longitudinal – circular – longitudinal respectively. Innermost layer is made up of transitional ___. It helps bladder to stretch.

iv. Urethra: It is a ___ structure arising from urinary bladder and opening to the exterior of the body.
There are ___ urethral sphincters between urinary bladder and urethra.
a. Internal sphincter: Made up of ___ muscles, involuntary in nature.
b. External sphincter: Made up of ___ muscles, voluntary in nature.
Answer:

i. Kidney: A pair of bean shaped kidneys are present on either side of the backbone from 12^th thoracic to 3^rd lumbar vertebra. Kidneys are present behind peritoneum. Hence are called retroperitoneal. Dimensions of each kidney are 10 × 5 × 4 cms. Average weight is 150 g in males and 135 g in females. Outer surface is convex and inner is concave. Notch on the inner concave surface is called hilum. Renal artery enters and renal vein as well as ureter leave the kidney through hilus. Each kidney has almost 1 million functional units called nephron.

ii. Ureters: A pair of ureters arise from hilum of each kidney. Each ureter is a long muscular tube 25 – 30 cm in length. Ureters open into urinary bladder by separate openings, which are not guarded by valves. They pass obliquely through the wall of urinary bladder. This helps in prevention of backward flow of urine due to compression of ureters while bladder is filled.

iii. Urinary bladder: It is a median pear-shaped sac. A hollow muscular organ, the bladder is situated in pelvic cavity posterior to pubic symphysis. At the base of the urinary bladder there is a small inverted triangular area called trigone. At the apex of this triangle is opening of urethra. At the two points of the base of the triangle are openings of ureters. Urinary bladder is covered externally by peritoneum. Inner to peritoneum is muscular layer. It is formed by detrusor muscles which consist of three layers of smooth muscles. Longitudinal – circular – longitudinal respectively. Innermost layer is made up of transitional epithelial tissue. It helps bladder to stretch.

iv. Urethra: It is a fibromuscular tube-like structure arising from urinary bladder and opening to the exterior of the body. There are two urethral sphincters between urinary bladder and urethra.
a. Internal sphincter: Made up of detrusor muscles, involuntary in nature.
b. External sphincter: Made up of striated muscles, voluntary in nature.
If this valve is not functioning properly during inflammation of bladder, it can lead to kidney infection.

Internet is my friend. (Textbook Page No. 179)

Question 1.
Find out what is floating kidney.
Answer:

1. Floating kidney or nephroptosis, is an inferior displacement or dropping of the kidney.
2. This condition occurs when the kidney slips from its normal position because it is not held securely in place by the adjacent organs or its fat covering.
3. It generally develops in extremely thin people whose adipose capsule or renal fascia is deficient.
4. It may result in twisting of the ureter and cause blockage of urine flow. The resulting backup of urine would put pressure on the kidney and damage the tissues.
5. Twisting of the ureter may also cause pain and discomfort.
6. This condition is more common in females than males and happens commonly among one in four people.
7. Weakening of the fibrous bands that hold the kidney in place can predispose to floating kidney.

Can you recall? (Textbook Page No. 179)

Question 1.
Observe the figure carefully and label various regions of L.S. of kidney.

Answer:

Can you tell? (Textbook Page No.182)

Question 1.
Why are kidneys called ‘retroperitoneal’?
Answer:
Kidneys are located in abdomen. Kidneys are not surrounded by peritoneum instead they are located posterior to it. Thus, kidneys are called retroperitoneal.

Question 2.
Why urinary tract infections are more common in females than males?
Answer:

• The urethra in women (4 cm) is much shorter than that of males (20 cm).
• This allows easy passage of bacteria into the urinary bladder.

Hence, urinary tract infections are more common in females than males.

Question 3.
What is nephron? Which are its main parts? Why are they important?
Answer:
Nephron is the structural and functional unit of kidney.
Structure of nephron:
A nephron (uriniferous tubule) is a thin walled, coiled duct, lined by a single layer of epithelial cells. Each nephron is divided into two main parts:

i. Malpighian body
ii. Renal tubule

i. Malpighian body: Each Malpighian body is about 200pm in diameter and consists of a Bowman’s capsule and glomerulus.

a. Glomerulus:
Glomerulus is a bunch of fine blood capillaries located in the cavity of Bowman’s capsule.
A small terminal branch of the renal artery, called as afferent arteriole enters the cup cavity (Bowman capsule) and undergoes extensive fine branching to form network of several capillaries. This bunch is called as glomerulus.
The capillary wall is fenestrated (perforated).

All capillaries reunite and form an efferent arteriole that leaves the cup cavity.
The diameter of the afferent arteriole is greater than the efferent arteriole. This creates a high hydrostatic pressure essential for ultrafiltration, in the glomerulus.

b. Bowman’s capsule:
It is a cup-like structure having double walls composed of squamous epithelium.
The outer wall is called as parietal wall and the inner wall is called as visceral wall.
The parietal wall is thin consisting of simple squamous epithelium.
There is a space called as capsular space / urinary space in between two walls.
Visceral wall consists of special type of squamous cells called podocytes having a foot-like pedicel. These podocytes are in close contact with the walls of capillaries of glomerulus.
There are small slits called as filtration slits in between adjacent podocytes.

ii. Renal tubule:

a. Neck:
The Bowman’s capsule continues into the neck. The wall of neck is made up of ciliated epithelium. The lumen of the neck is called the urinary pole. The neck leads to proximal convoluted tubule.

b. Proximal Convoluted Tubule :
This is highly coiled part of nephron which is lined by cuboidal cells with brush border (microvilli) and surrounded by peritubular capillaries. Selective reabsorption occurs in PCT. Due to convolutions (coiling), filtrate flows slowly and remains in the PCT for longer duration, ensuring that maximum amount of useful molecules are reabsorbed.

c. Loop of Henle :
This is ‘U’ shaped tube consisting of descending and ascending limb.
The descending limb is thin walled and permeable to water and lined with simple squamous epithelium.
The ascending limb is thick walled and impermeable to water and is lined with simple cuboidal epithelium.
The LoH is surrounded by capillaries called vasa recta.
Its function is to operate counter current system – a mechanism for osmoregulation.
The ascending limb of Henle’s loop leads to DCT.

d. Distal convoluted tubule:
This is another coiled part of the nephron.
Its wall consists of simple cuboidal epithelium.
DCT performs tubular secretion / augmentation / active secretion in which, wastes are taken up from surrounding capillaries and secreted into passing urine.
DCT helps in water reabsorption and regulation of pH of body fluids.

e. Collecting tubule:
This is a short, straight part of the DCT which reabsorbs water and secretes protons.
The collecting tubule opens into the collecting duct.

Think about ¡t. (Textbook Page No. 182)

Question 1.
How much blood ¡s supplied to kidney?
Answer:
Around 600 ml of blood passes through each kidney per minute.

Do this. (Textbook Page No. 183)

Question 1.
Check blood reports of patients and comment about possibility of glucosuria.
Answer:
Glucosuria is the presence of glucose sugar in urine. High glucose in urine is usually indicative of diabetes mellitus.

Condition	Glucose range in urine
Normal	0 to 15 mg/dL (0 to 0.8 mmol/L)
Prediabetes	100 to 125 mg/dL (5.6 to 6.9 mmol/L)
Diabetes	126 mg/dL (7 mmol/L)

[Students can get access to sample reports on the internet and refer the above table to comment on blood reports of patients on their own.]

Use your brain power. (Textbook Page No. 185)

Question 1.
In which regions of nephron the filtrate will he isotonic to blood?
Answer:
Filtrate leasing the proximal convoluted tubule (PCT) is isotonic to the blood plasma.

Can you tell? (Textbook Page No. 185)

Question 1.
Explain the process of urine formation in details.
Answer:
Process of urine formation is completed in three steps, namely;

i. Ultrafiltration/ Glomerular filtration,
ii. Selective reabsorption,
iii. Tubular secretion / Augmentation

i. Ultrafiltration / Glomerular filtration :
Diameter of afferent arteriole is greater than the efferent arteriole. The diameter of capillaries is still smaller than both arterioles. Due to the difference in diameter, blood flows with greater pressure through the glomerulus. This is called as glomerular hydrostatic pressure (GHP) and normally, it is about 55 mmHg. GIIP is opposed by osmotic pressure of blood (normally, about 30 mm Hg) and capsular pressure (normally, about 15 mm Hg).

Hence net / effective filtration pressure (EFP) is 10 mm Hg.
EFP = Hydrostatic pressure in glomerulus – (Osmotic pressure of blood + Filtrate Hydrostatic pressure)
= 55 – (30 + 15)
= 10 mm Hg

Under the effect of high pressure, the thin walls of the capillary become permeable to major components of blood (except blood cells and macromolecules like protein).
Thus, plasma except proteins oozes out through wall of capillaries.
About 600 ml blood passes through each kidney per minute.

The blood (plasma) flowing through kidney (glomeruli) is filtered as glomerular filtrate, at a rate of 125 ml / min. (180 L/d).
Glomerular filtrate / deproteinized plasma / primary urine is alkaline, contains urea, amino acids, glucose, pigments, and inorganic ions.
Glomerular filtrate passes through filtration slits into capsular space and then reaches the proximal convoluted tubule.

ii. Selective reabsorption :
Selective reabsorption occurs in proximal convoluted tubule (PCT). It is highly coiled so that glomerular filtrate passes through it very slowly. Columnar cells of PCT are provided with microvilli due to which absorptive area increases enormously.
This makes the process of reabsorption very effective.
These cells perform active (ATP mediated) and passive (simple diffusion) reabsorption.

Substances with considerable importance (high threshold) like – glucose, amino acids, vitamin C, Ca⁺⁺, K⁺, Na⁺, Cl^– are absorbed actively, against the concentration gradient. Low threshold substances like water, sulphates, nitrates, etc., are absorbed passively.
In this way, about 99% of glomerular filtrate is reabsorbed in PCT and DCT.

iii. Tubular secretion / Augmentation :
Finally filtrate reaches the distal convoluted tubule via loop of Henle. Peritubular capillaries surround DCT. Cells of distal convoluted tubule and collecting tubule actively absorb the wastes like creatinine and ions like K⁺, H⁺ from peritubular capillaries and secrete them into the lumen of DCT and CT, thereby augmenting the concentration of urine and changing its pH from alkaline to acidic.
Secretion of H⁺ ions in DCT and CT is an important homeostatic mechanism for pH regulation of blood. Tubular secretion is the only process of excretion in marine bony fishes and desert amphibians.

Question 2.
How does counter current mechanism help concentration of urine?
Answer:
Under the conditions like low water intake or high water loss due to sweating, humans can produce concentrated urine. This urine can be concentrated around four times i.e. 1200 mOsm/L than the blood (300 mOsm/L). Hence, a mechanism called countercurrent mechanism is operated in the human kidneys. The countercurrent mechanism operating in the Limbs of Henle’s loop of juxtamedullary nephrons and vasa recta is as follows:

1. It involves the passage of fluid from descending to ascending limb of Henle’s loop.
2. This mechanism is called countercurrent mechanism, since the flow of tubular fluid is in opposite direction through both limbs.
3. In case of the vasa recta, blood flows from ascending to descending parts of itself.
4. Wall of descending limb is thin and permeable to water, hence, water diffuses from tubular fluid into tissue fluid due to which, tubular fluid becomes concentrated.
5. The ascending limb is thick and impermeable to water. Its cells can reabsorb Na⁺ and Cl^– from tubular fluid and release into tissue fluid.
6. Due to this, tissue fluid around descending limb becomes concentrated. This makes more water to move out from descending limb into tissue fluid by osmosis.
7. Thus, as tubular fluid passes down through descending limb, its osmolarity (concentration) increases gradually due to water loss and on the other hand, progressively decreases due to Na⁺ and Cl^– secretion as it flows up through ascending limb.
8. Whenever retention of water is necessary, the pituitary secretes ADH. ADH makes the cells in the wall of collecting ducts permeable to water.
9. Due to this, water moves from tubular fluid into tissue fluid, making the urine concentrated.
10. Cells in the wall of deep medullar part of collecting ducts are permeable to urea. As concentrated urine flows through it, urea diffuses from urine into tissue fluid and from tissue fluid into the tubular fluid flowing through thin ascending limb of Henle’s loop.
11. This urea cannot pass out from tubular fluid while flowing through thick segment of ascending limb, DCT and cortical portion of collecting duct due to impermeability for it in these regions.
12. However, while flowing through collecting duct, water reabsorption is operated under the influence of ADII. Due to this, urea concentration increases in the tubular fluid and same urea again diffuses into tissue fluid in deep medullar region.
13.  Thus, same urea is transferred between segments of renal tubule and tissue fluid of inner medulla. This is called urea recycling; operated for more and more water reabsorption from tubular fluid and thereby excreting small volumes of concentrated urine.
14. Osmotic gradient is essential in the renal medulla for water reabsorption by counter current multiplier system.
15. This osmotic gradient is maintained by vasa recta by operating counter current exchange system.
16. Vasa recta also have descending and ascending limbs. Blood that enters the descending limb of the vasa recta has normal osmolarity of about 300 mOsm/L.
17.  As it flows down in the region of renal medulla where tissue fluid becomes increasingly concentrated, Na⁺, Cl^– and urea molecules diffuse from tissue fluid into blood and water diffuse from blood into tissue fluid.
18. Due to this, blood becomes more concentrated which now flows through ascending part of vasa recta. This part runs through such region of medulla where tissue fluid is less concentrated.
19. Due to this, Na⁺, Cl^– and urea molecules diffuse from blood to tissue fluid and water from tissue fluid to blood. This mechanism helps to maintain the osmotic gradient.
    

Try this. (Textbook Page No. 185)

Question 1.
Read the given urine report and prepare a note on composition of normal urine.

Answer:
The composition of normal urine is as follows:

1. A volume of 1 – 2 litres of urine in 24 hours is normal. This volume can however vary considerably as it depends on fluid intake, physical activity, temperature, etc.
2. The colour of normal urine is generally pale yellow due to urochrome (pigment produced by breakdown of bile). The colour of urine may vary slightly due to urochrome concentration and diet.
3. The appearance of urine is generally clear and transparent.
4. Any form of deposits (sediments/ crystals) is generally absent in normal urine.
5. The pH of normal urine is acidic and is generally around 6.0 (Range: 4.6 to 8.0). The pH varies considerably with the diet of a person.
6. The specific gravity of urine is an average of 1.02 ( Range : 1.001 to 1.035).
7. Albumin, sugar, bile salts bile pigments, ketone bodies and casts are absent in normal urine.
8. Occult blood is generally not seen in normal urine.

Think (Textbook Page No. 185)

Question 1.
What would happen if ADH secretion decreases due to any reason?
Answer:
In absence of ADH, diuresis (dilution of urine) takes place and person tends to excrete large amount of dilute urine. This condition called as diabetes insipidus. Frequent excretion of large amount of dilute urine may cause a person to feel thirsty.

Think and appreciate. (Textbook Page No. 185)

Question 1.
How do kidneys bring about homeostasis? Is there any role of neuro endocrine system in it?
Answer:
The composition of urine depends upon food and fluid consumed by an individual. There are two ways in which it the composition is regulated. They are as follows:

i. Regulating water reabsorption through ADH
ii. Electrolyte reabsorption though RAAS
iii. Atrial Natriuretic Peptide

i. Regulating water reabsorption through ADH:
Hypothalamus in the midbrain has special receptors called osmoreceptors which can detect change in osmolarity (measure of total number of dissolved particles per liter of solution) of blood.
If osmolarity of blood increases due to water loss from the body (after eating namkeen or due to sweating), osmoreceptors trigger release of Antidiuretic hormone (ADH) from neurohypophysis (posterior pituitary). ADH stimulates reabsorption of water from last part of DCT and entire collecting duct by increasing the permeability of cells.

This leads to reduction in urine volume and decrease in osmolarity of blood.
Once the osmolarity of blood comes to normal, activity of osmoreceptor cells decreases leading to decrease in ADH secretion. This is called negative feedback.
In case of hemorrhage or severe dehydration too, osmoreceptors stimulate ADH secretion. ADH is important in regulating water balance through kidneys.

In absence of ADH, diuresis (dilution of urine) takes place and person tends to excrete large amount of dilute urine. This condition called as diabetes insipidus.
[Note: Hypothalamus is a part of forebrain]

ii. Electrolyte reabsorption through RAAS:
Another regulatory mechanism is RAAS (Renin Angiotensin Aldosterone System) by Juxta Glomerular Apparatus (JGA).
Whenever blood supply (due to change in blood pressure or blood volume) to afferent arteriole decreases (e.g. low BP/dehydration), JGA cells release Renin. Renin converts angiotensinogen secreted by hepatocytes in liver to Angiotensin I. ‘Angiotensin converting enzyme’ further modifies Angiotensin I to Angiotensin II, the active form of hormone. It stimulates adrenal cortex to release another hormone called aldosterone that stimulates DCT and collecting ducts to reabsorb more Na and water, thereby increasing blood volume and pressure.

iii. Atrial natriuretic peptide (ANP): A large increase in blood volume and pressure stimulates atrial wall to produce atrial natriuretic peptide (ANP). ANP inhibits Na⁺ and Cl^– reabsorption from collecting ducts inhibits release of renin, reduces aldosterone and ADH release too. This leads to a condition called Natriuresis (increased excretion of Na⁺ in urine) and diuresis.

No. Both ADH and RAAS are essential for homeostasis.

1. Only ADH can lower blood Na⁺ concentration by way of water reabsorption in DCI and collecting duct. whereas RAAS stimulates Na⁺ reabsorption and maintains osmolarity of body fluid.
2. Action of ADH and RAAS leads to increase in blood volume and osmolarity.
3. For mechanism of Atrial natriuretic peptide:

Atrial natriuretic peptide (ANP): A large increase in blood volume and pressure stimulates atrial wall to produce atrial natriuretic peptide (ANP). ANP inhibits Na⁺ and Cl^– reabsorption from collecting ducts inhibits release of renin, reduces aldosterone and ADH release too. This leads to a condition called Natriuresis (increased excretion of Na⁺ in urine) and diuresis.

ADH is produced by the hypothalamus and is stored and released by the posterior pituitary or the neurohypophysis in response to appropriate trigger. Hence, there is a role of the neuroendocrine system in homeostasis.

Use your brain power. (Textbook Page No. 186)

Question 1.
Can we use this knowledge in treatment of high blood pressure? Why high BP medicines are many a times diuretics?
Answer:

1. Yes, the knowledge of homoeostasis is used in the treatment of high blood pressure.
2. Some commonly used theories for treatment of high blood pressure are as follows:
   • Angiotensin II receptor blockers (ARBs) are used as medications to treat high blood pressure. These medications block the action of angiotensin II by preventing angiotensin II from binding to angiotensin II receptors on the muscles surrounding blood vessels. As a result, blood vessels enlarge (dilate), and blood pressure is reduced.
   • Another method is the use of ‘Angiotensin converting enzyme’ ACE blockers. These inhibitors inhibit activity of ACE and therefore decrease the production of angiotensin II. As a result, these medications cause the blood vessels to enlarge or dilate, and this reduces blood pressure.
3. Vasodilation reduces arterial pressure. Reduced angiotensin II leads to natriuresis (increased excretion of Na⁺ in urine) and diuresis, thereby reducing blood pressure.
4. Too much salt can cause extra fluid to build up in the blood vessels, raising blood pressure. Diuretics are substances that slow renal absorption of water and thereby cause diuresis (elevated urine flow rate) which in turn reduces blood volume and blood pressure by flushing out salt and extra fluid. Hence, high BP medicines are many a times diuretics.

Can you tell? (Textbook page no. 186)
How do skin and lungs help in excretion?
OR
Can you tell? (Textbook page no. 187)
Explain role of lungs and skin in excretion.
Answer:
Yes, various organs other than the kidney participate in excretion. They are as follows:

i. Skin:

Skin acts as an accessory excretory organ. The skin of many organisms is thin and permeable. It helps in diffusion of waste products like ammonia.
Human skin however is thick and impermeable. It shows presence of two types of glands namely, sweat glands and sebaceous glands.

• Sweat glands are distributed all over the skin. They are abundant in the palm and facial regions.
  These simple, unbranched, coiled, tubular glands open on the surface of the skin through an opening called sweat pore. Sweat is primarily produced for thermoregulation but it also excretes substances like water, NaCl, lactic acid and urea.
• Sebaceous glands are present at the neck of hair follicles. They secrete oily substance called sebum.
  It forms a lubricating layer on skin making it softer. It protects skin from infection and injury.

ii. Lungs:

Lungs are the accessory excretory organs. They help in excretion of volatile substances like CO₂ and water vapour produced during cellular respiration. Along with CO₂, lungs also remove excess of H₂O in the form of vapours during expiration. They also excrete volatile substances present in spices and other food stuff.

Can you tell? (Textbook Page No. 187)

Question 1.
When does kidney produce renin? Where is it produced in kidney?
Answer:
Kidney produces renin whenever blood supply (due to change in blood pressure or blood volume) to afferent arteriole decreases (e.g. low BP/dehydration).
The juxtaglomerular Apparatus (JGA) cells secrete renin.

Question 2.
Explain how electrolyte balance of blood plasma maintained.
Answer:
The composition of urine depends upon food and fluid consumed by an individual. There are two ways in which it the composition is regulated. They are as follows:

i. Regulating water reabsorption through ADH
ii. Electrolyte reabsorption though RAAS
iii. Atrial Natriuretic Peptide

i. Regulating water reabsorption through ADH:
Hypothalamus in the midbrain has special receptors called osmoreceptors which can detect change in osmolarity (measure of total number of dissolved particles per liter of solution) of blood.

If osmolarity of blood increases due to water loss from the body (after eating namkeen or due to sweating), osmoreceptors trigger release of Antidiuretic hormone (ADH) from neurohypophysis (posterior pituitary). ADH stimulates reabsorption of water from last part of DCT and entire collecting duct by increasing the permeability of cells.

This leads to reduction in urine volume and decrease in osmolarity of blood.
Once the osmolarity of blood comes to normal, activity of osmoreceptor cells decreases leading to decrease in ADH secretion. This is called negative feedback.
In case of hemorrhage or severe dehydration too, osmoreceptors stimulate ADH secretion. ADH is important in regulating water balance through kidneys.

In absence of ADH, diuresis (dilution of urine) takes place and person tends to excrete large amount of dilute urine. This condition called as diabetes insipidus.
[Note: Hypothalamus is a part of forebrain]

ii. Electrolyte reabsorption through RAAS:
Another regulatory mechanism is RAAS (Renin Angiotensin Aldosterone System) by Juxta Glomerular Apparatus (JGA).

Whenever blood supply (due to change in blood pressure or blood volume) to afferent arteriole decreases (e.g. low BP/dehydration), JGA cells release Renin. Renin converts angiotensinogen secreted by hepatocytes in liver to Angiotensin I. ‘Angiotensin converting enzyme’ further modifies Angiotensin I to Angiotensin II, the active form of hormone. It stimulates adrenal cortex to release another hormone called aldosterone that stimulates DCT and collecting ducts to reabsorb more Na and water, thereby increasing blood volume and pressure.

iii. Atrial natriuretic peptide (ANP): A large increase in blood volume and pressure stimulates atrial wall to produce atrial natriuretic peptide (ANP). ANP inhibits Na⁺ and Cl^– reabsorption from collecting ducts inhibits release of renin, reduces aldosterone and ADH release too. This leads to a condition called Natriuresis (increased excretion of Na⁺ in urine) and diuresis.

No. Both ADH and RAAS are essential for homeostasis.

1. Only ADH can lower blood Na⁺ concentration by way of water reabsorption in DCI and collecting duct. whereas RAAS stimulates Na⁺ reabsorption and maintains osmolarity of body fluid.
2. Action of ADH and RAAS leads to increase in blood volume and osmolarity.
3. For mechanism of Atrial natriuretic peptide:

Atrial natriuretic peptide (ANP): A large increase in blood volume and pressure stimulates atrial wall to produce atrial natriuretic peptide (ANP). ANP inhibits Na⁺ and Cl^– reabsorption from collecting ducts inhibits release of renin, reduces aldosterone and ADH release too. This leads to a condition called Natriuresis (increased excretion of Na⁺ in urine) and diuresis.

Can you tell? (Textbook Page No. 187)

Question 1.
What is the composition of sweat?
Answer:
Sweat is composed of water, NaCl, lactic acid and urea.

Internet my friend. (Textbook Page No. 189)

Question 1.
Treatments other than surgical removal of kidney stone like Lithotripsy. (Breaking down of kidney stones using shock waves).
Answer:
a. Cystoscopy and ureteroscopy:
During cystoscopy, the doctor uses a cystoscope to look inside the urethra and bladder to find a stone in the urethra or bladder.
During ureteroscopy, the doctor uses a ureteroscope, which is longer and thinner than a cystoscope, to see detailed images of the lining of the ureters and kidneys.

The doctor inserts the cystoscope or ureteroscope through the urethra to see the rest of the urinary tract. Once the stone is found, the doctor can remove it or break it into smaller pieces.
The doctor performs these procedures in the hospital with anesthesia.

b. Percutaneous nephrolithotomy:
The doctor uses a thin viewing tool, called a nephroscope, to locate and remove the kidney stone.
The doctor inserts the tool directly into your kidney through a small cut made in your back.
For larger kidney stones, the doctor also may use a laser to break the kidney stones into small pieces. The doctor performs percutaneous nephrolithotomy in a hospital with anesthesia.

c. Generally for smaller stones doctors recommend drinking lots of water, consuming pain relievers and consuming medicines like alpha blocker to relax the ureter muscles, and help pass the kidney stones more quickly and with less pain

[Students are expected to find more information using the internet.]

Question 2.
Dietary restrictions suggested for kidney patients.
Dietary restrictions for kidney patients include the following:

1. Drinking large amounts of water.
2. Reduce consumption of oxalate rich food like rhubarb, beets, okra, spinach, Swiss chard, sweet potatoes, nuts, tea, chocolate and soy products.
3. Follow a diet low in salt and animal protein.
4. Reduce consumption of calcium supplements (if any) but consume appropriate amount of calcium in food.

[Students are expected to find more information using the internet.]

Balbharti Maharashtra State Board Organisation of Commerce and Management 11th Textbook Solutions Chapter 8 Introduction to Management Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Organisation of Commerce and Management Solutions Chapter 8 Introduction to Management

1. (A) Select the correct option and rewrite the sentences

Question 1.
Management is ………………….
(a) dynamic
(b) static
(c) selective
Answer:
(a) dynamic

Question 2.
Management uses …………………. method of observation.
(a) unscientific
(b) artificial
(c) scientific
Answer:
(c) scientific

Question 3.
To help top level management in co-ordinating the activities is the function of …………………. level management.
(a) middle
(b) lower
(c) top
Answer:
(a) middle

Question 4.
To look after maintenance of machinery is the function of …………………. level of management.
(a) top
(b) middle
(c) lower
Answer:
(c) lower

Question 5.
Management is …………………. oriented action.
(a) policy
(b) profit
(c) goal
Answer:
(c) goal

1. (B) Match the pairs

Question 1.

Part A	Part B
(a) Management	(1) Need of creativity
(b) Management as a profession	(2) Can be seen.
(c) Top level management	(3) Administration
(d) Management as an art	(4) Intangible
(e) Lower level management	(5) Formal education
	(6) Policy framing.
	(7) Implementation of policies
	(8) Selective
	(9) Directing subordinates to complete the task
	(10) Cause effect relationship

Answer:

Part A	Part B
(a) Management	(4) Intangible
(b) Management as a profession	(5) Formal education
(c) Top level management	(6) Policy framing.
(d) Management as an art	(1) Need of creativity
(e) Lower level management	(9) Directing subordinates to complete the task

1. (C) Give one word / phrase / term

Question 1.
The art of getting things done through others.
Answer:
Management

Question 2.
The second level of management.
Answer:
Middle level

Question 3.
Systematic body of knowledge.
Answer:
Management as a Science

Question 4.
The level of management related to Managing Director.
Answer:
Top level management

Question 5.
Level of management that link top level and lower level management.
Answer:
Middle level management

1. (D) State True or False

Question 1.
Management is an ancient activity.
Answer:
True

Question 2.
Management is not objective oriented.
Answer:
False

Question 3.
In small organization also management is required.
Answer:
True

Question 4.
Middle level management forms policy of an organization.
Answer:
False

Question 5.
Various principles and techniques are used in management.
Answer:
True

1. (E) Find the odd one

Question 1.
Management is an – art, a science, a commerce, a profession.
Answer:
a commerce

Question 2.
Management as an Art – Singer, Actor, Dancer, Doctor.
Answer:
Doctor

Question 3.
Management as Profession – Doctor, Chartered Accountant, Lawyer, Singer.
Answer:
Singer

Question 4.
Level of Management – Lower Level, Middle Level, Top Level, High Level.
Answer:
High level

Question 5.
Top Level Management – Board of Directors, Presidents, Purchase Manager, Chief Executive Officer (CEO).
Answer:
Purchase manager

Question 6.
Middle Level Management – Departmental Head, Finance Manager, Production Manager, Supervisor.
Answer:
Supervisor

Question 7.
Lower Level Management – Supervisor, Foreman, Finance Manager, Office Superintendent.
Answer:
Finance Manager

1. (F) Complete the sentences

Question 1.
The English verb ‘manage’ came from Italian word ……………….
Answer:
Maneggiare

Question 2.
To manage is to forecast, to plan, to organize, to command, to co-ordinate and to ……………….
Answer:
control

Question 3.
Organization is a group of person who work together to achieve a ……………….
Answer:
common goal

Question 4.
Link between the top level and lower level of management is called as ……………….
Answer:
middle level management

Question 5.
Lower level management works under ……………….
Answer:
middle level management

Question 6.
Art is the bringing about a desired result through the application of ……………….
Answer:
skills

Question 7.
Any knowledge which is scientifically developed and consists of universally accepted principles is known as a ……………….
Answer:
science

Question 8.
Science is based on systematic experiments, on certain rules and on ……………….
Answer:
principles

Question 9.
Teacher, Doctor, C.A., Lawyer comes under the term ……………….
Answer:
profession

1. (G) Select the correct option

(Finance Manager, Art, CEO, Profession, Supervisors)

Part A	Part B
(1) Top level	—————-
(2) —————–	Lower level
(3) Middle level	—————-
(4) ————-	Doctor
(5) Singer	—————-

Answer:

Part A	Part B
(1) Top level	CEO
(2) Supervisors	Lower level
(3) Middle level	Finance Manager
(4) Profession	Doctor
(5) Singer	Art

1. (H) Answer in one sentence

Question 1.
What is called as management?
Answer:
Getting work done through others is called as management.

Question 2.
Name, the levels of management?
Answer:
Top level, middle level and lower level are the three levels of management.

Question 3.
What is top level management?
Answer:
The level of management that decides the long term objectives of organization and frame the plans and policies and take decisions is called as top level-management.

Question 4.
What do you mean by middle level management?
Answer:
The level of management which is concerned with implementation of plans and policies in the organization with the help of lower level management is known as middle level management.

Question 5.
What is lower level management?
Answer:
The level of management which is considered to do actual execution of plan is called as lower level management.

Question 6.
What do you mean by an art?
Answer:
An art means skill of individual, creativity, regular practice, personal abilities and innovativeness.

Question 7.
What is science?
Answer:
Science is a systematic body of knowledge and scientific methods.

Question 8.
What is the meaning of a profession?
Answer:
Profession means an individual having adequate knowledge and abilities which is helpful to handle different situations. They are trained people having formal education in management school and have a code of conduct.

1. (I) Arrange in proper order.

Question 1.
Middle level management, Top level management, Lower level management.
Answer:
Top level management, Middle level management, Lower level management.

Question 2.
Departmental Head, CEO, Supervisors, Managing Director.
Answer:
CEO, Managing Director, Departmental Head, Supervisors.

Question 3.
Office clerk, Shareholders, Managing Director, Board of Director.
Answer:
Shareholders, Board of Directors, Managing Director, Office clerk.

1. (J) Correct the underlined word and rewrite the following sentences

Question 1.
Management is only an art.
Answer:
Management is an art, a science and a profession.

Question 2.
There are five levels of management.
Answer:
There are three levels of management.

Question 3.
Lower level management works under top level management.
Answer:
Middle level management works under top level management.

Question 4.
Middle level management works under lower level management.
Answer:
Middle level management works under top-level management.

Question 5.
Management is pure science.
Answer:
Management is a social science.

Question 6.
The entry in any profession is not restricted.
Answer:
The entry in any profession is restricted.

Question 7.
Professionals need informal education.
Answer:
Professional need formal education.

2. Explain the following terms/concepts

Question 1.
Top level management.
Answer:

1. The top level management decides the long term objectives of organization, frames plans and policies and take decisions.
2. It consists of Board of Directors, Managing Director,CEO, President, etc.

Question 2.
Middle level management.
Answer:

1. The middle level management is a link between top level and lower level management.
2. It works under top level management.
3. This level is considered with implementation of plans and policies in the organization with the help of lower level management.
4. It consists of Head of Departments / Branch managers / Incharge, etc.

Question 3.
Lower level management.
Answer:

1. This is the last level of management.
2. It is a supervisory level of management.
3. It consists of superintendents, supervisors, foremen and junior executives.
4. Lower level management receives instructions from the authorities i.e. middle level management and direct subordinates wherever necessary.

3. Study the following case/situation and express your opinion

1. While working in a company, Pranav used to give order to his subordinates about what is to be done and which work is to be done? In this organization, various employees like Pratap are doing their work assigned by Pravin. In this company, Pravin submits his report of work completion to Pranav after completing the work done by the employees like Pratap.

Question 1.
Identify different levels of management in above mentioned company.
Answer:
There are two levels of management i.e. middle and lower level of management in above mentioned company.

Question 2.
Find the level of Pranav in management of company.
Answer:
Pranav works in the middle level management of company.

Question 3.
Explain the functions and role of Pratap in his company.
Answer:
Pratap is an employee in a company. He works at the lower level. He executes the work assigned by Pravin.

4. Distinguish between the following

Question 1.
Top level Management and Middle level Management.
Answer:

	Top Level Management	Middle level Management
(1) Meaning	Top level management refers to top position in the organization such as Board of Directors, CEO, President, Managing Director, etc.	Middle level management refers to middle positions in the organization such as Departmental head, Managers like Finance manager, Production manager, Sales manager, Marketing manager, etc.
(2) Nature of Work	It is concerned with framing plans and policies of the entire organization.	It is concerned with implementation of policies framed by top level management with the help of lower level management.
(3) Skills	It requires conceptual or decision making skills than technical skills.	It requires combination of conceptual and decision making skills than technical skills.
(4) Promotion Policy	Policy    At this level promotion is given on merit.	At this level promotion is based on merit as well as seniority basis.
(5) Time Frame	It considers longer period of time i.e. 5 to 20 years.	It considers medium period of time i.e. 1 to 5 years.
(6) Responsibility	Top level is responsible to shareholders, government and society.	Middle level is responsible to top level management.
(7) Flow of Order	Orders and instructions are passed to middle level.	Orders and instructions are passed to lower level and report of the work is submitted to top level.

Question 2.
Middle Level Management and Lower Level Management.
Answer:

	Middle Level Management	Lower Level Management
(1) Meaning	Middle level management refers to middle positions in the organization. E.g. departmental head, finance manager, etc.	Lower level management refers to lower / last positions in the organization. E.g. superintendents, supervisors, etc.
(2) Nature of Work	It is concerned with implementation of policies framed by top level with the help of lower level.	It is considered with actual execution of planning.
(3) Skills	It requires both conceptual and technical skills.	It requires more technical skills than other skills.
(4) Promotion Policy	Promotion is based on merit as well as seniority basis.	Promotion is based on seniority basis.
(5) Time Frame	It considers medium term i.e. 1 year to 5 years.	It considers very short period of time i.e. upto one year.
(6) Responsibility	It is responsible to top level management.	It is responsible to both top and middle level management.
(7) Flow of Order	Instructions and orders are passed to Lower level and report of the work is submitted to top level	Implementation is done as per the orders of middle level and reports are submitted to middle level.

Question 3.
Top Level Management and Lower Level Management.
Answer:

	Top Level Management	Lower Level Management
(1) Meaning	Top level refers to top positions in the organization. e.g. Board of directors, Managing Directors, CEO, President, etc.	Lower level management refers to lower positions in the organization. e.g. Junior Executives, Foremen, etc.
(2) Nature of Work	It is concerned with framing plans and policies of the entire organization.	It is concerned with execution of plans and policies.
(3) Skills	It requires conceptual or decision making skills than technical skills.	It requires more technical skills than conceptual skills.
(4) Promotion Policy	Promotion is based on merit basis.	Seniority is preferred for promotion.
(5) Time Frame	It considers for longer period of time i.e. 5 to 20 years.	Basically it is very short period of time i.e. upto 1 year.
(6) Responsibility	It is responsible to Shareholders, Government and Society.	It is responsible to middle as well as top level management.
(7) Flow of Order	Orders and instructions are passed to middle level.	It implements orders of middle level and also reports to the middle level.

5. Answer in brief

Question 1.
State any four features of Management.
Answer:
Following are the features of management:
(i) Management is Goal Oriented : Management activity is done to achieve pre-decided goals. The success of organization is calculated on the basis of achievement of desired goals. The success depends on proper management of all available resources. Management decides the goals before conducting the activity. The manager takes efforts to achieve these goals. So management is goal oriented.

(ii) Management is a Group Activity : “Management is an art of getting things done through others.” The above definition explains that management is done through / with the people. It is done by co-ordinating all resources effectively and efficiently. The activities are done under the guidance of manager by its team members. Also in organisation decisions are taken by the group of people such as Board of Directors, Shareholders, etc.

(iii) Management is Intangible : Management is the collective efforts of manager. It is a process of planning, organizing, decision making, directing and controlling. These cannot be seen physically. However, the results of best management can be experienced in the form of increased profits, attainments of pre-decided so management is tangible in nature.

(iv) Management is a Continuous Process : Management is essential throughout the life of organisation. It is necessary to start the activity, to run the activity in smooth manner and also to control the activity. It is endless process. It is required for survival of the organization continuously.

Question 2.
Write any four functions of Top Level Management.
Answer:
Following are the functions of Top Level Management:

1. To decide long term objectives of the organisation.
2. To frame plans and policies to achieve the set objectives.
3. To observe that policies are properly implemented.
4. To create various department and positions.

Question 3.
Write any four functions of Middle Level Management.
Answer:
Following are the functions of Middle Level Management:

1. To link the top level management and lower level management.
2. To understand plans and policies framed by top level management and prepare plan of action according to the nature of department.
3. To assign duties and responsibilities to the staff of the concerned department to achieve pre-defined goals.
4. To appoint lower level staff.

Question 4.
State any four functions of Lower Level Management.
Answer:
Following are the functions of Lower Level Management:

1. Do work under Middle Level Management.
2. To assign work to subordinates.
3. To give instruction to subordinates.
4. To direct the subordinates wherever necessary.

Question 5.
State any two points of relationship about management as an art.
Answer:
The relationship between art and management are:
(i) Personal Skills : In organization, every manager has to handle different situations, has to solve various problems, has to handle human resource, etc. with his own style. This style varies from manager to manager and so the result. This is because of the different skills and abilities of each person.

(ii) Creativity : Creativity refers to the ability to find something new and innovative. Manager finds out new ways to conduct organizational activities and to achieve desired goals. Manager develops new ideas according to the changes in business situation and in the business environment.

Question 6.
Explain any two points of relationship about management as a science.
Answer:
The following points shows there is a relationship in management as a science:
(i) Systematic Body of Knowledge : Pure science is based on systematic experiments, certain rules and principles. Management principles and functions are also based on the experiments which can be studied through various different management theories, techniques and approaches given by different management thinkers.

(ii) Universal Applicability of Principles : Scientific principles can be applied in all over the world in similar situation. Therefore, the same result can be drawn. Management principles are universal in nature. They can be applied everywhere, anytime and in any condition.

Question 7.
State hny two points of relationship about management as a profession.
Answer:
The relationship between profession and management can be seen with the help of following points:
(i) Expertise : A professional is an expert in his field. He has acquired specific knowledge through formal education. Also he has particular skill in practicing as a professional. Manager also becomes an expert by practicing regularly in the specific field and can acquire certain knowledge and skill.

(ii) Registration : Professionals need to register themselves with their associations. For example, a Chartered Accountant should register himself with Institute of Chartered Accountant of India i.e. ICAI. Without registration certificate he cannot practice. In case of managers, there is no such body to register themselves. However, managers can take the membership of ‘Chambers of Commerce’ which gives them benefits.

6. Justify the following statements

Question 1.
Management is essential in professional and non-professional activities.
Answer:
(i) As management is universal in nature it is applicable everywhere, either it is a professional activity or non-professional activity.

(ii) Every person performs certain activity to achieve pre-decided goals. E.g. earning of profit, winning games, competitions, to increase goodwill, etc.

(iii) Organisations either large, medium or small, profit making or non-profit making, government or non-government organisation, etc. to get success, to attain goals every time management plays a vital role. It’s application ensures smooth functioning of business.

(iv) A professional acquires certain specific formal education. He has specific skill to achieve his goals. He performs management functions like planning, organizing, controlling, etc. Similarly, while performing non- professional activities like plantation of tree, blood donation campaign, a person also requires to perform certain management functions.

(v) Thus, management is essential in performing professional as well as non-professional activities.

Question 2.
Level of management depends upon the size of organization.
Answer:
(i) Making different parts for the smooth functioning of business and to achieve decided goals is called levels of management.

(ii) The size of business is small, medium or large. Every business needs to manage the activities efficiently and’ effectively.

(iii) The levels of management depends on size of organization. If the size of organization is small the classification of levels becomes easy and simple. Whereas in large organization, it becomes difficult and complicated. The levels of management in such organisation may be classified into three parts i.e. Top level, Middle level and Lower level.

(iv) These levels of management are important in organization. These levels have their own functions. As per that they have to perform their duties and responsibilities to attain their set goals and objectives.

(v) Thus, levels of management depends on size of organisation.

Question 3.
Management is a group oriented action.
Answer:
(i) In the words of Mary Parker Follet, “management is an art of getting things done through others”. Management focuses on completing the work through human resource.

(ii) It is co-ordinating of all available resources effectively and efficiently. In organisation, the activities are done by the subordinates, employees under the guidance of a manager.

(iii) A manager, though he is competent, able, has capacity, skills to perform the organizational task, it is not possible for a single manager to perform all the activities related to the said task alone.

(iv) To fulfill or accomplish the objectives and goals, he needs the help of other people, as it should be divided among all the employees, as per their skill, qualification, interest, knowledge and experience.

(v) Thus, management is a group oriented activity.

Question 4.
Management is an art.
Answer:
(i) An art is the bringing about a desired result through the application of skills. It is an innovativeness and creativeness of a person. This is a natural and inborn quality of a person. But with the help of proper education and practice one can developed the quality or qualities. E.g. acting, dancing, cycling, magician, singing, drawing, skating, musician, sports etc. comes under art.

(ii) Art does not require any specific educational qualification or it not any compulsion. It is inborn quality or a skill.

(iii) As a manager, in organisation he has to handle various kinds of situations, employees, other resources effectively and efficiently with his own style. So he uses his skills which are either inborn or developed by practice and formal education.

(iv) A manger must be innovative. He should handle the conditions. He should not use same methods all the time. Like an artist, he has to develop his skills by regular practice. He should use different techniques and skills to get perfect result. This is possible due to regular practice.

(v) Every managers has different abilities and skills. They have to use their own abilities and skills to guide the people to achieve their goals.

(vi) Thus, manager need to be creative and skilled like an artist. Hence, management is an art.

Question 5.
Answer:
There is no need of registration for managers.

1. Every professional is registered with the concerned body or organisation. A professional is formally, educated and also a skilled person.
2. A manager, now-a-days, possess a diploma or degree in management studies like Indian Institute of Management (IIM), PUMBA (Savitribai Phule Pune University, MBA) etc. A person who wants to be a manager acquires the required knowledge of the profession and trains himself / herself for the managerial job.
3. As management is a profession, a manager is not practicing independently. He is an employee of the organisation.
4. Thus, there is no need of registration for managers.

Question 6.
Management is a continuous process.
Answer:

1. Management is a continuous process. It is an important feature of management.
2. A manager of an organisation has to perform managerial work to achieve pre-defined goals of the organisation. To achieve these goals, he as a manager applies various functions of management, principles of management and also techniques of management.
3. From the beginning till the achievement of success, continuous efforts of every individual is required.
   Continuity is an integral part of management. It is necessary through out the life of an organisation.
4. Therefore, management has become necessary to start the activity and to run the activity at the business smoothly.
5. Thus, management is a continuous activity. It is not a one time process, but it is never ending process.

Question 7.
Planning is a task of top level management.
Answer:

1. Planning means thinking before doing. It is thinking in advance. Before the action, there is a thinking process.
2. In the large organisation, management levels are divided into three parts i.e. top, middle and lower.
3. At the top level management, very few personnels are involves. They are Board of Directors, Managing Directors, President, Chief Executive Officer, etc.
4. The main function of top level management is to frame the plans and policies of the whole organisation to achieve its goals.
5. It decides the long term objectives of the organisation and also frames particular plans and policies to get them. Further, it also sees that the policies are properly implemented.
6. Framing of organizational objectives and accordingly plans and policies to achieve them becomes a tough task of top level management.
7. Thus, Planning is a task of top level management.

7. Attempt the following

Question 1.
State the definition of management.
Answer:
Following are some definitions given by management thinkers :

1. “Management is an art of getting things done through others.” – Mary Parker Follet
2. “To manage is to forecast, to plan, to organize, to command, to co-ordinate and to control”. – Henry Fayol
3. “Management is knowing exactly what is to be done and seeing that it is done in the best possible manner.”
4. “Management is the process consisting of planning, organizing, actuating and controlling, performed to determine and accomplish the objectives by the use of people and resources”. George Terry.

Question 2.
State the features of management.
Answer:
(i) Managements is Dynamic : Management is dynamic and not static in character. It deals with human efforts, different situations and different activities. It adopts changes in the form of technological, socio economical, political, environmental, etc.

(ii) Management is Intangible : Management is intangible in nature. It can’t be seen and touched. It is abstract. The principles, theories, techniques, concept are practically invisible. The result of best management can be experienced in the form of increased profit, smooth functioning of the organization and achievement.

(iii) Management is a Social Process : Management is a process because it sets objectives and takes steps to achieve them. It is said to be a social process because management is concerned with people. For example employees, shareholders, investors, customers, community, etc.

(iv) Management is a Continuous Process : Management is essential through the life of an organization. It is necessary to begin the activity, to run the activity smoothly and to control the activity. This is never ending process.

(v) Management is Different from Ownership : In an organization, management is different from ownership. Owners contribute capital which is taken care of by efficient and paid managers.

(vi) Management is an Inborn Quality : Management in the older days, was considered to be an inborn quality. But with the inception of various business schools, management is a quality which can be developed if a person has the interest and perseverance.

Question 3.
State management is as a science.
Answer:
1. Systematic Body of Knowledge : Management is not a science like physics, chemistry or biology. It is a science which deals with human beings. Management has a theories and concepts. Various principle of management like unity of command, division of labour, unity of direction are all universally accepted principle. Management follows procedures like collecting data, analyzing and drawing inferences from it. This makes management a science.

2. Use of Scientific Methods of Observation : In science the conclusions are drawn only on the basis of scientific methods of observation. Management also uses systematic methods of data collection, verification and analysis of that collected data and interpretation of data, then a manager takes decisions in the organisation.

3. Cause and Effect Relationship : Science is based on cause and effect relationship. After using specific formulae or their combination, we get particular result in science. Management theories are also based on relationships. Fair and regular salary along with good working conditions motivates people to work hard and to give best results. Whereas, old machineries, techniques, untrained employees may not give good / excepted output to the organization.

4. Universal Applicability of Principles : The management principles are universally accepted just like scientific principles are accepted all over the world. For e.g. Newton’s law, Archimedes principle, etc. are all accepted principles, similarly in management also the principle of authority and responsibility, unity of command, direction, etc. are all universally accepted and are used in all organization whether big or small. Management follows these principles in all kind of organizations, at all levels in the same organization. State the relationship between the management and art.

Question 4.
Following points clearly states the relation between management and art.
Answer:
1. Creativity: Managers must posses the quality of being creative. Every manager should be able to bring innovations in the business. This quality is developed with experience. The more experienced manager can bring about changes according to the situation.

2. Regular Practice: Practice makes a man perfect. This is the rule applied by a manager. The more the experience, the manager will be able to meet any challenge or difficult situation with ease. He is in a position to make the required changes according to the circumstances.

3. Personal Skill: Every manager being an individual is different. A good manager must posses good skills in dealing with human beings. He should use his skill in knowing the problem, allotment of work to employees, give judgment and understanding to deal with various situations in business.

4. Personal Abilities : Every person has different abilities and skills. Manager uses his own abilities to co-ordinate the activities, to guide the people and to achieve their goals.

Question 5.
State the importance of management
Answer:
1. Utilisation of Resources Optimally : Effective management is all about optimum use of available resources in the organization. Proper distribution of resources and work to right person results in higher output. It also reduces wastage and cost. Maximisation of output by reducing cost is the ultimate objective of every organization.

2. Better Relation : Different groups work at different levels in organization. The employer and the employees have different sets of demands from each other. Effective management ensures that employees fulfill their social responsibilities towards their employers. This helps in establishing good relationship among different groups in an organization.

3. Achievement of Goals : In an organization group of people performs various activities to achieve common goals of the organization. Effective management brings about harmony (cordial relationship) and good co-ordination in efforts of all group. A manager creates a feeling of team spirit among the members of the group.

4. Division of Work : Division of work means the entire work is divided into smaller lots. Each job is given to a particular employee as per his ability and skill to carry out the required activity. Division of work leads to specialization of work and completion of the work in the stipulated period of time.

Question 6.
State the similarities between management and profession.
Answer:
1. Formal Education : Professionals need formal education. Today’s managers are qualified with management diploma or degrees from various institutes like IIMS, PUMBA etc. They also take formal training from management schools which helps them to work professionally.

2. Code of Conduct : Code of conduct means set of rules and regulations to be followed by practicing in their field. Every profession has their own code of conduct to control the activities. It emphasis on ethical practice in that profession. E.g. Chartered Accountants have to work according to ICAI, Lawyers have to follow the code of conduct framed by Bar Council of India, etc. However for managers there is no such association of managers. Generally managers follow the code of conduct based on customs and traditions.

3. Expertise : A professional is an expert in his field. He has acquired specific knowledge through formal education. He has particular skill of practicing as professional. He becomes an expert due to regular practice in specific field.

4. Registration : Professionals need to register themselves with their associations. For example, Lawyers must get registered themselves with India Bar Council, Chartered Accountants with Institute of Chartered Accountant of India, where as for managers there is no need of such registration with any body. They can take membership of ‘Chambers of Commerce’ which give them membership benefits.

5. Restricted Entry : The entry in any profession is restricted. It needs formal education e.g. Lawyers, Architect, Interior designer, Doctors, Chartered Accountant, etc. However, such compulsion is not there for working as a manager.

8. Answer the Following

Question 1.
What is management? State the characteristics or features of management.
Answer:
Management and its features:
(A) Meaning:
The word management is derived from the Latin word ‘manus’ which means ‘hands.
It means management is handling some activity.
Many thinkers said that management means a fine consideration of co-ordinating and controlling the work either from direct individual or from a group.

(B) Definitions:
Some definitions of management given by different management thinkers as follows:

1. “Management traditionally can be defined as “an art of getting things done through others. ” – Mary Parker Follet
2. “To manage is to forecast, to plan, to organize, to command, to co-ordinate and to control”. – Henry Fayol
3. “Management is co-ordination of resources through the process of planning, organizing, directing and controlling in order to attain stated objectives.” – F.W. Taylor

Following are the characteristics features of management:
(i) Managements is Dynamic : Management is dynamic and not static in character. It deals with human efforts, different situations and different activities. It adopts changes in the form of technological, socio economical, political, environmental, etc.

(ii) Management is Intangible : Management is intangible in nature. It can’t be seen and touched. It is abstract. The principles, theories, techniques, concept are practically invisible. The result of best management can be experienced in the form of increased profit, smooth functioning of the organization and achievement.

(iii) Management is a Social Process : Management is a process because it sets objectives and takes steps to achieve them. It is said to be a social process because management is concerned with people. For example employees, shareholders, investors, customers, community, etc.

(iv) Management is a Continuous Process : Management is essential through the life of an organization. It is necessary to begin the activity, to run the activity smoothly and to control the activity. This is never ending process.

(v) Management is Different from Ownership : In an organization, management is different from ownership. Owners contribute capital which is taken care of by efficient and paid managers.

(vi) Management is an Inborn Quality : Management in the older days, was considered to be an inborn quality. But with the inception of various business schools, management is a quality which can be developed if a person has the interest and perseverance.

(vii) Management is Situational: Management helps to take decisions and also to handle specific situation. The decisions are based on the principles, theories and techniques. They are applicable in certain situation only.

(viii) Management is Goal Orientated : Every business organization either small, medium or large, profit making or non-profit making, Governmental or NGOs have their own goals. When the organization achieves its desired goals, it is called success of that organisation. This success depends on proper management of all available resources. A manager takes lots of efforts to achieve these goals. So management is goal oriented.

(ix) Management is Universal : Management principles and theories are equally applicable every where and in every field like business, profession, hospital, education, politics, government administration social activities. These principles are the guidelines to the manager. They are flexible and capable of adaptation to every type of organization.

(x) Management is Group Activity : Management is defined as “getting work done through others” or “with the people in formally organized groups”. The tasks set by the organization is complied by employees, all levels of managers and even with the help of outside parties.

Question 2.
Discuss whether management is an art science or profession.
Answer:
Art:
1. Creativity: Managers must posses the quality of being creative. Every manager should be able to bring innovations in the business. This quality is developed with experience. The more experienced manager can bring about changes according to the situation.

2. Regular Practice: Practice makes a man perfect. This is the rule applied by a manager. The more the experience, the manager will be able to meet any challenge or difficult situation with ease. He is in a position to make the required changes according to the circumstances.

3. Personal Skill: Every manager being an individual is different. A good manager must posses good skills in dealing with human beings. He should use his skill in knowing the problem, allotment of work to employees, give judgment and understanding to deal with various situations in business.

4. Personal Abilities : Every person has different abilities and skills. Manager uses his own abilities to co-ordinate the activities, to guide the people and to achieve their goals.

Science:
1. Systematic Body of Knowledge : Management is not a science like physics, chemistry or biology. It is a science which deals with human beings. Management has a theories and concepts. Various principle of management like unity of command, division of labour, unity of direction are all universally accepted principle. Management follows procedures like collecting data, analyzing and drawing inferences from it. This makes management a science.

2. Use of Scientific Methods of Observation : In science the conclusions are drawn only on the basis of scientific methods of observation. Management also uses systematic methods of data collection, verification and analysis of that collected data and interpretation of data, then a manager takes decisions in the organisation.

3. Cause and Effect Relationship : Science is based on cause and effect relationship. After using specific formulae or their combination, we get particular result in science. Management theories are also based on relationships. Fair and regular salary along with good working conditions motivates people to work hard and to give best results. Whereas, old machineries, techniques, untrained employees may not give good / excepted output to the organization.

Profession:
1. Formal Education : Professionals need formal education. Today’s managers are qualified with management diploma or degrees from various institutes like IIMS, PUMBA etc. They also take formal training from management schools which helps them to work professionally.

2. Code of Conduct : Code of conduct means set of rules and regulations to be followed by practicing in their field. Every profession has their own code of conduct to control the activities. It emphasis on ethical practice in that profession. E.g. Chartered Accountants have to work according to ICAI, Lawyers have to follow the code of conduct framed by Bar Council of India, etc. However for managers there is no such association of managers. Generally managers follow the code of conduct based on customs and traditions.

3. Expertise : A professional is an expert in his field. He has acquired specific knowledge through formal education. He has particular skill of practicing as professional. He becomes an expert due to regular practice in specific field.

4. Registration : Professionals need to register themselves with their associations. For example, Lawyers must get registered themselves with India Bar Council, Chartered Accountants with Institute of Chartered Accountant of India, where as for managers there is no need of such registration with any body. They can take membership of ‘Chambers of Commerce’ which give them membership benefits.

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 16 Skeleton and Movement Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Biology Solutions Chapter 16 Skeleton and Movement

1. Choose the correct option

Question (A).
The functional unit of striated muscle is …………..
a. cross bridges
b. myofibril
c. sarcomere
d. z-band
Answer:
c. sarcomere

Question (B).
A person slips from the staircase and breaks his ankle bone. Which bones are involved?
a. Carpals
b. Tarsal
c. Metacarpals
d. Metatarsals
Answer:
b. Tarsal

Question (C).
Muscle fatigue is due to accumulation of ……..
a. pyruvic acid
b. lactic acid
c. malic acid
d. succinic acid
Answer:
b. lactic acid

Question (D).
Which one of the following is NOT antagonistic muscle pair?
a. Flexo-extensor
b. Adductor-abductor
c. Levator-depressor
d. Sphinetro-suprinater
Answer:
d. Sphinetro-suprinater

Question (E).
Swelling of sprained foot is reduced by soaking in hot water containing a large amount of common salt,
a. due to osmosis
b. due to plasmolysis
c. due to electrolysis
d. due to photolysis
Answer:
a. due to osmosis

Question (F).
Role of calcium in muscle contraction is ……….
a. to break the cross bridges as a cofactor in the hydrolysis of ATP
b. to bind with troponin, changing its shape so that the actin filament is exposed
c. to transmit the action potential across the neuromuscular junction.
d. to re-establish the polarisation of the plasma membrane following an action potential
Answer:
b. to bind with troponin, changing its shape so that the actin filament is exposed

Question (G).
Hyper-secretion of parathormone can cause which of the following disorders?
a. Gout
b. Rheumatoid arthritis
c. Osteoporosis
d. Gull’s disease
Answer:
c. Osteoporosis

Question (H).
Select correct option between two nasal bones

Answer:
(c)

Question 2.
Answer the following questions

Question (A).
What kind of contraction occurs in your neck muscles while you are reading your class assignment?
Answer:

1. Isometric contractions occur in the neck muscles while reading class assignment.
2. These contractions are important for supporting objects in a fixed position.

Question (B).
Observe the diagram and enlist importance of ‘A’, ‘B’ and ‘C’.

Answer:

1. A – Posterior portion of vertebral foramen of atlas vertebrae; Importance – The spinal cord runs through this portion of vertebral foramen
2. B – Anterior portion of vertebral foramen of axis vertebrae; Importance – In this portion, the odontoid process of axis vertebrae forms ‘NO’ joint.
3. C – Inferior articular facet; Importance – It articulates with superior articular facet of axis and permits rotatory movement of head.

Question (C).
Raju intends to train biceps; while exercising using dumbbells, which joints should remain stationary and which should move?
Answer:
While performing exercise of biceps using dumbbells, the joint which should remain stationary are wrist joint or radiocarpal joint, ball and socket joint of shoulder. The only joint which should move is hinge joint of elbow.

Question (D).
In a road accident, Moses fractured his leg. One of the passers by, tied a wodden plank to the fractured leg while Moses
was rushed to the hospital Was this essential? Why?
Answer:

1. Fracture is a significant and traumatic injury which requires medical attention however, getting timely first aid is important.
2. If any bone is fractured, it is essential that the fractured part be immobilized to prevent further injury. It can be done with the help of any available wooden plank or batons or rulers. Thus, a wooden plank was tied to Moses’s fractured leg as a first aid for fracture.
3. A fractured bone is immobilized to prevent the sharp edges of the fractured bone from moving and cutting tissue, muscle, blood vessels and nerves. Immobilization can also help reduce pain or control shock.

Question (E).
Sprain is more painful than fracture. Why?
Answer:

1. A sprain is an injury that involves the ligaments (tissues that connect bones at joints), whereas a fracture is an injury that involves bones.
2. Sprains can be of three degree: 1^st degree: Mild with micro-tears, 2^nd degree: Partial with visible tear in ligament, 3^rd degree: Completely torn ligament.
3. If a sprain is 3^rd degree, it will be more painful than a fracture. It usually requires a surgery to fix this injury, while breaking a bone, most of the time does not require surgery.
4. Breaks or Fractures also vary greatly. Minor fractures (like stress/ hairline fractures) are much less painful than compound/ complex fractures in which the bone may be cracked into half.
5. Blood supply is essential for growth and regeneration. Bones are highly vascularized whereas, ligaments are not. This causes the bones to heal comparatively faster than severe sprains. Thus, the duration of enduring pain until the injury heals also differs.
6. Also, ligaments have a rich supply of sensory nerves, which may also be responsible for an elevated sense of pain during severe sprains.

[Note: 1^st and 2^nd degree sprains are not very serious and may be lesser painful than a fracture. Depending on the severity of the injury, intensity of pain will vary.]

Question (F).
Why a red muscle can work for a prolonged period whereas white muscle fibre suffers from fatigue after a shorter work? (Refer to chapter animal tissues.)
Answer:

1. Red muscle fibres contain large amount of myoglobin and mitochondria (site of aerobic respiration), whereas white muscles fibres contain lesser amount of myoglobin and mitochondria.
2. Myoglobin is an iron-containing pigment that carries oxygen molecules to muscle tissues. Abundance of these pigments in red muscle fibres supports higher rate of aerobic respiration, whereas white muscle fibres have less mitochondria and depend upon anaerobic respiration.
3. Anaerobic respiration in muscle white fibres leads to the production of lactic acid and accumulation of higher of levels lactic acid can result in fatigue in white muscle fibres.

Thus, red muscle fibres can perform prolonged work and show less fatigue due to accumulation of negligible amount loss or of lactic acid, whereas white muscle fibres suffer from fatigue after a shorter work due to accumulation of higher amount of lactic acid.

3. Answer the following questions in detail

Question (A).
How is the structure of sarcomere suitable for the contractility of the muscle? Explain its function according to sliding
filament theory. (Refer to chapter animal tissues.)
Answer:
i. Sarcomere is the functional unit of myofibril. It has specific arrangement of actin and myosin filaments. The components of sarcomere are organized into variety of bands and zones. Actin and myosin are referred as contractile proteins. Actin is called as thin filament whereas myosin in called as thick filament. The structure of sarcomere:

ii. ‘A’ band – dark bands present at the centre of sarcomere and contain myosin as well as actin.
‘H’ zone or Hensen’s zone – light area present at the centre of ‘A’ band
‘M’ line – present at the centre of ‘H’ zone
‘I’ band – light bands present on the either side of ‘A’ band containing only actin
Z’ line – adjacent ‘I’ bands are separated by ‘Z’ line.

iii. Sliding filament theory: It was put forth by H.E Huxley and A.F Huxley. It is also known as ‘Walk along theory’ or Ratchet theory.

• According to the sliding filament theory, the interaction between actin and myosin filaments is the basic cause of muscle contraction. The actin filaments are interdigitated with myosin filaments.
• The head of the myosin is joined to the actin backbone by a cross bridge forming a hinge joint. From this joint, myosin head cannot tilt forward or backward. This movement is an active process as it utilizes ATP.
• Myosin head contains ATPase activity. It can derive energy by the breakdown of ATP molecule. This energy can be used for the movement of myosin head.
• During contraction, the myosin head gets attached to the active site of actin filaments and pull them inwardly so that the actin filaments slide over the myosin filaments. This results in the contraction of muscle fibre.
  

Question (B).
Ragini, a 50 year old office goer, suffered hair-line cracks in her right and left foot in short intervals of time. She was worried about minor jerks leading to hair line cracks in bones. Doctor explained to her why it must be happening and prescribed medicines.

What must be the cause of Ragini’s problem? Why has it occurred? What precautions she should have taken earlier? What care she should take in future?
Answer:

1. Considering Ragini’s age, she may be undergoing menopause. After menopause, oestrogen level declines resulting in lower bone density.
2. Osteoporosis:
   • In this disorder, bones become porous and hence brittle. It is primarily age related disease and is more common in women than men.
   • Osteoporosis may be caused due to decreasing estrogen secretion after menopause, deficiency of vitamin D, low calcium diet, decreased secretion of sex hormones and thyrocalcitonin.
3. As age advances, bone resorption outpaces bone formation. Hence, the bones lose mass and become brittle. More calcium is lost in urine, sweat, etc., than it is gained through diet. Thus, prevention of disease is better than treatment by consuming adequate amount of calcium and exercise at young age.
4. A person with previous hairline fractures is more susceptible to reoccurrence of fractures. Hence, Ragini needs to take her medications and supplements properly, avoid jerky movements and maintain body weight.

Question (C).
How does structure of actin and myosin help muscle contraction?
Answer:
i. Myosin filament:

1. Each myosin filament is a polymerized protein.
   Many meromyosins (monomeric proteins) constitute one thick filament.
2. Myosin molecule consists of two heavy chains (heavy meromyosin / HMM) coiled around each other forming a double helix. One end of each of these chains is projected outwardly is known as cross bridge. This end folds to form a globular protein mass called myosin head.
3. Two light chains are associated with each head forming 4 light chains/light meromyosin / LMM.
4. Myosin head has a special ATPase activity. It can split ATP to produce energy.
5. Myosin contributes 55% of muscle proteins.
6. In sarcomere, myosin tails are arranged to point towards the centre of the sarcomere and the heads point to the sides of the myofilament band.
   

ii. Actin filament: It is a complex type of contractile protein. It is made up of three components:

1. F actin: It forms the backbone of actin filament. F actin is made up of two helical strands. Each strand is composed of polymerized G actin molecules. One ADP molecule is attached to G actin molecule.
2. Tropomyosin: The actin filament contains two additional protein strands that are polymers of tropomyosin molecules. Each strand is loosely attached to an F actin. In the resting stage, tropomyosin physically covers the active myosin-binding site of the actin strand.
3. Troponin: It is a complex of three globular proteins, is attached approx. 2/3^rd distance along each tropomyosin molecule. It has affinity for actin, tropomyosin and calcium ions. The troponin complex is believed to attach the tropomyosin to the actin. The strong affinity of troponin for calcium ions is believed to initiate the contraction process.
   

Question (D).
Justify the structure of atlas and axis vertebrae with respect to their position and function.
Answer:
i. Atlas vertebrae:

1. Atlas is the ring-like, 1^st cervical vertebrae. It has anterior, posterior arches and large lateral masses.
2. It lacks centrum and spinous process. The superior surfaces of the lateral masses are concave and are known as superior articular facets.
3. These facets articulate with the occipital condyles of the occipital bone thereby forming atlanto-occipital joints. This articulation permits ‘YES movement’ or nodding movement.
4. The inferior surfaces of the lateral masses known as inferior articular facets articulate with axis vertebrae,

ii. Axis vertebrae:

1. It is the 2^nd cervical vertebrae.
2. A peg-like process called odontoid process projects superiorly through the anterior portion of the vertebral foramen of the atlas.
3. The odontoid process forms a pivot on which the atlas and head rotate. This arrangement allows ‘NO movement’ or side to side movement of the head.
4. The articulation formed between the anterior arch of atlas, the odontoid process of the axis and between their articular facets is called as atlanto-axial joint.

Question (E).
Observe the blood report given below and diagnose the possible disorder.

Answer:
On observing Report D, it is clear that the level of uric acid is more than normal, thus the patient must be suffering from gouty arthritis.

Also, the elevated blood urea nitrogen (BUN) indicates dysfunctional liver and/ or kidneys. It generally occurs due to decrease in GFR, caused by renal disease or obstruction of urinary tract.

Question 4.
Write short notes on following points

Question (A).
Actin filament
Answer:
Actin filament: It is a complex type of contractile protein. It is made up of three components:

1. F actin: It forms the backbone of actin filament. F actin is made up of two helical strands. Each strand is composed of polymerized G actin molecules. One ADP molecule is attached to G actin molecule.
2. Tropomyosin: The actin filament contains two additional protein strands that are polymers of tropomyosin molecules. Each strand is loosely attached to an F actin. In the resting stage, tropomyosin physically covers the active myosin-binding site of the actin strand.
3. Troponin: It is a complex of three globular proteins, is attached approx. 2/3rd distance along each tropomyosin molecule. It has affinity for actin, tropomyosin and calcium ions. The troponin complex is believed to attach the tropomyosin to the actin. The strong affinity of troponin for calcium ions is believed to initiate the contraction process.
   

Question (B).
Myosin filament
Answer:
i. Myosin filament:

1. Each myosin filament is a polymerized protein.
   Many meromyosins (monomeric proteins) constitute one thick filament.
2. Myosin molecule consists of two heavy chains (heavy meromyosin / HMM) coiled around each other forming a double helix. One end of each of these chains is projected outwardly is known as cross bridge. This end folds to form a globular protein mass called myosin head.
3. Two light chains are associated with each head forming 4 light chains/light meromyosin / LMM.
4. Myosin head has a special ATPase activity. It can split ATP to produce energy.
5. Myosin contributes 55% of muscle proteins.
6. In sarcomere, myosin tails are arranged to point towards the centre of the sarcomere and the heads point to the sides of the myofilament band.
   

Question (C).
Role of calcium ions in contraction and relaxation of muscles.
Answer:
Calcium ions play a major role in contraction and relaxation of muscles.

1. Calcium ions are released from the sarcoplasm during muscle contraction and stored in sarcoplasmic reticulum during muscle relaxation.
2. When a skeletal muscle is excited and an action potential travels along the T tubule, the concentration of calcium ions increases.
3. These calcium ions bind to troponin which in turn undergoes a conformational change that causes tropomyosin to move away from the myosin-binding sites on actin. Once these binding sites are free, myosin heads bind to them to form cross-bridges and the muscle fiber contracts.
4. The decrease in calcium ion concentration in the sarcoplasmic reticulum causes tropomyosin to slide back and block the myosin binding sites on actin. This causes the muscle to relax.

Question 5.
Draw labelled diagrams

Question (A).
Synovial joint.
Answer:
i. Synovial joints / freely movable joints / diarthroses:

1. It is characterized by presence of a space called synovial cavity between articulating bones that renders free movement at the joint.
2. The articulating surfaces of bones at a synovial joint are covered by a layer of hyaline cartilage. It reduces friction during movement and helps to absorb shock.
3. Synovial cavity is lined by synovial membrane that forms synovial capsule. Synovial membrane secretes synovial fluid.
4. Synovial fluid is a clear, viscous, straw coloured fluid similar to lymph. It is viscous due to hyaluronic acid. The synovial fluid also contains nutrients, mucous and phagocytic cells to remove microbes.
   Synovial fluid lubricates the joint, absorbs shocks, nourishes the hyaline cartilage and removes waste materials from hyaline cartilage cells (as cartilage is avascular). Phagocytic cells destroy microbes and cellular debris formed by wear and tear of the joint.
5. If the joint is immobile for a while, the synovial fluid becomes viscous and as joint movement starts, it becomes less viscous.
6. The joint is provided with capsular ligament and numerous accessory ligaments. The fibrous capsule is attached to periosteum of articulating bones. The ligament helps in avoiding dislocation of joint.
7. The types of synovial joints are on follows:

1. Pivot joint: In this type of joint, the rounded or pointed surface of one bone articulates with a ring formed partly by another bone and partly by the ligament. Rotation only around its own longitudinal axis is possible. e.g. in joint between atlas and axis vertebrae, head turns side ways to form ‘NO’ joint.

2. Ball and socket joint: The ball like surface of one bone fits into cup like depression of another bone forming a movable joint. Multi-axial movements are possible. This type of joint allows movements along all three axes and in all directions. e.g. Shoulder and hip joint.

3. Hinge joint: In a hinge joint, convex surface of one bone fits into concave surface of another bone. In most hinge joints one bone remains stationary and other moves. The angular opening and closing motion (like hinge) is possible. In this joint only mono-axial movement takes place like flexion and extension. e.g. Elbow and knee joint.

4. Condyloid joint: It is an ellipsoid joint. The convex oval shaped projection of one bone fits into oval shaped depression in another bone. it is a biaxial joint because it permits movement along two axes viz, flexion, extension, abduction, adduction and circumduction is possible. e.g. Metacarpophalangeal joint.

5. Gliding joint: It is a planar joint, where the articulating surfaces of bones are flat or slightly curved. These joints are non-axial because the motion they allow does not occur along an axis or a plane. e.g. Intercarpal and intertarsal joints.

6. Saddle joint: This joint is a characteristic of Homo sapiens. Here the articular surface of one bone is saddle-shaped and that of other bone fits into saddle (each bone forming this joint have both concave and convex areas). It is a modified condyloid joint in which movement is somewhat more free. It is a biaxial joint that allows flexion, extension, abduction, adduction and circumduction.
e.g. Carpometacarpellar joint between carpal (trapezium) and metacarpal of thumb.

Question (B).
Different cartilagenous joints.
Answer:
Cartilaginous / slightly movable joints / amphiarthroses:
These joints are neither fixed nor freely movable. Articulating bones are held together by hyaline or fibrocartilages. They are further classified as

a. Synchondroses: The two bones are held together by hyaline cartilage. They are meant for growth. On completion of growth, the joint gets ossified, e.g. Epiphyseal plate found between epiphysis and diaphysis of a long bone, Rib – Sternum junction.

b. Symphysis: In this type of joint, broad flat disc of fibrocartilage connects two bones. It occurs in mid-line of the body. e.g. Intervertebral discs, manubrium and sternum, pubic symphysis.

Practical / Project :

Identify the following diagrams and demonstrate the concepts in classroom.

Answer:
The diagrams A, B and C represent Class I, Class II and Class III lever respectively.
For description:

1. Class I lever: The joint between the first vertebra and occipital condyle of skull is an example of Class I lever. The force is directed towards the joints (fulcrum); contraction of back muscle provides force while the part of head that is raised acts as
   
2. Class II lever: Human body raised on toes is an example of Class II lever. Toe acts as fulcrum, contracting calf muscles provide the force while raised body acts as resistance.
   
3. Class III lever: Flexion of forearm at elbow exhibit lever of class III. Elbow joint acts as fulcrum and radius, ulna provides resistance. Contracting bicep muscles provides force for the movement.
   

[Students are expected to perform the given activity on their own]

12th Biology Digest Chapter 16 Skeleton and Movement Intext Questions and Answers

Movements And Locomotion (Textbook Page No. 193)

Question 1.
Streaming of protoplasm, peristalsis, walking, running, etc. Which of the above-mentioned movements are internal? Which are external? Can you add few more examples?
Answer:

1. Streaming of protoplasm, peristalsis are internal movements. Walking and running are external movements.
2. Examples of internal movement: Contraction and relaxation heart, inspiration and expiration, contraction of blood vessels, etc.
3. Examples of external movement: Swimming; movement tongue, jaws, snout, tentacles, movement of ear pinna, etc.

Can you recall? (Textbook Page No. 193)

Question 1.
Which are different types of muscular tissues?
Answer:

1. Smooth / non-striated / visceral / involuntary muscles
2. Cardiac muscles
3. Skeletal / straited / voluntary muscles.

Question 2.
Name the type of muscles which bring about running and speaking.
Answer:
Skeletal muscles (Voluntary muscles)

Question 3.
Name the muscles which do not contract as per our will.
Answer:
Involuntary muscles (smooth muscles and cardiac muscles)

Question 4.
Which type of muscles show rhythmic contractions?
Answer:
Cardiac muscles

Question 5.
Which type of muscle is present in the diaphragm of the respiratory system?
Answer:
Skeletal muscle

Question 6.
State the functions of:

1. Smooth muscles
2. Cardiac muscles
3. Striated muscles

Answer:

1. Smooth muscles: They bring about involuntary movements like peristalsis in the alimentary canal, constriction and dilation of blood vessels.
2. Cardiac muscles: They bring about contraction and relaxation of the heart.
3. Striated muscles: They control voluntary movements of limbs, head, trunk, eyes, etc.

Can you recall? (Textbook Page No. 193)

Question 1.
Name the part of human skeleton situated along the vertical axis.
Answer:
Axial skeleton

Question 2.
Give an account of bones of human skull.
Answer:
Skull is made up of 22 bones. It is located at the superior end of vertebral column. The bones of skull are
joined by fixed or immovable joints except for jaw.

Skull consists of cranium or brain box and facial bones.

i. Cranium: It is made up of four median bones and two paired bones.

1. Frontal bone: It is median bone (unpaired) forming forehead, roof of orbit (eye socket) and the most anterior part of cranium. It is connected to two parietals, sphenoid and ethmoid bone.
2. Parietal bones: These paired bones form the roof of cranium and greater portion of sides of the cranium.
3. Temporal bones: These paired bones are situated laterally just above the ear on either side. Each temporal bone gives out zygomatic process that joins zygomatic bone to form zygomatic arch. Just at the base of zygomatic process is mandibular fossa, a depression for mandibles (lower jaw bone) that forms the only movable joint of the skull. This bone harbors the ear canal that directs sound waves into the ear. The processes of temporal bones provide points for attachment for various muscles of neck and tongue.
4. Occipital bone: It is a single bone present at the back of the head. It forms the posterior part and most of the base of cranium. The inferior part of this bone shows foramen magnum, the opening through which medulla oblongata connects with spinal cord. On the either sides of foramen magnum are two prominent protuberances called occipital condyles. These fit into the corresponding depressions present in 1^st vertebra.
5. Sphenoid bone: Median bone present at the base of the skull that articulates with all other cranial bones and holds them together. This butterfly shaped bone has a saddle shaped region called sella turcica. In this hypophyseal fossa, the pituitary gland is lodged.
6. Ethmoid bone: This median bone is spongy in appearance. It is located anterior to sphenoid and posterior to nasal bones. It contributes to formation of nasal septum and is major supporting structure of nasal cavity.

ii. Facial Bones: Fourteen facial bones give a characteristic shape to the face. The growth of face stops of the
age of 16.
Following bones comprise the facial bones:

1. Nasals: These are paired bones that form the bridge of nose.
2. Maxillae: These form the upper jaw bones. They are paired bones that join with all facial bones except mandible. Upper row of teeth are lodged maxillae.
3. Palatines: These are paired bones forming the roof of buccal cavity or floor of the nasal cavity.
   
4. Zygomatic bones: They are commonly called as cheek bones.
5. Lacrimal bones: These are the smallest amongst the facial bones.
   These bones form the medial wall of each orbit. They have lacrimal fossa that houses lacrimal sacs. These sacs gather tears and send them to nasal cavity.
6. Inferior nasal conchae: They form the part of lateral wall of nasal cavity. They help to swirl and filter air before it passes to lungs.
7. Vomer: The median, roughly triangular bone that forms the inferior portion of nasal septum.
8. Mandible: This median bone forms the lower jaw. It is the largest and strongest facial bone. It is the only movable bone of skull. It has curved horizontal body and two perpendicular branches i.e. rami. These help in attachment of muscles. It has lower row of teeth lodged in it.

Think about it. (Textbook Page No. 193)

Question 1.
Did you ever feel tickling in muscles?
Answer:
Yes, the tickling sensation in muscles can be felt and sometimes it is also accompanied by itching sensation.

Question 2.
What is locomotion?
Answer:
The change in locus of whole body of living organism from one place to another place is called locomotion.

Question 3.
State the four basic types of locomotory movements seen in animals.
Answer:
The four basic types locomotory movements seen in animals are:

1. Amoeboid movement: It is performed by pseudopodia, e.g. leucocytes.
2. Ciliary movement: It is performed by cilia, e.g. ciliated epithelium. In Paramoecium, cilia help in passage of food through cytopharynx.
3. Whirling movement: It is performed by flagella, e.g. sperms.
4. Muscular movement: It is performed by muscles, with the help of bones and joints.

Question 4.

Question 1.
Why do muscles show spasm after rigorous contraction?
Answer:

1. Rigorous contraction of muscles occurs during strenuous activities (swimming. running, cycling, aerobics. etc.)
2. Muscle contraction requires energy. Glucose in muscle cells breakdown during anerobic respiration resulting in accumulation of lactic acid.
3. This lactic acid buildup triggers muscle spasm around muscle cells.

Question 2.
Why do we shiver during winter?
Answer:

1. Humans are homeotherms as the can regulate their body temperature with respect to the surrounding temperature. During winter, when temperature falls, the thermoreceptors detect the change in temperature and send signals to the brain.
2. Shivering reflex i.e. rapid contraction of muscles is triggered by the brain to generate heat and raise the body temperature.

Can you tell? (Textbook Page No. 194)

Question 1.
Why are movement and locomotion necessary among animals?
Answer:

• Movement is one of the important characteristics of all the living organisms. Animals exhibit wide range of
  movements like rhythmic beating of heart, movement of diaphragm during respiration, ingestion of food,
  movement of eyeballs, etc.
• Locomotion results into change in place or location of an organism. Animals locomote in search of food, mate, shelter, breeding ground. while escaping from the enemy, etc.
• Thus, locomotion and movement are necessary to support the living of animals.

Question 2.
All locomotions are movements but all movements are not locomotion. Justify
Answer:
Locomotion occurs when body changes its position, however all movements may not result in locomotion. Thus, all locomotions are movements but all movements are not locomotion.

Question 3.
Kriti was diagnosed with knee tendon injury. She asked the doctor whether she will be able to walk due to the injury? If not then state the reason.
Answer:
Knee tendon injury affects the ability to walk. Kriti may not be able to walk freely as the tendons attached to the bones help in the movement of the parts of skeleton.

Question 4.
What are antagonistic muscles? Explain with example.
Answer:

1. The muscles that work in pairs and produce opposite action are known as antagonistic muscles, e.g. biceps and triceps of upper arm.
2. The biceps (flexors) bring flexion (folding) and triceps (extensors) bring extension of the elbow joint.
3. One member from a pair is capable of bending the joint by pulling of bones the other member is capable of straightening the same joint by pulling.
4. In antagonistic pair of muscles, one member is stronger than the other, e.g. The biceps are stronger than the triceps.

Question 5.
Describe the antagonistic muscles in detail.
Answer:
Following are the important antagonistic muscles:

1. Flexor and extensor: Flexor muscle on contraction results into bending or flexion of joint, e.g. Biceps. Extensor muscle on contraction results in straightening or extension of a joint, e.g. Triceps.
2. Abductor and adductor: Abductor muscle moves a body part away from the body axis. e.g. Deltoid muscle of shoulder moves the arm away from the body. Adductor muscle moves a body part towards the body axis, e.g. Latissimus dorsi of shoulder moves the arm near the body.
3. Pronator and supinator: Pronator turns the palm downwards and supinator turns the palm upward.
4. Levator and depressor: Levator raises a body part and the depressors lower the body part.
5. Protractor and retractor: Protractor moves forward, whereas the retractor moves backward.
6. Sphincters: Circular muscles present in the inner walls of anus, stomach, etc., for closure and opening.

Question 6.
Differentiate between:
i. Flexor and extensor muscles
Answer:

	Flexor Muscles		Extensor Muscles
a.	Flexor muscles contract and bring about the bending or flexion of joint.	a.	Extensor muscles contract and bring about the straightening or extension of joint.
b.	These muscles decrease the angle between the bones on two sides of a joint.	b.	These muscles increase the angle between the components of limb.
e.g.	Biceps	e.g.	Triceps

ii. Pronator and supinator: Pronator turns the palm downwards and supinator turns the palm upward.

Can you recall? (Textbook Page No. 194)

Question 1.
Comment on contraction of skeletal muscles.
Answer:
Skeletal muscles show quick and strong voluntary contractions. They bring about voluntary movements of the body. For mechanism of muscle contraction:

When the muscles are relaxed, the active sites remain covered with tropomyosin and troponin complex. Due to this, myosin cannot interact with active site of actin and thus contraction cannot occur.

1. When an impulse (action potential) comes to muscle through motor end plate, it spreads throughout the sarcolemma of the myofibril.
2. The transverse tubules of sarcoplasmic reticulum release large number of Ca⁺⁺ ions into sarcoplasm. These calcium ions interact with the troponin molecules and the interaction inactivates troponin-tropomyosin complex. This causes change in the structure of tropomyosin.
3. As a result, tropomyosin gets detached from the active site of actin (F actin) filament, exposing the active site for actin.
4. The myosin head cleaves the ATP to derive energy and gets attached to the uncovered active site of actin. This results into the formation of acto-myosin complex.
5. The myosin heads are now tilted backwards and pull the attached actin filament inwardly. This results in contraction of the muscle fibres.

Do You Know How (Textbook Page No. 195)

Question 1.
Do skeletal muscles contract and bring about movement and locomotion?
Answer:
When the muscles are relaxed, the active sites remain covered with tropomyosin and troponin complex. Due to this, myosin cannot interact with active site of actin and thus contraction cannot occur.

1. When an impulse (action potential) comes to muscle through motor end plate, it spreads throughout the sarcolemma of the myofibril.
2. The transverse tubules of sarcoplasmic reticulum release large number of Ca⁺⁺ ions into sarcoplasm. These calcium ions interact with the troponin molecules and the interaction inactivates troponin-tropomyosin complex. This causes change in the structure of tropomyosin.
3. As a result, tropomyosin gets detached from the active site of actin (F actin) filament, exposing the active site for actin.
4. The myosin head cleaves the ATP to derive energy and gets attached to the uncovered active site of actin. This results into the formation of acto-myosin complex.
5. The myosin heads are now tilted backwards and pull the attached actin filament inwardly. This results in contraction of the muscle fibres.

Internet my friend. (Textbook Page No. 196)

Question 1.
Collect information about‘T’ tubules of sarcoplasmic reticulum.
Answer:

1. T tubules or the transverse tubules are invaginations of the sarcolemma penetrating into the myocyte interior, forming a highly branched and interconnected network that makes junctions with the sarcoplasmic reticulum.
2. These tubules are selectively enriched with specific ion channels and proteins crucial in the development of calcium transients necessary in excitation-contraction coupling, thereby facilitating a fast-synchronous contraction of the entire cell volume.
3. They are unique to straited muscle cells.
   [Source: https://www. uniprot. org/locations]

Can you tell? (Textbook Page No. 197)

Question 1.
Explain the chemical changes taking place in muscle contraction.
Answer:
The muscle undergoes various chemical changes during contraction, they are as follows:

1. A nerve impulse arrives at the motor nerve. The neurotransmitter – acetylcholine is released at the neuromuscular junction (N-M junction or motor endplate) enters into the sarcomere.
2. This leads to inflow of Na+ inside the sarcomere and generates an action potential in the muscle fibre.
3. The action potential passes down the T tubules and activates calcium channels in the T tubular membrane. Activation of calcium channel allows calcium ions to pass into the sarcoplasm. These Ca⁺⁺ ions binds to the specific sites on troponin of actin filaments and a conformational change occurs in the troponin – tropomyosin complex, thereby removing the masking of active sites for myosin on the actin filament.
4. In the myosin head, the enzyme ATPase gets activated in the presence of Ca⁺⁺ and converts ATP into ADP and inorganic phosphate.
5. This energy from ATP hydrolysis is utilized by myosin bridges or myosin heads to bind with active sites of actin and form actomyosin complex pulling the actin filaments towards the centre of sarcomere. The myosin heads are now tilted backwards and pull the attached actin filament inwardly towards them. The actin filament slides over mysosin and contraction occurs.
6. Also, the ADP needs to be converted back to ATP immediately as they required for muscular contraction, This is achieved in the muscles by the presence of another high energy compound, creatine phosphate.
7. ADP combines with creatine phosphate to produce ATP and creatinine due to which the supply of ATP for muscle contraction is restored but the level of creatine phosphate keeps decreasing and the level of creatinine keeps on increasing.
8. The creatinine formed needs to be reconverted to creatine phosphate. This is done by ATP produced during oxidation of glycogen through glycolysis.

Question 2.
Why are muscle rich in creatine phosphate?
Answer:

1. Creatine phosphate or phosphocreatine is formed from ATP, when the muscle is in relaxed state. It is a phosphorylated form of creatine.
2. Muscle cells contain creatine phosphate which acts as energy reserve as this high energy compound acts as a phosphate donor for ATP formation.
3. ATP acts as an immediate source of energy for contraction

Question 3.
Explain the mechanism of muscle contraction and relaxation.
Answer:
Mechanism of muscle contraction:

When the muscles are relaxed, the active sites remain covered with tropomyosin and troponin complex. Due to this, myosin cannot interact with active site of actin and thus contraction cannot occur.

1. When an impulse (action potential) comes to muscle through motor end plate, it spreads throughout the sarcolemma of the myofibril.
2. The transverse tubules of sarcoplasmic reticulum release large number of Ca⁺⁺ ions into sarcoplasm. These calcium ions interact with the troponin molecules and the interaction inactivates troponin-tropomyosin complex. This causes change in the structure of tropomyosin.
3. As a result, tropomyosin gets detached from the active site of actin (F actin) filament, exposing the active site for actin.
4. The myosin head cleaves the ATP to derive energy and gets attached to the uncovered active site of actin. This results into the formation of actomyosin complex.
5. The myosin heads are now tilted backwards and pull the attached actin filament inwardly. This results in contraction of the muscle fibres.

Muscle relaxation: During relaxation, all the events occur in reverse direction as that of muscle contraction.

1. When the stimulation is terminated, the actomyosin complex is broken down and myosin head gets detached from actin filaments. This process utilizes ATP.
2. Also, the Ca⁺⁺ ions are pumped back into the sarcoplasmic reticulum. This process too is an energy dependent process and utilizes ATP.
   
3. As a result, the troponin-tropomyosin complex is restored again which covers the active sites of act in filament, due to disappearance of the Ca⁺⁺ ions. The interaction between actin and myosin ceases and the actin filaments return back to their original position.
4. This results in muscle relaxation.

Question 4.
What do you understand by muscle twitch?
Answer:
Single muscle twitch:

Single muscle twitch: It is a muscle contraction initiated by a single brief-stimulation. It occurs in 3 stages: a latent period of no contraction, a contraction period and a relaxation period.

1. The involuntary contraction of muscle fibers is known as muscle twitch.
2. Muscle twitch is also known as fasciculation.
3. It is caused due accumulation of lactic acid in muscles.

Do You Know How (Textbook Page No. 198)

Question 1.
Exoskeletal components change from lower to higher group of animals. These include chitinous structures, nails, horns, hooves, scales, hair, shell, plates, fur, muscular foot, tube feet, etc.

Question 1.
Do you know any of these exoskeletal structures help in movement and locomotion?
Answer:
Nails, hooves, scales, plates, muscular foot and tube feet help in movement and locomotion.

Question 2.
How do scales and plates help in movement and locomotion?
Answer:
Scales and plates in reptiles like snakes provide grip to move on rough edgy surfaces.

Question 3.
Are scales of a fish and that of snake similar?
Answer:
Fishes have dermal scales (bony scales), whereas reptiles like snakes have epidermal scales or scutes (horny, tough extensions of outer layer of skin i.e., stratum corneum).

Question 4.
Find out more information about exoskeletal structures and their role in movement and locomotion.
Answer:
Exoskeletal structures: Exoskeleton provide support, help in movement and also provides protection from predators. The exoskeletal structures vary from organism to organism. Echinoderms have tube feet for locomotion whereas molluscs (e.g. Chiton) have muscular foot for movement and locomotion

[Students are expected to find out more information about exoskeletal structures on their own.]

Question 2.
Name the tissues that form the structural framework of the body.
Answer:
Cartilage and bone

Do you remember? (Textbook Page No. 198)

Question 1.
What are the components of skeletal system?
Answer:
The components of skeletal system are bones, tendons, ligaments and joint.

Question 2.
What type of bones are present in our body?
Answer:
Long bones, short bones, flat bones, irregular bones and sesamoid bones.

Question 3.
How do bones help in various ways?
Answer:

1. Bones form the framework of our body and thus provide shape to the body.
2. They protect vital organs thus help in the smooth functioning of body.
3. The joints between the bones help in movement and locomotion.
4. They provide firm surface for attachment of muscles.
5. They are reservoirs of calcium and form important site for hemopoiesis.

Use your brain power. (Textbook Page No. 198)

Question 1.
Can you compare bone, muscle and joint which help in locomotion with any simple machines you have studied earlier?
Answer:
Bone, muscle and joint can be compared to the simple machines called levers. Joints act as fulcrum, respective muscle generates the force required to move the bone associated with joint.

Question 2.
Explain the three types of lever found in human body.
Answer:
The three types of lever are as follows:

1. Class I lever: The joint between the first vertebra and occipital condyle of skull is an example of Class I lever. The force is directed towards the joints (fulcrum); contraction of back muscle provides force while the part of head that is raised acts as resistance.
   
2. Class II lever: Human body raised on toes is an example of Class II lever. Toe acts as fulcrum, contracting calf muscles provide the force while raised body acts as resistance.
   
3. Class III lever: Flexion of forearm at elbow exhibit lever of class III. Elbow joint acts as fulcrum and radius, ulna provides resistance. Contracting bicep muscles provides force for the movement.
   

Use your brain power. (Textbook Page No. 199)

Question 1.
Why are long bones slightly bent and not straight?
Answer:

1. Long bones include tibia, fibula, femur, humerus, radius, ulna, etc.
2. They have greater length than width. They consist of a shaft and variable number of epiphysis.
3. They are slightly bent or curved to absorb the stress of the body’s weight and evenly distribute the body weight at several different points.
4. If long bones were straight, the weight of the body would be unevenly distributed and the bone would fracture more easily.

Identify and label. (Textbook Page No. 199)

Question 1.
Identify the different bones.
Answer:

Identify and label. (Textbook Page No. 200)

Question 1.
Name A, B, C and D from the given figure and discuss in group.

Answer:
A – Coronal suture,
B – Sagittal suture,
C – Lambdoidal suture,
D – Lateral / Squamous suture

Skull has many sutures (type of immovable joints) present, out of which four prominent ones are:

1. Coronal suture: Joins frontal bone with parietals.
2. Sagittal suture: Joins two parietal bones.
3. Lambdiodal suture: Joins two parietal bones with occipital bone.
4. Lateral/squamous sutures: Joins parietal and temporal bones on lateral side.

Can you tell? (Textbook Page No. 201)

Question 1.
Give schematic plan of human skeleton.
Answer:

[Note: Numbers in the bracket indicate the number of bones.]

Can you tell? (Textbook Page No. 201)

Question 1.
Enlist the bones of cranium.
Answer:
Cranium: It is made up of four median bones and two paired bones.

1. Frontal bone: It is median bone (unpaired) forming forehead, roof of orbit (eye socket) and the most anterior part of cranium. It is connected to two parietals, sphenoid and ethmoid bone.
2. Parietal bones: These paired bones form the roof of cranium and greater portion of sides of the cranium.
3. Temporal bones: These paired bones are situated laterally just above the ear on either side. Each temporal bone gives out zygomatic process that joins zygomatic bone to form zygomatic arch. Just at the base of zygomatic process is mandibular fossa, a depression for mandibles (lower jaw bone) that forms the only movable joint of the skull. This bone harbors the ear canal that directs sound waves into the ear. The processes of temporal bones provide points for attachment for various muscles of neck and tongue.
4. Occipital bone: It is a single bone present at the back of the head. It forms the posterior part and most of the base of cranium. The inferior part of this bone shows foramen magnum, the opening through which medulla oblongata connects with spinal cord. On the either sides of foramen magnum are two prominent protuberances called occipital condyles. These fit into the corresponding depressions present in 1st vertebra.
5. Sphenoid bone: Median bone present at the base of the skull that articulates with all other cranial bones and holds them together. This butterfly shaped bone has a saddle shaped region called sella turcica. In this hypophyseal fossa, the pituitary gland is lodged.
6. Ethmoid bone: This median bone is spongy in appearance. It is located anterior to sphenoid and posterior to nasal bones. It contributes to formation of nasal septum and is major supporting structure of nasal cavity.

Can you tell? (Textbook Page No. 201)

Question 1.
Write a note on structure and function of skull.
Answer:
i. Structure of skull:
Skull is made up of 22 bones. It is located at the superior end of vertebral column. The bones of skull are joined by fixed or immovable joints except for jaw.

Skull consists of cranium or brain box and facial bones.

i. Cranium: It is made up of four median bones and two paired bones.

1. Frontal bone: It is median bone (unpaired) forming forehead, roof of orbit (eye socket) and the most anterior part of cranium. It is connected to two parietals, sphenoid and ethmoid bone.
2. Parietal bones: These paired bones form the roof of cranium and greater portion of sides of the cranium.
3. Temporal bones: These paired bones are situated laterally just above the ear on either side. Each temporal bone gives out zygomatic process that joins zygomatic bone to form zygomatic arch. Just at the base of zygomatic process is mandibular fossa, a depression for mandibles (lower jaw bone) that forms the only movable joint of the skull. This bone harbors the ear canal that directs sound waves into the ear. The processes of temporal bones provide points for attachment for various muscles of neck and tongue.
4. Occipital bone: It is a single bone present at the back of the head. It forms the posterior part and most of the base of cranium. The inferior part of this bone shows foramen magnum, the opening through which medulla oblongata connects with spinal cord. On the either sides of foramen magnum are two prominent protuberances called occipital condyles. These fit into the corresponding depressions present in 1^st vertebra.
5. Sphenoid bone: Median bone present at the base of the skull that articulates with all other cranial bones and holds them together. This butterfly shaped bone has a saddle shaped region called sella turcica. In this hypophyseal fossa, the pituitary gland is lodged.
6. Ethmoid bone: This median bone is spongy in appearance. It is located anterior to sphenoid and posterior to nasal bones. It contributes to formation of nasal septum and is major supporting structure of nasal cavity.

ii. Facial Bones: Fourteen facial bones give a characteristic shape to the face. The growth of face stops of the age of 16.
Following bones comprise the facial bones:

1. Nasals: These are paired bones that form the bridge of nose.
2. Maxillae: These form the upper jaw bones. They are paired bones that join with all facial bones except mandible. Upper row of teeth are lodged maxillae.
   
3. Palatines: These are paired bones forming the roof of buccal cavity or floor of the nasal cavity.
4. Zygomatic bones: They are commonly called as cheek bones.
5. Lacrimal bones: These are the smallest amongst the facial bones. These bones form the medial wall of each orbit. They have lacrimal fossa that houses lacrimal sacs. These sacs gather tears and send them to nasal cavity.
6. Inferior nasal conchae: They form the part of lateral wall of nasal cavity. They help to swirl and filter air before it passes to lungs.
7. Vomer: The median, roughly triangular bone that forms the inferior portion of nasal septum.
8. Mandible: This median bone forms the lower jaw. It is the largest and strongest facial bone. It is the only movable bone of skull. It has curved horizontal body and two perpendicular branches i.e. rami. These help in attachment of muscles. It has lower row of teeth lodged in it.

ii. Functions of skull:

1. It protects the brain.
2. It provides sockets for ear, nasal chamber and eyes.
3. Mandible bone of the skull helps in opening and closing of the mouth.

Internet my friend. (Textbook Page No. 201)

Question 1.
Cleft palate and cleft lip
Answer:

1. Cleft palate and cleft lip are the birth defects that occur when a baby’s lip or mouth does develop properly.
2. Cleft palate happens when the tissue that forms the roof of the mouth does not join together completely during pregnancy.
3. Cleft lip happens when the tissue that makes up the lip does not join completely before birth. This leads to formation of an opening in the upper lip.

[Students are expected to find more information about Cleft Palate and lip on internet.]

Can you tell? (Textbook Page No. 202)

Question 1.
Why skull is important for us? Enlist few reasons.
Answer:
Functions of skull:

• It protects the brain.
• It provides sockets for ear, nasal chamber and eyes.
• Mandible bone of the skull helps in opening and closing of the mouth.

Internet my friend. (Textbook Page No. 202)

Question 1.
Find out information about sinuses present in skull, functions of skull and disorder ‘sinusitis’.
Answer:
Sinuses are the hollow cavities present in the skull. They humidify the air we breathe.

i. The four types of sinuses present in the skull:

• Frontal sinuses: They are located above each eye. There are right and left frontal sinuses.
• Maxillary sinuses: They are the largest among all sinuses, located just behind the cheekbones near to upper jaws.
• Sphenoid sinuses: These are present just behind the nose.
• Ethmoid sinuses: These are present between the eyes.

ii. Functions of skull:

Functions of skull:

• It protects the brain.
• It provides sockets for ear, nasal chamber and eyes.
• Mandible bone of the skull helps in opening and closing of the mouth.

iii. Sinusitis: It is the inflammation of tissue lining the sinuses. Healthy sinuses when get blocked with mucus and germs causing infection which may lead to sinusitis.

[Students are expected to find more information about sinusitis, using the internet.]

Something interesting. (Textbook Page No. 202)

Question 1.
If police suspect strangulation, they carefully inspect hyoid bone and cartilage of larynx. These get fractured during strangulation. V arious such investigations are done in case of suspicious death of an individual where ossification of sutures in skull, width of pelvic girdle, etc. are examined to find out approximate age of victim or gender of victim, etc. You may find out information about forensic science.
Answer:
Forensic science is an application of science which is used in the matter of criminal determination and civil law. It is generally used in investigation of crimes. Forensic scientists collect, preserve and analyze the evidence during the course of investigation.
[Students are expected to find more information about forensic science on internet.]

Try this. (Textbook Page No. 202)

Question 1.
Feel your spine (vertebral column). Is it straight or curved?
Answer:
Our spinc shows four slight curves which are visible when viewed from the sides.

Question 2.
Find information about slipped disc. (Textbook page no.202)
Answer:

1. The bones of vertebral column are supported by the intervertebral discs.
2. These intervertebral discs act as shock absorbers due to which they are constantly compressed.
3. The disc consists of two parts – soft gelatinous inner part (nucleus pulposus) and tough outer ring.

If the ligaments of the intervertebral discs become injured, the pressure developed in the nucleus pulposus protrudes posteriorly or into one of the adjacent vertebrae. This is known as slipped disc.

[Students are expected to find more information using the internet.]

Can you tell? (Textbook Page No. 204)

Question 1.
Write a note on curvatures of vertebral column and mention their importance.
Answer:

1. The four curvatures in human spine are cervical, lumbar, thoracic and sacral curvatures.
2. The cervical and lumbar curvatures are secondary and convex whereas the thoracic and sacral curvatures are
   primary and concave.
3. Importance: Curvatures help in maintaining balance in upright position. They absorb shocks while walking
   and also protect the vertebrae from fracture.

Question 2.
Explain the structure of typical vertebra.
Answer:

1. Each vertebra has prominent central body called centrum.
2. The centra of human vertebrae are flat in anterio-posterior aspect. Thus, human vertebrae are amphiplatyan.
3. From the either side of the centrum are two thick short processes which unite to form an arch like structure called neural arch, posterior to centrum.
4. Neural arch forms vertebral foramen which surrounds the spinal cord.
5. Vertebral foramina of all vertebrae form a continuous ‘neural canal’. Spinal cord along with blood vessels and protective fatty covering passes through neural canal.
6. The point where two processes of centrum meet, the neural arch is drawn into a spinous process called neural spine.
   
7. From the base of neural arch, two articulating processes called zygapophyses are given out on either side. The anterior is called superior zygapophyses and posterior called inferior zygapophyses.
8. In a stack of vertebrae, inferior zygapophyses of one vertebra articulates with superior zygapophyses of next vertebra. This allows slight movement of vertebrae without allowing them to fall.
9. At the junction of zygapophyses, a small opening is formed on either side of vertebra called intervertebral foramen that allows passage of spinal nerve.
10. From the base of neural arch, lateral processes are given out called transverse processes. Neural arch, neural spine and transverse processes are meant for attachment of muscles.

Question 2.
How will you identify a thoracic vertebra?
Answer:
Thoracic vertebrae can be identified on the basis of centrum, as the centrum of the thoracic vertebrae is heart shaped.

Can you recall? (Textbook Page No. 206)

Question 1.
How does humerus form ball and socket joint? Where is it located?
Answer:
The head of humerus fits into the glenoid cavity of scapula and forms ball and socket joint. it is located in shoulder and hips.

Can you tell? (Textbook Page No. 208)

Question 1.
Differentiate between the skeleton of palm and foot.
Answer:

	Skeleton of palm	Skeleton of foot
a.	It consists of metacarpals and phalanges	It consists of metatarsals and phalanges
b.	Saddle joints and condyloid joints are in the palm.	Condyloid or saddle joints are not present the foot.

Question 2.
Explain the longest bone in human body.
Answer:
Femur: The thigh bone is the longest bone in the body. The head is joined to shaft at an angle by a short neck. It forms ball and socket joint with acetabulum cavity of coxal bone. The lower one third region of shaft is triangular flattened area called popliteal surface. Distal end has two condyles that articulate with tibia and fibula.

Internet my friend. (Textbook Page No. 212)

Question 1.
Find out information about types of fractures and how they heal.
Answer:

1. Fractures are classified based on their severity, shape or position of the fracture line or the physician who first described them.
2. Types of fractures:
   1. Open fractures: The broken ends of the bone protrude through skin.
   2. Comminuted fractures: The bone is splintered, crushed or broken into pieces at the site of impact and smaller bone fragments lie between the two main fragments.
   3. Greenstick fractures: A partial fracture in which one side of the bone is broken and the other side bends.
   4. Impacted fractures: One end of the fractured bone is forcefully driven into the inferior of the other.
   5. Pott fractures: Fracture of the distal end lateral leg bone with serious injury of the distal tibial articulation.
   6. Codes fractures: Fracture of the distal end of the lateral forearm in which the distal fragment is displaced posteriorly.
3. A fractured bone heals in four phases viz, reactive phase, fibrocartilaginous formation phase, bony callus formation phase and bone remodeling phase.

[Source: Tortora, G., Derrickson, B. Principles of Anatomy and Physiology. 15^th Edition]

[Students are expected to find out more information about healing of fractures using the internet.]

Do you remember? (Textbook Page No. 208)

Question 1.
What are joints? What are the types?
Answer:
i. A point where two or more bones get articulated is called joint or articulation or athrosis.
They are classified based on degree of flexibility or movement they permit into lastly synovial or freely movable or diarthroses type of joints.

ii. Synarthroses / fibrous joints / movable joints:
In this joint, the articulating bones are held together by means of fibrous connective tissue. Bones do not exhibit movement. Hence, it is immovable or fixed type of joint. Synarthroses are further classified into sutures, syndesmoses and gomphoses.

• Sutures: It is composed of thin layer of a dense fibrous connective tissue. Sutures are places of growth. They remain open till growth is complete. On completion of growth, they tend to ossify. Sutures may permit some moulding during childhood. Sutures are further classified into butt joint, scarf joint, lap joint and serrate joint.
• Syndesmoses: It is present where there is greater distance between articulating bones. At such locations, fibrous connective tissue is arranged as a sheet or bundle, e.g. Distal tibiofibular joint, interosseous membrane between tibia and fibula and that between radius and ulna.
• Gomphoses: In this type of joint, a cone shaped bone fits into a socket provided by other bone,
  e. g. Tooth and jaw bones.
  

iii. Cartilaginous / slightly movable joints / amphiarthroses:
These joints are neither fixed nor freely movable. Articulating bones are held together by hyaline or
fibrocartilages. They are further classified as

• Synchondroses: The two bones are held together by hyaline cartilage. They are meant for growth.
  On completion of growth, the joint gets ossified, e.g. Epiphyseal plate found between epiphysis and diaphysis of a long bone, Rib – Sternum junction.
• Symphysis: In this type of joint, broad flat disc of fibrocartilage connects two bones. It occurs in mid-line of the body. e.g. Intervertebral discs, manubrium and sternum, pubic symphysis.
  

iv. Synovial joints / freely movable joints / diarthroses:

1. It is characterized by presence of a space called synovial cavity between articulating bones that renders free movement at the joint.
2. The articulating surfaces of bones at a synovial joint are covered by a layer of hyaline cartilage. It reduces friction during movement and helps to absorb shock.
   
3. Synovial cavity is lined by synovial membrane that forms synovial capsule. Synovial membrane secretes synovial fluid.
4. Synovial fluid is a clear, viscous, straw coloured fluid similar to lymph. It is viscous due to hyaluronic acid. The synovial fluid also contains nutrients, mucous and phagocytic cells to remove microbes. Synovial fluid lubricates the joint, absorbs shocks, nourishes the hyaline cartilage and removes waste materials from hyaline cartilage cells (as cartilage is avascular). Phagocytic cells destroy microbes and cellular debris formed by wear and tear of the joint.
5. If the joint is immobile for a while, the synovial fluid becomes viscous and as joint movement starts, it becomes less viscous.
6. The joint is provided with capsular ligament and numerous accessory ligaments. The fibrous capsule is attached to periosteum of articulating bones. The ligament helps in avoiding dislocation of joint.

g. The types of synovial joints are on follows:

1. Pivot joint: In this type of joint, the rounded or pointed surface of one bone articulates with a ring formed partly by another bone and partly by the ligament. Rotation only around its own longitudinal axis is possible. e.g. in joint between atlas and axis vertebrae, head turns side ways to form ‘NO’ joint.

2. BaIl and socket joint: The ball like surface of one bone fits into cup like depression of another bone forming a movable joint. Multi-axial movements are possible. This type of joint allows movements along all three axes and in all directions. e.g. Shoulder and hip joint.

3. Hinge joint: In a hinge joint, convex surface of one bone fits into concave surface of another bone. In most hinge joints one bone remains stationary and other moves. The angular opening and closing motion (like hinge) is possible. In this joint only mono-axial movement takes place like flexion and extension. e.g. Elbow and knee joint.

4. Condyloid joint: It is an ellipsoid joint. The convex oval shaped projection of one bone fits into oval shaped depression in another bone. It is a biaxial joint because it permits movement along two axes viz, flexion, extension, abduction, adduction and circumduction is possible. e.g. Metacarpophalangeal joint.

5. Gliding joint: It is a planar joint, where the articulating surfaces of bones are flat or slightly curved. These joints are non-axial because the motion they allow does not occur along an axis or a plane. e.g. Intercarpal and intertarsal joints.

Saddle joint: This joint is a characteristic of Homo sapiens. Here the articular surface of one bone is saddle-shaped and that of other bone fits into saddle (each bone forming this joint have both concave and convex areas). It is a modified condyloid joint in which movement is somewhat more free. It is a biaxial joint that allows flexion, extension, abduction, adduction and circumduction. e.g. Carpometacarpellar joint between carpal (trapezium) and metacarpal of thumb.

Imagine. (Textbook Page No. 208)

Question 1.
If your elbow joint would be a fixed type of joint and joint between teeth and gum would be freely movable.
Answer:

1. If the elbow joint would be fixed the flexion and extension of the forearm won’t be possible. Also, rotation of the forearm and wrist would not be not possible.
2. Gomphoses is the type of joint that holds the teeth in the jaw bone. If this joint would be freely movable, we would not be able to chew and all our teeth would fall out.

Use your brain power. (Textbook Page No. 210)

Question 1.
Why are warming up rounds essential before regular exercise?
Answer:

1. Warming up before exercise stimulates the production and secretion of synovial fluid which reduces the stress on joints during exercise.
2. Also, if a joint is immobile for a while, the synovial fluid becomes viscous and as joint movement starts, it becomes less viscous.
3. Warming up increases the blood circulation, loosening the joints and increasing the blood flow. It also prepares the muscles for physical activity and prevents injuries.

Can you tell? (Textbook Page No. 211)

Question 1.
Classify various types of joints found in human body. Present the information in the form of chart. Give example of each type.
Answer:

i. A point where two or more bones get articulated is called joint or articulation or athrosis.
They are classified based on degree of flexibility or movement they permit into lastly synovial or freely movable or diarthroses type of joints.

ii. Synarthroses / fibrous joints / movable joints:
In this joint, the articulating bones are held together by means of fibrous connective tissue. Bones do not exhibit movement. Hence, it is immovable or fixed type of joint. Synarthroses are further classified into sutures, syndesmoses and gomphoses.

1. Sutures: It is composed of thin layer of a dense fibrous connective tissue. Sutures are places of growth. They remain open till growth is complete. On completion of growth, they tend to ossify. Sutures may permit some moulding during childhood. Sutures are further classified into butt joint, scarf joint, lap joint and serrate joint.
2. Syndesmoses: It is present where there is greater distance between articulating bones. At such locations, fibrous connective tissue is arranged as a sheet or bundle, e.g. Distal tibiofibular joint, interosseous membrane between tibia and fibula and that between radius and ulna.
3. Gomphoses: In this type of joint, a cone shaped bone fits into a socket provided by other bone,
   e. g. Tooth and jaw bones.
   

iii. Cartilaginous / slightly movable joints / amphiarthroses:
These joints are neither fixed nor freely movable. Articulating bones are held together by hyaline or
fibrocartilages. They are further classified as

• Synchondroses: The two bones are held together by hyaline cartilage. They are meant for growth.
  On completion of growth, the joint gets ossified, e.g. Epiphyseal plate found between epiphysis and diaphysis of a long bone, Rib – Sternum junction.
• Symphysis: In this type of joint, broad flat disc of fibrocartilage connects two bones. It occurs in mid-line of the body. e.g. Intervertebral discs, manubrium and sternum, pubic symphysis.
  

iv. Synovial joints / freely movable joints / diarthroses:

1. It is characterized by presence of a space called synovial cavity between articulating bones that renders free movement at the joint.
2. The articulating surfaces of bones at a synovial joint are covered by a layer of hyaline cartilage. It reduces friction during movement and helps to absorb shock.
   
3. Synovial cavity is lined by synovial membrane that forms synovial capsule. Synovial membrane secretes synovial fluid.
4. Synovial fluid is a clear, viscous, straw coloured fluid similar to lymph. It is viscous due to hyaluronic acid. The synovial fluid also contains nutrients, mucous and phagocytic cells to remove microbes.
   Synovial fluid lubricates the joint, absorbs shocks, nourishes the hyaline cartilage and removes waste materials from hyaline cartilage cells (as cartilage is avascular). Phagocytic cells destroy microbes and cellular debris formed by wear and tear of the joint.
5. If the joint is immobile for a while, the synovial fluid becomes viscous and as joint movement starts, it becomes less viscous.
6. The joint is provided with capsular ligament and numerous accessory ligaments. The fibrous capsule is attached to periosteum of articulating bones. The ligament helps in avoiding dislocation of joint.

g. The types of synovial joints are on follows:

1. Pivot joint: In this type of joint, the rounded or pointed surface of one bone articulates with a ring formed partly by another bone and partly by the ligament. Rotation only around its own longitudinal axis is possible. e.g. in joint between atlas and axis vertebrae, head turns side ways to form ‘NO’ joint.

2. BaIl and socket joint: The ball like surface of one bone fits into cup like depression of another bone forming a movable joint. Multi-axial movements are possible. This type of joint allows movements along all three axes and in all directions. e.g. Shoulder and hip joint.

3. Hinge joint: In a hinge joint, convex surface of one bone fits into concave surface of another bone. In most hinge joints one bone remains stationary and other moves. The angular opening and closing motion (like hinge) is possible. In this joint only mono-axial movement takes place like flexion and extension. e.g. Elbow and knee joint.

4. Condyloid joint: It is an ellipsoid joint. The convex oval shaped projection of one bone fits into oval shaped depression in another bone. It is a biaxial joint because it permits movement along two axes viz, flexion, extension, abduction, adduction and circumduction is possible. e.g. Metacarpophalangeal joint.

5. Gliding joint: It is a planar joint, where the articulating surfaces of bones are flat or slightly curved. These joints are non-axial because the motion they allow does not occur along an axis or a plane. e.g. Intercarpal and intertarsal joints.

Saddle joint: This joint is a characteristic of Homo sapiens. Here the articular surface of one bone is saddle-shaped and that of other bone fits into saddle (each bone forming this joint have both concave and convex areas). It is a modified condyloid joint in which movement is somewhat more free. It is a biaxial joint that allows flexion, extension, abduction, adduction and circumduction. e.g. Carpometacarpellar joint between carpal (trapezium) and metacarpal of thumb.

[Students are expected to prepare a chart on their own.]

Can you tell? (Textbook Page No. 211)

Question 1.
Human beings can hold an object in a better manner than monkeys. Why?
Answer:

1. Humans and monkeys both have five fingers including thumb, however humans can hold an object in better manner than monkeys because humans have highly developed opposable thumbs. The opposable thumb allows better grip.
2. The saddle joint in thumb allows free and independent movement to the thumb the carpometacarpellar joint between carpal (trapezium) and metacarpal of thumb makes the thumb opposable. It allows biaxial movements, i.e. flexion – extension and adduction – abduction but not rotation.

[Note: Gorillas, chimpanzees, orangutans and some other variants of apes have opposable thumb.]

Internet my friend. (Textbook Page No. 211)

Question 92.
Now a days we hear from many elderly people that they are undergoing knee replacement surgery. Find out why one has to undergo knee replacement; how it is carried out and how it can be prevented.
Answer:
Knee replacement is done in following cases:

1. Osteoarthritis: The cartilage in the knee undergoes degradation. It is caused by many factors such as muscle weakness, aging, obesity, etc.
2. Rheumatoid arthritis: It is characterised by inflammation of the synovial membrane, where it starts secreting excess of synovial fluid in the joint. This fluid exerts extensive pressure on the joint and causes severe pain.
3. Post-traumatic arthritis: This is caused due to breakage of ligament or cartilage. The breakage can be due to severe injury or accident. It causes severe pain and requires knee replacement.
4. Procedure:
   The procedure involves removal of the damaged cartilage or ligament and replaces it with artificial implant made up of either metal, plastic or both. Metal or plastic knee caps are used to cover the knees. The implant is connected to the bone and an artificial knee joint is made between them.
5. Prevention: Maintaining body weight, exercising regularly, consuming appropriate medications and supplements, etc.

[Students are expected to find out more information about knee replacement on internet]

Find out. (Textbook Page No. 212)

Question 1.
You must have heard of Sachin Tendulkar suffering from ‘tennis elbow’, a cricketer suffering from a disorder named after another game. Can common people too suffer from this disorder? Find out more information about this disorder.
Answer:

1. Tennis elbow is caused due to inflammation of tendon which joins muscles of forearm to the bone of upper arm (humerus). It is known as lateral epicondylitis.
2. It causes severe pain in the elbow. It occurs due to extensive repetitive movement of hand. This damages the tendon and increases the tenderness of the elbow joint.
3. This disorder develops not only in athletes but also in other common people whose job involves extensive movement of hand such as carpenter, painter, plumber, etc.

[Students are expected to find more information about tennis elbow on their own.]

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 1 Living world Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Biology Solutions Chapter 1 Living World

1. Choose correct option

Question A.
Which is not a property of living being?
a. Metabolism
b. Decay
c. Growth
d. Reproduction
Answer:
b. Decay

Question B.
A particular plant is strictly seasonal plant. Which one of the following is best suited if it is to be studied in the laboratory?
a. Herbarium
b. Museum
c. Botanical garden
d. Flower exhibition
Answer:
a. Herbarium

Question C.
A group of students found two cockroaches in the classroom. They had a debate whether they are alive or dead. Which life property will help them to do so?
a. Metabolism
b. Growth
c. Irritability
d. Reproduction
Answer:
c. Irritability

Question 2.
Distinguish between botanical gardens, zoological park and biodiversity park with reference to characteristics.
Answer:

No.	Botanical Gardens	Zoological Parks	Biodiversity Parks
1.	Plants of different varieties collected from different parts of the world are grown in vivo in a scientific and systematic manner in a botanical garden.	Zoological parks are places where wild animals are kept in captivity.	It is an assemblage of species that form self-sustaining communities on degraded barren landscape.
2.	It is a type of ex situ conservation.	It is a type of ex situ conservation.	It is a type of in situ conservation.
3.	It is related to conservation of various	It is related to conservation of various fauna.	It is related to conservation of all biodiversity.

3. Answer the following questions

Question A.
Jijamata Udyan, the famous zoo in Mumbai has acclimatised humbolt penguins. Why should penguins be acclimatised when kept at a place away from their natural habitat?
Answer:

1. Zoological park (zoo) is a type of ex-situ conservation in which wild animals are kept in captivity.
2. Humboldt penguins are native to South America and the surrounding environment differs significantly at Jijamata Udyan (zoo) in Mumbai.
3. In order to ensure that these penguins survive longer and are healthy they need to be acclimatised (adjust) to their new environment slowly.
4. If they are not acclimatised or the facilities in the zoo where the penguins are kept are not optimal/ suitable, they may develop abnormal stress and exhibit unusual behaviours due to it.
5. These penguins may also be more prone to contracting certain diseases, since they are suited to living in a particular climatic condition.
6. The enclosure of these penguins consists of water pool, air handling units and a chiller system to maintain temperatures between 12 – 14°C, where the penguins were kept for around 8 to 10 days to get acclimatised to their new environment before allowing any visitors inside the zoo.

Hence, Humboldt penguins need to be acclimatised to their new surroundings, when kept at a place away from
their natural habitat.

Question B.
Riya found peculiar plant on her visit to Himachal Pradesh. What are the ways she can show it to her biology teacher and get information about it?
Answer:

1. Riya can press and mount the plant specimen on a herbarium sheet and preserve the dried plant material, until she returns back from her visit.
2. She can also write any available information regarding the collected specimen on the herbarium sheet, which can be useful for further studies with her biology teacher.
3.  Various taxonomical aids can be useful to get information about this peculiar plant.
   [Note: In order to conserve the local flora, Riya can collect photographs of plant and describe it’s structure to her teacher.]

Question C.
At Andaman, authorities do not allow tourists to collect shells from beaches. Why it must be so?
Answer:

1. Seashells are an important part of the coastal ecosystem and are crucial for the survival of various marine creatures.
2. They provide material for building nests of birds and also act as a substratum for attachment of algae, sea grass, sponges and various microbes.
3. Fishes use shells for hiding from predators, whereas hermit crabs use shells as temporary shelters.
4. Removal of seashells from seashores may also indirectly affect the rate of shoreline erosion.
   Hence, in an attempt to protect the ecosystem, authorities in Andaman do not allow tourists to collect shells from beaches.

Question D.
Why do we have green house in botanical gardens?
Answer:

1. Greenhouse is a structure with suitable walls and a roof in which plants are grown under regulated climatic conditions.
2. Most botanical gardens exhibit ornamental plants which require stringent/ optimum climatic conditions for their growth and/or flowering.
3. The greenhouse associated with botanical gardens are also used to grow and propagate those plants that may not survive seasonal changes.

Hence, in order to provide optimum temperature for better growth and flowering and also to protect the plants from certain diseases, there are greenhouses in botanical gardens.

Question E.
What do you understand from terms like in situ and ex situ conservation?
Answer:

1. In situ conservation: It includes conservation of species in their natural habitats. Grazing, cultivation and collection of products from the forests is banned in such areas. Legally protected areas include national parks, wildlife sanctuaries and biosphere reserves.
2. Ex situ conservation: It includes conservation of species outside their natural habitats. Species are conserved in botanical gardens, culture collections and zoological parks.

4. Write short notes

Question A.
Role of human being in biodiversity conservation.
Answer:

1. Due to rapid increase in human population and industrialization, humans have over utilized natural resources; leading to degradation of the environment and hence only humans can help conserve the ecosystem.
2. Humans are capable of conserving and improving the quality of nature and thus, can play a major role in biodiversity conservation.
3. In order to conserve biodiversity and its environmental resources, humans must use the resources rationally and avoid excessive degradation of environment.
4. Human beings are stakeholders of the environment and need to come together to overcome pollution and improve the environment quality in order to conserve biodiversity. E.g. Ban or limit on use of harmful products (plastic, chemicals, etc.) that are toxic to various birds, animals, etc.
5. Human beings also play a role in conservation of biodiversity by establishment of various sites for in situ (national parks, wildlife sanctuaries and biosphere reserves) and ex situ (botanical gardens, culture collections and zoological parks) conservation.

Question B.
Importance of botanical garden.
Answer:
The importance of botanical gardens is as follows:

1. It is a place where there is an assemblage of living plants maintained for botanical teaching and research purpose.
2. Botanical gardens are important for their records of local flora.
3. Botanical gardens provide facilities for the collection of living plant materials for botanical studies.
4. Botanical gardens also supply seeds and material for botanical investigations.
5. The development of botanical gardens in any country is associated with its history of civilization, culture, heritage, science, art, literature and various other social and religious expressions.
6. Botanical gardens besides possessing an outdoor garden may contain herbaria, research laboratory, greenhouses and library.
7. Botanical gardens are not only important for botanical studies, but also to develop tourism in the country.

Question 5.
How can you, as an individual, prevent the loss of Biodiversity?
Answer:
As individuals, we can prevent loss of biodiversity in the following ways:

1. Increasing awareness about environmental issues. Making posters that provide more information about biodiversity conservation, to raise public awareness.
2. Increased support and/ or active participation in government policies and actions laid down for conservation of biodiversity.
3. Protect various plant and animal species in our surrounding.
4. Set up bird and bat houses wherever possible.
5. Prevent felling of trees especially native plants or trees in a particular area.
6. Reduce, recycle and reuse resources. Especially, reduce pollution and use of plastic bags and other materials that are potential threats for the environment.
7. Use environment friendly products, segregate and dispose garbage correctly.
8. Convince people about the importance of trees and the need to participate in tree plantation campaign.
9. Obey the rules that fall under Biodiversity Act.
   [Students can use the given points as reference and mention additional preventive measures on their own.]

Practical / Project :

Question 1.
Make herbarium under the guidance of your teacher.
Answer:
Students are expected to perform the given activity by themselves under the guidance of their teacher.

Question 2.
Find out information about any one sacred grove (devrai) in Maharashtra.
Answer:
Sacred groves in Maharashtra are located in districts like Ahmednagar, Bhandara, Chandrapur, Jalgaon, Kolhapur, Nashik, Pune, Raigad, Ratnagiri, Sangli, Satara, Sindhudurg, Thane, Yavatmal.
[Source: Data as per C.P.R. Environment Education Centre, Chennai.]
e. g. Sacred grove of Parinche valley, Pune district of Maharashtra:

The Parinche valley region is comprised of the inaccessible rear part of the Purandhar fort and its surrounding valley region and is situated about 63 km to the southeast of Pune city and 18 km from Saswad town. The total area of the valley region is about 132 sq. km. Parinche is the biggest village and a nodal place in the valley. The majority (12) of the documented groves are located in the Kaldari and Pangare zones. The size of the sacred groves has however reduced due to various human related activities that have taken place in recent years.

The biggest sacred grove in the Parinche valley belongs to Buvasaheb of Tonapewadi and spreads over an area of 4.80 hectares. The forest types are unique to the groves. Presence of key species in the sacred groves varies from region to region. Two key tree species, i.e. Terminalia bellerica and Ficus spp., are present in these sacred groves which have almost disappeared from the surrounding areas. Large buttressed trees are another important feature of well-preserved sacred groves. The presence of these tree species indicates the vegetation of the past and also the type of potential vegetation that can be regenerated in these regions.

[Source: Waghchaure, C. K., Tetali, P., Gunale, V. R., Antia, N. H., & Birdi, T. J. (2006). Sacred Groves of Parinche Valley of Pune District of Maharashtra, India and their Importance. Anthropology & Medicine, 13(1), 55-76]
[Students can refer the given answer and search for more information about other sacred groves on their own.]

11th Biology Digest Chapter 1 Living World Intext Questions and Answers

Can you recall? (Textbook Page No. 1)

Question 1.
Whether all organism are similar? Justify your answer.
Answer:
No, all organisms are not similar.

1. Organisms on the earth exhibit great diversity.
2. Organisms are grouped as microbes, plants (autotrophs), animals (heterotrophs) and decomposers.
3. Different microbes and decomposers have various shapes and sizes.
4. Plants can be further classified on their shape, size, structure, mode of reproduction, etc. Plants also differ greatly based on the locations in which they are found, e.g. Snowy, desert, forest, aquatic, etc.
5. Even animals show a high degree of variation. They are classified as unicellular, multicellular, invertebrates, vertebrates, etc. Also, based on the environment in which they live, they are classified as terrestrial, aerial, aquatic and amphibians.

Question 2.
What is the difference between living and non-living things?
Answer:

	Living Things	Non-living Things
a.	Living things show growth from within.	Non-living things show growth by accumulation of materials on their surface.
b.	They reproduce asexually or sexually, except mules, sterile worker bees, infertile males.	They do not reproduce.
c.	They perform metabolism in order to obtain energy.	No metabolic changes occur in non-living things.
d.	They show irritability and respond to changes in their surroundings.	They do not show irritability.
e.	They undergo ageing and eventually die.	Non-living things do not have a finite life span.

Question 3.
Enlist the characters of living organisms.
Answer:
The basic principles of life are as follows:

1. Metabolism: Metabolism is breaking of molecules (catabolism) and making of new molecules (anabolism). An organism performs metabolism in order to obtain energy and various chemical molecules essential for survival.
2. Growth and development: Organisms tend to grow and develop in a well-orchestrated process from birth onwards.
3. Ageing: It is the process during which molecules, organs and systems begin to lose their effective working and become old.
4. Reproduction: For continuity of race (species), organisms reproduce (asexually or sexually) to produce young ones like themselves. However, mules and worker bees do not reproduce, yet are living.
5. Death: As the body loses its capacity to perform metabolism, an organism dies.
6. Responsiveness: Living organisms respond to thermal, chemical or biological changes in their surroundings.

Can you tell? (Textbook Page No. 1)

Question 1.
Whether all organisms prepare their own food?
Answer:
No, all organisms do not prepare their own food. Organisms that prepare their own food are known as autotrophs (e.g. Green plants, certain microbes). These organisms prepare their own food in the presence of sunlight, water and carbon dioxide.

Question 2.
Which feature can be considered as all-inclusive characteristic of life? Why?
Answer:
Metabolism can be considered as an all-inclusive (defining) feature of life since it is exhibited by all living organisms and does not take place in non-living things.

Another all-inclusive characteristic of life is responsiveness or irritability. This is a unique property of living beings since all living beings are conscious of their surroundings.

Question 3.
How can we study large number of organisms at a glance?
Answer:
Systematic study of organisms with the help of taxonomical aids can be used to study a large number of organisms at a glance.

Can we call? (Textbook Page No. 1)

Question 1.
Reproduction as inclusive character of life?
Answer:
No, we cannot call reproduction as an inclusive character of life. Certain organisms like mules and worker bees do not reproduce and are still living. Thus, reproduction cannot be considered as an all inclusive defining characteristic of living organisms.

Think about it (Textbook Page No. 1)

Question 1.
Can metabolic reactions demonstrated in a test tube (called ‘in vitro’ tests) be called living?
Answer:

• The sum total of all the chemical reactions occurring in the body is known as metabolism and no non¬living object exhibits metabolism.
• However, metabolic reactions can be demonstrated outside the body in a test tube (cell-free medium).
• Thus, isolated metabolic reaction (s) outside the body of an organism, performed in a test tube is neither living nor non-living.
• Metabolic reactions occurring in vitro are living reactions but not living things.

Question 2.
Now a days patients are declared ‘brain dead’ and are on life support. They do not show any sign of self-consciousness. Are they living or non-living?
Answer:
The brain controls all life processes. Hence, when a patient is declared as ‘brain dead’, he does not carry out any of the inclusive defining characters of living things (e.g. metabolism, consciousness, etc.) and is completely dependent on machines. Since, such patients do not show any sign of self-consciousness, these patients cannot exactly be called as living.

Internet my friend (Textbook Page No. 2)

Question 1.
Collect information about Prof. Almeida, Prof. V. N. Naik, Dr. A. V. Sathe, Dr. P. G. Patwardhan with reference to their taxonomic work and biodiversity conservation.
Answer:
i. Prof. Almeida:
Prof. (Dr.) Marselin R. Almeida was a renowned Plant Taxonomist and Medicinal Plant Consultant of India. He was a curator at the Blatter Herbarium (Mumbai). He discovered four new species of pteridophytes from Bombay presidency. His work includes – Pteridophytes of Maharashtra and Flora of Mahabaleshwar. He has contributed to the Flora of Maharashtra, Sawantwadi and its adjoining areas along with Dr. S. M. Almeida.

ii. Prof. V. N. Naik:
Prof. V. N. Naik is a renowned ‘Angiosperms Taxonomist’ of India. He completed the Flora of Marathwada. He has produced 15 Ph.D., 110 research articles and 6 books. His book on ‘Taxonomy of Angiosperms’ (Tata McGraw-Hill Education, 1984) is widely used throughout the world. He is currently a faculty of Dr. Babasaheb Ambedkar Marathwada University, Aurangabad.
[Source: http://www. bamu. ac. in/dept-of-botany/Achievements, aspxj]

iii. Dr. A. V. Sathe:
Collection and taxonomic studies of mushrooms in Maharashtra started around 1974. Prof. A.V. Sathe and his team were amongst the first to begin these studies. They recorded 75 species distributed in 43 genera. These species were collected from Maharashtra, Karnataka and Kerala. The collection of these species was documented in the form of a Monograph on Agaricales.
[Source: Borkar P., Doshi A., Navathe D. (2015) Mushroom diversity of Konkan region of Maharashtra, India. Journal of Threatened Taxa. 7(10): 7625-7640]

iv. Dr. P. G. Patwardhan:
Dr. Patwardhan and his associates at the M.A.C.S. Research Institute, Pune-renamed as Agharkar Research Institute (ARI), Pune have performed detailed studies on lichens. His school is in possession of over 600 species of crustose lichens, obtained after intensive collection programmes. These specimens have been deposited in the Ajarekar Mycological Herbarium in the Department of Mycology and Plant Pathology at the M.A.C.S. Research Institute, Pune.
[Source: http://lib.unipune.ac.in:8080/xmlui/bitstreamfhandle/l23456789/7451/07_introduction.pdf? sequence=7&is Allowedly]
[Students are expected to find more information on their own.]

Can you tell? (Textbook Page No. 3)

Question 1.
What are the essentials of a good herbarium?
Answer: The essentials of a good herbarium are as follows:

1. It is essential to identify and label the collected specimen correctly.
2. Specimens should be stored in a dry place.
3. The plants are usually pressed and mounted on the sheet of paper known as herbarium sheets. Some plants are not suitable for pressing or mounting, like succulents, seeds, cones, etc. They need to be preserved in suitable liquid like formaldehyde, acetic alcohol, etc.
4. In order to preserve the specimen for longer durations, acid-free paper, special glues and inks must be used to mount the specimen so that the specimen does not deteriorate.
5. The specimens should be dried well before preparing a herbarium in order to prevent rotting of specimen.
6. It is also essential to record the date, place of collection along with detailed classification and highlighting with its ecological peculiarities, characters of the plant on a sheet. Local names of plant specimens and name of the collector may be added. This information is given at lower right comer of sheet and is called ‘label’.

Question 2.
Why does the loss of biodiversity matter?
Answer:

1. The loss of biodiversity is an moral and ethical issue.
2. Biodiversity helps to maintain stability in an ecosystem.
3. Humans share the environment with various other organisms and harm to these species can result in loss of biodiversity.
4. The loss of even one variety of organisms can affect the entire ecosystem.
   Hence, due to all these reasons, loss of biodiversity matters.

Question 3.
Why should we visit botanical gardens, museums and zoo?
Answer:

1. Botanical gardens, museums and zoos are taxonomical aids which can be used to study biodiversity.
2. Botanical gardens have a wide range of plant species that are protected and preserved which can be observed and studied.
3. Museums help gain information about various plants and animals that are preserved and may even be extinct. They act as reference hubs for biodiversity studies.
4. Zoos provide information about various animals. They also harbour certain endangered animals and help us understand the role of biodiversity conservation. They can also be visited to study the food habits and behaviour of animals.
   Hence, we should visit botanical gardens, museums and zoos.

Find out (Textbook Page No. 4)

Question 1.
Human being is at key position in maintaining biodiversity of earth. Find out more information about the following.
i. Laws to protect and conserve biodiversity in India.
Answer:
a. Forest (Conservation) Act, 1980
b. Biological Diversity Act, 2002
c. Wildlife (Protection) Act, 1972
d. Environment Protection Act, 1986
[Students can find out more laws to protect and conserve Biodiversity in India ]

ii. Environmental effects of ambitious projects like connecting rivers or connecting cities by constructing roads.
Answer:
Connecting rivers or connecting cities by constructing roads have the following environmental effects:
a. They form barriers to animals.
b. Construction of roads requires cutting down of trees and results in large scale deforestation.
c. They occupy large land resources resulting in loss of habitat of various species.
d. It can alter the water flow pattern and damage many ecosystems.
e. Increase in air, water, soil and noise pollution can disturb various animals and birds, thus affecting their behavioural pattern.

iii. Did bauxite mining in Western Ghats affect critically endangered species like – Black panther, different Ceropegia spp., Eriocaulon spp. ?
Answer:
a. The Western Ghats, is one of the global biodiversity hotspots and retains more than 30% of all plant, aquatic, reptile, amphibian and mammal species found in India.

b. Recently, this ecologically sensitive region has been subjected to various developmental activities that have adversely affected the flora and fauna of the region.

c. Bauxite mining is one such activity which has had significant negative impact on the local environment. To access bauxite ore deposits, the above-ground vegetation needs to be completely removed, causing large scale deforestation. The vegetation in the adjoining area is also affected due to dumping.

d. The major threats of this activity include vegetation loss, forest fragmentation and biodiversity loss.
e. Since most mines fall in Eco-Sensitive Zones (ESZ), it has seriously affected the flora and fauna of the Western Ghats.

f. Black panthers have frequently been spotted at various locations in the Western Ghats and mining in these areas can seriously affect their health and numbers.

g. Certain species of Ceropegia and Eriocaulon that are endemic in the area have been reported to be critically endangered.

[Source: Chandore A. (2015) Endemic and threatened flowering plants of Western Ghats with special reference to Konkan region of Maharashtra. Journal of Basic Sciences. 2 (21-25)]
Hence it is most likely that bauxite mining in Western Ghats has adversely affected the critically endangered species like – Black panther, different Ceropegia spp., Eriocaulon spp.

Internet my friend (Textbook Page No. 4)

Question 1.
i. Collect information about botanical gardens, zoological parks and biodiversity hotspots in India.
Answer:
a. Botanical gardens in India:

No.	Botanical Gardens of India	Location
1.	Acharya Jagadish Chandra Bose Indian Botanic Garden	Kolkata
2.	Lloyd Botanical Garden	Darjeeling
3.	National Botanical Research Institute	Lucknow
4.	Botanical Garden of the Forest Research Institute	Dehradun
5.	The State Botanical Garden	Odisha
6.	Botanical Garden	Saharanpur
7.	Government Botanical Garden	Ootacamund

b. Zoological Parks in India:

No.	Zoological parks	Location	Type of animals
1.	Rajiv Gandhi Zoological Park	Pune [Katraj]	Reptiles, mammals, birds. They have a snake park.
2.	Jijamata Udyan	Mumbai	Endangered species of animals and birds.
3.	Nehru Zoological Park	Hyderabad	3500 species of birds, animals and reptiles.
4.	Indira Gandhi Zoological Park	Vishakhapatanam	Primates, carnivores, small mammals, reptiles and birds.
5.	Padmja Naidu Himalayan Zoological Park	Darjeeling	Endangered animals like snow leopards, red pandas, gorals (mountain goat), Siberian tigers and a variety of endangered bird species.
6.	Allen Forest Zoo	Kanpur	Hyena, Bear, Rhinoceros, Hippopotamus, Langoor, Musk deer. Ostrich, Emu, Crane etc.
7.	Lucknow Zoo	Lucknow	Royal Bengal Tiger, White Tiger, Gibbon, Black Bear, Asiatic Elephant, Great pied, Horn Bill etc.
8.	Alipore Zoological Gardens	Kolkata	Royal Bengal Tiger, African Lion, Hippopotamus, Great Indian One-homed Rhinoceros.
9.	The Madras Crocodile Bank Trust	Chennai	Crocodiles and many species of turtles, snakes and lizards.
10.	Parassinikkadavu Snake Park	Kannur	Spectacled Cobra, King Cobra, Russell’s Viper, Krait and Pit Viper.

c. Biodiversity hotspots in India:

No.	Biodiversity Hotspots
1.	The Eastern Himalayas (Arunachal Pradesh, Bhutan, Eastern Nepal)
2.	Indo – Burma (Purvanchal Hills, Arakan Yoma, Eastern Bangladesh)
3.	The Western Ghats and Srilanka

ii. Collect information of endemic flora and fauna of India.
Answer:
a. Endemic flora:
Albizia sikharamensis (Mimosaceae), Argvreia arakuensis (Convolvulaceae), Arundinella setosa (Poaceae), Acacia diadenia (Mimosaceae), Citrus assamensis (Rutaceae), Magnolia bailloni (Magnoliaceae), etc.
[Source: http://www. bsienvis. nic. in/Database/E_3942. aspx]

b. Endemic fauna:
Bare Bellied Hedgehog (Paraechinus nudiventris), Andaman Shrew (Crocidura andamanensis), Aruanchal Macaque (Macaca munzala), Car Nicobar Rat (Rattus palmarum), Peter’s Tube-nosed Bat (Harpiola grisea) etc.
[Source: http://faunaofindia.nic.in/PDFVolumes/spb/056/index.pdf]
[Students are expected to use the given sources and find more information on their own.]