Bhagya

MP Board Class 11th Maths Solutions Chapter 11 शंकु परिच्छेद विविध प्रश्नावली

प्रश्न 1.
यदि एक परवलयाकार परावर्तक का व्यास 20 सेमी और गहराई 5 सेमी है, तो नाभि ज्ञात कीजिए।
हल:
परवलयाकार परावर्तक AOB का व्यास,
AB = 20 सेमी
तथा AM = 10 सेमी
परावर्तक की गहराई, OM= 5 सेमी

यदि OX, OY निर्देशांक अक्ष हो तो बिन्दु परवलय पर स्थित है।
माना परवलय का समीकरण, y² = 4ax
∴ 10² = 4a.5 या 100 = 20a
∴ a = 5
परवलय की नाभि (a, 0) या (5, 0) है।

प्रश्न 2.
एक मेहराब परवलय के आकार का है और इसका अक्ष ऊर्ध्वाधर है। मेहराब 10 मीटर ऊँचा है और आधार में 5 मीटर चौड़ा है। यह परवलय के दो मीटर की दूरी पर शीर्ष से कितना चौड़ा होगा?
हल:
इसका आकार परवलय की आकृति का है।
माना OX, OY इसके निर्देशांक अक्ष है, और समीकरण y² = 4ax है।

मेहराब की ऊँचाई, OL = 10 मीटर
चौड़ाई EF = 5 मीटर
LF = \(\frac{1}{2}\) EF = \(\frac{1}{2}\) × 5 = \(\frac{5}{2}\)

= 2.24 मीटर (लगभग)।

प्रश्न 3.
एक सर्वसम भारी झूलते पुल की केबिल (cable) परवलय के रूप में लटकी हुई है। सड़क पथ जो क्षैतिज है 100 मीटर लम्बा है तथा केबिल से जुड़े अर्ध्वाधर तारों पर टिका हुआ है, जिसमें सबसे लम्बा तार 30 मीटर और सबसे छोटा तार 6 मीटर है। मध्य से 18 मीटर दूर सड़क पथ से जुड़े समर्थक (supporting) तार की लंबाई ज्ञात कीजिए।
हल:
माना OX, OY निर्देशांक अक्ष हैं। AOB परवलय के रूप में केबिल है। इसका समीकरण x² = 4ay के रूप में होगा।
सबसे छोटे तार की लम्बाई OL = 6 मीटर
सबसे बड़े तार की लम्बाई BM = 30 मीटर
शीर्ष O से रेखा LM की दूरी OL
= 6 मीसा है
सड़क की लंबाई AB = 100 मीटर, यदि C मध्य बिन्दु हो तो
CB = \(\frac{1}{2}\)AB = \(\frac{1}{2}\) × 100 = 50 मीटर
OC = CL – OL = 30 – 6 = 24 मीटर

OY से 18 मीटर दूरी पर एक बिन्दु R लिया गया है। OX से R की दूरी b हो, तब
R के निर्देशांक (18, b)
यह परवलय x² = \(\frac{625}{6}\)y पर स्थित है
∴ 18 x 18 = \(\frac{625}{6}\)b
∴ RP = b = \(\frac{18 \times 18 \times 6}{625}\)
= \(\frac{1944}{625}\)
= 3.11 मीटर
आधार LM से R की दूरी = RS = RP + PS
= 3.11 + 6
= 9.11 मीटर
∴ तार की लंबाई = 9.11 मीटर।

प्रश्न 4.
एक मेहराब अर्ध-दीर्घवृत्ताकार रूप का है। यह 8 मीटर चौड़ा है और केंद्र से 2 मीटर ऊँचा है। एक सिरे से 1.5 मीटर दूर बिन्दु पर मेहराब की ऊँचाई ज्ञात कीजिए।
हल:
% आकृति में ELF एक मेहराब है जिसकी चौड़ाई EF = 8 मीटर और ऊंचाई = 2 मीटर है।
माना OX, OY निर्देशांक अक्ष है। ELF एक दीर्घवृत्त है जिसमें a = 4, b = 2

एक बिन्दु Q सिरे F से 1.5 मीटर की दूरी पर है।
∴ Q की 0 से दूरी = 4 – 1.5 = 2.5 मीटर
मान लीजिए बिन्दु Q पर मेहराब की ऊंचाई p है।
∴ P(2.5, p) दीर्घवृत्त पर स्थित है।

अत: Q बिन्दु पर मेहराब की ऊंचाई = 1.56 मीटर (लगभग) है।

प्रश्न 5.
एक 12 सेमी छड़ इस प्रकार चलती है कि इसके सिरे निर्देशांक्षों को स्पर्श करते हैं। छड़ के बिन्दु P का बिन्दुपथ ज्ञात कीजिए जो x-अक्ष के संपर्क वाले सिरे से 3 सेमी दूर है।
हल:
माना OX, OY निर्देशांक्ष हैं। इन अक्षों पर रेखा PQ = 12 सेमी चलती है।
∆ POQ में, PQ² = OP² + OQ²
12² = a² + b²
या a² + b² = 144 …(1)
जहाँ OA = a, OB = b अक्षों पर अंत:खण्ड हैं।

अतः L का बिन्दुपथ एक दीर्घवृत्त है। जिसका समीकरण \(\frac{x^{2}}{81}+\frac{y^{2}}{9}\) = 1 है।

प्रश्न 6.
त्रिभुज का क्षेत्रफल ज्ञात कीजिए जो परवलय x² = 12y के शीर्ष को इसकी नाभिलंब जीवा के सिरों को मिलाने वाली रेखाओं से बना है।
हल:
परवलय का समीकरण,
x² = 12y
नाभि के निर्देशांक (a, 0) या (3, 0) हैं।

प्रश्न 7.
एक व्यक्ति दौड़पथ पर दौड़ते हुए अंकित करता है कि उससे दो झंडा चौकियों की दूरियों का योग सदैव 10 मीटर रहता है। और झंडा चौकियों के बीच की दूरी 8 मीटर है। व्यक्ति द्वारा बनाए पथ का समीकरण ज्ञात कीजिए।
हल:

स्पष्ट है कि P का बिन्दुपथ एक दीर्घवृत्त है। .
PF₁ + PF₂ = 10 = 2a
∴ a = 5
F₁F₂ = 8 = 2c
∴ c = 4
c² = a² – b²
या 16 = 25 – b²
∴ b² = 9
दीर्घवृत्त का समीकरण,
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1
अर्थात् \(\frac{x^{2}}{25}+\frac{y^{2}}{9}\) = 1.

प्रश्न 8.
परवलय y² = 4ax के अंतर्गत एक समबाहु त्रिभुज है जिसका एक शीर्ष परवलय का शीर्ष है। त्रिभुज की भुजा की लंबाई ज्ञात कीजिए।
हल:
परवलय y² = 4ax, एक समबाहु त्रिभुज बनाई गई है।
मान लीजिए इसकी भुजा की लंबाई p है।

यह परवलय y² = 4ax पर स्थित है।

MP Board Class 11th Maths Solutions

MP Board Class 12th Biology Solutions Chapter 8 Human Health and Disease

Human Health and Disease NCERT Textbook Questions and Answers

Question 1.
What are the various public health measures which you would suggest as safeguard against infectious disease ?
Answer:
The common preventive measures are as follows :

• Education : People should be educated about communicable diseases to protect , themselves from such diseases.
• Vaccination : People should get vaccination on time to avoid infection.
• Sanitation : The sanitation condition should be improved to avoid infection from polluted water, contaminated food, etc.
• Eradication of vectors : The breeding places of vectors should be destroyed. adult vectors should be killed by suitable methods.
• Sterilisation : The patient’s surroundings and articles of use should be completely sterilised so as to reduce the chances of infection.

Question 2.
In which way has the study of biology helped us to control infectious diseases ?
Answer:
Study of biology helps us to diagnose the pathogen in following ways :

1. The life-cycle of many pathogens studied.
2. Alternate and reservoir hosts are known.
3. The mechanisms of transmission of disease is known.
4. The protective measures are suggested against disease and pathogens based an above studies.
5. Suitable medicines against infectious disease are suggested.
6. The preparation of vaccines against many pathogens also entitle the use and study of biology.

Question 3.
How does the transmission of each of the following diseases take place :
(a) Amoebiasis
(b) Malaria
(c) Ascariasis
(d) Pneumonia.
Answer:
(a) Amoebiasis : Through faecal-oral route.
(b) Malaria : Through the bite of female Anopheles mosquito.
(c) Ascariasis : Through taking contaminated food and water.
(d) Pneumonia : Droplets from the sputum of the patient.

Question 4.
What measure would you take to prevent water-borne diseases ?
Answer:
To prevent water borne diseases, following measures are required :

• Drinking water should be clean, free from contamination. This could be achieved by filtration, boiling or sedimentation and chemical treatment of water.
• Water resources/ reservoirs should be periodically de-contaminated / disinfected.
• Water should not be allowed to stand for long to become breeding pools.
• Standards practices of hygiene should be strictly maintained in public catering.

Question 5.
Discuss with your teacher what does ‘a suitable gene’ means, in the context
of DNA vaccines.
Answer:
The term ‘suitable gene’ refers to that specific segment of DNA which forms immunogenic protein, such genes can be cloned and then integrated with vector for introducing into an individual to be immunised for certain disorder producing a particular vaccine against the pathogens.

Question 6.
Name the primary and secondary lymphoid organs.
Answer:
Primary lymphoid organs are bone marrow and thymus. Secondary lymphoid organs are spleen, lymph nodes, tronsils, peyer’s patches of small intestine and appendix.

Question 7.
The following are some well known abbreviations, which have been used in this chapter. Expand each one to its full form :
(a) MALT
(b) CMI
(c) AIDS
(d) NACO
(e) HIV.
Answer:
(a) MALT : Mucosa Associated Lymphoid Tissue.
(b) CMI : Cell Mediated Immunity. . .
(c) AIDS : Acquired Immune Deficiency Syndrome.
(d) NACO : National AIDS Control Organisation.
(e) HIV : Human Immunodeficiency Virus.

Question 8.
Differentiate the following and give examples of each :
(a) Innate and Acquired immunity.
(b) Active and Passive immunity.
Answer:
(a) Differences between Innate and Acquired immunity :

(b) Differences between Active and Passive immunity:

Question 9.
Draw a well-labelled diagram of an antibody molecule.
Answer:

Question 10.
What are the various routes by which transmission of Human Immuno-deficiency Virus takes place ?
Answer:
Various routes of entry of ADDS virus are :

• Sexual contact with the infected person.
• Through placenta (from infected mother to foetus).
• Transfusion of infected blood or blood products.
• Sharing infected needles by drug abusers.

Question 11.
What is the mechanism by which the AIDS virus causes deficiency in the immune system of the infected person ?
Answer:
The virus enters macrophages after getting into the body of individual where RNA forms viral DNA by reverse transcription. The viral DNA gets incorporated in the host cell’s DNA and directs the infected cells to produce viral copies. The newly produced virus particles attack helper T-cells and thus, the number of T-cells decrease. Since, the helper T-cells are essential for functioning of immune system, the person suffers from various diseases due to dificient immune system.

Question 12.
How is a cancerous cell different from a normal cell?
Answer:
Cancer cell is different from normal cells in the sense that it :

• Looses the property of contact inhibition.
• Continues to grow and divide.
• Produces masses of cells called tumours.

Question 13.
Explain what is meant by metastasis ?
Answer:
Metastasis is the spread of cancerous cells through migration from one tissue to other tissue and organs resulting in formation of secondary tumour. Malignant tumour is a mass of proliferating cells called neoplastic cells. They grow rapidly and invade surrounding unaffected normal cells or tissues. Cells get sloughed off from such tumour and migrate to distant sites through blood. A new place of infection is thus, established and a new tumour is formed. This property is called metastasis.

Question 14.
List the harmful effects caused by alcohol/drug abuse.
Answer:
The adverse effects of drugs and alcohol abuse are :

• Low to moderate doses can cause reckless behaviour, vandalism and violence depression, fatigue, weight fluctuations, etc.
• Excessive doses of drugs may lead to coma and death due to respiratory failure, heart failure or cerebral haemorrhage.
• A combination of different drugs or alcohol mixed with drugs, results in overdosing and even death. ,

Question 15.
Do you think that friends can influence one to take alcohal / drugs? If yes, how may one protect himself / herself from such as influence?
Answer:
Yes, friends can influence one to take drugs and alcohol. A person can take the following steps for the prevention of themself against drug abuse :

• By avoiding under peer pressure as everyone has their own field of interest which should be respected by there teachers and family. One should not experiment with alcohal for curiosity and fun.
• Avoid the company of friends who take drugs.
• Seek help from parents and peers. A child should not pushed beyond his/her threshold
  limits.
• Take proper knowledge and counselling about drug abuse. Devolve your energy in other extra-curricular activities.
• Seek immediate professional and medical help from psychologists and psychiatrists if symptoms of depression and frustration become apparent.
• Get rid of the problem completely and lead, perfectly normal life by increasing their will power.

Question 16.
Why is that once a person starts taking alcohol or drugs, it is difficult to get rid of this habit ? Discuss it with your teacher.
Answer:
It is difficult to get rid of this habit, because these substances are addictive and one starts having unpleasant feelings or withdrawal symptoms like nausea, vomiting, diarrhoea, shivering, muscle twitching, excessive perspiration, muscular and abdominal cramps. Mind loses control and all one can think of is taking the addictive substance. That is why resisting the temptation/pressure for the first time is the only way to avoid getting into the addictive habit and committing a slow suicide.

Question 17.
In your view what motivates youngsters to take to alcohol or drugs and how can this be avoided?
Answer:
Probably the ‘motivation’ comes from :

• Curiosity to experience the effect.
• Foolishness to try to prove oneself in front of the peers.
• Wrongly taking it as an excuse to escape from reality.
• Wrong thinking that one time ‘try’ is not going to do any harm.

But youngsters who, are strong willed, who understand its ill effects and who are satisfied with their academic and other achievements and who don’t want to waste their precious life, don’t fall for this kind of ‘Motivation’.

Human Health and Disease Other Important Questions and Answers

Human Health and Disease Objective Type Questions

1. Choose the correct Answers :

Question 1.
Useful vaccines from the protection of Polio, Diphtheria and Tetanus :
(a)B.C.G.
(b) D.RT.
(c) M.M.R.
(d) S.T.D.
Answer:
(b) D.RT.

Question 2.
After the disease, immunity produced in the body is called:
(a) Active immunity
(b) Inactive immunity
(c) Both (a) and (b)
(d) None of these.
Answer:
(a) Active immunity

Question 3.
Cancer is related to :
(a) Irregular growth of tissues
(b) Ageing
(c) Irregular division of tissues
(d) None of these.
Answer:
(a) Irregular growth of tissues

Question 4.
Concept against small pox vaccination is:
(a) W.B.Cs. received from animals
(b) Antigen obtained from other animals
(c) Of Antigens
(d) To weak the small pox virus.
Answer:
(d) To weak the small pox virus.

Question 5.
Syphilis is a sexually transmitted disease which is caused by :
(a) Pastrulla
(b) Leptospira
(c) Treponema pallidum
(d) Vibrio.
Answer:
(c) Treponema pallidum

Question 6.
Antibody is a:
(a) A molecule which destroy a specific antigen
(b) W.B.Cs. which eats bacteria
(c) Secretion of mammalian R.B.Cs.
(d) Component of Nucleus.
Answer:
(a) A molecule which destroy a specific antigen

Question 7.
One of the form of allergy is :
(a) Asthama
(b) Yellow eyes
(c) Typhoid
(d) Mumps.
Answer:
(a) Asthama

Question 8.
AIDS virus enters in human by :
(a) Food
(b) Kiss
(c) Water
(d) Blood.
Answer:
(d) Blood.

Question 9.
Credit of vaccine goes to :
(a) Alexander Fleming
(b) Edward Jenner
(c) Louis Pasteur
(d) Robert Koch.
Answer:
(b) Edward Jenner

Question 10.
Blood cancer is:
(a) Carcinoma
(b) Sarcoma
(c) Lymphoma
(d) Leukemia.
Answer:
(d) Leukemia.

Question 11.
AIDS test is known by :
(a) ELISA
(b) Australian antigen
(c) HIV test
(d) None of these.
Answer:
(a) ELISA

Question 12.
Reason of liver cancer is:
(a) Alcohol
(b) Tobacco
(c) Both (a) and (b)
(d) None of these.
Answer:
(a) Alcohol

Question 13.
Probe is used in:
(a) Fingerprinting
(b) Segregation of gene
(c) Identification of disease
(d) All of these.
Answer:
(d) All of these.

Question 14.
Monoclonal antibodies are used in :
(a) Identification of blood group
(b) Identification of pathogen
(c) Identification of cancer
(d) All of these.
Answer:
(d) All of these.

Question 15.
Study of embryological abnormalities are called :
(a) Tricology
(b) Terentology
(c) Traumatology
(d) Termitology.
Answer:
(b) Terentology

Question 16.
The disease which occurs in embryonic stage of a child due to the use of the
Thalidomide by women:
(a) Vaginal carcinoma
(b) Microcephaly
(c) Phocomelia
(d) Viralism.
Answer:
(c) Phocomelia

Question 17.
A chemical substance that causes loss of sensation is called:
(a) Antianxiety
(b) Analgesic
(c) Anesthetic
(d) Exciting (Stimulating substance).
Answer:
(c) Anesthetic

Question 18.
In which part is opium found in plant:
(a) From leaves
(b) From fruits
(c) From seeds
(d) From bark.
Answer:
(b) From fruits

Question 19.
Lethal Hallucinogen is:
(a) Opium
(b) Morphine
(c) L.S.D.
(d) Heroin.
Answer:
(c) L.S.D.

Question 20.
Stimulant found in tea:
(a) Tannin
(b) Cocain
(c) Caffeine
(d) Freak.
Answer:
(c) Caffeine

Question 21.
Which of the following is narcotic opiate:
(a) Barbiturates
(b) Morphine
(c) Amphetamine
(d)L.S.D.
Answer:
(b) Morphine

Question 22.
Name the medicine which changes our mind:
(a) Psychotropic
(b) Hallucinogen
(c) Barbiturates
(d) Stimulants.
Answer:
(a) Psychotropic

2. Fill in the Blanks:

1. ………………….. prescribed the germ theory of disease.
2. ………………….. is the first human insulin which is made by genetic engineering.
3. AIDS is ………………….. transmitted disease.
4. Gonorrhoea is caused by …………………..
5. ………………….. genes which is responsible for cancer.
6. Blood cancer is known as …………………..
7. Hepatitis disease is caused by …………………..
8. Standard vaccine should not be …………………..
9. Pathogens are in living condition in …………………..
10. Cholera disease is caused by bacteria …………………..
11. ………………….. is the good technique to kill pathogens.
12. World Red Cross Day is celebrated on …………………..
13. The mixture of narcotic drugs are called …………………..
14. Benzodiazepine is ………………….. drug.
15. World Mental Health Day is celebrated on …………………..
16. L.S.D drug is obtained from ………………………… fungus.
17. The medicine which releases stress and edginess without sleeping is called …………………..

Answer:

1. Robert Koch
2. Humulin
3. Sexually
4. Neisseria gonorrhoeae
5. Oncogenes
6. Leukemia
7. Virus
8. Diseased and Poisonous
9. In living vaccine
10. Vibrio cholerae
11. Vaccination
12. 8th May
13. Cocktail
14. Psychotropic
15. 10th December
16. Claviceps purpurea
17. Tranquilizer.

3. Match the Following:
I.

Answer:

1. (e)
2. (f)
3. (d)
4. (a)
5. (b)
6. (c)

II.

Answer:

1. (d)
2. (a)
3. (e)
4. (b)
5. (c).

III.

Answer:

1. (d)
2. (e)
3. (a)
4. (b)
5. (c).

IV.

Answer:

1. (c)
2. (a)
3. (d)
4. (b)

4. Answer in One Word/Sentence :

1. Who has proposed germ theory of disease?
2. Name the vector of malarial parasite.
3. Name the genetically engineered human insulin.
4. Give full name of HIV.
5. Name the vaccines used for the vaccination of diphtheria, polio and whooping cough.
6. What is the source of opium?
7. Give two examples of stimulants.
8. Give the names of harmful chemical compounds of tobacco.
9. Give the name of source of LSD.
10. Give one example of sedative drugs.
11. Give one example of tranquilizer.
12. Name the drugs responsible for altering mood of man.

Answer:

1. Robert Koch
2. Plasmodium
3. Humulin
4. Human Immunodeficiency Virus
5. DPT vaccine
6. Papaver officinarum
7. Caffeine, cocaine
8. Nicotin
9. Claviceps purpurea (Fungus)
10. Benzodiazepines
11. Phenothiazines
12. Barbiturates.

Human Health and Disease Very Short Answer Type Questions

Question 1.
Name the carrier of malaria parasite.
Answer:
Female Anophelies mosquito.

Question 2.
Write the full form of HIV.
Answer:
Human Immunodeficiency Virus.

Question 3.
Name the vaccine which prevent Dephtheria, Polio and Tetanus.
Answer:
DPT and Polio vaccine.

Question 4.
Write the full form of ELISA.
Answer:
Enzyme-linked Immunosorbent Assay.

Question 5.
Which disease is cured by chemotherapy ?
Answer:
Cancer.

Question 6.
What is the source of opium ?
Answer:
Poppy.

Question 7.
Give three examples of stimulating substances.
Answer:
Caffeine, cocaine, amiphimine.

Question 8.
Name the harmful chemical component which is found in tobacco.
Answer:
Nicotine.

Question 9.
Write the source of LSD.
Answer:
Claviceps purpurea.

Question 10.
Give one example of tranquilizer’s.
Answer:
Phenothiazine.

Question 11.
What is the name of drug which converts thoughts and feelings of human and causes fear ?
Answer:
Psychedelic drugs.

Question 12.
Which disease is caused by transmission of HIV?
Answer:
AIDS.

Question 13.
Name the method which is used in the diagnosis of AIDS.
Answer:
ELISA Test.

Question 14.
In different parts of the country Chikungunya is clarified. Name the carrier of this disease.
Answer:
Chikungunya is spread through bites from A. aegypti mosquitoes.

Question 15.
Name the types of Immunity.
Answer:
Immunity is two types :

1. Innate immunity,
2. Acquired immunity.

Question 16.
Where is Wuchereria found ?
Answer:
Filarial worms Wuchereria is found in lymph vessels.

Question 17.
Name the types of tumours.
Answer:
Tumours are of two types :

1. Benign and
2. Malignant.

Question 18.
What harm does AIDS cause in humans?
Answer:
Immune power of the body is decreased by AIDS.

Question 19.
When inability to open the mouth because human chewing areca nuts in betel quid or its variants (gutkha) and jaw muscles become hard. Give the name of possible disease.
Answer:
Submucous fibrasis disease.

Question 20.
Name the full form of DPT.
Answer:
Diphtheria, Pertussis and Tetanus.

Question 21.
What is the side effects of cancer treatment ?
Answer:
Hair falling and Anemia.

Question 22.
What is the full form of NAC ?
Answer:
National Aids Control Organization.

Question 23.
Which plant extract is called marijuana ?
Answer:
Cannabis sativa.

Question 24.
Name combine form of the poppy, morphine, herein pethidine and methedrine.
Answer:
Meconium.

Question 25.
What is the second name of Mary Mallon ?
Answer:
The second name of Mary Mallon is Typhoid Mary.

Question 26.
Name the cells which are multiplied by HIV when it enters the human body.
Answer:
Macrophages and helper T-lymphocytes.

Question 27.
Which body organs are affected by Pneumonia?
Answer:
Lungs and Alveolies are affected.

Question 28.
What is the confirmation test of Typhoid?
Answer:
Typhoid fever can be confirmed by Widal test.

Question 29.
Write the full form of LSD.
Answer:
Lysergic Acid Diethylamide.

Question 30.
Which viruses have responsible for cancer ?
Answer:
Oncogenic viruses have cancer-causing viral oncogenes.

Question 31.
Name the two major groups of cells required in attaining specific-immunity.
Answer:
B-cells and T-cells.

Question 32.
What are primary lymphoid organs ?
Answer:
The primary lymphoid organs are those organs in which B-lymphocytes and T- lymphocytes undergo maturation, e.g., Bone marrow and thymus.

Human Health and Disease Short Answer Type Questions

Question 1.
Define the following :
(a) Immunity
(b) Vaccine
(c) Interferon
(d) Vaccination.
Answer:
(a) Immunity: The capacity of any organism to fight with the disease and casual organisms is known as Immunity. The Immunity is due to B and T cells.
(b) Vaccine : The chemical substances which protect our body by disease causing organisms are called vaccines.
(c) Interferon : Interferon is an antiviral protein which is produced within animal cells due to stimulus produced after viral infection and which prevents the infection and multiplication of other viruses.
(d) Vaccination: Vaccination is the process by which resistance against specific disease is created in any living organism.

Question 2.
Distinguish between Inborn and Acquired immunity.
Answer:
Differences between Inborn and Acquired immunity are :

Question 3.
What is auto-immunity ?
Answer:
This is an abnormality which sometimes develop in the immune system of the body. Instead of destroying foreign molecules, it starts attacking the body’s own cells leading to serious consequence. Such diseases are called autoimmune diseases.

Auto-immune diseases depend on the type of self-antigen involved. If the self-antigens are R.B.Cs. then the body destroys its own R.B.Cs. resulting in chronic anaemia. When the self-antigen is a muscle cell, it results in destruction of its own-muscles resulting in severe weakness, when the self-antigens are liver cells, it results in chronic hepatitis.

Question 4.
What is allergens? What cause the allergy is produce?
Answer:
Allergy is the condition of hypersensitivity of the body against certain substances or to a physical or chemical agent allergens. When antigen antibody reaction takes place in the body, it results allergy. Sufficient antibodies will not be produced in lacking of proper immune system. The allergens combine with the antibody bound mast cells, which causes the cell to burst releasing histamine. It results inflammatory responses in the body. Allergy may be caused by a medicine, cosmetics and other substances like, pollen grains, dust particles etc.

Question 5.
What is B-cells and T-cells ? Explain it
Answer:
B-lymphocytes or B-cells produce an army of protein called antibodies in blood. In response to pathogens, T-lymphocytes or T-cells help B-cells to produce antibodies immune response are produced by these two types of lymphocytes.

Question 6.
Which types of diseases are protected by D.P.T. vaccine? Write the name of pathogens for each disease.
Answer:
D.P.T. vaccine is protect three types of disease :

(i) Diphtheria
(ii) Pertussis
(iii) Tetanus.
Name of the Pathogens:

Question 7.
What is Eugenics ?
Answer:
Eugenics: The branch of biology which deals with the study of improvements of human race is called eugenics.
Importance:
1. Development of selective reproduction in similar species.
2. Transfer of genetic materials in various organisms.
3. Development of GM food and GM crops.
4. Gene cloning.
5. Gene therapy, etc.

Question 8.
Why vaccination is important in immunity ?
Answer:
Vaccination is the process of introduction of weakened or inactivated pathogens or proteins (vaccine) into a person so, provide protection against the disease. Vaccination provides immunisation after a time gap. Due to immunisation our body produces antibodies against the vaccine and develops the ability to neutralise pathogens during actual infection. Nowadays different types of vaccine which gives in children like polio, tetanus, deptheria, pertusis, small pox etc.

Question 9.
What is Drug addiction ? What are its causes ?
Answer:
Addiction is most common problem of our youths because they start to take different types of drugs and alcoholic beverages due to various reasons. It makes them habitual and dependent on them. Addiction is the habitual, psychological and physiological dependence on a substance or practice which is beyond voluntary control. A person who is habituated to a substance or a practice especially harmful one, is called drug addiction.
Its causes are :

• Curiosity
• Fun and stimulation
• Will of doing more work
• Feeling of freeding
• Temporary escape from the life problems.

Question 10.
Write down the differences between Sedative and Tranquilizer.
Answer:
Differences between Sedative and Tranquilizer:

Question 11.
Give the name of source of LSD. Give its effects also.
Answer:
LSD (Lysergic Acid Diethylamide): It is a crystalline amidated alkaloid which is obtained from the sclerotium of ergot fungus Claviceps purpurea, pathogen of ergot disease of rye.
LSD causes horrible dreams, hallucinations, chronic psychosis and damage the brain. LSD was tried as treatment for alcoholism, neurosis and cancer patients. LSD brings about chromosomal and foetal abnormalities. Pathological condition caused by LSD abuse or by eating grain affected by ergot is called ergotism.

Question 12.
Describe the effect of alcohol on human body and society.
Answer:
The effect of alcohol on human body and society are :

• It effects the central nervous system.
• Persons is incapable of differentiating what is right and what is wrong.
• Taking excess of alcohol weakens heart, lungs, liver and other parts of the body.
• Pupil of eye expands on taking excess of alcohol.
• This increases the criminal activities in human beings.

Question 13.
Write withdrawal symptoms of addiction.
Answer:
Withdrawal symptoms from Addiction : When a drug dependent person fails to get the drug, he feels severe physical and psychological disturbances. These are called withdrawal symptoms. It includes tremors, nausea, vomiting, weakness, insomnia, anxiety run fits, decreased appetite, restlessness, elevated blood pressure, rapid heartbeat and epilepsy. These symptoms indicates that the body of addict person is unable to further use of intoxicant and they should stop it immediately.

Question 14.
Describe about the psychotropic drugs.
Answer:
These drugs effect on thinking power, mental processes-, continuous use of these drugs, bring change in human behaviour, consciousness and perception. So, they are called mood altering drugs. The person becomes addict by the use of these drugs and cannot live without these drugs.

Question 15.
What are sedative ? What are its types ? Write its effect.
Answer:Sedative : Drugs which directly depress the brain and central nervous system are included in this group. They make the body free from anxiety and lethargic. The excessive use leads to sleep. Sleep causing drugs are also called Hypnotics.
It is of two types :

1. Barbiturates and
2. Benzodiazopines.

1.  Barbiturates : It is a sedative and tranquilizer. It supresses brain’s activity and creates a feeling of relaxation, drawsiness and sleepiness.
2. Benzodiazopines : These drugs are commonly prescribed to help relieve anxiety in people who have anxiety disorder or another mental illness where anxiety is a symptom. They are very addictive.

Question 16.
What is Interferon ?
Answer:
Interferon : Interferon is an antiviral protein which is produced within animal cells due to stimulus produced after viral infection and which prevents the infection and , multiplication of other viruses.

Human Health and Disease Long Answer Type Questions

Question 1.
What is cancer ? Write the names of two types of cancer and causes of cancer.
Answer:
Cancer: When tumours arCformed due to unorganized and uncontrolled division of cell, then it is called cancer.
Types of Cancer: Cancer is not a single disease but a complex of many diseases. Today about 200 distinct types of cancer have been recognized.

These are grouped into four main types:

1. Carcinomas
2. Sarcomas
3. Lymphomas and
4. Leukemias.

1. Carcinomas : The tumours which are made up of epithelial cells of ectodermal or endodermal origin are called carcinomas, e.g., solid tumours in nerve tissues and in tissues of body surfaces or their attached glands. It includes breast, skin, cervical and brain carcinomas.

2. Sarcomas : Hie tumours which are made up of connective tissue cells of mesodermal origin are called sarcomas. e.g., solid tumours growing from connective tissues, bones, cartilages and muscles. It constitutes only about 2% of human cancers.

3. Lymphomas : Cancers in which there is excessive production of lymphocytes by the lymph nodes and spleen are called lymphomas, e.g., Hodgkin’s disease. It constitute about 5 % of human cancers.

4. Leukemias : It is cancer of blood characterized by excessive number of W.B.Cs. (or leucocytes) in the blood (neoplastic growth). These cells invade into bone-marrow, lymph nodes and the spleen. It is more frequent in the children of age group 5 to 7 years but it can occur at any age. Acute leukemia causes death. It has no sure remedy. It constitute about 4% of human cancers.

Causes of Cancer : Physical and chemical factors which cause cancer are called as carcinogens. Main causes of cancer are as follows :

• Smoking causes mouth and lung cancer.
• Radiation, such as X-rays, ultraviolet rays and other ionizing radiations cause cancer.
• Viruses may cause cancer.
• Chemical substances such as Nicotine, caffeine, products of combustion of coal and oil, polycyclic hydrocarbons may cause cancer.

Symptoms of Cancer: There are some symptoms of cancer which must be kept in mind:

• Any lump or thickening in the tissue especially in breast, tongue or lip.
• A wound that is not healing.
• Any sudden change in mole or warts.
• Persistent indigestion and difficulty in swallowing things.
• Regular cough and hoarseness in sound.
• Unusual weight loss.
• A change in bowel habits.
• Any ulcer that does not get well.
• Bleeding in vagina at times other than the menstruation.
• Non-injury bleeding from the surface of the skin, mouth or any other opening of the body.

Question 2.
What is AIDS ? Write down its causes, transmission and symptoms.
Or
Write the name, two symptoms and important preventive measures of today’ dreadening disease spread through blbod transfusion.
Or
Where is AIDS reported for the first time ? Describe the causes, method of transmission and control measures of this disease.
Answer :
AIDS (Acquired Immune Deficiency Syndrome) disease is discovered for the first time in America in the year 1981.

Causes of Disease :  This disease is caused by the infection of a virus known as HIV (Human Immunodeficiency Virus).

Transmission : It is transmitted through sexual contact, homosexuality, contaminated needles, blood transfusion, drugs, artificial insemination and organ transplantation etc.

Symptoms of Disease : It is characterized by showing swollen lymph nodes, fever, loss of weight. The person loses the immunity against the infection. In this disease, the number of helper T-cells are reduced.

Treatment : No suitable drug is available so far against this disease. Only anti-viral cells may increase in the number by immune stimulation method.

Control : The following measures are advised to prevent spreading of disease :

• Providing health education and explain the hazardous effects of AIDS.
• Do not reutilize the used syringe. Throw it away or destroy it.
• The blood of donor person and organs of transplantation like kidney, liver, cornea of eyes, growth hormones would be carefully examined before use.
• Sexual contact with many people must be avoided.

Question 3.
What is immune system ? Describe various components of immune system of man and its role.
Answer:
The Immune system or Immunity (A Specific Defence Mechanism) : The ability of an organism to resist the pathogen or development of disease resistance is known as immunity and the study of immunity is called immunology, while the infected person with no disease is called immune. The most peculiar characteristic of immune system is that it can differentiate the self (body’s own cells) and non-self (foreign microbes).

Cells of the Immune System : Lymphocytes are the type of W.B.Cs. or Leucocytes which are chief cells of immune system of body. There are two types of lymphocytes that promote cellular immunity and humoral immunity. Both of these types of lymphocytes are derived from lymphocyte stem cells in the bone-marrow in the embryo.

1. T-cells of lymphocytes : These cells eventually migrate to the lymphoid tissue. Before doing so, the lymphocytes first migrate to thymus gland and are processed in the gland, hence called T-lymphocytes or T-cells. These are responsible for cellular immunity. There are following types of T-cells :

Killer T-cells : Killer T-cells or KT-cells directly attack and destroy antigen. In doing so, they move to the site of invasion and produce some chemical that attracts and stimulate phagocytes to feed more voraciously on antigen.
Helper T-cells : Helper T-cells or HT-cells stimulate B-cells to produce more antibody.
Suppressor T-cells: Suppressor T-cells or ST-cells keep entire immune system to attack on the own body cell.

Mode of action of T-cells to antigens : T-cells are antigen specific (each T-cell recognizes a specific antigen and different types of T-cells are stimulated by different types v of antigens). When a T-cell is stimulated by specific antigen, T-lymphocytes divide ripidly to form a clone of T-cells called lymphoblasts. T-cells live for 4-5 years or even longer. T- cells of a clone are morphologically similar but they perform different functions. According to their functions, their are three classes of T-cells, i.e., killer T-cells, helper T-cells and 1 suppressor T-cells.

2. B-cells of lymphocytes : The other population of lymphocytes which produces antibodies are processed in some unknown area, possibly liver or spleen. This population ‘ was firstly discovered in birds in which processing occurs in the bursa of fabricus, a structure not found in mammals. For this reason they are called as B-cells or B-lymphocytes. They are responsible for humoral immunity.

Mode of action B-cells to antigens : Once a B-cells is activated by the antigen, it multiplies very fast and forms a clone of plasma cells. Most of these produce antibody at a tremendous rate of 2,000 molecules per second. These antibodies circulate in the lymph to fight the antigen. So, forming humoral immune system. B-cells are short lived and are replaced by new cells from the bone-marrow after every few days.
The capacity of B-cells to produce specific antibodies is acquired during its development and maturation.

Question 4.
Describe the harmful effects of tobacco smoking.
Answer:
Effect of chewing and smoking Tobacco: The chewing and smoking of tobacco affects us in the following ways :

• It affects the route of the transfer of nerve impulse and central nervous system. In low concentration nicotine stimulates conduction of nerve impulse but long-term use reduces the activity of nervous system.
• Nicotine stimulates the release of adrenaline leading to high blood pressure. Increased blood pressure due to smoking enhances the risk of heart diseases.
• It retards foetal growth in pregnant women.
• The smoke of tobacco contains aromatic hydrocarbons and tar along with carbon monoxide (CO). Carbon monoxide reduces oxygen carrying capacity of the blood. Hydrocarbons induces cancer. Due to these reasons tobacco chewing person suffer from mouth cancer and smokers suffer from throat and lung cancers.
• The use of tobacco in any form stimulates the secretion of saliva and gastric juices due to which acidity is increased in stomach. It may cause ulcers in the wall of stomach. The absorption capacity of mucous membrane of stomach also decreases. Thus, person get suffered from hyponutrition, loss of appetite and constipation.
• Smoking also affects the activity of kidneys.
• Nicotin relaxes the muscular and skeletal tissues due to which person becomes weak.
• Long-term smoking may also cause diseases like bronchitis and Emphysema.
• Smoking reduces immunity of the body.
• Lips of smokers may become dark coloured. Teeth and fingers get stained. The breath becomes foul.

Question 5.
What are the causes of addiction ?
Answer:
Causes of addiction are : Young peoples take drugs and alcoholic beverages and other intoxicans substances because of the following reasons:

• Curiosity: It is one of the most important cause of addiction because a curiosity is generated among young peoples due to advertisement of drugs in news papers and other means of communications. So, that they take them for fust time and then became habitual.
• Fun and Stimulation: Some drugs are stimulative in nature. These drugs are used to increase body power but regular use of these drugs make them habitual. These drugs cause physical and mental weakness.
• Will of doing more work: It is believed that the use of intoxicating substances increases die capacity of the body and person is able to do more physical and mental work. But it is not true because these substances increases mental and physical fatigueness.
• Feeling of freedom.
• Inability to face problems of life.
• Encouragement or Pressure by friends.
• Temporary escape from the life problems.
• Mental relaxation.
• Relax from pain.
• Feeling of freedom, etc.

Question 6.
What is rehabilization ?
Answer:
Rehabilitation; The long-term treatment of addicts require behavioural training which enables the patient to make different and useful responses to die stimuli that led him/her to drug taking. This is called rehabilization of a drug dependent.

Rehabilization requires psychological and social therapy in the form of counselling by relatives, friends and physicians They educate the patients about ill effects of taking drug and motivate them total drug abstinence in a sympathetic manner. The counsellors should not criticise the patient for the past actions and should be affectionate and sympathetic to him/her. The measure taken to rehabilize a drug dependent are reffered to as psychological therapy.

Vitamin administration, proper nutrition, restoration of electrolyte balance and proper hydration are necessary during the rehabilization. These measures are aimed at process of restoring the health damaged by drug abuse. It has been noticed that the level of cyclic AMP suddenly rises in the brain of an addict who has denied the drug. It increases withdrawal symptoms. Vitamin C checks the rise in cAMP level and helps in preventing the withdrawal symptoms.

MP Board Class 12th Biology Solutions

MP Board Class 11th Maths Solutions Chapter 11 शंकु परिच्छेद Ex 11.4

प्रश्न 1 से 6 तक प्रत्येक में, अतिपरवलयों के शीर्षों, नाभियों के निर्देशांक, उत्केंद्रता और नाभिलंब जीवा की लंबाई ज्ञात कीजिए।
प्रश्न 1.
\(\frac{x^{2}}{16}-\frac{y^{2}}{9}\) = 1.
हल:
अतिपरवलय का समीकरण \(\frac{x^{2}}{16}-\frac{y^{2}}{9}\) = 1
अनुप्रस्थ अक्ष, x-अक्ष के अनुदिश है।
a² = 16, b² = 9
∴ c² = a² + b² = 16 + 9 = 25
∴ a = 4, b = 3, c = 5
शीर्षों के निर्देशांक (± a, 0) या (± 4,0).
नाभियों के निर्देशांक (± c, 0) या (± 5, 0)

प्रश्न 2.
\(\frac{y^{2}}{9}-\frac{x^{2}}{27}\) = 1
हल:
अतिपरवलय का समीकरण : \(\frac{y^{2}}{9}-\frac{x^{2}}{27}\) = 1
अनुप्रस्थ अक्ष, y-अक्ष के अनुदिश है
a² = 9, b² = 27
∴ c² = a² + b² = 9 + 27 = 36
∴ a = 3, b = 3\(\sqrt{3}\), c = 6
शीर्षों के निर्देशांक (0, ± a) या (0, ± 3)
नाभियों के निर्देशांक (0, ± c) या (0, ± 6)
उत्केंद्रता e = \(\frac{c}{a}=\frac{6}{3}=2\)
नाभिलंब जीवा की लंबाई = \(\frac{2 b^{2}}{a}=\frac{2 \times 27}{3}\) = 18.

प्रश्न 3.
9y² – 4x² = 36
हल:
अतिपरवलय का समीकरण 9y² – 4x² = 36
36 से भाग देने पर, \(\frac{y^{2}}{4}-\frac{x^{2}}{9}\) = 1
⇒ अनुप्रस्थ अक्ष, x-अक्ष के अनुदिश है।
∴ a² = 4, b² = 9
c² = a² + b² = 4 + 9 = 13
∴ a= 2, b = 3, c = \(\sqrt{13}\)
शीर्षों के निर्देशांक (0, ± a) या (0, ± 2)

प्रश्न 4.
16x² – 9y² = 576.
हल:
अतिपरवलय का समीकरण : 16x² – 9y² = 576
576 से भाग देने पर,

प्रश्न 5.
5y² – 9x² = 36.
हल:
अतिपरवलय का समीकरण : 5y² – 9x² = 36

प्रश्न 6.
49y² – 16x² = 784.
हल:
अतिपरवलय का समीकरण : 49y² – 16x² = 784
784 से भाग देने पर, \(\frac{y^{2}}{16}-\frac{x^{2}}{49}\) = 1
अनुप्रस्थ अक्ष, y-अक्ष के अनुदिश है।
a² = 16, b² = 49
∴ c² = a² + b² = 16 + 49 = 65
∴ a = 4, b = 7, c = \(\sqrt{65}\)
शीर्षों के निर्देशांक (0, ± a) या (0, ± 4).
नाभियों के निर्देशांक (0, ± c) या (0, ± \(\sqrt{65}\)).

निम्नलिखित प्रश्न 7 से 15 तक प्रत्येक में, दिए गए प्रतिबंधों को संतुष्ट करते हुए अतिपरवलय का समीकरण ज्ञात कीजिए।
प्रश्न 7.
शीर्ष (± 2, 0), नाभियाँ (± 3, 0).
हल:
अनुप्रस्थ अक्ष, x-अक्ष के अनुदिश है।
a = 2, c = 3, c² = a² + b²
या 9 = 4 + b²
∴ b² = 5

प्रश्न 8.
शीर्ष (0, ± 5), नाभियाँ (0, ± 8).
हल:
अनुप्रस्थ अक्ष, y-अक्ष के अनुदिश है।
a = 5, c = 8, c² = a² + b²
या 64 = 25 + b²
∴ b² = 64 – 25 = 39, a² = 25
अतिपरवलय का समीकरण : \(\frac{y^{2}}{25}-\frac{x^{2}}{39}\) = 1.

प्रश्न 9.
शीर्ष (0, ± 3), नाभियाँ (0, ± 5).
हल:
अनुप्रस्थ अक्ष, y-अक्ष के अनुदिश है।
शीर्ष (0, ± 3) ⇒ a = 3, a² = 9
नाभियाँ (0, ± 5) ⇒ c = 5
∴ c² = a² + b²
या 25 = 9 + b²
∴ b² = 16
∴ अतिपरवलय का समीकर: \(\frac{y^{2}}{9}-\frac{x^{2}}{16}\) = 1

प्रश्न 10.
नाभियाँ (± 5, 0), अनुप्रस्थ अक्ष की लम्बाई = 8.
हल:
अनुप्रस्थ अक्ष की लम्बाई = 2a = 8
⇒ a = 4
∴ a² = 16
नाभियाँ (± 5, 0)
⇒ c = 5, c² = a² + b²
या 25 = 16 + b²
∴ b² = 9
∴ अतिपरवलय का समीकरण : \(\frac{x^{2}}{16}-\frac{y^{2}}{9}\) = 1.

प्रश्न 11.
नाभियाँ (0, ± 13), संयुग्मी अक्ष की लम्बाई = 24.
हल:
नाभियाँ (0, ± 13)
⇒ अनुप्रस्थ अक्ष, y-अक्ष के अनुदिश है।
और C = 13, c² = 169
संयुग्मी अक्ष की लम्बाई, 2b = 24
∴ b = 12, b² = 144
c² = a² + b²
या 169 = a² + 144
∴ a² = 169 – 144 = 25
∴ अतिपरवलय का समीकरण \(\frac{y^{2}}{25}-\frac{x^{2}}{144}\) = 1.

प्रश्न 12.
नाभियाँ (± 3\(\sqrt{5}\), 0), नाभिलंब जीवा की लम्बाई = 8.
हल:

प्रश्न 13.
नाभियाँ (± 4,0), नाभिलंब जीवा की लम्बाई 12 है।
हल:
नाभियाँ (± 4,0)
⇒ अनुप्रस्थ अक्ष, x-अक्ष के अनुदिश है।
और c = 4 या c² = 16
या c² = a² + b², ∴ a² + b² = 16 …(1)
नाभिलंब जीवा की लम्बाई = \(\frac{2 b^{2}}{a}\) = 12
∴ b² = 6a …(2)
समी (1) और (2) से,
a² + 6a – 16 = 0
या (a + 8)(a – 2) = 0 .
a = – 8 या a = 2
परन्तु a ≠ – 8, ∴ a = 2, a² = 4
b² = 6a = 6 × 2 = 12
∴ अतिपरवलय का समीकरण, \(\frac{x^{2}}{4}-\frac{y^{2}}{12}\) = 1

प्रश्न 14.
शीर्ष (± 7, 0), e = \(\frac{4}{3}\).
हल:
शीर्ष (± 7, 0)
⇒ अनुप्रस्थ अक्ष, x-अक्ष के अनुदिश है।

∴ अतिपरवलय का समीकरण, \(\frac{x^{2}}{49}-\frac{y^{2}}{\frac{343}{9}}\) = 1
या \(\frac{x^{2}}{49}-\frac{9 y^{2}}{343}\) = 1.

प्रश्न 15.
नाभियाँ (0, ± \(\sqrt{10}\)) हैं तथा (2, 3) से होकर जाता है।
हल:
नाभियाँ (0, ± \(\sqrt{10}\))
⇒ अनुप्रस्थ अक्ष, y-अक्ष के अनुदिश है।
और c= \(\sqrt{10}\) या c² = 10 = a² + b²
∴ a² + b² = 10 …(1)
मान लीजिए अतिपरवलय का समीकरण

MP Board Class 11th Maths Solutions

MP Board Class 12th Biology Solutions Chapter 11 Biotechnology : Principles And Processes

Biotechnology : Principles And Processes NCERT Text Book Questions and Answers

Question 1.
Can you list 10 recombinant proteins which are used in medical practice? Find out where they are used as therapeutics (use the internet).
Answer:

Question 2.
Make a chart (with diagrammatic representation) showing a restriction enzyme, the substrate DNA on which it acts, the site at which it cuts DNA and the products it produces.
Answer:
The substrate DNA on which a restriction enzyme acts :

Question 3.
From what you have learnt, can you tell whether enzymes are bigger or DNA is bigger in molecular size ? How did you know ?
Answer:
DNA is bigger in molecular size than enzymes. Because DNA is a long double stranded molecule which can go up to a few meters in length when stretched end to end but enzymes although variable in size, would still be smaller than the DNA.

Question 4.
What would be the molar concentration of human DNA in a human cell? Consult your teacher.
Answer:
The average molecular weight of a nucleotide in human DNA is 130.86. The molecular weight of human DNA will therefore be 6 x 10⁹ nucleotides (based on the human genome project) x 130-86 = 784-56 x 109 gm/mol. The molar concentration of DNA can be calculated accordingly.
The molarity can be calculated as
Molar Concentration = \(\frac { No. of molecules }{ Molecular Weight }\)

Question 5.
Do eukaryotic cells have restriction endonucleases? Justify you answer.
Answer:
No. of Eukaryotic cells do not have restriction endonucleases. All the restriction endonucleases have been isolated from die various strains of bacteria and they are also turned according to the genus and species of prokaryotes. The first letter of the enzyme comes from the genus and the second two letters come from the species of the prokaryotic cell from which they have been isolated.

Question 6.
Besides better aeration and mixing properties, what other advantages do stirred tank bioreactors have over shake flasks?
Answer:
Shake flask are used for growing microbes and mixing the desired materials on a small scale in the laboratory. However, the large-scale production of a desired biotechnological product requires large stirred tank bioreactors.

Besides better aeration and mixing properties, bioreactors have the following advantages:

• It has an oxygen delivery system.
• It has a foam control, temperature and pH control system.
• Small volumes of culture can be withdrawn periodically.

Question 7.
Collect 5 examples of palindromic DNA sequences by consulting your teacher. Better try to create a palindromic sequence by following base-pair rules.
Answer:
Some palindromic DNA sequences and the restriction enzymes which act on them are:

Question 8.
Can you recall meiosis and indicate at what stage a recombinant DNA is made?
Answer:
A recombinant DNA is made in the pachytene stage of prophase I by crossing over during meosis cell division. Recombination nodules are visible in synaptonemal complex in pachytene sub stage. Crossing over occurs in this time between chromatids than recombinant DNA is formed.

Question 9.
Can you think and answer how a reporter enzyme can be used to monitor transformation of host cells by foreign DNA in addition to a selectable marker?
Answer:
Reporter enzyme can differentiate recombinants from non-recombinants on the basis of their ability to produce a specific colour in the presence of a chromogenic substrate. DNA is inserted within the coding sequence of the enzyme β-galactosidase. This result into inactivation of the enzyme which is referred to as insertional inactivation.

The presence of a chromogenic substrate gives blue-coloured colonies if the plasmid in the bacteria does not have an insert The presence of the insert results in insertional inactivation of β-galactosidase and the colonies do not produce any colour. These are identified as recombinant colonies.

Question 10.
Describe briefly the followings :
(a) Origin of replication
(b) Bioreactors
(c) Downstream processing.
Answer:
(a) Origin of replication : Origin of replication (ori) is a sequence on the chromosome, form where replication starts and any place of DNA when linked to this sequence can be made to replicate within the host cells.

This sequence also controls the copy number of the linked DNA.
So, if we want to recover many copies of the target DNA it should be linked to the ‘ori’ site and should be cloned in a vector whose origin supports high copy number.

(b) Bioreactors : Bioreactors are large vessels in which raw materials are biologically converted into specific products, individual enzymes, etc., using microbial, plant, animal or human cells.

A bioreactor provides the optimal conditions for achieving the desired production levels by providing optimum growth conditions of temperature, pH, substrate, salts, vitamins, oxygen, etc.

(c) Downstream processing : Downstream processes include separation and purification, formulation with suitable preservatives, etc, which are collectively referred to as downstream processing.

Such formulation has to undergo through clinical trials as in case of drugs.
Strict quality control testing for each product is also required. The downstream processing and quality control testing vary from product to product.

Question 11.
Explain briefly
(a) PCR
(b) Restriction enzymes and DNA
(c) Chitinase.
Answer:
(a) PCR : PCR stands for polymerase chain reaction; a method of amplifying fragments of DNA. This method can make multiple copies of even asingle DNA fragment or the gene of interest, in a test tube. The reaction mixture requires.

• Double-stranded DNA fragment (gene of interest).
• Primers-small chemically synthesized oligonucleotides that are complementary to the regions of this DNA.
• The special thermostable DNA polymerase (isolated from a bacterium, Thermus aquaticus), that does not denature and remain active even at high temperatures.

Unwinding of two strands of DNA by heating the sample at 92-94°C helps primers to get positioned on the exposed nucleotides as per base pairing rules. DNA polymerase recognizes primes as ‘start’ tags and begins to extend the primes using the free nucleotides provided in the reaction and the genomic DNA as template. With each round of reactions, the DNA doubles.

(b) Restriction enzymes and DNA : These enzymes are used in genetic engineering to cut the large DNA molecule into smaller fragments.

When DNA from two different sources are cut by the same restriction enzyme, the resultant DNA fragments have the same kind of ‘sticky-ends’ and these can be joined together (end-to end) using DNA ligases.

This new DNA created by joining fragments, from two different sources/genomes together is recombinant DNA.

(c) Chitinase : Chitinase is an enzyme that breaks down chitin, a component of fungal cell wall. It is useful for isolating the fungal cell DNA.

Question 12.
Discuss with your teacher and find out how to distinguish between :
(a) Plasmid DNA and Chromosomal DNA.
(b) RNA and DNA.
(c) Exonuclease and Endonuclease.
Answer:
(a) Distinguish between Plasmid DNA and Chromosomal DNA :

(b) Distinguish between DNA and RNA :

(c) Distinguish between Exonuclease and Endonucleases :
Exonucleases are nucleases which cut off the nucleotides from the 5′ or 3′ ends of a DNA molecule, whereas endonucleases are nucleases which cleave the DNA duplex at any point except at the ends.

Biotechnology : Principles And Processes Other Important Questions and Answers

Biotechnology : Principles And Processes Objective Type Questions

1. Choose the Correct Answers :

Question 1.
Exchange between genetic material in artificial is called:
(a) Gene recombination
(b) Gene transfer
(c) (a) and (b)
(d) None of these.
Answer:
(c) (a) and (b)

Question 2.
When was DNA recombinant technique discovered:
(a) In 1971
(b) In 1972
(c) In 1973
(d) In 1974.
Answer:
(b) In 1972

Question 3.
Which recombinant technique was given by H. Harris and J.F. Watkins:
(a) Transformation
(b) Transduction
(c) Cloning
(d) Protoplast recombination.
Answer:
(d) Protoplast recombination.

Question 4.
An artificial gene is made by scientists of California which capacity was:
(a) For making of Insulin
(b) For making of artificial gene
(c) For protection of pests
(d) For production of food nutrients.
Answer:
(a) For making of Insulin

Question 5.
The process in which a gene of interest is located and copied out of DNA extracted is called:
(a) Animal cloning
(b) Gene cloning
(c) DNA cloning
(d) RNA cloning.
Answer:
(b) Gene cloning

Question 6.
An organism, produced a sexually from one ancestor are called:
(a) Callus
(b) Ovum
(c) Clone
(d) None of these.
Answer:
(c) Clone

Question 7.
Which is used as molecular scissors in genetic engineering :
(a) DNA Polymerase
(b) DNA Ligase
(c) Helicase
(d) Restriction endonuclease.
Answer:
(d) Restriction endonuclease.

Question 8.
Transgenic expression of transgene in target tissue is determined by :
(a) Enhancer
(b) Reporter
(c) Promoter
(d) Transgene.
Answer:
(b) Reporter

Question 9.
First restriction endonuclease is :
(a) EcoRI
(b) Hind II
(c) Hind III
(d) Taq I.
Answer:
(b) Hind II

Question 10.
Cloning vector pBR³²² showing restriction site by which :
(a) Ampicillin
(b) Tetracycline
(c) Both (a) and (b)
(d) None of these.
Answer:
(c) Both (a) and (b)

Question 11.
DNA segment (T-DNA) of Agrobacterium tumefaciens is caused the disease in plant cells:
(a) Cancer
(b) Decomposition
(c) Turner
(d) None of these.
Answer:
(c) Turner

2. Fill in the Blanks :

1. ………………………. is the process by which information from a gene is used in the synthesis of a functional gene product.
2. Man made insulin is …………………………….
3. ……………………….. in RNA replaces thymine in DNA.
4. Formation of m-RNA from DNA is called …………………………
5. Gene that effects more than one character is called …………………………

Answers:

1. Gene expression
2. Humulin
3. Uracil
4. Transcription
5. Polygene.

3. Match the Following:

Answers:

1. (c)
2. (d)
3. (a)
4. (b).

4. Answer in One Word/Sentence :

1. Name the plant in which sequence of nucleotide in DNA was read for the first time.
2. Stickyudy of structural and functional aspects of genome.
3. Name the scientist who has discovered DNA finger- printing technique.
4. DNA having similar nucleotide sequence.
5. Organism in which gene of other organism is inserted.
6. Where does CCMB situated?
7. Name of the first cloned animal.
8. Medical system by which disturbed genes can be replaced by normal genes.
9. Biomolecule that destroy viruses and produce immunity in human beings.
10. Source of Ti plasmid.

Answers:

1. Arabidopsis
2. Genomics
3. Alec Jeffrey
4. Repetitive DNA
5. Transgenic
6. Hyderabad
7. Dolly
8. Gene therapy
9. Interferon
10. Agrobacterium tumefaciens.

Biotechnology : Principles And Processes Very Short Answer Type Questions

Question 1.
Which enzyme is used for isolation of target gene ?
Answer:
Restriction endonuclease is used for isolation of target gene.

Question. 2.
Which DNA polymerase is active in high tempereature ?
Answer:
Taq DNA polymerase is active in high temperature.

Question. 3.
Name three Restriction endonuclease enzyme.
Answer:

1. EcoRI
2. Hindi II
3. Hind III.

Question. 4.
Write full form of PCR. Which enzyme is used in ?
Answer:
Polymerase Chain Reaction (PCR).
Taq DNA is used in PCR.

Question. 5.
What is Bacteriophage ?
Answer:
Viruses they are infected to bacteria are called Bacteriophage.

Question. 6.
First recombinant DNA was formed in.
Answer:
In Bacteria Salmonella typhiurium.

Question. 7.
What is molecular scissors ?
Answer:
Restriction enzyme is called molecular scissors.

Question. 8.
Where does Hind II cuts the DNA molecule ?
Answer:
It cuts the DNA molecule at a specific 6 base pair seQuestionuence.

Question. 9.
What is the funtions of sticky ends ?
Answer:
It helps enzyme DNA ligase.

Question. 10.
Which type of charge found in DNA ?
Answer:
Negative charge.

Question. 11.
How does ethidium bromide cause DNA to fluorence in electrophoresis ?
Answer:
The most commonly used stain for detecting DNA is ethidium bromide.

Question. 12.
What is electrophoresis ?
Answer:
Electrophoresis is a techniQuestionue used in laboratories in order to separate micromolecules based on size.

Question. 13.
Give the function Of circular DNA which is found in bacterial cell.
Answer:
It is works as a vector.

Question. 14.
Name the technique in which we should be isolate DNA segment.
Answer:
Electrophoresis.

Question. 15.
Name two antibiotic restriction gene which found in plamid pBR³²²
Answer:
Ampicillin and Tetracycline.

Biotechnology : Principles And Processes Short Answer Type Questions

Question. 1.
What is genetic engineering ?
Answer:
It is the process of attaining the DNA in a organism’s genome. Genetic engineering is used by scientists to enhance the characterstics of an individual organism.

Question. 2.
Give any two purpose of genetic engineering.
Answer:

1. It is a set of technologies used to change the genetic make up of cells.
2. Transfer of genes within and across species boundaries to produce improved organisms.

Question. 3.
What is bacteriophage ?
Answer:
Bacteriophages are composed of proteins that encapsulate a DNA genome and may have relatively simple structure.

Question. 4.
What is restriction endonuclease ?
Answer:
An enzyme produced chiefly by certain bacteria, that has the property of cleaving DNA molecules at or near a specific sequence of basis.

Question. 5.
What is vector ?
Answer:
A vector is a quantity or phenomenon that has two independent properties. It used to transfer gentic material to a target cell.

Question. 6.
Write the four characters of vector.
Answer:

1. The plasmid DNA act as vectors to transfer the piece of DNA.
2. Have compatible restriction site for insertion of DNA molecule.
3. It should be capable of self reptication.
4. It is not dergadate in host cell.

Question. 7.
What is plasmid ?
Answer:
A plasmid is a small DNA molecule within a cell that is physically separated from a chromosomal DNA and can replicate independently.

Question. 8.
What is DNA ligase ?
Answer:
DNA ligase is a specific type of enzyme, a ligase that facilitates the joining of DNA strands togather by catalyzing the formation of a phosphodiesterbond.

Question. 9.
What is C-DNA?
Answer:
C-DNA is a single stranded RNA. C-DNA is often used to clone eukaryotic genes in prokaryotes.

Question. 10.
What is Ti-plasmid ?
Answer:
Ti or tumour inducing plasmid is a plasmid that is a part of the genetic equipment that Agrobacterium tumifaciens use to transduce their genetic material to plants.

Question. 11.
Name of the first clone of the world.
Answer:
Dolly was a female domestic sheep and the first mammal cloned.

Question. 12.
Name the scientist who discovered artificial DNA synthesizing method.
Answer:
The Nobel prize in physiology 1968 was awarded jointly to Robert W. Holley, Hargobind Khorana and M. Nirenberg for their interpretation of the genetic code.

Question. 13.
What is gene manipulation or genetic engineering ? Explain it.
Answer:
Genetic engineering, also called genetic modification is the direct manipulation of an organisms genes using biotechnology. It is a set of technologies used to change the genetic makeup of cells, including the transfer of genes and across species boundaries to produce improved organisms. New DNA is obtained by either isolating and copying the genetic material of interest using recombinant DNA methods or by artificially synthesising the DNA.

It may also mean extracting DNA from another organisms genome and combining it with the DNA of that individual. Genetic engineering is used by scientists to enhance or modify the characterstics of an individual organisms genetic engineering can be used to produce plants that have a higher nutritional value or can tolerate exposure to herbicides.

Question. 14.
Write application of genetic engineering to crop improvement.
Answer:
Genetic engineering has placed an important role in improvement of plant production. There are following applications of genetics in plant improvement.

• Production of polyploidy crops.
• Hybridization : Hybridization is used to produce plants with desirable traits.
• Transgenic plants : It helps a lot in improving the yield and Questionualify of crops.
• Insect and herbicides resistant plants are engineered.

Question. 15.
Write the uses of animal cloning.
Answer:
Applications of Animal cloning:

• By this techniQuestionue desired genotypes of any organism can be conserved.
• It produces organisms with better characters.
• Endangered plant and animal species can be conserved by this techniQuestionue.
• Animals with good Questionuality of milk and protein can be produced.

Question. 16.
Write applications of genetic engineering in medical field.
Answer:

• The hereditary diseases like colour-blindness, haemophilia which are caused by recessive genes therapy. .
• Substances like vitamins, hormones, amino acids and antibiotics can be synthesized in bacteria by introducing the genes which code these substances.
• Production of insulin: It is a medicine used for the treatment of diabetes. It is produced by gene splicing.
• Hepatitis – B vaccine: Hepatitis-B is a viral disease of liver, today this vaccine is prepared with the help of genetic engineering.

Question. 17.
What do you understand by gene bank? What are its significances?
Answer:
Gene bank : The institution which conserves the genes of the organisms is called as gene bank. The genetic material (DNA) found in the cells of organisms are conserved in gene banks. The best measure of conserving genes is to conserve endangered organisms. The tissues or cells of organisms is also conserved in gene banks.

Significance : Genes stored in gene bank are used for the production of improved varieties of species and for scientific tests.

Question. 18.
What do you understand by gene cloning ? What are its significance?
Or
What is gene cloning ? Write its importance.
Answer:
Gene cloning : It is a process in which DNA of an organism is cut into smaller DNA fragments by the use of restriction endonuclease enzymes. Each DNA fragment is introduced into a bacterial, yeast, insect, plant or animal cell. The cells are grown on a suitable medium under suitable conditions. Each ceil containing a particular DNA fragment multiplies to give rise a group of cells, all containing the same foreign DNA.

These groups of cells are known as clone of cells. These copies of DNA resulting from the multiplication or recombinant DNA are called as cloned DNA and the process is known as gene cloning.
Significance:

• Useful hereditary characters are obtained by this process.
• Many diseases are cured by this process.
• Many medicines are synthesized with the help of this process.
• This process should also be used in eugenics.

Biotechnology : Principles And Processes Long Answers Type Questions

Question. 1.
Describe the useful and harmful effects of genetic engineering.
Or
Describe the utility of genetic engineering.
Or
Describe the Genetic engineering. Write the importance of it in human life.
Answer:
The branch of molecular genetics in which we can manipulate or transplant the genes or the genetic material or DNA according to our will is called gene manipulation or genetic engineering. The main objective of genetic engineering is to synthesize recombinant DNA (formed of the DNA segments of two different organisms).
The useful and harmful effects of genetic engineering are as follows :
(A) Useful effects or Utility :

1. Industrial uses : Various types of substances such as vitamins, hormones and antibiotics can be synthesized in bacteria by introducing genes that code these substances. In this way, bacteria can function as living factories for the synthesis of these substances. Humulin (human insulin) is synthesized by this method.

2. Treatment of diseases : A new system of medicines, gene therapy may develop to cure several genetic disorders such as haemophilia, colour-blindness, etc. Also many inborn metabolic disorders due to defective genes such as alkaptonuria, phenylketonuria, etc. can be cured.

3. Use in agriculture : The genes for N₂ fixation found in symbiotic bacteria Rhizobium leguminosarum or blue-green algae may be transferred to the major food crops, increases food production without using expensive fertilizers. Thus, we can save millions of rupees spent otherwise on fertilizers and manures to boost food production.

4. Changes in the structure and expression of genes : We can obtain new plants, animals having traits tailored according to our will.

(B) Harmful effects:

• Normal harmless bacteria can be transformed into cancer causing forms thus ushering a new era of biological warfare.
• During experiments, it is Questionuite possible to obtain super viruses for which we might have no defence.
• By the use of recombinant DNA, the bacteria may be made resistant to antibiotics.

Question. 2.
Explain the mechanism of recombinant DNA technology in genetic engineering by using plasmid as carrier of genes.
Answer:
Mechanism of Recombinant DNA Technology: Mechanism of recom-binant DNA technology involves the following steps

1. Isolation of desired gene or fun¬ctional DNA segment : From the eukaryotic cell desired DNA segment is isolated with the help of enzyme restriction endo¬nuclease. Now this segment of DNA is known as foreign DNA.

2. Transfer of DNA segment from one organism to other : Plasmid is an extra chromosomal circular DNA found mostly in bacteria over and above the main genome. When bacteria multiplies the plasmid DNA also multiplies along with the chromosomal DNA. These plasmids can be easily isolated from the bacterial cell with the help of restriction endonu¬cleases. Plasmid serves as a vector for transferring the foreign DNA into a suitable recipient.

Foreign DNA and plasmid sliced with the help of endonucleases has free sticky ends through which they join each other with complementary base pairing with the help of enzyme DNA ligase. Thus, a recombinant DNA is formed,

Such a recombinant DNA when introduced into a recipient bacterium (transformation), it replicates and expresses itself, within the bacterial cell, the recombinant DNA molecule replicates along with the endogenous DNA of the host cell and produces copies of cloned DNA. This process is known as gene cloning. The cloned recombinant DNA produced in large Questionuantities can be isolated, purified and analysed.

Question. 3.
Describe the role of genentic engineering in artificial synthesis of human insulin.
Answer:

• Aldric and his supporter prepared two DNA seQuestionuences corresponding to the A and B chains of human insulin.
• Sticky ends were produced in the E.coli plasmid and the insulin gene by treating them both with the same restriction endonucleases.
• These two are then joined together by the enzyme DNA ligase.
• The bacteria are then grown in sterilised bioreactors in the appropriate growth medium.
• The chain A and B are produced separately, extracted and purified.

Question. 4.
What is clone ? Give its preparation, extracted and purified.
Answer:
An organism or cell or group of organisms, produced a sexually from an ancestor, to which they are genetical.

1. Gene cloning : Following steps are used by gene cloning:

1. Preparation of gene : DNA extracted from an organism, with the gene of interest is cut into gene size pieces with restriction enzyme.

2. Insertion into vector : Bacterial plasmid are cut with the same restriction enzyme. Plasmids are small circles of DNA in bacterial cells that are naturally present in addition to the bacterial other DNA.

3. Transformation of host cells : The recombinant plasmids are then transferred into bacteria using either electrophOration. The plasmid are small enough to pass through the holes into the cells. However rather than using electricity to create holes in the bacterium, it is done by alternating the temperature between hot and cold.

The bacteria is grown on a culture dish and allowed to grow into colonies. All the colonies on all the plates are called a gene library.

2. Plant cloning : Plant tissue culture is a method of propagation that has been sprouting in popularity throught as an alternative to cloning.

Plant can be cloned artificially using tissue culture. Vegetative propagation works because the end of the cutting forms a mass of non specialized cells called a callus, the callus will grow divide and form various specialized cells eventually forming a new plant.

MP Board Class 12th Biology Solutions

MP Board Class 12th Biology Solutions Chapter 6 Molecular Basis of Inheritance

Molecular Basis of Inheritance NCERT Textbook Questions and answers

Question 1.
Group of the following as nitrogenous bases and nucleosides:
Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.
Answer:
Adenine, Guanosine, Thymine, Uracil and Cytosine are nitrogenous bases. (Adenine and Guanosine → Purine, Thymine, Uracil and Cytosine → Pyrimidine) Cytidine is a nucleoside.

Question 2.
If a double stranded DN A has 20 % of cytosine, calculate the percent of adenine in the DNA.
Answer:
According to Chargaff’s rule, the DNA molecule should has an equal ratio
Cytosine = 20% therefore, Guanine = 20%
A+T = 100 – (G + C)
A + T = 100 – 40 since, both Adenine and Thymine are in equal amounts.
Thymine = Adenine = \(\frac { 60 }{ 2 }\) = 30%
So, quantity of Adenine is 30% in DNA helix.

Question 3.
If the sequence of one strand of DNA is written as follows:
5′ -ATG CATGCA TGC ATG CAT GCA TGCATGC – 3′
Write down the sequence of complementary strand in 5′-3’direction.
Answer:
In 3’→ 5′ direction, 3′ –
TACGTACGTACGTACGTACGTACGTACG-5′
In 5’→ 3′ 5′ direction, 5-
GCATGCATGCATGCATGCATGCATGCAT-3′.

Question 4.
If the sequence of the coding strand in a transcription unit is written as follows:
5’ – ATG CAT GCA TGC ATG CAT GCA TGC ATGC – 3’
Write down the sequence of mRNA.
Answer:
5’ – AUGCAUGCAUGC AUGCAUGCAUGCAUG-3’

Question 5.
Which property of DNA double helix led Watson and Crick to hypothesis semiconservative mode of DNA replication? Explain.
Answer:
The two strands of DNA show complementary base pairing. This property of DNA led Watson and Crick to suggest a semi-conservative mechanism of DNA replication in which one strand of parent is conserved while the other complementary strand formed is new.

Question 6.
Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acid synthesized from it (DNA or RNA) list the types of nucleic acid polymerases.
Answer:
These are two different types of nucleic acid polymerases :

1. DNA – dependent DNA polymerases
2. DNA – dependent RNA polymerases

The DNA dependent DNA polymerases use a DNA template for synthesizing a new strand of DNA, whereas DNA dependent RNA polymerases use a DNA template strand for synthesizing RNA.

Question 7.
How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material ?
Answer:
Hershey and Chase experiment :

• They grew some bacteriophages on a medium that contained radioactive phosphorus and some in another medium that contained radioactive sulphur.
• Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein as phosphorus is present only in DNA.
• Viruses grown on radioactive sulphur contained radioactive protein but not radioactive DNA because DNA does not contain sulphur.
• It was found that bacteria which were infected with bacteriophages that had radioactive DNA were radioactive, indicating that DNA was the material that passed from the virus to the bacteria.
• Bacteria that were infected with viruses that had radioactive proteins were not radioactive. This indicated that proteins did not enter the bacteria from the viruses.
  
  (vi) This was a clear cut proof that DNA is die genetic material that is passed from virus to bacteria.

Question 8.
Differentiate between the followings:
(a) Repetitive DNA and Satellite DNA
(b) mRNA and tRNA
(c) Template strand and Coding strand.
Answer:
(a) Differentiate between Repetitive DNA and Satellite DNA:

(b) Differentiate between mRNA and fRNA:


(c) Differentiate between Template strand and Coding strand:

Question 9.
List two essential roles of ribosome during translation.
Answer:
Two essential roles of ribosome during translation are:

1. One of the RNA acts as apeptidyl transferase ribozyme for formation of peptide bonds.
2. Ribosome provides sites for attachment of mRNA and charged tRNA for polypetide synthesis.

Question 10.
In the medium where E. coli was growing, lactose was added, which induced the lac operon. Then, why does lac operon shut down sometime after addition of lactose in the medium?
Answer:
Lactose regulates switching on and off of the lac operon. If lactose is provided in the growth medium of the bacteria, it is transported into the cells through the action of permease.

The lactose then induces the operon in the following manner :

• The repressor of the operon is synthesized all the time from the is gene.
• It binds the repressor protein which binds to the operator region of the operon and prevents RNA polymerase from transcribing the operon.

• In the presence of an inducer, such as lactose or allolactose, the repressor is inactivated by interaction with the inducer.
• This allows RNA polymerase access to the promoter and transcription proceeds.

Question 11.
Explain (in one or two lines) the function of the followings:
(a) Promoter
(b) tRNA
(c) Exons.
Answer:
(a) Promoter is an essential component of the transcription unit. It is located at the beginning of 5′ -end.
It provides a site for the attachment of transcription factors and RNA polymerase.
(b) tRNA is a small sized RNA molecule that takes part in transcription. It physically picks up activated amino acids from the cytoplasm and carries (transfers) them to ribosomes, where they join together through peptide bonds and leave the tRNA to fetch more amino acids.
(c) Exons are the coding sequences of DNA that are transcribed and translated.

Question 12.
Why is the human genome project called a mega project ?
Answer:
Human genome project is called a mega project because :

• Its aim was to determine the nucleotide sequence of complete human genome which was a task of enormous magnitude.
• A total of 3 x 109 base pairs were to be sequenced and the cost was about 9 billion US dollars.
• It required bioinformatics data base techniques and other contemporary devices for the analysis, storage and retrieval of information.
• May countries worked jointly to complete this timed project.

Question 13.
What is DNA Fingerprinting ? Mention its application.
Answer:
DNA Fingerprinting : Every human individual is characterised by unique print at the fingertips. The study of fingers, palm and sole print is called dermatogly phics’.

Like prints of the fingertips, each individual has unique DNA fingerprint. Unlike the prints of finger, the DNA fingerprints can not be altered by surgery. The latter is exactly similar in all the cells and tissues of an individual. It can not be changed by medical treatment. The distinction of individuals on the basis of DNA fingerprint is due to sequence of nucleotides in whole genomic DNA. The technique to identify a person on the basis of his/her DNA specificity is called DNA fingerprinting. This was invented by Sir Alec Jeffreys in 1984 at Leicester University U.K. In India, Dr. V. K. Kashyap and Dr. Lalji Singh started this technique at CCMB, Hyderabad.

DNA fingerprinting involves following steps:

The DNA of the organism to be tested is isolated, it is called host DNA.
Host DNA is cleaved with the help of specific restriction enzymes into several fragments.
Double stranded DNA fragments are denatured to produce single stranded DNA by alkali treatment.
DNA segments are separated by electrophoresis.

Question 14.
Briefly describe the following:
(a) Transcription
(b) Polymorphism
(c) Translation
(d) Bioinformatics.
Answer:
(a) Transcription : It is the formation of RNA over the template of DNA. It forms single-stranded RNA which has a coded information similar to the sense or coding strand of DNA with the exception that thymine is replaced by uracil. One strand of DNA is used as template strand for the synthesis of a complementary strand of RNA called wRNA.

(b) Polymorphism : Genetic polymorphism means occurrence of genetic material in more than one form. It is of three major types, i. e„ allelic, SNP and RFLP.

Allelic polymorphism : Allelic polymorphism occurs due to multiple alleles of a gene. Allele possess different mutations which alter the structure and function of a protein formed by them as a result, change in phenotype may occur.

SNP or single nucleotide polymorphism : Over 1-4 million single base DNA differences have been observed in human beings. According to SNP, every human being is unique. SNP is very useful for locating alleles, identifying disease-associated sequence and tracing human history.

(c) Translation : It is the process during which the genetic information which is stored in the sequence of nucleotides in an mRNA molecule is converted, following dictations of the genetic code, into the sequence of amino acids in the polypeptide. It takes place in cytoplasm in both eukaryotes and prokaryotes.

(d) Bioinformatics : The science which deals with handling storing of huge information of genomics as databases, analysing, modeling and providing various aspects of biological information, especially the molecules connected with genomics and proteomics is called bioinformatics.

Molecular Basis of Inheritance Other Important Questions

Molecular Basis of Inheritance Objective Type Questions

1. Choose the Correct Answer:

Question 1.
Which component they have free amino acid and carboxylic group:
(a) Glucose
(b) Nucleotide
(c) Amino acid
(d) None of these.
Answer:
(c) Amino acid

Question 2.
Nucleotide which are participate in energy transfer:
(a) NAD
(b) FAD
(c) FMN
(d) ATP.
Answer:
(d) ATP.

Question 3.
What is the unit of protein:
(a) Fatty acid
(b) Monosaccharides
(c) Amino acid
(d) Glycerol.
Answer:
(c) Amino acid

Question 4.
Nucleic acids are polymers of:
(a) Amino acid
(b) Nucleoside
(c) Nucleotide
(d) Globulin.
Answer:
(c) Nucleotide

Question 5.
Peptide bonds are found in:
(a) Protein
(b) Fat
(c) Nucleic acid
(d) Carbohydrate.
Answer:
(a) Protein

Question 6.
Glycosidic bonds are found in:
(a) Nucleic acid
(b) Protein
(c) Polysaccharides
(c) Monosaccharides.
Answer:
(c) Monosaccharides.

Question 7.
By which the control and coordinate of heredity:
(a) By DNA
(b) By RNA
(c) Mostly DNA but some organisms by RNA
(d) None of these.
Answer:
(c) Mostly DNA but some organisms by RNA

Question 8.
It is not a protein:
(a) Myosin
(b)Actin
(c) Haematind
(d) Albumin.
Answer:
(c) Haematind

Question 9.
A source which give immediate energy:
(a) Glucose
(b) NADH
(c) AIP
(d) Pyruvic acid.
Answer:
(c) AIP

Question 10.
Who discovered ATP:
(a) Karl Lohmann
(b)Lipman
(c) Bowman
(d) Blackman.
Answer:
(a) Karl Lohmann

Question 11.
Which nitrogenous base is found in only RNA:
(a) Cytosine
(b) Adenine
(c) Uracil
(d) Guanine.
Answer:
(c) Uracil

Question 12.
Molecule which shows very difference from other molecules in the cell:
(a) Mineral salt
(b) Lipids
(c) Proteins
(d) Carbohydrate.
Answer:
(c) Proteins

Question 13.
Who prescribe the double helix structure of DNA:
(a) Nirenberg
(b) Komberg
(c) Holley and Nirenberg
(d) Watson and Crick.
Answer:
(d) Watson and Crick.

Question 14.
Which is non-essential for the plants:
(a) Ca
(b) Zn
(c) Cu
(b) Na.
Answer:
(b) Na.

Question 15.
Number of nucleotides are found in one whorl of DNA:
(a) 9
(b) 10
(c) 11
(d) 12.
Answer:
(b) 10

Question 16.
Which are microelements:
(a) Ca
(b) N
(c) Mg
(d) Mn.
Answer:
(d) Mn.

Question 17.
Which simillarities is found in DNA and RNA:
(a) Same type of pyrimidine is found in both
(b) Thymine is present in both
(c) Some sugar is found in both
(d) Both are polymers of nucleotide.
Answer:
(d) Both are polymers of nucleotide.

Question 18.
Cholesterol is a:
(a) Simple lipids
(b) Complex lipids
(c) Derivatives lipids
(d) Protein.
Answer:
(c) Derivatives lipids

2. Fill in the Blanks:

1. Cut up and join of polynucleoid chain is called …………………….
2. Changes ……………………. done by genetic engineering.
3. UAA, UAG and UGA are …………………….codon.
4. …………………….code is universal and non-ambiguous.
5. Transcription of DNA information is the form of …………………….
6. ……………………. enzyme is involved in transcription.

Answer:

1. Genetic engineering
2. Characters of organisms
3. Termination
4. Codon
5. mRNA
6. RNA Polymerase.

3. Match the Following:

I.

Answer:

1. (c)
2. (e)
3. (b)
4. (a)
5. (d)
6. (f)

II.

Answer:

1. (c)
2. (d)
3. (a)
4. (b).

III.

Answer:

1. (c)
2. (a)
3. (b)
4. (e)
5. (d).

4. Answer in One Word/Sentence:

1. Who discovered nucleic acid for the first time ?
2. Name the Indian-American scientist who is well known for chemical synthesis of gene.
3. Name the small-RNA which is synthesized before DNA replication.
4. Which codon is known as the starting codon ?
5. Name the RNA which function as enzyme.
6. Who proposed the operon model of gene regulation ?
7. What chemical is produced by regulator gene of Lac-operon ?
8. What term is used for those genes which are active in all the cells and tissues of an individual ?
9. Name the operon which regulate the metabolism of lactose in E. coli.
10. Name the organism which contain single stranded DNA.

Answer:

1. Friedrich Miescher
2. H.G. Khorana
3. RNA-primer
4. AUG or GUG
5. Ribozyme
6. Jacob and Monod
7. Repressor
8. Housekeeping genes
9. Lac-operon
10. Φ × 174phage.

Molecular Basis of Inheritance Very Short Answer Type Questions

Question 1.
Structure formed by regulation + structural + operator+ promoter gene.
Answer:
Operon.

Question 2.
What are the animals that have a foreign gene deliberately inserted into their genome ?
Answer:
Transgenic animals.

Question 3.
What are the group of cells or organisms which have same hereditary characters ?
Answer:
Clone.

Question 4.
By which the instructions of our DNA are converted into a functional product ?
Answer:
Gene expression.

Question 5.
Write the name of sugar found in RNA.
Answer:
Ribose sugar.

Question 6.
Which codon is AUG?
Answer:
Anticodon.

Question 7.
Name the enzyme which takes part in transcription.
Answer:
RNA Polymerase.

Question 8.
Who tell that DNA is a heredity material ?
Answer:
Alfred Hershey and Martha Chase. ,

Question 9.
Which bond is make in DNA when join the sugar and phosphoric acid ?
Answer:
Phosphodiester bond.

Question 10.
Name the segment in which any nucleotide sequence within a gene that is removed by RNA splicing during maturation of the final RNA products.
Answer:
Intron.

Question 11.
Who gave the Operon model ?
Answer:
Jacob and Monod.

Question 12.
What do you mean by commaless due to genetic code ?
Answer:
Between two codon has no internal punctuation.

Question 13.
Write the full name of Sn RNP.
Answer:
Small nuclear Ribonucleo Proteins.

Molecular Basis of Inheritance Short Answer Type Questions

Question 1.
What is genetic code ? What do you know about the discovery of genetic code ?
Answer:
Genetic code is that sequence of three nitrogenous bases of mRNA in which genetic information for the synthesis of one amino acid is coded.

The triplet codons of the genetic codes are discovered for the first time by M.W. Nirenberg in 1950. He synthesized a RNA by the use of repetitive sequence of Uracil which is called as polyuracil (UUUUU …………). They added synthesized mRNA to a cell free extract containing protein synthesizing enzymes and ribosome from E. coli together with a mixture of 20 amino acids. The only molecules synthesized in a polypeptide chain were phenyl alanine and their number in chain was one third of the Uracil base on poly-U-m-RNA. This confirmed triplet nature of genetic code.

Question 2.
What are oncogenes ?
Answer:
Genes which are responsible for production of cancer in host by uncontrolled mitotic cell division are called as oncogenes.

Question 3.
What are Okazaki fragments and leadings strands?
Answer:
Okazaki fragments : On second parental DNA template new complementary DNA strands are formed in smaller fragments starting from RNA primer. These short fragments are called Okazaki fragments.
Leading strands : Second strand is formed on 5’→ 3’ strand of parental DNA in a continuous stretch in reverse direction 3’→ 5’ and is called as leading strand.

Question 4.
DNA nucleotide is formed which molecule ?
Answer:
Components are DNA Nucleotides:

Question 5.
Explain the Watson and Crick model of DNA.
Answer:
The structure of DNA was proposed by Watson and Crick. It is twisted ladder like structure. It has got two coiled polynucleotides which are joined together by nitrogen bases with hydrogen bond in the centre. The longitudinal strands of DNA are made of sugars and phosphates of nucleotides. The horizontally placed nitrogen bases are of two types, purine and pyrimidine. Purines are adenine and guanine whereas pyrimidines are cytosine and thymine.

Question 6.
What is RNA primer ? Why it is necessary from DNA synthesis ?
Answer:
A primer is a short strand of RNA or DNA. RNA primer serves as a starting point for DNA polymerase, which builds complementary DNA. It is reqiured for DNA replication because the enzymes that catalyze this process. DNA polymerase can only add new nucleotide to an existing strand of DNA.

Question 7.
What is peptide bond ?
Answer:
The bond formed between the carboxylic group (- COOH) of one amino acid and amino group (- NH₂) of another amino acid is called as peptide bond. A molecule of water is released during the formation of peptide bond.

Question 8.
Define Codon and Anticodon.
Answer:
Codon: A specific sequence of three consecutive nucleotides that is a part of the genetic code and that specifies a paticular amino acid in a protein or starts or stops protein synthesis e.g., AUG codon which is situated on die mRNA, code methionine amino acid.

Anticodon : A sequence of three adjacent nucleotides located on one end of transfer RNA. It bounds to the complementary coding triplet of nucleotides in mRNA during translation phase of protein synthesis.
For example, the anticodon for Glycine is ccc that binds to the codon (which is GGE) of mRNA.

Molecular Basis of Inheritance Long Answer Type Questions

Question 1.
Explain DNA duplication in short
Answer:
Watson and Crick after giving the double helix model of DNA, also postulated the mechanism of DNA duplication, also known as replication. According to them, during duplication, the weak hydrogen bonds between the nitrogenous base of the nucleotides get separated, so that two polynucleotide chains of DNA also separate and uncoil. The chains thus, separated are complementary to one another. These strands act as template and because of the specificity of base pairing each nucleotide of separated chain attracts its complementary nucleotide from the cell cyto-plasm. Once the nucleotides are attached by their hydrogen bonds their sugar radicals write through their Fig. DNA duplication, phosphate components completing the formation of a new polynucleotide chain. This results in the formation of two double helixes of DNA where in each molecule has one old strand contributed by parent DNA and one synthesized new. This method of DNA duplication is known as semi-conservative method.

Question 2.
Describe the functions of nucleic acids.
Or
Explain the utility of nucleic acids.
Answer:
Utility of Nucleic acids:

• Nucleic acids are the hereditary materials of organisms which involve in the transfer of hereditary characters
• from one generation to the next.
• DNA controls the synthesis of enzymes which control the various activities of the body.
• Nucleic acids also control protein synthesis.
• Nucleic acids form maximum portion of chromatin network.
• It causes mutation in living beings.
• They form enzymes.

Question 3.
Explain the structure of RNA.
Answer:
RNA molecules are single stranded nucleic acids composed of nucleotides. Four types of bases are present in RNA. These nitrogenous bases joint in different manner and form the ribonucleoside. Ribonucleoside joins together and make a polyribonucleotide chain.
AH four types of nucleoside and nucleotide are as follows :

Question 4.
Elaborate the term RNA. Also describe the types and functions of RNA.
Or
Write location and kinds of RNA in the cell.
Answer:
RNA(Ribonucleic acid): RNA located in the nucleus, cytoplasm, ribosomes and in some other cell organelles.
Types of RNA and their functions: RNA are of three types:

1. Messenger RNA (mRNA): It makes a small fraction 5-10%. This RNAdirects the sequence of amino acids in protein synthesis after joining with ribosomes. It carries the genetic information contained in DNA. It is short lived and has rapid turnover. It is formed of 700-1500 nucleotides and has a molecular weight from 5,00,000 to 20,00,000. The sequence of three nitrogenous bases of mRNA forms a codon which is responsible for coding of one amino acid.

2. Ribosomal RNA (rRNA): It makes 80% of total cell RNA. It is the most stable type of RNA and is associated with ribosomes.

3. Transfer RNA (tRNA): It makes a small fraction (10-15%) of RNA. These are smallest molecules formed of 73-93 nucleotides with molecular weight ranging between 25,000 to 30,000. rRNA works as adaptor molecules for carrying amino acids to the mRNA template during protein synthesis.

Question 5.
Describe the structure of nucleotides.
Answer:
Structure of Nucleotides : Nucleotides are the basic unit of nucleic acids and, they are also involved in the energy transfer reactions of the body. A nucleotide consists of a nitrogenous base, a pentose sugar and a phosphate group. A nucleotide possessing two groups of nitrogenous bases :

1. Purines and
2. Pyrimidines.

Purines are double ringed nitrogenous bases, e.g., Adenine and Guanine. Pyrimidines are single ring nitrogenous bases like cytosine, thymine and uracil.

The pentose sugar of nucleotides is also of two types :

(i) Ribose sugar [CH₂OH(CHOH)₂.CHOH.CHO].
(ii) Deoxyribose sugar [CH₂OH(CHOH₂.CH₂CHO].

A nucleotide may have one, two or three phosphates to form a nucleotides. A combination of a base and a sugar is called nucleoside and combination of a base, a sugar and phosphate group is known as nucleotide. As there are five bases so, five different kinds of nucleosides and nucleotides are known. These are listed in table given below :

Question 6.
Give the functions of nucleotides.
Answer:
Functions of Nucleotides :

• It works as a activated precursors of DNA and RNA.
• They are perform the storage and conduction of energy to the form of ATP.
• It required for activation of intermediates in many biosynthetic pathway.
• It works as Carrier of methyl group in the form of SAM.
• It’s Components of co-enzyme: NAD, FAD and Co A.
• Some functions are as a vitamin. ,
• They are control and coordinates different activities in our body.

Question 7.
Write four features of genetic code.
Answer:
According to Nirenberg, Khorana and Holley, genetic code is that sequence of nitrogenous bases of DNA in which genetic informations for the synthesis of protein are coded.

Characteristic features of genetic code :

1. The code is triplet: The codon is a specific sequence of three nitrogenous bases of mRNA.

2. The code is commaless: The sequence of bases read in blocks of three at a time form a particular position. There is no gap between two subsequent codons.

3. Code is degenerating: Presence of more than one codon for one amino acid is called as degeneracy of codons, e.g., Serine having three codons UCU, UCA, AGU.

4. Codes are universal: Codons are similar in all organisms, e.g., serine is coded by UCU codon in all the living beings.

5. Codes are non-ambiguous : The position of genetic code in cellular medium is nonambiguous because a codon always codes only one amino acid. Sometimes a codon codes more than one amino acid, e.g., in E. coli. UUV codon generally code phenylalanine, after treatment of their ribosome with streptomycin. It can also code isoleucine, leucine and serine.

6. Initiation and termination codon: Codons responsible for the initiation of polypeptide chain are called as initiation codon, e.g., AUG. Likewise, codons responsible for the termination of polypeptide chain are called as chain termination codon, e.g., UAA, UAG, UGA.

Question 8.
Write five characters of gene hypothesis.
Answer:
Sutton, Bridges, Muller and Morgan suggest these theory. The character’s of gene of this theory are as follows :

• Genes are situated on the chromosome.
• They make the physiological character of organisms.
• These are called functional unit of specific characters.
• Genes have the capacity of self-transcription.
• They perform mutation.
• Characters are goes to one generation to other by parents.

Question 9.
What is gene expression ? Explain by different methods of gene expression in animals.
Answer:
The mechanism at molecular level by which a gene is able to express itself in the phenotype of an organism is called gene expression.
Different methods of gene expression in animals are:

1. Transduction : It is the process in which bacteriophages pick up pieces of DNA from one bacterial cell and transfer the same to another on infection.
2. Transformation : It is the process by which DNA isolated from one type of cell when introduced into another, is able to bestow some of the properties of the former to the latter.

Question 10.
What is proof reading and repair of DNA ?
Answer:
Variety of environmental factors such as radiation, chemicals etc. may cause damage in DNA of a cell. The bacterial DNA polymerase III can do proofreading, in the sense that it can go back and remove the wrong base before it proceeds to add new bases in the 5′ → 3′ direction. It is called proof reading. Obviously, the survival of the cell depends on its availability of damages:

(i) Monoadduct: Which involve alterations in a single nitrogenous base.
(ii) Diadducts : They are the alterations involving more than one nitrogenous base. Number of nucleases have been found to be involved in repair replication such as Exonucleases (defined as phosphodiesterases which require a terminus for hydrolysis and cut off terminal nucleotides), Endonucleases (which are also phosphodiesterases which do not require a terminus for hydrolysis and break internal bonds). The endonucleases which act on the damaged DNA and cause repair or correction of this molecule are referred to as correctional nucleases. The following steps are said to be involved in the repair repl’iGatipn i.e., Incision, Excision, Reinsertion and joining of newly formed strands.

Question 11.
Write any four differences between DNA and RNA.
Answer:
Differences between DNA and RNA:

Question 12.
Write the names of enzymes used in DNA replication.
Ans.
The names of enzymes used in DNA replication are as follows :

• DNA helicase : For unwinding of two strands.
• DNA gyrase : For relieving tension.
• Primase : For formation of primer.
• DNA polymerase : For DNA synthesis.
• RNA primer : For initiation of the synthesis of DNA segments.
• DNA ligase : For joining of DNA Okazaki segments.

Question 13.
What is transcription ? Name the enzyme catalysing it
Answer:
Transcription : Formation of mRNA from DNA in the presence of enzyme is called transcription. It is the first stage of protein synthesis which is catalysed, by RNA polymerase enzyme. The process of transcription involves in the following steps :

1. Exposing of the bases of DNA : The two strands of DNA are separated due to presence of an unwinding protein and thus, their bases are exposed. The exposed chain of DNA functions as template for the synthesis of mRNA in the presence of RNA polymerase enzyme.

2. Base pairing: The ribonucleotides are jointed in a definite fashion on the exposed strand of DNA. G is bonded with ‘C’, ‘C’ bonded with ‘G’, ‘T’ bonded with ‘A’ and ‘A’ bonded with‘‘T’respectively.

3. Synthesis of RNA chain : The new ribonucleotide bonded on DNA template are jointed with the help of RNA polymerase and thus, forming a new chain of RNA. Then this mRNA is separated from DNA and reaches the cytoplasm. Where it combines with ribosomes and thus, initiating the synthesis of protein.

Question 14.
What is translation ? Explain it
Answer:
Translation : The translation step of protein synthesis involves translation of the language of nucleic acids (available in the form of mRNA) into language of protein. The sequence of bases in mRNA, decides the sequence of amino acids in proteins. Each amino acid is programmed by a triplet code. It consists of a sequence of three bases in the DNA and the complementary bases in mRNA. The synthesis of protein occurs in three steps, initiation, elongation and termination. After the final step i.e., termination, the proteins are transported out of the cell or translocated within the cell. Thus, the transformation of nucleotides chain of RNA into polypeptide chain of protein is called as translation.

It is completed in following steps :

1. Activation of amino acids.
2. Binding of activated amino acids with rRNA.
3. Binding of mRNA with smaller unit of ribosome.
4. Initiation of polypeptide chain.
5. Elongation of polypeptide chain.
6. Termination of polypeptide chain.

Question 15.
Describe the evidence given by Griffith in support of DNA as genetic material. Explain it along with suitable diagram.
Answer:
Griffith had done transformation experiments in mice to prove that DNA is the genetic material. He took virulent strain of Diplococcus pneumonae (S-III) which causes pneumonia in mice and injected it into mice which resulted in the production of pneumonia in mice. He also injected a non-virulent strain of that bacteria in the body of mice and found that all the mice were unaffected.

In third experiment he injected heat killed (S-III) strain and non-virulent strain R-II strain together in the body of mice and found that all the mice suffered from pneumonia and became dead. After analysis it was found that these mice contained both the strains of Diplococcus pneumonae. Thus, this experiment proved that any substance of S-III strain is transferred into R-II strain due to which R-II strain become virulent. Later, McLeod, Avery and McCarty observed that DNA molecules are transferred from S-III to R-II strain and make virulent. Thus, it is proved that DNA is the genetic material.

MP Board Class 12th Biology Solutions

MP Board Class 12th Biology Solutions Chapter 5 Principles of Inheritance and Variation

Principles of Inheritance and Variation NCERT Textbook Questions and Answers

Question 1.
Mention the advantage of selecting pea plant for experiment by MendeL
Answer:
Advantages of selecting pea plant:

• Pea plant showed visible contrasting characters, e.g., seed colour, texture, plant height, flower colour, etc, it was easy to track the passing on of these characters in the progeny.
• The pea flower remains closed and stamens and carpel are both present on the same flower. Therefore, it undergoes self-pollination and breeds true for traits (no mixing of gene pool).
• It was easy to manipulate cross-pollination.
• Many offspring’s produced in one generation.
• Pea plant has short life cycle.
• It is easy to grow.

Question 2.
Differentiate between the foifowing:
(a) Dominance and Recessive
(b) Homozygous and Heterozygous
(c) Monohybrid andDihybrid.
Answer:
(a) Dominance and Recessive:
When a cross is made between single pair of contrasting characters then the character one which expresses itself phenotypically in F₁ generation is known as dominant whereas other allele or trait which fails to express itself is recessive, e.g., when cross is made between two plants of pea, one having pure red flowers (RR) and other having pure white flowers (rr). In F₁ generation only red-flowered has appeared.

(b) Differences between Homozygous and Heterozygous:

(c) Differences between Monohybrid cross and Dihybrid cross:

Question 3.
A diploid organism is heterozygous for 4 loci, how many types of gametes can be produced?
Answer:
Here, we apply the formula 2ⁿ where, n = number of loci. The organism is heterozygous for 4 loci
n=4
So, 2ⁿ = 2⁴ = 2 × 2 × 2 × 2 = 16
The organism will produce 16 types of gametes.

Question 4.
Explain the law of Dominance using a monohybrid cross.
Answer:

Monohybrid cross : Crossing between single pair of contrasting character, is known as monohybrid cross, e.g., hybridization between tall and dwarf plant of pea. All offspring of F₁ generation of this cross will be tall. In F₂ generation tall, hybrid tall and dwarf plants are produced in the ratio of 1:2:1. Thus, the genotypic ratio will be 1 TT: 2 Tt: 1 tt (1:2:1). The phenotypic ratio of F₂ generation will be 3 : 1 (3TTand 1tt).

Question 5.
Define and design a test-cross.
Answer:
Crossing of F₁ individual having dominant phenotype with its homozygous recessive parent is called test cross. The test cross is used to determine whether the individuals exhibiting dominant character are homozygous or heterozygous.

Question 6.
Using a punnett square, workout the distribution of phenotypic features in the first filial generation after a cross between a homozygous female and a heterozygous male for a single locus. .
Answer:
A cross between a homozygous female and heterozygous male follow two conditions :

Condition (1) : Homozygous female – TT (Tall)
Heterozygous male – Tt (Tall)

Condition (2) : Homozygous female – tt (dwarf)
Heterozygous male – Tt (Tall)

Question 7.
When a cross is made between tall plant with yellow seed (TtYy) and tall plant with green seed (Ttyy), what proportion of phenotype in the offspring could be expected to be:
(a) Tall and green
(b) Dwarf and green.
Answer:

Question 8.
Two heterozygous parents are crossed. If the two loci are linked what would be the distribution of phenotypic features in F₁ generation for a dihybrid cross.
Answer:
In case two heterozygous parents, showing linkage result will be :
Parents BbLl × Bb Ll
Genotype : Blue long Blue long
Phenotype In F₁ all the combinations may show parental characters as the genes are completely linked. With all possible genotypes in Fj progeny may exhibit blue long type of phenotype in above case. However in case of incomplete linkage parental combinations will be less in number.

Question 9.
Briefly mention the contribution of T.H. Morgan in genetics.
Answer:
T.H. Morgan (1866-1945) was given the Nobel prize. His contributions are:

• Morgan worked on fruit fly Drosophila melanogaster and proposed the chromsomal theory of linkage.
• He stated and established that genes are located on the chromosome.
• He established the principle of linkage, crossing over, sex-linked inheritance and discovered the relationship between gene and chromosome.
• He established the technique of chromosome mapping.
• He observed and worked on mutation.

Question 10.
What is pedigree analysis ? Suggest how such an analysis can be useful.
Answer:
1. Pedigree analysis is an important method to study human genetics because :

• Human beings cannot be crossed at will.
• The generation time is very long (about 20 years).
• The number of offspring produced is small.

2. For pedigree analysis, information about the family’s history for a particular trait is first collected.
3. The expression of the trait is constructed in the form of a family tree.
4. In a pedigree, by convention :

• Circles denote females.
• Squares denote males.
• Solid symbols represent the trait being studied.
• Open symbols denote normal form.
• Symbol with a cross line indicates a carrier for a recessive trait.
• Parents are joined by horizontal lines.
• The offsprings are connected to a horizontal line below the parents in the order of birth and the line is
• connected to the parental line by a vertical line.

5. Pedigree analysis can yield valuable information about the possible genetic make up of a person for a trait.

Functions : From pedigree analysis, we can find out, what kind of characters a new born baby is having. If this practice is done before marriage, we can be safe from many problems. It helps in finding the solution of hereditary problems.

Question 11.
How is sex-determined in human beings?
Answer:
Each cell of human beings contains 23 pairs of chromosomes. Out of 23 pairs, 22 pairs are similar and called as autosomes whereas 23^rd pair is different from autosomes which are called as sex-chromosomes because they play important role in the determination of sex. In the cells of man, there are two types of sex chromosomes (XY) present while the cells of women have two similar sex chromosomes (XX). After spermatogenesis two types of sperms are formed, (i) One having 22 + X chromosomes and (ii) Other having 22 + Y chromosomes. But in woman there is only one type of ovum formed which contains 22 + X chromosomes only. At the time of fertilization when a sperm having 22 + X chromosomes is fused with ovum they produce a female child having two ‘XX’ chromosomes. But when a sperm having 22 + Y chromosomes is fused with ovum then resulting offspring will contain ‘XY’ sex chromosomes hence, it is male child.

Question 12.
A child has blood group ‘O’. If the father has blood group ‘A’ and mother blood group ‘B’. Workout the genotypes of the parents and the possible genotypes of the other off springs.
Answer:

Question 13.
Explain the following terms with example:
(a) Co-dominance
(b) Incomplete dominance.
Answer:
(a) Co-dominance : When the F₁ generation resembles both the parents and both the parental characters are expressed simultaneously, then a phenomenon is called co-dominance.
For example: ‘AB’ type blood group is possible when allele ‘A’ and ‘B’ come together and since, both the alleles are expressing their effects in F₁ generation, they are co-dominants.

(b) Incomplete dominance : It is a condition which occurs between dominant alleles in which one may be slightly more dominant than the other.
Example : As in Mirabilis jalapa if a red flowered plant is crossed with white flowered plant, then the resulting F₁ hybrid bears pink coloured flowers only. This intermediate inheritance is called incomplete dominance. The F₁ pink flowered plants when self-pollinated, give F₂ progeny which contains red, pink and white flowers in the ratio of 1 : 2 : 1. The phenotypic and genotypic ratio is same in case of incomplete dominance.

Question 14.
What is point mutation ? Give one example.
Answer:
Point mutation is a gene mutation that arises due to change in a single base pair of DNA.
Example : Sickle-cell anaemia.
A substitution of a single nitrogen base at the sixth codon of the β- globin chain of haemoglobin molecule causes the change in the shape of the R.B.C. from biconcave disc to elongated shaped, structure which results is sickle cell anaemia.

Question 15.
Who has proposed the chromosomal theory of inheritance?
Answer:
In 1902 Walter Sutton and Theodor Boveri proposed the ‘Chromosomal theory of inheritance’.

Question 16.
Mention any two autosomal genetic disorders with their symptoms.
Answer:
Phenylketonuria : This inborn error of metabolism is also inherited as the autosomal recessive trait. The affected individual lacks an enzme that converts the amino acid phenylalanine into tyrosine. As a result, this phenylalanine is accumulated and converted into phenyl pyruvic acid and other derivatives. Accumulation of these in brain results in mental retardation.

Sickle cell anaemia : This is an autosome liked recessive trait that can be transmitted from parents to the offspring when both the partners are carrier for the gene of heterozygous.

Principles of Inheritance and Variation Other Important Questions and Answers

Principles of Inheritance and Variation Objective Type Questions

Question
1. Choose the Correct Answer
Question 1.
What will the percent of ab types of gamete by the Aa Bb parents:
(a) 75%
(b) 50%
(c) 25%
(d) 12-5%.
Answer:
(d) 12-5%.

Question 2.
Who evolved DNA segregation and refinement techniques:
(a) Beadle and Tatum
(b) Carl Correns
(c) Watson and Crick
(d) Sutton and Boveri.
Answer:
(c) Watson and Crick

Question 3.
Who prescribed the word chromosome:
(a) Johannsen
(b) Waldeyer
(c) Benda
(d) de Duve.
Answer:
(b) Waldeyer

Question 4.
Two types of proteins which found in eukaryotic chromosome:
(a) Conjugated and Complex protein
(b) Histone and Non histone protein
(c) DNA and RNA
(d) Histone and DNA.
Answer:
(b) Histone and Non histone protein

Question 5.
Heredity material which is found in cytoplasm:
(a) Genome
(b) Plasmone
(c) Nucleosome
(d) Chromatid.
Answer:
(b) Plasmone

Question 6.
The main functions of the chromosome:
(a) Transfers of characters from parents to children
(b) Growth
(c) Respiration
(d) Reproduction.
Answer:
(a) Transfers of characters from parents to children

Question 7.
Bacteriophage is a heredity material:
(a) Single helix RNA
(b) Single helix DNA
(c) Double helix DNA
(d) Single helix RNA and Double helix DNA.
Answer:
(d) Single helix RNA and Double helix DNA.

Question 8.
Word gene is indicated to:
(a) Part of DNA which is coded to polypeptide
(b) A part of RNA
(c) Linkage group
(d) Sequence of amino acid of protein.
Answer:
(a) Part of DNA which is coded to polypeptide

Question 9.
What method of arrangement of genes on the chromosome is:
(a) Linear
(b) Oval
(c) Scattered
(d) Spiral.
Answer:
(a) Linear

Question 10.
The features does not appear in the first generation called:
(a) Dominant
(b) Recessive
(c) Special
(d) General.
Answer:
(b) Recessive

Question 11.
Who described the heredity of chlorophyll in first:
(a) Correns
(b) Mendel
(c) Watson
(d) Sutton and Boveri.
Answer:
(c) Watson

Question 12.
Who observed the jumping genes of Maize:
(a) Jacob and Monod
(b) Beadle and Tatum
(c) Khurana
(d) Barbara Me Clintock.
Answer:
(d)Barbara Me Clintock

Question 13.
Genetic information which goes to offsprings from parents:
(a) Cytoplasmic heredity
(b) Nucleus heredity
(c) Genetic code
(d) All of these.
Answer:
(b) Nucleus heredity

Question 14.
What is the cytoplasmic unit of parental inheritance:
(a) Hormogon
(b) Plasmagene
(c) Genome
(d) None of these.
Answer:
(d) None of these.

Question 15.
All parental characters which are inherited by cytoplasm are called:
(a) Plasmone
(b) Caryotype
(c) Ideogram
(d) Phenotype.
Answer:
(a) Plasmone

Question 16.
Which are found in cytoplasmic control and sterility:
(a) Maize
(b) Wheat
(c) Gram
(d) Rice.
Answer:
(b) Wheat

Question 17.
The model of sex determination in honeybees are called:
(a) Female monoploidy
(b) Single diploidy
(c) Gamete diploidy
(d) Gametogenesis.
Answer:
(a) Female monoploidy

Question 18.
Which are found in unfertilized egg of human:
(a) Single Y Chromosome
(b) X and Y Chromosome
(c) XX Chromosome
(d) Single X Chromosome.
Answer:
(d) Single X Chromosome.

Question 19.
A hemophilic male and a normal female are married then offsprings are:
(a) Allhaemophilic
(b) Haemophilic female
(c) Haemophilic male
(d) All the normal.
Answer:
(d) All the normal.

Question 20.
Amniocentesis is a technique which is used in:
(a) Determine any disease in heart
(b) Know about any disease in brain
(c) Determine any hereditary disease in embryo
(d) All of these.
Answer:
(c) Determine any hereditary disease in embryo

Question 21.
Which base is not found in DNA:
(a) Gaunine
(b) Cytosine
(c) Uracil
(d) Adenine.
Answer:
(c) Uracil

Question 22.
Plasmids are:
(a) Extra chromosome
(b) Extra nuclei
(c) Extra metabolic
(d) All of these.
Answer:
(a) Extra chromosome

2. Fill in the Blanks :

1. Genes which are found in some chromosomes are called …………………….
2. Chromosomes are the carrier of …………………….
3. ……………………. is the unit of mutation.
4. ……………………. model is prescribed by Komberg.
5. Without any changes, characters are transfered to one generation to another this process is called …………………….
6. A ……………………. map is graphic representation of the relative distance of genes.
7. Sex-determination in human occurs by ……………………. chromosome.
8. Number of chromosomes in man’s sperm is …………………….

Answer:

1. Linked chromosome
2. Hereditary characters
3. Muton
4. Nucleosome
5. Complete linkage
6. Chromosome map
7. Y
8. Two.

3. Match the Following :
I.

Answer:

1. (c)
2. (d)
3. (a)
4. (b)

II.

Answer:

1. (c)
2. (d)
3. (a)
4. (b)

III.

Answer:

1. (c)
2. (d)
3. (a)
4. (b)

IV.

Answer:

1. (d)
2. (e)
3. (a)
4. (c)
5. (b)

V.

Answer:

1. (c)
2. (e)
3. (a)
4. (b)
5. (d)

Question 4.
Answer in One Word/Sentence:

1. Give an example of multiple allelism.
2. Removal of anther from a floral bud.
3. Name the set of two crosses in which two plants are used as parents of apposite-Sex one by one.
4. The external appearance of an individual.
5. What is the ratio of a dihybrid test cross?
6. What is the genotypic ratio of a monohybrid cross?
7. The allelic gene interaction in which homozygous dominant individual does not survive.
8. Name the plant in which the inheritance of flower colour shows incomplete dominance.
9. Who discovered the chromosomes ?
10. Name a chromosome having no centromere.
11. What will be the sex of an individual (Human) having 44 + XX chromosomes?
12. The chromatin part which stains lightly on staining with basic dyes.
13. Who proposed nucleosome model of chromosomal structure ?
14. Who discovered Y-chromosome?
15. What is another name of Bleeder’s disease?
16. Name a plant in which genic sex determination is found.
17. What type of male children are produced from a carrier mother and a normal father (For colour blindness)?
18. Which type of sex determination is found in Coccinia indica?
19. Name the syndrome having XXY sex chromosomes.
20. Name the twin’s developmenting from two different fertilized ova.
21. Which law is useful in the study of frequency distribution of a gene in a population?
22. Name the pigment lack of which cause albinism in human beings.
23. Name the group of characteristics that identified a particular set of chromosomes.

Answer:

1. Blood group in humans
2. Emasculation
3. Reciprocal crosses
4. Phenotype,
5. 1:1:1:1,
6. 1:2:1
7. Lafliality
8. Mirabilis jalapa
9. Strasburger
10. Acentric
11. Female
12. Euchromatin
13. Kornberg
14. Stevens
15. Haemophilia
16. Asparagus
17. 50% normal and 50% colour blind
18. XX female and XY male mechanism
19. Klinefelter’s syndrome
20. Heterozygotic,
21. Hardy-Weinberg law
22. Melanin
23. Karyotype.

Principles of Inheritance and Variation Very Short Answer Type Questions

Question 1.
Name the method in which removal of male reproductive organ from any bisexual flower.
Answer:
Emasculation.

Question 2.
What is known as the alternative forms of a gene that can occupy the same locus on a particular chromosome and that control the same character ?
Answer:
Allelomorph.

Question 3.
What is one, two or more genes that when present together produce effect qualitatively distinct from the separate effect of any one of them ?
Answer:
Complementary gene.

Question 4.
What do you mean by heredity material which are found in outer part of the chromosome?
Answer:
Plasmogene.

Question 5.
What do you mean by the act or process of mating organisms of different varieties or species to create a hybrid?
Answer:
Hybridization.

Question 6.
What do you say when a gene shows dominant effect?
Answer:
Dominant gene.

Question 7.
What do you mean by a factor which shows the single character in the cell?
Answer:
Allele.

Question 8.
What is meant by true breeding?
Answer:
True breeding means that the parents will also pass-down a specific phenotypic trait to their offspring.

Question 9.
Which type of true breeding varities are selected by Mendel ?
Answer:
14 types of true breeding verities are selected by mendel.

Question 10.
What do you mean by Fj in mendels law?
Answer:
First fertile generation.

Question 11.
What is the phenotypic and genotypic ratio in dihybrid test cross?
Answer:
There are 1:1:1:1 ratio found in genotype and phenotype.

Question 12.
Who gave the name ’X’ body? Which genes are select sex determination in human beings?
Answer:
Herman Henking suggested the name and occur.es the sex determination in human by XX and XY chromosome.

Question 13.
What is the reason of frame shift mutation?
Answer:
A frame shift mutation is a genetic mutation caused by a deletion or insertion in a DNA sequence.

Question 14.
Give one example of Co-dominancy.
Answer:
‘A’, ‘B’ and ‘O’ genes are co-dominant which found in blood group.

Question 15.
What is point mutation?
Answer:
A point mutation is a genetic mutation where a single nucleotide base is changed.

Question 16.
What do you mean by diagrammatic representation of hereditary characters which shows specific characters from generation-to-generation?
Answer:
Pedigree analysis.

Question 17.
Name the method in which breakdown of an segment of chromosome.
Answer:
Segregation.

Question 18.
Give the name ofany allopolyploid species which is originate by natural way.
Answer:
Wheat (Triticum astivum).

Question 19.
When a third sex chromosome is added to the normal two in mammals, arfe called?
Answer:
Trisomic.
Question 20. When an individual exhibiting morphological characteristics of both sexes in drosophila called this method ?
Answer:
Gynandromorphy.

Question 21.
Disease in which blood clotting is not occur when injured any part
Answer:
Haemophilia.

Question 22.
Names a specific DNA segment which are functioned as a unit of heredity.
Answer:
Gene.

Question 23.
What is the name of cell chromosome except sex chromosome?
Answer:
Autosomes.

Question 24.
What is the name of the arms of chromosome?
Answer:
Chromatid.

Question 25.
Write one example of sex-linked inheritance.
Answer:
Colour blindness.

Question 26.
Which type of sex determination found in human and Drosophila?
Answer:
X, Y type.

Question 27.
Which disease caused by point mutation?
Answer:
Sickle cell anaemia.

Question 28.
Name the disease caused by inadiquate or defective nutrition.
Answer:
Dystrophy.

Question 29.
What is mutation?
Answer:
A mutation occurs when a DNA gene is damaged or changed in such a way as to alter the genetic message carried by that gene.

Question 30.
What is the exception of Mendel’s Independent assortment theory?
Answer:
Linkage.

Principles of Inheritance and Variation Short Answer Type Questions

Question 1.
Give reasons for Mendel’s success.
Answer:
Reasons for Mendel’s success are as follows :

• Mendel selected seven pairs of traits in garden pea and used all the seven pairs of different characters individually.
• He provided correct information of quantitative method.
• He explained correct ratio of phenotype.
• Mendel did his experiment by correcting the previous work of scientists.
• Experiment was done systematically and explained result clearly.

Question 2.
Explain test cross and back cross.
Answer:
Crossing of F₁ hybrid with either of two parents is called as back cross. For example, when tall plant (TT) was crossed with dwarf plant (tt) then in Fi generation all plants are long (Tt) and when this F₁ plants are crossed with any individual then called back cross. Whereas crossing of F₁ individual with recessive parent is called test cross because it is used to test whether parents are homozygous or heterozygous.

Question 3.
What are plasmagenes?
Answer:
Genetic material presents outside the nucleus is called as plasmon or plasmagene e.g., Genetic material (DNA and RNA) present within mitochondria and chloroplast belonging to this category.

Question 4.
Explain the ratio of genotype and phenotype. (In non-hybrid cross).
Answer:
Monohybrid cross :

Question 5.
Why did Mendel choose garden pea plants as experimental material?
Answer:
Mendel selected garden pea as experimental material because of the following reasons :

• The life cycle of pea plant is short and is completed in few months and gives results in short-time.
• Plants are easily cross-pollinated.
• Presence of 7 pairs of contrasting characters.
• The flowers of pea are bisexual and naturally self-pollinated.
• The hybrids obtained from cross were fertile.
• Emasculation can be easily done when androecium (anthers) are removed before maturity, then the plant
• exhibits like a unisexual plant.

Question 6.
When a colour-blind man is married with a normal wpman, then work-out the progeny of these parents with the help of diagram only.
Answer:

Question 7.
What is Down’s Syndrome?
Answer:
Down’s Syndrome: It is a common human karyotype in which trisomy occur at 21st chromosome. This type of person contains 47 chromosomes. It was first studied by Down (1866) and was popularly known as Mongolian idiocy because the facial features of persons suffering from Down’s syndrome resembled Mongolians. They have broad forehead, flat hands, short neck, projecting lips, stubby fingers and long extending tongue. The Mongoloid individuals are feeble minded and their mental age never exceeds to those of six to seven year’s old children.

Question 8.
Write Differences between Genotype and Phenotype.
Answer:
Differences between Genotype and Phenotype:

Question 9.
What are plasmids ? Describe its characters.
Answer:
Plasmids : Plasmids are extrachromosomal, extranuclear, self-replicating, covalently closed DNA molecules. These plasmids also play an important role in the transmission of hereditary characters. Plasmids are generally found in bacterial cell in addition to bacterial chromosome. The discoverer of plasmid are William Hays and Lederberg.

Characters of plasmids:

• Plasmids are extranuclear DNA molecules.
• These are smaller than chromosomes.
• These possess the capacity of self-replication.
• They can transfer the genes from donor to recipient cells, they act as the vector molecule of DNA.
• They have a capacity of integration with host DNA hence, it is used in genetic engineering.

Question 10.
Give the structure of E. coli chromosome.
Answer:
Isolated E. coli chromosomes display many individually super coiled loops emanating from a central region. These super coiled loops were hypothesized to be topological domains. These domains are sufficient to generate the observed precision of E. coli chromosome structure. Its chromosomal DNA has been completely sequenced by lab researchers. E. coli has a single chromosome with about 4,600 kb, about 4,300 potential coding sequences and only about 1,800 known E. coli proteins.

Question 11.
What do you understand by reciprocal cross ?
Answer:
Reciprocal cross : A set of two reciprocal crosses means that the same two parents are used in two experiment in such a way that:

1. In one experiment, ‘A’ is used as the female parent and ‘B’ is used as the male parent.
2. In the other experiment, ‘A’ will be the male parent and ‘B’ the female parent.

Question 12.
What is modifier gene?
Answer:
Modifier genes, instead of making the effects of another gene. A gene can modify the expression of a second gene. In mice, coat colour is controlled by the B gene. THeB allele conditions black coat colour and is dominant to the b allele that produces a brown coat.

Question 13.
What do you understand by Multiple allele? Explain it with example.
Answer:
Multiple allele : There are two alternative forms of each trait for the seven pairs of contrasting characters studied by Mendel. It means that there are two alleles for each trait. The first allele will be dominant and other will be recessive. But further studies show that, there are two or more alternative forms of a gene or allele, which are known as multiple alleles.

Example : (i) The skin colour of rabbit is due to the presence of four alleles of a gene.
(ii) A well known example of multiple allele is ABO blood type in human beings. The four human blood groups ‘A’, ‘B’ ‘AB’ and ‘O’ are the four phenotypes for this trait. Persons having ‘A’ blood group have ‘A’ type of glycoprotein (antigen) coating the red blood cells and persons having ‘B’ blood group have ‘B’ type of glycoprotein (antigen) coating the red blood cells. Persons having ‘AB’ blood group have both type of glycoproteins i.e., ‘A’ and ‘B’ antigens coating the red blood cells, while persons having ‘O’ blood group do not have any glycoprotein on R.B.Cs.

Question 14.
Describe the causes of hereditary variation.
Or
Why does hereditary variations originate ?
Answer:
Any change in the structure of gene of an organism will produce heritable variations. The chief causes of these variations are as follows :

(i) Genetic recombination : The exchange of chromosomal segments of genes takes place during reduction division which causes recombination of genes. These recombinant organisms fuse to form zygote. The genetic structure of these zygotes is different from their parents. Adults formed from this zygote will show variations.

(ii) Changes in the number of chromosomes : Changes in the number of chromosomes will also cause genetic variations, e.g., modern varieties of wheat contain 16,21 and 42 chromosomes. These varieties are produced from its parents having only 7 chromosomes.

(iii) Changes in the structure of genes : Changes in the structure of gene will also cause genetic variations, e.g., 30 variety of wheat is the best example of genetic variations. Like this, if any changes in the genes of human being which are related with pigment production, will inhibit the production of these pigments and thus man become albino.

Question 15.
Explain recessive epistasis with example.
Answer:
Recessive epistasis (9: 3 :4): In case of recessive epistasis, the epistatic gene is recessive to its own allele. Thus, in this condition, the epistatic gene can have its inhibiting influence only when it is in homozygous condition.

The inheritance of body colour in mice is an example of recessive epistasis. It may have three body colours i.e., agouti, coloured and albino. The agouti body colour is controlled by a dominant gene ‘A’. Another dominant gene ‘C’ gives coloured body only in the absence of dominant ‘A’ gene. The expression of agouti body colour by gene ‘C’ is inhibited by recessive gene ‘C’. Therefore, even in the presence of ‘A’ gene, the mice develop albino body colour if recessive genes ‘CC’ are present.

Question 16.
Why are the women normally carrier of sex-linked diseases ? Write with example.
Or
Men are generally suffering from colour-blindness but women are only carriers pf this disease. Explain the reason.
Answer:
The gene for colour-blindness is recessive and carried by X-chromosomes. A colour-blind man has a single recessive gene (X^CY). The gene for normal vision is dominant. When sex chromosomes (X) of a man contains genes for colour-blindness (X^C), they express itself and produce colour-blindness in them. A carrier woman contains one recessive gene which remains suppressed due to presence of the dominant gene of normal vision on the other sex chromosomes (X^CX). These women play important role in the conduction of genes of colour-blindness. Hence, they are carrier (X^CX). Women become colour-blind only when its both chromosomes contain the genes of colour-blindness (X^CX^C).

Question 17.
What is Turner’s Syndrome?
Answer:
Turner’s Syndrome: It is a disorder arising from chromosomal abnormalities. The cells of a person suffering from this disease contain 45 chromosomes (44 A + X) only. Cytologicaily, they can be recognized by presence of only one X-chromosome (XO female). They are female in general appearance with underdeveloped breasts, broad chest, webbed neck, low set ears, poorly developed ovaries. They are sterile females, often short and of subnormal intelligence.

Question 18.
When a haemophilic man is married with a normal woman then workout the progeny of these parents with the help of diagram only.
Answer:
When a haemophilic man is married with a normal woman, all sons will be normal whereas daughters will be carrier.

Fig. Offspring of haemophilic man and normal woman.

Question 19.
Why human males are generally suffering from baldness but woman are not suffering from it ?
Answer:
The genes responsible for baldness are found in autosomes but are expressed by sex-linked genes. The baldness is also caused due to irradiation, abnormal functioning of thyroid gland and abnormality in heredity. Genetical baldness depends upon an autosomal allelomorph (Bb). When ‘BB’ is occurred in the form of dominant homozygous then man and woman both suffer from this disease whereas it occurs in heterozygous stage (Bb) in man only because a male hormone is required for their development, hence woman does not suffer from baldness. Baldness is not expressed in recessive homozygous stage (bb).

Question 20.
Describe the inheritance of colour-blindness in the offspring of a colour-blind man and carrier woman with the help of suitable ray diagram.
Answer:
When a colour-blind man is married with a carrier woman, 25% boys will be colour-blind and 25% boys will be normal whereas 25% girls will be colour-blind and 25% girls offspring become carrier of colour-blindness. In other words, we can say that 50% offspring would be colour-blind, 25% normal and remaining 25% offspring will be carrier of colour-blindness.

Question 21.
Explain the inheritance of haemophilia when a haemophilic woman is married with normal man with the help of diagram.
Answer:
When a haemophilic woman is married with a normal man, all the boys offspring will be haemophilic whereas all the girls offspring will be carrier of haemophilia. In other words, 50% offsprings will be haemophilic and 50% offsprings will be carrier.

Question 22.
What is Klinefelter’s syndrome ? Also describe their symptoms.
Answer:
Klinefelter’s syndrome: It is a sex chromosomal syndrome. It is found only in men. They possess 47 chromosomes in their cells (22 pairs autosomes + XXY sex chromosomes). One extra X chromosome is found in them, thus, there is Trisomy in sex-chromosome.

Principles of Inheritance and Variation Long Answer Type Questions

Question 1.
Write down the law of segregation.
Or
Describe Mendel’s law of segregation with example.
Answer:
According to law of segregation, two genes of a character separate and get distributed randomly to different gametes and then to different offspring as per the law of probability. In other words, genes for each trait segregate without mixing.

When a cross is made between pure tall pea plant and dwarf pea plant, it gives rise to tall plant in the F₁ generation. On self-breeding both tall and dwarf plants appear in F₂ generation in the ratio of 3 : 1. This ratio can be achieved only if the two factors separate at the time of gamete formation and only one factor passes into gamete. Therefore, this law is also called as law of purity of gametes.

Question 2.
Explain Complementary gene (9:7) with example.
Answer:
Complementary genes (9:7): The complementary genes are the two pairs of non-allelic genes which are present on separate loci and interact to produce only one phenotypic trait but either of them if present alone produces the phenotypic trait in the absence of other.

The example of complementary gene interaction was provided by Bateson and Punnet in sweet pea (Lathyrus odoratus). He demonstrated that two dominant genes ‘C’ and ‘P’ are responsible for the development of purple coloured flowers in a plant of sweet pea. The dominant gene ‘P’ or ‘C’ in homozygous (PP and CC) and heterozygous (P_p and C_c) state are unable to produce their effect if they are present alone. They observed that when two white flowered plants are crossed with each other, the F₁have coloured flowers.

This is due to the fact that crossing between two white flowered plants brought together two non-allelic dominant genes in Ft. On intercrossing, these Fi progenies produced coloured and white flowered plants in 9:7 ratio (Fig.).

Question 3.
Explain the Law of Independent Assortment of Characters.
Answer:
The gametes and zygotes usually carry several chromosomes which are different in appearance and perhaps in contents. The genotype consists of many pairs of genes. Although chromosomes were not known to Mendel yet he felt the need of finding out how different characters would behave in relation to each other in their passing from generation to generation. This also could be demonstrated with two pairs of opposite characters. Such a cross which involves two character differences separable in inheritance is termed as a dihybrid cross. For example, taking parents round yellow pea and a wrinkled green pea (Fig.).

Round, yellow (RY) : Dominant characters Wrinkled,
green (ry) : Recessive characters.

Question 4.
What is chromosomal theory of inheritance ?
Answer:
In 1902, Sutton and Boveri proposed a theory to explain inheritance of characters. This theory is known as chromosomal theory of inheritance.

The main points of this theory are as follows :

• Chromosomes and factors are found in pairs in diploid cells.
• Chromosomes and factors get’separated during the formation of gametes.
• During transmission of characters only one chromosome and factor is transmitted to offspring.
• Pairing between homologous chromosomes and factor occurs after fertilization.

Question 5.
What is crossing over ?
Or
Explain crossing over.
Or
Write the process of crossing over with diagram that occurs during meiosis.
Answer:
Exchange of chromosomal segments between two homologous chromosomes during diplotene of meiosis I division is called as crossing over. When homologous chromosomes tend to start their separation due to repulsion during diplotene, they are bounded

In some places. These places are called Chiasmata. Due to the process of terminalization chiasmata start to move along the length of the chromosome from the centromere. The terminalization of chiasmata will result in the formation of ‘X’-shaped configuration in which breakage of chromosomal segments takes place, then these segments are reunited in a manner in which the segments of non-sister chromosomes are exchanged. This exchange of chromosomal segments or genes of homologous chromosomes is termed as crossing over. This process is very important because it originates new characters and causes variations in the organisms.

Question 6.
Explain sex-linkage by giving example of haemophilia.
Or
What is sex-linkage? Describe the inheritance of haemophilia on the basis qf sex linkage.
What is sex-linkage? Explain with suitable example.
Answer:
Genes that are formed on sex chromosomes and are inherited together are called linked genes and the process is called sex-linked inheritance of sex-linkage.

Inheritance of haemophilia in man : Haemophilia is a sex-linked disease in which patient losses the capacity of blood clotting during bleeding. This disease is caused by a recessive gene present on the X-chromosomes (X^h). A man possessing a single recessive gene suffers from the disease because the Y-chromosome does not contain any gene (X^hY). A carrier women shall have only one gene of the disease (XX^h). She does not suffer from the disease because of the presence of the normal gene on the second sex chromosomes.

We can understand the process of inheritance of haemophilia by the following examples:

1. Inheritance of haemophilia in offspring of a normal father and carrier mother: 50%
children (25% boys + 25% girls) of these parents are normal, 25% boys are haemophilic and 25% girls are carrier of this disease.

2. Inheritance of haemophilia in the offspring of haemophilic man and normal woman : All the boys of these parents will be normal while all the girls will be carrier of the disease.

3. Inheritance of haemophilia in the offspring of normal man and haemophilic woman : All die boys of these parents will be haemophilic and all the girls will be carrier of the disease.

4. Inheritance ofhaemophilia in the offspring ofhaemophilic man and carrier woman : The 25% boys of this parents will be haemophilic and 25% boys will be normal while 25% girls will be haemophilic and remaining 25% girls will be carrier of the disease.

Question 7.
What is linkage? It is of how many types?
Answer:
Linkage : The tendency of genes inherited together, without independent assortment is called linkage and such genes are linked genes. Such genes are closely situated showing a strong attraction between them. Linkage always reduces the changes of crossing over. All those genes which are located in the single chromosome from one linkage
group. The linkage groups correspond to the haploid number of chromosomes of a species or number of chromosomes pairs. Thus, the linkage groups in maize would be 10,7 in pea, 4 in Drosophila, 23 in man etc.

Types of Linkages :

T.H. Morgan along with Castle working on Drosophila discovered the following types of linkage:

1. Complete linkage : It is the phenomenon in which parental combination of characters appear together for two or more generations continuously.
Example: Genes for bent wings and shaven bristles in Drosophila melanogaster.

2. Incomplete linkage : In this phenomenon the linked genes may remain together for one generation only and have chances of separation in subsequent generation, because of crossing over.
Example : Genes for grey body and long wings characters.

Question 8.
What is gene mutation ? Explain the causes of gene mutation.
Answer:
Gene mutation : Genes are responsible for the transmission of hereditary characters from generation to generation. Sometimes few sudden changes in the structure and arrangement of genes take place which result in the production of new characters. These sudden changes in genes are called as gene mutation.

Causes of gene mutation :

• According to Beadle and Tatum, mutations occurs due to the physical and chemical changes in genes during reproduction.
• According to Rasowsky, X-rays also cause mutation.
• According to Muller, radioactive radiations also cause mutations.
• According to Guber and Smith, some antibiotics also cause mutation.
• Sudden changes during reproduction also cause mutation.
• Changes in the number of chromosomes.

Question 9.
What is incomplete linkage and complete linkage ?
Answer:
When there are chances of separation of linked genes, it is called incomplete linkage. In this type of linkage, genes are situated at distance. During crossing over in meiosis, probability of separation of genes increases. In maize incomplete linkage is found.

When there are no chances of separation of linked genes, it is called complete linkage. Complete linkage takes place due to the non-break in the gene combination, situated on any chromosome. It is found is Drosophila and other insects.

Question 10.
Explain the types of sex chromosomes.
Answer:
Mainly sex chromosomes are of two types :
(A) Autosomes
(B) Sex chromosomes or Allosomes.

(A) Autosomes : That carry genes for body characters and general physiological activities. The two members of each homologous pair are similar in shape and size.
(B) Sex chromosomes : That carry genes for sex. A pair of them determines the sex. They are of following types :

1. XX and XY method of sex determination: In majority of animals female have XX sex-chromosome and male have XY. The females are homogametic and produce ova of one type of X chromosome. Male are heterogametic, they produce two types of sperms X and Y chromosomes. It means sex chromosomes in female are homomorphic and of male are heteromorphic.
Examples: Drosophila and man.

2. XX-XO method of sex determination: In this case female have two homomorphic sex chromosomes XX. They produce all ovas with X chromosomes. The males have only one X chromosomes, there is no Y chromosome. Example: Birds, Butterflies, Fishes and etc.

Question 11.
Explain the chromosomal theory of linkage.
Answer:
Chromosomal theory of linkage was produced by Morgan and Castle in 1911. The main points of this hypothesis are :

1. Linked genes are found on a chromosome.
2. The strength of linkage depends upon the distance between genes.
3. Two nearest genes exhibit linkage while genes exhibit crossing over.
4. All linked genes are found on a particular place on the chromosomes and are arranged in a linear fashion.

Question 12.
Explain the inheritance of colour-blindness when the man is blind and woman is normal.
Answer:
Colour blindness : It is a genetic disease. It is an inability to distinguish red from green. The recessive gene C, for the trait is carried by ‘X’ chromosomes, ‘Y’ chromosome has no corresponding gene. A single gene C, on ‘X’ chromosome is sufficient to express itself in the male. Female are generally carrier of the disease in such cases.
Example : The inheritance of colour-blindness in the offspring of a normal woman and colour blind man are as follow:

Ratio : 50% offsprings are carrier and 50% offsprings are normal. In other words all the girls will be carrier of colour blindness whereas all the boys will be normal.

MP Board Class 12th Biology Solutions

MP Board Class 11th Maths Solutions Chapter 11 शंकु परिच्छेद Ex 11.3

निम्नलिखित प्रश्नों 1 से 9 तक प्रत्येक दीर्घवृत्त में नाभियों और शीर्षों के निर्देशांक, दीर्घ और लघु अक्ष की लंबाइयाँ, उत्केंद्रता तथा नाभिलंब जीवा की लम्बाई ज्ञात कीजिए।
प्रश्न 1.
\(\frac{x^{2}}{36}+\frac{y^{2}}{16}\) = 1
हल:

प्रश्न 2.
\(\frac{x^{2}}{4}+\frac{y^{2}}{25}\) = 1
हल:

दीर्घ अक्ष की लंबाई = 2a = 2 × 5 = 10
लघु अक्ष की लंबाई = 2b = 2 × 2 = 4
उत्केंद्रता = e = \(\frac{c}{a}=\frac{\sqrt{21}}{5}\)
नाभिलंब जीवा की लंबाई = \(\frac{2 b^{2}}{a}=\frac{2 \times 4}{5}=\frac{8}{5}\).

प्रश्न 3.
\(\frac{x^{2}}{16}+\frac{y^{2}}{9}\) = 1
हल:

प्रश्न 4.
\(\frac{x^{2}}{25}+\frac{y^{2}}{100}\) = 1
हल:
दीर्घवृत्त का समीकरण \(\frac{x^{2}}{25}+\frac{y^{2}}{100}\) = 1
∴ a² = 100, b² = 25
∴ a = 10, b = 5
∴ c² = a² – b² = 100 – 25 = 75
∴ c = 5\(\sqrt{3}\)
नाभि के निर्देशांक (0, ± c) या (0, ± 5\(\sqrt{3}\))
शीर्षों के निर्देशांक (0, ± a) या (0, ± 10)
दीर्घ अक्ष की लंबाई = 2a = 2 × 10 = 20
लघु अक्ष की लंबाई = 2b = 2 × 5 = 10

प्रश्न 5.
\(\frac{x^{2}}{49}+\frac{y^{2}}{36}\) = 1
हल:
दीर्घवृत्त का समीकरण \(\frac{x^{2}}{49}+\frac{y^{2}}{36}\) = 1
∴ a² = 49, b² = 36
∴ a = 7, b = 6
दीर्घ अक्ष, x-अक्ष के अनुदिश है
c² = a² – b² = 49 – 36 = 13
c = \(\sqrt{13}\)
नाभियों के निर्देशांक (± c, 0) या (± \(\sqrt{13}\), 0)
शीर्षों के निर्देशांक (± a, 0) या (± 7, 0)
दीर्घ अक्ष की लंबाई = 2a = 2 × 7 = 14
लघु अक्ष की लंबाई = 2b = 2 × 6 = 12

प्रश्न 6.
\(\frac{x^{2}}{100}+\frac{y^{2}}{400}\) = 1
हल:
दीर्घवृत्त का समीकरण \(\frac{x^{2}}{100}+\frac{y^{2}}{400}\) = 1
∴ a² = 400, b² = 100
∴ a = 20, b = 10
c² = a² – b² = 400 – 100 = 300
∴ c = 10\(\sqrt{3}\)
दीर्घ अक्ष, y- अक्ष के अनुदिश है
नाभियों के निर्देशांक (0, ± c) या (0, ± 10\(\sqrt{3}\))
शीर्षों के निर्देशांक (0, ± a) या (0, ± 20)
दीर्घ अक्ष की लंबाई = 2a = 2 × 20 = 40
लघु अक्ष की लंबाई = 2b = 2 × 10 = 20

प्रश्न 7.
36x² + 4y² = 144.
हल:
दीर्घवृत्त का समीकरण 36x² + 4y² = 144
या \(\frac{x^{2}}{4}+\frac{y^{2}}{36}\) = 1
∴ a² = 36, b² = 4
∴ a = 6, b = 2
∴ c² = a² – b² = 36 – 4 = 32
∴ c = 4\(\sqrt{2}\)
दीर्घवृत्त का अक्ष, y-अक्ष के अनुदिश है
नाभियों के निर्देशांक (0, ± c) या (0, ± 4\(\sqrt{2}\))
शीर्षों के निर्देशांक (0, ± a) या (0, ± 6)
दीर्घ अक्ष की लंबाई = 2a = 2 × 6 = 12
लघु अक्ष की लंबाई = 2b = 2 × 2 = 4

प्रश्न 8.
16x² + y² = 16.
हल:
दीर्घवृत्त का समीकरण 16x² + y² = 16
या \(\frac{x^{2}}{1}+\frac{y^{2}}{16}\) = 1
∴ दीर्घवृत्त का अक्ष, y-अक्ष के अनुदिश है।
a² = 16, b² = 1
∴ a = 4, b = 1
c² = a – b² = 16 – 1 = 15
∴ c = \(\sqrt{15}\)
नाभियों के निर्देशांक (0, ± c) या (0, ± \(\sqrt{15}\))
शीर्षों के निर्देशांक (0, ± a) या (0, ± 4)
दीर्घ अक्ष की लंबाई = 2a = 2 × 4 = 8
लघु अक्ष की लंबाई = 2b = 2 × 1 = 2

प्रश्न 9.
4x² + 9y² = 36.
हल:
दीर्घवृत्त का समीकरण 4x² + 9y² = 36
या \(\frac{x^{2}}{9}+\frac{y^{2}}{4}\) = 1
दीर्घ अक्ष, x-अक्ष के अनुदिश है।
∴ a² = 9, b² = 4
∴ a= 3, b = 2
c² = a² – b² = 9 – 4 = 5
∴ c = \(\sqrt{5}\)
नाभियों के निर्देशांक (± c, 0) या (± \(\sqrt{5}\), 0)
शीर्षों के निर्देशांक (± a, 0) या (± 3,0)
दीर्घ अक्ष की लंबाई = 2a = 2 × 3 = 6
लघु अक्ष की लंबाई = 2b = 2 × 2 = 4

निम्नलिखित प्रश्नों 10 से 20 तक प्रत्येक में, दिए प्रतिबंधों को संतुष्ट करते हुए दीर्घवृत्त का समीकरण ज्ञात कीजिए।
प्रश्न 10.
शीर्षों (± 5, 0), नाभियाँ (± 4, 0).
हल:
a = 5, c = 4, c² = a² – b²
या 16 = 25 – b²
∴ b² = 25 – 16 = 9
और a² = 25
दीर्घवृत्त का समीकरण,
\(\frac{x^{2}}{25}+\frac{y^{2}}{9}\) = 1

प्रश्न 11.
शीर्षों (0, ± 13), नाभियाँ (0, ± 5).
हल:
दीर्घ अक्ष, y-अक्ष के अनुदिश है।
∴ c = 5, a = 13, c² = a² – b²
∴ 25 = 169 – b²
∴ b² = 169 – 25 = 144,
और a² = 13² = 169
∴ दीर्घवृत्त का समीकरण
\(\frac{x^{2}}{144}+\frac{y^{2}}{169}\) = 1

प्रश्न 12.
शीर्ष (± 6, 0), नाभियाँ (± 4, 0).
हल:
दीर्घ अक्ष x-अक्ष के अनुदिश है।
a= 6, ∴ a² = 36, c = 4
c² = a² – b² या 16 = 36 – b²
∴ b² = 36 – 16 = 20
∴ दीर्घवृत्त का समीकरण
\(\frac{x^{2}}{36}+\frac{y^{2}}{20}\) = 1.

प्रश्न 13.
दीर्घ अक्ष के अंत्य बिन्दु (± 3, 0), लघु अक्ष के अंत्य बिन्दु (0, ± 2).
हल:
दीर्घ अक्ष x-अक्ष के अनुदिश है।
a = 3, b = 2, ∴ a² = 9, b² = 4
दीर्घवृत्त का समीकरण,
\(\frac{x^{2}}{9}+\frac{y^{2}}{4}\) = 1

प्रश्न 14.
दीर्घ अक्ष के अंत्य बिन्दु (0, ± \(\sqrt{5}\)), लघु अक्ष के अंत्य बिन्दु (± 1, 0).
हल:
दीर्घ अक्ष, y-अक्ष के अनुदिश है।
a = \(\sqrt{5}\), b = 1, ∴ a² = 5, b² = 1
दीर्घवृत्त का समीकरण,
\(\frac{x^{2}}{1}+\frac{y^{2}}{5}\) = 1

प्रश्न 15.
दीर्घ अक्ष की लंबाई = 26, नाभियाँ (45, 0).
हल:
दीर्घ अक्ष, x-अक्ष के अनुदिश है।
और 2b = 26, ∴ b = 13 या a² = 169,
c = 5, c² = 25 = a² – b² = 169 – b²
∴ b² = 169 – 25 = 144
अतः a² = 169, b² = 144
∴ दीर्घवृत्त का समीकरण,
\(\frac{x^{2}}{169}+\frac{y^{2}}{144}\) = 1.

प्रश्न 16.
दीर्घ अक्ष की लंबाई = 16, नाभियाँ (0, ± 6).
हल:
दीर्घ अक्ष, y-अक्ष के अनुदिश है।
2b = 16, ∴ b = 8 या b² = 64,
c = 6, c² = 36,
c² = a² – b²
या 36 = a² – 64
∴ a² = 64 + 36 = 100
∴ दीर्घवृत्त का समीकरण
\(\frac{x^{2}}{64}+\frac{y^{2}}{100}\) = 1

प्रश्न 17.
नाभियाँ (± 3, 0), a = 4.
हल:
दीर्घ अक्ष, x-अक्ष के अनुदिश है।
∴ c = 3, a = 4
अब c² = a² – b²
या 9 = 16 – b²
∴ b² = 16 – 9 = 7
∴ दीर्घवृत्त का समीकरण,
\(\frac{x^{2}}{16}+\frac{y^{2}}{7}\) = 1

प्रश्न 18.
b = 3, c = 4, केन्द्र मूल बिन्दु पर, नाभियाँ x-अक्ष पर है।
हल:
दीर्घ अक्ष, x-अक्ष के अनुदिश है
c² = a² – b²
16 = a² – 9
a² = 16 +9 = 25
∴ दीर्घवृत्त का समीकरण,
\(\frac{x^{2}}{25}+\frac{y^{2}}{9}\) = 1

प्रश्न 19.
केंद्र (0, 0) पर, दीर्घ अक्ष y-अक्ष पर और बिन्दुओं (3, 2) और (1, 6) से जाता है।
हल:

प्रश्न 20.
दीर्घ अक्ष,x-अक्ष पर और बिन्दुओं (4, 3), (6, 2) से जाता है।
हल:
मान लीजिए दीर्घवृत्त का समीकरण \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1

∴ दीर्घवृत्त का समीकरण,
\(\frac{x^{2}}{52}+\frac{y^{2}}{13}\) = 1

MP Board Class 11th Maths Solutions

MP Board Class 12th Biology Solutions Chapter 12 Biotechnology And its Application

Biotechnology And its Application NCERT Text Book Questions and Answers

Question 1.
Crystals of Bt toxin produced by some bacteria do not kill the bacteria themselves because:
(a) Bacteria are resistant to the toxin,
(b) Toxin is immature
(c) Toxin is inactive
(d) Bacteria enclose toxin in a special sac
Answer:
(c) Toxin is inactive: In bacteria, the toxin is present in an inactive form called prototoxin. This gets converted into the active form when it enters the salivary gland of insects having alkaline medium.

Question 2.
What are transgenic bacteria? Illustrate using any one example.
Answer:
The becteria whose DNA is manipulated to carry and express a foreign DNA is called transgenic bacteria. These microbes are used for producing important bio-chemicals. They have been synthesizing alcohol, enzymes, steroids and antibiotics. Example: Bacillus thuringiensis for Bt cotton, hirudin from transgenic Brassica napus seed. Hirudin is a protein which prevents blood clotting. Its gene was chemically synthesized and introduced in Brassica napus, in which hirudin accumulates in the seed from where it is extracted, purified and used as a medicine.

Question 3.
Compare and contrast the advantages and disadvantages of production of genetically modified crops.
Answer:
Advantages of GM crops :

• Genetic modification has made crops more tolerant to abiotic stresses (cold, drought, heat, salt.)
• Viral resistance can be introduced.
• Over ripening losses can be reduced. Example: Flavr Savr Tomato.
• Enhanced nutritional value of food. Example: Golden Rice.
• Reduced reliance on chemical pesticides.

Disadvantages of GM crops :

• Transgenes in crop plants can endanger native species. Example: The gene for Bt toxin expressed in pollen may end natural pollinators such as honeybees.
• Weeds also become resistant.
• Products of transgenes may be allergic or toxic.
• They cause damage to the natural environment.

Question 4.
What are Cry proteins? Name an organism which produces it How has man exploited this protein to his benefit?
Answer:
Cry proteins are toxic proteins (insecticidal proteins) secreted by Bacillus thuringiensis in crystal form during a particular phase of their growth. The toxin is coded by a gene called cry.

The genes encoding cry proteins called Bt toxin genes were isolated from B. thuringiensis and incorporated into several crop plants such as Bt cotton, Bt com etc. to provide resistance against insect pests.

Question 5.
What is gene therapy? Illustrate using the example of adenosine deaminase (ADA) deficiency.
Answer:
It is a collection of methods which allows correction of a gene defect that has been diagnosed in a child or embryo. In gene therapy, normal genes are inserted into a person’s cells or tissues to treat a hereditary defect. Gene therapy is being tried for sickle cell anaemia and Severe Combined Immuno Deficiency (SCID).

In some children, ADA deficiency can be cured by bone marrow transplantation. In others, it can be treated by enzyme replacement therapy, in which functional ADA is given to the patient by injection. However, both of these approaches are not completely curative.

In gene therapy, lymphocytes from the blood of the patient are grown in culture outside the body. A functional ADA cDNA (using a retroviral vector) is then introduced into these lymphocytes, which are subsequently returned to the patient. Because these cells are not immortal, the patient requires periodic infusion of such genetically engineered lymphocytes. However, if the gene isolated from marrow cells producing ADA is introduced into cells at early embryonic stages, the disease could be cured permanently.

Question 6.
Diagrammatically represent the experimental steps in cloning and expressing a human gene (say the gene for growth hormone) into a bacterium like E. coli.
Answer:

Question 7.
Can you suggest a method to remove oil (hydrocarbon) from seeds based on your understanding of rDNA technology and chemistry of oil?
Answer:
To remove oil from seeds using recombinat DNA technology would involve:

• Identifying the genes that code for oil production.
• Deleting these genes from the seed genome.
• Splicing back together the remaining DNA.
• Putting it back into the cell.

It will be not be very easy because the oils are made up of fatty acids and glycerol. Since, fatty acids are important components of cell membrane system, deleting or switching off of its genes might affect the cell structure itself.

Question 8.
Find out from internet what is golden rice.
Answer:
Golden rice is a variety of rice produced through genetic engineering to biosynthesize β -carotene, a precursor of vitamin ‘A’ in the edible parts of rice. It is intended to produce a fortified food to be grown and consumed in areas with a shortage of dietary vitamin ‘A’.

Question 9.
Does our blood have proteases and nucleases?
Answer:
No.

Question 10.
Consult internet and find out how to make orally active protein pharmaceutical. What is the major problem to be encountered?
Answer:
For making any oral drug or nutritional supplement, the action of digestive enzymes has to be taken into account. Most of the antibiotics and vitamin supplements are made in capsule form to prevent the action of HC1 in the stomach. For protein preparation, the major source is groundnut shells. The protein extracted from the source is predigested so, as to make it absorbable by the digestive system.

Biotechnology And its Application Other Important Questions and Answers

Biotechnology And its Application Objective Type Questions

1. Choose the Correct Answer:

Question 1.
Triticum aestivum wheat is:
(a) Haploid
(b) Diploid
(c) Tetraploid
(d) Hexaploid.
Answer:
(c) Tetraploid

Question 2.
Man-made cereal is:
(a) Potato
(b) Triticale
(c) Triticum
(d) Sugarcane.
Answer:
(b) Triticale

Question 3.
Wheat grain is a:
(a) Fruit
(b) Seed
(c) Embryo
(d) Glume.
Answer:
(b) Seed

Question 4.
Removal of stamens from the flower during hybridization is called:
(a) Cutting
(b) Self-fertilization
(c) Emusculation
(d) Topnin.
Answer:
(c) Emusculation

Question 5.
New crop is:
(a) Triticale
(b) Rye
(c) Winged bean
(d) Wheat.
Answer:
(a) Triticale

Question 6.
Wheat used in bread is:
(a) Triticum aestivum
(b) Triticale
(c) All species of triticum
(d) Secale.
Answer:
(a) Triticum aestivum

Question 7.
Sonera-64 and Lerma roja-64 A are the varieties of:
(a) Wheat
(b) Rice
(c) Pea
(d) Maize.
Answer:
(a) Wheat

Question 8.
Haploid male plants can be produced by the culturing of:
(a) Filament
(b) Pollen grains
(c) Stamens
(d) Androecium.
Answer:
(b) Pollen grains

Question 9.
Ti plasmid which is used in genetic engineering, is found in:
(a) Escherichia coli
(b) Bacillus thuringiensis
(c) Agrobacterium rhizogenes
(d) Agrobacterium tumefaciens.
Answer:
(d) Agrobacterium tumefaciens.

Question10.
Bt-Toxin is:
(a) Intercellutar lipid
(b) Intercellular crystal protein
(c) Extracellular crystal protein
(d) Lipid.
Answer:
(c) Extracellular crystal protein

Question 11.
The function of Bacillus thuringiensis is:
(a) Bio-metallurgy technique
(b) Bio-insecticides plant
(c) Bio-fertilizers
(d) Bio-mineralization process.
Answer:
(b) Bio-insecticides plant

Question 12.
Non-toxic crystal of Bt are made by bacteria but bacterias are not killed by own because:
(a) Non-toxic is immature
(b) Bacteria is immune resistant
(c) Non-toxic is inactive
(d) Bacteria has non-toxic sac.
Answer:
(c) Non-toxic is inactive

Question 13.
Genetic transfer through viruses is called:
(a) Sexduction
(b) Transduction
(c) Conjugation
(d) Transformation.
Answer:
(b) Transduction

Question 14.
Biopiracy is realated with:
(a) Discovery of biomolecules and genes
(b) Cultural knowledge
(c) Bio-research
(d) All of these.
Answer:
(d) All of these.

Question 15.
Golden rice is rich in which vitamin:
(a) Vitamin A
(b) Vitamin B
(c) Vitamin C
(d) Vitamin D.
Answer:
(a) Vitamin A

2. Fill in the Blanks:

1. …………………………. is the appropriation of another’s knowledge of use of biological resources.
2. A ……………………………. is a patent granted by the government to the inventor for biological entities.
3. ………………………………….. is a loosely used term for molecules that are present in organisms.
4. …………………………………… is a group of standards which is used in control of relations between our work and bio-diversity.
5. The production of product, its extraction and process is called ……………………………….

Answer:

1. Biopiracy
2. Biopatent
3. Bio-molecules
4. Bio-code of conduct
5. Downstream.

3. Match the Following:

Answer:

1. (b)
2. (d)
3. (a)
4. (c).

4. Answer in One Word/Sentence:

1. Give the name of first transgenic crop.
2. What is the name of insect resistant protein which is transferred in Bt Cotton?
3. What is the name of first man-made insulin?
4. In which organism nif-genes are found?
5. Give the name of an antiviral protein.

Answer:

1. Tobacco
2. Cry protein
3. Humulin
4. Rhizobium
5. Interferon.

Biotechnology And its Application Very Short Answer Type Questions

Question 1.
Name the organism whose genetic material has been altered using genetic engineering techniques.
Answer:
Genetically Modified Organisms (GMO).

Question 2.
Name the crops which are prepaired with the help of biotechnology.
Answer:
Bt cotton, Bt maize, paddy, tomato, potato and soya been.

Question 3.
Through whom Bt toxin protein originates?
Answer:
By Bacillus thuringiensis.

Question 4.
By which Bt toxin is coded?
Answer:
By cry genes Bt toxin is coded.

Question 5.
Full form of RNAi.
Answer:
RNA interference (RNAi).

Question 6.
Name the therapy which is help of missing of defective ones in order to correct genetic disorders.
Answer:
Gene therapy.

Question 7.
Name the scientific name of bacteria in which be form organism toxin.
Answer:
Bacillus thuringiensis.

Biotechnology And its Application Short Answer Type Questions

Question 1.
What is genetically modified (GM) food ? Give two examples.
Answer:
Genetically modified food (GM food): The food substances produced from ge-netically modified crops or transgenic crops is called GM food. This food differ from conventionally developed varieties in the following aspects :

• GM food contains antibiotic resistance gene itself.
• It contains protein produced by transgene, e.g. Cry protein in insect resistance varieties.
• These GM foods contain enzyme produced by the antibiotic resistance gene that was used during gene transfer by recombinant DNA technology.

Examples of GM Crops, Food and Fruits:

1. Flavr Savr Tomato : It is the first food containing genetically engineered DNA. . These tomatoes contain genes for antibiotic resistance for kanamycin.

2. Maize : GM maize has a bacterial gene which increases its resistance to pests and
diseases. It also has a gene for ampicillin resistance which is harmful for us, therefore introduction of GM maize is opposed by many European countries. ,

3. Rape oil seed : It is a new type of plant that contain genes for resistance to the herbicide Basta. It has for more potential, dangers and can become a weed and would be impossible to control with Basta. It could cross fertilize with relatives such as wild mustard, thus, spreading the resistance to wild plants. Such type of environmental risks could occur with genetically modified rapeseed crop. They might also effect food chains in unpredictable ways.

Question 2.
Write down the advantage of GM Crops.
Answer:

Advantages of GM crops :

• Genetic modification has made crops more tolerant to abiotic stresses (cold, drought, heat, salt.)
• Viral resistance can be introduced.
• Over ripening losses can be reduced. Example: Flavr Savr Tomato.
• Enhanced nutritional value of food. Example: Golden Rice.
• Reduced reliance on chemical pesticides.

Disadvantages of GM crops :

• Transgenes in crop plants can endanger native species. Example: The gene for Bt toxin expressed in pollen may end natural pollinators such as honeybees.
• Weeds also become resistant.
• Products of transgenes may be allergic or toxic.
• They cause damage to the natural environment.

Question 3.
What is perfect agriculture? How is this method better than traditional method? Explain.
Answer:
Perfect agriculture is a method of agriculture which is sustainable, perfect and harmless. Green revolution and there after the production of agricultural crops has definitely increased due to use of new and high yielding varieties, development of irrigation facilities, increased irrigated area, use of fertilizers etc. but it results many problems such as loss of soil fertility, pollution of food and water and diseases. The resistance power of plants and human heings falls slowly. Food and water borne diseases affecting the health of human beings and animals.

All of these conditions and events taking place due to the modem commercial agriculture. Therefore, it would become necessary to develop a method of agriculture which would be free from above mentioned demerits. This kind of agriculture is called to be as perfect or sustainable agriculture. Organic agriculture is the best example of perfect agriculture.

Question 4.
What is organic cropping? What are its basis?
Answer:
Organic agriculture: Organic agriculture is a method of agriculture which does not allow the use of synthetic fertilizers, pesticides, insecticides, weedicides, plant growth regulators, substances of animal origin and genetically modified bacteria. In this method biofertilizers, crop rotation methods are used to increase crop production and biopesticides are used to control insects and weeds.

Thus, organic farming is a holistic way of agriculture which tries to bridge the widening gap between man and nature. If has the commitment of meeting production needs on one hand and sustaining resources tand ecosystem function on the other hand. Thus, organic farming is an alternative agriculture production system which avoid or largely excludes the use of synthetic chemicals, fertilizers, pesticides and growth regulating hormones and live stock additives.

Basics of Organic agriculture :

• Organic agriculture is based on improvement of soil, plants, animals, man and global scinery and make it sustainable.
• Organic agriculture is based on those ecosystems and biocycles which utilize that organisms which would be promoted.
• It is based on the principle which are related with making pollution free environment and possibilities of life.
• It is also based on saving environment and health of present and future generations.

Question 5.
What is gene library?
Answer:
Gene library: Several clones of cells, each clone containing one or a few foreign genes representing almost all the genes of an organism is referred to as genes library. From this gene library it is possible to identify a clone containing gene of interest. In order to obtain gene library of an organism, its genome is first cut into smaller DNA fragments containing one or a few genes such as fragments can be cloned into a cell which may brfthat of bacteria, yeast, insects, plant or animal cell.

When such a cell multiplies to form a group of cells, all cells will contain the same foreign DNA fragment which was introduced initially. These cells which have similar foreign DNA fragment are referred to as a clone of cells. Several clones of cells each clone containing one or a few foreign genes are finally obtained and is called gene library.

Biotechnology And its Application Long Answer Type Questions

Question 1.
What is eugenics? Write importance of eugenics.
Answer:
Eugenics: The branch of biology which deals with the study of improvements of human race is called eugenics.
Importance:

• Development of selective reproduction in similar species.
• Transfer of genetic materials in various organisms.
• Development of GM food and GM crops.
• Gene cloning
• Gene therapy, etc.

Question 2.
Explain the following in brief:

1. Biopiracy,
2. Biopatent.

Answer:
1. Biopiracy : Some organisations and multinational companies exploid biological resources and genetical resources indegenous to a country without proper authorisation. This is called biopiracy. In fact it is illegal removal of biological material. The process of biopyracy involves collection of samples of biological sources, which can be done unnoticed. This biological material is then subjected to product development for use on a commercial scale.

Today a range of biological resources are facing biopiracy. It includes plants and animals, micro-organisms genetic materials etc. Western companies are getting great benefits from using the knowledge and biological resources of the third world communities.While the companies stand to make huge revenue from this process, the local communities are unrewarded and infact, may have to buy the products of these companies at high prices.
To check illegal exploitation of biological resources Government of India has signed the General Agreement on Tariffs and Trade (GATT), which opens country natural resources for foreign exploitation.

2. Biopatent : The protection given by government to an inventor of biological material to secure him for a specific time the exclusive right of manufacturing, exploiting, using and selling of an invention is called biopotent.

Today manufacturing companies are being granted patents for products and technologies that make of biological resources, such as plants and animals, genetic materials which was identified developed and used by farmers and indegenous peoples.

There is growing worldwide opposition to the granting of patents on biological materials such as genes, plants, animals and human. Farmers and indegenous peoples are outaged that’ plants that they developed are being ‘hijacked’ by companies. Groups are diverse as religious leaders, parliamentarians and environment NGOs are intensifying campaign against corporate patenting of living things.

Question 3.
Describe the application of genetic engineering in the field of Agriculture and Medicine.
Answer:
(A) Application of Genetic engineering or Biotechnology in Agriculture : Genetic engineering is found to be very beneficial in agriculture. Its important use in agriculture are:
1. Increase in photosynthetic efficiency: An increase in photosynthetic efficiency of crop plants can be achieved by introducing suitable Carbon dioxide Fixation Gene (cfx) from any plant into the crop plants.

2. Transfer of nitrogen fixing ability: Number of symbiotic and non-symbiotic micro-organism have capacity of fixing atmospheric nitrogen. Nitrogen fixers are found to possess nitrogen fixing gene (nif genes) which are located on chromosomes or plasmids. Introduction of nif gene in crop plants results in ability in crop plants to fix atmospheric nitrogen and reduction in the use of chemical nitrogen fertilizers.

3. Disease resistance in crop plants : Plant breeders at present are developing high yield varieties by transferring gene for disease resistance through conventional breeding.

4. Plant tissue in crop improvement: Some of the areas of plant improvement where tissue culture has been applied with success are as follows :

• Rescuing hybrids through embryo culture.
• Multiplication of germplasm.
• Production of disease free plants.
• Production of haploid through another culture.
• Somaclonal variation.
• Somatic hybridization.
• Cryopreservation of germplasm.

5. VAM (Vesicular-Arbuscular Mycorrhiza) fungi with Rhizobium can boost the yields: Recently there has been a new dimension to this farm practice by the way of increasing Rhizobium inoculation effect by simultaneous inoculating seeds with VAM as well as Rhizobium culture.VAM are structural modification of hyphae helping in absorption and storage of phosphorus.

(B) Application of Genetic engineering in Medical field :
1. The hereditary diseases like colour-blindness, haemophilia which are caused by recessive genes and also many inborn metabolic disorders due to defective genes as alkaptonuria, phenylketonuria can be cured with the gene therapy.

2. Substances like vitamins, hormones, amino acids and antibodies can be synthesized in bacteria by introducing the genes which code these substances. In this way bacteria can be used as biofactories for the synthesis of these substances.

3. Production of insulin: Insulin is medicine used for the treatment of diabetes. Initially it is derived from animals (pig and cows) but today it is produced by gene splicing.

4. Hepatitis-B vaccine: Hepatitis-B is a viral disease of liver. Today this vaccine is prepared with the help of genetic engineering.

MP Board Class 12th Biology Solutions

MP Board Class 12th Biology Solutions Chapter 4 जनन स्वास्थ्य

जनन स्वास्थ्य NCERT प्रश्नोत्तर

प्रश्न 1.
समाज में जनन स्वास्थ्य के महत्व के बारे में अपने विचार प्रकट कीजिए।
उत्तर
जनन स्वास्थ्य का तात्पर्य जनन के सभी पहलुओं जैसे शारीरिक, भावनात्मक, व्यावहारिक तथा सामाजिक स्वास्थ्य से है। दुनिया में भारत पहला ऐसा देश है जिसने राष्ट्रीय स्तर पर जननात्मक स्वस्थ समाज को प्राप्त करने की कार्य योजनाएँ बनाई हैं। इन कार्यक्रमों को परिवार कल्याण के नाम से जाना जाता है। इनकी शुरूआत 1951 में (परिवार नियोजन के नाम से) हुई थी। जनन संबंधित और आवधिक क्षेत्रों को इसमें सम्मिलित करते हुए बहुत उन्नत व व्यापक कार्यक्रम फिलहाल ‘जनन एवं बाल स्वास्थ्य सेवा कार्यक्रम (आ.सी.एच.)’ के नाम से प्रसिद्ध है। इन कार्यक्रमों के अंतर्गत् जनन संबंधी विभिन्न पहलुओं के बारे में लोगों में जागरूकता पैदा करते हुए और जननात्मक रूप से संबंध समाज तैयार करने के लिए अनेक सुविधाएँ एवं प्रोत्साहन दिए जा रहे हैं।

प्रश्न 2.
जनन स्वास्थ्य के उन पहलुओं को समझाइए, जिस पर आज के परिदृश्य में विशेष ध्यान देने की जरूरत है
उत्तर
जनन स्वास्थ्य प्राप्ति के लिए विभिन्न कार्य-योजनाओं के सफलतापूर्वक क्रियान्वयन के लिए मजबूत संरचनात्मक सुविधाओं, व्यावसायिक विशेषज्ञता तथा भरपूर भौतिक सहारों की आवश्यकता होती है। लोगों को जनन संबंधी समस्याओं जैसे-सगर्भता (Pregnancy), प्रसव (Parturition), यौन संचारित रोगों, गर्भपात (Abortion), गर्भ निरोधकों, आर्तव चक्र संबंधी समस्याओं, बांझपन (Infertility) आदि के बारे में चिकित्सकीय सहायता देने के लिए बेहतर तकनीकों एवं नई कार्य योजनाओं को क्रियान्वित करने की भी आवश्यकता है ताकि लोगों की अधिक सुचारु रूप से देखभाल और सहायता की जा सके।

बढ़ती “मादा भ्रूण हत्या” की कानूनी रोक के लिए ऐम्नियोसेंटेसिस (Amniocentesis) जैसे लिंग परीक्षण पर वैधानिक प्रतिबंध तथा व्यापक बाल प्रतिरक्षीकरण (टीकाकरण) आदि कुछ महत्वपूर्ण कार्यक्रमों को शामिल किया गया है। जनन संबंधी विभिन्न अनुसंधानों को बढ़ावा देने के लिए हमारे देश की सरकारी एजेंसियाँ सतत् क्रियाशील हैं। लखनऊ स्थित केन्द्रीय औषध अनुसंधान संस्थान (Central Drug Research Institue CDRI) ने “सहेली” नामक गर्भनिरोधक गोली का निर्माण किया है। यौन संचारित रोगों की सही जाँच तथा देखभाल और लगभग सभी जनन स्वास्थ्य समस्याओं हेतु विकसित चिकित्सा सुविधाओं के होने से बेहतर समाज एवं जनन स्वास्थ्य के संकेत प्राप्त हो रहे हैं।

प्रश्न 3.
क्या विद्यालयों में यौन शिक्षा आवश्यक है ? यदि हाँ तो क्यों ?
उत्तर
हाँ, विद्यालयों में यौन शिक्षा आवश्यक है, ताकि छात्र/छात्राओं को यौन संबंधी विभिन्न पहलुओं के बारे में फैली हुई भ्रान्तियों एवं यौन संबंधी गलत धारणाओं से छुटकारा मिल सके। बच्चों को जनन अंगों, किशोरावस्था एवं उससे संबंधित परिवर्तनों, सुरक्षित और स्वच्छ यौन क्रियाओं, यौन संचारित रोगों एवं एड्स की जानकारी देना विशेष रूप से किशोर आयु वर्ग में जनन संबंधी स्वस्थ जीवन बिताने में सहायक होती है।

प्रश्न 4.
क्या आप मानते हैं कि पिछले 50 वर्षों के दौरान हमारे देश के जनन स्वास्थ्य में सुधार हुआ है? यदि हाँ तो इस प्रकार के सुधार वाले कुछ क्षेत्रों का वर्णन कीजिए।
उत्तर
जी हाँ, पिछले 50 वर्षों के दौरान हमारे देश के जनन स्वास्थ्य क्षेत्र में सुधार हुआ है। निम्न क्षेत्रों में हुए सुधार को आसानी से देखा जा सकता है

1. यौन संबंधी मामलों के बारे में बेहतर जागरुकता आयी है।
2. चिकित्सकीय देख-रेख में होने वाले प्रसवों की संख्या में वृद्धि हुई है।
3. प्रसवोत्तर देखभाल बेहतर हुई है।
4. मातृ मृत्यु-दर (MMR) में गिरावट आई है।
5. शिशु मृत्यु-दर (IMR) में गिरावट आई है।
6. यौन संचारित रोगों की सही जाँच-पड़ताल तथा देखभाल व उपचार बेहतर हुआ है।
7. जनसंख्या वृद्धि दर पर रोक लगी है।
   कुल मिलाकर सभी यौन समस्याओं हेतु बढ़ी हुई चिकित्सा सुविधाओं का होना समाज के बेहतर जनन स्वास्थ्य की ओर संकेत देता है।

प्रश्न 5.
जनसंख्या विस्फोट के कौन-से कारण हैं ?
उत्तर
शीघ्र एवं अनियमित जनसंख्या वृद्धि को जनसंख्या विस्फोट कहा जाता है।इसके प्रमुख कारण निम्नलिखित हैं

1. रोगों व महामारियों पर नियंत्रण
2. कृषि का विकास
3. संचार व आवागमन के साधन
4. उच्च जन्मदर
5. निम्न मृत्युदर
6. मनुष्य का वर्ष भर मैथुनकाल होना।

प्रश्न 6.
क्या गर्भ-निरोधकों का उपयोग न्यायोचित है ? कारण बताइए।
उत्तर
भारत में जनसंख्या वृद्धि-दर अत्यधिक होने के कारण राष्ट्रीय संकट उत्पन्न हो गया है। अतः गर्भ निरोधकों का उपयोग न्यायोचित है। इसके उपयोग से परिवार को सीमित किया जा सकता है एवं उनकी सुविधाओं में वृद्धि की जा सकती है। इन गर्भ निरोधकों के उपयोग से यौन संचारित रोगों (STDs) से बचा जा सकता है। इसके साथ ही दो संतानों के बीच अन्तराल भी रखा जा सकता है।जनसंख्या वृद्धि को कम करके परिवार, समाज व देश की समृद्धि में सहयोग कर सकते हैं।

प्रश्न 7.
जनन ग्रंथि को हटाना, गर्भ निरोधकों का विकल्प नहीं माना जा सकता है, क्यों?
उत्तर
जनन ग्रंथियाँ सन्तान उत्पन्न करने वाले अंग हैं । वृषण में शुक्राणुओं तथा अण्डाशय में अण्डों का निर्माण होता है। गर्भ निरोधक के लिए स्वस्थ अंगों को शरीर से हटाना उचित नहीं है। इससे मानसिक, शारीरिक स्वास्थ्य संबंधी समस्याएँ उत्पन्न हो सकती हैं। जबकि गर्भ निरोधक विधियाँ सुलभ होती हैं एवं उनका कोई बुरा प्रभाव भी शरीर पर नहीं पड़ता है। इनका आसानी से प्रयोग भी किया जा सकता है। जरूरत पड़ने पर इसे हटाया भी जा सकता है। जनन ग्रंथि को हटाना एक जटिल प्रक्रिया है, कानून हमें इसकी आज्ञा भी नहीं देता है।

प्रश्न 8.
उल्बवेधन एक घातक लिंग निर्धारण (जाँच) प्रक्रिया है, जो हमारे देश में निषेधित है। क्या यह आवश्यक होना चाहिए ? टिप्पणी कीजिए।
उत्तर
बढ़ती मादा भ्रूण हत्या की कानूनी रोक के लिए उल्बवेधन (Amniocentesis) जाँच (भ्रूणीय लिंग निर्धारण), लिंग परीक्षण पर वैधानिक प्रतिबंध उचित है क्योंकि यह एक खतरनाक प्रवृत्ति है। कई बार ऐसा देखा जाता है कि यह पता चलने पर कि भ्रूण मादा (लड़की) है, चिकित्सीय सगर्भता समापन (Medical termination of pregnancy) कराया जाता है, सामान्य भाषा में इसे गर्भपात (Abortion) कहा जाता है। यह पूरी तरह गैर-कानूनी है। इस प्रकार की प्रवृत्ति से बचना चाहिए क्योंकि यह माता एवं बच्चा (भ्रूण) दोनों के लिए खतरनाक है । इससे समाज में पुरुष एवं महिलाओं की संख्या का अनुपात भी बिगड़ सकता है। जिससे वैवाहिक तथा स्वास्थ्य संबंधी समस्याएँ भी उत्पन्न हो सकती हैं।

प्रश्न 9.
बन्ध्य दम्पत्तियों को संतान पाने हेतु सहायता देने वाली कुछ विधियाँ बताइए।
उत्तर
यदि दंपत्ति बच्चा पैदा करने में अक्षम है तथा ऐसे दोष को ठीक करने का इलाज संभव न हो तो कुछ विशेष तकनीकों के द्वारा उन्हें बच्चा पैदा करने में सहायता की जा सकती है, ये तकनीकें सहायक जनन प्रौद्योगिकी (ART) कहलाती हैं।
1. पात्रे निषेचन (Invitro fertilization IVF)-
इस विधि में अण्डाणु का निषेचन शरीर से बाहर लगभग शरीर के भीतर जैसी स्थिति में कराया जाता है। इसके पश्चात् स्त्री में भ्रूण स्थानान्तरण (Embryo transfer, ET) कराया जाता है।

• युग्मनज डिंबवाहिनी स्थानान्तरण (Zygote intrafallopian transfer ZIFT)-इस प्रक्रिया में प्रारंभिक भ्रूण को 8 ब्लास्टोमियर तक की अवस्था में स्त्री की डिंबवाहिनी या फैलोपियन नलिका में अग्रिम परिवर्धन के लिए स्थानांतरित किया जाता है।
• आन्तर गर्भाशयी स्थानान्तरण (Intrauterine transfer IUT)-जब भ्रूण को 8 ब्लास्टोमियर से अधिक अवस्था पर परिवर्तन हेतु स्त्री के गर्भाशय में स्थानान्तरित किया जाता है तो इसे IUT कहते हैं । यदि किसी स्त्री में गर्भधारण नहीं हो पाता है तो उसकी सहायता के लिए पात्रे निषेचन अर्थात् अन्य स्त्री के भीतर ही निषेचन कराने के बाद भ्रूण को उस स्त्री में स्थानान्तरित किया जा सकता है।
• टेस्ट ट्यूब बेबी (Test tube baby)-इस विधि में दाता स्त्री के अण्डे को दाता पुरुष शुक्राणु से प्रयोगशाला में परखनली के भीतर (स्त्री शरीर के बाहर) निषेचन कराया जाता है तथा निश्चित समय तक अनुरूपी परिस्थितियों में परिवर्धन के बाद अग्रिम परिवर्धन हेतु स्त्री के गर्भाशय में स्थानान्तरित कर दिया जाता है। इस विधि द्वारा जन्मे बच्चे को “टेस्ट ट्यूब बेबी” कहते हैं।

2. युग्मक आन्तर फैलोपीयन स्थानान्तरण (Gamete intrafallopian transfer GIFT)-
जिन स्त्रियों में अण्डाणु उत्पन्न नहीं होता है किन्तु इनमें निषेचन और भ्रूण परिवर्धन के लिए उचित वातावरण होता है उनमें यह विधि अपनायी जाती है। इसमें एक दाता स्त्री से अण्डाणु लिया जाता है तथा उस स्त्री को फैलोपीयन नलिका में स्थानांतरित कर दिया जाता है।

3. आन्तर कोशिकीय शुक्राणु निक्षेपण (Intra-cytoplasmic sperm injection)-
प्रयोगशाला में भ्रूण बनाने के लिए इस प्रक्रिया में शुक्राणु को सीधे ही अण्डाणु में अंत:क्षेपित कर दिया जाता है।

4. कृत्रिम वीर्य सेचन (Artificial insemination AI)-
यदि पुरुष-स्त्री को वीर्य सेचित करने में अक्षम हो अथवा उसके वीर्य में शुक्राणुओं की संख्या कम हो तो ऐसी अवस्था में यह तकनीक अपनायी जाती है। इस विधि में पति अथवा एक स्वस्थ दाता का वीर्य कृत्रिम रूप से स्त्री की योनि में प्रविष्ट कर दिया जाता है। जब कृत्रिम विधि से वीर्य को गर्भाशय में प्रविष्ट किया जाता है तो इसे गर्भाशय वीर्य सेचन (Intra-uterine insemination IUI) कहते हैं।

5. परपोषी मातृत्व (Host mothering)-
इस विधि में भ्रूण को प्राकृतिक माता से निकालकर एक अन्य स्त्री धात्रेय माता (Foster mother) में रोपित कर दिया जाता है। धात्रेय माता में यह भ्रूण जन्म तक अथवा अस्थायी रूप से निश्चित समय तक वर्धित होता है। जिसके बाद इन्हें पुनः मूल माता में या अन्य किसी स्त्री में रोपित कर दिया जाता है। यह तकनीक ऐसी स्त्रियों के लिए लाभदायक है जिनमें भ्रूण बन तो जाता है किन्तु पूर्ण परिवर्धन के समय तक ये इसे रख नहीं पाती हैं।

प्रश्न 10.
किसी व्यक्ति को यौन संचारित रोगों की चपेट में आने से बचने के लिए कौन-से उपाय अपनाने चाहिए?
उत्तर
यौन संचारित रोगों से बचाव (Protection from Sexually Transmitted Diseases)यौन संचारित रोगों से बचाव हेतु निम्न बातों का ध्यान रखना चाहिए

1. किसी अनजान व्यक्ति या बहुत से व्यक्तियों के साथ यौन संबंध न रखें।
2. मैथुन के समय हमेशा कंडोम का इस्तेमाल करें।
3. संक्रमित व्यक्ति का रुधिर किसी अन्य व्यक्ति को न चढ़ाया जावे।
4. पहले से उपयोग किये गये इंजेक्शन व सुईयों का प्रयोग न करें।
5. समलैंगिकता से बचें।
6. यदि कोई आशंका है तो तुंरत ही प्रारंभिक जाँच के लिए किसी योग्य चिकित्सक से मिलें और रोग की पहचान होने पर पूरा इलाज करावें।

प्रश्न 11.
निम्न वाक्य सही है या गलत, व्याख्या सहित बताइए

(क) गर्भपात स्वतः भी हो सकता है। (सही/ गलत)
उत्तर
गलत, सामान्य परिस्थितियों में गर्भपात नहीं होता। किसी दुर्घटनावश या स्वैच्छिक रूप से गर्भसमापन (गर्भपात) होता है।

(ख) बंध्यता को जीवनक्षम संतति न पैदा कर पाने की अयोग्यता के रूप में परिभाषित किया गया है और यह सदैव स्त्री की असामान्यताओं/दोषों के कारण होती है।(सही/गलत)
उत्तर
गलत, बंध्यता हमेशा स्त्री की असामान्यताओं/ दोषों के कारण नहीं होती, कभी-कभी पुरुष भी बंध्यता के लिए दोषी होता है।

(ग) एक प्राकृतिक गर्भ निरोधक उपाय के रूप में शिशु को पूर्णरूप से स्तनपान कराना सहायक होता है। (सही/गलत)
उत्तर
सही, शिशु को पूर्णरूप से स्तनपान कराने से अण्डोत्सर्ग नहीं होता है। अतः आर्तव चक्र (Menstrual cycle) भी नहीं होता है, जिसके कारण गर्भ की संभावनाएँ समाप्त हो जाती हैं। किन्तु यह विधि शिशु के जन्म के अधिकतम 6 माह तक कारगर है।

(घ) लोगों के जनन स्वास्थ्य के सुधार हेतु यौन संबंधित पहलुओं के बारे में जागरुकता पैदा करना एक प्रभावी उपाय है।(सही/गलत)
उत्तर
सही, क्योंकि ऐसा करने से लोगों की जनन स्वास्थ्य की समस्याएँ समाप्त अथवा कमतर हो जाती हैं।

प्रश्न 12.
निम्नलिखित प्रश्नों को सही कीजिए

(क) गर्भ निरोधक के शल्य क्रियात्मक उपाय युग्मक बनने को रोकते हैं।
उत्तर
युग्मक बनने से नहीं, बल्कि युग्मकों के परिवहन को रोकते हैं।

(ख) सभी प्रकार के यौन संचारित रोग पूरी तरह उपचार योग्य हैं।
उत्तर
हिपेटाइटिस-B, एड्स एवं जननिक हर्पिस का उपचार नहीं होता।

(ग) ग्रामीण महिलाओं के बीच गर्भ निरोधक के रूप में गोलियाँ (पिल्स) बहुत अधिक लोकप्रिय
उत्तर
ग्रामीण महिलाओं में गर्भ निरोधक के रूप में गोलियाँ (पिल्स)लोकप्रिय नहीं हैं । ग्रामीण महिलाओं को यौन शिक्षा की आवश्यकता है।

(घ) ई. टी. तकनीकों में भ्रूण को सदैव गर्भाशय में स्थानान्तरित किया जाता है।
उत्तर
ई. टी. तकनीकों में 8 ब्लास्टोमियर से ज्यादा अवस्था वाले भ्रूण को गर्भाशय में स्थानान्तरित किया जाता है। जबकि 8 ब्लास्टोमियर से कम अवस्था वाले भ्रूण को अण्डवाहिनी में स्थानान्तरित किया जाता है।

जनन स्वास्थ्य अन्य महत्वपूर्ण प्रश्नोत्तर

जनन स्वास्थ्य वस्तुनिष्ठ प्रश्न

1. सही विकल्प चुनकर लिखिए-

प्रश्न 1.
स्त्रियों के शरीर से फैलोपियन ट्यूब को अलग करना कहलाता है
(a) वैसेक्टोमी
(b) ट्यूबेक्टोमी
(c) ओवरीक्टोमी
(d) कैस्ट्रेशन।
उत्तर
(b) ट्यूबेक्टोमी

प्रश्न 2.
आबादी की सर्वाधिक वृद्धि का प्रमुख कारण है
(a) कम मृत्यु दर
(b) जन्मदर में वृद्धि
(c) अकाल न पड़ना
(d) युद्ध कम होना।
उत्तर
(a) कम मृत्यु दर

प्रश्न 3.
बड़े शहरों में अधिक जनसंख्या का कारण है
(a) शिक्षा के अवसर
(b) उपलब्ध भौतिक सुविधाएँ
(c) अधिक आय के स्रोत
(d) उपर्युक्त सभी।
उत्तर
(d) उपर्युक्त सभी।

प्रश्न 4.
जनसंख्या घनत्व अधिक है____
(a) यू.एस.ए. में
(b) भारत में
(c) चीन में
(d) जापान में।
उत्तर
(c) चीन में

प्रश्न 5.
AIDS रोग फैलता है
(a) बैक्टीरिया से
(b) प्रोटोजोआ से
(c) वाइरस से
(d) फंगस से।
उत्तर
(c) वाइरस से

प्रश्न 6.
यौन संक्रमित रोगों के कारक होते हैं
(a) विषाणु
(b) जीवाणु
(c) प्रोटोजोआ
(d) उपर्युक्त सभी।
उत्तर
(a) विषाणु

प्रश्न 7.
भारत में जनसंख्या की गणना की गई थी
(a) सन् 1891
(b) सन् 1947
(c) सन् 1950
(d) सन् 1961.
उत्तर
(a) सन् 1891

प्रश्न 8.
निम्न में से जन्मदर को नियंत्रित करने की विधि
(a) TUD
(b) GIFT
(c) MIF
(d) IVEET
उत्तर
(a) TUD

प्रश्न 9.
एल्कोहॉल से सर्वाधिक प्रभावित अंग है
(a) यकृत
(b) सेरीब्रम
(c) सेरीबेलम
(d) हृदय।
उत्तर
(a) यकृत

प्रश्न 10.
विश्व की जनसंख्या वृद्धि दर है
(a) 2.4%
(b) 2%
(c) 3%
(d) 4%.
उत्तर
(b) 2%

2. रिक्त स्थानों की पूर्ति कीजिए

1. मानव जनसंख्या का सांख्यिकीय अध्ययन ………………. कहलाता है।
2. माल्थस के अनुसार जनसंख्या में ………………. वृद्धि होती है जबकि खाद्य उत्पादन में ……………….. वृद्धि होती है।
3. किसी समष्टि में एक निश्चित अवधि में जीवों का आना और बाहर जाना ………………. कहलाता
4. शुक्राणु नली को काटकर बन्द करना ………………. कहलाता है।
5. स्थलीय जीवों के जनसंख्या घनत्व के लिये स्थान के क्षेत्रफल को ……………… से प्रदर्शित करते हैं।
6. जन्मदर एवं अप्रवासन जनसंख्या घनत्व में ………………. करते हैं।
7. संशोधित जन्मदर ……………… जन्मदर से अधिक होती है।
8. जनसंख्या आकार को नियन्त्रित करने वाले कारक को …………….. कहते हैं।
9. जैव सूचकांक = ( ………………. / मृत्युदर) x 1000.
10. बालिकाओं में बालकों की अपेक्षा ……………. परिपक्वता आती है।
11. भारत पहला देश है जिसने …………………… स्वास्थ्य देखभाल को ………………….में सामाजिक लक्ष्य के रूप में लाया।
12. RCH भारत में .. ……………. में लाया गया।
13. एम्नियोसेन्टेसिस में लिंग निर्धारण …………. गुणसूत्र के निरीक्षण द्वारा लाया गया।
14. प्रसव के ठीक बाद महिलाएँ …… अनार्तव (मासिक धर्म amoeriorrhoea) का अनुभव करती हैं।
15. LNG-20 ……………….. स्रावी IUD है।
उत्तर

1. डेमोग्राफी
2. बीजगणितीय, अंकगणितीय
3. प्रवासन
4. वैसेक्टोमी
5. द्विविमा (m²)
6. वृद्धि,
7. अशोधित
8. धनात्मक अवरोधक
9. जन्मदर
10. जल्दी
11. प्रजनन, सन् 1951
12. सन् 1997,
13. लिंग
14. स्तनपान संबंधी
15. हॉर्मोन।

3. सही जोड़ी बनाइए

‘A’ – ‘B’

1. S-आकृति वक्र – (a) जन्म नियंत्रण विधि
2. kसीमा स्थिरांक – (b) जनसंख्या
3. शुक्राणु नाशक रसायन – (c) लॉजिस्टिक वक्र
4. माल्थस 2009 – (d) D/Px1000
5. वास्तविक मृत्युदर – (e) संतृप्ति अवस्था।
उत्तर
1.(c), 2.(e), 3.(a), 4. (b), 5.(d).

‘A’ – ‘B’

1. माल्थस – (a) कॉपर-टी
2 वैसेक्टोमी – (b) लिंग परीक्षण
3. ट्यूबेक्टोमी – (c) सेन्सस
4. जनगणना – (d) जनसंख्या पर निबंध
5. एम्निओसेण्टेसिस – (e) पुरुष
6. आई.ग.सी.डी. – (f) स्त्री या जनसंख्या नियंत्रण।
उत्तर
1.(d),2.(e),3. (1), 4.(c), 5.(b), 6.(a).

4. एक शब्द में उत्तर दीजिए

1. दो ऐसे कारण दीजिए जिनसे जनसंख्या अनियन्त्रित रूप से बढ़ जाती है।
2. IUCD का पूरा नाम लिखिए।
3. मानव जनसंख्या का अध्ययन क्या कहलाता है?
4. माला D और N में कौन-सा रासायनिक गर्भ-निरोधक पाया जाता है ?
उत्तर

1. (i) कम उम्र में विवाह, (ii) शिक्षा
2. इन्ट्रायूरिक कन्ट्रासेप्टिव डिवाइसेस
3. डेमोग्राफी (Demography)
4. एस्ट्रोजन एवं प्रोजेस्टीरॉन हॉमोन।

जनन स्वास्थ्य अति लघु उत्तरीय प्रश्न

प्रश्न 1.
परिवार नियोजन को अब किस नाम से जाना जाता है ?
उत्तर
परिवार नियोजन को अब “परिवार कल्याण” के नाम से जाना जाता है।

प्रश्न 2.
सेन्ट्रल ड्रग रिसर्च इन्स्टीट्यूट (CDRI) कहाँ अवस्थित है।
उत्तर
लखनऊ (उत्तरप्रदेश) में स्थित है।

प्रश्न 3.
कंडोम के उपयोग के कोई एक लाभ लिखिए।
उत्तर
कंडोम यौन संचरित रोगों से बचाव करता है।

प्रश्न 4.
हफ्ते में एक बार लेने वाली गर्भनिरोधक गोली का नाम लिखिए।
उत्तर
“सहेली”

प्रश्न 5.
IUCD काशब्द विस्तार बताइए।
उत्तर
इन्ट्रा यूटेराइन कॉन्ट्रासेप्टिव डिवाइस।

प्रश्न 6.
STD का सम्पूर्ण रूप लिखिये।
उत्तर
यौन संचारित रोग (Sexually Transmitted Diseases)।

प्रश्न 7.
विवाह की वैधानिक आयु स्त्री और पुरुष के लिए क्या सुनिश्चित है ?
उत्तर
स्त्री की आयु 18 वर्ष तथा पुरुष के लिए 21 वर्ष सुनिश्चित है।

प्रश्न 8.
दो STD के नाम लिखिए जो संदूषित रक्त से संचारित होते हैं।
उत्तर
(a) एड्स
(b) हिपेटाइटिस-BI

प्रश्न 9.
लैंगिक संपर्क से होने वाले कोई दो रोग बताइए।
उत्तर

• सूजाक (Gonorrhoea)
• सिफलिस (Syphilis)

प्रश्न 10.
HIV एवं AIDS का सम्पूर्ण रूप लिखिए।
उत्तर

• HIV—ह्यूमन इम्यूनोडेफीसिएन्सी वाइरस।
• AIDS- एक्वायर्ड इम्यूनोडेफीसिएन्सी सिन्ड्रोम।

प्रश्न 11.
ZIFT का पूरा नाम लिखिए।
उत्तर
युग्मनज अन्तः डिम्बवाहिनी स्थानान्तरण (Zygote Intra-fallopian Transfer)।

प्रश्न 12
पुरुष में नसबंदी की शल्य क्रिया का क्या नाम है ?
उत्तर
वैसेक्टोमी (Vasectomy)।

प्रश्न 13.
दो ऐसे कारण दीजिये जिनसे जनसंख्या अनियन्त्रित रूप से बढ़ जाती है।
उत्तर
रोगों का नियंत्रण, कृषि का विकास।

प्रश्न 14.
मानव जनसंख्या का अध्ययन क्या कहलाता है ?
उत्तर
डेमोग्राफी।

प्रश्न 15.
माला D और N में कौन-सा रासायनिक गर्भ निरोधक पाया जाता है ?
उत्तर
एस्ट्रोजन एवं प्रोजेस्ट्रॉन।

प्रश्न 16.
एम्नियोसेण्टेसिस विधि द्वारा किसका परीक्षण किया जाता है ?
उत्तर
भ्रूण का लिंग परीक्षण।

प्रश्न 17.
किसी इकाई क्षेत्रफल में पाई जाने वाली मानव संख्या कहलाती है ?
उत्तर
जनसंख्या घनत्व।

जनन स्वास्थ्य लघु उत्तरीय प्रश्न

प्रश्न 1.
परखनली शिशु कैसे प्राप्त किये जा सकते हैं ?
उत्तर
स्त्रियाँ जब गर्भधारण करने योग्य नहीं होती हैं, उस स्थिति में परखनली शिशु तकनीक का उपयोग किया जाता है। स्त्रियों के अनिषेचित अण्डाणु को ऐण्टिसेप्टिक स्थिति में निकाला जाता है। इस अण्डाणु को परखनली में लेकर शुक्राणु द्वारा निषेचन (Fertilization) की क्रिया कराते हैं । निषेचित अण्डाणु से विदलन क्रिया द्वारा 32 कोशिकीय अवस्था वाला भ्रूण बनता है, जिसे ब्लास्टोसिस्ट (Blastocyst) कहते हैं। इस 32 कोशिकीय भ्रूण का रोपण स्त्रियों के गर्भाशय में कर दिया जाता है जहाँ पर इस भ्रूण का अगला विकास होता है और शिशु बनता है जिसे ‘टेस्ट ट्यूब बेबी’ कहते हैं।

प्रश्न 2.
गिफ्ट (GIFT) क्या है ?
उत्तर
GIFT-इसका पूरा नाम गैमीट इन्ट्रा फैलोपियन ट्रान्सफर (Gamete Intra Fallopian Transfer) है । यह गर्भ निरोधन की आधुनिक तकनीक है। इस तकनीक में शुक्राणु कैथेटर (Catheter) में संगृहीत कर लिया जाता है। वाश स्विम तकनीक (Wash swim up technique) द्वारा शुक्राणुओं को अण्डाणु के पीछे कैथेटर में संगृहीत किया जाता है। यह सम्पूर्ण क्रिया स्त्रियों की फैलोपियन नलिका में होती है। निषेचन क्रिया शुक्राणुओं की संख्या कम होने से पूर्ण नहीं हो पाती है।

प्रश्न 3.
ट्यूबेक्टोमी और वैसेक्टोमी को समझाइये।
उत्तर
ट्यूबेक्टोमी-इसमें स्त्री की अण्डवाहिनी को काटकर बाँध दिया जाता है। जिससे अण्डाणु गर्भाशय में नहीं आ पाते और निषेचन नहीं हो पाता। इसे ट्यूबेक्टोमी कहते हैं। यह परिवार नियोजन की एक विधि है। वैसेक्टोमी-यह भी परिवार नियोजन की एक विधि है जो पुरुषों द्वारा अपनाई जाती है। इसमें पुरुष की शुक्राणु नलिका को काटकर बाँध दिया जाता है जिससे शुक्राणु स्खलन के समय नहीं निकलते और निषेचन नहीं हो पाता। यह क्रिया वैसेक्टोमी कहलाती है।

प्रश्न 4.
एम्नियोसेण्टेसिस क्या है ?
उत्तर
एम्नियोसेण्टेसिस भ्रूण परीक्षण की एक तकनीक है जिसमें सर्जिकल सुई द्वारा मादा के गर्भाशय से एम्नियोटिक द्रव को शरीर से बाहर निकाला जाता है और एम्नियोटिक द्रव में उपस्थित फोयटस कोशा का संवर्धन किया जाता है और इसका गुणसूत्रीय परीक्षण करके निम्न बातों का पता लगाया जाता है

• गुणसूत्रीय असामान्यता, जैसे-डाउन सिण्ड्रोम, फिलोडेल्फिया सिण्ड्रोम एवं एडवर्ड सिण्ड्राम।
• उपापचयी अनियमितताएँ, जैसे- PKU, क्रिटेनिज्म, एल्केप्टोन्यूरिया।
• लिंग भ्रूण के परीक्षण में इसका उपयोग किया जाता है।

प्रश्न 5.
मानव जनसंख्या वृद्धि के सामाजिक कारण समझाइये।
उत्तर
मानव जनसंख्या वृद्धि के प्रमुख सामाजिक कारण निम्नलिखित हैं

• समाज का अशिक्षित होना ।
• निम्न सामाजिक स्तर का होना।
• समाज में विभिन्न प्रकार की कुरीतियाँ होना।
• विभिन्न प्रकार की सामाजिक मान्यताएँ ।
• कम उम्र में विवाह करना, क्योंकि उम्र के पूर्वार्द्ध में प्रजनन क्षमता अधिक होती है।
• निम्न सामाजिक स्तर के कारण मनोरंजन का अभाव होना।
• सामाजिक मान्यताएँ जैसे-पुत्र रत्न की प्राप्ति से मोक्ष मिलता है
• सामाजिक पिछड़ापन।

MP Board Class 12th Biology Solutions

MP Board Class 12th Biology Solutions Chapter 7 Evolution

Evolution NCERT Textbook Questions and Answers

Question 1.
Explain antibiotic resistance observed in bacteria in light of Darwinism selection theory.
Answer:
Darwinism theory of natural selection states that environment selects organisms with favourable variations and these organisms thus, survive and reproduce. It is observed when bacterial populations are exposed to certain antibiotic, the sensitive bacteria could not tolerate and hence, died due to the adverse environment. Whereas some bacteria that developed mutation became resistant to the particular antibiotic and survived. As a result such resistant bacteria survive and multiply quickly as compared to other sensitive bacteria. So, the whole population is regained by multiplication of resistant variety and antibiotic resis¬tant gene becomes widespread in the bacterial population.

Question 2.
Find out from newspapers and popular science articles any new fossil dis-coveries or controversies about evolution.
Answer:
Fossils of dinosaurs have revealed the evolution of reptiles in Jurassic period. As a result of this evolution of other animals such as birds and mammals has also been discovered. However, two unusual fossils recently unearthed in China have ignited a controversy over the evolution of birds confiiciusomis is one such genus of primitive birds that were crow sized and lived during the Cretaceous period in China.

Question 3.
Attempt giving a clear definition of the term species.
Answer:
A species generally includes similar organism. Members of this group can show interbreeding. Similar group of genes are found in the members of same species and this group has capacity to produce new species. Every species has some cause of isolation which intruped the interbreeding with nearest reactional species which is refer as reproductively isolated.

Question 4.
Try to trace the various components of human evolution (Hint: Brain size and function, skeletal structure, dietary preference, etc.)
Answer:
The various components of human evolution are as follows:

1. Brain capacity.
2. Posture
3. Food/Dietary preference and their important features.

Name brain capacity, posture and food features :

• Dryopithecus Africans : Knuckle walker, walked similar to Gorillas and Chim-panzees (was more apelike) soft fruit and leaves; canines large, arm and legs are of equal size.
• Ramapithecus: Semi-erect (more manlike) seeds, nuts canines were small while molars were large.
• Australopithecus africanus: Australopithecus africanus 450 cm3 full erect prosture, height (1 -05m) Herbivorous (ats fruits). Hunted with stone weapons, lived as trees, canines and incisors were small.
• Homo habilus: Homo habilus 735 cm3 fully erect posture, height (1 -5m) carnivo¬rous canines were small. They were first tool makers.
• Homo erectus: Homo erectus 800-1100 cm3 fully erect posture, height (1-5- l-8m) omnivorous. They used stone and bone tools for hunting games.
• Homo neanderihalensis : Homo neanderthalensis 1300 – 1600 cm3 fully erect posture, height (1-5 – l-6m) omnivorous cave dwellers, used hiles to protect their bodies and buried their dead.
• Homo sapiens fossils : Homo sapiens fossils 1650 cm3 fully erect posture with height (l -8m) omnivorous. They had strong jaw with teeth close together. They were cave dwellers, made painting and carvings in the caves. They developed a culture and were called first modem man.
• Homo sapiens sapiens : Homo sapiens sapiens 1200 – 1600 cm3 fully erect pos¬ture, height (1-5 – l-8m) omnivorous. They are the living modem men with high intelligence. They developed art, culture, language, speech, etc. They cultivated crops and domes¬ticated animals.

Question 5.
Find out through internet and popular science articles whether animals other than man has self-consciousness.
Answer:
There are many animals other than humans, which have self-consciousness. An example of an animal being self-conscious is dolphins. They are highly intelligent. They have a sense of self and, they also recognize others among themselves and others. They communicate with each other by whistles, tail-slapping, and other body movements, not dolphins, there are certain other animals such as Crow, Parrot, Chimpanzee, Gorilla, Orangutan, etc., which exhibit self-consciousness.

Question 6.
List 10 modern day animals and using the internet resources link it to a corresponding ancient fossil. Name both.
Answer:
Modern and Ancient corresponding animals :

Question 7.
Practise drawing various animals and plants.
Answer:

Question 8.
Describe one example of adaptive radiation.
Answer:
Darwin’s finches in the Galapagos Island once had a common ancestor but with evolution, they modified into different types according to their food habitat.

Question 9.
Can we call human evolution as adaptive evolution ?
Answer:
No, we can not be called human evolution as adaptive evolution.

Question 10.
Using various resources such as your school Library or the internet and discussions with your teacher, trace the evolutionary stages of any one animal, say horse.
Answer:
The evolution of horse started with Eohippus during Eocene period. It involved the following evolutionary stages :

• Gradual increase in body size
• Elongation of head and neck region
• Increase in the length of limbs and feet
• Gradual reduction of lateral digits
• Enlargement of third functional toe
• Strengthening of the back
• Development of brain and sensory organs
• Increase in the complexity of teeth for feeding on gras

• Eohippus : It had a short head and neck. It had four functional toes and a splint of 1 and 5 on each hind limb and a splint of 1 and 3 in each forelimb. The molars were short crowned that were adapted for grinding the plant diet.
• Mesohippus : It was slightly taller than Eohippus. It had three toes in each foot.
• Merychippus : It had the size of approximately 100 cm. Although it still had three toes in each foot, but it could run on one toe. The side toe did not touch the ground. The molars were adapted for chewing the grass.
• Pliohippus : It resembled the modern horse and was around 108 cm tall. It had a single functional toe with splint of 2nd and 4th in each limb.

Equus : Pliohippus gave rise to Equus or the modern horse with one toe in each foot. They have incisors for cutting grass and molars for grinding food.

Evolution Other Important Questions and Answers

Evolution Objective Type Questions

1. Choose the Correct Answer.

Question 1.
These are found at the atmosphere of earth before evolution:
(a) Water vapour, CH₄, NH₃ and Oxygen
(b) CO₂, NH₃, H₂ and Water vapour
(c) CH₄, NH₃, H₂ and Water vapour
(d) CH₄, O₃, O₂ and Water vapour.
Answer:
(c) CH₄, NH₃, H₂ and Water vapour

Question 2.
Which gas is absent at the atmosphere of earth at the time of origin of earth:
(a) NH₃
(b) H₂
(c) O₃
(d) CH₄
Answer:
(c) O₃

Question 3.
Which gaseous mixture are used and found the amino acid by Miller:
(a) Methane, Ammonia, Hydrogen and Water vapour
(b) Methane, Ammonia, Nitrogen and Water vapour
(c) Methane, Nitrogen, Hydrogen and Water vapour
(d) Ammonia, Carbon dioxide, Nitrogen and Water vapour.
Answer:
(a) Methane, Ammonia, Hydrogen and Water vapour

Question 4.
Who gave the theory “Survival of the fittest” :
(a) Charles Darwin
(b) Herbert Spencer
(c) Lamarck
(d) Hugo de Vries.
Answer:
(b) Herbert Spencer

Question 5.
Who gave the theory of inheritance of acquired, characters :
(a) Charles Darwin
(b) Lamarck
(c) Valles
(d) de Vries.
Answer:
(b) Lamarck

Question 6.
Who wrote origin of species :
(a) Oparin
(b) Beajman
(c) Lamarck
(d) Darwin.
Answer:
(d) Darwin.

Question 7.
What is the name of Darwin’s ship:
(a) Gangotri
(b) Beagle
(c) Atlantic
(d) Seagull.
Answer:
(b) Beagle

Question 8.
‘Atmosphere is a factor of variation’ who gave this theory :
(a) Mendel
(b) Darwin
(c) Lamarck
(d) Laplace.
Answer:
(c) Lamarck

Question 9.
Darwinism explain it:
(a) Characters are grown by heredity
(b) Species are changed into structural form with time .
(c) Nature selected the animals which are adapted
(d) Origin of evolution due to effect of environment.
Answer:
(c) Nature selected the animals which are adapted

Question 10.
Which antibiotic is used to Replica planting experiment of Ladderberg:
(a) Penicillin
(b) Streptomycin
(c) Erythromycin
(d) Neomycin.
Answer:
(a) Penicillin

Question 11.
Which is the basic unit of natural selection:
(a) Species
(b) Community
(c) Genous
(d) Solitary organism.
Answer:
(c) Genous

Question 12.
Who gave the theory of natural selection :
(a) Lamarck
(b) de Vries
(c) Darwin
(d) Mendel.
Answer:
(d) Mendel.

Question 13.
Darwinism is based on:
(a) Segregation
(b) Independent assortment
(c) Quantitative heredity
(d) Natural selection.
Answer:
(b) Independent assortment

Question 14.
Who discover the uses and experiment of organs:
(a) Valles
(b) Lamarck
(c) Darwin
(d) de Vries
Answer:
(a) Valles

Question 15.
The unit of natural selection :
(a) Solitary animal
(b) Family
(c) Community
(d) Species.
Answer:
(d) Species.

Question 16.
Mule is a product of:
(a) Mutation
(b) Reproduction
(c) Inter-species hybridization
(d) Intra-species hybridization
Answer:
(b) Reproduction

Question 17.
Who use first the word ‘species’:
(a) Linnaeus
(b) John
(c) Aristotle
(d) Darwin
Answer:
(b) John

Question 18.
Homologus organs are :
(a) Similar in origin
(b) Similar in function
(c) Similar in evolution
(d) Similar in behaviour.
Answer:
(a) Similar in origin

Question 19.
Which era are called as golden period of Reptilia or Dinosaurs :
(a) Mesozoic
(b) Cenozoic
(c) Palaeozoic
(d) Cenozoic.
Answer:
(a) Mesozoic

Question 20.
Dinosaurs are distinct in this era:
(a) Jurassic
(b) Triassic
(c) Cretaceous
(d) Permian.
Answer:
(c) Cretaceous

Question 21.
Vestigial organs of human’ are:
(a) Wisdom teeth, Coccyx, Nail, Eyelids, Vermiform appendix
(b) Wisdom teeth, Coccyx, Vermiform appendix, Pancreas
(c) Wisdom teeth, Coccyx, Vermiform appendix, Nictating membrane
(d) Coccyx, Wisdom teeth, Nail, Auricular muscle.
Answer:
(c) Wisdom teeth, Coccyx, Vermiform appendix, Nictating membrane

Question 22.
How made fossils :
(a) Animals are naturally buried in land after death
(b) Animals are decomposed by decomposers
(c) Animals are eaten by their hunter species
(d) Animals are destroyed due to environmental conditions.
Answer:
(d) Animals are destroyed due to environmental conditions.

Question 23.
Species which are similar in shape and different in reproduction are called :
(a) Sub-species
(b) Sibling
(c) Isotopic
(d) Allopetric species.
Answer:
(c) Isotopic

Question 24.
Name the ship which had been used by Darwin for journey :
(a) Ciloge
(b) Beagle
(c) Seagull
(d) Atlantic.
Answer:
(b) Beagle

Question 25.
Life is not in this era :
(a) Mesozoic
(b) Palaeozoic
(c) Cenozoic
(d) Azoic.
Answer:
(d) Azoic.

Question 26.
Which factor is responsible for evolution due to New-darwinism theory :
(a) Mutation
(b) Usefull variation
(c) Hybridization
(d) Mutation and Natural selection.
Answer:
(d) Mutation and Natural selection.

Question 27.
Found the age of fossils by which :
(a) By quantity of calcium ions
(b) By quantity of organic radioactive components
(c) By struggle from other mammals
(d) By structure of bones.
Answer:
(d) By structure of bones.

Question 28.
Darwin finches are related to which of the following evidences :
(a) Fossils
(b) Embryology
(c) Anatomy
(d) Geographical distribution.
Answer:
(d) Geographical distribution.

Question 29.
Who gave Recapitulation theory :
(a)Weizmann
(b) Haeckel
(c) Darwin
(d) Malthus.
Answer:
(b) Haeckel

Question 30.
Which definition of organic evolution is correct:
(a) Evolutionary history of a species
(b) History of species with the variations is the species
(c) Embryonal history of species
(d) Development of species.
Answer:
(a) Evolutionary history of a species

Question 31.
What is the perfect sequence of development of human:
(a) Peking man, Heidelberg man, Neanderthal Cro-Magnon
(b) Peking man, Homo sapiens, Cro-Magnon, Neanderthal
(c) Peking man, Neanderthal,.Homo sapiens, Heidelberg
(d) Peking man, Cro-Magnon, Homo sapiens, Neanderthal.
Answer:
(a) Peking man, Heidelberg man, Neanderthal Cro-Magnon

Question 32.
Which of the following is the most primitive ancestor of man :
(a.) Australopithecus
(b) Ramapithecus
(c) Homo habilis
(d) Homo neanderthalensis.
Answer:
(b) Ramapithecus

Question 33.
Ancestral amphibians were tetrapods that evolved during :
(a) Jurassic period
(b) Cretaceous period
(c) Devonian period
(d) Carboniferous period.
Answer:
(c) Devonian period

2. Fill in the Blanks :

1. Earth is a member of …………….
2. ……………. wrote ‘Origin of life’.
3. Oxygen is find the earth because photo-synthetic organisms are present on the earth, this phenomenon is called …………….
4. ………….. is the connecting link between Reptiles and Aves.
5. …………………. is the ancesters of horse.
6. The evolution of ……………. molecule set the stage for evolution of autotrophs.
7. Evolution of birds and mammals were ……………. era.
8. Golden era of Dinosaurs is ……………. period.
9. ……………. fossil man has been known from Shivalik hills in India.
10. Book of Charles Darwin “Origin of life” has been explained of …………….
11. ……………. be change of hareditary characters.

Answer:

1. Solar system
2. Oparin
3. Oxygen revolution
4. Archaeopteryx
5. Eohippus
6. Chlorophyll
7. Jurassic
8. Mesozoic
9. Ramapithecus
10. Natural selection
11. Mutation.

Question 3.
Match the Following :

Answer:

1. (d)
2. (a)
3. (e)
4. (c)
5. (b)

Answer:

1. (d)
2. (c)
3. (e)
4. (a)
5. (b)
6. (f).

4. Answer in One Word / Sentence:

1. The matter which gave rise to the universe 15 billion years ago.
2. The matter which is a link between acellular and cellular system.
3. The island at which Darwjn studies the organisms for his theory of evolution.
4. Origin of two or more unlike species from a common ancestor.
5. The whole process of development at changes from embryo to adult organism.
6. The process of changes in the course of origin of a new species.
7. The theory of organic evolution proposed by Darwin.
8. The permanent, heritable and spontaneous changes in organisms.
9. The remains of ancient organisms.
10. The organs which are different in their origin but have similar function.

Answer:

1. Ylem
2. Coacervates
3. Galopegos
4. Adaptive radiation
5. Ontogeny
6. Phytogeny
7. Natural selection
8. Mutation
9. Fossils
10. Homologous organs.

Evolution Very Short Answer Type Questions

Question 1.
Name the non-cellular organism.
Answer:
Virus.

Question 2.
Name the theory which tell about the milky way and formation of star.
Answer:
Big Bang.

Question 3.
Releasing of 02 by the evolution of photosynthesis an ancient time is called?
Answer:
Oxygen revolution.

Question 4.
Which planet has the possibility of origin of life other than earth?
Answer:
ars.

Question 5.
Where did life originate ?
Answer:
In water.

Question 6.
What is the art of limitation, which is practiced the world over from annoying siblings or plant resembles another creature to gain other advantages?
Answer:
Mimicry.

Question 7.
Name the method of the changing of the structure of a gene resulting in a variant form that may be transmitted to.
Answer:
Mutation.

Question 8.
What is the collection of different genes within an inter breading population?
Answer:
Gene pool.

Question 9.
Who wrote origin of species?
Answer:
Charles Darwin.

Question 10.
Who proposed the theory of natural selection?
Answer:
Darwin.

Question 11.
Who proposed the Recapitulation theory or Biogenetic law?
Answer:
Haeckel.

Question 12.
Astronomical distance measured in?
Answer:
Astronomical distance measured in Light years.

Question 13.
Name the theory by which earth is said to originate.
Answer:
The big bang theory.

Question 14.
What is fossil ?
Answer:
Fossils are the remains or impressions of ancient organisms preserved in sedimentary rocks or other media.

Question 15.
What is mutation ?
Answer:
New species originate due to changes of hereditary characters are called mutation.

Question 16.
Name the scientist who tell the spontaneous theory is wrong.
Answer:
Louis Pasteur’s.

Question 17.
Which era is called golden period of Dinosaurs?
Answer:
Mesozoic period is called golden period of Dinosaurs.

Question 18.
In which ship Darwin studied the nature ?
Answer:
Beagle.

Question 19.
Name any two vertibrates body organ which are homologus organs of human forelimb.
Answer:
(1) Flipper of Whale,
(2) Wing of birds.

Question 20.
What is the scientific name of modern man ?
Answer:
Homo sapiens.

Question 21.
Who is the early man of the modern human?
Answer:
Cro-Magnon peoples are early human of modern human.

Question 22.
Who are the early human and sub-human.
Answer:
Ramapithecus is early human and Australopithecus is early human.

Question 23.
Which human form first started to walk on two legs?
Answer:
Australopithecus form first started to walk on two legs.

Question 24.
Which type of human was ‘Cro-Magnon’ on the basis of food in take?
Answer:
Cro-Magnon was carnivorous.

Question 25.
On the basis of evolution which human had brain size of 1400cc?
Answer:
Neanderthal.

Question 26.
Give the name of apelike ancestors of humans
Answer:
Apelike ancestors of human’s are Dryopithecus.

Question 27.
Differentiate between Dryopithecus and Ramapithecus.
Answer:
Dryopithecus were apelike but Ramapithecus were mostly human like.

Evolution Short Answer Type Questions

Question 1.
What is virus ? Why is it treated as a link between living and non-living ?
Answer:
Viruses are simplest organism of the earth, which consists of nucleic acid (DNA or RNA) surrounded by protein cover. It shows characteristics of living as well as non-living organisms.
(A) Living characters of virus:

• Virus shows structural differentiations.
• They contain hereditary material.
• They exhibit mutation.
• They spread plant and animal diseases.
• Growth and development present.
• They exhibit adaptaion.
• They possess sensitivity.

(B) Non-living characters of virus:

• Lack protoplasm and cell organelles.
• Can be crystallized.
• No metabolic activities seen.
• Cannot reproduce outside living cells.
• They lack enzymes.

Due to above reason, viruses are considered as link between living and non-living organisms, thus, it is the first life originated in the earth.

Question 2.
What is oxygen revolution? Explain.
Answer:
Oxygen revolution : Evolution of 02 in photosynthesis during primitive environmental conditions is very important because, it is required in the evolution of organism and conversion of reducing environment into oxidizing environment hence, it is called oxygen revolutioa Oxygen evolution should cause the following changes in the environment:

• Oxygen evolution should cause the conversion of reducing environment into oxidizing environment
• Ozone layer is formed 15 miles above from the earth which absorbs the ultraviolet light of the sunlight and thus, prevents the entry of uv light in the atmosphere.
• O₂ present in the environment dissociates methane (CH₄) into CO₂ and O₂. This CO₂ is used in photosynthesis.
• NH₃ of the primitive environment is dissociated into H₂O.and nitrogen.
  CH₄ + 2O₂2 → CO₂ + 2H₂O
  4NH₃ + 3O₂ → 2N₂ + 6H₂O.

Question 3.
Explain the origin of the earth.
Answer:
Origin of the earth:

• Earth was formed 4-5 billion years back.
• Initially, die surface was covered with water vapOur, methane, C02 and NH3.
• The UV rays of the sun broke water into hydrogen and oxygen.
• Hydrogen escaped and oxygen combined with NH3 and CH4 to form water, C02 and other gases, also forming die ozone layer.
• Cooling of water vapour led to rain which filled the depressions on earth’s surface, forming water bodies.

Question 4.
Mention the names of discoveries and principles given by the following scientists:

1. Louis Pasteur
2. A. I. Oparin
3. Urey and Miller
4. Francesco Redi
5. Faux.

Answer:

1. Louis Pasteur : He proved that air contains spores of microorganism and Biogenesis theory was supported by him.
2. A. I. Oparin: He presented the biochemical explanation of origin of life in his book “The Origin of Life on Earth”.
3. Urey and Miller : They supported the evidence of Oparin-Haldane theory of Origin of life.
4. Francesco Redi: He by conducting experiments proved that abiogenesis cannot exist but bioginesis theory can exist i.e., Life arises from pre-existing life.
5. Faux: He has been experimentally supporting the organic substances as described by Oparin.

Question 5.
Write down the difference between :

1. Ozone and Oxygen
2. Micro-molecules and Macro-molecules.

Answer:
1. Differences between Ozone and Oxygen :

2. Differences between Micro-molecules and Macro-molecules :

Question 6.
What are homologous organs ?
Or
What is homology ?
Answer:
Organs which are similar in structure and origin but different in appearance and functions are called homologous organs and the phenomenon is called homology.
Examples : Forelimbs of bat, wings of bat, hands of man, forelimbs of horse. These are the examples of homologous organs because, they are made up of similar bones, hu¬merus, radius-ulna, carpals, metacarpals and fingers.

Question 7.
What do you mean by analogous ?
Answer:
Analogous organs: Organs which are different in origin and structure but per¬forming similar functions are known as analogous organs and the phenomenon is called as analogy. Analogous organs do not indicate phylogeny.
Examples : Wings of butterflies are made up of chitin, wings of birds made by production of feathers on forelimbs and skin present between the fingers of bat are the examples of analogous organs.

Question 8.
What is the difference between homologous and analogous organs ? Give two examples of each of them.
Answer:
Differences between Homologous and Analogous organs :

Question 9.
What do you mean by vestigial organs ?
Or
Write the two names of vestigial organs of man.
Or
What are vestigial organs ? Explain. Write four vestigial organs of the human body.
Answer:
Vestigial organs : Organs that are reduced and have become functionless in an organism but were functional in their ancestors are called as vestigial organs :
Examples :

• Vermiform appendix
• Coccygial vertebrae
• Nictitating membrane in the eyes of human
• Muscles of external ear (Pinna).

Question 10.
What is connecting link ? Also explain the significance of connecting links.
Answer:
Connecting link : Certain organisms which share characters with two different groups. Such organisms are called connecting links.
Examples : Some fossils present in nature like Archaeopteryx is known as fossils connecting link and it is the connecting link of birds and reptiles. Neopilina, Platyptus, Protopterus, etc. Neopilina is a mollusc and is the connecting link of mollusca and annelida.
Significance of Connecting links : It proves the process of organic evolution and helps in the identification of organism closely related.

Question 11.
Write down the demerits of Darwinism.
Answer:
Some of the demerits of Darwinism are :

• Darwinism stresses upon small fluctuating variations which has no role in evolution.
• Does not satisfactorily explain effect of use and disused and presence of vestigial organs:
• It did not differentiate somatic and germinal variations.
• It explains survival of the fittest but not arrival of the fittest. .

Question 12.
Differentiate between Lamarckism and Darwinism.
Answer:
Differences between Lamarckism and Darwinism:

Question 13.
Explain Lamarckism in short
Or
Explain Lamarckism of organic evolution in brief.
A ns. Lamarckism : In 1809, Lamarck has proposed a theory to explain organic evolution, which is known, as the “Theory of inheritance of acquired characters.” According to Lamarck, organisms acquire certain characters during their lifetime due to changes in environment and these acquired characters are heritable. According to this theory, new species are originated as follows :

• New requirements and wills are produced in the organisms due to the effect of changed environment
• New requirements and wills of organisms resulting in the production of new habits.
• Changes in the habit bringing about modifications of the organ.
• New habits resulting in the use or disuse of the organs:
• Use of organ resulting in the development of acquired characters.
• These acquired characters are heritable.
• Inheritance of acquired characters resulting in the development of new species.

Evolution Long Answer Type Questions

Question 1.
Draw a well-labelled diagram of Miller and Urey’s experiment
Answer:

Experimental evidence of Chemical evolution or Miller’s experiment:

• Experiment was performed by S.L. Miller and H.C. Urey in 1953.
• Experimental set-up: In a closed flask containing CH4, H2, NH3 and water vapoiy at 800°C, electric discharge was create. The conditions were similar to those in primitfve atmosphere.
• Observations: After a week, they observed presence of amino acids and complex molecules like sugars, nitrogen bases, pigments and fats in the flask.
• Conclusions : (i) It provides experimental evidence for the theory of chemical origin.
  (ii) It showed that the first non-cellular form of life was created about 3 billion years ago.
  (iii) It showed that non-cellular biomolecules exist in the form of DNA, RNA, polysaccharides and protein.

Question 2.
Write an essay on modern concepts of origin of life.
Or
Explain the role of non-living in origin of life.
Answer:
Modern concept of origin of life : The modem concept of origin of life was postulated by a Russian biochemist A.I. Oparin in 1936. According to this theory, after the formation of earth various chemicals played important role in the formation of atmosphere. Life originated and first organism came into existence from certain molecules when atmospheric conditions became suitable. According to Oparin, life originated in the following steps:

1. Formation of earth and its atmosphere: Earth is believed to be originated some 4,500 million years ago by the condensation and cooling of the clouds of cosmic dust and gases called ylem. The heavier elements collected at the core and lighter elements around the core. Outermost layer contains H, C, O and N. Oxygen was found only in combination of other elements. These four elements reacted with each other forming H₂, H₂O, CH₄, NH₃, CO₂ and HCN.

2. Formation of small organic molecules : The mixture of methane, ammonia, water and hydrogen comes in contact of solar energy. Cosmic rays and electric discharge could produce some simple organic compounds. These simple organic compounds formed in such a way and accumulated in primitive atmosphere and oceans were responsible for synthesis of complex micro molecules as follows :

3. Formation of polymers : It is clearly understood from the above description that a large number of micro molecules such as hydrocarbons, amino acids, fatty acids, purine and pyrimidines and simple sugars accumulated in the oceans. When atmospheric water condensed on further cooling, the inorganic precursors collided, reacted and aggregated to form new molecules of increasing size and complexity. Thus, by polymerization macromolecules were formed. The chemical reactions for the formation of macro-molecules can be summarized as follows:

(a) Sugar + Sugar → Polysaccharides
(b) Fatty acid + Glycerine → Lipids
(c) Amino acid + Amino acid → Protein
(d) Nitrogenous base (Adenine) + Sugar + Phosphate → Adenosine phosphate
(e) Nitrogenous base + Sugar → Nucleoside
(f) Nucleoside + Phosphate → Nucleotide
(g) Nucleotide + Nucleotide → Nucleic acid.

4. Formation of molecular aggregates and primitive cells : Over a vast of time, these molecules became associated with one temporary complex. Ultimately, it leads to the formation of a coacervate. A coacervate is a solution of high molecular weight of chemicals, i.e., proteins and carbohydrates, which become bounded by lipid membrane, which is selectively permeable. The coacervate grows by absorbing molecules from their environment. The substances which got accumulated in the coacervates underwent reactions and resulted in the molecular reorganization of some proteins into enzymes. A coacervate having nucleoprotein surrounded by various nutritive organic substances and covered by surface membrane is considered to be the precell, which got later transformed into first living cell. The coacervate can reproduce by budding.

5. Evolution of complex biochemical reactions : Primitive organism utilize chemi¬cal substances present in the environment as food hence, they are:

(a) Heterotrophic, chemosynthetic organisms appeared due to mutation and natural selection in heterotrophs.
(b) Blue-green algae evolved from chemosynthetic organisms by mutation and natural selection.
(c) The liberation of free oxygen into the atmosphere produced by the blue-green algae due to the process of photosynthesis. It finally changed the reducing atmosphere into an oxidizing one and therefore, all possibilities of further chemical evolution were finished.
Free living eukaryotes originated in the ocean from blue-green algae.

6. The origin of well-developed organisms : From the simple eukaryotes which were like unicellular organisms of today various forms of life evolved during passage of time.

Question 3.
Write the process of formation of organic molecules in sea water on earth with the help of Miller and Urey’s experiment
Answer:
The work of A.I. Oparin (1938-1965), HLUrey and Stanley Miller (1959) pro¬vided evidences in the favour of biochemical origin of life. They had prepared the atmos¬phere like that of primitive earth and as described by Oparin, they made the synthesis of organic compounds by the following methods:

(i) Four elements H, C, O (not free O₂) and N react .with each other to form H₂O, CH₄, NH₃, CO₂ and HCN on primitive earth.

(ii) From these four elements following organic molecules were formed in sea water of earth:
(a)

(b) CH₄, H₂O, NH₃ → Amino acids.
(c) CH₄, HCN, H₂O, NH₃ → Nitrogenous bases.

(iii) Macro-molecules of organic compounds were synthesized by these above pre¬pared organic compounds.

(a) Sugar + Sugar → Polysaccharides (Carbohydrate).
(b) Fatty acids + Glycerol → Fats.
(c) Amino acid + Amino acid → Protein.
(d) Nitrogenous base + Sugar → Nucleoside.
(e) Nucleoside + Phosphoric acid → Nucleotide.
(f) Nucleotide + nucleotide → Nucleic acid.
(g) Nitrogenous base (Adenine) + Sugar + Phosphate →Adenosine phosphate.
(iv) The above organic compounds and salts together constituted the first living being.

Question 4.
Name connecting link of reptiles and birds. Also write their characters.
Answer:
Archaeopteryx is the connecting link of birds and reptiles. Archaeopteryx was a bird. It is regarded as the connecting link between reptiles and birds, which suggests the path of evolution of the latter from the former. It is found as fossils. They are found during Jurassic period 140 million years. Archaeopteryx exhibits both reptiles and birds like characters.

1. Reptiles like characters:

• Bones were similar to that of reptiles in which air sacs were absent.
• Tail bearing vertebra.
• It had teeth in jaws, scales were present on the body.
• Metacarpals were free.
• Pelvic girdle recombines with the pelvic girdle of reptiles.

2. Birds like characters :

• Presence of feathers on body. .
• Forelimbs were modified in wings.
• Skull large and monocondylar.
• Jaws were modified into beak.
• Hallux was backwarded and pointed.

Question 5.
Give a detailed account of theory of natural selection.
Or
Describe the Darwin’s opinion about the origin of new species of organisms.
Answer:
Charles Darwin (1809-1882) explained the theory of evolution in his book “Origin of species by natural selection.” Darwin undertook a long voyage for five years in the capacity of a naturalist on a British warship ‘Beagle’. He travelled to islands of Galapagos and collected evidences to explain evolution.
To explain origin of species he gave theory of natural selection as a mechanism for evolution.

The main points of Darwinism are given below:

1. Over production of offsprings : Every living being has an inherent tendency to produce more offspring than that can survive.

2. Struggle for existence : Though the offsprings are produced in large number yet their production remain almost constant. This is because of struggle for existence. There is struggle for food, space, breeding, etc. Moreover death of individuals due to diseases an

3. Survival of the fittest: Darwin believed that any individual is successful in struggle for existence if it survives long enough to produces offsprings. Individuals who are fit in a particular environment can only survive.

4. Variation: Due to constant struggle, the organisms change themselves in accordance with the new needs.

5. Natural selection : In the struggle, for existence organisms having variations favourable to the environment would have more chance to survive and reproduce their own kind. Those which do not possess favourable variations would die or fail to reproduce.

6. Origin of species : Any changes in environmental conditions cause natural selection to act upon the population and select the well adapted individuals. It results in changes of characters of the populations. By the inheritance of these changes in successive generations, new species are formed.

Question 6.
Organic Evolution is a continued process, explain it in favour of it giving any three evidences.
Answer:

Evolution is a complex phenomenon accounting for the present day diversity among organisms. But it has clearly maintained the basic unity among them since it occurred over a period of millions of years, no one would have seen/recorded evolution and hence scientists have provided various evidences to prove evolution.

Some of the evidences of organic evolution are described below :

I. Evidences from Embryology :

• Important activities that occur various animals are:
• For the survival, all animals get energy and various substances from environment.
• In all organisms energy is produced from ATP.
• In all organisms, the duplication of DNA is similar.
• In all organisms, protein synthesis is same and it is produced from ribosomes.
• In all organisms respiration and steps of respiration is same.
• All organisms multiplicate and reproduce, due to which they have basic similarities.
• All organisms conduct hereditary characters on similar principles.

II. Evidence from Anatomy: Anatomy of living organisms will be explained with different examples:

• Homologous organs: Organs which are similar in structure and origin but diffeifent in function and appearance is known as homologous organs.
• Analogous organs : Organs which are different in origin and structure but performing similar functions are known as analogous organs.
• Vestigial organs : Organs that are reduced and have become functionless in an organism, but were functional in their ancestors are called vestigial organ.

III. Evidences from vestigial organs: Organs of the body which are non-functional but they are functional in some other organisms are called vestigial organs:

Morphological evidence of evolution is provided by the presence of vestigial organs of body which are often undesired, degenerated and non-functional. These might have been large and functional in some other animals or in ancestors of those which now possess it in rudimentary forms. e.g„ vermiform appendix in man, muscles of external ear (pinna) in man, nictitating membrane or plica, semilunaris in human eye, wisdom teeth (third pair of molars) tail bone (coccyx) in man, wings of ostrich, hindlimbs in snakes, etc.

Question 7.
What do you mean by organic evolution? How do fossils exhibit evidence to prove organic evolution?
Answer:
Descent with modification in organism is known as organic evolution.

Evidences of organic evolution from fossils record : Fossils are treated as significant evidence of organic evolution. Fossils are the remains or impressions of ancient organisms preserved in the layers of rock and soil. Fossils only do not prove the theory of organic evolution, yet it evidently prove that gradually complexity increased in body organization. The complexity in the body of organization can be noticed as we study the upper layers. Thus, it can be concluded from above observations :

• The crust of the earth and the organisms living on it underwent change in the course of time.
• The organisms with simple structural organization originated earlier than the complex ones.
• Some of the organisms lived on the earth for short time and became extinct. This was a result of drastic changes in the climate on the earth.

Hence, forth fossils produce bonafide record of such plants and animals which had shown their existence once upon a time and now are extinct or not present exactly in the same form, thus, producing strong evidence in favour of organic evolution.

MP Board Class 12th Biology Solutions