RBSE Solutions for Class 12 Biology

RBSE Solutions for Class 12 Biology Chapter 28 Man-Reproductive System

August 14, 2019 by Prasanna Leave a Comment

Rajasthan Board RBSE Class 12 Biology Chapter 28 Man-Reproductive System

RBSE Class 12 Biology Chapter 28 Multiple Choice Questions

Question 1.
Graafian follicles are found in:
(a) Thyroid gland of a man
(b) A prostate gland of a man
(c) Ovaries of woman
(d) Testes of man
Answer:
(c) Ovaries of woman

Question 2.
Which gland secretes an alkaline fluid that helps in smoothening of the vagina at the time of mating?
(a) Prostate
(b) Rectal
(c) Cowper
(d) Perineal
Answer:
(c) Cowper

Question 3.
After how many days menstrual cycle occurs in the women.
(a) 14 days
(b) 20 days
(c) 32 days
(d) 28 days
Answer:
(d) 28 days

RBSE Class 12 Biology Chapter 28 Very Short Answer Type Questions

Question 1.
Write the function of Corpus luteum.
Answer:
Secretion of two hormones-Progesterone & Relaxin.

Question 2.
Which of the cells function as endocrine gland in the testes? Write the name of the hormone secreted by these cells.
Answer:
Leydig’s cells: Androgen hormone.

Question 3.
Write the name and function of the hormone secreted by Leydig’s cells.
Answer:
Androgen hormone.
It controls spermatogenesis & provides male secondary sexual characters.

Question 4.
Write the names of primary sex organs found in male and female human beings.
Answer:
Male – Testes
Female – Ovaries

Question 5.
Write the name of the mucous membrane found internally in the Uterus.
Answer:
Endometrium.

Question 6.
Write two secondary sexual characters of a woman.
Answer:
Well developed mammary glands.
The high pitch of voice.

RBSE Class 12 Biology Chapter 28 Short Answer Type Questions

Question 1.
What are the changes visible at the puberty in male and female human beings?
Answer:
Changes in Male and Female at Puberty:

Male	Female
Growth in the penis, scrotal sacs, prostate gland and seminal vesicle.	Growth is uterus, vagina, oviducts and clitoris.
Beginning of spermatogenesis.	Growth of mammary glands and beginning of a menstrual cycle.
The low pitch of voice.	The high pitch of voice.
Growth of hair on face (moustache), thorax and pelvic region.	Absence of hair on the body except for pelvic region.
Growth of the body as a whole.	Pelvic broadening, growth of mammary glands and fat deposition in the body.
Secretion of testosterone, FSH, LH hormones increases.	Secretion of Progesterone, estrogen, LH hormones increases.
Psychological attraction towards a male.	Psychological attraction towards female.

Question 2.
During the winter season, the scrotal sac shrinks and become small. Explain it.
Answer:
Spermatogenesis needs a temperature of 34 to 35° C. During winters, the outside temperature is low. Hence, the scrotal sacs shrink to come closure to the body to get more heat.

Question 3.
What is Cryptorchidism?
Answer:
The arrest of spermatogenesis due to high temperature is called cryptorchidism.

Question 4.
What is andropause?
Answer:
Male menopause is called andropause. there is decreased secretion of testosterone. It occurs normally after 60 years.

Question 5.
What do you mean by Sertoli cells? What are their functions?
Answer:
They are special cells which are found inside the seminiferous tubeless of testes. They provide support & nutrition to the sperms.

Question 6.
Draw a neat diagram of mammary glands of a woman and explain how it is helpful in reproductive activity.
Answer:
The mammary glands are female sexual character & are meant to secrete milk to feed young ones.

Question 7.
Draw labelled diagram of a transverse section of the testis.
Answer:

RBSE Class 12 Biology Chapter 28 Essay Type Questions

Question 1.
Describe the male reproductive system of a man with suitable diagram.
Answer:
Male Reproduction System:

The male reproductive system consists of primary reproductive and accessory reproductive organs. Primary sex organs are a pair of testes or male gonads. Accessory reproductive organs are mainly scrotal sac, epididymis, vas deferens, penis, prostate gland and Cowper’s glands.

Testes:

An adult man has a pair of testes (oval bodies suspended outs the body in the scrotum).
The wall of tests is an elastic covering of thin and hairy and thick inner subcutaneous layer of unstriated muscle fibres which is known as Dartos muscle.
In each half of scrotal sac, a tuft of rod-shaped striated muscles fibres are found. It connects the subcutaneous layer with abdominal subcutaneous muscle and known as Cremaster muscle.
Each cavity of the scrotal sac is connected to the abdominal cavity with the help of a white Inguinal canal. In adult mammals, testes are ex-abdominal which are found outside the abdominal cavity in the testes sacs or scrotal sacs.
The spermatogenesis in the testes: needs less temperature and the temperature of the scrotal sacs is 33 to 35°C which is 2 to 4o C less than the body temperature.
The testis descends into the scrotal sac through an inguinal canal. The testicular artery, testicular vein & the testicular nerve also descend into the scrotal sac in the form of a spermatic cord.
Each testis is attached in the scrotal sac with the help of a gubernaculum. Both the spermatic cord & the gubernaculum are provided with elastin fibres.
If the testes fail to descend into the scrotal sacs, the spermatogenesis gets arrested due to a high temperature which called cryptorchidism.

Question 2.
Explain the structure of ovary.
Answer:
Structure of Ovary:

Ovaries are paired structures found in the abdominal cavity.
Size of each ovary is 1.5 – 3 cm long and 8 mm thick and shape is like an almond.
It is found behind kidneys in the pelvic region.
Ovary forms ovum and secretes a female hormone, estrogen and progesterone.

Question 3.
Explain the role of various accessory glands in male and female helping in the reproduction process.
Answer:
(1) Accessory Glands in Male:

Three types of accessory glands are found in man which secretes their secretions into the Urethra.
These secretions are essential for viability and motility of spermatozoa. The secretions of accessory glands, epididymis, and sperms are collectively formed semen.

Following accessory reproductive glands are found in man –
1. Prostate Glands:

It is situated around the anterior end of the urethra and opens into the urethra.
The prostate gland secretes a white fluid which forms 25 – 30%part of semen.
It is alkaline secretion and neutralizes the acidic medium of the Vagina.
This fluid contains Phosphatase, citrate, lysozyme, fibrinolysin, spermin etc.
It’s secretions activate the sperms and prevent the semen to coagulate.
In aged men, the size of the prostate may be enlarged which create problem in urine discharge.
It consists of Four lobes –
1. Ventral lobe – 1
2. Dorsal lobe – 1
3. Lateral lobes – 2

2. Seminal Vesicles:

It is also called as uterus masculine.
There is a pair of seminal vesicles in man.
The secretions of the seminal vesicle form 70% part of the seminal fluid.
The secretions of the seminal vesicle are alkaline, slimy & its pH is 7.4. It consists of fructose, ascorbic acid, prostaglandin & enzymes.
The fructose provides energy to the species.

3. Cowper’s glands:

There is a pair of Cowper’s glands which are also called as Bulbo-urethral glands.
They are small, round & yellow in colour and open into the middle part of the urethra.
They secrete transparent & alkaline fluid having a pH 7.5 to 8. This fluid makes the urethra alkaline before the copulation

(2) Various Accessory Glands in Female:
There are two types of sex accessory glands –

Prostate Glands:
- A pair of prostate glands open into the anterior part of the vagina. Their secretions make the vagina moist before copulation.
Bartholin Glands:
- On both sides of the vaginal opening, a pair of bean-shaped glands are situated. These are Bartholin’s glands.
- These glands secrete an alkaline and lubricating fluid, which keeps vulva moist and facilitate sexual intercourse.

Question 4.
Write short notes on:

Ovary
Penis
Epididymis
Secondary sexual characters
Corpus luteum

Answer:
1. Ovary:

Ovaries are paired structures found in the abdominal cavity.
Size of each ovary is 1.5 – 3 cm long and 8 mm thick and shape is like an almond.
It is found behind kidneys in the pelvic region.
Ovary forms ovum and secretes a female hormone, estrogen and progesterone.

2. Penis:

In man, the penis is suspended between the legs along with the scrotal sacs attached to the abdomen. It is cylindrical, erectile and highly vascularized copulatory organ. It is covered by skin.
Body of the penis is made up of three blood sinuses composed of filamentous, muscular connective tissues. Two large are dorsolateral and one is ventrally around Urethra. Dorsolateral sinuses are called corpora cavernosa and ventral sinus as corpus sporangium.
The apex of the penis is swollen and form glans. Glans penis is a form of only corpus sporangium. Covering of skin over glans is a cap-like & called as prepuce. It gets folded on the base of glans during copulation.
In normal condition blood, sinuses remain empty and muscles are in contraction stage, at this time only urine can flow outside through urinogenital aperture.

At the time of erection or during erect position for mating, blood moves in the sinuses by penial artery, muscles relaxed and glans of the penis becomes hard and swollen as maximum as possible.
Glans become nacked from prepuce and it is now in the erection state. The penis is ready for deep penetration in the vagina of the female. The semen ejaculated by penis into the vagina.

3. Epididymis:

It is thin, highly coiled and comma-shaped tube about 6 meters long that leads into a vas deferens. Its coiled rings are adhered by connective tissues.
Its outer side is covered with thick muscular layer and inner side is lined with stratified epithelium.
It is attached to the inner surface of the testis & has 3 parts –
1. Caput or globulus major
2. Corpus
3. Cauda or globulus minor
Globulus major is a large cap-like structure which receives the vasa efferentia.
Globulus minor is a small structure which opens into vas deferens.
The epididymis stores the sperms & it is the site for physiological maturation of the sperms.

4. Secondary sexual characters:

Male	Female
Growth in the penis, scrotal sacs, prostate gland and seminal vesicle.	Growth is uterus, vagina, oviducts and clitoris.
Beginning of spermatogenesis.	Growth of mammary glands and beginning of a menstrual cycle.
The low pitch of voice.	The high pitch of voice.
Growth of hair on face (moustache), thorax and pelvic region.	Absence of hair on the body except for pelvic region.
Growth of the body as a whole.	Pelvic broadening, growth of mammary glands and fat deposition in the body.
Secretion of testosterone, FSH, LH hormones increases.	Secretion of Progesterone, estrogen, LH hormones increases.
Psychological attraction towards a male.	Psychological attraction towards female.

5. Corpus luteum:

The follicular cells of the burst follicles reorganize under the control of LH to form corpus luteum.
It has yellow coloured luteal cells containing lutein yellow pigments.
This structure performs endocrine function & secretes progesterone & inhibin hormones.
There is a blood clot in the centre of the corpus luteum which is called corpus haemorrhagic.

Question 5.
Explain with diagram the female reproductive system.
Answer:
Female Reproductive Organs:

One pair of ovaries are found in a woman as a Primary reproductive organ.
Accessory reproductive organs found in a female are oviducts, uterus, vagina and vulva.
Reproductive glands and mammary glands are also included as accessory organs.

(1) Ovary:

Ovaries are paired structures found in the abdominal cavity.
Size of each ovary is 1.5 – 3 cm long and 8 mm thick and shape is like an almond.
It is found behind kidneys in the pelvic region.
Ovary forms ovum and secretes the female hormone. estrogen and progesterone.

(2) Histology of Ovary:

There is a pair of ovaries which are attached dorsally in the abdominal cavity with the help of mesovarium. The mesovarium originates from the visceral peritoneum.
The ovary has germinal epithelium inside the mesovarium which is made up of cuboidal germ cells.
The ovary has stroma made up of fibrous connective tissues. The peripheral part of the stroma is condensed which is called cortex & the inner less dense part is called the medulla.
The cortical part of the ovary has ovarian follicles which are as follows –
1. Primary follicles:
  - They are formed during embryonic stage & begin to grow at puberty.
  - It has an oogonium surrounded by one laver of follicular cells.
2. Secondary follicles:
  - It has a primary oocyte surrounded by 2 layers of follicular cells.
3. Tertiary follicles:
  - It has primary oocyte surrounded by 3 layers of the follicular cells.
4. Mature follicles or Graaffian follicles:
  - It is surrounded by a membrane granulosa which consists of two layers v.z., theca external & theca internal.
  - It has a fluid-filled cavity, the antrum.
  - The mature follicle has a secondary oocyte which is attached to the membrane granulosa by a stalk.
  - This stalk is called as germ hill or discuss proliferous or cumulus oophoritis.
  - The Gryphon follicle gradually migrates to the periphery & gets protruded.
  - The mature follicles bursts at the surface to release the secondary oocyte outside. It is called ovulation.
5. Atretic follicle:
  - Some times, the Graffian follicle fails to ovulate & degenerates gradually. The degenerating Graffian follicle is called as atretic follicle and the process is called as atresia.
  - It is due to deficiency of vitamin E & by hormonal imbalance.
6. Corpus luteum:
  - The follicular cells of the burst follicles reorganize under the control of LH to form corpus luteum.
  - It has yellow coloured luteal cells containing lutein yellow pigments.
  - This structure performs endocrine function & secretes progesterone & inhibin hormones.
  - There is a blood clot in the centre of the corpus luteum which is called corpus haemorrhagic.
7. Corpus Albicans:
  - If there is no fertilization, the corpus luteum begins to degenerate. The degenerating corpus luteum is called as corpus Albicans.
  - It is colourless & non-functional.

3. Oviduct:

There is a pair of oviducts which originate from the Mullerian duct.
The anterior end of the oviduct modified to form ciliated fimbriated funnel or infundibulum.
Its mouth is funnelshaped which is called as ostium.
The ovum enters into the oviduct through the oviducal funnel.
The middle part of the oviduct is called a Fallopian tube.
Its wall hås unstriated muscles & ciliated internally.
The Fallopian tube is the site for secondary maturation division, fertilization & embryonic development.

4. Uterus:

The posterior part of the oviduct modifies to for uterus.
The uterus is simplex in a woman.
A uterine wall has three layers –
1. Epimedium:
  - Outer & made up of visceral peritoneum.
2. Myometrium:
  - It is the middle layer which is made up of smooth muscles. It contains the longest smooth muscle.
3. Endometrium:
  - It is the innermost layer which has 2 layers viz., stratum functional & stratum basale.

5. Vagina:

The large, elastic & muscular tube that runs from the cervix (terminal part of the uterus) to outside is called Vagina.
The Vaginal opening is partially closed by a thin membrane called Hymen.
This hymen may be torn off due to physical labour, sexual contact and exercise etc.
Vagina provides a path for menstrual discharge beside copulatory organ of female.
It serves as a birth canal during the birth of the child. Two folds of tissue called vulva protect the vaginal and urethral openings.
Vaginal wall stores glycogen. The Lactobacilli bacteria present in the vagina makes mucous acidic by fermentation.
This prevents the vagina from infection.

6. Vulva:

The external genitalia of the woman is called vulva. It is situated just above the perineum is the pelvic area.
Following structures are included in vulva:
1. Mons pubis or Mons veneris:
  - It is swollen fatty tissue covered by skin and situated above the pubic symphysis.
2. Labia Majora:
  - A pair of transverse folds expanded from the mons pubis to the lower side and till back. It is tissue covered by skin and hair are found on its surface.
3. Labia Minora:
  - There is a pair of small folds present inside the labia majora surrounding vestibule.
4. Clitoris:
  - It is a sensory and erectile organ situated at the anterior corner of labia minora and below the mons pubis. This is homologous to glans of the penis (intromittent organ of male).
5. Vestibule:
  - “A fissure like structure in the middle of labia minora is called vestibule”.
  - The urinary orifice and orifice of the vagina are found. Below the clitoris is urinary orifice or meatus is found as a small aperture. Below this is orifice of the vagina is situated.

7. Sex Accessory Glands:
There are two types of sex accessory glands –

Prostate Glands:
- A pair of prostate glands open into the anterior part of the vagina. Their secretions make the vagina moist before copulation.
Bartholin Glands:
- On both sides of the vaginal opening, a pair of bean-shaped glands are situated. These are Bartholin’s glands.
- These glands secrete an alkaline and lubricating fluid, which keeps vulva moist and facilitate sexual intercourse.

8. Breasts:

Breasts begin to develop at the age of puberty and finally become Mammary glands as accessory reproductive organs of the female.
One pair of mammary glands found in women. These are found on the pectoral muscles on the ventral side.
Each gland composed of connective tissues, 15 – 20 tubular partitioned lobules. Fatty tissue is found between these structures.
Each lobule has a proliferation of glandular tissues caused by the action of oestrogen and progesterone.
Glands secrete milk as the nutrition of newborn. Many small ductules unite and form lactiferous duct in each lobule.
Such many lactiferous tubules open on the nipples independently. The nipple is a knob-like pigmented structure on each mammary gland.
The area around the nipple is darker (dark pigmented) and called as areola mammae.
Each nipple has 15 to 25 pores for milk ejection. Growth and function of mammary glands are controlled by somatotropin, prolactin, estrogen, progesterone and oxytocin hormones.
Milk contains fat, lactose, casein, lysozyme, calcium, vitamin and important immunoglobins.
Mother’s milk is a complete food for the neonatal and it develops immunity in the child.

RBSE Solutions for Class 12 Biology

RBSE Solutions for Class 12 Biology Chapter 36 Man-Chromosomal Aberrations

August 14, 2019 by Prasanna Leave a Comment

Rajasthan Board RBSE Class 12 Biology Chapter 36 Man-Chromosomal Aberrations

RBSE Class 12 Biology Chapter 36 Multiple Choice Questions

Question 1.
Number of autosomes found in man are:
(a) 42
(b) 44
(c) 46
(d) 48
Answer:
(b) 44

Question 2.
Which of the following disease is caused by numerical changes in autosomes?
(a) Turner’s syndrome
(b) Klinefelter’s syndrome
(c) Triple female
(d) Down syndrome
Answer:
(d) Down syndrome

Question 3.
Progeny of normal female and colourblind father will be:
(a) Normal vision in all children and no carrier
(b) Colour blind boys and normal girls
(c) Girls are a carrier but boys normal
(d) All children colourblind
Answer:
(c) Girls are a carrier but boys normal

Question 4.
A number of chromosomes in Turner’s syndrome:
(a) 44
(b) 45
(c) 46
(d) 47
Answer:
(b) 45

Question 5.
Which of the following disease is sex-linked?
(a) Diarrhoea
(b) Edward syndrome
(c) Mongolism
(d) Haemophilia
Answer:
(d) Haemophilia

Question 6.
Haemophilic father and disease carrier mother will produce offsprings in which:
(a) Half normal sons but half haemophilic sons
(b) All haemophilic sons
(c) All haemophilic daughters
(d) Half daughters normal and other half carriers
Answer:
(a) Half normal sons but half haemophilic sons

Question 7.
When linkage is found in a group of genes than they:
(a) Cannot show independent assortment
(b) Stimulate cell division
(c) Did not show chromosomal mapping
(d) Show recombination during meiosis
Answer:
(a) Cannot show independent assortment

Question 8.
Hereditary recombination is found in higher animals as the result of crossing over, In which of the following it occurs:
(a) Between sister chromatids of any bivalent
(b) Between non-sister chromatids of any bivalent
(c) Between two daughter nuclei
(d) Between two different bivalents
Answer:
(b) Between non-sister chromatids of any bivalent

Question 9.
Which of the following enzyme cannot be synthesized in phenylketonuria disease?
(a) Phenylalanine hydroxylase
(b) Phenylalanine dehydrogenase
(c) Phenyl oxygenase
(d) Phenyl hydroxylases
Answer:
(a) Phenylalanine hydroxylase

Question 10.
Sickle cell anaemia is caused by:
(a) Mutation of the autosomal gene
(b) Polyploidy of sex chromosome
(c) Aneuploidy of sex chromosome
(d) None of the above
Answer:
(a) Mutation of the autosomal gene

RBSE Class 12 Biology Chapter 36 Very Short Answer Questions

Question 1.
What is Down syndrome?
Answer:
It is an autosomal abnormality which is caused due to one extra 21st chromosome (45 + 2 = 47). It is also called as 21st terisomy or mongolism.

Question 2.
What is the function of Phenylalanine hydroxylase?
Answer:
This enzyme catabolizes phenylalanine into tyrosine.

Question 3.
Which type of mutation occurs in the haemoglobin of sickle cell anaemia?
Answer:
Point mutation: At 6th position of B chain of haemoglobin glutamic acid is replaced by valine amino acid.

Question 4.
How many chromosomes may be present in Klinefelter’s syndrome?
Answer:
47 or 48 or 49

Question 5.
When and in which type of cell division crossing over occurs?
Answer:
Meiosis, during Pachytene phase of Meiosis – 1.

Question 6.
What is a linkage group?
Answer:
The group of genes found on one chromosome which has a tendency to be inherited together is called a linkage group.

Question 7.
What do you mean by haemophilia?
Answer:
It is a sex-linked hereditary disease. There is delayed or no blood clotting. Its recessive defective gene is found on the X chromosome.

Question 8.
Write a number of chromosomes found in a man and woman.
Answer:
Man – 44 + XY.
Woman – 44 + XX.

RBSE Class 12 Biology Chapter 36 Short Answer Questions

Question 1.
What is colourblindness?
Answer:
Colour blindness:

It is a hereditary disease. In this disease, the person is unable to differentiate different colours.
The colourblindness is of two types –
- Red-Green colourblindness
- Blue colourblindness
This disease is mainly found in men because males have only one X chromosome. If X – chromosome contains recessive allele of pigment formation gene than no cone formation takes place and man suffers from colour blindness. While two X – chromosomes are found in a woman and hence woman may be a carrier of this disease or may suffer from this disease.

Question 2.
Express the inheritance of disease in the progeny of carrier of colourblind mother and a normal father.
Answer:
Conclusion: 50% daughter carrier; 50% daughter normal; 50% son colourblind; 50% son normal.

Question 3.
What is Turner’s syndrome?
Answer:
Turner’s Syndrome:

A person suffering from this disease is always a woman. Such a woman has only one X chromosome instead of two and hence a total number of chromosomes found are only 45 (44 + X 0). It is called Turner’s Syndrome.
One girl born per 3000 as turner syndrome.
Symptoms:
- Such women are of short stature.
- Broad chest and lack of secondary sexual characteristics and sterile.
- Short neck, underdeveloped mammary glands.
- Mental – retardedness

Question 4.
Explain the experiment of Bateson and Punnet.
Answer:
An experiment of Bateson and Punnet:

Bateson and Punnet (1906) did a dihybrid cross on plants of blue flowers and long pollens with red flower and round pollen plant and found the following results:
[Abbreviations: Blue flower = B, Red flower = b, Long pollens = L and round pollens = L.]
A cross between Blue flower and long pollens (BBLL), and red flower and round pollens yielded blue flowers and long pollens in the F₁generation.
Now they made a test cross between heterozygous F₁ offsprings (B bLl) with homozygous recessive parent (bbll). By the concept of linked genes and independent assortment BbLl should give four kinds of gametes and hence four phenotypes in equal proportions 1:1:1:1 were expected. But Bateson and Punnet obtained the following results –
The above results in 7:1:1:7 ratio were obtained. This suggests that new combinations or recombinants were less and a number of parental plants were more.
It means the selected genes for the study have not followed the law of independent assortment, but their nature showed linked inheritance or they show Linkage.
The genes present on the same chromosome show their nature to inherit together are called Linked genes and the incidence is known as Linkage.
Sometimes a group of genes found on a single chromosome are inherited together, is called Linkage group.

Question 5.
Explain various types of linkage.
Answer:
Various Types of Linkage:
Linkage is of two types –

Complete linkage:
When certain genes located so nearly on the same chromosome and tend to remain linked together while passing from one generation to another without new combinations are termed complete linkage.
Incomplete linkage:
Linked genes are not inherited always in their combination. In the homologous chromosomes or at the time of meiosis exchange of such genes may be found due to crossing over. The genes located at distance on the chromosome are also incompletely linked because of more chances of isolation by crossing over in these.

Question 6.
What are sex-linked character and their inheritance?
Answer:
Sex-linked Inheritance:

The human female has XX chromosomes and the human male has XY chromosome along with 44 autosomes in both sexes.
The genes present on the X chromosome may express in both male and female gender and hence it is called crisscross heredity. While genes present on Y – chromosome found only in the male.
Genes of approximately 20 characters found on the X chromosome of human are sex-linked characters and a father can not transmit sex-linked traits (of the X chromosome) to his sons, it can be passed on to the daughters only. Some of the
sex-linked hereditary diseases found in human beings are as follows:
- Colour Blindness
- Haemophilia

RBSE Class 12 Biology Chapter 36 Essay Type Questions

Question 1.
What do you mean by sex-linked inheritance? Discuss it with reference to colourblindness and haemophilia diseases.
Answer:
1. Sex-linked Inheritance:

The human female has XX chromosomes and the human male has XY chromosome along with 44 autosomes in both sexes.
The genes present on the X chromosome may express in both male and female gender and hence it is called crisscross heredity. While genes present on Y- chromosome found only in the male.
Genes of approximately 20 characters found on the X chromosome of human are sex-linked characters and a father can not transmit sex-linked traits (of the X chromosome) to his sons, it can be passed on to the daughters only.

Some of the sex-linked hereditary diseases found in human beings are as follows:
Colour blindness:

It is a hereditary disease. In this disease, the person is unable to differentiate different colours.
The colourblindness is of two types –
1. Red-Green colourblindness
2. Blue colourblindness
This disease is mainly found in men because males have only one X chromosome. If X – chromosome contains recessive allele of pigment formation gene than no cone formation takes place and man suffers from colour blindness. While two X – chromosomes are found in a woman and hence woman may be a carrier of this disease or may suffer from this disease.

Red-Green colourblindness:

It is caused due to the absence of red/green cones and it is also called as protanopia/Deuteranopia.
The person is unable to distinguish red, green, yellow and orange colours. These colours appear to red in the absence of green cones and green in the absence of red cones.

Blue colourblindness:

It is caused due to the absence of blue cones.
Such a person is unable to distinguish violet, Indico, Blue and Green colours. They all appear to be green.

2. Inheritance of colourblindness:

Example 1: Cross between a normal woman and a colourblind man.
If a colour blind male marries with a normal female the possibilities would be – All children (son and daughters) would be normal, but the daughters would be carriers. Because they are heterozygous, they will not be colourblind.
The woman acts as a carrier of colourblindness.
Example 2: Cross between a carrier woman and a colourblind man.
In this cross, 50% of girls will be colourblind and 50% of girls will be a carrier.
Similarly, 50% of boys will be normal and 50% of boys will be colourblind.
Conclusion: Following are conclusions regarding heredity of colour blindness:
1. Identification of colours in the vision is a sex-linked character. Its gene is located on the X chromosome and its allele is not found in males.
2. Males are more sufferer from this disease. Because if the normal gene for colour differentiation is absent on X chromosome man than the colour blindness the disease is found.
3. Women have two X-chromosomes and both X chromosomes have the gene for colour identification and expression.
4. While only one chromosome in man has a gene for colour identification and expression. If only one X chromosome carries the recessive gene for colourblindness in woman than the normal woman becomes the carrier of the disease.
5. Gene for the identification of colours is dominant. If both X chromosomes of woman carry this recessive gene then a woman will be colour-blind.
6. Father and son of a colourblind woman are also colour blind. If the husband of a colour blind woman is also colour blind than their daughters will be colour blind. Daughter (normal) of a colour blind father produce 50%sons of normal vision and 50% sons as colourblind.

3. Haemophilia or Bleeder’s disease:

It is a sex-linked hereditary disease which results in delayed or no blood clotting. It was first discovered by John Otto.
Haldane discovered haemophilia in the family of Queen Victoria. It is caused due to one recessive gene on the X – chromosome of man.
It is also called as Royal disease or bleeder’s disease. It is expressed only in men The women act as a carrier of this disease. Recessive genes on both the X – chromosomes cause haemophilia in females but they do not survive.
Father of haemophilia female is always haemophilic.
Haemophilia is of 2 types –
1. Haemophilia – B or Christmas disease: It is caused due to the absence of factor – IX. It is a weak type of haemophilia.
2. Haemophilia – A: It is caused due to the absence of factor – VIII (AHG).
This disease is also due to the recessive gene located on the X chromosome. Haemophilic man and normal woman produce 50% carrier daughters and 50% normal sons.
Carrier woman and normal man produce 25% carrier daughters, 25% haemophilic sons, 25% normal daughters and 25% normal sons.
This disease is found only in males as males have only one X chromosome. If a woman has both recessive genes on their X chromosomes than such woman cannot survive, because blood flow during menstruation does not stop and hence it is fatal for them.
When females contain only one recessive gene of haemophilia on the X chromosome and other X chromosomes without such gene, they are a carrier of haemophilia.

Question 2.
Differentiate between linkage and crossing over.
Answer:
1. Linkage:

Mendel’s principle of independent assortment is truly applicable for both genes and chromosomes. Genes control various characteristics of the organisms and gene is a basic unit of heredity.
First-time German scientist Sutton studied the interrelationship between factors and chromosomes. He discovered linkage.
According to him in each pair of chromosome one chromosome is from each parent, (paternal and maternal member).
It is the reason by which the genes present on separate chromosomes are independently assorted, but all genes present on the same chromosome are said to be linked and should be inherited together.
A single chromosome may carry a large number of genes.
All the genes located on the same chromosome are said to be linked genes. The characters controlled by these genes are said to be Linked characters.
These linked genes have a tendency to be inherited together but independently. Genes present on a single chromosome are not independently assorted.
Bateson and Punnet (1906) explained about linked genes in his experiment on sweet pea (Lathyrus odoratus).

An experiment of Bateson and Punnet:

Bateson and Punnet (1906) did a dihybrid cross on plants of blue flowers and long pollens with red flower and round pollen plant and found the following results:
[Abbreviations: Blue flower = B, Red flower = b, Long pollens = L and round pollens = L.]
A cross between Blue flower and long pollens (BBLL), and red flower and round pollens yielded blue flowers and long pollens in the F₁ generation.
Now they made a test cross between heterozygous F₁ offsprings (BbLl) with homozygous recessive parent (bbll). By the concept of linked genes and independent assortment BbLl should give four kinds of gametes and hence four phenotypes in equal proportions 1:1:1:1 were expected. But Bateson and Punnet obtained the following results –
The above results in 7:1:1:7 ratio were obtained. This suggests that new combinations or recombinants were less and a number of parental plants were more.
It means the selected genes for the study have not followed the law of independent assortment, but their nature showed linked inheritance or they show Linkage.
The genes present on the same chromosome show their nature to inherit together are called Linked genes and the incidence is known as Linkage.
Sometimes a group of genes found on a single chromosome are inherited together, is called Linkage group.

Various Types of Linkage:
Linkage is of two types –

Complete linkage:
When certain genes located so nearly on the same chromosome and tend to remain linked together while passing from one generation to another without new combinations is termed complete linkage.
Incomplete linkage:
Linked genes are not inherited always in their combination. In the homologous chromosomes or at the time of meiosis exchange of such genes may be found due to crossing over. The genes located at distance on the chromosome are also incompletely linked because of more chances of isolation by crossing over in these.

2. Crossing Over:

Bateson and Punnet obtained some recombinant type plants during experiments. It is because of a special process called crossing over.
In crossing over an exchange of genetic material takes place between non-sister chromatids of homologous chromosomes. It results in the formation of new combinations.

During prophase I of meiosis there is an exchange of parts of chromatids between homologous chromosomes. The homologous chromosomes become intertwined and the chromatids of homologous chromosomes cross over at certain point called Chiasmata.
The Chromatids break at this point and then rejoin. Crossing over results in the exchange of corresponding segments at the same locus between the paternal and maternal chromosomes (means non-sister chromatids of homologous chromosomes). Linked genes separate from each other by crossing over and new combinations appear.

Types of Crossing over:

On the basis of the number of Chiasma formation in the chromosomes, the three types of crossing over are found –
- Single Crossing over When only one chromatid of homologous chromosomes participate in crossing over and only one chiasma is formed. It is called Single crossing over.
- Double-crossing over: When two, three or all four chromatids of homologous, chromosomes participate in crossing over and two chiasmata are formed. It is known as Double-crossing over.
- Multiple Crossing over When more than two chiasmata are formed between two non-sister chromatids. It is called multiple crossing over.

Significance of Crossing Over:

This process occurs during the pachytene stage of prophase – I of meiosis.
It occurs between non-sister chromatids of homologous chromosomes.
Exchange of segments or genes between non-sister chromatids is known as Crossing over.
Numbers of points of a chromosome where crossing over takes place at a single time depends on the length of the chromosome. Percentage of crossing over increases with the increase in the length of a chromosome.
Distance between the genes in a chromosome affects the probability of crossing over. More distance between genes increase the probability of crossing over and if the distance is less than there is a decrease in the probability of crossing over.

Question 3.
Describe various chromosomal aberrations in man.
Answer:
Human Chromosomal Disorders(Abnormalities):

Chromosomal disorders occur due to a change in the number of chromosomes or change in their structure.

These abnormalities are of two types:
1. Autosomal Abnormalities:
1.1. Mongolism or Down’s Syndrome:

This abnormality was first described by Langdon Down in 1866.
The person suffering from. Down syndrome has 47 chromosomes.
This is due to one extra 21st chromosome. Hence, it is also called 21st Trisomy. Down syndrome is found in one out of 600 children.
Such defect generally occurs in a child which is born by the woman of more than 40 years ago. It is due to one extra 21st more chromosome the ovum at the time of meiotic division (expulsion of the polar body), so instead of 23 chromosomes, there are 24 chromosomes in the ovum.
A child develope from such ovum contains 47 chromosomes and abnormality develops is Down syndrome.

Salient Features:

They have protruding round heads, webbed neck.
A furrowed tongue that causes the mouth to remain partially open and short.
Broad hands with blunt fingers and characteristic fingerprint patterns.

Physical, psychomotor and mental development is retarded. Life expectancy is shortened.

1.2. Edward’s Syndrome:

It is caused due to one extra 18th chromosome (45 + 2 = 47). Hence, it is also called as 18th Trisomy.
It occurs in one out of 6000 persons.
Symptoms: Mentally retarded, congenital heart disease, Polydactylous, Maximum life of 6 months.

1.3. Cry du Chat Syndrome:

It is caused due to loos of small arm (p – arm) of the 5th chromosome.
It is also called the cat’s cry syndrome.
Symptoms: Microcephaly, Infant voice resembles the cat’s cry.

2. Allosomal Abnormalities:

Following are main autosomal abnormalities.

2.1. Turner’s Syndrome:

A person suffering from this disease is always a woman.
Such a woman has only one X chromosome instead of two and hence a total number of chromosomes found are only 45 (44 + X 0). It is called Turner’s Syndrome.
One girl born per 3000 as turner syndrome.
Symptoms:
1. Such women are of short stature.
2. Broad chest and lack of secondary sexual characteristics and sterile.
3. Short neck, underdeveloped mammary glands
4. Mental – retardedness

2.2. Klein felters Syndrome:

A person suffering from this disease is a man.
Such man has 47 or 48 or 49 total chromosomes instead of 46 in their cells. This additional number is either of X or Y chromosome. Klinefelter’s syndrome in man may have chromosomes as:
1. 44 + XXY (One additional X chromosome) = 47
2. 44 + XXXY (two additional X chromosome) = 48
3. 44 + XXXXY (three additional X chromosome) = 49
4. 44 + XXYY (addition one X and one Y chromosome) = 48
5. 44 + XXXYY (Additional two X and one Y chromosome) = 49
Such males with an abnormal number of chromosomes have feminine characters. Specific features of a Klinefelter’s syndrome are:

In 2A + XXX condition (47), the female has usually normal genitalia, limited fertility and slight mental retardedness.
Higher doses of X chromosomes may lead to more pronounced effects. In the male, the mammary glands are like female and this, the condition is called Gynaecomastia (gina = woman and massage = male with mammae). Hands and Legs are long as to compare to the normal one.

2.3. Super Female:

In women sometimes a total number of chromosomes may be up to 47 to 49. This increased number is due to additional X chromosomes –
- 44 + XXX ⇒ 47
- 44 + XXXX ⇒ 48
- 44 + XXXXX ⇒ 49
Symptoms:
- Such women are of low mental ability.
- Delayed sexual development.
- Highly beautiful.
- More emotional.

2.4. Jacob’s syndrome:

It is also called as a criminal syndrome.
It is caused due to one extra Y-chromosome in Man (44 + XYY = 47).
Symptoms:
- Abnormal development of sex organs.
- Exceptionally tall and stout body.
- Low mental level.
- The criminal tendency since birth.

RBSE Solutions for Class 12 Biology

RBSE Solutions for Class 12 Biology Chapter 35 Mendel’s Law of Inheritance

August 13, 2019 by Prasanna Leave a Comment

Rajasthan Board RBSE Class 12 Biology Chapter 35 Mendel’s Law of Inheritance

RBSE Class 12 Biology Chapter 35 Multiple Choice Questions

Question 1.
The main reason for the success of Mendel was that he:
(a) Selected pea plant
(b) Studied one character at a time in hybrid
(c) Kept record of pedigree
(d) All of the above
Answer:
(d) All of the above

Question 2.
Law of Independent assortment is proved by:
(a) All offspring of the F₁ generation is long
(b) Long and dwarf plants produce in 3:1 ratio
(c) Expression of plants having smooth & wrinkled seeds in the F₂ generation.
(d) Expression of Long & Dwarf plants in the F₂ generation.
Answer:
(c) Expression of plants having smooth & wrinkled seeds in the F₂ generation.

Question 3.
A monohybrid cross gives an F₂ generation is a characteristic phenotypic ratio of:
(a) 9:3:3:1
(b) 3:1
(c) 1:1
(d) 2:1
Answer:
(b) 3:1

Question 4.
Red and White cross give pink progeny. In this R gene represents:
(a) Hybrid
(b) Recessive
(c) Incomplete dominance
(d) Mutation
Answer:
(c) Incomplete dominance

Question 5.
The effect of the genotype of group AB in human being shows:
(a) Dominant-Recessive
(b) Incomplete dominance
(c) Co-dominance
(d) Complimentary
Answer:
(c) Co-dominance

Question 6.
What is the genotype ratio in a lethal gene?
(a) 1:2:1
(b) 3:1
(c) 2:1
(d) 9:3:3:1
Answer:
(c) 2:1

RBSE Class 12 Biology Chapter 35 Very Short Answer Type Questions

Question 1.
Who is called the father of heredity and why?
Answer:
Gregor Johann Mendel (1822 to 1884) was first to explain the mechanism of transmission of characters from generation to generation. He is, therefore, called as “father of genetics”.

Question 2.
What are the monohybrid and dihybrid phenotypic ratio of the F₂ generation?
Answer:
Monohybrid ⇒ 3:1 (Dominant: Recessive)
Dihybrid ⇒ 9:3:3:1

Question 3.
What are the multiple gene characters?
Answer:

Normally, phenotype expression is controlled by a pair of alleles. But when 3 or more alleles are responsible for one character, it is called multiple alleles.
Example: ABO blood system.

Question 4.
Where and in which organization Mendel read the research paper of his experiments?
Answer:
Brunn Society of Natural History from 8th Feb. to 8th March 1865.

Question 5.
In which language the original research paper of Mendel was published and what was its title?
Answer:
German
Title: Versuche Uber Pflanzenhybriden.

Question 6.
Who rediscovered Mendel’s work?
Answer:
Hugo De Vries (Holland), E.V. Tschermak (Austria) and K. Correns (Germany)

RBSE Class 12 Biology Chapter 35 Short Answer Type Questions

Question 1.
Differentiate the following:

Homozygous and Heterozygous
Dominant and Recessive trait
Genotype and Phenotype
Monohybrid Cross and Dihybrid Cross

Answer:
1. Homozygous and Heterozygous:

There are two genes for every character. If the two genes for a particular character are identical, is said to be homozygous (G homos ⇒ same; zygotes ⇒ a pair). For example, tall plants (TT) & dwarf plant (tt) are homozygous.
If the two genes for a particular character are unlike, it is said to be heterozygous (G. heteros ⇒ other, zygotes ⇒ a pair). For example, Tt is heterozygous tall.

Homozygous	Heterozygous
Formation of only one type of gametes	Formation of 2 types of gametes
Breed true to the particular character	Do not breed true to a particular character
It possesses two alleles of a gene	It possesses two identical alleles of a contrasting gene

2. Dominant and Recessive trait:

Dominant character:
- Amongst two alleles of a character, the character express in a heterozygous condition is called Dominant character. The incidence is known as Dominance and the gene responsible for it is dominant allele.
- For Example, Tt containing pea plant has T for tallness and t for dwarfness and T is the dominant character.
Recessive character:
- Amongst Alleles of a character in the F₁ generation, which is not expressed is called recessive character. It expresses only in the homozygous condition of character.
- For Example, tt condition express dwarfness of a plant.

3. Genotype and Phenotype:

Genotype:
- Genetic expression of a feature, which obtained from parents.
- For Example Pure round seed producing parent plants has genotype R R.
Phenotype:
- The visible expression of the hereditary character possessed by an organism.
- For example, around or wrinkled shape of seed is its phenotype.

4. Monohybrid Cross and Dihybrid Cross:

Monohybrid cross:
- Across in which inheritance of only one pair of contrasting character is studied.
Dihybrid cross:
- Across in which inheritance of two pairs of contrasting characters is studied simultaneously.

Question 2.
Define:

Allele
Codominance
Polygenes
Lethal gene

Answer:
1. Allele:

Alternative forms of a gene for eg. blue colour and black colour of the eye are two alleles of the colour of the eye gene. An allele is a short form of Allelomorph. Similarly, for the seed of the pea plant, the shape of the seed Round (R) and Wrinkled (r) are two alleles.

2. Codominance:

The dominant and recessive factor of an allele equally expresses their characters in F₁ generation is called codominance. The skin colour in castles is red (RR) and white (RR) is found. The cross of these gives a roan colour of skin in the F₁ generation.
Then cross between Fı generation gives red, roan and white skin castles in the F₂ generation. The ratio of red, roan and white coloured skin animals are found 1:2:1 (RR: Rr: RR). The results do not follow Mendelian ratio with regard to colours.

3. Multiple gene or Polygenes:

The phenotype traits which are governed by three or more genes are called polygenic traits. The polygenic traits show a wide range of phenotypes. Each gene of the polygenic trait contributes to the phenotype but to a small degree.
Presence of more than one dominant gene makes the phenotype more prominent. All the dominant genes add up their effects to produce a full phenotype. Therefore, the polygenic traits are also called as quantitative traits.
Inheritance of such traits is called polygenic or quantitative inheritance.
The polygenes may occupy two or more different loci on the same homologous chromosome pair.
It was first studied by Galton (1883) in man.
The well-known examples of a polygenic trait are human skin colour & kernel colour in wheat.
The other examples of polygenic traits in human beings are – eye colour & body height.
Example: 1 – Human skin colour
The human skin colour is a polygenic effect and controlled by three pairs of genes Aa, Bb, Cc.
These genes are located on different chromosomes and are inherited independently.
Each gene contributes to a unit of darkness due to incomplete dominance.
The skin colour varies from very dark to very light and has many shades of intermediate colours. Hence, a total of 64 phenotypic combinations is possible for skin colours.

4. Lethal gene:

Some genes besides controlling external characteristics also affect the viability of organisms.
L. Cuenot in 1905 presented results of his experiment on the body colour of rats. These results were different from Mendel’s law of segregation.
According to his experiments, the yellow colour of the skin is due to gene Y and grey colour is due to gene y in which Y is dominant over y (grey colour).
When yellow colour rats were crossed then yellow and grey colour rats in 2:1 ratio produced.
Genotypically, yellow rats were heterozygous (Yy) and grey rats were homozygous (yy).
The Yy (heterozygous) yellow colour rats were unable to survive. Hence Y gene is responsible for the yellow colour of rats and it also affects survival capacity.
The yellow homozygous (YY) rats can not survive, hence homozygous rats are not found at all.
Hence YY genes are called lethal genes and this incidence is known as Lethality.
Some lethal genes are lethal only in homozygous recessive condition and called as recessive lethal gene, while dominant lethal gene may be lethal even in heterozygous condition.

Question 3.
What are the reasons for Mendel’s success?
Answer:
Reasons for Mendel’s Law:

Mendel with his experiments discarded all the theories given by earlier scientists, as these theories were hypothetical and were not based on scientific experiments.
Initially, Mendel’s work was also not recognized by other scientists. But it was accepted after 34 years of the death of
Mendel, when three scientists Hugo de Vries, Karl corners and Eric Von Tschermak obtained the same results as of Mendel.
Mendel is known as “Father of Genetics” for his valuable contribution in the area of heredity.

Achievements of Mendel’s work are mentioned below:

Scientist discovered dominant and recessive characters in living beings on the basis of Mendel’s work.
Beneficial characters of different genera are brought together in a genus.
By using Mendel’s law, the discovery of immunity against diseases, continuity in adverse conditions, flowers and fruits of good quality and in quantity are developed.
Similarly, Breed of Cow, buffalo and hen are improved.

Question 4.
Describe Incomplete dominance with examples?
Answer:
Incomplete Dominance:

It was discovered by corners (1903).
It is also called as blended or partial or mosaic or ‘intermediate inheritance’.
It is an exception to be an outcome of Mendel’s monohybrid cross.
It is found in both plants & animals.
In Mendel’s pea experiments, dominance was essentially complete & there was no difference between the homozygous & heterozygous plants in the expression of dominant character.
However, there are characters or alleles that are neither dominant nor recessive. In such cases, both the alleles of the contrasting conditions of a character express as a blend.
As a result, in F₁ generation the hybrid produced by crossing two pure individuals does not resemble either of them but is midway between them.
This expression of an intermediate trait of two pure parents in the F₁ hybrids is called as incomplete dominance.
Example 1: Four O’clock plant (Mirabilis jalapa) & snapdragon plant (Antirrhinum majus) Incomplete dominance in Snapdragon Here the phenotypic and genotypic ratios are same. The ratio 1:2:1 is the characteristics of Incomplete Dominance.

Question 5.
Why Mendel selected Pea plant in his experiments? Explain.
Answer:
Selection of Garden Pea by Mendel for his experiments:

Mendel carried out his work on garden pea Pisum sativum. He selected this plant because of the following reasons:
- It was easy to grow the plants in open ground and in pots.
- The plant had a short life span and it completes in one season so the study of many generations is possible in a few years.
- The pea plant is small, easy to crossbreed artificially.
- The plant flower being bisexual, and exhibits self-fertilization in nature so that purity of characters can be maintained until many generations but can be easily cross-pollinated experimentally.
- A large number of true-breeding varieties of peas are available or we can say many contrasting characters are found in pea plants.

RBSE Class 12 Biology Chapter 35 Essay Type Questions

Question 1.
Mendel studied which hereditary traits of a Pea. in his experiments, explain.
Answer:
Selection of Characters by Mendel:

Mendel conducted a breeding experiment on the garden pea plant, Pisurnsativum. He studied the inheritance of seven pairs of contrasting characters in the garden pea & he restricted his experiments to one or two pairs of contrasting traits in each experiment.
The seven pairs of contrasting characters selected by Mendel are as follows:

Character	Recessive	Dominant
Height of Plant	Dwarf	Tall
Position of Flower	Terminal	Auxillary
The shape of Pod	Constricted	Inflated
Colour of Pod	Yellow	Green
The shape of a seed	Wrinkled	Round
Colour of Seed Coat	White	Grey
Colour of Cotyledon	Green	Yellow

Some authors follow the colour of flowers – White and Purple character as a 7th character instead of the colour of cotyledons.
Mendel found all the above 7 traits to be pure genetically in experiments.

Question 2.
Explain Mendel’s laws in detail.
Answer:
Mendel’s Breeding Experiments:
Mendel did the following experiments to cross plants using one or more contrasting characters at one time –

Monohybrid cross
Dihybrid cross

1. Monohybrid cross:

In this cross, he used one contrasting character at a time.

Experiment:

He made a cross between pure tall and pure dwarf plants –
- Tall plant: Height 6 feet
- Dwarf plant: Height 1 feet
Mendel used plants of F₁ generation for self-pollination and seeds obtained were sowed. The resulting plants of second filial (F₂) generation were both tall and dwarf. in 3:1 ratio. This 3:1 ratio is called monohybrid phenotypic ratio.
Out of these of dwarf plants were self-pollinated and their seeds again gave all dwarf plants. While the tall plants were self-pollinated and the seeds obtained were sowed they gave both tall and dwarf plants in the 3:1 ratio, phenotypically.
The genotype of these plants was found to be 1:2:1 (1TT; 2Tt: 1tt).
The result shows that tall plants homozygous TT and heterozygous Tt while dwarf plants were tt homozygous.

Mendel’s Explanation:

Hence monohybrid cross produces phenotype ratio 3:1 and genotype ratio 1:2:1 in the F₂ generation. The logical conclusion arrived at by Mendel from these points was that Fı plants must have carried a ‘hidden factor’ for dwarfness. Thus
The second conclusion arrived at by Mendel was that if the F₁ plants contain two ‘factors’ for height (the one responsible for tallness which showed up and one for dwarfness which remain hidden).
Hence pure breed of the tall plant contains two factors (TT) for tallness and dwarf plant contain two factors (tt) for dwarfness.
These factors are presently referred to as a gene. Out of them one factor from father and another factor from mother came into the offspring.
The factor which expresses its character is termed Dominant and which could not express in the presence of dominant factor is called Recessive factor.
Example 1: Seed Shape in Pea.

2. Dihybrid cross:

In this experiment, Mendel selected two pairs of contrasting characters i.e. Mendel studied the inheritance of two characters at a time. He crossed pea plant with round & yellow seeds (RRYY) with the plant having wrinkled & green seeds.
In the F₁ generation, all pea plants produce round & yellow seeds (Rr Yy), which indicates that round is dominant over the wrinkled shape and yellow is dominant over green colour.
In F₂ generation (resulting by self-pollination of F₁ plants), all the four characters were assorted out independent to each other which indicates that seed colour is independent of seed shape.
The F₁ hybrids form 4 types of gametes viz. RY, Ry, rY & ry They on random mating produced & types of the generation having a phenotypic ratio of 9:3:3:1.
Example: Shape and colour of seeds of a pea plant

Result of F₂ generation:
- yellow Round ⇒ 9 (1, 2, 3, 4, 5, 7, 9, 10, 13)
- yellow Wrinkled ⇒ 3 (11, 12, 15)
- Green Round ⇒ 3 (6, 8, 14)
- Wrinkled green ⇒ 1 (16)

Question 3.
Explain the deviations of principles of Mendel’s laws.
Answer:
principles of Mendel’s laws:
Incomplete Dominance:

It was discovered by corners (1903).
It is also called as blended or partial or mosaic or ‘intermediate inheritance’.
It is an exception to be an outcome of Mendel’s monohybrid cross.
It is found in both plants & animals.
In Mendel’s pea experiments, dominance was essentially complete & there was no difference between the homozygous & heterozygous plants in the expression of dominant character.
However, there are characters or alleles that are neither dominant nor recessive. In such cases, both the alleles of the contrasting conditions of a character express as a blend.
As a result, in F₁ generation the hybrid produced by crossing two pure individuals does not resemble either of them but is midway between them.
This expression of the intermediate trait of two pure parents in the F₁ hybrids is called as incomplete dominance.
Example 1: Four O’clock plant (Mirabilis jalapa) & snapdragon plant (Antirrhinum majus) Incomplete dominance in Snapdragon Here the phenotypic and genotypic ratios are same. The ratio 1:2:1 is the characteristics of Incomplete Dominance.

Codominance:

A dominant and recessive factor of an allele equally express their characters in F₁ generation is called codominance. The skin colour in cattles is red (RR) and white (rr) is found. The cross of these gives a roan colour of skin in the F₁ generation.
Then cross between F₁ generation gives red, roan and white skin cattles in the F₂ generation. The ratio of red, roan and white coloured skin animals are found 1:2:1 (RR:Rr:rr). The results do not follow Mendelian ratio with regard to colours.

Differences between Co-dominance and Incomplete dominance:

Co-dominance	Incomplete dominance
The heterozygous shows ratio of both alleles in the F1 generation. For example blood group AB has both the antigens A as well as B.	The heterozygous is of the intermediate type and shows none of the original parental forms. Example Pink colour flower of snapdragon.
Both alleles are equally dominant. Both expressed fully and equally.	There is partial dominance of one and hence intermediate type id formed.
Examples: ABO blood group and skin colour of cattle.	Examples: the colour of snapdragon flowers and flowers of Four O clock plant.

RBSE Solutions for Class 12 Biology

RBSE Solutions for Class 12 Biology Chapter 37 Mutations

August 13, 2019 by Prasanna Leave a Comment

Rajasthan Board RBSE Class 12 Biology Chapter 37 Mutations

RBSE Class 12 Biology Chapter 37 Multiple Choice Questions

Question 1.
What is a Genome?
(a) Sum total of a haploid set of chromosome & extra chromosomal material of the cell
(b) Total No. of a chromosome is an organism
(c) diploid sets of chromosomes
(d) number of chromosomes found in the zygote
Answer:
(a) Sum total of a haploid set of chromosome & extra chromosomal material of the cell

Question 2.
How many numbers of nucleotides pairs found in human?
(a) 3 lac
(b) 30 lac
(c) 3 crores
(d) 3 Billion
Answer:
(d) 3 Billion

Question 3.
What is the basis of DNA fingerprinting?
(a) Self copy of DNA
(b) Fingerprint of DNA
(c) DNA fingerprints of two individual are not the same
(d) none of the above
Answer:
(c) DNA fingerprints of two individual are not the same

Question 4.
Dolly was produced by the help of which technique:
(a) From normal hybridization
(b) From normal sexual reproduction
(c) From cloning
(d) From Tissue Culture
Answer:
(c) From cloning

Question 5.
What is a Mutation?
(a) Change in the genetic material of DNA
(b) Permanent and Hereditary change in the DNA
(c) Change in the cytoplasm of the cell
(d) any kind of variations
Answer:
(b) Permanent and Hereditary change in the DNA

Question 6.
If adenine is replaced from guanine than mutation is –
(a) Frameshift mutation
(b) Transcription
(c) Transition
(d) Transversion
Answer:
(c) Transition

Question 7.
Genetic code includes –
(a) 3 nucleotide bases, 64 codons
(b) 3 nucleotide bases, 18 codons
(c) 2 bases, 32 codons
(d) 2 bases, 64 codons
Answer:
(a) 3 nucleotide bases, 64 codons

Question 8.
Polyploidy can be produced by –
(a) Colchicines
(b) X rays
(c) Gamma Rays
(d) None of Above
Answer:
(a) Colchicines

RBSE Class 12 Biology Chapter 37 Very Short Answer Type Questions

Question 1.
Human Genome Project was started by which international Agencies.
Answer:
National Institute of Health and Department of Energy, America.

Question 2.
What are VNTRs?
Answer:
A variable number of Tandem Repeat are small segments of nucleotides which are repetitive.

Question 3.
The stage through which change in a group of chromosomes numbers occurs is called?
Answer:
Polyploidy.

Question 4.
What is Non-disjunction?
Answer:
Failure of a pair of homologous chromosomes to separate in meiosis – I, failure of sister chromatids to separate during meiosis – II is called non-disjunction.

Question 5.
What are mutagens?
Answer:
The factors which produce mutations are called mutagens.

Question 6.
Define Probe.
Answer:
The probe is a fragment of DNA or RNA of variable length which can be radioactively or fluorescently labelled.

Question 7.
What is the Honolulu Technique? Who developed this technique and where.
Answer:
It was developed by Teruhiko Wakayama and Ryuzo Vanagimachi at University of Hawaii (1998). It is used to produce an animal clone.

RBSE Class 12 Biology Chapter 37 Short Answer Type Questions

Question 1.
Define Gene Mutation.
Answer:
Some time mismatch combination of nitrogenous bases occurs in daughter DNA. If this change occurs at gene level it is known as Gene Mutation.

Question 2.
What is duplication?
Answer:
Duplication:
Sometimes, some genes are present more than once in a chromosome i.e. one, segment repeats twice in a chromosome. It is called duplication.

Question 3.
Define point mutation.
Answer:
It occurs at a particular locus of the chromosomes so it is also known as a Point mutation.

Question 4.
Define genome.
Answer:
Genome:

It is a scientific project which was aimed to study the complete sequence of the human genome. Genes are the key which decides how we appear and work. Moreover, the genes indicate our history and future.
The DNA sequence is different in two individuals at a certain location, this promotes the scientists to completely sequence a human genome.
The haploid set of chromosomes in a cell is called a Genome.

Question 5.
What is Euploidy?
Answer:
Euploidy or Polyploidy:
The total set of a number of chromosomes in cell or organism is called monoploidy. The gaining of one or more complete sets of chromosomes is called euploidy. This can be of two types –

Autopolyploidy
Allopolyploidy

Question 6.
What is a silent mutation?
Answer:
Silent Mutation:
Silent mutations are base substitutions that result in no change of the amino acid or amino acid functionality when the altered messenger RNA (mRNA) is translated. For example, if the codon AAA is altered to become AAG, the same amino acid – lysine – will be incorporated into the peptide chain.

Question 7.
What is gene cloning?
Answer:
Gene Cloning:

The production of a large population of a DNA fragment in pure form is called as gene cloning.
It involves the formation of recombinant DNA and its introduction into the mother bacterial cell.
The recombinant DNA is formed by combining DNAs from two different organisms.
This process is called genetic engineering. A DNA fragment of one organism (called as Passenger DNA) is transferred into another organism by combining it with another DNA(called as Vehicle DNA).
Normally plasmid or bacteriophage DNA is used as Vehicle DNA. Both the passenger and vehicle DNA are cleaved by using restriction endonuclease enzyme and they are mixed under suitable conditions.
Their complementary sticky ends pair and their ends are sealed by ligase enzyme. The resulting DNA is called recombinant DNA.
The recombinant DNA is added to the culture containing growing host bacteria.
These bacteria take the recombinant DNA along with the nutrients.
This DNA replicates when the bacteria divide and produce a large clone of recombinant DNA.

Question 8.
Define Wobble Hypothesis.
Answer:
The genetic codes are ambiguity in nature. It means a particular amino acid can be coded by more than one codons. It was explained by Crick who gives Wobble hypothesis. According to this theory, if the amino acid is coded by more than one code than initial two nucleotides of the codon are same only 3rd nitrogenous base are different.

RBSE Class 12 Biology Chapter 37 Essay Type Questions

Question 1.
What is the mutation? Write the features of mutation.
Answer:
Mutation:

The sudden changes in the hereditary material i.e. Chromosomes are termed as mutations.
Hugo De Vries (1901) propounded “theory of mutation”.
It is based on the results of experiments on a plant, the evening primrose.
De Vries explained that mutations are sudden changes and are the sole agent for evolution.
Mutations are always irreversible.

Features of Mutation:

Mutations occur from time to time among the individuals of a naturally breeding species.
The mutant organisms are markedly different from their parents.
Mutations are large, sudden & are heritable.
Mutations lead to the formation of new species.
Mutations are both harmful & beneficial.
The animals with harmful mutations are destroyed by natural selection and the beneficial mutations get established in the animals.

Question 2.
Describe the structural changes in chromosomal mutation.
Answer:
Chromosomal mutation or Chromosomal aberration or Numerical changes in
chromosomes:
Any change in gene arrangement or insertion of a gene or deletion of a gene then these changes are known as Chromosomal Mutation. They are of two types:

Change in the number of chromosomes
Change in the structure of chromosomes

1. Change in the number of chromosomes.
It is of two types:
(1) Euploidy or Polyploidy:

The total set of a number of chromosomes in cell or organism is called monoploidy. The gaining of one or more complete sets of chromosomes is called euploidy.

This can be of two types –
(a) Autopolyploidy:

It is a type of euploidy, where the additional set of chromosomes is derived from a parent or identical parental species.
For example Triploid, Tetraploid, Pentaploid etc. Colchicine which is isolated from the corn of Colchicum autumnale has an ability to destroy spindle fibres during cell division so that the chromosomes do not move to the respective poles and ultimately chromosome number gets double.

(b) Allopolyploidy:

In this type, more than two haploid sets of chromosomes that are dissimilar and derived from different species are added.
Example: Raphnobraccia (2n = 36). Russian scientist GD. Carpichako(1927) make a cross between Raphanus sativus (Radish) and Brassica oleracea (Cauliflower);. It is sterile. Triticale is a cross product of Triticum (wheat) and Secale (Rai) is also used as man-made cereal “Triticale” and is commercially exploded all over.

(2) Aneuploidy:

Aneuploidy is the addition or loss of one or more chromosomes to the complete diploid chromosome complement of an organism. It is caused by nondisjunction of the chromosome during segregation of chromosomes.

It is of two types –
(a) Hypoploidy:

Hypoploidy is the loss of one or more chromosomes from the diploid genome. If there is a loss of one chromosome (2n – 1), it is called monosomy and if the loss of pair of chromosome (2n – 2), it is called Nullisomy.

(b) Hyperploidy:

Addition of one or more chromosome it is called Hyperploidy.
Excess of one chromosome is called trisomy.
Example: Down’s syndrome or 21st trisomy or Mongolism or Mongoloid Idiocy. (2n + 1)
Excess of two chromosomes is called as tetrasomy (2n + 2).

2. Change in the structure of chromosomes:

These mutations result in structural changes in the chromosome. Hence, the original structure of the gene is changed.
They involve changes either in the total number of genes or gene loci in a chromosome or their rearrangement.

They are of 4 types –
(a) Deletion:

It is the simplest kind of chromosomal aberration.
The deletion is loss of chromosomal segment.
Terminal deletion refers to the loss of a segment from one or the other end of the chromosome.
Intercalary of Interstitial deletion involves an intercalary segment of the chromosome with the reunion of the terminal segment.

(b) Translocation:

It is a type of chromosomal aberration.
A chromosomal segment goes to another chromosome (Unilateral) or normally, exchange of chromosomal segment (genes) between two non-homologous chromosomes (Bilateral).
Sometimes there is a change in the expression of the gene due to Position effect.

Some times the number of genes in a chromosome does not change but the sequence of genes is altered by the rotation of the chromosomal segment within the chromosomes by 180°. Due to these change, genotypes remains unchanged but phenotypic characters change.

(d) Duplication:

Sometimes, some genes are present more than once in a chromosome i.e. one, segment repeats twice in a chromosome. It is called duplication.

Question 3.
Describe the Human Genome Project.
Answer:
Human Genome Project:

The human genome project (HGP) is a project to map and sequence the 3 billion nucleotides contained in the human genome and to identify all the genes present in it.
It is under the control of an international body, the Human Genome Organization (HUGO). This project was started in 1988 by the US Department of Energy (DOE). The DOE and National Institute of Health (NIH) hand in hand started this project in 1990.
250 laboratories from 8 countries were involved in this project James D. Watson was the leader of the project and it was started under the leadership of Francis Collins. This is also called Mega Project & it was completed in 2003.

Aims of the Human Genome Project:

To identify one lakh genes in human DNA.
To determine the base sequence of 3 billion chemical base pairs that make up human DNA.
To store this information in the database.
To improve tools for data analysis.
To address the ethical, legal and social issues that may arise from the project.
To determine the sequence of several modal organisms like E. coli, yeast, Drosophila, rat (Mus musculus)

The technique used in the Human Genome Project:
There are two types of approaches for analyzing the genome.

Identify all the genes that are expressed as RNA – Expressed sequence tags or ESTs.
Sequencing the whole genome (both coding and non-coding regions) and later assigning the different regions with functions – Sequence annotation.

The second methodology was used in the HGP which involves the following steps:

The whole DNA of the cell is isolated and broken randomly into fragments.
They are inserted into specialised vectors like BAC (bacterial artificial chromosomes) and YAC (yeast artificial chromosome).
The fragments are cloned in suitable hosts like bacteria and yeast. PCR (polymerase chain reaction) can also be used for cloning or making copies of DNA fragments.
The fragments are sequenced as annotated DNA sequences (an offshoot of the methodology developed by double Nobel laureate, Frederick Sanger), Sanger used Di-deoxy Chain Termination method through advanced DNA sequences.
The sequences were then arranged on the basis of some overlapping regions. It necessitated the generation of overlapping fragments for sequencing.
For physical and genetic mapping of sequence recognition sites of restriction endonuclease are used which are located on microsatellite of the chromosomes. Sequencing of the overlapping end is done through a special sequence program through the computer.

Salient Features of Human Genome:

The human genome has 3.1657 Billions of nucleotide base pairs.
According to a presumption of the human gene from 80,000 – 140000 but HGP revealed the Number of genes has 30,000.
The average gene size is 3000 base pairs. Dystrophin is the biggest gene, contains 2.4 crores base pairs.
Human Genome project revealed Single Nucleotide polymorphism (SNP). At 104 million places this SNP is found. It helps in identifying sequence for disease on the chromosome.
The function of 50% of genes has been identified.
Less than 2% of genes are involved in protein synthesis.
Chromosome Number 1 has the highest gene number which is 2968 genes and the lowest number of genes are present in the Y chromosome which only 231 genes.
99.9% of the nucleotide bases are exactly similar in all human beings.
Only 0.1% of the human genome with some 3.2 million nucleotides represents the variability observed in human beings.
Repeated or repetitive sequences make up a large portion of the human genome. There are some 30,000 minisatellite loci, each having 11 – 60 bp repeated randomly up to a thousand times. These are about 2,00,000 microsatellites, each with up to 10 basic pair repeated 10 – 100 times.
Repetitive sequences are nucleotide sequences that are repeated many times, sometimes hundred to thousand times. They do not have direct coding function but provide information as to chromosome structure, dynamics and evolution.

Question 4.
Explain in detail about DNA fingerprinting technique and write about its applications.
Answer:
DNA Fingerprinting:

Human is identified by their names and their characteristic. No two organisms have similar characteristics. (except identical twins).
There are certain characteristics such as skin colour, hair colour, the colour of pupil of eye, height, voice, a pattern of walking, sitting, way of talking, way of living through which we can differentiate or identify two individuals.
Every individual differs in fingerprints. There is a difference in imprints and pristine we observe is called fingerprint.
This is actually an official way of identification of an individual. It was developed by Francis-Galton, which is still in use.
Alec Jeffreys & colleagues (1985) were investigating a DNA fragment from the gene for a human blood protein, globin.
They discovered that this fragment contained a sequence of bases repeated several times.
This repeated DNA is called as minisatellite. Further, the same minisatellite sequence was found in other places in the human genome repeated several times.
More interestingly, the individuals differ in the pattern of repeats of the base sequence & they differ enough that two individuals have only a remote chance of having exactly the same pattern.
That means that these patterns are like fingerprints & are called by Jeffreys as DNA fingerprints.

Principle:

Several small nucleotides are found repetitively in DNA. They are called VNTRs (Variable Number of Tandem Repeats).
The length to VNTR of two persons may be of the same length but the sequence will be different.
In every child half of the VNTRs come from mother and other half comes from father.

DNA fingerprinting Technique:

DNA is isolated from the source material like a drop of blood, saliva, semen, teeth etc. by the help of high-speed Refrigerated centrifuge.
If DNA is less in the amount it should be amplified using Polymerase Chain Reaction (PCR).
DNA is cut into fragments using restriction endonuclease enzyme for Restriction Fragment Length Polymorphism (RFLP).
These fragments are poured into separate wells of an electrophoresis gel and electrophoresed. These fragments are stained with fluorescent dyes like Ethidium Bromide and observed under UV light.
Alkaline solutions changed this double-strand DNA into single-strand DNA. This process is known as Denaturation.
The separated DNA fragments are transferred to a nitrocellulose membrane by placing it over the gel. This process is called Southern Blotting.

Application of DNA Fingerprinting:

In Criminal cases and Family affairs:
- DNA fingerprinting is commonly used in forensic cases to undercover key evidence for criminal trials. In crime scenes, bodily fluids such as blood are taken as samples and tested in a lab to help identify the DNA of individuals who might have contributed or been the victim of the crime. For this technique biological wastes such as blood may be dried, semen, hair, tooth, a piece of bone, anybody tissues etc. is obtained at the site of crime.
- This technique is also used in case of disputed parents where DNA of probable parents are tested against child DNA and help in the identification of actual parents.
- This technique is also helpful in Murder, rape, crime, property nomination, divorce cases. Where the identification of parent is very important. Every individual has received DNA from its parents and can be identified by the help of VNTRs.
Medical Field:
Several hereditary diseases can be detected at the early stage of pregnancy. It can be managed at a certain level through DNA fingerprinting.
It is used to assess the pattern of migration of ancient populations.
It is also employed in the genetic analysis of various strains of agricultural crops and animals.
It is also employed to understand evolution.

Question 5.
What do you mean by cloning? Explain how first animal clone was produced?
Answer:
Cloning:

A group of cells or organisms which originate from one parent cell/organism and which are similar physically and genetically to the parent are called a clone. The process of formation of a clone is called cloning. (G. Klon > cutting used for propagation).
In simple organisms, the cloning is found in a natural way. But it is lacking in higher animals.
Identical Twins are also like clones as they are formed by the first division of a single zygote so these two divided cells grow individually and develop into a complete separate embryo.
In plants, tissue culture technique is used to make plants. In tissue culture technique a single cell from the desired plant is cultured in a suitable medium

Process of Cloning of Doly:

Ian Wilmut and coworkers in (1995) prepared Magon and Morgan clones of sheep by nuclear transplantation from sheep embryo into sheep ovum at Roslin Institute, Scotland. In Feb 1997, Ian Wilmut has produced a clone of an adult lamb named “Dolly”.
Further, in July 1997, Ian Wilmut and coworkers prepared sheep clones called Polly and Molly.
These clones were created from a single cell taken from the udder of an adult sheep. The complete procedure is shown below.

Question 6.
Explain in detail changes in chromosomal numbers.
Answer:
Change in the number of chromosomes.
It is of two types:
(1) Euploidy or Polyploidy:

The total set of a number of chromosomes in cell or organism is called monoploidy. The gaining of one or more complete sets of chromosomes is called euploidy.

This can be of two types –
(a) Autopolyploidy:

It is a type of euploidy, where the additional set of chromosomes is derived from a parent or identical parental species.
For example Triploid, Tetraploid, Pentaploid etc. Colchicine which is isolated from the corn of Colchicum autumnale has an ability to destroy spindle fibres during cell division so that the chromosomes do not move to the respective poles and ultimately chromosome number gets double.

(b) Allopolyploidy:

In this type, more than two haploid sets of chromosomes that are dissimilar and derived from different species are added.
Example: Raphnobraccia (2n = 36). Russian scientist GD. Carpichako(1927) make a cross between Raphanus sativus (Radish) and Brassica oleracea (Cauliflower);. It is sterile. Triticale is a cross product of Triticum (wheat) and Secale (Rai) is also used as man-made cereal “Triticale” and is commercially exploded all over.

(2) Aneuploidy:

Aneuploidy is the addition or loss of one or more chromosomes to the complete diploid chromosome complement of an organism. It is caused by nondisjunction of the chromosome during segregation of chromosomes.

It is of two types –
(a) Hypoploidy:

Hypoploidy is the loss of one or more chromosomes from the diploid genome. If there is a loss of one chromosome (2n – 1), it is called monosomy and if the loss of pair of chromosome (2n – 2), it is called Nullisomy.

(b) Hyperploidy:

Addition of one or more chromosome it is called Hyperploidy.
Excess of one chromosome is called trisomy.
Example: Down’s syndrome or 21st trisomy or Mongolism or Mongoloid Idiocy. (2n + 1)
Excess of two chromosomes is called as tetrasomy (2n + 2).

RBSE Solutions for Class 12 Biology

RBSE Solutions for Class 12 Biology Chapter 29 Man-Chemical Coordination

August 9, 2019 by Prasanna Leave a Comment

Rajasthan Board RBSE Class 12 Biology Chapter 29 Man-Chemical Coordination

RBSE Class 12 Biology Chapter 29 Multiple Choice Questions

Question 1.
Secretions of endocrine glands are called:
(a) Pheromone
(b) Enzyme
(c) Hormone
(d) Mucous
Answer:
(c) Hormone

Question 2.
Father of Endocrinology is:
(a) Claude Bernard
(b) Thomas Edison
(c) Balis and Starling
(d) Banting and Best
Answer:
(b) Thomas Edison

Question 3.
Endocrine glands are:
(a) Glands with ducts
(b) Acidic glands
(c) Alkaline glands
(d) Ductless glands
Answer:
(d) Ductless glands

Question 4.
Hyposecretion of which hormone develops:
(a) Thyroxin
(b) Vasopressin
(c) Oxytocin
(d) Calcitonin
Answer:
(b) Vasopressin

Question 5.
A hormone to control calcium and phosphorus metabolism is produced by the organ:
(a) Pancreas
(b) Thymus
(c) Thyroid
(d) Parathyroid
Answer:
(d) Parathyroid

Question 6.
Alfa cells of Islets of Langerhans secrete a hormone which is:
(a) Insulin
(b) Glucagon
(c) Melatonin
(d) Somatostatin
Answer:
(b) Glucagon

Question 7.
The gland which stimulates fear, fight & flight during an emergency is:
(a) Adrenal
(b) Thyroid
(c) Pituitary
(d) Thymus
Answer:
(a) Adrenal

Question 8.
Name of the hormone releases by corpus luteum is:
(a) Androgen
(b) Progesterone
(c) Estrogen
(d) Testosterone
Answer:
(b) Progesterone

Question 9.
In human a disease caused by iodine deficiency is:
(a) Diabetes
(b) Goitre
(c) Sterile
(d) Addison’s disease
Answer:
(b) Goitre

Question 10.
Androgen hormone is secreted by:
(a) Ovaries
(b) Pituitary
(c) Thyroid
(d) Testes
Answer:
(d) Testes

RBSE Class 12 Biology Chapter 29 Very Short Answer Type Questions

Question 1.
Who is the father of endocrinology?
Answer:
Thomas Edison.

Question 2.
Where pituitary gland is located?
Answer:
It is attached below the hypothalamus with the help of an infundibulum.

Question 3.
If there is ADH deficiency, what will effect on the body?
Answer:
Diabetes insipid us.

Question 4.
Who is supreme commander of the endocrine system?
Answer:
Hypothalamus.

Question 5.
Which hormone is secreted by parathyroid?
Answer:
Parathornone (PTH).

Question 6.
Which endocrine gland secretes life-saving hormone?
Answer:
Adrenal Cortex.

Question 7.
Write names of hormone secreted by the Thymus gland.
Answer:
Thymosin.

Question 8.
Write the name of the gland which acts as a sexual biological clock.
Answer:
Pineal body.

Question 9.
Where islets of Langerhan’s are found?
Answer:
Pancreas.

Question 10.
Write the name of anyone hormone secreted by Adrenal cortex.
Answer:
Aldosterone.

Question 11.
Who discovered Seretin hormone?
Answer:
H. Starling.

Question 12.
Which hormone secreted by thyroid gland?
Answer:
T₃ & T₄ (Thyroxine).

Question 13.
Which hormone is secreted by Beta cells of islets of Langerhans?
Answer:
Insulin.

Question 14.
Which disease is caused by the deficiency of Insulin hormone?
Answer:
Diabetes mellitus.

Question 15.
Write the name of hormones secreted by Graafian follicles.
Answer:
Estrogen.

RBSE Class 12 Biology Chapter 29 Short Answer Type Questions

Question 1.
Differentiate endocrine and exocrine glands with examples.
Answer:

Endocrine Gland	Exocrine Gland
They are ductless	They have ducts
Their secretions are called hormones	Not called
Example: Pituitary	Example: Liver

Question 2.
Explain the functions of thyroxin hormone.
Answer:
Functions of Thyroxine hormone:

It increases oxidative metabolism so the rate of energy production increases our speed of life increases.
It increases protein synthesis, gluconeogenesis, the temperature of body and functions of nerves.
It increases the glucose absorption, by intestine oxygen consumption and basic metabolic rate (BIR).
It increases the heartbeat.
It plays a significant role in the metamorphosis of tadpole larva into a frog.
It plays a role in osmotic regulation in cold-blooded vertebrates. It also increases the activity of neurosecretion adrenalin and non-adrenalin.

Question 3.
Write the name of hormones secreted by islets of Langerhan’s.
Answer:

α-cells-Secrete glucagon.
β-cells-Secrete insulin.
δ-cells-Secrete somatostatin.

Question 4.
Write a source of thyroxin hormone.
Answer:
Cells of thyroid follicles.

Question 5.
Write the name of releasing and inhibiting hormones secreted by the hypothalamus.
Answer:

Question 6.
Explain in brief the diseases caused by irregular secretions of Adrenal gland.
Answer:
Diseases:

Addison’s disease: It is caused due to deficiency of mineralocorticoids.
Symptoms: Reduced hunger, vomiting, reduced BMR & body temperature. dehydration & bronzing of the skin.
Conn’s disease: It is caused due to excess secretion of mineralocorticoids.
Symptoms: Muscle weakness, high BP, malfunctioning of kidneys etc.
Cushing’s disease: It is caused due to hypersecretion of corticosteroids. It is caused due to hypersecretion of corticosteroids which is due to excess of ACTH.
Symptoms: Virilism, oedema, hyperglycaemia etc. In virilism, opposite secondary sexual characters appear and it is more common in women. The virilism in a female is also called as hirsutism.

Question 7.
“Goiter disease frequently occurs in the persons living on the hills”. Explain with reasons.
Answer:
The drinking water at high attitude is without dissolved iodine. Moreover, there is a lack of aquatic food at hills which is a good source of iodine.

Question 8.
Write names of hormones secreted by Adenohypophysis.
Answer:

Growth Hormone
Thyroid-stimulating Hormone
Adrenocorticotrophic Hormone
Follicle-stimulating Hormone
Luteinizing Hormone
Prolactin
Melanophore Stimulating Hormone

Question 9.
If thyroidectomy performs in man, what will happen?
Answer:
It will cause hypothyroidism in the body which in turn will result in cretinism or Goitre or Myxaedena.

Question 10.
Write symptoms of Myxedema disease.
Answer:
Puffiness of skin on face & hands, Reduced blood pressure, BMR & heartbeats, increased obesity, Rough & dry skin, Loss of Hair etc.

RBSE Class 12 Biology Chapter 29 Essay Type Questions

Question 1.
Draw a labelled diagram of the pituitary gland and Describe the hormones secreted by
Neurohypophysis.
Answer:
Hormones of Neurohypophysis:

The pars nervosa stores & releases two hormones which are secreted by supraoptic nucleus & paraventricular nucleus viz., oxytocin & vasopressin respectively.
Both the oxytocin & vasopressin are octapeptide & each has 8 amino acids.

1. Oxytocin:

It is found only in females & it is also called as Pitocin.
It helps in parturition by stimulating contraction of the uterine muscles.
It also helps in the release of the milk from the mammary glands after parturition.

2. Vasopressin:

It is also called an Antidiuretic hormone (ADH) or Pitressin.
It controls reabsorption of water in the DCT.
Deficiency of ADH causes diabetes insipid us. Its main symptom is diuresis or polyurea.
In this disease, the amount of urine is increased 15 to 25 litre per day from the normal.

Question 2.
Describe in brief the hormones secreted by the thyroid gland and diseases caused by their irregular secretions.
Answer:
The thyroid secrets two hormones.

Thyroxine
Calcitonin

Disorders Related to Thyroxine Hormone:
(1) Hypothyroidism:

Hypothyroidism deficiency of thyroid hormone may be due to malfunctioning of the thyroid gland or deficiency of iodine in the diet.
It leads to the following diseases.

1. Cretinism:

It occurs in children due to deficiency of thyroxine. It leads to dwarfism with serious mental deficiencies. Such individuals are called Cretins.
Retarded body growth – Child looks as pot-bellied, protruding tongue etc. Blood pressure, heart rate and Basal metabolic rate (BMR) are below normal. Such individuals may be sterile.

2. Myxedema:

It occurs in adults due to deficiency of thyroxine. Puffiness of skin on the face and hands.
The person becomes dull, lack intelligence and alertness, lack of memory.
Blood pressure, heart rate and BMR are below normal. Skin becomes yellowish and decreases in reproductive ability. It can be cured by treating the patient with thyroxin tablets orally.

3. Goitre:

This disease occurs due to a lack of iodine in the food. The thyroid gland enlarges and neck of such patient becomes swollen. Less production of thyroxine due to unavailability of iodine, which is a major constituent of thyroxine.
This disease is more prevalent in people living at high altitude as their water lacks in iodine.

4. Hashimoto’s Disease:

It is caused due to the formation of antibodies against the medicines given to cure diseases caused by hyposecretion of the thyroid. These antibodies destroy the thyroid gland. It is also called as Suicide of thyroid gland. It is an autoimmune disorder.

(2) Hyperthyroidism:

Increases activity of thyroid gland produces an excess of hormone in the body which increases the rate of metabolic activities, rate of heartbeat change in behaviour more sweat etc. It also causes oscillations in hands and feet. Hypersecretion of thyroxin results in the following diseases.

1. Exophthalmic Goiter:

The over secretion of the gland produces swelling in the neck region, but in this disease, goitre is accompanied by bulging of eyeballs and vision of such a person looks very fearing and eyes look staring. This disease is called Grave’s disease.

2. Plumer’s disease:

At various places formation of nodes/tumour in the thyroid gland is called more plum’s disease.

Question 3.
Describe briefly the various hormones secreted by Adrenal gland.
Answer:
Adrenal cortex:
Structure:

It is the outer & light yellow coloured part which originates from the mesoderm.
Its weight is about 4 gm & it forms 80 to 85% of the total gland.
It secretes about many hormones which are collectively called as corticosteroids. They are under the control of ACTH.
They are also called as life-saving hormones.

The adrenal cortex has 3 parts:
1. Zona glomerulus:

It is the outer part of the cortex which consists of small & cuboidal cells.
It secretes mineralocorticoids eg. Aldosterone. These hormones are called as salt retaining hormones.
These hormones maintain a definite quantity of water, Na⁺, K⁺ & Cl⁺ in the ECF. These hormones stimulate the absorption of Na⁺¹ & Cl^– & excretion of K⁺ in the kidneys.

2. Zona fasciculate:

It is the middle part of the cortex which consists of large & polyhedral cells.
This part secretes glucocorticoids eg. cortisol, corticosterone etc.
These hormones are used to treat allergy & help in organ transplantation.
These hormones act as anti-inflammatory & help in the treatment of arthritis.
These hormones stimulate glycogenesis & gluconeogenesis.

3. Zona reticularis:

It is the inner zone of the cortex which has cords of polyhedral cells. It also has modified glandular cells.
This part secretes androgen, estrogen & progesterone in small quantity. These hormones affect sexual behaviour, hair growth, bones & muscles.
The androgen is secreted in the form of dihydroxy-epi-androsterone.

Diseases:

Addison’s disease: It is caused due to deficiency of mineralocorticoids.
Symptoms: Reduced hunger, vomiting, reduced BMR & body temperature. dehydration & bronzing of the skin.
Conn’s disease: It is caused due to excess secretion of mineralocorticoids.
Symptoms: Muscle weakness, high BP, malfunctioning of kidneys etc.
Cushing’s disease: It is caused due to hypersecretion of corticosteroids. It is caused due to hypersecretion of corticosteroids which is due to excess of ACTH.
Symptoms: Virilism, oedema, hyperglycaemia etc. In virilism, opposite secondary sexual characters appear and it is more common in women. The virilism in a female is also called as hirsutism.

Medulla:

It is the core part of the adrenal gland which is light brown in colour.
It originates from neural ectoderm.
This part is not controlled by any pituitary hormone.
Its chromaffin cells secrete two hormones viz.,
1. Adrenalin or Epinephrin
2. Nor-adrenalin or Nor-epinephrin.
These hormones are collectively called as Catecholamines (Amine acid derivatives)
Their molecular structure was studied by Axelrod.
These hormones are secreted in the 4:1 ratio.
They are also called as 3F hormones (Fear, Fight & Flight) and emergency hormones.
They also provide a chemical defence mechanism by protecting the body from external & internal dangers.
These hormones increase BP, heartbeats, breathing rate, BMR, blood glucose, lipolysis, glycogenolysis etc.
They act as neurotransmitters.

Question 4.
Which hormones are secreted by ovaries? Describe any two of them.
Answer:
Ovaries:

They are female gonads but also act as an endocrine gland. They secrete three types of hormones.
Its follicular cells & corpus luteum perform an endocrine function.
In woman, the ovaries activate as an age of 12 years. It is called as beginning of puberty.
The follicular cells under the control of FSH secrete estrogen hormone which is also called as estradiol or feminizing hormone.
It is a steroid hormone and performs the following functions:
It controls oogenesis
It provides secondary sexual characters
It controls menstrual and oestrous cycles.
It stimulates the development of mammary glands.
It controls female sexual accessory organs.
It controls sexual behaviour
The corpus luteum of the ovaries secrete progesterone under the control of LH. The progesterone is a steroid hormone which is also called a pregnancy hormone. It performs the following functions:
It prepares the uterus for embryonic development. The uterine wall becomes more muscular, glandular & vascular.
It helps in implantation of the embryo.
It inhibits oxytocin & ovulation during pregnancy.
In woman, the corpus luteum also secretes relaxin hormone at the time of completion of embryonic development. This hormone helps in parturition by relaxing pubic symphysis.

RBSE Solutions for Class 12 Biology

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RBSE Solutions for Class 12 Biology

Rajasthan Board RBSE Class 12 Biology Chapter 28 Man-Reproductive System

RBSE Class 12 Biology Chapter 28 Multiple Choice Questions

RBSE Class 12 Biology Chapter 28 Very Short Answer Type Questions

RBSE Class 12 Biology Chapter 28 Short Answer Type Questions

RBSE Class 12 Biology Chapter 28 Essay Type Questions

Rajasthan Board RBSE Class 12 Biology Chapter 36 Man-Chromosomal Aberrations

RBSE Class 12 Biology Chapter 36 Multiple Choice Questions

RBSE Class 12 Biology Chapter 36 Very Short Answer Questions

RBSE Class 12 Biology Chapter 36 Short Answer Questions

RBSE Class 12 Biology Chapter 36 Essay Type Questions

Rajasthan Board RBSE Class 12 Biology Chapter 35 Mendel’s Law of Inheritance

RBSE Class 12 Biology Chapter 35 Multiple Choice Questions

RBSE Class 12 Biology Chapter 35 Very Short Answer Type Questions

RBSE Class 12 Biology Chapter 35 Short Answer Type Questions

RBSE Class 12 Biology Chapter 35 Essay Type Questions

Rajasthan Board RBSE Class 12 Biology Chapter 37 Mutations

RBSE Class 12 Biology Chapter 37 Multiple Choice Questions

RBSE Class 12 Biology Chapter 37 Very Short Answer Type Questions

RBSE Class 12 Biology Chapter 37 Short Answer Type Questions

RBSE Class 12 Biology Chapter 37 Essay Type Questions

Rajasthan Board RBSE Class 12 Biology Chapter 29 Man-Chemical Coordination

RBSE Class 12 Biology Chapter 29 Multiple Choice Questions

RBSE Class 12 Biology Chapter 29 Very Short Answer Type Questions

RBSE Class 12 Biology Chapter 29 Short Answer Type Questions

RBSE Class 12 Biology Chapter 29 Essay Type Questions

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