CUET UG Exam 2024 : Biology Important Chapter-wise MCQs Question with Answers

Practice chapter-wise Biology MCQs suitable for various entrance exams including CUET. Biology MCQ questions are very useful for various competitive exams.

Biology MCQ questions are based on the latest CUET 2024 exam pattern and other entrance exams, so that you can get a fair idea about the exam trends adopted for various entrance exams.

Exposure to different types of questions increases understanding of types of questions, difficulty level and optimal approach. The increased exposure allows students to refine their strategies, ultimately improving their performance in the CUET UG examination.

1. Reproduction in Organisms

Multiple Choice Questions

1. Choose the correct statement from amongst the following—

(a) Dioecious (hermaphrodite) organisms are found only in animals.
(b) Dioecious organisms are found only in plants.
(c) Dioecious organisms are found in both plants and animals.
(d) Dioecious organisms are found only in vertebrates.

Ans. (c) Dioecious organisms are found in both plants and animals.

Explanation: Dioecious organisms produce either male or female reproductive organs and gametes but never both at the same time. Animals are generally dioecious organisms and some plants such as Spinach, Mulberry, Ginkgo, etc. are also examples of dioecious plants.

2. There is no natural death in single celled organisms like Amoeba and Bacteria because— 

(a) they cannot reproduce sexually.
(b) they reproduce by binary fission.
(c) parental body is distributed among the offspring.
(d) they are microscopic.

Ans. (c) parental body is distributed among the offspring.

Explanation: During mitotic division, the cell divides into two equal-sized daughter cells. In this method, two similar individuals are produced from a single parent cell of Amobea or bacteria. This means that parental genetic material never dies even though their identity is changed.

3. There are various types of reproduction. The type of reproduction adopted by an organism depends upon—

(a) the habitat and morphology of the organism.
(b) morphology of the organism.
(c) morphology and physiology of the organism
(d) the organism’s habitat, physiology and genetic makeup.

Ans. (d) the organism’s habitat, physiology and genetic makeup.

Explanation: Habitat determines whether the organism will undergo self- pollination, cross-fertilisation, sexual reproduction. Physiology plays an important role both in asexual and sexual reproduction. Somatic cell favours asexual reproduction and germ cell favours sexual reproduction.

4. Identify the incorrect statement—

(a) In asexual reproduction, the offsprings produced are morphologically and genetically identical to the parent.
(b) Zoospores are sexual reproductive structures.
(c) In asexual reproduction, a single parent produces offspring with or without the formation of gametes.
(d) Conidia are asexual structures in Penicillium.

Ans. (b) Zoospores are sexual reproductive structures.

Explanation: Zoospores are motile asexual spores having flagellum used for locomotion.

5. Which of the following is a post-fertilisation event in flowering plants?

(a) Transfer of pollen grains
(b) Embryo development
(c) Formation of flower
(d) Formation of pollen grains

Ans. (b) Embryo development

Explanation: Embryo development is the post-fertilisation event in flowering plants. Transfer and formation of pollen grains, formation of flower are prefertilisation events.

6. The number of chromosomes in the shoot tip cells of maize plant is 20. The number of chromosomes in the microspore mother cells of the same plant shall be—

(a) 20
(b) 10
(c) 40
(d) 15

Ans. (a) 20

Explanation: Shoot tip cells and microspore mother cells both are diploid in maize plants. So, number of chromosoms in microspore mother cells in maize will be 20.

7. Vegetative propagation in Pistia occurs by—

(a) Stolon
(b) Offset
(c) Runner
(d) Sucker

Ans. (b) Offset

Explanation: The vegetative propagation in Pistia occurs by offset where one internode long runner grows horizontally along the soil surface and gives rise to new plants either from axillary or terminal buds.

8. Which one of the following processes results in the formation of clone of bacteria?

(a) Transformation
(b) Transduction
(c) Binary fission
(d) Conjugation

Ans. (c) Binary fission

Explanation: Clone means the cell or organism that is genetically identical to the original cell or organism from which it is derived. In bacteria, reproduction occurs through binary fission which results in the formation of clones i.e., the original cell is divided into two identical cells.

9. Microsporangia is a technique—

(a) for the production of true to type plants.
(b) for production of haploid plants.
(c) for production of somatic hybrid.
(d) for production of somaclonal plants.

Ans. (a) for the production of true to type plants.

Explanation: Microsporangia occur in all vascular plants that have heterosporic life cycles and are true type plants. Whereas haploid plants are produced from anther culture. Protoplast fusion produces a somatic hybrid and Somaclonal plants can be produced by plant tissue culture.

10. Example of Corm is—

(a) Ginger
(b) Colocasia
(c) Onion
(d) Potato

Ans. (b) Colocasia

Explanation: Corms are vertical fleshy underground stems that act as a food storage structure. Colocasia has a corm that stores starches to fuel the growth of the plant. Onion is a modified bud and called bulb. Potatoes are stem tuber whereas ginger is a rhizome, not a tuber.

11. What is common between negative reproduction and apomixis?

(a) Both are applicable to dicot plants only.
(b) Both bypass the flowering phase.
(c) Both occur round the year.
(d) Both produces progeny identical to the parent.

Ans. (d) Both produces progeny identical to the parent.

Explanation: Negative reproduction and Apomixis both produces progeny identical to the parents. None occurs round the year but in a certain condition, it could happen. Both can happen in dicot and monocot plants.

12. Which part would be most suitable for raising virus free plants for micropropagation?

(a) Bark
(b) Vascular tissue
(c) Meristem
(d) None of these

Ans. (c) Meristem

Explanation: In meristem, cells are divided at a very fast rate and such tissues are devoid of any infection.

13. Which one of the following is common to multicellular fungi, filamentous algae and protonema of mosses?

(a) Diplontic life cycle
(b) Members of kingdom Plantae
(c) Mode of nutrition
(d) Multiplication by fragmentation

Ans. (d) Multiplication by fragmentation

Explanation: In algae we can find a diplontic life cycle but not in moss or fungi. Kingdom Plantae does not consist of fungi and algae. The mode of nutrition possesses by mosses and algae is autotrophic whereas fungi are heterotrophic. The only thing they are common in is multiplication by fragmentation.

14. Banana is vegetatively propagated by—

(a) Tubers
(b) Rhizomes
(c) Bulbs
(d) Suckers

Ans. (b) Rhizomes

Explanation: Banana is vegetatively propagated by rootstock rhizomes. Suckers and bulbs are the underground base of the shoot and bud respectively and are used by mint and onions respectively. Root tubers originate from roots that have been modified to store nutrients.

15. A scion is grafted to stock. The quality of fruits produce will be determined by genotype of—

(a) Stock
(b) Scion
(c) Both stock and scion
(d) Neither stock nor scion

Ans. (b) Scion

Explanation: Scion is selected for good quality of fruit, flower and good resistance to diseases.

2. Sexual Reproduction in Flowering Plants

Multiple Choice Questions

1. Which one of the following parts of the plant when put into the soil is likely to produce new offspring?

(a) Part of an internode
(b) A stem cutting with a node
(c) Part of a primary root
(d) A flower

Ans. (b) A stem cutting with a node

Explanation: Vegetative plant propagation is often done with stems. Sections of stems that contain nodes and internodes are used in an effective way to propagate many ornamental plants.

2. Among the terms listed below, those that of are not technically correct names for a floral whorl are: [NCERT Exemplar]

(i) Androecium
(ii) Carpel
(iii) Corolla
(iv) Sepal

(a) (i) and (iv)
(b) (iii) and (iv)
(c) (ii) and (iv)
(d) (i) and (ii)

Ans. (c) (ii) and (iv)

Explanation: The technically correct terms for the floral whorls are (from outermost to innermost) calyx, corolla, androecium and gynoecium. They are made up of sepals, petals, stamens and carpels respectively.

3. Filiform apparatus in the embryo sac of an angiosperm is present at the micropylar tip of :

(a) Central cell
(b) Egg cell
(c) Synergids
(d) Antipodals

Ans. (c) Synergids

Explanation: The synergids have special cellular thickening at the micropylar tip called filiform apparatus, which plays an important role in guiding the pollen tube into the synergids.

4. Embryo sac occurs in—

(a) embryo
(b) axis part of embryo
(c) ovule
(d) stamen

Ans. (c) ovule

Explanation: Embryo sac occurs in ovule. Ovule is integumented megasporangium. It consists of nucleus covered by one or two integuments, mounted on a funicle, chalaza and micropyle. The ovule is vascularised.

5. One of the most resistant biological material is—

(a) lignin
(b) cutin
(c) sporopollenin
(d) cellulose

Ans. (c) sporopollenin

Explanation: Outer layer (exine) of pollen grain is made of a highly resistant substance called sporopollenin. Sporopollenin is not degraded by any enzyme because of sporopollenin, pollen grains are well preserved as fossils.

6. In angiosperms________lead to the formation of a mature male gametophyte from a pollen mother cell.

(a) two meiotic divisions
(b) three mitotic division
(c) two mitotic and one meiotic division
(d) a single mitotic division

Ans. (c) two mitotic and one meiotic division

Explanation: Meiosis produces pollen grain. Its cell divides mitotically to form generative nucleus and tube cell. The generative nucleus undergoes another mitosis to form two male gametes.

7. Micropyle occurs in—

(a) ovary
(b) ovule
(c) seed
(d) both (b) and (c)

Ans. (d) both (b) and (c)

Explanation: Micropyle is the small minute pore which is differentiated from the surface of the egg. It is formed by the projection of integuments into which the male gamete through pollen tube enters into the egg of the ovule. It is usually located at the top of the seed or ovule.

8. Embryo sac is also called—

(a) microspore
(b) megaspore
(c) megagametophyte
(d) microgametophyte

Ans. (c) megagametophyte

Explanation: Megagametophyte or the female gametophyte is the embryo sac that develops from the megaspore through megagametogenesis. Female gamete is the egg cell which upon fusion with male gamete forms a diploid zygote. Hence, embryo sac is also called megagametophyte.

9. Egg apparatus comprises of __________.

(a) Polar nuclei
(b) Antipodal cells
(c) Egg cell and synergids
(d) Male gametes

Ans. (c) Egg cell and synergids

Explanation: Egg apparatus is present at the micropylar end. It consists of two synergids and one egg cell.

10. Which one of the following is not found in a female gametophyte of an angiosperm ?

(a) Germ pore
(b) Synergids
(c) Filiform apparatus
(d) Central cell

Ans. (a) Germ pore

Explanation: Germ pore is not found in the female gametophyte of an angiosperm. A germ pore is a small pore in the outer wall of a fungal spore through which the germ tube exits upon germination. The main work of a germ pore is to absorb water for seed germination.

11. The gamete mother cell is known as: 

(a) Diploid
(b) Meiocytes
(c) Haploid
(d) Isogamete

Ans. (b) Meiocytes

Explanation: Gamete mother cell in diploids are specialised cells called meiocytes.

12. Pollen grains are well preserved as fossils because of the presence of _____.

(a) Exine
(b) Intine
(c) Germ pores
(d) Sporopollenin

Ans. (d) Sporopollenin

Explanation: Outer layer (exine) of pollen grain is made of a highly resistant substance called sporopollenin. Sporopollenin is not degraded by any enzyme. It is not affected by high temperature, strong acid or strong alkali. Because of sporopollenin, pollen grains are well preserved as fossils.

13. The outer layer pollen grain is called ____A____. This is made up of ____B____ which is absent on the ____C____. A B C

(a) intine cellulose micropyle
(b) exine sporopollenin germ pores
(c) intine sporopollenin germ pores
(d) exine cellulose micropyle

Ans. (b) exine, sporopollenin, germ pores

Explanation: Pollen grain has a two-layered wall i.e., exine and intine. The hard outer layer of pollens, named exine, is made of sporopollenin. It has prominent apertures called germ pores where sporopollenin is absent.

14. Embryo sac is to ovule as _______ is to an anther.

(a) Stamen
(b) Filament
(c) Pollen grain
(d) Androecium

Ans. (c) Pollen grain

Explanation: Embryo sac is the female gametophyte which is an oval structure in the nucellus of the ovule of flowering plants and the ovule is the megasporangium. Similarly, the pollen grain is the male gametophyte and the anther microsporangium.

15. Which is the innermost wall layer of microsporangium?

(a) Tapetum
(b) Epidermis
(c) Endothecium
(d) Endodermis

Ans. (a) Tapetum

Explanation: The wall layers of a microsporangium from outermost to innermost are: epidermis, endothecium, middle layers and tapetum. The first three layers generally provide protection and help in dehiscence of anther. Tapetum performs nutritive function for pollen grains.

3. Human Reproduction

Multiple Choice Questions

1. Urethral meatus refers to the— [NCERT Exemplar] 

(a) Urinogenital duct
(b) Opening of vas deferens into urethra
(c) External opening of the urinogenital duct
(d) Muscles surrounding the urinogenital duct

Ans. (c) External opening of the urinogenital duct

Explanation: Urethral meatus refers to the external opening of urinogenital duct, through which, in males, urine and semen both are expelled out.

2. Clitoris in females is—

(a) homologous to penis
(b) analogous to penis
(c) functional penis in female
(d) non-functional

Ans. (a) homologous to penis

Explanation: The clitoris is structurally and functionally homologous to the penis of the male reproductive system, except that the clitoris does not contain the urethra and plays no role in urination.

3. Prostate gland is present—

(a) on ureter
(b) on kidney
(c) on testis
(d) around urethra

Ans. (d) around urethra

Explanation: The prostate is located just below the bladder and in front of the rectum. It is about the size of a walnut and surrounds the urethra. The primary function of the prostate is to produce the fluid that nourishes and transports sperm (seminal fluid).

4. Corpus luteum develops from—

(a) Oocyte
(b) Nephrostome
(c) Ruptured Graafian follicle
(d) None of the above

Ans. (c) Ruptured Graafian follicle

Explanation: Corpus luteum is the structure which is formed by the follicles after the egg is released from the Graafian follicle. This is a yellow coloured structure which is present in the ovary.

5. Select the feature of human female.
(a) Well-developed mammary glands
(b) High-pitched voice
(c) Strong muscles
(d) Both (a) and (b)

Ans. (d) Both (a) and (b)

Explanation: Females have well-developed mammary glands and voice is pitched higher than males.

6. Match the structures of male reproductive system given in column I with their features given in column II and select the correct match from the options given below.

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

Ans. (d) A-(ii), B-(iv), C-(iii), D-(i)

Explanation: Rete testes carries sperms from the seminiferous tubules (where sperms are produced through meiosis) of the testes into the vasa efferentia. Leydig cells synthesise and secrete testicular hormones called androgens. The penis is the male external genitalia that facilitates insemination.

7. The given figure depicts a diagrammatic sectional view of the human female reproductive system. Select the option with correctly identified parts.

(a) A – Ovary, G – Vagina, D – Fimbriae
(b) B – Isthmus, I – Vagina, F–Perimetrium
(c) L – Endometrium, H – Cervical canal, C–Ampulla
(d) E – Infundibulum, J – Endometrium, K – Myometrium

Ans. (c) L – Endometrium, H – Cervical canal, C–Ampulla

Explanation: In the given figure, A is uterus, B, C, and D are the parts of the oviducts, i.e., isthmus, ampulla, and infundibulum, respectively, E is the fingerlike projections called fimbriae at the edges of infundibulum, G is the cervix, H is the cervical canal, I is the vagina, and J, K, and L are the three layers of uterus from outer to inner namely, perimetrium, myometrium, and endometrium respectively.

8. Egg is liberated from ovary in—

(a) secondary oocyte stage
(b) primary oocyte stage
(c) oogonial stage
(d) mature ovum stage

Ans. (a) secondary oocyte stage

Explanation: During the process of ovulation, the release of the egg occurs at the secondary oocyte stage in which meiosis-I have been completed.

9. Location and secretion of leydig cells are—

(a) Liver-cholesterol
(b) Ovary-estrogen
(c) Testes-testosterone
(d) Pancreas-glucagon

Ans. (c) Testes-testosterone

Explanation: The Leydig cells are the interstitial cells which are located adjacent to the seminiferous tubules in the testes. The important function known of the leydig cells is to produce the androgen and testosterone.

10. Which of the following is primary sex organ?

(a) Scrotum
(b) Penis
(c) Testes
(d) Prostate

Ans. (c) Testes

Explanation: The testes are the primary male reproductive organ and are responsible for testosterone and sperm production.

11. Corpus luteum secretes—

(a) LH
(b) Estrogen
(c) Progesterone
(d) FSH

Ans. (c) Progesterone

Explanation: The corpus luteum secretes large amounts of progesterone which is essential for maintenance of the endometrium. Such an endometrium is necessary for implantation of the fertilised ovum and other events of pregnancy.

12. Spermatozoa are nourished during their development by—

(a) Sertoli cells
(b) Interstitial cells
(c) Connective tissue cells
(d) None of the above

Ans. (a) Sertoli cells

Explanation: Sertoli cells provides nourishment to the developing sperm cells during spermatogenesis therefore, they are also named as ‘mother or nurse cell’.

13. Sperms produce an enzymatic substance for dissolving egg membrane called—

(a) Hyaluronic acid
(b) Hyaluronidase
(c) Androgen
(d) Estrogen

Ans. (b) Hyaluronidase

Explanation: The acrosome of the sperm contains digestive enzymes like hyaluronidase. These enzymes break down the outer membrane of the ovum, called as the zona pellucida.

14. The Leydig cells found in the human body are the secretory source of—

(a) Progesterone
(b) Intestinal mucus
(c) Glucagon
(d) Androgens

Ans. (d) Androgens

Explanation: The regions outside the seminiferous tubules called interstitial spaces contain small blood vessels and interstitial cells or leydig cells. It synthesises and secretes testicular hormones called androgens.

15. Seminal plasma, the fluid part of semen, is contributed by— [NCERT Exemplar]

(i) Seminal vesicle
(ii) Prostate gland
(iii) Urethra
(iv) Bulbourethral gland

(a) (i) and (ii)
(b) (i), (ii) and (iv)
(c) (ii), (iii) and (iv)
(d) (i) and (iv)

Ans. (b) (i), (ii) and (iv)

Explanation: Secretion of seminal vesicle (paired), prostate gland (unpaired) and bulbourethral glands or Cowper’s glands (paired) constitute the seminal plasma. It contains various proteins and fructose as energy suppliers for sperm motility and is also responsible for making the largest proportion of the alkaline buffer. Fructose, citric acid and supplementary nutrients are secreted by seminal vesicles.

4. Reproductive Health

Multiple Choice Questions

1. From the sexually transmitted diseases mentioned below, identify the one which does not specifically affect the sex organs– [NCERT Exemplar]

(a) Syphilis
(b) AIDS
(c) Gonorrhea
(d) Genital warts

Ans. (b) AIDS

Explanation: AIDS is the sexually transmitted disease which does not specifically affect the sex organs.

2. Which one of the following groups includes all sexually transmitted disease?

(a) AIDS, Syphilis, Cholera
(b) HIV, Malaria, Trichomoniasis
(c) Gonorrhea, Hepatitis-B, Chlamydiasis
(d) Hepatitis-B, Haemophilia, AIDS

Ans. (c) Gonorrhea, Hepatitis-B, Chlamydiasis

Explanation: All of the diseases like AIDS, Gonorrhoea, hepatitis-B, chlamydiasis, and syphilis are sexually transmitted diseases. While cholera, malaria, haemophilia are water-borne, parasitic and Mendelian disorders respectively.

3. Which of the following does not belong to STDs?

(a) Gonorrhea
(b) Syphilis
(c) Trichomoniasis
(d) Dengue

Ans. (d) Dengue

Explanation: Dengue is a mosquito-borne viral infection, found in tropical and subtropical climates worldwide, mostly in urban and semi-urban areas. The virus responsible for causing dengue is called dengue virus.

4. Reproductive health in society can be improved by–

(i) introduction of sex education in schools.
(ii) increased medical assistance.
(iii) awareness about contraception and STDs.
(iv) equal opportunities to male and female child.
(v) encouraging myths and misconceptions.

(a) (i), (ii), (iv), and (v)
(b) (i), (ii), (iii), and (iv)
(c) (ii) and (v)
(d) (i), (ii), (iii), (iv), and (v)

Ans. (b) (i), (ii), (iii), and (iv)

Explanation: Introduction of sex education in school should be encouraged to provide right information to the young so as to discourage children from believing in myths and having misconceptions about sex related aspects.

5. Which of the following is correct regarding HIV, hepatitis-B, gonorrhoea, trichomoniasis?

(a) Hepatitis-B is eradicated completely whereas others are not.
(b) HIV is a pathogen whereas others are diseases.
(c) Gonorrhoea is a viral disease whereas others are bacterial.
(d) Trichomoniasis is an STD whereas others are not.

Ans. (b) HIV is a pathogen whereas others are diseases. 

Explanation: HIV is known as a retrovirus (pathogen) that causes AIDS whereas hepatitis B, gonorrhoea, and trichomoniasis are categorised as a disease. HIV (Human immunodeficiency virus) is a pathogen that attacks the body’s immune system.

6. Lactational amenorrhoea is effective only up to a maximum period of–

(a) 6 months before conception.
(b) 6 months after conception.
(c) 1 year after parturition.
(d) 6 months after parturition.

Ans. (d) 6 months after parturition.

Explanation: Lactational amenorrhoea (absence of menstruation) method is based on the fact that cycle does not occur during the period of intense lactation following parturition. This method has been reported to be effective only up to a maximum period of 6 months following parturition.

7. The other name for STDs is–

(a) reproductive tract infections.
(b) venereal diseases.
(c) non-communicable diseases.
(d) both (a) and (b)

Ans. (d) both (a) and (b)

Explanation: Sexually Transmitted Diseases (STDs) are the diseases or infections which are transmitted through sexual intercourse. STDs are also called Venereal Diseases (VD) or Reproductive Tract Infections (RTI).

8. Select the correct statement regarding sexually transmitted diseases.

(a) Use of condoms does not protect the user from contracting STDs.
(b) Gonorrhoea is transmitted from an infected mother to the foetus through placenta.
(c) The chances of contracting STDs are very high among persons in the age group of 12–25 years.
(d) Infected females may often be asymptomatic and hence, may remain undetected for long.

Ans. (d) Infected females may often be asymptomatic and hence, may remain undetected for long.

Explanation: Use of condoms protects the user from contracting STDs. Gonorrhoea is transmitted from an infected mother to the foetus during delivery.

9. Diaphragms are contraceptive device used by females. Choose the correct option from the statements given below:

(i) They are introduced into the uterus.
(ii) They are placed to cover the cervical region.
(iii) They act as physical barrier for sperm entry.
(iv) They act as spermicidal agents.

(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iii)
(d) (iii) and (iv)

Ans. (c) (ii) and (iii)

Explanation: A diaphragm or cap is a barrier method of contraception. It fits inside vagina and prevents sperm from passing through the cervix (the entrance of womb). It needs to use it with a gel that kills sperm (spermicide).

10. Cu2+ ions released from copper releasing Intra Uterine Device (IUDs)—

(a) Prevent ovulation
(b) Make uterus unsuitable for implantation
(c) Increase phagocytosis sperms
(d) Suppress sperm motility

Ans. (d) Suppress sperm motility

Explanation: Cu2+ ions released by copper releasing intra uterine devices suppress sperm motility. Intra-uterine devices are inserted by doctors in the uterus through vagina. They are available as the non-medicated IUDs, copper releasing IUDs and hormone releasing IUDs.

11. Cu-T prevents pregnancy by preventing—

(a) fertilisation
(b) ovulation
(c) implantation of fertilised egg
(d) none of the above

Ans. (a) fertilisation

Explanation: The copper-coated IUD prevents pregnancy by not allowing the sperm to fertilise the egg. It may also make it harder for a fertilised egg to implant in the uterus.

12. Oral contraceptive pills help in the birth control by—

(a) killing sperms
(b) killing ova
(c) preventing ovulation
(d) forming barrier between sperm and ova

Ans. (c) preventing ovulation

Explanation: Oral contraceptives (birth-control pills) are used to prevent pregnancy. Estrogen and progesterone are two female sex hormones. Combinations of estrogen and progesterone work by preventing ovulation (the release of eggs from the ovaries).

13. Emergency contraceptives are effective if used within– [NCERT Exemplar]

(a) 72 hrs of coitus
(b) 72 hrs of ovulation
(c) 72 hrs of menstruation
(d) 72 hrs of implantation
Ans. (a) 72 hrs of coitus

Explanation: Of all the contraceptive methods, IUDs are widely accepted methods of contraception in India. There are different kinds of IUDs like copper releasing and hormones releasing.

14. Medical Termination of Pregnancy (MTP) is considered safe upto how many weeks to avoid pregnancy?

(a) 8 weeks
(b) 12 weeks
(c) 18 weeks
(d) 6 weeks

Ans. (b) 12 weeks

Explanation: Medical termination of pregnancy (MTP) or medical abortion is the use of abortion pills for terminating a pregnancy. MTP is feasible only up to 12 weeks of pregnancy, and after that surgical termination takes over. MTP is one of the safest methods of terminating an unwanted pregnancy.

15. A national level approach to build up a reproductively healthy society was taken up in our country in: [NCERT Exemplar]

(a) 1950s
(b) 1960s
(c) 1980s
(d) 1990s

Ans. (a) 1950s

Explanation: India was among the first countries in the world to initiate action planes and programmes at a national level to attain total reproductive health as a social goal. These programmes called, ‘family planning’ were initiated in 1951 and periodically assessed over parts decades.

5. Heredity and Variation

Multiple Choice Questions

1. G. J. Mendel was a—

(a) British monk
(b) Australian monk
(c) Austrian monk
(d) German scientist

Ans. (c) Austrian monk

Explanation: Gregor Mendel was an Austrian monk, teacher, and Augustinian prelate who lived in the 1800s. He experimented on garden pea hybrids while living at a monastery and is known as the father of modern genetics.

2. Match the column I with column II and select the correct option.

Column I	Column II
A) Johannsen	(i) Crossing over in Drosophila
B) Mendel	(ii) Coined the term gene
C) T.H. Morgan	(iii) Law of segregation

(a) A-(ii), B-(i), C-(iii)
(b) A-(iii), B-(i), C-(ii)
(c) A-(ii), B-(iii), C-(i)
(d) A-(iii), B-(ii), C-(i)

Ans. (c) A-(ii), B-(iii), C-(i)

Explanation: Johannsen coined the term gene. Mendel introduced law of independent assortment. T.H Morgan performed experiments on Drosophila.

3. In a dihybrid cross, if you get 9:3:3:1 ratio it denotes that: [NCERT Exemplar] 

(a) the alleles of two genes are interacting with each other
(b) it is a multigenic inheritance
(c) it is a case of multiple allelism
(d) the alleles of two genes are segregating independently.

Ans. (d) the alleles of two genes are segregating independently.

Explanation: Cross involving two contrasting characters is called a dihybrid cross. The two flowers of each trait assort at random and independent of their traits and get randomly as well as independently rearranged in the offspring.

4. A monohybrid cross produced tall and dwarf pea plants in ratio of 3 : 1. Their genotypes would be—

(a) TT × Tt
(b) Tt × Tt
(c) TT × tt
(d) Tt × tt

Ans. (b) Tt × Tt

Explanation: Crossing of homozygous tall TT with a homozygous dwarf (tt) is called Monohybrid cross. In this, all F1 progeny are heterozygous tall (Tt).

5. Common test to find genotype of hybrid is by—

(a) studying sexual behaviour of F1 progeny
(b) crossing F1 individuals with recessive parents
(c) crossing one F2 progeny with male parent
(d) crossing one F2 progeny with female parent

Ans. (b) crossing F1 individuals with recessive parents

Explanation: To find the genotype of a hybrid, test cross is performed in which an organism showing dominant phenotype is crossed with the recessive parent instead of selfing. The progenies of such cross can easily be analysed to predict the genotype of the test organism.

6. Which one is a test cross? 

(a) Ww × Ww
(b) Ww × ww
(c) ww × ww
(d) WW × Ww

Ans. (b) Ww × ww

Explanation: A test cross is obtained by the crossing of an F1 progeny with a homozygous recessive parental progeny. It is done in order to determine whether the progeny is homozygous or heterozygous for a character under study.

7. Mating of an organism to a double recessive in order to determine whether it is homozygous or heterozygous for a character under consideration is called—

(a) Reciprocal cross
(b) Test cross
(c) Dihybrid cross
(d) Back cross

Ans. (b) Test cross

Explanation: A test cross is performed to determine the genotype of a dominant parent if it is a heterozygous- or homozygous-dominant. For the purpose, a dominant parent is crossed with the homozygous recessive parent.

8. A cross between two tall plants resulted in offspring having few dwarf plants. What would be the genotypes of both the parents? [NCERT Exemplar]

(a) TT and Tt
(b) Tt and Tt
(c) TT and TT
(d) Tt and tt

Ans. (b) Tt and Tt

Explanation: In the case of TT and Tt; all offspring would be tall (TT, Tt). In the case of option ‘c’ no gene for a dwarf is present, so all offspring will be tall. In the case of option ‘d’, one of the parent plants is dwarf, so it is incorrect. In the case of option ‘b’, most of the offspring will be tall and a few will be a dwarf (TT, Tt, tt).

9. The term genetics was proposed by—

(a) Mendel
(b) Bateson
(c) Morgan
(d) Johannsen

Ans. (b) Bateson

Explanation: The word genetics was introduced in 1905 by English biologist William Bateson, who was one of the discoverers of Mendel’s work and who became a champion of Mendel’s principles of inheritance.

10. The F2 genotypic ratio of monohybrid cross is—

(a) 1 : 1
(b) 1 : 2 : 1
(c) 2 : 1 : 2
(d) 9 : 3 : 3 : 1

Ans. (b) 1 : 2 : 1

Explanation: In F2 generation, we see two types of plants, tall and dwarf that appear phenotypically in the ratio 3:1. But genotypically, the tall plants are of two types, homozygous tall (TT) and heterozygous tall (Tt), therefore, the genotypic ratio comes out to be 1:2:1.

11. A dihybrid for qualitative trait is crossed with homozygous recessive individual of its type, the phenotype ratio is— 

(a) 1 : 2 : 1
(b) 3 : 1
(c) 1 : 1 : 1 : 1
(d) 9 : 7

Ans. (c) 1 : 1 : 1 : 1

Explanation: The given cross is a test cross. The ratio for a test cross is always 1:1 for a monohybrid cross and 1:1:1:1 for a dihybrid cross.

12. In order to find out the different types of gametes produced by a pea plant having genotype AaBb, it should be crossed with a plant with the genotype—

(a) AABB
(b) AaBb
(c) AABb
(d) aabb

Ans. (d) aabb

Explanation: Test cross is performed always between the F1 heterozygous plants and pure recessive (homozygous) parent plant. So, in the given case AaBb should be crossed with aabb.

13. In Mirabilis jalapa, the number of F2 red flowered plants in a cross of red flowered and white flowered would be—

(a) 1
(b) 2
(c) 8
(d) 6

Ans. (a) 1

Explanation: A red flowered Mirabilis jalapa (RR) is crossed to white flowered Mirabilis jalapa (rr). The F1 individuals are Pink flowered (Rr). When F1 individuals are self-crossed, in the F2 generation, 1 plant with red flowers (RR), 2 plants with pink flowers (Rr), 1 plant is with white flowers (rr) are formed.

14. In pea plants, yellow seeds are dominant to green. If a heterozygous yellow seeded plant is crossed with a green-seeded plant, what ratio of yellow and green seed plants would you expect in F1 generation?

(a) 9 : 1
(b) 3 : 1
(c) 1 : 3
(d) 50 : 50

Ans. (d) 50 : 50

Explanation: The genotype of the heterozygous yellow seeded plant will be “Yy” and that for green seeded plant will be “yy” (as the recessive allele is expressed only in homozygous conditions). The cross between heterozygous yellow seeded plant and green seeded plant will produce 50% yellow seeded plants and 50% green seeded plants.

15. Which of the following is best suited for codominance?

(a) Both are recessive
(b) Both are dominant
(c) One is recessive
(d) One is dominant

Ans. (b) Both are dominant

Explanation: Codominance occurs when a heterozygous genotype expresses both recessive and dominant genotype. This means that none of the factors is recessive, but they can both manifest themselves regardless of whether they are homozygous or heterozygous.

6. Molecular Basis of Inheritance

Multiple Choice Questions

1. In a DNA strand, the nucleotides are linked together by— 

(a) glycosidic bonds
(b) phosphodiester bonds
(c) peptide bonds
(d) hydrogen bonds

Ans. (b) phosphodiester bonds

Explanation: In DNA molecule, the nucleotides are linked by phosphodiester bonds whereas the nitrogen bases are held by hydrogen bonds.

2. Khorana was awarded nobel prize for—

(a) discovering DNA
(b) discovering RNA
(c) chemical synthesis of gene
(d) discovering DNA polymerase

Ans. (c) chemical synthesis of gene

Explanation: Har Gobind Khorana was awarded Nobel Prize in Physiology or Medicine for their interpretation of the genetic code and its function in protein synthesis.

3. Who discovered DNA polymerase?

(a) Okazaki
(b) Kornberg
(c) Messelson and Stahl
(d) Watson and Crick

Ans. (b) Kornberg

Explanation: The deoxynucleotides precursors of DNA were unknown at the time of Watson and Crick’s proposal of the double helical DNA structure, and many people thought DNA synthesis was a “vital” activity. Kornberg disproved them by discovering DNA polymerase, the need for a DNA template in addition to deoxynucleotides and primers, and 3′→5′ error correcting mechanism of DNA polymerase I.

4. While analysing the DNA of an organism a total number of 5386 nucleotides were found out of which the proportion of different bases were: Adenine = 29%, Guanine = 17%, Cytosine = 32%, Thymine = 17%. Considering the Chargaff’s rule it can be concluded that: [NCERT Exemplar]

(a) it is a double stranded circular DNA
(b) it is single stranded DNA
(c) it is a double stranded linear DNA
(d) no conclusion can be drawn

Ans. (b) it is single stranded DNA

Explanation: According to Chargaff’s rule, the ratio of adenine to thymine and that of guanine to cytosine is always equal to one. In the given organism, DNA is not following the Chargaff’s rule, hence it can be concluded that it is a single-stranded DNA not a double-stranded DNA.

5. The protein of DNA, which contains information for an entire polypeptide is called as—

(a) Cistron
(b) Muton
(c) Recon
(d) Operon

Ans. (a) Cistron

Explanation: Cistron is a section of DNA or RNA molecule that codes for a specific polypeptide in protein synthesis.

6. Discontinuous synthesis of DNA occurs on one strand because—

(a) DNA molecule being synthesised is very long
(b) DNA dependent DNA polymerase catalyses polymerisation only in one direction (5’ → 3’)
(c) it is a more efficient process
(d) DNA ligase has to play some role 

Ans. (b) DNA dependent DNA polymerase catalyses polymerisation only in one direction (5’ → 3’)

Explanation: DNA dependent DNA polymerase catalyses polymerisation only in one direction (5’→ 3’). Consequently the replication on template with polarity 3’ to 5’ is continuous while the other template with polarity 5’ to 3’ is discontinuous and is synthesised in Okazaki fragments which later join by DNA ligase.

7. Choose the incorrect pair—

(a) Negative charged DNA wrapped around positive charged DNA– Nucleosome
(b) Thread-like, colourless unit of structure–Chromatin in nucleus
(c) Unit of 8 molecules in histones–Histone octamer
(d) Basic amino residues in histones–Lysines and arginines

Ans. (a) Negative charged DNA wrapped around positive charged DNA– Nucleosome

Explanation: Negatively charged DNA wrapped around positively charged histone octamer is called as nucleosome.

8. Hershey and Chase used 35S and 32P to prove that DNA is the genetic material. Their experiments proved that DNA is genetic material because—

(a) loss of 35S in progeny viruses indicated that proteins were not passed on.
(b) progeny viruses retained 32P but not S35.
(c) retention of P32 in progeny viruses indicated that DNA was passed on.
(d) all of the above

Ans. (b) progeny viruses retained 32P but not S35.

Explanation: In Hershey and Chase experiment they used isotopes of phosphorus and sulphur but the progeny only retained isotopic phosphorus proving that DNA is the genetic material as it contains phosphorus.

9. Read the following statements and select the correct option—
(i) Loosely packed and lightly stained regions of chromatin are called heterochromatin.
(ii) Densely packed and dark stained regions of chromatin are called euchromatin.
(iii) A typical nucleosome contains 200 bp of DNA helix.

(a) (i) and (ii)
(b) only (iii)
(c) (ii) and (iii)
(d) (i), (ii), and (iii)

Ans. (b) only (iii)

Explanation: Heterochromatin is densely packed and darkly stained whereas the euchromatin is loosely packed and lightly stained. A typical nucleosome contains 200 base pairs of DNA helix .

10. DNA replication enzymes are given below. Select their correct sequence in DNA replication— 

(i) Helicase
(ii) Primase
(iii) SSB
(iv) DNA ligase
(v) DNA polymerase

(a) (i) → (ii) → (iii) → (v) → (iv)
(b) (iv) → (i) → (iii) → (v) → (ii)
(c) (iii) → (ii) → (i) → (v) → (iv)
(d) (i) → (iii) → (ii) → (v) → (iv)

Ans. (d) (i) → (iii) → (ii) → (v) → (iv)

Explanation: Helicase unwinds the two strands of DNA. SSB enzyme prevents the strands from coiling back. Primase synthesises primers. DNA polymerase helps to create duplicate strands. DNA ligase finally joins the DNA fragments to form two daughter DNAs.

11. The Okazaki fragments in DNA chain—

(a) polymerise in the 5᾽→ 3᾽ direction and explain 3᾽→ 5᾽ DNA replication.
(b) prove semi-conservative nature of DNA replication.
(c) result in transcription.
(d) polymerise in the 3᾽→ 5᾽ direction and form replication fork.

Ans. (a) polymerise in the 5᾽→ 3᾽ direction and explain 3᾽→ 5᾽ DNA replication.

Explanation: On DNA template strand with 5᾽→3᾽ orientation, DNA polymerase synthesises short pairs on new DNA (about l000 nucleotides long) in 5᾽→3᾽ direction and then joins these pieces together. These small fragments are called Okazaki fragments and new DNA strand made in this discontinuous manner is called lagging strand. Okazaki fragments are joined by means of DNA ligase.

12. Escherichia coli fully labelled with 15N is allowed to grow in 14N medium. The two strands of DNA molecule of the first generation bacteria have—

(a) same density but do not resemble with their parent DNA.
(b) same density and resemble with their parent DNA.
(c) different density and do not resemble with their parent DNA.
(d) different density but resemble with their parent DNA.

Ans. (c) different density and do not resemble with their parent DNA.

Explanation: The two strands of first generation DNA have different density since one strand has N 15 while other has N 14 hence they will not resemble their parent DNA.

13. Nitrogenous bases are linked to sugar by— (a) phosphodiester bond
(b) N-glycosidic bond
(c) O-glycosidic bond
(d) hydrogen bond 

Ans. (b) N-glycosidic bond

Explanation: A nitrogenous base is linked to the pentose sugar through N-glycosidic linkage to form a nucleoside, such as adenosine or deoxyadenosine, guanosine or deoxyguanosine, cytidine or deoxycytidine and uridine or deoxythymidine.

14. What does A and B represent in the given representation?


Ans. (a) Nucleoside Nucleotide

Explanation: A nucleotide has three components, a nitrogenous base, a pentose sugar, and a phosphate group. A nitrogenous base is linked to the pentose sugar through N-glycosidic linkage to form a nucleoside, such as adenosine or deoxyadenosine, guanosine or deoxyguanosine, cytidine or deoxycytidine and uridine or deoxythymidine.

15. In Meselson and Stahl’s experiments, heavy DNA was distinguished from normal DNA by centrifugation in–

(a) CsOH gradient
(b) CsCl gradient
(c) 15NH2Cl
(d) 14NH2Cl

Ans. (b) CsCl gradient

Explanation: In Meselson and Stahl’s experiments, heavy DNA was distinguished from normal DNA by centrifugation in CsCl gradient. When DNA is mixed with caesium chloride, it will settle down at a particular height in centrifugation and heavier one higher up.

7. Evolution

Multiple Choice Questions

1. Coacervates were experimentally produced by—
(a) Oparin and Sidney Fox
(b) Fischer and Huxley
(c) Jacob and Monod
(d) Urey and Miller

Ans. (a) Oparin and Sidney Fox

Explanation: Coacervates were non-living colloidal aggregates of organic molecules that formed by the intermolecular attraction force between the organic molecules synthesised abiotically by Oparin and Sydney Fox.

2. Which of the following pairs of structures is homologous?

(a) Wings of grasshopper and forelimbs of flying squirrel
(b) Tentacles of Hydra and arms of starfish
(c) Forelimbs of bat and forelegs of horse
(d) Wings of birds and wings of moth

Ans. (a) Wings of grasshopper and forelimbs of flying squirrel

Explanation: Wings of grasshopper and forelimbs of flying squirrel are homologous as they are anatomically same in structure. None of the other combinations has anatomically same structure.

3. Theory of chemical origin of life was prepared by—

(a) Miller and Fox
(b) Oparin and Haldane
(c) Miller and Watson
(d) Watson and Melvin

Ans. (b) Oparin and Haldane

Explanation: According to the theory of the chemical origin of life, pre-biological changes slowly transform simple atoms and molecules into the more complex chemical needed to produce life and it is proposed by Oparin and Haldane in 1992.

4. Non-directional alteration in Hardy-Weinberg equilibrium is—

(a) Gene flow
(b) Mutation
(c) Genetic drift
(d) Gene recombination

Ans. (c) Genetic drift

Explanation: It is genetic drift. It is the mechanism of evolution in which allele frequencies of a population change over generations due to chance. It's nondirectional and non-directive.

5. The earliest hominids evolved about 4 million years ago were—

(a) Homo erectus
(b) Neanderthal man
(c) Cro-magnon man
(d) Australo pithecus

Ans. (d) Australo pithecus

Explanation: The first early hominid evolved from Africa and known as Australopithecus lived 4 million ago. Neanderthal and Cro-magnon man have lived 40,000 years ago. Homo erectus have lived 2 million years ago.

6. A baby has a small tail. It is a case of—

(a) Mutation
(b) Metamorphosis
(c) Atavism
(d) Retrogressive evolution

Ans. (c) Atavism

Explanation: Atavism is a modification of a biological structure whereby an ancestral genetic trait reappears after having been lost through an evolutionary change in previous  generations. So, a baby that has a small tail is an example of atavism.

7. Industrial melanism is related to—  

(a) skin darkening due to smoke
(b) drug resistance
(c) defence against radiation
(d) protective resemblance to surroundings

Ans. (d) protective resemblance to surroundings

Explanation: Industrial melanism is a change in the colour of skin, fur or feathers acquired by a population of animals living in an industrial region where the environment is soot-darkened. This protective resemblance to surroundings helps members to survive and reproduce.

8. Darwin’s finches are good example of— 

(a) Connecting link
(b) Adaptive radiation
(c) Convergent evolution
(d) Industrial melanism

Ans. (b) Adaptive radiation

Explanation: Adaptive radiation is a rapid increase in the number of species with a common ancestor, characterised by great ecological and morphological diversity. Darwin's finches are a classical example of adaptive radiation. Their common ancestor arrived on the Galapagos about two million years ago.

9. Analogous organs arise due to— 

(a) Divergent evolution
(b) Artificial selection
(c) Genetic drift
(d) Convergent evolution

Ans. (d) Convergent evolution 

Explanation: Convergent evolution is the process whereby organisms not closely related (not monophyletic), independently evolve similar traits as a result of having to adapt to similar environments or ecological niches, bat and mosquitoes wings are example of such.

10. The theory of use and disuse of organ was proposed by—

(a) Darwin
(b) Lamarck
(c) de Vries
(d) Hooker

Ans. (b) Lamarck

Explanation: The use or disuse theory explains that the parts of an organism that the organism uses most will undergo hypertrophy and will become more developed and was proposed by Lamarck.

11. Birbal Sahni Institute of Palaeobotany is located at—

(a) Delhi
(b) Lucknow
(c) Dehradun
(d) Kolkata

Ans. (b) Lucknow

Explanation: Birbal Sahni Institute of Palaeobotany is situated in Lucknow, Uttar Pradesh and established in 1946. Birbal Sahni is also known as father of Palaeobotany in India.

12. A connecting link between reptiles and birds is— 

(a) Dimetrodon
(b) Dodo
(c) Archaeopteryx
(d) Sphenodon

Ans. (c) Archaeopteryx 

Explanation: Archaeopteryx is the connecting link between reptiles and aves. It had teeth, long bony tail and three claws in its wings as reptiles. It also retains a wishbone, a breastbone, hollow thin-walled bones, air sacs in the backbones, and feathers, which are also found in the nonavian coelurosaurian relatives of birds.

13. Genetic drift operates only in— 

(a) island population
(b) smaller population
(c) larger population
(d) mendelian population 

Ans. (b) smaller population 

Explanation: Genetic drift occurs mainly in a small isolated population. The reason is the irregular phenomenon of alleles getting lost with higher probability. When this loss of alleles starts then the genetic drift continues until that specific allele is completely lost within the entire population.

14. Species occuring in different geographical areas are called—

(a) Sibling
(b) Neopatric
(c) Sympatric
(d) Allopatric

Ans. (d) Allopatric

Explanation: Siblings: Generate from same parents. Neopatric, Sympatric species are developed due to reproductive isolation and occur in overlapping or same area of geographical distribution. Allopatric are species of different geographical regions.

15. The abiogenesis occurred about____________billion years ago.

(a) 1.2
(b) 1.5
(c) 2.5
(d) 3.5

Ans. (d) 3.5

Explanation: Abiogenesis or spontaneous creation or autobiogenesis was proposed by Von Helmont. The stated that life originated abiogenetically from non living material by spontaneous generation about 3.5 billion year ago.

8. Health and Diseases

Multiple Choice Questions

1. Which of the following is not a hereditary disease? 

(a) Cretinism
(b) Cystic fibrosis
(c) Thalassemia
(d) Haemophilia

Ans. (a) Cretinism

Explanation: Cretinism is not a hereditary disease out of the given options. T3 and T4 are thyroid hormones; As both T3 and T4 require iodine for their synthesis, therefore when deficiency of iodine occurs, low production of these hormones results in cretinism.

2. The disease tetanus is also called–

(a) Bleeder’s disease
(b) Lockjaw
(c) Whooping cough
(d) Gangrene

Ans. (b) Lockjaw

Explanation: Tetanus is an infection caused by the bacterium called Clostridium tetani. When the bacteria invade the body, they produce a poison (toxin) that causes painful muscle contractions. Another name for tetanus is “lockjaw”.

3. Health is not defined by–

(a) physical well-being
(b) mental well-being
(c) social well-being
(d) genetic disorders

Ans. (d) genetic disorders

Explanation: Health is defined as a state of complete physical, mental and social wellbeing. On the other hand, genetic disorders are the deficiencies or defects which the child inherits from parents by birth, which affect the health of an individual.

4. Which of the following is not a biological agent that causes disease?

(a) Mycoplasma
(b) Virus
(c) Radiations
(d) Fungi

Ans. (c) Radiations

Explanation: Radiations are physical agents that cause disease along with electricity, pressure, heat, cold, etc. Mycoplasma, virus and fungi are biological agents or pathogens that cause various diseases.

5. Which characteristic is not exhibited by a disease-causing pathogen?

(a) Invasiveness
(b) Toxigenicity
(c) Virulence
(d) Co-operation

Ans. (d) Co-operation

Explanation: Co-operation is a characteristic not exhibited by a pathogen. Invasiveness of a pathogen is its ability to gain entry into a host and grow. Virulence is the ability of a pathogen to produce disease. Toxigenicity is the power of a pathogen to form toxins capable of damaging host cells.

6. What is the first sign or symptom of tetanus? 

(a) Stiffness of neck
(b) Swelling in neck
(c) Paralysis
(d) Numbness in fingers

Ans. (a) Stiffness of neck

Explanation: Stiffness of neck is the first symptom of tetanus. Then there is difficulty in swallowing and chewing. Subsequently, spasms of muscles of the jaw and face take place. There is severe pain. It is often a fatal disease.

7. What is the full form of ATS injection?

(a) Anti Tuberculosis Serum
(b) Artificial Typhoid Serum
(c) Anti Tetanus Serum
(d) Alpha Typhoid Serum

Ans. (c) Anti Tetanus Serum

Explanation: ATS stands for Anti Tetanus Serum injection. It should be administered within 48 hours of injury if the person has not had a tetanus shot in the past five years because Clostridium tetani enter in the blood through any wound or cut.

8. The organisms which cause diseases in plants and animals are called — [NCERT Exemplar]

(a) pathogens
(b) vectors
(c) insects
(d) worms

Ans. (a) pathogens

Explanation: A wide range of organisms including bacteria, viruses, fungi, protozoans, helminths, etc., cause diseases in plants and animals. Such disease causing organisms are called pathogens. While vectors are the carriers of pathogens which may be insects or worms.

9. Diseases are broadly grouped into infectious and non-infectious diseases. In the list given below, identify the infectious diseases. [NCERT Exemplar]

I. Cancer
II. Influenza
III. Allergy
IV. Smallpox

(a) I and II
(b) II and III
(c) III and IV
(d) II and IV

Ans. (d) II and IV

Explanation: Influenza, is an infectious disease caused by influenza viruses. Cancer is a non-infectious disease. Smallpox is a serious, highly contagious disease. It is caused by the Variola Virus. Allergy is the exaggerated response of the immune system to certain antigens. It is a non-infectious response.

10. A non-communicable disease is–

(a) Diabetes
(b) Measles
(c) Influenza
(d) Diphtheria

Ans. (a) Diabetes

Explanation: Diabetes is one of the four major types of non-communicable diseases (cardiovascular disease, diabetes, cancer and chronic respiratory diseases).

11. Which of the following is the most fatal non-infectious disease?

(a) AIDS
(b) Cancer
(c) Diabetes
(d) Obesity 

Ans. (b) Cancer

Explanation: Cancer is the most fatal non-infectious disease. In 2017, 9.6 million people were estimated to have died from various forms of cancer. While AIDS is the most fatal infectious disease affecting nearly 36.9 million people worldwide. Diabetes and obesity are also non-infectious diseases but not the most fatal.

12. How are infectious diseases transmitted?

(a) By droplet infection
(b) They are inherited
(c) By seeing patients
(d) By eating an unhealthy diet

Ans. (a) By droplet infection

Explanation: Infectious diseases are transmitted from one person to another by droplet infection, direct contact with an infected persons, animal bite, contact with soil or by transplacental transmission.

13. A person with sickle-cell anaemia is— [NCERT Exemplar]

(a) more prone to malaria
(b) more prone to typhoid
(c) less prone to malaria
(d) less prone to typhoid

Ans. (c) less prone to malaria

Explanation: Sickle-cell anaemia is related to malaria, not to typhoid and person suffering from sickle cell anaemia are resistant to malarial parasite as RBCs of sickle-cell anaemic patients is distorted in shape and is not affected by Plasmodium sp.

14. Haemozoin is– [NCERT Exemplar]

(a) precursor of haemoglobin
(b) toxin from Streptococcus
(c) toxin from Plasmodium species
(d) toxin from Haemophilus species

Ans. (c) toxin from Plasmodium species

Explanation: Haemozoin is a toxin released by Plasmodium species, which is responsible for the chills and high fever recurring every three to four days. The rupture of RBCs during life cycle or plasmodium is associated with release of a toxic substance, haemozoin.

15. The sporozoites that cause infection when a female Anopheles mosquito bites a human being are formed in– [NCERT Exemplar]

(a) liver of human
(b) RBCs of mosquito
(c) salivary glands of mosquito
(d) intestine of human

Ans. (c) salivary glands of mosquito

Explanation: Sporozoites enter the female Anopheles mosquito when they bite an infected person where these sporozoites fertilise and multiply in the stomach wall of the female Anopheles and get stored in the salivary gland of mosquito till they are again transferred to the human body by a mosquito bite.

9. Improvement in Food Production

Multiple Choice Questions

1. ‘Jaya’ and ‘Ratna’ developed for green revolution in India are the varieties of—

(a) Maize
(b) Rice
(c) Wheat
(d) Bajra

Ans. (b) Rice

Explanation: “Jaya” and “Ratna” are the Indian varieties of rice. Indian varieties of Maize are P-3501, NK-6240, P-3396 etc. Indian varieties of bajra are GHB-558, Pusa- 605, Nandi-32 etc. and Indian varieties of wheat are UP-2338, PBW-502, etc.

2. ‘Himgiri’ developed by hybridisation and selection for disease resistance against rust pathogen is a variety of—

(a) Chilli
(b) Maize
(c) Sugarcane
(d) Wheat

Ans. (d) Wheat

Explanation: Himgiri is a disease-resistant variety of wheat.

3. Which one of the following is not a biofertiliser?

(a) Agrobacterium
(b) Rhizobium
(c) Nostoc
(d) Mycorrhiza

Ans. (a) Agrobacterium

Explanation: Agrobacterium is a genus of Gram-negative bacteria, that causes tumours in plants.

4. Which one of the following species of bees are used for the commercial production of honey?

(a) Apis dorsata
(b) Apis indica
(c) Apis florea
(d) Apis mellifera

Ans. (d) Apis mellifera

Explanation: Apis mellifera has more ability to protect itself from enemies so they are used for the commercial production of honey.

5. Which one of the following is a breed of cattle? 

(a) Ayrshire
(b) Ghagus
(c) Kadakanath
(d) Scampi

Ans. (a) Ayrshire

Explanation: Ayrshire is an efficient grazer and noted for her vigour and efficiency of milk production. Ghagus is a chicken breed of Karnataka. Kadakanath is also a chicken breed of India. Scampi is an edible lobster.

6. Single cell protein refers to—

(a) a specific protein extracted from pure culture of a single type of cell.
(b) sources of mixed proteins extracted from pure or mixed culture of organism from a single cell.
(c) protein extracted from a single cell.
(d) a specific protein extracted from a single cell.

Ans. (b) sources of mixed proteins extracted from pure or mixed culture of organism from a single cell.

Explanation: Single-cell protein or SCP is a source of mixed protein extracted from pure or mixed culture of single-cell organisms such as yeast, bacteria, fungi, algae etc.

7. Micropropagation is a technique—

(a) for production of thousand of plants
(b) for production of haploid plants
(c) for production of somatic hybrids
(d) for production of somaclonal plants.

Ans. (a) for production of thousand of plants

Explanation: Micropropagation is a technique for the production of thousands of plants. Haploid plants are obtained from the cells of the gametes produced on the anther or inside an ovary.

8. Root development is promoted by—

(a) Abscisic acid
(b) Auxin
(c) Ethylene
(d) Gibberellins

Ans. (b) Auxin

Explanation:

Native auxin is transported down the stem to the roots and the overall development of the roots is stimulated.

9. The sugar present in milk is—

(a) Glucose
(b) Lactose
(c) Fructose
(d) Sucrose

Ans. (b) Lactose

Explanation: Lactose is a sugar found in milk.

10. One of the commonly used plant growth hormones in tea plantation is —

(a) Ethylene
(b) Abscisic acid
(c) Zeatin
(d) Indole-3- acetic acid and IBA

Ans. (d) Indole-3- acetic acid and IBA

Explanation: Indole-3-acetic acid is the most common naturally occurring plant auxin class hormone but can be artificially synthesised. It is the best known of the auxins which help in growth. IDA is indole butyric acid and synthetic plant hormone used for plant easy growth.

11. The term ‘totipotency’ refers to the capacity of a—

(a) cell to generate whole plant
(b) bud to generate whole plant
(c) seed to germinate
(d) cell to enlarge in size

Ans. (a) cell to generate whole plant

Explanation: Totipotency is the ability of a single cell to divide and produce all the differentiated cells in an organism, including extraembryonic tissues. Totipotent cells formed during sexual and asexual reproduction include spores and zygotes to generate whole plants.

12. An explant is —

(a) dead plant
(b) part of the plant
(c) part of the plant used in tissue culture
(d) part of the plant that expresses a specific character

Ans. (c) part of the plant used in tissue culture

Explanation: Explant is the living tissue removed from an organism and placed in a medium for tissue culture.

13. 33% of India’s gross domestic product comes from—

(a) industry
(b) agriculture
(c) export
(d) small scale cottage industry 

Ans. (b) agriculture 

Explanation: One-third of Indian gross domestic products comes from agriculture. About 8% of GDP comes from industry and about 20% of GDP comes from MSMEs. Rest comes from service sectors.

14. In the hexaploid wheat, the haploid (n) and basic (x) number of chromosomes are—
(a) n = 21 and x = 21
(b) n = 21 and x = 14
(c) n = 21 and x = 7
(d) n = 7 and x = 21

Ans. (c) n = 21 and x = 7 

Explanation: In a hexaploid wheat 2n = 6x = 42. So, the haploid chromosome number of hexaploid is 42 ÷ 2 = 21 and the basic number of chromosomes is 42 ÷ 6 = 7.

15. The scientific process of which crop plants are enriched with certain desirable nutrient is called—

(a) Crop protection
(b) Breeding
(c) Biofortification
(d) Bioremediation

Ans. (c) Biofortification

Explanation: Biofortification is the process of increasing the micronutrient content of a food crop through selective breeding, genetic modification.

10. Microbes in Human Welfare

Multiple Choice Questions

1. Lactic acid is formed by the process of:

(a) Fermentation
(b) Glycolysis
(c) HMP pathway
(d) None of these

Ans. (a) Fermentation

Explanation : In lactic acid fermentation, glucose is converted into pyruvic acid, which is then reduced to lactic acid:

2. Large-holes in ‘Swiss-Cheese’ are due to:

(a) Propionibacterium shermanii
(b) Saccharomyces cerevisiae
(c) Penicillium chrysogenum
(d) Acetobacter aceti

Ans. (a) Propionibacterium shermanii

Explanation : Propionibacterium shermanii is the bacterium used for the production of a large amount of CO2 for making swiss cheese that has large holes.

3. Which of the following is also known as baker’s yeast?

(a) Saccharomyces cerevisiae
(b) Streptococcus pneumoniae
(c) Leuconostoc mesenteroides
(d) Streptococcus faecalis

Ans. (a) Saccharomyces cerevisiae

Explanation : Saccharomyces cerevisiae is also known as baker’s yeast. This yeast is used for fermenting wheat flour after it is kneaded. Similarly, the dough which is used for making bread is also fermented using baker’s yeast.

4. Which of the following gas is responsible for the puffed-up appearance of the dough?

(a) NH3
(b) CO2
(c) O2
(d) CH4

Ans. (b) CO2

Explanation : The dough which is used for making foods such as dosa and idli is fermented by using bacteria like Saccharomyces. The puffed-up appearance of this dough is due to the production of CO2 gas.

5. Which of the following household products is not made from Soybean?

(a) Tempeh
(b) Yoghurt
(c) Tofu
(d) Soy sauce

Ans. (b) Yoghurt

Explanation : Soybean products include Tempeh, Tofu and Soy sauce. Tofu is cheese-like product of soybean obtained after fermentation with Mucor species. While tempeh is an Indonesian food formed by fermenting, drying, salting and frying of soybean.

6. Which of the following statements is incorrect regarding Yoghurt?

(a) It is fermented milk.
(b) Mixed with an inoculum containing Streptococcus faecalis.
(c) Preparation takes approximately 4-5 hours.
(d) The original flavour of yoghurt is of acetaldehyde.

Ans. (b) Mixed with an inoculum containing Streptococcus faecalis.

Explanation : Yoghurt is fermented milk. The milk is heated to 80-90°C for half an hour and then it is cooled to 40-43°C and mixed with an inoculum containing Streptococcus. Preparation of yoghurt takes approximately 4-5 hours. The original flavour of yoghurt is of acetaldehyde.

7. Which of the following organisms is responsible for the production of citric acid? 

(a) Bacteria
(b) Archaebacteria
(c) Protozoa
(d) Yeast

Ans. (d) Yeast

Explanation : Species of yeast like Mucor species, Aspergillus niger, can ferment sugar to produce citric acid. Citric acid is a preservative and flavouring agent.

8. Which of the following is a function of pectinases?

(a) Clearing of fruit juices
(b) Degumming of silk
(c) Cleaning of hides
(d) Manufacturing of soap

Ans. (a) Clearing of fruit juices

Explanation : Pectinases are obtained from fungi grown on pectin containing media. Pectinases are used in enhancing juice extraction from fruits, clearing of fruit juices, removing bitterness, retting of fibres and fermentation of green coffee.

9. Which of the following is not a function of amylases?

(a) Manufacturing of soap
(b) Sweetening of bread
(c) Separation and desising of fibres
(d) Clearing starch related turbidity in fruit juices

Ans. (a) Manufacturing of soap

Explanation : Amylases are starch digesting enzymes that are used in softening and sweetening of bread, separation and desising of fibres, clearing of starch related turbidity in juices and clearing starch related stains on utensils and clothes.

10. The vitamin whose content increases following the conversion of milk into curd by lactic acid bacteria is: [NCERT Exemplar]

(a) Vitamin C
(b) Vitamin D
(c) Vitamin B12
(d) Vitamin E

Ans. (c) Vitamin B12

Explanation : A small amount of curd added to the fresh milk as inoculum or starter contains millions of LAB, which at suitable temperature multiply, thus converting milk to curd, which also improves its nutritional quality by increasing vitamin-B12.

11. Match the following columns of bacteria and their commercially important products.

Codes:
A B C D
(a) 2 3 4 1
(b) 2 4 3 1
(c) 4 3 2 1
(d) 4 1 3 2

Ans. (c) 4 3 2 1

Explanation : Following is the list of bacteria and their commercially important products.

12. Match the following columns of bioactive substances and their roles. [NCERT Exemplar]

Choose the correct match:
Codes:

A B C D
(a) 2 3 1 4
(b) 4 2 1 3
(c) 4 1 3 2
(d) 3 4 2 1

Ans. (d) 3 4 2 1

Explanation : Statins produced by the yeast Monascus purpureus have been commercialised as blood cholesterol lowering agents. Cyclosporin A is used as an immunosuppressive agent in organ transplant patients. Streptokinase is used as a ‘clot buster’ for removing clots from the blood attack. Lipases are used in detergent formulations and are helpful in removing only stains from the laundry.

13. Ethanol is commercially produced through a particular species of:

(a) Saccharomyces
(b) Clostridium
(c) Trichoderma
(d) Aspergillus

Ans. (a) Saccharomyces

Explanation : Ethanol is commercially produced through a particular species of Saccharomyces. Ethanol is an attractive renewable biofuel. Increasing the availability of this alternative energy source requires ethanologenic yeasts that can produce ethanol more efficiently.

14. Depending on the type of raw material used for fermentation and the type of processing, different types of alcoholic drinks are obtained.

(a) True
(b) False
(c) Depends on conditions
(d) Can't be determined

Ans. (a) True

Explanation : We can obtain different types of alcoholic drinks by using different raw materials for fermentation and by using different techniques for processing like processing alcoholic drinks with distillation or without distillation.

15. Which of the following is the function of lipases?

(a) Manufacturing of soaps
(b) Clearing of hides
(c) Sweetening of bread
(d) Removing bitterness

Ans. (a) Manufacturing of soaps

Explanation : Lipases are lipid digesting enzymes used in flavouring cheese, hydrolysing oils for manufacture of soap and in detergent formulations for removing fatbased stains.

11. Principles and Processes of Biotechnology.

Multiple Choice Questions

1. The ability of a cell to grow into a complete plant is called:

(a) Protoplasmic fusion
(b) Tissue culture
(c) Cellular totipotency
(d) Somaclonal hybridisation

Ans. (c) Cellular totipotency

Explanation : Totipotency is the ability of a single cell to divide and produce all the differentiated cells in an organism. For example, spores and zygotes are examples of totipotent cells.

2. Reproducing new plants by cells instead of seeds is known as:

(a) Biofertiliser
(b) Mutation
(c) Antibiotics
(d) Tissue culture

Ans. (d) Tissue culture

Explanation : Plant tissue culture is a collection of techniques used to maintain or grow plant cells, tissues or organs under sterile conditions on a nutrient culture medium of known composition. Plant tissue culture is widely used to produce clones of a plant in a method known as micropropagation.

3. The most common bacterium used in genetic engineering is:

(a) E. coli
(b) Clostridium
(c) Bacillus thuringiensis
(d) Salmonella

Ans. (a) E. coli

Explanation : E. coli is a preferred host for gene cloning and protein production with in cell. Thus, it is widely used in genetic engineering.

4. Bt cotton is resistant to:

(a) insects
(b) herbicides
(c) salt
(d) drought

Ans. (a) insects

Explanation : Bacillus thuringiensis is a soil bacterium which produces certain proteins that have ability to kill some insects. If Bt gene from this bacteria is incorporated in the cotton then it resist insects.

5. Bt toxin is:

(a) intracellular lipid
(b) polysaccharide
(c) intracellular crystalline protein
(d) extracellular crystalline protein

Ans. (d) extracellular crystalline protein

Explanation : Bacillus thuringiensis (Bt) is a gram-positive, spore-forming bacterium that synthesises Bt toxin. It secretes toxin extracellularly, which is a crystalline protein that blocks insect gut proteins upon ingestion.

6. Plasmid is a:

(a) fungus
(b) plastid
(c) extra-chromosomal DNA
(d) part of plasma membrane

Ans. (c) extra-chromosomal DNA

Explanation : Plasmids are self-replicating extra-chromosomal double stranded circular DNA molecules found in gram-negative and gram-positive bacteria as well as in some yeast and other fungi.

7. Transfer of any gene into a completely different organism can be done through:

(a) genetic engineering
(b) tissue culture
(c) transformation
(d) none of the above

Ans. (a) genetic engineering

Explanation : Genetic engineering is a biological method in which genes of one organism are being inserted into another organism includes cloning. Tissue culture - Technique to grow in an artificial medium from living tissues. Transformation is the process by which an organism acquires exogenous DNA.

8. PCR is required for:

(a) DNA synthesis
(b) DNA amplification
(c) Protein synthesis
(d) Amino acid synthesis

Ans. (b) DNA amplification

Explanation : Polymerase chain reaction (PCR) is a laboratory technique used to amplify DNA sequences. The method involves using short DNA sequences called primers to select the portion of the genome to be amplified.

9. Gel electrophoresis is used for:

(a) construction of recombinant DNA by joining with cloning vectors
(b) isolation of DNA molecule
(c) cutting of DNA into fragments
(d) separation of DNA fragments according to their size

Ans. (d) separation of DNA fragments according to their size

Explanation : Gel electrophoresis is a method that is widely used in laboratories to separate the molecules such as DNA, RNA, or proteins, based on their size, under the influence of an electric field.

10. Polyethylene glycol method is used for:

(a) gene transfer without a vector
(b) bio-diesel production
(c) seedless fruit production
(d) all of the above

Ans. (a) gene transfer without a vector

Explanation : Polyethylene glycol is used in chemical mediated gene transfer wherein it induces DNA uptake into plant protoplasts.

11. The genetic defect Adenosine Deaminase (ADA) deficiency may be cured permanently by:

(a) administering adenosine deaminase activators
(b) periodic infusion of genetically engineered lymphocytes having functional ADA cDNA
(c) enzyme replacement therapy
(d) introducing bone marrow cells producing ADA into cells at early embryonic stages

Ans. (d) introducing bone marrow cells producing ADA into cells at early embryonic stages
Explanation : Adenosine deaminase deficiency is caused due to deletion of a gene responsible for coding the enzyme adenosine deaminase (ADA). Bone marrow transplant and enzyme replacement therapy are not permanent cures. Genetically engineered lymphocytes are to be periodically infused. If bone marrow cells producing ADA are introduced at an early embryonic stage then it can be a permanent cure for the disease.

12. Product of biotechnology is:

(a) Humulin
(b) Bt cotton
(c) Biofertilisers
(d) All of these

Ans. (d) All of these

Explanation : Biotechnology is a set of technologies used to change the genetic make-up of the cells, including the transfer of genes within and across species to producing novel organisms like transgenic crop, biofertiliser, humulin, etc.

13. Which among the following is regarded as molecular scissors?

(a) Reverse transcriptase
(b) Restriction endonuclease
(c) Exonucleases
(d) Polymerases

Ans. (b) Restriction endonuclease

Explanation : Restriction enzymes are also called “molecular scissors” as they cleave DNA at or near specific recognition sequences known as restriction sites. These enzymes make one incision on each of the two strands of DNA and are also called restriction endonucleases.

14. Enzyme used in PCR technology is:

(a) Taq polymerase
(b) Polymerase
(c) Exonuclease
(d) Helicase

Ans. (a) Taq polymerase

Explanation : Taq DNA Polymerase is a highly thermostable recombinant DNA polymerase. PCR relies upon Taq Polymerase, to produce numerous copies of a specific DNA region in vitro.

15. In Bt cotton, a transgenic plant, Bt refers to:

(a) Botanical
(b) Beta
(c) Biotechnology
(d) Bacillus thuringiensis

Ans. (d) Bacillus thuringiensis

Explanation : Bt cotton is a genetically modified plant produced by injecting the gene of Bacillus thuringiensis, a bacterium that produces a toxin which is insecticidal in nature.

12. Application of Biotechnology in Health and Agriculture

Multiple Choice Questions

1. Name the drug used in cancer treatment produced by using biotechnology:

(a) Terramycin
(b) HGH
(c) TSH
(d) Interferon

Ans. (d) Interferon

Explanation : Interferons are a family of naturally-occurring proteins that are made and secreted by cells of the immune system. Interferons can promote or hinder the ability of some cells to differentiate.

2. What does the second critical research area of biotechnology provide?

(a) Optimum conditions for the catalyst.
(b) The pure form of catalyst.
(c) An impure form of catalyst.
(d) Purification of products.

Ans. (a) Optimum conditions for the catalyst.

Explanation : There are three critical research areas in biotechnology. They are:
(a) It provides the best catalyst in the form of an improved organism (transformants) usually a microbe or pure enzyme.
(b) It creates optimum conditions for the catalyst by the method of genetic engineering.
(c) Purification of products by downstream processing.

3. The first antibiotic was discovered by ______.

(a) Louis Pasteur
(b) R. Koch
(c) W. Fleming
(d) A. Fleming

Ans. (d) A. Fleming

Explanation : Scottish physician Alexander Fleming was the first to experimentally discover that a Penicillium mould secretes an antibacterial substance, and was first to concentrate the active substance involved, which he named penicillin in 1928.

4. Human insulin is being commercially produced from a transgenic species of:

(a) Rhizobium
(b) Saccharomyces
(c) Escherichia
(d) Mycobacterium

Ans. (c) Escherichia

Explanation : Human insulin is produced from genetically engineered E.coli. By using genetic engineering or recombinant DNA technology. In 1983, an American pharmaceutical firm Eli Lilly synthesised artificial insulin (Humulin) with the help of plasmids of Escherichia coli. Since then, genetically engineered E. coli bacteria are being used to produce human insulin.

5. Genetically engineered bacteria are being employed for production of:

(a) Thyroxine
(b) Human insulin
(c) Cortisol
(d) Epinephrine

Ans. (b) Human insulin

Explanation : Human insulin is made by genetically engineered microbes in a pure form. Genetically-engineered bacteria are grown in large stainless steel fermentation vessels. The vessel contains all the nutrients needed for growth. When the fermentation is complete, the mixture containing the bacteria is harvested. The bacteria are filtered off and broken open to release the insulin they have produced. It is then purified and packaged into bottles for distribution.

6. From which animals were insulin obtained in the early days?

(a) Insects
(b) Lizard and snakes
(c) Cats and dogs
(d) Cattle and pigs

Ans. (d) Cattle and pigs

Explanation : In early days, the insulin was obtained from the pancreas of slaughtered cattle and pigs. Insulin obtained this way had many disadvantages.

7. How many recombinant therapeutics have been approved for humans to date?

(a) 30
(b) 20
(c) 13
(d) 25

Ans. (a) 30

Explanation : Recombinant therapeutics are the type of medicines obtained from recombinant organisms. These are mostly protein in nature. Till date, 30 recombinant therapeutics have been approved for human use which includes insulin, human factor IX, etc.

8. The polypeptide chains present in insulin is connected by ___________ bonds.

(a) ionic
(b) covalent
(c) disulphide
(d) hydrophobic interactions

Ans. (c) disulphide

Explanation : The polypeptide chains present in insulin is connected by disulphide bonds. These disulphide bonds are formed between two cysteine residues. In total, insulin consists of 3 disulphide bonds.

9. Which organ secretes insulin?

(a) Stomach
(b) Pancreas
(c) Thyroid
(d) Intestine

Ans. (b) Pancreas

Explanation : Insulin is secreted by the pancreas. The stomach secretes various digestive enzymes like pepsin and gastric juice. The thyroid secretes thyroid hormones while small intestine secretes trypsin.

10. Insulin in pro-hormone form contains an extra stretch called ___________.
(a) B-peptide
(b) G-peptide
(c) C-peptide
(d) S-peptide

Ans. (c) C-peptide

Explanation : The pro-hormone form of insulin consists of an extra stretch called Cpeptide. The C peptide is a connecting peptide having approximately 3 amino acids. It connects A and B chain of pro-insulin.

11. ___________ in 1983 used recombinant DNA technology to produce insulin.

(a) Eli Lilly
(b) Emily Lilly
(c) Lilly Rose
(d) Amy Sanger

Ans. (a) Eli Lilly

Explanation : In 1983, Lilly Company of America that was founded by Eli Lilly used recombinant DNA technology to produce insulin.

12. ___________ organism was used to produce recombinant insulin.

(a) Cyanobacteria
(b) E.coli
(c) Saccharomyces cerevisiae
(d) B. subtilis

Ans. (b) E.coli

Explanation :The first recombinant insulin was constructed by using E.coli. It was developed by Lilly Company.

13. C-peptide is removed during ___________ phase of insulin.

(a) initiation
(b) maturation
(c) termination
(d) elongation

Ans. (b) maturation

Explanation : C-peptide is removed during the maturation phase of insulin production. It is released as a by-product of insulin formation.

14. C-peptide of human insulin is: [NCERT Exemplar]
(a) a part of mature insulin molecule.
(b) responsible for the formation of disulphide bridges.
(c) removed during maturation of pro-insulin to insulin.
(d) responsible for its biological activity.

Ans. (c) removed during maturation of pro-insulin to insulin

Explanation : The connecting peptide or C-peptide is a short protein containing 31 amino acids. It connects the A and B chain of proinsulin molecule. After the processing of proinsulin molecule, C-peptide is removed leaving behind A and B chains which bound together by disulphide bonds to constitute a insulin molecule.

15. The first clinical gene therapy was given for treating:

(a) Diabetes mellitus
(b) Chicken pox
(c) Rheumatoid arthritis
(d) Adenosine Deaminase deficiency

Ans. (d) Adenosine Deaminase deficiency

Explanation : The first clinical gene therapy was given in 1990 to a 4-year old girl with Adenosine Deaminase (ADA) deficiency. This enzyme is crucial for the immune system to function. The disorder is caused due to the deletion of the gene for Adenosine Deaminase.

13. Organisms and Environment 

Multiple Choice Questions

1. The physiological capacity of a species to produce their progeny in absence of predators, diseases and other inhibiting factors is called:

(a) carrying capacity
(b) biotic potential
(c) environmental resistance
(d) population explosion

Ans. (b) biotic potential

Explanation : The ability of a population of a particular species to propagate under ideal environmental conditions such as sufficient food supply, no diseases, and no predators, is called biotic potential.

2. Zone of atmosphere near the Earth surface is:

(a) Stratosphere
(b) Mesosphere
(c) Troposphere
(d) Thermosphere

Ans. (c) Troposphere

Explanation : The troposphere is the lowest layer of our atmosphere. Starting at ground level, it extends upward to about 10 km (6.2 miles or about 33,000 feet) above sea level. After troposphere, stratosphere (up to 50 km), mesosphere (up to 85 km) and thermosphere (between 500 km – 1000 km) are found.

3. Photosynthetically active radiation (PAR) represent the following range of wavelength:

(a) 450–950 nm
(b) 340–450 nm
(c) 400–700 nm
(d) 500–600 nm

Ans. (c) 400–700 nm

Explanation : Photosynthetically active radiation, often abbreviated PAR, designates the spectral range (wave band) of solar radiation from 400 to 700 nanometers.

4. Niche overlap indicates:

(a) mutualism between two species.
(b) active cooperation between two species.
(c) two different parasites on the same host.
(d) sharing of one or more resources between the two species.

Ans. (d) sharing of one or more resources between the two species.

Explanation : Organisms or populations in the competition have a niche overlap of a limited resource for which they compete. Both owl and cat feed on shrews and mice it means two or more species sharing of one or more resources.

5. Who is referred as Father of Ecology in India?

(a) Ramdeo Misra
(b) Raja Ram Mohan Roy
(c) Verghese Kurien
(d) M S Swaminathan

Ans. (a) Ramdeo Misra

Explanation : The Father of Ecology in India is Ramdeo Misra. Raja Ram Mohan Roy was a freedom fighter. Verghese Kurien is the Father of the White Revolution of India while MS Swaminathan is the Father of Green Revolution of India.

6. What is a group of individuals belonging to the same species within an ecosystem called?

(a) Species
(b) Population
(c) Biomass
(d) Manipulation

Ans. (b) Population

Explanation : A group of individuals belonging to the same species within an ecosystem is called population while biomass is plant based renewable organic material and manipulation is a skill full handling work.

7. What is the place where a particular organism lives called?

(a) Factors
(b) Niche
(c) Habitat
(d) Environment

Ans. (c) Habitat

Explanation : The place where a particular organism lives is called a habitat. It depends on both biological and physical factors present within it while niche is the unique place of  species in an ecosystem and environment is our surroundings.

8. Biosphere is: [NCERT Exemplar]

(a) a component in the ecosystem.
(b) composed of the plants present in the soil.
(c) life in outer space.
(d) composed of all living organisms present on Earth which interact with the physical environment.

Ans. (d) composed of all living organisms present on Earth which interact with the physical environment.

Explanation : Biosphere is composed of all living organisms present on Earth which interact with their physical environment. In other words a biosphere or ecosphere term is collectively used for all the ecosystems of world. Ecosphere or biosphere can not be defined by other three options, as they represent only a part of ecosystem.

9. Ecological niche is: [NCERT Exemplar]

(a) the surface area of the ocean.
(b) the surface area of forests.
(c) the physical position and functional role of a species within the community.
(d) formed of all plants and animals living at the bottom of a lake.

Ans. (c) the physical position and functional role of a species within the community.

Explanation : Ecological niche is determined by factors such as by its range of tolerance i.e., the physical position and functional role of a species within the community. etc while rest of the options are incorrect.

10. Salt concentration (salinity) of the sea measured in parts per thousand is: [NCERT Exemplar]

(a) 10-15
(b) 30-70
(c) 0-5
(d) 30-35

Ans. (d) 30-35

Explanation : The salt concentration (salinity) of sea is measured in parts per thousand is 30-35.

11. Formation of tropical forests needs mean annual temperature and mean annual precipitation as: [NCERT Exemplar]

(a) 18-25 °C and 150-400 cm
(b) 5-15 °C and 50-100 cm
(c) 30-50 °C and 100-150 cm
(d) 5-15 °C and 100-200 cm

Ans. (a) 18-25 °C and 150-400 cm

Explanation : Formation of tropical forest needs annual temperature 18-25°C and annual rainfall (precipitation) above 140 cm, usually between 150-400 cm and reach upto 1000 cm/year. Because tropical forest (tropical rain or evergreen forest mainly occurs in equatorial or sub-equatorial region like Amazon, Central America and Orinoco and Congo river basins of South America and Africa respectively.

12. Which of the following forest plants controls the light conditions at the ground? [NCERT Exemplar]

(a) Lianas and climbers
(b) Shrubs
(c) Tall trees
(d) Herbs

Ans. (c) Tall trees

Explanation : In forest, intensity, duration and quality of light at ground is controlled by tall trees, which have higher productivity than shrubs and herbs growing underneath. Lianas and climbers are woody vines which make commercialism association with tall trees. Herbs and shrubs occupy lower strata of forest.

13. What will happen to a well growing herbaceous plant in the forest if it is transplanted outside the forest in a park? [NCERT Exemplar]

(a) It will grow normally.
(b) It will grow well because it is planted in the same locality.
(c) It may not survive because of change in its micro climate.
(d) It grows very well because the plant gets more sunlight.

Ans. (c) It may not survive because of change in its micro climate.

Explanation : A well growing herbaceous plant in forest receives less intensity, duration and quality of light, but when it is transplanted in a park outside the forest, it will face microclimate charge and may not survive due to uninterrupted receiving of light. Rest of the other options are incorrect to depict the fate of plant.

14. Lichens are well known combination of an algae and fungus where fungus is:

(a) an epiphytic relationship between algae.
(b) a parasitic relationship with the algae.
(c) a symbiotic relationship with the algae.
(d) a saprophytic relationship with the algae.

Ans. (c) a symbiotic relationship with the algae.

Explanation : Lichens are a well-known combination of an algae and a fungus where the fungus has a symbiotic relationship with the algae. The fungus depends on the algae for food, as fungi cannot perform photosynthesis. In return, they provide shelter to the algae.

15. Pneumatophores are present or common in:

(a) xerophytes
(b) hydrophytes
(c) mesophytes
(d) halophytes

Ans. (d) halophytes

Explanation : Pneumatophore roots are specialised aerial roots in halophytes which help the plants to breathe air in waterlogged soil. Halophytic plants possess the negatively geotropic vertical roots.

14. Ecosystem

Multiple Choice Questions

1. Identify the possible link “A” in the following food chain:

Plant → Insect → Frog → “A” → Eagle.

(a) Rabbit
(b) Wolf
(c) Cobra
(d) Parrot

Ans. (c) Cobra

Explanation: Frog is prey to Cobra. Neither rabbit, parrot nor wolf eats frog.

2. Both hydrarch and xerarch successions lead to—

(a) medium water conditions
(b) xeric conditions
(c) highly dry conditions
(d) excessive wet conditions

Ans. (a) medium water conditions

Explanation: Both Hydrarch and Xerarch succession lead to medium water conditions. Xerarch succession takes place in dry areas and the series progress from xeric to mesic (medium water) condition.

3. Which one of the following is not a gaseous biogeochemical cycle in ecosystem?

(a) Sulphur cycle
(b) Phosphorus cycle
(c) Nitrogen cycle
(d) Carbon cycle

Ans. (b) Phosphorus cycle

Explanation: Phosphorus does not follow the gaseous biogeochemical cycle in the ecosystem.

4. Which one of the following animals may occupy more than one trophic levels in the same ecosystem at the same time?

(a) Sparrow
(b) Lion
(c) Goat
(d) Frog

Ans. (a) Sparrow

Explanation: Sparrow may occupy more than one tropic level in the same ecosystem at the same time. Sparrow is a primary consumer when it eats seeds, fruits, peas and a secondary consumer when it eats insects and worms.

5. The upright pyramid of number is absent in—

(a) pond
(b) forest
(c) lake
(d) grassland

Ans. (b) forest

Explanation: Pyramid of numbers represents the number of individuals per unit area at various trophic levels. It is always upright in grassland, pond and lake ecosystems. But in a forest, it is partly upright because producers are lesser in number and support a greater number of herbivores which in turn support a fewer number of carnivores.

6. Which one of the following is not a functional unit of an ecosystem?

(a) Energy flow
(b) Decomposition
(c) Productivity
(d) Stratification

Ans. (d) Stratification

Explanation: Stratification is not the functional unit of an ecosystem. It is seen in different communities and is different for different types of habitats.

7. Identify the likely organism (1), (2), (3) and (4) in the food web shown below:

(1) (2) (3) (4)
a) deer rabbit frog rat
(b) dog squirrel bat deer
(c) rat dog tortoise crow
(d) squirrel cat rat pigeon

Ans. (a) deer rabbit frog rat

Explanation: Deer is prey to lion. Rabbit is prey to foxes. Snake eats a frog. Hawk preys to rat. So, option (a) is correct.

8. Which of the following representations shows the pyramid of numbers in a forest ecosystem?

(a) D
(b) A
(c) B
(d) C

Ans. (d) C

Explanation: In forest ecosystem, it is a partly upright or spindle shaped because producers are lesser in number and support a greater number of herbivores and which in turn support a fewer number of carnivores. After that the number goes down at each successive level.

9. In Primary succession on rocks, the pioneer species are usually—

(a) algae
(b) fungi
(c) lichens
(d) bryophytes

Ans. (c) lichens

Explanation: Lichens are the first organisms to colonise bare rocks.

10. The ‘10 percent law’ is related to—

(a) mendelian genetics
(b) non-Mendelian genetics
(c) energy transfer from lower trophic level to higher trophic level
(d) energy consumption during photosynthesis in C4 plants

Ans. (c) energy transfer from lower trophic level to higher trophic level

Explanation: 10 per cent law is related to the energy transfer from a lower trophic level to a higher level. When the energy is passed on from one trophic level to another, only 10 per cent of the energy is passed on to the next trophic level.

11. Chipko movement was launched for the protection of—

(a) wetlands
(b) grasslands
(c) forest
(d) livestock

Ans. (c) forest

Explanation: Chipko movement originated in 1973 in the Himalayan region of Uttarakhand to protect the forest. It’s the first forest conservation movement in India.

12. Pyramid of energy in the river ecosystem is—

(a) always upright
(b) always inverted
(c) constant
(d) declining

Ans. (a) always upright

Explanation: The pyramid of energy in the river ecosystem is always upright as the transfer of energy from one trophic level to the next as per 10% law.

13. Climax community is in a state of—

(a) non-equilibrium
(b) equilibrium
(c) disorder
(d) constant change

Ans. (b) equilibrium

Explanation: Climax community is in a state of near-equilibrium or equilibrium. It’s in equilibrium because the population remains relatively unchanged until destroyed by an event such as fire or human interference.

14. Energy transferred from one trophic level to another is—

(a) 5%
(b) 10%
(c) 15%
(d) 20%

Ans. (b) 10%

Explanation: According to the 10% law, 10% energy transferred from one trophic level to another.

15. Which ecosystem has the highest primary productivity?

(a) Pond
(b) Ocean
(c) Desert
(d) Forest

Ans. (d) Forest

Explanation: Forests have the highest primary productivity because they generate highest amount of organic matter each year as trees are large and their leaves can trap and use a great deal of sunligh.

15. Biodiversity and Its Conservation

Multiple Choice Questions

1. If the Bengal tiger becomes extinct:

(a) hyenas and wolves will become scarce.
(b) the wild areas will be safe for man and domestic animals.
(c) its gene pool will be lost for ever.
(d) the population of beautiful animals like deers will get stabilised.

Ans. (c) its gene pool will be lost for ever.

Explanation : When the Bengal tiger becomes extinct, its gene pool will be lost until the end of time forever. In the event that the Bengal tiger becomes extinct, the total species is lost for ever since it is the genes and chromosomes that determine the species. Hyenas and wolves populations will increase as the Bengal tiger will not eat them.

2. The world biodiversity day is celebrated annually on:

(a) 5th June
(b) 22nd May
(c) 22nd April
(d) 16th September

Ans. (b) 22nd May

Explanation : The International Day for Biological Diversity is celebrated every year on May 22.

3. Extinction of a species in a food chain is compensated by: 

(a) food chain
(b) ecological pyramid
(c) food web
(d) none of the above

Ans. (c) food web

Explanation : When a particular species in a food chain becomes extinct, it is compensated by the remaining generalised species at the same trophic level in a food web.

4. World Summit on sustainable development (2002) was held in:

(a) Argentina
(b) South Africa
(c) Brazil
(d) Sweden

Ans. (b) South Africa

Explanation : The ‘World Summit on Sustainable Development’ 2002 took place in Johannesburg, South Africa, from 26th August to 4th September, 2002.

5. What does IUCN stands for?

(a) International Union for Control of Disease
(b) International Union for Conservation of Nature
(c) International Unit of Nucleic acids
(d) International Union of Carbohydrates and Nucleic acids

Ans. (b) International Union for Conservation of Nature

Explanation : International Union for Conservation of Nature (IUCN) was founded on 5 October 1948 in France by Julian Huxley.

6. What is the global species diversity according to Robert May?

(a) 70 million
(b) 7 million
(c) 2 million
(d) 20 million

Ans. (b) 7 million

Explanation : Robert McCredie May is a theoretical ecologist and promoter of science. He estimated that there are almost 7 million species globally.

7. Of all the animal species recorded, what percent do insects comprise of?

(a) 7 %
(b) 70 %
(c) 50 %
(d) 2 %

Ans. (b) 70 %

Explanation : Out of all known animal species, 70 percent is comprised of insects.

8. Which of the following species is native of India?

(a) Rhodes grass
(b) Jackalberry tree
(c) Ebony tree
(d) Umbrella thorn acacia

Ans. (c) Ebony tree

Explanation : Ebony tree (Diospyros celibica) is the native of India. It is a threatened species found in Karnataka.

9. Which community should show less variation in productivity from year to year?

(a) Dynamic community
(b) Stable community
(c) Unstable community
(d) Reproductive community

Ans. (b) Stable community

Explanation : Less variation in productivity from year to year is a feature of a stable community. While dynamic and resproductive community show more variations due to environmental disturbances and genetic combinations.

10. Which of the following is a feature of a stable community?

(a) Resistant or resilient to occasional disturbances.
(b) More variation in productivity.
(c) Susceptible to occasional disturbances.
(d) Susceptible to invasions by alien species.

Ans. (a) Resistant or resilient to occasional disturbances.

Explanation : Resistant or resilient to occasional disturbances is a feature of a stable community.

11. For what reason is rich biodiversity important?

(a) Community issues
(b) Ecosystem health
(c) Ecological issues
(d) Community problems

Ans. (b) Ecosystem health

Explanation : Throughout the world, the diversity of plants and animals is not uniform that is uneven distribution is observed. For ecosystem health, rich biodiversity is essential.

12. What is considered as the rivet in the ‘Rivet popper hypothesis’?

(a) Ecosystem
(b) Community
(c) Species
(d) Individual

Ans. (c) Species

Explanation : Paul Ehrlich, a Stanford ecologist gave the ‘Rivet popper hypothesis’. In this hypothesis, he considered species as a rivet in an ecosystem. Rivet popper hypothesis states the importance of richness of species in maintenance of ecosystem.

13. What does describes ‘The Evil Quartet’?

(a) An increase in the number of species.
(b) The decrease in the number of species.
(c) Mass extinction.
(d) Formation of the universe.

Ans. (b) The decrease in the number of species.

Explanation : ‘The Evil Quartet’ is the sobriquet used to describe the reasons for the decline in the number of species. The four reasons include : Loss of habitat and fragmentation, Over-exploitation of species, Invasion of alien (new) species and Co- extinctions.

14. What are the excess and the unsustainable use of resources called?

(a) Loss of habitat and fragmentation
(b) Co-extinctions
(c) Invasion of alien species
(d) Over-exploitation

Ans. (d) Over-exploitation

Explanation : The excess and unsustainable use of resources is known as overexploitation. It is also known as overharvesting.

15. Which utilitarian states that humans derive countless direct economic benefits from nature?

(a) Big utilitarian
(b) Broadly utilitarian
(c) Narrowly utilitarian
(d) Small utilitarian

Ans. (c) Narrowly utilitarian

Explanation : The narrowly utilitarian arguments for conserving biodiversity are obvious, human derive countless direct economic benefits from nature.

16. Environmental Issues

Multiple Choice Questions

1. In 1984, the Bhopal gas tragedy took place because methyl isocyanate—

(a) reacted with ammonia
(b) reacted with CO2
(c) reacted with water
(d) reacted with DDT

Ans. (c) reacted with water

Explanation: Bhopal gas tragedy was happened due to leakage of MIC [methyl isocyanate] gas which reacts with water, giving off heat and producing methylamine and carbon dioxide.

2. Lead concentration in blood is considered alarming if it is—

(a) 30 µg/100 ml
(b) 4–6 µg/100 ml
(c) 10 µg/100ml
(d) 20 µg/100ml

Ans. (a) 30 µg/100 ml

Explanation: The presence of lead in blood is considered alarming when it is 30µg/100ml. Exposure to high level of lead may cause anemia, weakness and kidney and brain damage.

3. Common indicator organism of water pollution is—

(a) Eichhornia crassipes
(b) Escherichia coli
(c) Entamoeba histolytica
(d) Lemna paucicostata

Ans. (b) Escherichia coli

Explanation: Common indicator organism of water pollution is Escherichia coli it indicates faecal contamination. The presence of Lichens indicates pollution of air.

4. The World Environment Day is celebrated on—

(a) 6th of June
(b) 5th of June
(c) 6th of August
(d) 5th of May

Ans. (b) 5th of June

Explanation: 5 th June is celebrated as World Environment Day every year.

5. Expand BOD—

(a) Biological Oxygen Demand
(b) Biosynthetic Oxygen Demand
(c) Biogeochemical Oxygen Destroyer
(d) Biogeochemical Oxygen Dimension

Ans. (a) Biological Oxygen Demand

Explanation: BOD means Biological Oxygen Demand. It generally represents how much oxygen is needed to break down organic matter in water.

6. The term ‘Bio-magnification’ refers to the—

(a) growth of organism due to food consumption.
(b) increase in population size.
(c) blowing up of environmental issue by man.
(d) increase in concentration of non-degradable pollutants as they pass through food chain.
Ans. (d) increase in concentration of non-degradable pollutants as they pass through food chain.

Explanation: Biomagnification refers to the increase in the concentration of nondegradable pollutants as they pass through the food chain.

7. Effect of pollution is observed first on—

(a) micro-organism
(b) food crop
(c) green vegetable
(d) herbivores

Ans. (c) green vegetable

Explanation: The effect of pollution is first observed in green vegetables. As air is absorbed by the green vegetables and water is also absorbed by their root so the first effect is found in it.

8. Which is always present in photochemical smog?

(a) Ozone
(b) SO2
(c) NO2
(d) CH4

Ans. (c) NO2

Explanation: NO2 creates photochemical smog. Photochemical smog is a mixture of pollutants that are formed when nitrogen oxides and volatile organic compounds react to sunlight, creating a brown haze above cities.

9. Greenhouse effect is due to—

(a) CO2
(b) CO
(c) NO
(d) PO4

Ans. (a) CO2

Explanation: CO2 is a greenhouse gas as its molecules in the atmosphere absorb and trap long-wavelength infrared energy (heat) from the Earth.

10. Ozone hole results in—

(a) UV radiation reaching the earth
(b) cataract
(c) increase in skin cancer
(d) all of the above

Ans. (d) all of the above

Explanation: Ozone hole results in UV radiation-related skin cancer and also causes cataract.

11. Checking of re-radiating heat by atmospheric dust O3, CO2, and water vapour is—

(a) green house
(b) solar effect
(c) ozone layer effect
(d) radioactive effect

Ans. (a) green house

Explanation: Checking or trapping re-radiating heat by atmospheric dust O3, CO2 and water vapour causes greenhouse effect.

12. Which of the following is biogas?

(a) CO2
(b) N2O
(c) CH4
(d) N2

Ans. (c) CH4

Explanation: Methane (CH4) is a biogas. Raw biogas typically consists of methane (50 – 75%), carbon dioxide (25 – 50%), and smaller amounts of nitrogen (2 – 8%). As contained methane is maximum so it's known as biogas only.

13. Which of the following is secondary pollutant?

(a) NO
(b) NO2
(c) SO2
(d) PAN

Ans. (d) PAN

Explanation: PAN i.e., Peroxyacyl nitrates is called a secondary pollutant. Secondary pollutants are pollutants that form in the atmosphere rather than released from the emission of vehicles. 

14. The blue baby syndrome results from—

(a) excess of TDS (Total Dissolved Solid)
(b) excess of chlorides
(c) methemoglobinemia
(d) excess of dissolved oxygen

Ans. (c) methemoglobinemia

Explanation: Blue baby syndrome is also known as methemoglobinemia. The blue baby syndrome is typically caused by abnormalities in the heart, lungs, or blood.

15. Photochemical smog pollution does not contain—

(a) PAN (Peroxy Acyl Nitrate)
(b) Ozone
(c) Nitrogen dioxide
(d) Carbon dioxide

Ans. (d) Carbon dioxide

Explanation: Photochemical smog pollution does not contain carbon dioxide. It is a mixture of pollutants that are formed when nitrogen oxides and volatice organic compounds react to sunlight, creating a brown haze above cities.