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

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