Important Questions

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance

Very short answer type questions

Question 1.
A charge ‘q’ is moved from a point A above a dipole of dipole moment ‘p’ to a point B below the dipole in equatorial plane without acceleration. Find the work done in the process.

Answer:
Work done in the process is zero. Because, equatorial plane of a dipole is equipotential surface and work done in moving charge oh equipotential surface is zero.
W = qV_AB = q × 0 = 0

Question 2.
A point charge Q is placed at point O as shown in the figure. Is the potential difference (V_A – V_B) positive, negative or zero if Q is
(i) positive
(ii) negative

Answer:
Let the distance of points A and B from charge Q be r_A and r_B, respectively.
∴ Potential difference between points A and B
V_A – V_B = \(\frac{q}{4 \pi \varepsilon_{0}}\left[\frac{1}{r_{A}}-\frac{1}{r_{B}}\right]\)
As r_A = OA, r_B = r_B = OB
and r_A < r_B, \(\frac{1}{r_{A}}>\frac{1}{r_{B}}\)
There,[latex]\frac{1}{r_{A}}-\frac{1}{r_{B}}[/latex] has positive value.
(V_A – V_B) depends on the nature of charge q.
(V_A – V_B) is positive when Q > 0
(V_A – V_B ) is negative when Q < 0

Question 3.
A hollow metal sphere of radius 5 cm is charged such that potential on its surface is 10 V. What is the potential at the centre of the sphere?
Answer:
The electric potential at every point inside the charged spherical shell is same and equal to the electric potential on its surface.
The electric potential at the centre of sphere is 10 V.

Question 4.
Can two equipotential surfaces intersect each other? Justify your answer.
No, two equipotential surfaces cannot intersect each other because

1. Two normals can be drawn at intersecting point on two surfaces which gives two directions of E at the same point which is impossible.
2. Also two values of potential at the same point is not possible.

Question 5.
Why electrostatic potential is constant throughout the volume of the conductor and has the same value as on its surface?
Since, electric field intensity inside the conductor is zero. So, electrostatic potential is a constant.
But, E = –\(\frac{\Delta V}{\Delta r}\)
∴ E = 0, ΔV = 0
or V₂ – V₁ =0, V₂ – V₁
The potential at every point inside the conductor remains same.

Question 6.
In a certain 0.5 cm³ of space, electric potential is found to be 7 V throughout. What is the electric field in this region?
Answer:
Zero, because electric potential is same throughout as
E = \(\frac{d V}{d r}\)

Question 7.
Distinguish between a dielectric and a conductor.
Answer:
Dielectrics are non-conductors and do not have free electrons at all. While conductor has free electrons in its any volume which makes it able to pass the electricity through it.

Question 8.
The given graph shows the variation of charge q versus potential difference V for two capacitors C₁ and C₂. Both the capacitors have same plate separation but plate area of C₂ is greater than that C₁. Which line (A or B) corresponds to C₁ and why?

Answer:
Line B corresponds to Q because slope (q/V) of B is less than slope of A.

Question 9.
Do free electrons travel to region of higher potential or lower potential? (NCERT Exemplar)
Answer:
Free electrons would travel to regions of higher potentials as they are negatively charged.

Question 10.
Can there be a potential difference between two adjacent conductors carrying the same charge? (NCERT Exemplar)
Answer:
Yes, if the sizes are different.

Question 11.
Prove that, if an insulated, uncharged, conductor is placed near a charged conductor and no other conductors are present, the uncharged body must intermediate in potential between that of the charged body and that of infinity. (NCERT Exemplar)
Answer:
Let us take any path from the charged conductor to the uncharged conductor along the direction of electric field. Therefore, the electric potential decrease along this path.
Now, another path from the uncharged conductor to infinity will again continually lower the potential further. This ensures that the uncharged body must be intermediate in potential between that of the charged body and that of infinity.

Question 12.
A test charge q is made to move in the electric field of a point charge Q along two different closed paths [figure]. First path has sections along and perpendicular to lines of electric field.
Second path is a rectangular loop of the same area as the first loop. How does the work done compare in the two cases? (NCERT Exemplar)

Answer:
As electric field is conservative, work done will be zero in both the cases.

Short answer type questions

Question 1.
What is electrostatic shielding? How is this property used in actual practice? Is the potential in the cavity of a charged conductor zero?
Answer:
Whatever be the charge and field configuration outside, any cavity in a conductor remains shielded from outside electric influence. The field inside a conductor is zero. This is known as electrostatic shielding.

• Sensitive instruments are shielded from outside electrical influences by enclosing them in a hollow conductor.
• During lightning it is safest to sit inside a car, rather than near a tree. The metallic body of a car becomes an electrostatic shielding from lightening.

Potential inside the cavity is not zero. Potential is constant.

Question 2.
Two capacitors of unknown capacitances C₁ and C₂ are connected first in series and then in parallel across a battery of 100 V. If the energy stored in the two combinations is 0.045 J and 0.25 J respectively, determine the value of C₁ and C₂. Also calculate the charge on each capacitor in parallel combination.
Answer:
Energy stored in a capacitor,

Q₁ = C₁V = 38.2 × 10^-6 × 100 = 38.2 × 10^-4 C
Q₂ = C₂V = 11.8 × 10^-6 × 100 = 11.8 × 10^-4C

Question 3.
Calculate the equivalent capacitance between points A and B in the circuit below. If a battery of 10 V is connected across A and B, calculate the charge drawn from the battery by the circuit.

Answer:

Question 4.
A parallel plate capacitor is charged by a battery to a potential. The battery is disconnected and a dielectric slab is inserted to completely fill the space between the plates.
How will
(a) its capacitance
(b) electric field between the plates and
(c) energy stored in the capacitor be affected? Justify your answer giving necessary mathematical expression for each case.
Answer:
On introduction of dielectric slab in a isolated charged capacitor.
(a) The capacitance (C’) becomes K times of original capacitor as
C = \(\frac{\varepsilon_{0} A}{d}\)
and C ‘ = \(\frac{K \varepsilon_{0} A}{d}\)

(b) The total charge on the capacitor remains conserved on introduction of dielectric slab. Also, the capacitance of capacitor increases to K times of original values.
∴ CV = C’V’
CV = (KC)V’
⇒ V’ = \(\frac{V}{K}\)
∴ New electrical field,
E’ = \(\frac{V^{\prime}}{d}=\left(\frac{V / K}{d}\right)=\left(\frac{V}{d}\right) \frac{1}{K}=\frac{E}{K}\)
∴ On introduction of dielectric medium new electric field E’ becomes \(\) times of its original value.
(c) Energy stored initially,
U = \(\frac{q^{2}}{2 C}\)
Energy stored later,
U’ = \(\frac{q^{2}}{2(K C)}\) [< C’ = KG]
where, K = dielectric constant of medium
⇒ U’ = \(\frac{1}{K}\left(\frac{q^{2}}{2 C}\right)\)
⇒ U’ = \(\frac{1}{K}\) × U
The energy stored in the capacitor decreases and becomes \(\frac{1}{K}\) times of original energy.

Question 5.
Two identical parallel plate capacitors A and B are connected to a battery of V volts with the switch S closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant K. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric.

Answer:
Let C be the capacitance of each capacitor.
With switch S closed, the two capacitor are in parallel.
∴ Equivalent capacitance is 2 C.
∴ Energy stored = \(\frac{1}{2}\)(2C)V²
U₁ = CV² …………… (1)

Now, when switch is opened and then free space of capacitors are filled with dielectric, the capacitance of each capacitor will be KC. For capacitor B, the charge will remain as before and for A, the potential difference will remain same.

Charge on each capacitor in the previous case will be CV.
∴ Energy stored in capacitor A in circuit case is

Question 6.
Consider two conducting spheres of radii R₁ and R₂ with R₁ > R₂. If the two are at the same potential, the larger sphere has more charge than the smaller sphere. State whether the charge density of the smaller sphere is more or less than that of the larger one. (NCERT Exemplar)
Answer:

Question 7.
Calculate potential on the axis of a ring due to charge Q uniformly distributed along the ring of radius R.
(NCERT Exemplar)
Answer:
Let us take point P to be at a distance x from the centre of the ring, as shown in figure. The charge element dq is at a distance x from point P. Therefore, V can be written as

Long answer type questions

Question 1.
(i) Explain using suitable diagrams, the difference in the behaviour of a
(a) conductor and
(b) dielectric in the presence of external electric field. Define the terms polarisation of a dielectric and write its relation with susceptibility.

(ii) A thin metallic spherical shell of radius R carries a charge Q on its surface. A point charge \(\frac{Q}{2}\) is placed at its centre C and an other charge +2Q is placed outside the shell at a distance x from the centre as shown in the figure. Find (a) the force on the charge at the centre of shell and at the point A, (b) the electric flux through the shell.

Answer:
(i) When a capacitor is placed in an external electric field, the free charges present inside the conductor redistribute themselves in such a manner that the electric field due to induced charges opposes the external field within the conductor. This happens until a static- situation is achieved, i.e., when the two fields cancels each other and the net electrostatic field in the conductor becomes zero.

In contrast to conductors, dielectrics are non- conducting substances, i.e., they have no charge carriers. Thus, in a dielectric, free movement of charges is not possible. It turns out that the external field induces dipole moment by stretching molecules of the dielectric. The collective effect of all the molecular dipole moments is the net charge on the surface of the dielectric which produces a field that opposes the external field. However, the opposing field is so induced does not exactly cancel the external field, ft only reduces it. The extent of the effect depends on the nature of dielectric.

Both polar and non-polar dielectric develop net dipole moment in the presence of an external field. The dipole moment per unit volume is called polarisation and is denoted by P for linear isotropic dielectrics.
P = %E

(a) At point C, inside the shell.
The electric field inside a spherical shell is zero. Thus, the force experienced by charge at the centre C will also be zero.

Question 2.
(a) Find the ratio of the potential differences that must he applied across the parallel and series combination of two capacitors C ₁and C₂ with their capacitances in the ratio 1:2 so that the energy stored in the two cases becomes the same.

(b) Show that the potential energy of a dipole making angle θ with the direction of the field is given by u(θ) = – p \(\). Hence find out the amount of work done in rotating it from the position of unstable equilibrium to the stable equilibrium.

(b) As charges +q and -q traverse equal distance under equal and opposite forces; therefore, net work done in bringing the dipole in the region of electric field perpendicular to field-direction will be zero, i.e.,W₁ = 0.

Now, the dipole is rotated and brought to orientation making an angle θ with the field direction (i.e., θ₀ = 90° and θ₁ = θ°), therefore, work done
W₂ = p_E (cosθ₀ – cosθ ₁)
= p_E (cos 90° – cos θ) = -pE cos θ

∴Total work done in bringing the electric dipole from infinity, i.e., electric potential energy of electric dipole. Thus, work done by external torque in rotating a dipole in uniform electric field is stored as the potential energy of the system.

U = W₁ + W₂ = 0 – pEcosθ
= -pE cosθ
In vector form
U = – \(\vec{p} \cdot \vec{E}\)
For rotating dipole from position of unstable equilibrium (θ₀ =180°) to the stable equilibrium (θ = 0°)
∴W = pE (cos 180° – cos0°)
= pE (-1 -1) = -2 pE

Punjab State Board PSEB 12th Class Biology Important Questions Chapter 4 Reproductive Health Important Questions and Answers.

PSEB 12th Class Biology Important Questions Chapter 4 Reproductive Health

Very short answer type questions

Question 1.
Reproductive health refers only to healthy reproductive functions. Comment. [NCERT Exemplar]
Answer:
Reproductive health refers to the total well-being in all aspects of reproduction, i.e., physical, behavioural, psychological and social.

Question 2.
The present population growth rate in India is alarming. Suggest ways to check it. [NCERT Exemplar]
Answer:

• By increasing marriageable age.
• By promoting use of birth control measure.
• By educating people about consequences of un- controlled population growth.

Question 3.
Why do intensely lactating mothers not generally conceive? [NCERT Exemplar]
Answer:
Due to suppression of gonadotropins, ovulation and menstrual cycle do not take place.

Question 4.
Name an IUD that you would recommend to promote the cervix hostility to sperms.
Answer:
The hormone releasing IUD’s, e.g. progestasert, LNG-20 are , recommended to promote the cervix hostility to sperms.

Question 5.
Mention any tvgo events that are exhibited by the intake of oral contraceptive pills to prevent pregnancy in humans.
Answer:
Two events that are exhibited by the intake of oral contraceptive pills to prevent pregnancy in humans are ovulation and implantation.

Question 6.
Why is tubectomy considered a contraceptive method?
Answer:
Tubectomy involves cutting a piece of the fallopian tube and tying its ends. This way, the sperms are not able to reach the egg and it acts as a contraceptive method.

Question 7.
Mention the primary aim of the ‘Assisted Reproductive Technology’ (ART) programme. [NCERT Exemplar]
Answer:
‘Assisted Reproductive Technology’ (ART) is the collection of certain I1 special techniques. The primary aim of the ART programmes is to assist infertile couples to have children through certain special techniques (like ZIFT, IUT, GIFT, ICSI, AI, etc.), when corrective treatment for infertility problems is not possible.

Question 8.
Expand GIFT and ICSI.
Answer:
GIFT: Gamete Intra Fallopian Transfer.
ICSI: Intra Cytoplasmic Sperm Injection.

Short answer type questions

Question 1.
Comment on the RCH programme of the government to improve the reproductive health of the people. [NCERT Exemplar]
Answer:
The basic aims of the RCH programmes are creating public awareness f regarding reproduction-related aspects and providing facilities to build up a healthy society with added emphasis on the health of mother and child.

Question 2.
(a) List any four characteristics of an ideal contraceptive.
(b) Name two intrauterine contraceptive devices that effect the motility of sperms.
Answer:
(a) An ideal contraceptive should be:

• user friendly,
• easily available,
• effective,
• reversible with no or least side-effects,
• non-interfering with the sexual drive/desire and/or the sexual act of the user, (any four)

(b) Intrauterine Devices (IUDs): Lippes loop, copper releasing IUDs (Cu-T, Cu-7, multiload 375).

Question 3.
Name two hormones that are constituents of contraceptive pills. Why do they have high and effective contraceptive value? Name a commonly prescribed non-steroidal oral pill.
Answer:
Hormonal preparations (progestogens or progesterons and estrogens) are highly effective contraceptive because they inhibit ovulation and implantation, e.g., Mala-D, Mala-L. Morning after pills are used as emergency contraceptives, to avoid pregnancy due to rape or casual unprotected intercourse.

“Saheli”, a new oral pill is used “once-a-week” a non-steroidal preparation With very less side effects and high contraceptive value developed by CDRI in Lucknow, India.

Question 4.
Name and explain the surgical method advised to human males and females as a means of birth control. Mention its one advantage and one disadvantage.
Answer:
In the males-Vasectomy. In this method, a small part of the vas deferens is removed or tied up through a small incision on the scrotum. In the females-Tubectomy. In this method, a small part of the fallopian tube is removed or tied up through a small incision in the abdomen or through vagina.
Advantage: Highly effective
Disadvantage: Reversibility is very poor

Question 5.
A childless couple has agreed for a test tube baby programme. List only the basic steps of the procedure would involve to conceive the baby.
Answer:
IVF is the technique used in the case of childless couple. In IVF or In Vitro fertilisation, fertilisation is carried out in a glass container outside the body of the mother. Purified semen is poured over the mature retrieved oocytes. The fertilised eggs are separated and allowed to remain in culture medium, maintained in incubator for 48-72 hours. During the period, the fertilised egg undergoes cleavage and reach 4-8 celled stage, 2-3 fertilised 4-8 celled embryos are transferred or implanted .into the uterus of the recipient surrogate mother, for further development up to delivery.
Note: Excess fertilised oocytes are cryopreserved for use in case of implantation failure.

Question 6.
Why is medical termination of pregnancy (MTP) carried out?
Answer:
MTP is carried out to get rid of unwanted pregnancies. It is also essential when the foetus is suffering from an incurable disease or when continuation of the pregnancy could be harmful or even fatal to the mother and/or foetus.

Long answer type questions

Question 1.
Your school has been selected by the Department of Education to organise and host an interschool seminar on “Reproductive Health-Problems and Practices.” However, many parents are reluctant to permit their wards to attend it. Their argument is that the topic is “too embarrassing.”
Put forth four arguments with appropriate reasons and explanation to justify the topic to be very essential and timely.
Answer:
The selection of your school, to host a seminar on “Reproductive Health-Problems and Practices” is a matter great pride for the prestigious institute. The students will have an opportunity to listen to the diverse ideas, suggested by the learned speakers.

It is sad that many parents are reluctant to permit their wards to attend the seminar assuming that the topic is too embarrassing. The following arguments will justify the relevance of the topic in the present time:
(i) Introduction of sex education and the proper information about reproductive organs, adolescence and related changes will protect the youth from social evils like sex-abuse and sex-related crimes.

(ii) Right information about safe, healthy and hygienic sexual practices, sexually transmitted diseases (STDs) would help the people to lead a reproductive healthy life.

(iii) Decline in sex-ratio is a matter of great concern. The Govt, has put a I statutory ban on female foeticide. Both girls and boys have equal rights and equal opportunities in all spheres of life.

(iv) India is facing another problem of population explosion. It is eating, almost all the benefits of overall development. The benefits of development are not trickling down to the poor at lower strata. There is need for family planning, socially conscious healthy families of desired size i.e., Hum Do Humare Do.

Punjab State Board PSEB 12th Class Biology Important Questions Chapter 3 Human Reproduction Important Questions and Answers.

PSEB 12th Class Biology Important Questions Chapter 3 Human Reproduction

Very short answer type questions

Question 1.
Why are the human testes located outside the abdominal cavity? Name the pouch in which they are present.
Answer:
The human testes need lower temperature, 2-2.5°C less than the body temperature, for the formation of sperms which is provided outside the body. Testes are present in scrotal sac or scrotum.

Question 2.
Write the location and function of the sertoli cells in humans.
Answer:
Sertoli cells are present in seminiferous tubules. They provide nutrition to the germ cells or sperms.

Question 3.
Mention the location and the function of leydig cells in humans.
Answer:
Leydig cells are present in seminiferous tubules. They synthesise and secrete androgens.

Question 4.
The path of sperm transport is given below. Provide the missing steps in blank boxes. [NCERT Exemplar]

Answer:
Vasa efferentia, Vas deferens.

Question 5.
Female reproductive organs and associated functions are given below in column A and B. Fill in the blank boxes. [NCERT Exemplar]

Column A	Column B
Ovaries	Ovulation
Oviduct	A
B	Pregnancy
Vagina	Birth

Answer:
A – Fertilisation
B – Uterus

Question 6.
What is the role of cervix of the human female system in reproduction? [NCERT Exemplar]
Answer:
Cervix helps in regulating the passage of sperms into the uterus and forms the birth canal to facilitate parturition.

Question 7.
When do the oogenesis and the spermatogenesis initiate in human females and males respectively?
Answer:
In females; oogenesis initiates during foetal life. In males, spermatogenesis begins at the time of puberty.

Question 8.
Mention the importance of LH surge during menstrual cycle. [NCERT Exemplar]
Answer:
LH surge is essential for the events leading to ovulation.

Question 9.
Menstrual cycles are absent during pregnancy. Why? [NCERT Exemplar]
Answer:
The high levels of progesterone and estrogens during pregnancy suppress the release of gonadotropins required for the development of new follicles. Therefore, new cycle cannot be initiated.

Question 10.
How does the sperm penetrate through the zona pellucida in [ human ovum?
Answer:
The sperm penetrates through zona pellucida with the help of secretions from acrosome.

Question 11.
Mention the function of trophoblast in human embryo.
Answer:
Trophoblast is the outer layer of blastocyst which helps in the attachment of blastocyst to the endometrium of the uterus.

Question 12.
Explain the function of umbilical cord.
Answer:
It transports nutrients and respiratory gases and metabolic wastes to and from mother and foetus.

Question 13.
Given below are the stages in human reproduction. Write them ‘ in correct sequential order.
Insemination, Gametogenesis, Fertilisation, Parturition, Gestation, Implantation. [NCERT Exemplar]
Answer:
Gametogenesis, Insemination, Fertilisation, Implantation, Gestation, Parturition.

Question 14.
What stimulates pituitary to release the hormone responsible for parturition? Name the hormone.
Answer:
The signal from the fully developed foetus and placenta or the foetal , ejection reflex induces mild uterine contraction. The hormone released is oxytocin.

Question 15.
Name the important mammary gland secretions that help in resistance of the new bom baby. [NCERT Exemplar]
Answer:
Colostrum

Question 16.
Mention the function of mitochondria in sperm.
Answer:
It provide energy for the movement of sperm tail.

Short answer type questions

Question 1.
Write the function of each of the following:
(a) Middle piece in human sperm.
(b) Luteinising hormone in human males.
Answer:
(a) Provides energy for movement.
(b) Stimulates synthesis and secretion of androgens or male hormones for spermatogenesis.

Question 2.
Differentiate between mayor structural changes in the human ovary during the follicular and luteal phase of the menstrual cycle.
Answer:

Follicular phase	Luteal phase
1. During this, primary follicles grow to become fully mature Graafian follicle.	During this, remaining part of Graafian follicle transforms into corpus luteum.
2. Endometrium regenerates through proliferation.	Endometrium further thickens secreting progesterone for implantation after fertilisation. If fertilisation does not occur, corpus luteum degenerates.

Question 3.
Explain the events in a normal woman during her menstrual cycle on the following days :
(a) Pituitary hormone levels from 8 to 12 days.
(b) Uterine events from 13 to 15 days.
(c) Ovarian events from 16 to 23 days.
Answer:
(a) The level of LH and FSH secreted by anterior lobe of pituitary, stimulated by GnRH, increases.

(b) The endometrium of the uterus regenerates through proliferation. It grows and become thickened. There is repair of ruptured blood vessels and new blood capillaries develop. Uterine glands elongate.

(c) The remnant of Graafian follicle forms corpus luteum which secretes large amount of progesterone essential for maintenance of endometrium for implantation and for pregnancy.

Question 4.
What happens to corpus luteum in human female if the ovum is (a) fertilised, (b) not fertilised?
Answer:
(a) In case the ovum is fertilised, the corpus luteum persists and secretes a large amount of progesterone. The progesterone is essential for maintenance of endometrium, a necessity for implantation and for pregnancy.

(b) In the absence of fertilisation, corpus luteum degenerates. The level of LH and progesterone decreases very low. The absence of hormonal support, leads to the disintegration of endometrium and results in menstrual flow.

Question 5.
A large number of married couples the world over are childless. It is shocking to know that in India the female partner is often blamed for the couple being childless.
(a) Why in your opinion the female partner is often blamed for such situations in India? Mention any two values that you as a biology student can promote to check this Social evil.
(b) State any two reasons responsible for the cause of infertility.
(c) Suggest a technique that can help the couple to have a child where the problem is with the male partner.
Answer:
(a) The female partner is wrongly blamed for not bearing the child. It is due to the lack of proper information and knowledge about the reproduction and reproductive organs. Being a biology student I will advise and explain the couple as well as their other family members. Both male and female are equally responsible for bearing child. The two value, promoted are (i) concern for others (ii) scientific
temperament.

(b) Infertility is the inability to produce children inspite of unprotected sex and sexual co-habitation. It may be due to (i) physical/congenital disease (ii) immuno- logical or even physiological reason.

(c) In case, the problem is with male partner, artificial insemination (AI) is adopted. Semen collected either from the husband or a healthy donor is artificially introduced in the vagina or into the uterus of the female.

Question 6.
Name and explain the role of inner and middle walls of the human uterus.
Answer:
The inner wall of the uterus is called endometrium. It supports foetal growth and helps in placenta formation after implantation.
The middle wall of the uterus is called myometrium, exhibits strong contraction during delivery of baby.

Question 7.
Write a brief account of the structure and functions of placenta. [NCERT Exemplar]
Answer:
Placenta connects the foetus to the uterus through an umbilical chord. Both the foetal and the maternal tissues contribute to its formation. The foetal part is the chorionic villi and the maternal part is the uterine mucosa.

Long answer type questions

Question 1.
(a) Briefly explain the events of fertilisation and implantation in an adult human female.
(b) Comment on the role of placenta as an endocrine gland.
Or
Name the stage of human embryo of which it gets implanted.
Explain the process of implantation.
Or Draw a labelled diagram of a human blastocyst. How does it get implanted in the uterus?
Answer:
(a) (i) Fertilisation : Fertilisation occurs, if the ovum and sperms are transported simultaneously to the ampullary-isthmic junction and involve fusion of sperm with an ovum.
Secretions of acrosome of sperm help it to enter into the cytoplasm of ovum through zona pellucida and the plasma membrane. It induces meiotic division-II to form haploid ovum (ootid) and secondary polar body. The fusion of sperm with ovum to form diploid zygote is called fertilisation.

(ii) Implantation: Zygote undergoes cleavage to form a solid mass of 16 cells-morula, with daughter cells called blastomeres. Morula develops into a embryo with about 64 cells and with a cavity called blastocoel and the embryo is termed as blastocyst. It consists of outer layer of cells-trophoblast and inner cell mass.

The trophoblast gets attached to the endometrium- uterine wall of mother, after 7 days of fertilisation by a process called implantation leading to pregnancy. The uterine cells divide rapidly and cover blastocyst. The blastocyst gets embedded in the endometrium. Inner cell mass forms embryo.

(b) Placenta beside providing nutrients to the foetus also act as an endocrine gland. Placenta secretes human chorionic gonadotrophin (hCG), human placental lactogen (hPL), estrogen, progesterone etc., and later relaxin is secreted by ovary which facilitates parturition. Increased level of hormones like cortisol, prolactin and thyroxine etc., help in foetal growth, metabolic changes in the mother and maintenance of pregnancy.

Punjab State Board PSEB 12th Class Biology Important Questions Chapter 15 Biodiversity and Conservation Important Questions and Answers.

PSEB 12th Class Biology Important Questions Chapter 15 Biodiversity and Conservation

Very short answer type questions

Question 1.
Why is genetic variation important in the plant Rauwolfia vomitoria? [NCERT Exemplar]
Answer:
Genetic variation affects the variation in potency and concentration of me drug reserpine in the medicinal plant Rauwolfia vomitoria.

Question 2.
Name the type of biodiversity represented by the following:
(i) 50,000 different strains of rice in India
(ii) Estuaries and alpine meadows in India.
Answer:
(i) Genetic diversity
(ii) Ecological diversity

Question 3.
Identify ‘a’ and ‘b’ in the figure given below representing proportionate number of major vertebrate taxa.

Answer:
(a) Mammals
(b) Amphibians

Question 4.
Suggest two practices giving one example by each, that help protect rare or threatened species.
Answer:

1. By using cryopreservation (preservation at -196°C) technique, sperms, eggs, tissues, and embryos can be stored for long period in gene banks, seed banks etc.
2. Plants are propagated in vitro using tissue culture methods.

Question 5.
According to David Tilman, greater the diversity greater is the primary productivity. Can you think of a very low diversity man-made ecosystem that has high productivity? [NCERT Exemplar]
Answer:
Agricultural fields like wheat field or paddy field which are also examples of monoculture practices.

Question 6.
What is Red Data Book? [NCERT Exemplar]
Answer:
The Red Data Book is a compilation of data on species threatened with extinction and is maintained by IUCN. ‘

Question 7.
What is the expanded form of IUCN? [NCERT Exemplar]
Answer:
International Union for Conservation of Nature.

Question 8.
Why Western Ghats in India have been declared as biological hotspots?
Answer:
Western Ghats are biological hotspots because they have species richness and species evenness.

Question 9.
What is a national park?
Answer:
It is a protected area reserved for wildlife where human activities are not permitted.

Question 10.
What is the difference between endemic and exotic species? [NCERT Exemplar]
Answer:
Endemic species are native species restricted to a particular geographical region. Exotic species are species which are introduced from other geographical regions into an area.

Question 11.
How is the presently occurring species extinction different from the earlier mass extinctions? [NCERT Exemplar]
Answer:
Species extinction occurring at present is due to anthropogenic or man-made causes whereas the earlier extinction was due to natural causes.

Question 12.
Differentiate between in situ and ex situ approaches of conservation of biodiversity.
Answer:

In situ approach	Ex-situ approach
1. It involves protection of endangered species of plants and animals.	It involves protection of endangered species by removing them from the natural habitat.
2. This is done by protecting the natural habitat or ecosystem.	This is done by placing the species under special care.

Short answer type questions

Question 1.
What is biodiversity? Why is it a matter of concern now?
Answer:
Biodiversity is the occurrence of different types of genes, gene pools, species, habitats and ecosystems at a particular place and various parts of earth. It is a matter of concern because species are continuously lost, limiting the diversity and this will affect our survival and well-being on earth due to the changes in environment.

Question 2.
Where would you expect more species biodiversity-in tropics or in polar regions? Give reasons in support of your answer.
Answer:
More biodiversity is found in the tropics. This is because tropical regions remain undisturbed from frequent glaciations as in polar regions. Also, the tropics are less seasonal/more constant.

Question 3.
Is it true that there is more solar energy available in the tropics? Explain briefly. [NCERT Exemplar]
Answer:
As one moves from the equator to the polar regions, the length of the day decreases and the length of the night increases. The length of day and night are same at the equator. Therefore, it is true that there is more solar energy available in the tropics.

Question 4.
The given graph alongside shows species-area relationship. Write the equation of the curve ‘a’ and explain.
Answer:

The equation of the curve ‘a’ is S = CA^Z.
(i) Within a region, species richness increases with increasing explored area but only up to a limit.
(ii) Relationship between species richness and area for a wide variety of taxa turns out to be a rectangular hyperbola.

Question 5.
Explain ‘rivet popper hypothesis. Name the ecologist who proposed it.
Answer:
Paul Ehrlich proposed the rivet popper hypothesis. This hypothesis states that in an airplane (ecosystem) all parts are joined together using thousands of rivets (species). If every passenger traveling in it starts popping a rivet to take home (causing a species to become extinct), it may not affect flight safety (proper functioning of the ecosystem) initially but as more and more rivets are removed, the plane becomes dangerously weak over a period of time. Also, which rivet is removed may also be critical like loss of rivets on the wings (key species) is more serious threat to flight safety than loss of few rivets on the seats or windows inside the plane.

Question 6.
How do human activities cause desertification?
Answer:
Human activities like over-cultivation, unrestricted grazing, deforestation, and poor irrigation practices result in arid patches of land. The fertile topsoil that may take centuries to develop is eroded due to these activities. When large barren patches extend and meet over time, a desert is created. Increased urbanization is also one of the causes of desertification.

Question 7.
Why are conventional methods not suitable for the assessment of biodiversity of bacteria? [NCERT Exemplar]
Answer:
Many bacteria are not culturable under normal conditions in the laboratory. This becomes a problem in studying their morphological, biochemical, and other characterizations which are useful for their assessment.

Question 8.
List any four techniques where the principle of ex-situ conservation of biodiversity has been employed.
Answer:
Ex-situ Conservation (off-site conservation)

1. Zoological parks and botanical gardens.
2. Wildlife safari parks, aquaria.
3. Preservation of germplasm-seed gene banks, tissue culture, cryopreservation.
4. Sacred plants grown in homes, villages, and religious places.

Long answer type questions

Question 1.
(a) Why should we conserve biodiversity? How can we do it?
(b) Explain the importance of biodiversity hotspots and sacred groves.
Or
Why should biodiversity be conserved? List any two ethical arguments in its support. ‘
Answer:
(a) Need for Conservation of Biodiversity: Reasons for conservation of biodiversity can be grouped into three categories :

1. narrowly utilitarian
2. broadly utilitarian and
3. ethical.

(i) Narrowly Utilitarian: The reasons for conserving biodiversity are obvious because of their :
(a) direct economic benefits such as

• food (cereals, pulses, fruits)
• firewood
• fiber
• construction material
• products of medicinal importance
• industrial products (tannins, gums, lubricants, dyes, resins, perfumes).

(b) More than 25% of the drugs are derived from plants.

(c) 25,000 species of plants are used as traditional medicines by native people.

(ii) Broadly Utilitarian: Biodiversity plays a major role in providing ecosystem services that nature provides and which cannot be given a price tag are :

• Production of Oxygen
• Pollination of flowers by bees, bumblebees, birds, and bats, etc.
• Resulting in the formation of fruits and seeds.
• Aesthetic pleasures like bird watching, walking through the thick forests, waking up to bulbul’s song, etc.

(iii) Ethical :
(a) We share this planet with millions of plants, animals, and microbe species. Every species has an intrinsic value even if it is not of current or any economic value to us.

(b) We have an essential duty to care for their well-being and pass on the biological legacy in a proper form to our future generations. We can conserve biodiversity by two major approaches

• In situ conservation (on site/ conservation)
• Ex situ conservation (off-site conservation).

(c) Hotspots are the areas identified by conservationists for the very high level of species richness and high degree of endemism (species confined to a particular area and not found anywhere else). Hotspots help in protection of certain biodiversity-rich regions.

Sacred groves are the tracts of forest set aside where all the trees and wildlife within are given religious sanctity and total protection. Such sacred groves are found in Khasi and Jaintia hills in Meghalaya, Aravalli Hills of Rajasthan etc.

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 10 Cell Cycle and Cell Division Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 10 Cell Cycle and Cell Division

Very short answer type questions

Question 1.
During which phase of mitotic cell division, chromosomes gets separated?
Answer:
During anaphase.

Question 2.
Does mitosis occurs before or after the interphase?
Answer:
Yes, mitosis occurs before or after the interphase, as dividing phase (meiosis or mitosis) and interphase are considered only as the major phases of a cell cycle.

Question 3.
Mitosis cell division helps in regeneration process. How?
Answer:
Mitosis helps in regeneration by keeping all the somatic cells of an organism genetically similar, so that they are able to regenerate part or whole of the organism.

Question 4.
Given that average duplication time of E. coli is 20 minutes. How much time will two E. coli cells take to become 32 cells?
Answer:
2 hours (2ⁿ = 2⁵ = 2 × 2 × 2 × 2 × 2 = 32 generations).

Question 5.
If a tissue has 1024 cells at a given time, how many cycles of mitosis had the original parental single cell undergone?
[NCERT Exemplar]
Answer:
10 (2ⁿ, where n =10 generations).

Question 6.
Two key events take place during S-phase in animal cells, i.e., DNA replication and duplication of centriole. In which parts of the cell do these events occur? [NCERT Exemplar]
Answer:
DNA replication in the nucleus. Centriole duplication in the cytoplasm.

Question 7.
At what stage of meiosis, formation of tetrads occurs? Name it.
Answer:
Tetrads are formed during pachytene of prophase-I (meiosis-I).

Question 8.
Meiosis is essential in sexually reproducing organisms. How?
Answer:
Meiosis is essential in sexually reproducing organisms because it keeps the chromosome number constant.

Question 9.
Which cells of our body do not divide?
Answer:
Neuron cells stops dividing soon after the birth of a child.

Question 10.
The diagram shows a bivalent at prophase-I of meiosis. Which of the four chromatids can cross over? [NCERT Exemplar]
Answer:
The non-sister chromatids of homologous pair of chromosome undergo meiosis.

Short answer type questions

Question 1.
Imagine a situation if there was no meiosis. Then what would have happened to the next generation?
Answer:
In the absence of meiosis the next generation would have double the number of chromosomes after fusion of gametes. This would have resulted in the birth of an altogether new species. The maintenance of characters set would have been possible only through asexual reproduction.

Question 2.
Give a description of metaphase I of meiosis.
Answer:
Metaphase I: The bivalent chromosomes align on the equatorial plate. The microtubules from the opposite poles of the spindle attach to the pair of homologous chromosomes.

Question 3.
Describe telophase I of meiosis.
Answer:
Telophase I

• The nuclear membrane and nucleolus reappear, cytokinesis follows and this is called as diad of cells.
• Although in many cases the chromosomes do undergo some dispersion, they do not reach the extremely extended state of the interphase nucleus.
• The stage between the two meiotic divisions is called interkinesis and is generally short lived. Interkinesis is followed by prophase II, a much simpler prophase than prophase I.

Question 4.
What is the process of cell division in prokaryotes?
Answer:
Prokaryotes do not have nucleus. So, there is no elaborate karyokinesis, as seen in eukaryotes. In prokaryotes the replication of DNA starts the process of cell division. Once genetic material is replicated, it is followed by division of cytoplasm. The process is known as binary fission.

Question 5.
How does meiosis facilitate creation of offsprings, with distinct characters?
Answer:
Meiosis happens during gametogenesis and as a result gametes have half the number of chromosomes. During fertilization, when gametes fuse together two different sets of chromosomes make a new set. This results in an offspring, who has distinct characters, compared to parents.

Question 6.
What is the significance of mitosis?
Answer:
Significance of Mitosis

• In multicellular organisms, body growth is by mitotic divisions of the cells.
• Replacement of worn out tissues/cells (e.g., blood cells, skin cells) and repair of the injured tissues is by mitosis.
• In unicellular organisms, mitosis are involved in asexual reproduction
  (multiplication of cells).
• In plants, vegetative propagation involves only mitotic divisions and genetically identical individuals are produced.
• Uncontrolled cell divisions in certain tissues/organ (cancer) result in tumours.

Long answer type questions

Question 1.
Briefly describe the significance of cell division.
Answer:
Cell division is significant in the following ways :

• Cell Multiplication: Cell division is a means of cell multiplication or formation of new cells from pre-existing cells.
• Continuity: It maintains continuity of living matter generation after generation.
• Multicellular Organisms: The body of a multicellular organism is formed of innumerable cells. They are formed by repeated divisions of a single cell or zygote. As the number of cells increases, many of them begin to differentiate, form tissues and organisms.
• Cell Size: Cell division helps in maintenance of a particular cell size which is essential for efficiency and control of cell activities.
• Genetic Similarity: The common type of cell division or mitosis maintains genetic similarity of all the cells in an individual despite being different, i.e., structurally and functionally.

Question 2.
Explain meiosis-II in an animal cell.
Answer:
All these happen in the two haploid nuclei simultaneously.

• Prophase-II: It takes short time. Spindle formation begins and the chromosomes become short. Two chromatids, are joined to a single centromere. Nuclear membrane and nucleolus disintegrate.
• Metaphase-II: At the equator, the chromosomes lie and spindle is formed. The centromere of every chromosomes is joined to the spindle fibre and centromere also divides.
• Anaphase-II: The daughter chromosomes are formed. Chromatids move towards their poles with the spindle fibres.
• Telophase-II: Reaching at the poles, chromosomes form nuclei which are haploid (n) daughter nuclei. Again nuclear membrane is constructed. Nucleolus now becomes clearly visible.
• Cytokinesis: It occurs and four daughter cells are formed which are haploid (n). It may occur once or twice (i.e., in meiosis-I and II) or only after the meiosis-II cell division.

Question 3.
Describe briefly the phases of meiotic division.
Answer:
Meiotic division takes place in germ cells. The number of chromosomes is reduced to half in daughter cells.

Meiotic cell division is divided into two phases, i.e., meiotic-I and II.
In the meiotic-I division, the homologous chromosomes pair to form bivalent. Exchange of genetic material takes place. The chromosomes now separate and get distributed into daughter cells.

Long prophase-I is divided into five sub-stages, i.e., leptotene, zygotene, pachytene, diplotene and diakinesis. During metaphase-I, the bivalents arrange themselves on equatorial plate with their arms on the plate but the centromere is directed towards opposite pole. It is followed by anaphase-I. Now, the homologous chromosomes repel each other, move to the opposite poles with both their chromatids. In this way each pole gets half the chromosomes number of the parent cell.

In telophase-I, the nuclear envelope and nucleolus again appear. The centromere of each chromosome breaks, separating the chromatids, one each to a daughter cell. The meiotic cell division maintains chromosome number of a species.

As a result of meiotic division, the four daughter cells are formed with half the chromosome number (haploid) in each cell.

Punjab State Board PSEB 12th Class Biology Important Questions Chapter 2 Sexual Reproduction in Flowering Plants Important Questions and Answers.

PSEB 12th Class Biology Important Questions Chapter 2 Sexual Reproduction in Flowering Plants

Very short answer type questions

Question 1.
An anther with malfunctioning tapetum often fails to produce viable male gametophytes. Give any one reason.
Answer:
A malfunctioning tapetum does not provide enough nourishment to the developing male gametophytes and thus fail to produce viable male gametophytes.

Question 2.
Complete the following flow chart
Pollen mother cell → Pollen tetrad

[NCERT Exampler]
Answer:
Generative cell.

Question 3.
Gynoecium of a flower may be apocarpous or syncarpous. Explain with the help of an example each.
Answer:
Gynoecium of a flower may be apocarpous means the carpels are free from one another and there is no fusion of any part e.g., Ranunculus, Rose. Gynoecium of a flower is syncarpous, means the carpels are fused by their ovaries. The number of fusing carpels may vary from 2 (Petunia) to ∞ (Hibiscus).

Question 4.
Name the parts of the gynoecium which develop into fruit and seeds. [NCERT Exemplar]
Answer:
Ovary develops into fruit and ovules develop into seeds.

Question 5.
How many haploid cells are present in a mature female gametophyte of a flowering plant? Name them.
Answer:
One dikaryotic polar cell with two haploid nuclei and six haploid cells, viz, 3 antipodal, 2 synergids and 1 egg.

Question 6.
Name the type of pollination in self-incompatible plants. [NCERT Exemplar]
Answer:
Xenogamy.

Question 7.
How do flowers of Vallisneria get pollinated?
Answer:
In Vallisneria, the female flower stalk is coiled to reach the water surface to receive the pollen grains carried by water currents.

Question 8.
What is pollen-pistil interaction and how is it mediated?
Answer:
The pistil accepts the right type (compatible) of pollen and promotes fertilisation and rejects the pollen of other species and incompatible pollen of the same species. It is the result of interaction between the chemicals of the pollen and those of stigma.

Question 9.
State the function of filiform apparatus found in mature embryo sac of an angiosperm.
Answer:
Filiform apparatus plays an important role in guiding the path of pollen tubes into the synergids.

Question 10.
Normally one embryo develops in one seed but when an orange seed is squeezed many embryos of different shapes and sizes are seen. Mention how it has happened?
Answer:
An orange seed has many embryos because of polyembryony.

Question 11.
Name the component cells of the ‘egg apparatus’ in an embryo sac. [NCERT Exemplar]
Answer:
Two synergids and an egg.

Question 12.
Name the common function that cotyledons and nucellus perform. [NCERT Exemplar]
Answer:
Cotyledons and nucellus provide nourishment.

Question 13.
In the embryos of a typical dicot and a grass, which are the true homologous structures? [NCERT Exemplar]
Answer:
Cotyledons and scutellum.

Question 14.
In a case of polyembryony, if an embryo develops from the synergid and *another from the nucellus, which is haploid and which is diploid? [NCERT Exemplar]
Answer:
Synergid embryo is haploid and nucellar embryo is diploid.

Short answer type questions

Question 1.
Name the organic materials the exine and intine of an angiosperm pollen grains are made up of. Explain the role of exine.
Answer:
Exine is made up of sporopollenin and intine is made up of cellulose and pectin.
Due to the sporopollenin, exine can withstand high temperature and strong acids. It is also not affected by enzymes. It is because of this reason that pollen grains are well preserved as fossils.

Question 2.
Differentiate between geitonogamy and xenogamy in plants. Which one between the two will lead to inbreeding depression and why?
Answer:

Geitonogamy	Xenogamy
1. It is transfer of pollen grains from the anther to the stigma of another flower of same plant.	It is transfer of pollen grains from the anther to the stigma of a different plant.
2. The pollen grains are genetically similar to the plant.	The pollen grains are genetically different from the plant.

Geitonogamy will lead to inbreeding depression because the pollen grains are genetically similar, which results in inbreeding. Continued inbreeding will thus reduce fertility and productivity.

Question 3.
Double fertilisation is reported in plants of both, castor and groundnut. However, the mature seeds of groundnut are non-albuminous and castor are albuminous. Explain the post fertilisation events that are responsible for it.
Answer:
The development of endosperm (preceding the embryo) takes place from primary endosperm nucleus (PEN) in both, castor and groundnut. The developing embryo derives nutrition from endosperm.

PEN undergoes repeated division to give free nuclei. Subsequently cell wall is formed and endosperm becomes cellular. At this stage endosperm is retained in castor or is not fully consumed but in groundnut endosperm is consumed by growing embryo.

Question 4.
Differentiate between albuminous and non-albuminous seeds, giving one example of each.
Answer:

• Albuminous seeds have residual endosperm ip them. For example, maize.
• Non-albuminous seeds do not have any residual endosperm. For example, pea.

Question 5.
A non biology person is quite shocked to know that apple is a false fruit, mango is a true fruit and banana is a seedless fruit. As a biology student how would you satisfy this person?
Answer:
In apple only the thalamus (along with ovary) portion contributes to fruit. Therefore, it is a false fruit. Mango develops only from the ovary, therefore it is a true fruit.
Banana develops from ovary but without fertilisation. The method is known as parthenocarpy. Since there is no fertilisation, no seeds are formed.

Question 6.
Why are some seeds referred to as apomictic seeds? Mention one advantage and one disadvantage to a farmer who uses them.
Answer:
Seeds produced without fertilisation are referred to as apomictic.
Advantage: Desired characters are retained in offspring (progeny) as there is no segregation of characters in offspring (progeny). Seed production is assured in absence of pollinators.

Disadvantage: Cannot control accumulation of deleterious genetic mutation. These are usually restricted to narrow ecological niches and lack ability to adapt to changing environment.

Long answer type questions

Question 1.
A flower of brinjal plant following the process of sexual reproduction produces 360 viable seeds.
Answer the following questions giving reasons:
(a) How many ovules are minimally involved?
(b) How many megaspore mother cells are involved?
(c) What is the minimum number of pollen grains that must land on stigma for pollination?
(d) How many male gametes are involved in the above case?
(e) How many microspore mother cells must have undergone reduction division prior to dehiscence of another in the above case?
Answer:
(a) 360 ovules are involved. One ovule after fertilisation forms one seed.
(b) 360 megaspore mother cells are involved. Each megaspore mother cell forms four megaspores out of which only one remains functional.
(c) 360 pollen grains. One pollen grain participates in fertilisation of one ovule.
(d) 720 male gametes are involved. Each pollen grain carries two male gametes (which participate in double fertilisation) (360 × 2 = 720).
(e) 90 microspore mother cells undergo reduction division. Each microspore mother cell meiotically divides to form four pollen grains (360/4 = 90).

Punjab State Board PSEB 12th Class Biology Important Questions Chapter 16 Environmental Issues Important Questions and Answers.

PSEB 12th Class Biology Important Questions Chapter 16 Environmental Issues

Very short answer type questions

Question 1.
Why are lichens regarded as pollution indicators?
Answer:
Lichens are regarded as pollution indicators because they do not grow in areas that are polluted. So their presence indicates no pollution in that area and their absence indicates that the area is polluted.

Question 2.
Why is the use of unleaded petrol recommended for motor vehicles equipped with catalytic converters?
Or
Why are owners of motor vehicles equipped with catalytic converters advised to use unleaded petrol?
Or
Why is it desirable to use unleaded petrol in vehicles fitted with catalytic converters?
Answer:
Vehicles fitted with catalytic converters need to use unleaded petrol because lead inactivates the catalyst in the catalytic converter and increases hydrocarbon emission, thereby harming the environment.

Question 3.
In which year was the Air (Prevention and Control of Air Pollution) Act amended to include noise as air pollutant? [NCERT Exemplar]
Answer:
The Air (Prevention and Control of Air Pollution) Act 1981 was amended in 1987 to include noise as an air pollutant.

Question 4.
Name an industry which can cause air pollution, thermal pollution, and eutrophication. [NCERT Exemplar]
Answer:
Fertilizer factory.

Question 5.
Mention two advantages for preferring CNG over diesel as an automobile fuel.
Answer:
Advantages of CNG :

1. It burns more efficiently than petrol or diesel.
2. It is cheaper than petrol or diesel and cannot be siphoned off by thieves or adulterated.

Question 6.
What is an algal bloom? [NCERT Exemplar]
Answer:
The excessive growth of algae (free-floating) that causes coloration of water bodies is called algal bloom.

Question 7.
Write the name of the organism that is referred to as the ‘Terror of Bengal’.
Or
Why is Eichhomia crassipes nicknamed as* ‘Terror of Bengal’?
Answer:
Eichhornia crassipes is nicknamed as ‘Terror of Bengal’ because it grows very fast in the water body and depletes the dissolved oxygen. Hence, disturbing the ecosystem dynamics of the water body.

Question 8.
What is the raw material for polyblend? [NCERT Exemplar]
Answer:
Polyblends are natural man-made fibres, made by the mixture of two or more polymers, especially plastic waste products.

Question 9.
Mention the effect of UV rays on DNA and proteins in living organisms.
Answer:
The high energy of UV rays breaks the chemical bonds within DNA and protein molecules.

Question 10.
Write the unit used for measuring ozone thickness.
Answer:
Dobson unit.

Question 11.
State the purpose of signing the Montreal Protocol.
Answer:
Montreal Protocol was signed at Montreal, in 1987 to curb the emission of ozone-depleting substances.

Question 12.
What is reforestation? [NCERT Exemplar]
Answer:
Reforestation is the process of restoring a forest that once existed but was removed at some point of time in the past.

Short answer type questions

Question 1.
It is a common practice to plant trees and shrubs near the boundary walls of buildings. What purpose do they serve?
[NCERT Exemplar]
Answer:
The practice of growing and maintaining trees and shrubs near the boundary wall of residential or official buildings is a common practice. This is because it acts as a barrier for sound and check noise pollution. This green belt of trees and shrubs also acts as an effective measure to check primary air pollutants like dust, fly ash, etc.

Question 2.
‘Determination of Biological Oxygen Demand (BOD) can help in suggesting the quality of a water body. Explain.
Answer:
Biochemical Oxygen Demand (BOD) is the amount of dissolved oxygen required for microbial breakdown of biodegradable organic matter. Aerobic organisms use a lot of oxygen and as a result, there is a sharp decline in Dissolved Oxygen (DO) in the water body. This Can cause death of fishes and other aquatic species. The relationship between BOD and sewage can be understood from the graph given below :

Determination of BOD is thus an important parameter in suggesting the quality of a water body. The presence of more organic waste increases the microbial activity thus decreasing the DO. BOD is higher in polluted water and lesser in clean water.

Question 3.
Explain the process of secondary treatment given to the primary effluent up to the point it shows a significant change in the level of biological oxygen demand (BOD) in it.
Answer:
The primary effluent is passed into large aeration tanks where it is constantly agitated. Air is pumped into it mechanically. This allows vigorous growth of useful aerobic microbes into floes. These microbes consume the major part of organic matter in the effluent (this significantly reduces the BOD of the effluent).

Question 4.
Excessive nutrients in a freshwater body cause fish mortality. Give two reasons.
Answer:
Excessive nutrients in the water body result in the excessive and rapid growth of plants and animals, resulting in formation of increased organic matter, death of plants and animals increased the organic matter at the bottom which decomposes, increased BOD deplete the oxygen content, resulting in fish mortality.

Question 5.
Is it true that, if the dissolved oxygen level drops to zero, the water will become septic? Give an example which could lower the dissolved oxygen content of an aquatic body. [NCERT Exemplar]
Answer:
Yes, it is true. In case of zero level of Dissolved Oxygen (DO), the water becomes septic. Organic pollutants like fertilizer in aquatic bodies are responsible for lowering (up to zero) the level of dissolved oxygen.

Question 6.
Explain giving reasons why thermal power plants are not considered eco-friendly?
Answer:
Thermal power plants release particulate and gaseous pollutants in the environment. Inhalation of these pollutants can cause breathing or respiratory symptoms^irritation, inflammation, damage to lungs, and even premature death.

Question 7.
Blend of polyblend and bitumen, when used, helps to increase road life by a factor of three. What is the reason? [NCERT Exemplar]
Answer:
Polyblend is a fine powder of recycled modified plastic. The binding property of plastic makes the road last longer besides giving added strength to withstand more loads. This is because: plastic increases the melting point of the bitumen which would prevent it from melting in India’s hot and extremely humid climate, where temperature frequently crosses 50°C. rainwater will not seep through because of the plastic in the blend.

Question 8.
What is the main idea behind ‘Joint Forest Management Concept’ introduced by the Government of India?
[NCERT Exemplar]
Answer:
The main idea behind joint forest management concept introduced by the Government of India was involving the local communities in forest conservation. This concept was adopted considering the extraordinary courage and dedication the local people showed in protecting the wildlife through the movements like Bishnoi’s movement in Jodhpur and Chipko Movement in Garhwal Himalayas.

Long answer type questions

Question 1.
Explain giving reasons the cause of appearance of peaks ‘a’ and ‘b’ in the graph shown below:
img
Answer:
‘a’-High BOD due to sewage discharge.
‘b’-Increase in dissolved oxygen due to sewage decomposition. Micro-organisms involved in biodegradation of organic matter consume a lot of oxygen, therefore, there is a sharp decline in dissolved oxygen. When the sewage is completely degraded, oxygen concentration again increases.

Question 2.
Why is the concentration of toxins found to be more in the organisms occupying the highest trophic level in the food chain in a polluted* water body? Explain with the help of a suitable example.
Answer:
The concentration of toxic materials like heavy metals and pesticides increase at each trophic level of a food chain and is more in organisms of highest trophic level due to their accumulation at each trophic level For example, when DDT was used to control mosquitoes in a lake of USA, 800 times more DDT was found in the phytoplanktons than in the water of the lake. Zooplanktons had about 13 times more DDT than phytoplanktons. It was also observed that the fishes population had 9-40 times more DDT than zooplanktons and fish-eating birds had 25 times more DDT than fish.

Question 3.
Refrigerants arc considered to be a necessity in modem living but are said to be responsible for ozone holes detected in Antarctica. Justify.
Answer:
The widely used refrigerants are CFCs or. chlorofluorocarbons. CFCs discharged in the lower part of atmosphere move upwards to the stratosphere. Here, the UV rays act on them and release chlorine atoms. These free chlorine atoms react with ozone to release molecular oxygen. Chlorine atoms are not consumed in this reaction and hence, these continuously degrade ozone and have resulted in ozone holes.

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 14 Respiration in Plants Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 14 Respiration in Plants

Very short answer type questions

Question 1.
Why does anaerobic respiration/fermentation yields less energy than aerobic respiration?
Answer:
It happens due to incomplete oxidation of the substrate.

Question 2.
Write the overall equation of respiration.
Answer:
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + Energy

Question 3.
Name two openings in plants through which exchange of gases takes place?
Answer:
Stomata and lenticels.

Question 4.
Write the reaction where substrate-level phosphorylation takes place in glycolysis.
Answer:
Substrate level phosphorylation takes place during the following reaction

• When 1, 3-bisphosphoglycerate is converted into 3-phosphoglycerate.
• When phosphoenolpyruvate is converted into pyruvic acid.

Question 5.
List two instances where lactic acid is formed by fermentation.
Answer:
Instances, where lactic acid is formed by fermentation, are given below:

1. During fermentation by lactic acid bacteria.
2. During strenuous exercise, in the striated muscles in humans.

Question 6.
Mention the step of citric acid cycle, which is not mediated by dehydrogenase enzyme.
Answer:
Conversion of oxaloacetic acid to citric acid is not mediated by dehydrogenase enzyme.

Question 7.
At which step of respiration, hydrogen of NADH₂ is used?
Answer:
The hydrogen atoms accepted by NADH₂ during glycolysis are introduced to route I of ETS. In this route 3 ATP molecules are produced.

Question 8.
Mention the number of protons that passes through complex V for the synthesis of 2 molecules of ATP.
Answer:
Two pairs of protons (i.e., 4) passes through complex V for the synthesis of two molecules of ATP.

Question 9.
Name the inhibitor of oxidative phosphorylation.
Answer:
Oligomycin.

Question 10.
Name the unit of oxidative phosphorylation.
Answer:
Oxysomes.

Question 11.
F₀ – F₁ particles participate in the synthesis of ……………………. [NCERT Exemplar]
Answer:
ATP (Adenosine Triphosphate), the energy currency of the cell.

Question 12.
What do you mean by respiratory balance sheet?
Answer:
Respiratory Balance Sheet
The calculations of the net gain of ATP for every glucose molecule oxidized, can be made only on certain assumptions.
But this kind of assumptions are not valid in a living system for the following reasons :

• All pathways work simultaneously and do not take place one after the other.
• Substrates keep entering the pathways and are also withdrawn from the pathways.
• ATP is utilized as and when needed,
• Rates of enzyme actions are controlled by multiple means.

Short answer type questions

Question 1.
Why ATP is called an energy currency?
Answer:
The energy produced during various steps of cellular respiration is stored in the form of ATP. This is later utilized on an SOS basis. So, ATP is also called as energy currency.

Question 2.
There is no special respiratory organ in plants, yet plants efficiently manage exchange of gases. Justify.
Answer:
Every part of plant manages its gas exchange need. There is no exchange of gases between different organs. So unlike animals plants do not need special respiratory organs to facilitate exchange of gases. In leaves the exchange of gases takes place through stomata, while in stems it takes place through lenticels.

Question 3.
How glycolysis takes place in anaerobic environment?
Answer:
Glycolysis is the breakdown of glucose into pyruvic acid and it does not need oxygen. So in all living beings, irrespective of them being either aerobic or anaerobic glycolysis takes place. In fact glycolysis is the first step towards oxidation of glucose and oxidation takes place either during anaerobic respiration or during aerobic respiration.

Question 4.
The maximum concentration of alcohol produced by natural fermentation is 13%. But most of the alcoholic preparations for human consumption contain more than this percentage. How this higher percentage is achieved?
Answer:
The higher percentage of alcohol is achieved through distillation of the liquid, which gives pure alcohol as well. The boiling involved in distillation helps evaporate the liquid part and higher concentration is achieved.

Long answer type questions

Question 1.
Oxygen is critical for aerobic respiration. Explain its role with respect to ETS.
Answer:

• Oxygen is an essential requirement for aerobic respiration because an element of strong electronegativity to pull the electrons down the chain is needed.
• It ensures that protons are pumped into the outer lumen of the mitochondria, where they can come down their concentration gradient through ATP-synthase making ATP.
• The oxygen picks up electrons and protons, thus forming water. As the electrons in the ETS are used to do work, the electrons lose energy and reach a point at the end of the ETS, where they have to be gotten rid of.
• The scheme the cell uses to do this is to combine the electrons with hydrogen ions and oxygen to produce water.

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 9 Biomolecules Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 9 Biomolecules

Very short answer type questions

Question 1.
Write the name of any one amino acid, sugar, nucleotide and fatty acid. [NCERT Exemplar]
Answer:
Alanine is an amino acid, glucose is a sugar, adenylic acid is a nucleotide and linolenic acid is a fatty acid.

Question 2.
Mention four essential major elements of life.
Answer:
Oxygen, carbon, hydrogen and nitrogen are the four essential elements of life.

Question 3.
Name one element invariably found in proteins but not in all carbohydrates and lipids.
Answer:
Nitrogen is found invariably in proteins, but not in all carbohydrates and lipids.

Question 4.
What is the name given to a polysaccharide composed of two different monomers? Also give example for this.
Answer:
The name given is heteropolysaccharide to the type of polysaccharide, which is composed of different types of monomers, e.g., Pectin.

Question 5.
One of the homopolysaccharide is also known as animal starch. Name it.
Answer:
Glycogen is also known as animal starch.

Question 6.
The macromolecules that forms the hereditary determinants of the living organism. Name it.
Answer:
Nucleic acid.

Question 7.
A nitrogenous base is present in RNA but absent in DNA. Name it. Also give example in which it exists.
Answer:
Uracil (U), is the nitrogenous base present only in RNA, e.g., viruses like hepatitis C.

Question 8.
How many hydrogen bonds are formed between:
(i) Guanine and cytosine
(ii) Adenine and thymine, respectively?
Answer:
(i) Three hydrogen bonds between guanine and cytosine.
(ii) Two hydrogen bonds between adenine and thymine.

Question 9.
Reaction given below is catalysed by oxidoreductase between two substrates A and A, complete the reaction. [NCERT Exemplar] A reduced + A oxidised →
Answer:
A reduced + A’ oxidised → A oxidised + A’ reduced + A’ reduced

Question 10.
Name two physical factors which can affect the enzyme activity?
Answer:
Temperature and pH are the two physical factors that affects activity of an enzyme.

Question 11.
The enzyme that works only in the presence of a co-factor or coenzyme called
Answer:
Apoenzyme works only in the presence of a co-factor or coenzyme.

Question 12.
What do you mean by living state?
Answer:
The living state is a non-equilibrium steady-state to be able to perform work.

Short answer type questions

Question 1.
Give a tabular representation of different constituents of a living cell.
Answer:

Component	% of Total Cellular Mass
Water	70-90%
Proteins	10-15%
Carbohydrates	3%
Lipids	2%
Nucleic Acids	5-7%
Ions	1%

Question 2.
What are polysachharides?
Answer:
Polysaccharides are long chains of sugars. They are threads containing different monosaccharides as building blocks.

In a polysaccharide chain (say glycogen), the right end is called the reducing end and the left end is called the non-reducing end. Starch forms helical secondary structures. In fact, starch can hold 12 molecules in the helical portion.
Examples: Cellulose, chitin

Question 3.
Give a brief description of nucleic acid.
Answer:
For nucleic acids, the building block is a nucleotide. A nucleotide has three chemically distinct components. One is a heterocyclic compound, the second is a monosaccharide and the third is a phosphoric acid or phosphate.

The heterocyclic compounds in nucleic acids are the nitrogenous bases named adenine, guanine, uracil, cytosine, and thymine. Adenine and Guanine are substituted purines while the rest are substituted pyrimidines. The skeletal heterocyclic ring is called as purine and pyrimidine respectively. The sugar found in polynucleotides is either ribose (a monosaccharide pentose) or 2’ deoxyribose. A nucleic acid containing deoxyribose is called deoxyribonucleic acid (DNA) while that which contains ribose is called ribonucleic acid (RNA).

Question 4.
What is the difference between primary and secondary metabolites?
Answer:
Primary metabolites are found in both, animal cells and plant cells. Secondary metabolites are found only in plant cells.
Functions of primary metabolites are known to scientists, while functions of secondary metabolites are not known yet.

Question 5.
Explain the basic structure of an amino acid.
Answer:

Amino acid is an organic compound, which has an amino group and an acidic group, present
as substituents on the same carbon ; i.e., the a-carbon. Because of this amino acids are also called α-amino acids. On four valency positions there are four substituent groups. They are hydrogen, carboxyl group, amino group and a variable group. The variable group is called the ‘R’ group. The nature of R group decides the type of an amino acid.

Question 6.
Describe the classification and nomenclature of enzymes.
Answer:
Classification and Nomenclature of Enzymes: The International Union of Biochemists (IUB) has classified all the enzymes into the following six classes:

(a) Class 1: Oxidoreductases: These enzymes catalyse the oxidation (by adding oxygen or removal of hydrogen or removal of electrons) or reduction (by adding hydrogen or adding electrons to a substrate) of a substance.
S reduced + S’ oxidised → S oxidised + S’ reduced

(b) Class 2: Transferases: These enzymes catalyse the transfer of specific groups from one substrate to another. S – G + S’ → S + S’ – G.

(c) Class 3: Hydrolases: These enzymes catalyse the breakdown of larger molecules into smaller molecules with the addition of water.

(d) Class 4: Lyases: These enzymes catalyse the cleavage of specific covalent bonds and removal of specific group (s), without the use of water.

(e) Class 5: Isomerases: These enzymes catalyse the rearrangement of atoms in a molecule to form isomers.

(f) Class 6: Ligases: These enzymes catalyse covalent bonding (of C-0, C-S, C-N, P-O etc.) between two substrates to form a large molecule, mostly involving utilisation of energy by hydrolysis of ATP.

Long answer type questions

Question 1.
Enumerate the difference between a nucleotide and nucleoside. Give two examples of each with their structure. [NCERT Exemplar]
Answer:
Differences between nucleotide and nucleoside are given below:

Nucleotide	Nucleoside
A nucleotide consists of a nitrogenous base, a sugar (ribose or deoxyribose) and one to three phosphate groups, i. e., sugar + base + phosphate.	A nucleoside consists of a nitrogenous base t covalently attached to a sugar (ribose or deoxyribose), but without the phosphate group, i. e., sugar + base

Question 2.
What is the concept of metabolism? What are the metabolic basis for living?
Answer:
The continuous process of breakdown and synthesis of biomolecules through chemical reactions occurring in the living cells is called metabolism.

• Each of the metabolic reaction results in a transformation of biomolecules.
• Most of these metabolic reactions do not occur in isolation but are always linked with some other reactions.
• In these reactions, the metabolites are converted into another metabolite in a series of linked reactions called metabolic pathways.
• Each metabolite has a definite rate and direction during the flow through a metabolic pathways called the dynamic state.

In living systems, metabolism involves two following types of pathways:
(a) The anabolic pathway is called biosynthetic pathway. It leads to a more complex structure from a simpler structure, e.g., The pathway involving the conversion of acetic acid into cholesterol. These pathways consume energy.

(b) The catabolic pathways lead to simpler structure from a complex structure, e g., The pathway involving conversion of glucose into lactic acid in our skeletal muscles. This pathway lead to the release of energy, e.g., Energy is liberated when glucose is degraded to lactic acid in our skeletal muscles.

Question 3.
Formation of Enzyme-Substrate complex (ES) is the first step in catalysed reactions. Describe the other steps till the formation of product.
Answer:
Mechanism of Enzymatic Action: The catalytic cycle of an enzyme action can he described in the following steps :

• First, the substrate binds to the active site of the enzyme, fitting into the active site.
• The binding of the substrate induces the enzyme to alter its shape, fitting more tightly around the substrate. The formation of the ES complex is essential for catalysis.
  E + S → ES → EP → E + P
• The active site of the enzyme, now in close proximity of the substrate breaks the chemical bonds of the substrate and the new enzyme product complex is formed.
• The enzyme releases the products of the reaction and the free enzyme is ready to bind to another molecule of the substrate and run through the catalytic cycle one again.

Punjab State Board PSEB 12th Class Biology Important Questions Chapter 1 Reproduction in Organisms Important Questions and Answers.

PSEB 12th Class Biology Important Questions Chapter 1 Reproduction in Organisms

Very short answer type questions

Question 1.
What is life span?
Answer:
The period between birth and the natural death of an organism represents its life span.

Question 2.
On what factors does the type of reproduction adopted by an organism depend on? [NCERT Exemplar]
Answer:
The organism’s habitat, physiology and genetic make-up determines the type of reproduction adopted by it.

Question 3.
Name an organism where cell division is itself a mode of reproduction.
Answer:
Protists/Monerans/Amoeba/Paramecium.

Question 4.
Mention two inherent characteristics of Amoeba and Yeast that enable them to reproduce asexually. [NCERT Exemplar]
Answer:

1. They are unicellular organisms.
2. They have a very simple body structure.

Question 5.
What is conidia?
Answer:
The asexual, non-motile spores produced externally/exogenously in some fungi are called conidia, e.g., Penicillium.

Question 6.
Define gemmules.
Answer:
Internal asexual reproductive units or buds in sponges are called gemmules.

Question 7.
Name the vegetative propagules in the following
(a) Agave
(b) Bryophyllum
Answer:
(a) Agave – Bulbil
(b) Bryophyllum – Leaf buds/adventitious buds.

Question 8.
Mention the unique feature with respect to flowering and fruiting in bamboo species.
Answer:
Bamboo flowers once in its life time generally after 50-100 yrs of vegetative growth. It produces large number of fruits and dies.

Question 9.
Is Marchantia monoecious or dioecious? Where are the sex 1 organs borne in this plant? [NCERT Exemplar]
Answer:
Marchantia is dioecious. The male sex organs, antheridia, are borne on j the antheridiophores and female sex organs called archegonia are borne on archegoniophores.

Question 10.
Suggest a possible explanation why the seeds in a pea are arranged in a row, whereas those in tomato are scattered in the juicy pulp. [NCERT Exemplar]
Answer:
The ovary of pea plant is monocarpellary and the ovules are arranged along one margin whereas in tomato the ovary is tricarpellary with axile placentation.

Question 11.
In a developing embryo, analyse the consequences if cell divisions are not followed by cell differentiation.
[NCERT Exemplar]
Answer:
If differentiation does not follow division, embryo will not develop and this will not develop into a new organism.

Question 12.
Name the phenomenon and one bird where the female gamete directly develops into a new organism.
Answer:
The phenomenon is called parthenogenesis. Turkey.

Question 13.
Mention the site where syngamy occurs in amphibians and reptiles, respectively.
Answer:
In amphibians, external fertilisation occurs hence, syngamy occurs in the medium of water. In reptiles, internal fertilisation occurs hence, syngamy occurs within the body of female parent.

Question 14.
Name the group of organisms that produce non-motile gametes. How do they reach the female gamete for fertilisation?
Answer:
Angiosperms produce non-moule gametes. They reach the female gamete with the help of air or water.

Question 15.
The number of taxa exhibiting asexual reproduction is drastically reduced in the higher plants (angiosperms) and higher animals (vertebrates) as compared with lower groups of plants and animals. Analyse the possible reasons for this situation. [NCERT Exemplar]
Answer:
Both angiosperms and vertebrates have a more complex structural organisation. They have evolved very efficient mechanism of sexual reproduction. Since asexual reproduction does not create new genetic pools in the offspring and consequently hampers their adaptability to external conditions, these groups have resorted to reproduction by the sexual method.

Short answer type questions

Question 1.
The cell division involved in gamete formation is not of the same type in different organisms. Justify.
Answer:
The parents may be haploid or diploid but the gametes always have to be haploid, biploid parents undergo meiosis to produce haploid gametes, whereas haploid parents undergo mitosis to produce haploid gametes.

Question 2.
Although sexual reproduction is a long drawn, energy-intensive complex form of reproduction, many groups of organisms in Kingdom Animalia and Plantae prefer this mode of reproduction. Give at least three reasons for this. [NCERT Exemplar]
Answer:

• Sexual reproduction brings about variation in the offspring.
• Since gamete formation is preceded by meiosis, genetic recombination occurring during crossing over (meiosis-I), leads to a great deal of variation in the DNA of gametes.
• The organism has better chance of survival in a changing environment.

Question 3.
Explain the importance of syngamy and meiosis in a sexual life cycle of an organism.
Answer:
Syngamy and meiosis play an important role in sexual life cycle of any diploid organisms, may be a plant or animal syngamy, i.e. fusion of haploid gametes/sex cells (n) (fertilisation) to form diploid egg cell/zygote (2n). Zygote divides repeatedly by mitotic divisons to form an embryo which develop into a new diploid organisms. After attaining sexual maturity, the organisms undergo special type of meiotic divisions – spermatogenesis/ microsporogenesis and megasporogenesis/ oogenesis to form haploid male sex cells and female sex cells. These sex cells (n) again fuse to form diploid zygote.

Question 4.
Why are mosses and liverworts unable to complete their sexual mode of reproduction in dry conditions? Give reasons.
Answer:
For sexual reproduction to take place in mosses and liverworts the motile male gametophytes, antherozoids, have to swim on the water surface to fertilise the immotile female gametophytes, egg. In dry conditions, the antherozoids do not reach the egg and hence fertilisation cannot occur.

Question 5.
Why do algae and fungi shift to sexual mode of reproduction just before the onset of adverse conditions?
Answer:
Algae and fungi shift to sexual mode of reproduction for survival during unfavourable conditions. Fusion of gametes helps to pool their resources for survival. The zygote develops a thick wall that is resistant to dessication and damage which undergoes a period of rest before germination.

Long answer type questions

Question 1.
What is gametogenesis? Describe the different types of gametes and draw labelled diagrams. ,
Answer:
The process of formation of two types of gametes – male and female inside the gametangia is called gametogenesis.

Depending upon the size and motility, the gametes are of following types :
1. Isogametes or Homogamets: The gametes are similar in shape, size, structure arid function. Their fusion is called isogamy e.g., Cladophora. However, when the isogametes are physiologically different and the gametes produced by one parent do not fuse with each other, the gametes belonging to different mating types can only fuse and their fusion is called physiological anisogamy, e.g., Ulothrvc.

2. Anisogametes or Heterogametes: The fusing gametes are different in form, size, structure and behaviour. The larger, non-motile, food-laden gamete is called ovum/egg/oosphere/ macrogamete. The smaller, motile, active gamete is called sperm/male gamete/antherozoid. Such gametes are called anisogametes or heterogametes and their fusion is termed as anisogamy or heterogamy e.g., Cladophora.

Question 2.
Write a note on sexuality in organisms.
Answer:
Sexuality in Organisms: Sexual reproduction in organisms generally involves the fusion of gametes from two different individuals. But this is not always true.

Sexuality in Plants: Plants may have both male and female reproductive structures in the same plant (bisexual) e.g., rose or on different plants (unisexual) e.g., papaya. In several fungi and plants, terms such as homothallic and monoecious are used to denote the bisexual condition and heterothallic and dioecious are the terms used to describe unisexual condition.

In flowering plants, the unisexual male flower is staminate i.e., bearing stamens, while the female is pistillate or bearing pistils. In some flowering plants, both male and female flowers may be present on the same individual (monoecious) or on separate individuals (dioecious). Some examples of monoecious plants are cucurbits and coconuts and of dioecious plants are papaya, mulberry and date palm.

Sexuality in Animals: There are species which possess both the reproductive organs (bisexual). Earthworms, sponge, tapeworm and leech are typical examples of bisexual animals that possess both male and female reproductive organs i.e., they are hermaphrodites. There are large number of animal species which are either male or female and are called as unisexual organisms, e.g., frog, lizard, crow, dog, cat, rabbit, human beings etc.