Biology

Amniocentesis is a process in which Medical Termination of Pregnancy
amniotic fluid containing foetal cells is collected (Amendment) Act 2017 under section 3 of the
using a hollow needle inserted into the uterus MTP Act 1971 was enacted by Government
under ultrasound guidance. The chromosomes of India. The intention of MTP Act is to
are studied to see the abnormalities in the reduce the incidence of illegal abortion and
developing foetus. But the dangerous trend consequent maternal mortality. As per the
is the misuse of amniocentesis to determine provisions of the MTP Act, only the consent of
the sex of the unborn child. Frequently, if the woman whose pregnancy is being terminated
foetus is found to be female, it is aborted which is required. According to MTP Act pregnancy
is totally illegal. So the Government of India may be terminated : 1. Within first 12 weeks
has legalised MTP Act in 1971, with strict 2. More than 12 weeks but lesser than 20
conditions to avoid its misuse. weeks. The registered medical practitioner’s
opinion is mandatory stating the continuation
Amniocentesis : Used to extract foetal cells for genetic analysis. of the pregnancy would involve a risk to the
life of the pregnant woman or grave abnormal
1. Ultrasound used to determine the physical or mental health or is substantial risk
position of the foetus in the uterus to the child.

Placenta 2. Needle inserted through the Internet my friend
abdominal and uterine wall
What are the effects of alcohol drinking
3. Amnionic fluid containing and smoking on foetus in pregnant women?
foetal cells extracted
2.13 Sexually Transmitted Diseases (STDs) :
4. Centrifuge of Diseases or infections which are

extracted fluid 5. Cells used in karyotype transmitted through sexual intercourse are
and foetal cells collectively called Sexually Transmitted
Diseases (STDs) or Venereal Diseases (VDs) or
Amnionic Karyotype Reproductive Tract Infections (RTI). The major
fluid venereal diseases are syphilis and gonorrhoea.

Risks associated with

amniocentesis
1. Miscarriage
2. Needle injury to foetus
3. Leaking amniotic fluid
4. Infection

Fig. 2.23 : Amniocentesis process

Always Remember

Act of MTP : The Medical Termination Internet my friend
of Pregnancy (MTP) Act 1971 provides the
legal framework for MTP Act 2017. Collect information about other sexually
transmitted diseases.

Activity :

In a sonography clinic, we observe a
board saying ‘Sex selection and detection is
NOT done in this centre and is punishable
under PC-PNDT Act;

Find out what is PC-PNDT Act. Why
do you think such a mandate is essential?

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Table 2.24 : Sexually Transmitted Diseases (STDs)

Name of Disease Syphilis Gonorrhoea
Causative agent Treponema pallidum (Bacteria) Neisseria gonorrhoeae (Bacteria)

Incubation 3-4 weeks Male – 2 to 14 days
period Female – 7 to 21 days
Infection site
Symptoms Mucous membrane in genital, Mucous membrane of urino-genital tract,

Preventive rectal and oral region. rectum, throat and eye.
measures
Treatment Primary lesion called chancre at In male, partial blockage of urethra and

the site of infection. Chancre is reproductive ducts, pus from penis, pain

formed on external genitalia, skin and burning sensation during urination,

rashes and mild fever, inflamed arthritis, etc.

joints, loss of hair. Paralysis, In female, pelvic inflammation of urinary

Degenerative changes occur in the tract, sterility, arthritis, the children born

heart and brain. to affected mother suffer from gonococcal

opthalmia and gonococcal vulvovaginitis

of girls before puberty.

Education about sex practices, Sex hygiene, using condom during coitus,
sex hygiene, avoiding sex with avoiding sex with unknown partner or
unknown partner or multipartners, multipartners.
using condom during coitus.

Antibiotic-Penicillin Antibiotic-Cefixime

2.14 Infertility : IVF (In Vitro Fertization) :
Infertility is defined as the inability It is a process of fertization where an

to conceive naturally after (one year of) egg is combined with sperm outside the body
regular unprotected intercourse. The causes in a test tube or glass plate to form a zygote
of infertility could be physical, congenital, under simulated conditions in the laboratory.
diseases, immunological or even psychological. The zygote or early embryos (with up to 8
The common physical causes in females are blastomeres) could be then transferred into the
polycystic ovary syndrome (PCOS), hormonal fallopian tube for further development.
imbalance, endometriosis while in male, it is
less sperm count and small size of penis. ZIFT (Zygote Intrafallopian Transfer) :
ZIFT is an infertility treatment used
Prior to 1978, infertile couple had two
options, adopt or be childless. Today infertile when there is a blockage in the fallopian tubes
couples have many options to have a child such which prevents the fertilization of egg by the
as fertility drugs, test tube babies, artificial sperm.
insemination, IUI, surrogate motherhood, etc.
The couple could be assisted to have child / In this method, egg is removed from
children through certain special techniques woman’s ovary. Fertilization of the egg with
commonly known as Assisted Reproductive sperms is brought about outside the body under
Technologies (ART). sterile conditions to form zygote by the process
called in vitro fertilization (IVF). The zygote
is then transferred to fallopian tube for further
development.

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Try This Sperm bank / Semen bank:
A sperm bank or semen bank is a place
IVF centres : Make a list of IVF centres
in Maharashtra. which collects, stores and provides human
sperms / semen. The semen is provided by
GIFT (Gamete Intrafallopian Transfer) : healthy males called sperm donors. The sperms
Transfer of an ovum collected from a are stored in sperm bank by cryopreservation
method (at low temperature).
donor into the fallopian tube of another female
who can provide suitable environment for its Can you recall?
fertilization and development.
Surrogate mother :
This technique called gamete intrafallopian Some women have problem in
transfer (GIFT) has been developed for the
cases in which only the entrance to the oviducts implantation of embryo in uterus. Such
or the upper segment of the oviducts in blocked. woman can take help of the modern remedial
In this procedure ova and sperms are directly technique called surrogacy. In this, embryo
injected into regions of the oviduct, where is implanted in surrogate mother, who is not
fertilization produces a blastocyst, which the biological mother.
enters the uterus via the normal route. GIFT
has a success rate of about 30 percent. Adoption :
Adoption is a legal process by which
ICSI (Intra Cytoplasmic Sperm Injection) :
ICSI is an in vitro fertilization procedure a couple or a single parent gets legal
rights, privileges and responsibilities that
in which a single sperm cell is injected directly are associated to a biological child for the
into cytoplasm of an ovum in the laboratory. upbringing of the adopted child.
Here the sperm has to naturally penetrate the
egg. An adoptive parent should be medically fit
and financially able to take care of the adopted
Artificial Insemination (AI) : child. A person wishing to adopt a child must
In some infertility cases, the male partner be at least 21 years old but there is no legal
upper age limit for adoption.
is unable to inseminate the female due to a very
low sperm count. This problem can be solved Always Remember
by artificial insemination. In this technique,
the sperms are collected from the male and Tobacco, marijuana and other drugs
artificially introduced into the cervix of female, smoking may cause infertility in both men
for the purpose of achieving a pregnancy and women. Nicotine blocks the production
through in vivo fertilization (inside the body). of sperm and decreases the size of testicles.
Alcoholism by men interferes with the
IUI (Intra Uterine Insemination) : synthesis of testosterone and has an impact
In this technique the process is somewhat on sperm count. Use of cocaine or marijuana
may temporarily reduce the number and
like that of artificial insemination, the only quality of sperm.
difference is that the sperms are introduced into
the uterine cavity instead of cervix.

44

Always Remember Always Remember

• Jayesh a young, married man of 26 yrs is 1. Cells of trophoblast do not form any part
suffering from T. B. for the last 2 years. of the embryo proper.
He and his wife are desirous of a child
but unable to have one, what could be 2. They form ectoderm of the chorion
the possible reason? Explain. (extra embryonic membrane).

• Neeta is 45 years old and the doctor 3. They play important role in formation of
has advised her not to go for such a late placenta.
pregnancy. She however wants to be the
biological mother of a child, without
herself getting pregnant. is this possible
and how?

Activity :
1. Prepare concept map on information of male reproductive system.

2. Prepare concept map on information of female reproductive system.

45

Activity :
3. Prepare concept map on information of menstrual cycle.

4. Prepare concept maps on information of gametogenesis.

Spermatogenesis Oogenesis

5. Prepare concept map on information of fertilization.

46

Exercise

Q. 1 Multiple choice questions. 8. Presence of beard in boys is a ............
1. The number of nuclei present in a zygote a. primary sex organ
is …… b. secondary sexual character
a. two b.one c. four c.eight c. secondary sex organ
d. primary sexual character
2. Which of these is the male reproductive
Q. 2 Answer in one sentence.
organ in human? 1. What is the difference between a foetus
and an embryo?
a. sperm b. seminal fluid 2. Outline the path of sperm upto the
urethra.
c. testes d. ovary 3. Which glands contribute fluids to the
semen?
3. Attachment of embryo to the wall of the 4. Name the endocrine glands involved in
maintaining the sex characteristics of
uterus is known as…….. males.
5. Where does fertilization and
a. fertilization b. gestation implantation occur?
6. Enlist the external genital organs in
c. cleavage d. implantation female.
7. Give two differences between blastula
4. Rupturing of follicles and discharge of and gastrula.
8. What is the difference between embryo
ova is known as .............. and zygote?

a. capacitation b. gestation Q. 3 Fill in the blanks:
1. The primary sex organ in human male
c. ovulation d. copulation is ……….

5. In human female, the fertilized egg gets 2. The………. is also called the womb.
implanted in uterus ......................
a. After about7 days of fertilization 3. Sperm fertilizes ovum in the
b. After about 30 days of fertilization ………….. of fallopian tube.
c. After about two months of fertilization
d. After about 3 weeks of fertilization 4. The disc like structure which helps in
the transfer of substances to and from
6. Test tube baby technique is called......... the foetus’s body is called…………
a. In vivo fertilization
b. In situ fertilization 5. Gonorrhoea is caused by ……………
c. In vitro fertilization bacteria.
d. Artificial insemination
6. The hormone produced by the testis is
7. The given figure shows a human sperm. ………………..
Various parts of it are labelled as A, B,
C, and D .Which labelled part represents
acrosome ?

A
B
C
D

a. B b. C c. D d. A

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Q. 4 Short answer questions. Q. 6 Long answer questions.
1. Write a note on budding in Hydra.
2. Explain the different methods of 1. Explain the following parts of male
reproduction occuring in sponges. reproductive system along with
3. Write a note on IVF labelled diagram showing these parts-
4. Comment on any two mechanical Testis, vasa deferentia, epididymis,
contraceptive methods. seminal vesicle, proastate gland and
5. Write a note on tubectomy penis.
6. Give the name of causal organism of
syphilis and write on its symptoms. 2. Describe female reproductive system
7. What is colostrum? of human.

Q. 5 Answer the following questions. 3. Describe the process of fertilization.
1. Describe the phases of menstrual cycle
and their hormonal control. 4. Explain the process by which zygote
2. Explain the steps of parturition. divides and redivides to form the
3. Explain the histological structure of testis. morula.
4. Describe the structure of blastula.
5. Explain the histological structure of ovary Project :
in human. Prepare a chart showing information about
6. Describe the various methods of birth other STDs, mentioning causal organisms,
control to avoid pregnancy. symptoms and control measures.
7. What are the goals of RCH programme.
8. Which hormones are involved in
parturition?
9. Which as the function of male accessory
glands?

10. What is capacitation? Give it’s importance.

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3 Inheritance and Variation

Can you recall? studied individually one at a time or in
combination of two or three character at a
1. Is there a similarity between the parents time. He processed the data mathematically
and offsprings? and statistically.
• Mendel postulated the principles of heredity
2. What are the roles of chromosomes in which then became the fundamental laws
living organisms? of heredity, as proposed by Correns (1900).
He visualized that the traits as such are
3. How are hybrid seeds produced?
4. Which are the chromosomal disorders?

3.1 Chromosomes and Mechanism of not inherited physically but by ‘something’
inheritance. present inside the gametic cell. To this
• The transmission of genetic information ‘something’, he coined term ‘factors’ that
are responsible for expression of a particular
from generation to generation is known as trait/ character. He proposed that factors
heredity or inheritance. The mechanism are particulate in nature. The Mendelian
of inheritance was successfuly investigated factors are now termed as ‘genes’. These
before chromosomes had been observed or factors occur in pairs in the parents and
genes were known. segregate from each other during gamete
• Gregor Mendel, son of the peasant farmer, formation without blending/ mixing.
was born in Moravia in 1822. Gregor

Mendel first gave the accurate explanation Reasons for Mendel’s Success :
for the mechanism of inheritance by using • His experiments were carefully planned
hybridization technique.
and involved large sample.
• Inheritance of the seven traits in garden • He carefully recorded the number of plants
pea (Refer the diagram below) plant were
of each type and expressed his results as

ratios.

1. Do you know ?

Seven pairs of contransting visible characters in pea plant (Pisum sativum)

Pea Plant Traits

Seed shape Seed color Pod shape Pod color Flower Flower Plant height
Green color location Tall
Round Yellow Inflated
Purple Axial

Dominant

Wrinkled Green Constricted Yellow White Terminal Short
(Dwarf)
Recessive

49

• In the pea plant, contrasting characters can Phenotype : The external apperance of an
be easily recognized. individual for any trait is called phenotype for
that trait. It is observable and is determined by
• The seven different charaters in pea different combinations of alleles. e.g. In pea,
plant were controlled by a single factor for the height of stem (plant) tall and dwarf are
each. The factors are located on separate the two phenotypes (Tall is determined by TT
chromosomes and these factors are or Tt and dwarf by tt).
transmitted from generation to generation. Genotype : Genetic constitution or genetic
make up of an organism with respect to a
• He introduced the concepts of dominance particular trait. It is representation of the genetic
and recessiveness. constitution of an individual with respect to
Before learning about Mendel’s a single character or a set of characters. e.g.
pea tall plants can have genotype TT or Tt and
experiments let us get acquainted with genetic dwarf has tt.
terms and symbols. Homozygous (pure) : An individual possessing
identical allels for a particular trait, is called
3.2 Genetic Terminology : homozygous or pure for that trait.
Character : It is a specific feature of an
organism e.g. height of stem. Homozygous breeds true to the trait and
produces only one type of gametes e.g., tall
Trait : An inherited character and its detectable with TT and dwarf with tt.
variant e.g. Tall or dwarf. Heterozygous : An individual possessing
contrasting alleles for a particular trait, is
Factor : It is a unit of heredity, a particle called heterozygous. Heterozygous does not
present in the organism which is responsible for breed true for that trait and produces two types
the inheritance and expression of a character. of gametes e.g. F1 generation hybrids (Tt).
(factor is passed from one generation to the Heterozygous individual is also called hybrid.
next through gametes). Factor determines a Pure line : An individual or a group of
genetical (biological) character of an organism. individuals (population) which is homozygous
or true breeding for one or more traits,
Gene : It is a particular segment of DNA which constitutes pure line i.e. plant which breeds true
is responsible for the inheritance and expression for a particular character. It is a descendent of
of that character. a single homozygous parent produced after self
Alleles or Allelomorphs : The two or more fertilization.
alternative forms of a given gene (factor) Monohybrid : It is heterozygous for one trait
are called alleles of each other. They occupy and is produced from a cross between two pure
identical loci (positions) on homologous parents differing in single pair of contrasting
chromosomes. Allele is a short form of characters e.g. Hybrid tall produced in a cross
Allelomorph. between pure tall and pure dwarf parents. It is a
Dominant : It is an allele that expresses its trait heterozygote for a single pair of alleles.
even in the presence of an alternative allele i.e. F1 generation : It refers to the first filial
in heterozygous condition only. Alternatively, generation. It consists of all off-springs
the allele that expresses in F1 is called produced from a parental cross. Alternatively, it
dominant. (It is an allele of a pair that masks is first generation from a given mating between
the expression of other allele in F1 generation.) pure parents having contrasting characters.
Recessive : This allele is not expressed
in the presence of an alternative allele (in
heterozygous condition). It expresses only in
the presence of another identical allele. It is an
allele that does not express in F1 hybrid.

50

F2 generation : The second generation Activity :
(progeny) produced by selfing (inbreeding) of
F1 generation offsprings is called second filial Complete the following chart :
generation. e.g. Progeny produced from a cross
between two F1 individuals (e.g. Tt × Tt). Phenotype Tall X Dwarf
Punnett square/checker board : It is a
probability table representing different Genotype TT
permutations and combination of fertilization
between gametes of the opposite mating types. Gametes T t
In short, it is a diagrammatic representation of
a particular cross to predict the progeny of a First filial Tt
cross. Tall
Homologous Chromosomes : The Generation (F1) Tall × Tall
morphologically, genetically and structurally Selfing of F1 Tt Tt
essentially identical chromosomes present Genotype
in a diploid cell, are called homologous
chromosomes. Such chromosomes synapse Gametes
during meiosis.
Second filial Male T t
Back cross : It is a cross of F1 progeny with Generation (F2) gametes
any of the parents (e.g. F1 tall × pure tall; F1 tall Female
× pure dwarf i.e. Tt × TT/tt). gametes

Test cross : It is a cross of F1 progeny with T
homozygous recessive parent (e.g. F1 tall ×
pure dwarf i.e. Tt × tt). It is used to test the t
homozygous/ heterozygous nature of hybrid. It
is a kind of back cross. Dihybrid cross :
A cross between parents differing in two

heritable traits, is called dihybrid cross e.g.
cross of pure tall, round seeded plant with
dwarf, wrinkled seeded plant. Mendel also
performed the dihybrid cross between pea
plants that differed in two pairs of contrasting
characters.

Phenotypic ratio : It is the ratio of the offsprings Activity :
produced in F2 and subsequent generation with
respect to their physical appearence e.g. 3Tall : Complete the following chart :
1 dwarf, is F2 ‘Phenotypic ratio’ in monohybrid
cross. Parental generation

Genotypic ratio : It is the ratio of the offsprings Phenotype Round, yellow Wrinkled, green
produced in the F2 and subsequent generation seeds X seeds
with respect to their genetic make up e.g. 1 TT
: 2Tt : 1 tt, is F2 genotypic ratio in monohybrid Genotype RRYY rryy
cross.
Monohybrid cross : Gametes RY ry

A cross between parents differeing in First filial RrYy
only one heritable trait is called monohybrid generation Round, yellow seeds
cross. e.g. cross of pure tall and pure dwarf (f1)
plants. Mendel performed the monohybrid Selfing of f1 Round, yellow × Round,
cross between two pea plants with only one
pair of contransting character. Genotype seeds yellow seeds
Gametes
RrYy RrYy

RY, Ry, rY, ry RY, Ry, rY, ry

51

F2 Generation : 2. Law of segregation (Law of purity of
Male gametes) :

gametes This law is based on the fact that the
RY Ry rY ry alleles do not show any blending/ mixing and
both the alleles (characters) are recovered as
Female such in the F2 generation, though one of these
gametes is not seen at the F1 stage. During formation of
gametes, these two alleles (factors) obviously
RY separate or segregate, otherwise recessive type
will not appear in F2.
Ry
The gametes which are formed are
rY always pure for a particular character (trait). A
gamete may carry either dominant or recessive
ry factor but not both. That’s why it is also called
as law of purity of gametes.
Use your brain power
There are 16 possible individuals in F2 Statement of Law of Segregation : The law
generation. Try to find out the phenotypes as states that “When hybrid (F1) forms gametes,
well as the genotypic and phenotypic ratios. the alleles segregate from each other and enter
in different gametes”. The gametes formed are
Can you tell? pure in that they carry only one allele each
Why are farmers and gardeners (either dominant allele or recessive allele).
advised to buy new F1 hybrid seeds every Hence, this law is also described as “Law of
year? purity of gametes”.

3.3 Mendel’s Laws of Inheritance : 3. Law of Independent Assortment :
Mendel proposed three basic postulates on This law is based on dihybrid cross.

the basis of which three laws were formulated. It is basic principle of genetics developed by
These are described below: a Mendel. It describes how different genes
1. Law of Dominance : or alleles present on separate chromosomes
independently separate from each other, during
In monohybrid and dihybrid crosses, the formation of gametes. These alleles are then
phenotypic characters are controlled by discrete randomly united in fertilization. In dihybrid
units, called factors. In a dissimilar pair of cross, F2 phenotypic ratio 9:3:3:1 indicates
factors, one member of the pair dominates that the two pairs of characters behave
(i.e. dominant) over the other (i.e. recessive). independent of each other. It can be concluded
The law of dominance is used to explain that the two characters under consideration are
the expression of only one of the parental assorted independently giving rise to different
characters of a monohybrid cross in F1 and the combinations.
expression of both in F2.
Statement of Law of Dominance : “When Statement of Law of Independent
two homozygous individuals with one or more Assortment: The law states that “When hybrid
sets of contrasting characters are crossed, possessing two (or more) pairs of contrasting
the alleles (characters) that appear in F1 are factors (alleles) forms gametes, the factors in
dominant and those which do not appear in F1 each pair segregate independently of the other
are recessive”. pair”.

52

Try This conclude that plant is heterozygous. Test cross
is also used to introduce useful recessive traits
Find the ratio of dihybrid test cross by in the hybrids of self pollinated plants during
using punnett square. rapid crop improvement programs.

3.4 Back Cross and Test Cross : Following is the graphic representation of
a. Back cross : The F1 individuals obtained in test cross (Fig. 3.1). Recessive parent is crossed
a cross are usually selfed to get the F2 progeny. to find out unknown genotype.
They can also be crossed with one of the two
parents from which they were derived (either 3.5 Deviations from Mendel’s findings :
recessive or dominant). Such a cross is known Few generalizations were arrived at by
as back cross.
Mendel, on the basis of his experiments of
b. Test cross : The cross of F1 hybrid with the garden pea plant- such as,
homozygous recessive parent is known as a test i. Single trait Single gene Two alleles.
cross. It is used to test whether an individual is ii. Two alleles show interaction in which one
homozygous (pure) or heterozygous(hybrid).
Test cross is easy, simple, repeatable and is completely dominant.
predictable. iii. Factors (genes) for different traits

Test cross can be used to find out genotype present on different chromosomes assort
of any plant with dominant expression. But it is independently.
not known whether it is homozygous (pure) or With the passage of time, number of
heterozygous for that trait. For example, A pea deviations were observed/ identified in the
plant having violet (purple) flowers is crossed post-Mendelian era, that gave additional
with a pea plant with white flowers. If all information on the patterns of inheritance.
flowers produced are violet, we can conclude These deviations are then described as
that plant is pure or homozygous and if we get Neo-Mendelism.
violet and white flowers in 1:1 ratio, we can
Internet my friend

Find out the principle involved in
heredity of sheep/ coat colour

Homozygous X Homozygous
recessive recessive

rr rr

rr rr

R RR Rr R
Rr Rr Dominant phenotype Rr Rr
(Genotype unknown)
R r
Rr Rr Result rr rr
All flowers are violet Interpretation
Half of the flowers are violet and
Unknown flower is half of the flowers are white.
homozygous dominant
Unknown flower is heterozygous

Fig. 3.1 : Graphical representation of test cross
53

Do you know ? Parents : red flowers X white flowers
RR rr

The deviations are :

a. Single trait Single gene Gametes : R r

two alleles showing interactions like Rr
Pink flowers
codominance and incomplete dominance. F1 - hybrids :

b. Single trait Single gene more F2 Generation : Selfing of F1
R
than two alleles (multiple allelism).

c. Single trait more than one genes r

(Polygenic inheritance) showing different

epistatic interactions or additive effect. R RR Rr
red pink
d. Single gene influencing many traits
(pleiotropy). r Rr rr
pink white

It was observed that the phenotypic Result :
expression of a gene can be modified or Genotypic ratio - 1RR : 2Rr : 1rr
influenced by the other gene. These gene Phenotypic ratio - 1Red : 2 Pink :1 White
interactions are of two types.
b. Co-dominance :
i. Intragenic interactions : Occur between In co-dominance, both the alleles (genes)
the alleles of same gene e.g. incomplete
dominance and co-dominance. It also of an allelomorphic pair express themselves
occurs between the multiple allele series of equally in F1 hybrids. Such alleles which are able
a gene. to express themselves equally independently in
hybrids, are called co-dominant alleles. Thus
ii. Intergenic (non-allelic) interactions: in co-dominance both alleles are expressed.
Occur between the allels of different-
genes present on the same or different- Phenotype Red X White
chromosomes. e.g. pleiotropy, polygenes, Genotype RR WW
epistasis, supplermetary and complementry Parent
genes, etc. Some of these interactions are generation R W
discussed below : Roan Roan
Gametes RW RW
a. Incomplete dominance:
In the incomplete dominance, both the F1 RW RW
Generation
alleles (genes) of an allelomorphic pair express Gametes
themselves partially. One allele (gene) cannot
supress the expression of the other allele F2
(gene) completely. In such case, there is an Generation
intermediate expression in the F1 hybrid. A
well-known example is the flower colour of Phenotypic RR RW RW WW
Mirabilis jalapa. If a red-flowered (RR) plant ration : Red Roan White
is crossed with a white-flowered (rr) plant, then 2
F1 offsprings have pink (Rr) flowers. Genotypic 1 1
ration : RR : RW : WW
1:2:1

Fig. 3.2 : Representation of co-dominance

in cattle

54

Classic example of co-dominance is coat Table 3.3 : Few phenotypes and genotypes
colour in cattle. There are two types one with
red coat (with red colour hair) and other with in Drosophila
white coat (with white hair). When red cattles
(RR) are crossed with white cattles (WW), F1 Phenotype Genotype
hybrids (RW) are roan.
Normal wings vg+
Roans have the mixture of red and white
colour hair. Thus both the traits are expressed Nicked wings vgni
equally. In F2 generation red (RR), roans (RW)
and white (WW) are produced in the ratio Notched wings vgno
1:2:1. Thus in Co-dominance, the genotypic
and phenotypic ratios are identical. Strap wings vgst

c. Multiple alleles : Vestigial wings vg
More than two alternative forms (alleles)
Curiosity Box
of a gene in a population accupying the same 1. What is qualitative and quantitative
locus on a chromosome or its homologue, are
known as multiple alleles. Multiple alleles arise inheritance?
by mutations of the wild type of gene. A gene 2. Find out the traits of quantitative
can mutate several times producing a series of
alternative expression. Different alleles in a inheritance in humans.
series show dominant- recessive relation or may
show co-dominance or incomplete dominance Another good example of multiple alleles
among themselves. Wild type is dominant over is A, B, O blood grouping in human beings.
all other mutant alleles.
d. Pleiotropy :
In Drosophila, a large number of multiple When a single gene controls two (or more)
alleles are known. e.g. The size of wings from
normal wings to vestigial (no) wings, i.e., just different traits, it is called pleiotropic gene
stumps, is due to one allele (vg) in homozygous and the phenomenon is called pleiotropy
condition. The normal wing is wild type while or pleiotropism. The phenotypic ratio is
vestigial wing is recessive type. 1:2 instead of 3:1 because of the death of
recessive homozygote. The disease, sickle-cell
anaemia, is caused by a gene Hbs. Normal or
healthy gene HbA is dominant. The carriers
(heterozygotes HbA/Hbs) show signs of mild
anaemia as their RBCs become sickle-shaped
i.e. half- moon- shaped only under abnormally
low O2 concentration.

PARENTS Phenotype Sickle-cell carrier × Sickle-cell carrier
GAMETES Genotype
HbA HbS HbA HbS
HbA HbS HbA HbS

Genotype HbA HbA HbA HbS HbA HbS HbS HbS

OFFSPRINGS Phenotypes Normal Sickle-cell carrier Sickle-cell anaemic

1 2 1 (dies)

Fig. 3.4 : Representation of Pleiotropy

55

The homozygotes with recessive gene Hbs Walter Sutton along with Theodor Boveri
however, die of total anaemia. Thus, the gene (1903) studied the parallel behaviour of
for sickle- cell anaemia is lethal in homozygous Mendel’s factors (genes) and behaviour of
condition and produces sickle cell trait in chromosomes, at the time of meiosis.
heterozygous carrier. Two different expressions
are produced by a single gene. Based on these observations, chromosomal
theory of inheritance was put forth by Sutton
A marriage between two carriers will and Boveri. This theory identifies chromosomes
produce normal, carriers and sickle-cell as the carriers of genetic material.
anaemic children in 1:2:1 ratio. Sickle cell
anaemics die leaving carriers and normals in This theory states that the chromosomes
the ratio 1:2. The heterozygotes or carriers can are present in pairs in somatic cells. During
be identified by microscopic examination of gamete formation homologous chromosomes
blood. pair, segregate and assort independently during
meiosis. Thus, each gamete contains only one
Internet my friend chromosome from a pair.

1. Find out the inheritance pattern in blood Nucleus of gametes contains chromosomes,
groups of human beings. which carry all hereditary traits. Male and
female gametes (sperms and eggs) carry all
2. Find out more about pleiotropy-(Sickle- the hereditary traits.They are the link between
cell anaemia) and Polygenic inheritance parents and offsprings. The fusion of haploid
- (human skin colour) male gamete and haplaid female gamete,
restores the diploid number of chromosomes of
3.6 Chromosomal Theory of Inheritance : the species.
Gregor Johann Mendel published his work
Activity :
on inheritance of traits in 1866 but for some
reasons, it remained unnoticed or unrecognised Observe the following diagram and
till 1900, as communication was not easy in those answer the questions given below -
days. His work was not widely recognized. His
approach of using mathematics and statistics A Aa a
to explain biological phenomenon was totally
new and unacceptable to the then biologists. Meiosis I
As continuous variations were observed in
nature, Mendel’s concept of factors (genes) as AA aa
stable and discrete unit which controlled the AA
expression of characters, and that a pair of Meiosis
alleles did not “blend” with each other, was not II
accepted by his contemporaries. He also did
not know the physical location of the ‘factors’ aa
(genes) in the gametic cell.
Questions :
In 1900, three scientists Hugo de Vries, 1. What is homologous chromosome?
Correns and von Tschermak, independently 2. In which phase of meiosis-I, homologous
rediscovered Mendel’s work on the inheritance
of traits. Due to advancements in microcopy, chromosomes segregate?
scientists were able to observe cell division 3. Where are genes located?
and the structure of chromosomes under 4. Do genes and chromosomes have similar
microscope.
behaviour? Justify.

56

Always Remember Function :
Chromosomes are carriers of heredity.
1. Genes and chromosomes occur always in
pairs in diploid organism. Number of chromosomes :
The number of chromosomes is specific and
2. Alleles located on chromosome segregate
along with chromosome during gamete constant for a particular species, therefore it is
formation. of great importance in the study of phylogeny
and taxonomy of the species.
Can you recall?
The term Ploidy speaks for the degree
1. What is chromosome? of repetition of the primary basic number of
2. How many chromosomes are present in chromosomes (i.e. ‘x’) in a cell. When the
chromosome number in a cell is the exact
human somatic and reproductive cell? multiple of the primary basic number, then it is
called euploidy. Euploids include monoploid/
3.7 Chromosomes : haploid (with one set of chromosomes where
Chromosomes are filamentous bodies x=n), diploids (2n-two sets of chromosomes),
triploids (3n-three sets of chromosomes),
present in the eukaryotic nucleus. The term tetraploid (4n-four sets of chromosomes) and
chromosomes (Gr., Chromo = colour, soma so on. When the chromosome number is not the
= body) was coined by W. Waldeyer (1888). exact multiple of the haploid set, it is described
The size of chromosome varies from species to as Aneuploidy. Aneuploidy is either addition
species. Each metaphase chromosome varies or deletion of one or more chromosome (s) to
from 0.1 to 33 mm in length and 0.2 to 2 mm in the total number of chromosomes in a cell (see
thickness. Chromosomes are visible during cell the chart 3.5).
division. They are capable of self replication
and play vital role in heredity, mutation, Structure of chromosome :
variation, and evolutionary devlopment of Chromosomes are best visible under
eukaryotic species. Chemically eukaryotic
chromosomes are made of DNA, histone and microscope, when the cell is at metaphase
non-histone proteins. stage. It is because at this stage chromosomes
are highly condensed. Typical chromosome

Ploidy

Euploidy Aneuploidy

Monoploidy Diploidy Polyploidy
(n) (2n) (3n, 4n, 5n, etc)

Hypoploidy Hyperploidy

Monosomy Nullisomy Trisomy Tetrasomy
(2n+2)
(2n-1) (2n-2) (2n+1)

Chart 3.5 : Variation in chromosome number (ploidy)

57

consists of two chromatides joind together at Depending upon the position of centromere
centromere or primary constriction. Primary there are four types (shapes) of chromosomes
constriction constists of a disk shape plate called viz. Acrocentric (j shaped), Telocentric
kinetochore. It is at the kinetochore, spindle (i shaped), Submetacentric (L shaped)
fibres get attached during cell division. Besides and Metacentric (V shaped). The ends of
primary constriction, some few chromosomes chromosome (i.e. chromatids) are known as
possess additional one or two constrictions telomeres.
called secondary constriction. At secondary Sex Chromosomes :
constriction I, nucleolus becomes organized
during interphase. A satellite body (SAT body) The chromosomes which are responsible
is attached at secondary constriction II, in very for the determination of sex are known as sex
few chromosomes. Each chromatid in turn chromosomes (Allosomes). Human being and
contains a long, unbranched, slender, highly other mammals have X and Y Chromosomes
coiled DNA thread, called Chromonema, as sex chromosomes.
extending through the length of chromatid.
X chromosome is straight, rod like and
Chromatid consists a double stranded longer than Y chromosome. X chromosome
DNA molecule which extends from one end of is metacentric, while Y chromosome is
chromosomes to other. acrocentric. X chromosome has large amount
of euchromatin (extended region) and small
Chromomeres Telomere amount of heterochromatin (highly condensed
Chromonemata region). Euchromatin has large amount of
Secondary DNA material, hence genetically active. Y
constriction (ii) chromosome has small amount of euchromatin
and large amount of heterochromatin, hence it
Centromere Primary is genetically less active or inert. Both X and
Matrix constriction Y chromosome show homologous and non-
(centromere) homologous regions. Homologous regions
Nucleolus show similar genes while non-homologous
Chromonemata regions show dissimilar genes.

Satellite Secondary Non-Homologous regions
constriction (i) or
Centromere
A B nucleolar organizer

A : Parts of chromosomes B : Showing
secondary constrictions and details
Fig. 3.6 : Structure of Chromosome

Activity :

Study the types of chromosome
according to position of centromere. Observe

and complete the following table.

Types of Name of Position of Homologous region

Chromosome Chromosome Contromere Fig. 3.7 : Structure of X and Y chromosomes
(in humans)
Metacentric -
Crossing over occurs only between
-- homologous regions of X and Y chromosomes.
Non-homologous region of X chromosome is
Acrocentric - longer and contains more genes than that of

- At one end

58

non-homologous region of Y chromosome. The number of linkage groups of a particular
X-linked genes are persent on non-homologous species corresponds to its haploid number of
region of X-chromosome while Y linked genes chromosomes. e.g. Drosophila melanogaster
are persent on non-homologous region of has 4 linkage groups that corrrspond to the
Y-chromosome. 4 pairs of chromosomes. Garden pea has 7
linkage groups and 7 pairs of chromosomes.
Can you tell?
Sex-linkage :
1. What are allosomes? The transmission (inheritance) of X - linked
2. Compare X and Y chromosomes.
3. In which region of chromosomes does and Y-linked genes from parents to offspring,
is called sex-linked inheritance. Sex-linked
crossing over take place? inheritance is of three types viz. X-linked,
Y-linked and XY-linked. Sex linkage is of two
3.8 Linkage and Crossing Over : kinds :
Linkage :
a. Complete sex linkage : It is exhibited by
It is a known fact that several genes are genes located on non-homologous regions of
present on the chromosome. As chromosomes X and Y chromosomes. They inherit together
are carriers of heredity, these genes have because crossing over does not occur in this
tendency to be inherited together. Such genes region.
are called linked genes. This tendency of two or
more genes present on the same chromosomes Examples of X-linked traits are
that are inherited together is known as linkage. haemophilia, red-green colour blindness,
Linkage was discovered in plants by Bateson myopia (near sightedness) and for Y-linked are
and Punnett and in animals by T. H. Morgan. hypertrichosis, Ichthyosis, etc.
Linkage is of two kinds - complete and
incomplete linkage: b. Incomplate sex linkage : It is exhibited by
genes located on homologous regions of X and
I. Complete linkage : The linked genes which Y chromosomes. They do not inherit together
are closely located on the chromosome do not because crossing over occurs in this region.
separate (no crossing over) and inherit together. Examples of X-Y linked traits are total colour
They are called completely linked (strongly blindess, nephritis, retinitis pigmentosa, etc.
linked) genes and the phenomenon of their
inheritance is called complete linkage. Thus Crossing Over :
the parental traits are inherited in offsprings. Crossing over is a process that produces
e.g. X chromosome of Drosophila males- show
complete linkage. new combinations (recombinations) of genes by
interchanging and exhanging of corresponding
II. Incomplete linkage : The linked genes segments between non-sister chromatids of
which are distantly located on the same homologous chromosomes. It occurs during
chromosome and have chances of separation pachytene of prophase I of meiosis. The
by crossing over, are known as incompletly term crossing over was coined by Morgan.
linked (weakly linked) genes. The phenomenon The mechanism of crossing over consists
of their inheritance, is called incomplete four sequential steps such as synapsis, tetrad
linkage. Thus, new traits occur in offsprings. formation, crossing over and terminalization.
e.g. In Zea mays - colour and shape of grain This you have already studied in the chapter
show incomplete linkage. on cell division in class XI. The phenomenon
of crossing over is universal and it is necessary
Linkage Groups : for the natural selection, because it increases
All the linked genes in a particular
the chances of variation.
chromosome, constitute a linkage group.

59

Use your brain power How many linkage groups are in human being and maize?

Know the Scientist : • Morgan’s Experiments showing linkage
and crossing over :
Thomas Hunt Morgan
was an American biologist. Morgan used Drosophila melanogaster
He used fruit fly (Drosophila (fruit fly) for his experiments because,
melanogaster) in genetic Drosophila can easily be cultured in laboratory.
reserch and established the It’s life span is short, about two weeks. More
chromosomal theory of over, it has high rate of reproduction.
T. H. Margan heredity. He also discovered
(1866-1945) the principle of linkage, sex Morgan carried out several dihybrid cross
linkage and crossing over. experiments in fruit fly to study genes that are
Margan’s work played key role in the sex-linked. The crosses were similar to dihybrid
field of genetics. He was awarded a Nobel crosses, as carried out by Mendel in Pea.
Prize in 1933, in Physiology and Medicine. For example, Morgan and his group crossed
yellow-bodied, white eyed female to the wild
Cross I type with brown-bodied, red eyed males and
intercrossed their F1 progeny.

Cross II

yw y+ w+ wm w+ m+
Wild type
Parental x x

Yellow, White Wild type wm
White, Miniature

yw yw wm wm

F1 generation y+ w+ Yellow, White w+ m+
Wild type Recombinant Wild type
types (1.3%) White, Miniature
Parental type y+ w
(98.7%) Parental Recombinant
type (62.8%) types (37.2%)
y+ w+
w+ m+ w+ m

Wild type Brown, White Wild type White, Miniature
yw y w+ wm w m+

F2 generation Yellow, White Yellow, Red White, Miniature White, Normal
y+ w+ y+ w
w+ m+ w+ m

yw yw wm wm
Wild type Brown, White Wild type Miniature

yw y w+ wm w m+

yw yw wm wm
Yellow, White Yellow, Red
White, Miniature White, Normal

Fig : 3.8 : Linkage and crossing over

60

He observed that the two genes did not Some characters are influenced by
segregate independently of each other and F2 dominant genes while some other are by
ratio deviated very signiticantly from 9:3:3:1 recessive genes, present on autosomes. For
ratio. example,
• Autosomal dominant traits like Widow’s
Morgan and his group knew that the genes
were located on X chromosome and stated that peak and Huntington’s disease, etc.
when two genes in a dihybrid cross are situated • Autosomal recessive traits like Phenyl
on the same chromosome, then the proportion
of parental combination is much higher than ketonuria (PKU), Cystic fibrosis and Sickle
non-parental type. This occurs due to physical cell anaemia.
association or linkage of the two genes. He also
found that, when genes are grouped on the same a. Widow’s peak :
chromosome, some genes are strongly linked. A prominent “V” shaped hairline on
They show very few recombinations (1.3 %).
When genes are loosely linked i.e. present far forehead is described as widow’s peak. It is
away from each other on chromosome, they determined by autosomal dominant gene.
show more (higher) recombinations (37.2 %). Widow’s peak occurs in homozygous dominant
(WW) and also heterozygous (Ww) individuals.
For example, the genes for yellow body
and white eye were strongly linked and showed Individuals with homozygous recessive
only 1.3 percent recombination (in cross-I). (ww) genotype have a straight hair line (no
White bodied and miniature wings showed widows peak). Both males and females have
37.2 percent recombination (in cross-II). equal chance of inheritance.
Cross I shows crossing over between genes y
and w. Cross II shows crossing over between Widow’s peak No widow’s peak
genes white (w) and miniature wing (m). Here
dominant wild type alleles are represented with
(+) sign.

Always Remember Fig. 3.9 : Widow’s peak and straight hair
line

Parental combinations occur more b. Phenylketonuria (PKU):-
due to linkage and new combinations less due It is an inborn metabolic disorder caused
to crossing over.
due to recessive autosomal genes. When
3.9 Autosomal Inheritance : recessive genes are present in homozygous
Human somatic (2n) cell contains condition, phenylalanine hydroxylase enzyme
is not produced. This enzyme is essential
23 pairs of chromosomes. They can be for conversion of amino acid phenylalanine
divided functionally as autosomes and sex into tyrosine. Due to absence of this enzyme,
chromosomes. A single pair of chromosomes is phenylalanine is not converted into tyrosine.
involved in sex determination and remaining 22 Hence, phenylalanine and its derivatives
pairs are called autosomes. Autosomes control are accumulated in blood and cerebrospinal
a variety of traits other than sex. These traits fluid (CSF). It affects development of brain
are called autosome linked traits. Transmission and causes mental retardation. Excess
of body characters other than the sex linked phenylalanine is excreted in urine, hence this
traits from parents to their offsprings through disease is called phenylketonuria.
autosomes, is called autosomal inheritance.

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Autosomal recessive traits appear in both as hypertrichosis is responsible for excessive
sexes with equal frequency. These traits tend to development of hair on pinna of ear. This
skip generations. charater is transmitted directly from father to
son.
3.10 Sex Linked Inheritance :
Genes located on non-homologous region Internet my friend

of sex chromosomes, are called sex-linked Collect information of Ishihara’s Test for
genes. The traits that are determined by sex colour blindess.
linked genes, are called sex-linked traits.
Colour blindness :
The inheritance of sex linked genes from Colour blindness is X-linked recessive
parents to their offsprings, is called sex linked
inheritance. There are two types of sex-linked disorder where person is unable to distinguish
genes as X-linked genes and Y-linked genes. between red and green colours as both the
colours appear grey. It is caused due to recessive
a. X-linked (sex linked) genes : X-linked genes (XC) which prevents formation
The X linked genes are located on non of colour sensitive cells, the cones, in the retina
of eye.
homologous region of X chromosome and
these gene do not have corresponding alleles The homozygous recessive females (Xc
on Y chromosome. Xc) and hemizygous recessive male (XcY) are
unable to distinguish between red and green
Female has two X chromosomes. In female colours. The frequency of colour blind women
two recessive sex linked genes are required is much less than colour blind men. Dominant
for expression of sex linked traits. If one X X linked gene (XC) is necessary for formation of
chromosome carries a recessive gene for sex- colour sensitive cells in the retina of eye. Thus,
linked trait (defect) its effect is suppressed genotypes of male and female individuals can
by the dominant gene present on other X be represented as follows-
chromosome. The females with one recessive
gene are carriers. The carrier female is Sex Normal Colourblind Carrier
physically normal as she does not suffer from
the disease (disorder). Male XCY XcY

Male has only one X-chromosome. If Female XCXC XcXc XCXc
X chromosome carries X-linked recessive
gene for sex linked trait, then it is expressed The inheritance of colourblindness can
phenotypically, because there is no dominant be studied in the following two types of
gene on Y chromosome to suppress its marriages:-
effect. Therefore, sex-linked / X-linked traits
appear more frequently in males than in the 1. Marriage between colour blind male with
females. Examples of X-linked traits include normal female, will produce normal visioned
haemophilia, colour blindness, night blindness, male and female offspring in F1. The sons have
myopia, muscular dystrophy, etc. normal vision but daughter will be carrier for
the disease.
b. Y-linked (Holandric) genes :
Genes located on non-homologous region

of Y chromosome, are called Y linked genes.
The Y-linked genes are inherited directly from
male to male. In man, the Y-linked genes such

62

Parents: Colour blind X Normal are dominant over the recessive genes for
female haemophilia. The person having recessive gene
male for haemophilia is deficient in clotting factors
XCXC (VIII or IX) in blood. Even minor injuries cause
Genotype XcY continuous bleeding, hence haemophilia is also
called as bleeder’s disease.
Gametes : Xc Y XC XC
The recessive gene for haemophilia is
F1 : XCXc XCXc XCY XCY located on non homologous region of X
chromosome.As there is no corresponding allele
Carrier daughters Normal sons on Y chromosome to suppress its expression, so
( 50% ) ( 50% ) men suffer from this disease. Women suffers
only when both X chromosomes have recessive
Fig. 3.10 : Sex linked inheritance genes (alleles).
(colour blindness)
The genotype of male and female
2. Marriage between carrier female (daughter) individuals can be represented as follow-
and normal male will produce female offsprings
with normal vision but half of them will be Sex Normal Haemophilic Carrier
carriers for the disease. Half of male offsprings
will be normal while remaining half will be Male XHY XhY
colour blind.
Female XHXH XhXh XHXh

Carrier Normal Like colour blindness, haemophilia also
Parents: Daughter X shows criss-cross inheritance. The inheritance
Male of haemophilia can be studied with the help of
following examples -
Genotype XCXc XCY

Gametes : XC Xc XC Y 1. Marriage between the Haemophilic male
and normal female.

F1 : XCXC XCY XCXc XcY Parents: Haemophic Normal
Genotype male X female
Normal
Female XhY XHXH
XH XH
25% Normal Carrier Colourblind Gametes : Xh Y
Male Female Male
25% 25%
25%

Fig. 3.11 : Sex linked inheritance (colour F1 : XHXh XHXh XHY XHY
blindness)
Carrier daughters Normal sons
From above example, it is clear that the ( 50% ) ( 50% )
X linked recessive gene for colour blindness
is inherited from colourblind father to his Fig. 3.12 : Sex linked inheritance
grandson through his daughter. This type of (Haemophilia)
inheritance is called as cris-cross inheritance.

Haemophilia (Bleeder’s disease) :
Haemophilia is X-linked recessive disorder

in which blood fails to clot or coagulates
very slowly. The genes for normal clotting

63

2. Marriage between carrier female (daughter) On the other hand, some species in which
and normal male. the organism has either male or female
reproductive organs, is said to be dioecious or
Carrier Normal unisexual. Humans are dioecious.
Parents: Daughter X
Male German biologist, Henking in 1891,
while studying spermatogenesis of the squash
Genotype XHXh XHY bug (Anasa tristis), noted that 50% of sperms
receive the unpaired chromosomes while other
Gametes : XH Xh XH Y 50% sperm do not receive it. Henking gave
a name to this structure as the x-body but he
F1 : XHXH XHY XHXh XhY could not explain its role in sex determination.
Further investigations by other scientists led
Normal Normal Carrier Colourblind to conclusion that the “x-body” of Henking
Female Male Female Male was infact a chromosome and gave the name
25% 25% ‘X-Chromosome’.
25% 25%
a. Sex Determination in human beings :
Fig. 3.13 : Sex linked inheritance
(Haemophilia)

Do you know ? The chromosomal mechanism of sex
determination in human beings is XX-XY type.
Haemophilia is also referred as “The In human beings, the nucleus of each somatic
royal disease”, because it affected the royal cell contains 46 chromosomes or 23 pairs
families of England, Germany, Russia and of chromosomes. Out of these, 22 pairs are
Spain in the 19th and 20th centuries. Queen autosomes and one pair of sex chromosomes.
Victoria of England, who ruled from 1837-
1901, was belived to have been the carrier of Human female has a pair of XX,
haemophilia. She passed the trait on to her homomorphic sex chromosomes while male
three of nine children. has XY, heteromorphic sex chromosomes.

Thus genotype of :

Female = 44 Autosomes + XX

Find out Male = 44 Autosomes + XY

Aarya shows normal blood clotting During gamete formation in male,
but her mother is haemophilic. Ramesh the diploid germ cells in testis undergo
shows normal blood clotting but his father is spermatogenesis to produce two types of haplaid
haemophilic. If Ramesh and Aarya were to sperms, 50% sperms contain 22 autosomes and
marry, then find out the posible phenotypes of X chromosome while, 50% sperms contain 22
their offsprings. autosomes and Y chromosome.

3.11 Sex Determination : In Female, the diploid germ cells in
The mechanism by which sex is established ovaries undergo oogenesis to produce only one
type of egg. All eggs contain 22 autosomes
is termed as sex determination. The term sex and X chromosome. Thus human male is
refers to sexual phenotype. In some species, heterogametic and female is homogametic.
both male and female reproductive organs are
present in same organism. It is discribed as If sperm containing X chromosome
bisexual or hermaphrodite or monoecious. fertilizes egg (ovum), then diploid zygote is
formed, that grows into a female child. If

64

P generation Female Parents : Male X Female
Male Genotype
2n+ZZ 2n+ZW
Gametes : ↓ ↓ n+W

n+Z n+Z n+Z

XY Meiosis XX F1 : 2n+ZZ 2n+ZW
Gametes X Y XX
Male bird Female bird
Fertilization

F1 generation X Sperm Y Fig. 3.15 : Sex determination in birds
XX XY
Something Interesting
X Male
Eggs Female XY In Bonellia viridis, the environmental
factors determine the sex of individual.
XX The sex of worm Bonellia viridis depends on
which location the Bonellia larva gets settled.
X Male
Female The marine female Bonellia worm has
about 10 cm long body. She has a proboscis
Conclusion : 1:1 sex ratio is produced that can extend over a meter in length. If
a Bonellia larva settles on the seafloor, it
Fig. 3.14 : Sex determination in becomes a female.
human beings
However when, a larva lands on a
sperm containing Y chromosome fertilizes the female’s proboscis and enters the female’s
egg, then diploid zygote is formed that grows mouth, it migrates into her uterus and
into a male child. differentiates into a male. Male lives as
parasite in uterus of female fertilizing her
This indicates that the sex of a child eggs.
depends on the type of sperm fertilizing the
egg and hence the father is responsible for
determination of sex of child and not the
mother. Due to lack of knowledge, women are
often blamed for giving birth to female child.

b. Sex Determination in birds : c. Sex Determination in honey bees :
In birds, the chromosomal mechanism of In honey bees, chromosomal mechanism

sex determination is ZW-ZZ type. In this type of sex determination is haplo-diploid type. In
females are heterogametic and produce two this type, sex of individual is determined by
types of eggs; 50% eggs carry Z- chromosome, the number of set of chromosomes received.
while 50% eggs carry W- chromosome. Females are diploid (2n=32) and males are
haploid (n=16). The female produces haploid
Males are homogametic and produce eggs (n=16) by meiosis and male produces
one type of sperms. Each sperm carries a Z- haplaid sperms (n=16) by mitosis. If the egg
chromosome. Thus sex of individual depends is fertilized by sperm, the zygote develops into
on the kind of egg (ova) fertilized by the sperm.

65

a diploid female (2n=32) (queen and worker) Depending upon which chain of
and unfertilised egg develops into haploid male haemoglobin is affected, thalassemia is
(n=16) (Drone) by way of parthenogenesis. classified as alpha-thalassemia and beta-
thalassemia. It is caused due to deletion or
The diploid female gets differentiated mutation of gene which codes for alpha (a) and
into either worker or queen depending on the beta (b) globin chains that result in abnormal
food they consume during their development. synthesis of haemoglobin. In Thalassemia,
Diploid larvae which get royal jelly as food person shows symptoms like anaemia, pale
develops into queen (fertile female) and other yellow skin, change in size and shape of RBCs,
develops into workers (sterile females). slow growth and development, dark urine, etc.
Massive blood transfusion is needed to these
Parents Female Male patients. Thalassemia differs from sickle-cell
2n=32 n=16 anaemia. The former is a qualitative problem
of synthesising few globin molecule, while the
Meiosis Mitosis latter is a qualitative problem of synthesising
an incorrectly functional globin.
Gametes n=16 n=16 n=16 Down’s Syndrome (21st trisomy) :

without 2n=32 Down’s syndrome is named after the
(Diploid) physician John Langdon Down who first
fertilization Female described this autosomal chromosomal disorder
in 1866.
(parthenogenesis)

F1 n=16
(Haploid)

Male

Fig. 3.16 : Sex determination in honey bee

3.12 Genetic Disorders : Fig. 3.17 : Down’s Syndrome
Genetic Disorders are broadly grouped
This Syndrome is caused due to an extra
into two categories as, Mendelian disorders copy of chromosome number 21st. It shows
and chromosomal disorders, Mendelian presence of three copies of 21st chromosome
discorders are mainly caused due to alteration instead of homologous pair. These individuals
or mutation in the gene. e.g. thalassemia, sickle- will have 47 chromosomes instead of the
cell anaemia, colourblindness, haemophilia, normal number 46. 21st Trisomy occurs due
phenylketonuria, etc. On the other hand, to non-disjuction or failure of separation of
chromosomal disorders are caused due to chromosomes (autosomes) during gamete
absence or excess of one or more chromosomes formation. The incidence of non-disjunction
or their abnormal arrangment. For eg, Down’s is distinctly higher in mothers who are over 45
syndrome, Turner’s syndrome, Klinefelter’s years old.
syndrome etc.
Thalassemia :

Thalassemia is an autosomal, inherited
recessive disease. Haemoglobin molecule is
made of four polypeptide chains- 2 alpha (a)
and 2 beta (b) chains. The synthesis of alpha
chains are controlled by two closely linked
genes (HBA1 and HBA2) on chromosome
16 while the synthesis of beta chain is controlled
by a single gene (HBB) on chromosome 11.

66

These patients have mild or moderate Turner’s Syndrome :
mental retardation and skeletal development (X monosomy / XO females)
is poor. Distinct facial features like small It is sex chromosomal disorder caused
head, ears and mouth, face is typically flat
and rounded with flat nose, open mouth and due to non-disjunction of chromosome during
protruding tongue, eyes slant up and out with gamete formation. Individual born with
internal epicanthal folds, flat hands and stubby Turner’s syndrome has 44 autosomes with XO.
fingers and palm is broad with single palmer They are phenotypically female. They have a
crease. short stature (height) and webbed neck, lower
posterior hair line, broad shield-shaped chest,
Down Syndrome - Trisomy 21 poorly developed ovaries and breast, and low
intelligence.
1 2 34 5 6 78
Klinefelter’s syndrome (XXY males) :
9 10 11 12 13 14 15 16 It is chromosomal disorder caused due to
Trisomy
or extra X chromosome in males. Thus genotype
of individuals is 44 + XXY. They are described
17 18 19 20 21 22 XX XY as feminized males. Extra chromosome is a
result of non-disjunction of X-chromosome
Fig. 3.18 : Karyotype of Down’s syndrome during meiosis. Individual is male and has
over all masculine development. Voice pitch
is harsh and have under developed testis. They
are tall with long arms, feminine development
(development of breast i.e. Gynaecomastia)
and no spermatogenesis, therefore, individuals
are sterile.

Activity :

Study the complementary and supplementary interaction (digenic interactions) -
both in plants and animals

67

Exercise

Q. 1 Multiple choice questions. 8. Find the mis match pair :-
a. Down’s syndrome = 44 + XY
1. Phenotypic ratio of incomplete b. Turner’s syndrome = 44 + XO
c. Klinefelter syndrome = 44 + XXY
dominance in Mirabilis jalapa. d. Super female = 44 + XXX

a. 2 : 1 : 1 b. 1 : 2 : 1

c. 3 : 1 d. 2 : 2

2. In dihybrid cross, F2 generation offsprings 9. A colourblind man marries a woman,
show four different phenotypes while the

genotypes are ................ who is homozygous for normal colour

a. six b. nine vision ,the probability of their son being

c. eight d. sixteen colourblind is –

3. A cross between an individual with a. 0% b. 25%

unknown genotype for a trait with c. 50% d. 100%

recessive plant for that trait is .............. Q. 2 Very Short Answer Questions.

a. back cross b. reciprocal cross 1. Explain the statements :

c. monohybrid cross d. test cross a. Test cross is back cross but back cross

4. When phenotypic and genotypic ratios is not necessarily a test cross.

are the same, then it is an example of b. Law of dominance is not universal.

............ 2. Define the following terms :

a. incomplete dominance a. Dihybrid cross b. Homozygous

b. complete dominance c. Heterozygous d. Test cross

c. Multiple alleles 3. What is allosome?

d. cytoplasmic inheritance 4. What is crossing over?

5. If the centromere is situated near the end 5. Give one example of autosomal recessive

of the chromosome, the chromosome is disorder.

called ................ 6. What are X-linked genes?

a. Metacentric b. Acrocentric 7. What are holandric traits?

c. Sub-Metacentric d. Telocentric 8. Give an example of chromosomal

6. Chromosomal theory of inheritance was disorder caused due to non-disjunction

proposed by ................ of autosomes.

a. Sutton and Boveri 9. Give one example of complete sex

b. Watson and Crick linkage?

c. Miller and Urey Q. 3 Short Answer Questions.

d. Oparin and Halden 1. Enlist seven traits of pea plant selected/
studied by Mendel.
7. If the genes are located in a chromosome
2. Why law of segregation is also called the
as p-q-r-s-t, which of the following gene law of purity of gametes?

pairs will have least probability of being 3. Write a note on pleiotropy.
4. What are the reasons of Mendel success?
inherited together?

a. p and q b. r and s

c. s and t d. p and s

68

5. “Father is responsible for determination Q. 5 Long answer type questions.
of sex of child and not the mother”. 1. What is dihybrid cross? Explain with
Justify. suitable example and checker board
method.
6. What is linkage? How many linkage 2. Explain with suitable example an
groups do occur in human being? independent assotrment.
3. Define test cross and explain its
7. Write note on –PKU significance.
8. Compare - X-chromosome and Y- 4. What is parthenogenesis? Explain
the haplo-diploid method of sex
chromosome. determination in Honey bee.
9. Explain the chromosomal theory of 5. In the answer for inheritance of X-linked
genes, Madhav had shown carrier male.
inheritance. His answer was marked incorrect.
10. Observe the given pedigree chart and Madhav was wondering why his marks
were cut. Explain the reason.
answer the following questions.
6. With the help of neat labelled diagram,
a. Identify whether the trait is sex describe the structure of chromosome.
linked or autosomal.
7. What is cris-cross inheritance? Explain
b. Give an example of a trait in human with suitable example.
beings which shows such a pattern
of inheritance. 8. Describe the different types of
choromosomes.
Q. 4 Match the column-I with column-II
and re-write the matching pairs. Project :
Study the genetic traits like Rolling of tongue
Column-I Column-II or Widow’s peak in your class and write your
own observations.
1. 21 trisomy a. Turner’s syndrome

2. X-monosomy b. Klinefelter’s
syndrome

3. Holandric traits c. Down's syndrome

4. Feminized male d. Hypertrichosis

69

4 Molecular Basis of Inheritance

Can you recall? On the other hand DNA thought to be
small, simple molecule whose composition
1. What is nucleic acid? varied little among species. Over the time,
2. What are the types of nucleic acid? these ideas about DNA were shown to be
3. What are the functions of nucleic acid? wrong. In fact DNA molecules are large and
4. What is the difference between DNA in vary tremendously within and among species.
Variations in the DNA molecules are different
prokaryotes and eukaryotes? than the variation in shape, electrical charge
and function shown by proteins so it is not
4.1 The Discovery of DNA: surprising that most researchers initially
Modern understanding of DNA has evolved favored proteins as the genetic material.

from the discovery of nucleic acid to the Over a period of roughly 25 years (1928-
development of the double-helix model. In 1869, 1952), geneticists became convinced that DNA
Friedrich Miescher began working with white and not protein, was the genetic material. Let
blood cells which are the major component of us study three important contributions that
pus from infections. He collected a lot of pus helped cause this shift of opinion.
from bandages at the local hospital. He used a
salt solution to wash the pus off the bandages. Griffith’s experiments :
When he added a weak alkaline solution to the In 1928, a British medical officer Frederick
cells, the cells lysed and nuclei precipitated
out of the solution. From the cell nuclei, he Griffith performed an experiment on bacterium
isolated a unique chemical substance to Streptococcus pneumoniae that causes
which he called nuclein. Chemically, nuclein pneumonia in humans and other mammals.
has high phosphorus content. Moreover it Griffith used two strains or two genetic varieties
showed acidic properties. Hence it was named of Streptococcus to find a cure for pneumonia,
as nucleic acid. which was a common cause of death at that
time. The two strains used were :
By the early 1900s, we knew that i. Virulent, smooth, pathogenic and
Miescher's nuclein was a mix (mixture) of
proteins and nucleic acids. There are two kinds encapsulated S type.
of nucleic acids. DNA (deoxyribonucleic ii. Non-virulent, rough, non-pathogenic and
acid) and RNA (ribonucleic acid).
non-capsulated R type.
4.2 The Genetic Material is a DNA:
By the early 1900s, geneticist knew that Griffith conducted four experiments on
these bacteria. First, when he injected bacteria
genes control the inheritance of traits, that of strain R to mice, the mice survived because
genes are located on chromosome and that it did not develop pneumonia. Second, when he
chemically chromosomes are mainly composed injected bacteria of strain S to mice, the mice
of DNA and proteins. Initially, most geneticists developed pneumonia and died. In the third
thought that protein are large, complex experiment, he injected heat-killed strain S
molecules and store information needed to bacteria to mice, once again the mice survived.
govern cell metabolism. Hence it was assumed In fourth experiment, he mixed heat-killed
that proteins caused the variations observed S bacteria with live bacteria of strain R and
within species. injected to mice. The mice died and Griffith

70

recovered large numbers of live strain S They also discovered that protein –digesting
enzymes (proteases), RNA-digesting enzyme
bacteria from the blood of the dead mice. (RNAases) did not affect transformation, so the
transforming substance was neither a protein
In these four experiments, something had nor RNA. DNA digested with DNAse did
inhibit the transformation, suggesting that DNA
caused harmless strain R bacterium to change caused the transformation. These experiments
proved that the transforming principle is DNA
into deadly S strain bacterium. Griffith showed but all biologists were not convinced.

that the change was genetic. He suggested Rough nonvirulent
(strain R)
that genetic material from heat-killed strain + Protease

S bacterium had somehow changed the living Mouse dead

strain R bacterium into strain S bacterium.

Griffith concluded that the R-strain

bacterium must have taken up, to what he

called a "transforming principle" from the heat-

killed S bacterium, which allowed R strain to

get transformed into smooth-coated bacterium

and become virulent. + DNase

Rough strain Heat-killed smooth
virulent (strain S)
Rough strain Smooth Heat-killed and Heat-killed

(II-R) Strain (III-S) Smooth Strain Smooth Strain Mouse alive

Fig. 4.2 : DNA transforms bacteria

Finally, Alfred Hershey and Martha Chase
(1952) proved that DNA is the genetic material
and not proteins, by using bacteriophages.

Mouse alive Mouse dead Mouse alive Mouse dead Hershey - Chase Experiment:
Hershey and Chase worked with viruses
Fig. 4.1 : Griffith’s experiment
that infect bacteria i.e. bacteriophages, which
Avery, McCarty and MacLeod’s experiment: are composed of DNA and protein. They used
In 1944, after some 10 years of research radioactive phosphorous 32P in the medium for
some viruses and radioactive sulphur 35S for
and experimentation, U. S. microbiologists some others.
Oswald T. Avery, Colin M. MacLeod and
Maclyn McCarty (all at Rockefeller University They grew some viruses on a medium that
in New York) first evidenced to prove the DNA contained radioactive phosphorus and some
is a genetic material (transforming principle), others on medium that contained radioactive
through the experiments. They purified DNA, sulphur. Viruses grown in the presence of
RNA, Proteins (enzymes) and other materials radioactive phosphorus contained radioactive
from cell free extract of S cells/ strain and mixed DNA (labelled DNA), but not radioactive
with heat killed S strain and R cells seperately proteins because DNA contains phosphorus
to confirm which one could transform living (labelled DNA) but proteins do not. Similarly,
R cells into S cells. Only DNA was able viruses grown on radioactive sulphur contained
to transform harmless strain R into deadly radioactive protein but not radioactive DNA
strain S. because DNA does not contain sulphur.

Radioactive phages were allowed to
infect E.coli bacteria grown on the medium
containing normal ‘P’ and ‘S’. Then, as

71

the infection proceeded, the viral coats Can you recall?
were removed with the help of centrifuge.
Bacteria which were infected by viruses with 1. What are the chemical components of
radioactive DNA, were radioactive, indicating DNA?
that DNA was the material that passed from
the viruses to the bacteria. Bacteria which were 2. What is a chromosome?
infected by viruses having radioactive sulphur
(protein) were not radioactive. This indicates 4.3 DNA packaging :
that proteins from the viruses, did not enter Length of DNA double helix molecule, in
the bacteria. DNA is, therefore, the genetic
material that is passed from virus to bacteria a typical mammalian cell is approximately
(fig. 4.3). 2.2 meters. (This can be worked out simply by
multiplying the total number of base pairs with
In other words, sometime after infection, distance between the consecutive base pairs).
radioactivity for ‘P’ and ‘S’ was tested. Only Approximate size of a typical nucleus is 10-6
radioactive ‘P’ was found inside the bacterial m. How this long DNA molecule can be then
cell, indicating that DNAis the genetic material. accommodated in such a small nucleus? It,
therefore, must be condensed, coiled and super
Bacteriophage Phosphorus coiled to fit inside such small nucleus.
labeled DNA
Sulfur labeled Packaging in Prokaryotes:
protein capsule core In prokaryotes like E. coli, cell size is almost

1. Infection 2-3m long. They do not have well organized
nucleus. It is without nuclear membrane and
2. Blending nucleolus. The nucleoid is small, circular,
highly folded, naked ring of DNA which is
3. Centrifugation 1100m long in perimeter, containing about 4.6
million base pairs.
No sulfur detected in cells Phosphorus detected in cells
+ + The 1100m long (approximately 1.1 mm, if
cut and stretched out) nucleoid is to be fitted
Sulfur detected in No phosphorus or packaged into a cell which is hardly 2-3m
supernatant long. Hence the negatively charged DNA
detected in supernatant becomes circular, reducing the size to 350mm
in diameter. This is further reduced to 30mm
Fig. 4.3 : Hershey - Chase Experiment in diameter because of folding/ looping. 40-50
domains (loops) are formed. Formation of loops
Do you know ? is assisted by RNA connectors. Each domain is
further coiled and supercoiled, thereby reducing
1. Multiple forms of DNA and their the size down to 2m in diameter. This coiling
differences. (packaging) is assisted by positively charged
HU (Histone like DNA binding proteins)
2. Significance of different forms of DNA. proteins and enzymes like DNA gyrase and
DNA topoisomerase I, for maintaining super
coiled state.

72

RNA cleaved

e. Partially unfolded
chromosome

350 m 30 m 2 m Partial
RNA Each loop is DNase
independently digestion
supercoiled
Partial
a. Circular, unfolded b. Folded chromosome c. Supercoiled, Nicked DNA RNase
folded digestion
chromosome actually 40 to 50 loops
chromosome

d. Partially uncoiled chromosome

Fig. 4.4 : DNA Packaging in Prokaryotes

Packaging in Eukaryotes: The negatively charged helical DNA
Eukaryotes show well organized nucleus is wrapped around the positively charged
histone octamer, forming a structure known
containing nuclear membrane, nucleolus as nucleosome. The nucleosome core is made
and thread-like material in the form of up of two molecules of each of four types of
chromosomes. In the chromosomes, DNA is histone proteins viz. H2A, H2B, H3and H4. H1
associated with histone and non-histone proteins protein binds the DNA thread where it enters
as was reported by R. Kornberg in 1974. The (arrives) and leaves the nucleosome.
organization of DNA is much more complex in
eukaryotes. Depending upon the abundance of One nucleosome approximately contains
amino acid residues with charged side chains, 200 base pair long DNA helix wound around
a protein acquires its charge. Histones are the it (fig. 4.5). About 146 base pair long segment
proteins that are rich in lysine and arginine of DNA remains present in each nucleosome.
residues. Both these amino acid residues are Nucleosomes are the repeating units of
basic amino acids and carry positive charges chromatin, which are thread-like, stained
with them. So, histones are a set of positively (coloured) bodies present in nucleus. These look
charged, basic proteins (histones + protamine). like ‘beads-on-string’, when observed under
These histones organize themselves to make a an electron microscope. DNA helix of 200 bps
unit of 8 molecules known as histone octamer. wraps around the histone octamer by 1¾ turns.
Six such nucleosomes get coiled and then form
H2A H4 solenoid that looks like coiled telephone wire.
The chromatin is packed to form a solenoid
H2B H3 H1 Histone structure of 30 nm diameter (300A0) and further
supercoiling tends to form a looped structure
DNA called chromatin fiber, which further coils
Octamer of Histones and condense at metaphase stage to form the
chromosomes. The packaging of chromatin at
Fig. 4.5 : Nucleosome higher levels, need additional set of proteins
that are called Non-Histone Chromosomal
proteins (NHC).

73

Nucleosomes Non-Histone Chromosomal Proteins
(NHC) : These are additional sets of proteins
0.5 mm that contribute to the packaging of chromatin
at a higher level.
Fig. 4.6 : Chromatin showing beads-on- Heterochromatin and Euchromatin :
string 1. Heterochromatin: In eukaryotic cells, some

Nucleosome segments of chromonema/ chromosome
Linker DNA during interphase and early prophase
remain in a condensed state. These region
DNA wound around histone proteins constitute heterochromatin. This term
was proposed by Heitz. These regions are
A : Beads-on-string magnified localized near centromere, telomeres and
are also intercalated. It is genetically mostly
111 Å Nucleosomes inactive. It stains strongly and appears
60 Å dark. Heterochromatin is 2 to 3 times more
Chromatin fibers “stretched” rich in DNA than in the euchromatin.
during preparation for electron
2. Euchromatin: The regions of chromonema
microscopy revealing linker which are in non-condensed state, constitute
DNA between nucleosome cores euchromatin. Euchromatic regions stain
light. Euchromatin is genetically very
much active and fast replicating.
Euchromatin is transcriptionally active,
while heterochromatin is transcriptionally
almost inactive.

Linker DNA, Nucleosome core, Can you recall?

varying in 146 nucleotide pairs 1. What is the backbone of the DNA
structure?
length from 8 to of DNA wrapped as
2. Name the nitrogen bases of DNA.
114 nucleotide ¾ turns around an 3. What are Nucleoside and Nucleotide?
4. Is the double helix right or left handed?
pairs octamer of histones

B : Chain of nucleosomes forming
10 to 11 nm thick fibre

300 Å110 Å fiber Find out

300Å What is the key difference between
metaphase DNA in prokaryotic and eukaryotic cells?

fiber or 4.4 DNA Replication :
“solenoid” The DNA molecule regulates and controls

110 Å all the activities of the cell. Because of its
unique structure, it is able to control the
C : Solenoid forming 30 nm thick fibre synthesis of other molecules of the cell. At
Fig. 4.7 : DNA packaging the same time when the cell reproduces, the
DNA also should duplicate itself to distribute

74

equally to the daughter cells. As a carrier of At the point ‘O’, enzyme endonuclease
genetic information, DNA has to perform two nicks one of the strands of DNA, temporarily.
important functions : The nick occurs in the sugar-phosphate back
bone or the phosphodiester bond.
a. Heterocatalytic function : When DNA
directs the synthesis of chemical molecules 3. Unwinding of DNA molecule:
other than itself, then such functions of
DNA are called heterocatalytic functions. Now enzyme DNA helicase operates by
Eg. Synthesis of RNA (transcription),
synthesis of protein (Translation), etc. breaking weak hydrogen bonds in the vicinity of

b. Autocatalytic function : When DNA ‘O. The strands of DNA separate and unwind.
directs the synthesis of DNA itself, then
such function of DNAis called autocatalytic This unwinding is bidirectional and continues
function. Eg. Replication.
The process by which DNA duplicates as ‘Y’ shaped replication fork. Each separated

itself is called replication. Through replication, strand acts as template.
it forms two copies that are identical to it.
The two separated strands are prevented
In eukaryotic organisms, replication of
DNA takes place only once in the cell cycle. from recoiling (rejoining) by SSBP (Single
It occurs in the S- phase of interphase in the
cell cycle. strand binding proteins). SSB proteins remain

DNA replicates through Semiconservative attached to both the separated strands so as
mode of replication.
to facilate synthesis of new polynucleotide
The model for Semiconservative replication
was proposed by Watson and Crick, on the basis strands. Lagging Okazaki
of antiparallel and complementary nature of template RNA primer fragments
DNA strands. The process of semicoservative Replication
replication is as below: fork 5’
3’
1. Activation of Nucleotides: 3’ 3’La5g(d’giisncgonsttriannodus)
The four types of nucleotides of DNA i.e. 5’
3’
dAMP, dGMP, dCMP and dTMP are present
in the nucleoplasm. They are activated by ATP 5’ Leading Leadi(ncgonsttirnaonuds)
in presence of an enzyme phosphorylase. This template
results in the formation of deoxyribonucleotide
triphosphates i.e. dATP, dGTP, dCTP and dTTP. 3’ A=T 5’ 5’
The process is known as Phosphorylation. 3’

2. Point of Origin or Initiation point: T=A Parental strand
It begins at specific point ‘O’ -origin and
C=G
terminates at point ‘T’. Origin is flanked by
‘T’ sites. The unit of DNA in which replication A=T
occurs, is called replicon. In prokaryotes, there
is noly one replicon however in eukaryotes, G=C
there are several replicons in tandem.
Lagging T=A Leading
template template
A=TT=CA=GA=GT=TC=A 3’
T=AG=CA=TC=GT=AA=T
5’

RNA primer

55’ ’ Okazaki fragments 3’
New

3’ strands 5’
Fig. 4.8 : Semiconservative Replication of

DNA

75

4. Replicating fork: Okazaki). Okazaki fragments are joined by
enzyme DNA ligase.
The point formed due to unwinding and
separation of two strands appear like aY-shaped RNA primers are removed by DNA
fork, called replicating/ replication fork. The polymerase and replaced by DNAsequence with
unwinding of strands imposes strain which is the help of DNA polymerase-I in prokaryotes
relieved by super-helix relaxing enzyme. and DNA polymerase-α in eukaryotes.

5. Synthesis of new strands: Finally, DNA gyrase (topoisomerase)
enzyme forms double helix to form daughter
Each separated strand acts as mould or DNA molecules.
template for the synthesis of new complementary
strand. It begins with the help of a small RNA 7. Formation of daughter DNA molecules:
molecule, called RNA primer. RNA primer get
associated with the 3’ end of template strand At the end of the replication, two daughter
and attracts complementary nucleotides from DNA molecules are formed. In each daughter
surrounding nucleoplasm. These nucleotides DNA, one strand is parental and the other
molecules bind to the complementary one is totally newly synthesized. Thus, 50%
nucleotides on the template strand by forming is contributed by mother DNA. Hence, it is
hydrogen bonds (i.e. A=T or T=A; G = C or described as semiconservative replication.
C = G). The newly bound nucleotides get
interconnected by phosphodiester bonds, Experimental confirmation :
forming a polynucleotide strand. The synthesis
of new complementary strand is catalyzed Semiconservative Replication : In newly
by enzyme DNA polymerase. The new formed DNA molecule, one strand is old
complementary strand is always formed in 5’- (i.e. conserved) and other strand is newly
3’ direction. synthesized. Thus, it is called Semiconservative
mode of replication.
6. Leading and Lagging strand:
It is experimentally proved by Matthew
The template strand with free 3’end is called Meselson and Franklin Stahl (1958) by
leading template and with free 5’ end is called using equilibrium - density - gradient -
lagging template. The process of replication centrifugation technique.
always starts at C-3 end of template strand and
proceeds towards C-5 end. As both the strands 1. Meselson and Stahl in 1958 performed
of the parental DNA are antiparallel, new an experiment to prove semiconservative
strands are always formed in 5’ → 3’ direction. nature (mode) of replication.

One of the newly synthesized strand 2. They cultured bacteria E.coli in the medium
develops continously towards replicating containing 14N (light nitrogen) and obtained
fork is called leading strand. Another new equilibrium density gradient band by using
strand develop discontinuously away from the 6M CsCl2. The position of this band is
replicating fork is called lagging strand. recorded.

Maturation of Okazaki fragments : 3. E. coli cells were then tranferred to 15N
medium (heavy isotopic nitrogen) and
DNA synthesis on lagging template takes allowed to replicate for several generations.
place in the form of small fragments, called At equilibrium point density gradient band
Okazaki fragments (named after scientist was obtained, by using 6M CsCl2. The
position of this band is recorded.

76

Generation I Generation II
15N-DNA 14N-DNA
14N-DNA
15N-DNA 14N15N-DNA
14N15N-DNA
20 min 40 min 14N-DNA

15N 15N 14N 15N 14N 14N 14N 15N

Heavy Hybrid Light Hybrid

Fig. 4.9 : Interpretation of results of Meselson’s experiment on the separation of DNA by

equilibrium density gradient centrifugation

4. The heavy DNA (15N) molecule can 4.5 Protein synthesis :
be distinguished from normal DNA by Proteins are very important biomolecules.
centrifugation in a 6M Cesium chloride
(CsCl2) density gradient. The density They serve as structural components, enzymes
gradient value of 6M CsCl2 and 15N DNA is and hormones. The cell needs to synsthesize
almost same. Therefore, at the equilibrium new protein molecules. The process of
point 15N DNA will form a band. In this protein synthesis includes transcription and
both the strands of DNA are labelled with translation. The process of copying of genetic
15N. information from one (template) strand of DNA
into a single stranded RNA transcript, is termed
5. Such E. coli cells were they transferred to as transcription. During this process, synthesis
another medium containing 14N i.e. normal of complementary strand of RNA takes place
(light) nitrogen. After first generation, (Except that the Adenine nitrogen base pairs
the density gradient band for 14N 15N was with the Uracil base instead of Thymine).
obtained and its position was recorded.
After second generation, two density Central Dogma :
gradient bands were obtained - one at 14N
15N position and other at 14N position. Double stranded DNA molecule gives
rise to mRNA which acts as a messenger to
6. The position of bands after two generations programme the synthesis of a polypeptide
clearly proved that DNA replication is chain (protein). This type of unidirectional
Semiconservative. flow of information from DNA to RNA to
protein/ proteins is referred as central dogma

of molecular biology. It was postulated by

F.H.C. Crick in 1958.

Use your brain power Transcription Translation

1. List as many different enzyme activities DNA mRNA Polypeptide
required during DNA synthesis as you
can. The present concept of central dogma in
retroviruses or riboviruses is given by Temin
2. This type of replication is called semi- (1970) and Baltimore (1970):
conservative replication. Considering
the meaning of these words, why DNA Transcription Translation
replication is called semiconservative DNA mRNA Polypeptide
replication?
Reverse Transcription

77

Accordingly enzyme RNA dependent DNA polymerase catalyses polymerisation in
polymerase, synthesizes DNA from RNA. 5’→3’ direction. So the DNA strand having
3’→5’ polarity acts as template strand. The
Can you recall? other strand of DNA having 5’→3’ polarity
is complementary to template strand. The
1. What is transcription? sequence of bases in this strand, is same
2. How many nucleotides are present in a as in RNA (where Thymine is replaced by
Uracil). It is the actual coding strand. The
codon? information on this strand of DNA is copied
3. Name the molecule which carries on mRNA. This is called sense strand.
iii. The terminator is located at 3’ end of coding
anticodon? strand i.e. downstream. It defines the end of
4. What is mutation? the transcription process.

A. Transcription: 3’ Template strand 5’
During transcription, information of only
5’ Structural gene 3’
one strand of DNA is copied into RNA. This Promoter Coding strand Terminator
strand of DNA acts as template. Enzyme RNA
polymerase catalyses the formation of RNA Fig. 4.10 : Transcription unit
transcript.
After binding to promoter, RNApolymerase
DNA is located in the nucleoid of
Prokaryotes and in nucleus of Eukaryotes. moves along the DNA and causes local
DNA transcription takes place in nucleus
in eukaryotes whereas translation occurs unwinding of DNA duplex into two chains in
in cytoplasm. DNA transfers information
to m-RNA which then moves to ribosomes. the region of the gene. Exposed ATCG bases
Transcription occurs in the nucleus during
G1 and G2 phases of cell cycle. DNA has project into nucleoplasm. Only one strand
promotor and terminator sites. Transcription
starts at promotor site and stops at terminator functions as template (antisense strand) and
site. Actually the process of transcription, in
both Prokaryotes and Eukaryotes, involves the other strand is complementary which is
three stages viz. Initiation, Elongation and
Termination. actiually a coding strand (sense strand). The

Transcription Unit: ribonucleoside tri phosphates join to bases of
Each transcribed segment of DNA is called
DNA template chain. As transcription proceeds,
transcription unit. It consists of i. Promotor,
ii. The structural gene, iii. A terminator. Two the hybrid DNA-RNA molecule dissociates and
strands of DNA in the structural gene show
following features : makes mRNA molecule free.
i. The promotor is located towards 5’ end of
A 5′ 3′ T A 5′
structural gene i.e. upstream. It is a DNA
sequence that provides binding site for G CG
enzyme RNApolymerase. RNApolymerase
binds to specific Promotor. In prokaryotes, U AT
the enzyme recognizes the promotor by its
sigma factor sub unit. C GC
ii. Structureal genes - two strands of DNAhave
opposite polarity. DNA dependent RNA A TA

G CG

U AT

A TA

C GC

A TA

U AT

G 3′ 5′ C G 3′
m-RNA Template strand Coding strand

of DNA of DNA

(antisense strand) (sense strand)

Fig. 4.11 : Formation of Template and

Coding strand of DNA

78

When RNA polymerase reaches the Transcription unit and the gene:
terminator signal on the DNA, it leaves DNA The DNA sequence coding for m-RNA/
and fully formed mRNA (primary transcript) is
released. t-RNA or r- RNA is defined as a gene. Cistron
is a segment of DNA coding for a polypeptide.
As the mRNA grows, the transcribed region A single structural gene in transcription unit
of DNA molecule becomes spirally coiled and is said to be monocistronic where as a long
attains (regains) double helical form. segment of DNAhaving set of various structural
genes in one transcription unit is referred as
In bacteria, m-RNA does not require polycistronic. Structral genes in eukaryotes
any processing because it has no introns. have interrupted non-coding sequences
Prokaryotes posses only one type of RNA (introns). The coding sequences or express-
polymerase. In eukaryotes, there are three sequences are defined are exons. Only exons
types RNA polymerases. RNA polymerase-I appear in procesed mRNA in Eukaryotes.
transcribes r-RNA. RNA polymerase-II
transcribes m-RNA (primary transcript) and Processing of hnRNA:
heterogeneous nuclear RNA (or hnRNA). RNA In eukaryotes, forms of RNA transcribed
polymerase-III is responsible for transcription
of t-RNA and small nuclear RNA (snRNA). from DNA are called primary transcripts.
Such transcripts undergo changes called
Do you know ? processing or maturation before becoming
functional. Primary transcript is non functional
1. Many viruses contain RNA as genetic and contains both exons and introns. During
material and replicate by synthesizing processing only introns are removed by the
first the DNA and then form RNA. This process called splicing.
process is called reverse transcription.
Such viruses are known as Retroviruses. 5’ 5’
3’ 3’
2. e.g. Human immuno deficiency virus
(HIV) is responsible for causing AIDS. Capping Intron 3’ mRNA
Cap Exon RNA splicing Polyadenylation
3. In some cases like E.coli, a chain
terminating protein, the rho factor stops m5G’ PPP
the synthesis of mRNA.
m5G’ PPP 3’
4. The process of transcription as well as Poly A tail
translation involves 3 stages - initiation, m5G’ PPP
elongation and termination. m5G’ PPP 3’

Always Remember Messenger RNA (mRNA)

Fig. 4.12 : Transcription and Processing of
hnRNA to mRNA in Eukaryotes

Types of RNA and process of transcription:
In bacteria, there are three types of RNAs : m- RNA provides the encoded message, t- RNA
brings specific amino acid, to the site of translation. r- RNA plays role in providing binding site
to mRNA.
There is single DNA dependent RNA polymerase that catalyses transcription of all 3 types
of RNA in bacteria. RNA polymerase binds to promotor and initiates transcription (initiation).

79

Exons are joined in a definite sequence there must be combination of three consecutive
(order) by DNA ligase enzyme. Heterogeneous nitrogen bases that will be sufficient to specify
nuclear RNA, undergoes the process of capping 20 different types of amino acids.
and tailing. In capping, methylated guanosine
tri phosphate is added to 5’ end of hnRNA. In Thus, there would be 64 different codons
tailing, polyadenylation take place at 3’end. (code words) in the dictionary of genetic code
It is the fully processed hnRNA, now called and that each code word has to be a triplet
m-RNA. For translation m-RNA is transported codon. Every three consecutive nucleotides in
out of the nucleus through nuclear pore. DNA will constitute a triplet codon. Genetic
code is a triplet code, was evidenced first by
Genetic Code: Crick (1961) using “frame- shift mutation”.
However, M. Nirenberg and Matthaei were
It is already known that DNA is a master able to synthesize artificial m-RNA which
molecule of a cell that initiates, guides, regulates contained only one type nitrogenous base
and controls the process of protein synthesis. i.e. Uracil (Homopolymer). This synthetic
To perform this complicated function, it must poly-U sequence was transferred to protein
carry the requisite information for the synthesis synthesizing enzymes. A small polypeptide
of proteins. Obviously this information has to molecule was produced/ formed by the linking
be verily located in the DNA itself. The site for of phenylalanine molecules. This explains that
storing this information lies in the sequence of UUU codes for phenyl alanine. Later different
nucleotides (i.e. nitrogen bases), as evidenced homopolymer codons were deciphered. Codons
by Yanofski and Sarabhai (1964). formed by two or more bases were also tried.

About, 20 different types of amino acids are Dr. Har Gobind Khorana : He devised a
involved in the process of synthesis of proteins. technique for artificially synthesizing m- RNA
DNA molecule has 4 types of nitrogen bases to with repeated sequences of known nucleotides.
identify these 20 different types of amino acids. By using synthetic DNA, Dr. Khorana prepared
Question arises then, how is it possible that 20 chains of polyribonucleotides with known
types of amino acids are encoded by 4 types of repeated sequences of two or three nucleotides.
nitrogen bases? eg. CUC UCU CUC UCU.

According to F.H.C. Crick, this This resulted in formation of polypeptide
information is stored in the form of coded chain having two different amino acids placed
language (cryptogram) called genetic code, alternately (Leucine and Serine). Similarly
that contains code words (codons) each one polynucleotide chain with three- nitrogen
specifying (representing) specific amino acid. base repeats gave polypeptide chain with only
Genetic code, therefore, is a collection of base one amino acids. Eg. CUA CUA CUA CUA
sequences that correspond to each amino acid. (leucine). Later, Severo Ochoa established that
the enzyme (polynucleotide phosphorylase)
A single nitrogen base in a codon (singlet was also helpful in polymerising RNA with
codon) will encode for only four different types defined sequences in a template- independent
of amino acids. A combination of two nitrogen manner (i.e. enzymatic synthesis of RNA).
bases (doublet codon) will specify only 16
different types of amino acids. A combination Finally Nirenberg, Matthaei and Ochoa
of three nitrogen bases (triplet codon) will deciphered all the 64 codons in the dictionary
specify 64 different types of amino acids. Hence of genetic code.
G. Gamov (1954) suggested that in a codon,

80

Second Letter

U CA G

UUU Phe UCU UAU Tyr UGU Cys U
Leu UCC Stop UGC Stop C
U UUC UCA Ser UAC Stop UGA Trp A
UUA Leu UCG UAA His UGG G
CCU Gln CGU Arg U
UUG Ile CCC UAG CGC C
Met CCA Asn CGA Ser A
CUU Val CCG CAU Lys CGG Arg G
ACU AGU U
First Letter C CUC ACC Pro CAC Asp AGC Gly C Third Letter
CUA ACA CAA Glu AGA A
ACG AGG G
CUG GCU CAG GGU U
GCC GGC C
AUU GCA AAU GGA A
GCG GGG G
A AUC Thr AAC
AUA AAA

AUG AAG

GUU GAU

G GUC Ala GAC
GUA GAA

GUG GAG

Fig. 4.13 : Dictionary of genetic code

During replication and transcription, a i. Genetic code is a triplet code: Sequence
nucleic acid is copied to form another nucleic of three consecutive bases constitute
acid. These two processes are based on codon, which specifies one particular
complementarity principle. During translation, amino acid. Base sequence in a codon
genetic information is transferred from a is always in 5’ 3’ direction. In every
polymer of nucleotides to a polymer of amino living organism genetic code is a triplet
acids. Here, complementarity principle does code.
not exist.
ii. Genetic code has distinct polarity :
It is evident that change in nucleic acid Genetic code shows definite polarity i.e.
(genetic material) results in the change in direction. It, therefore, is always read in
amino acids of proteins. This clearly explains 5’ 3’ direction and not in 3’ 5’
that genetic code directs the sequence of amino direction. Otherwise message will change
acids during synthesis of proteins. e.g. 5’ AUG 3’.

Find ouTthe iii. Genetic code is non-overlapping : Code
What is the amino acid sequence is non overlapping i.e. each single base is
encoded by base sequence UCA UUU UCC a part of only one codon. Adjacent codons
GGG AGU of an m- RNA segment? do not overlap. If non-overlapping, then
with 6 consequtive bases only two amino
Characterestic of Genetic code: acid molecules will be in the chain. Had it
been overlapping type, with 6 bases, there
Genetic code of DNA has certain would be 4 amino acid molecules in a
fundamental characteristics – chain. Experimental evidence is in favour
of non-overlapping nature.

81

iv. Genetic code is commaless : There is Activity :
no gap or punctuation mark between
successive/ consecutive codons. 5′ 3′

v. Genetic code has degeneracy : Usually A UG U C G A C G C C C U A A
single amino acid is encoded by single
codon. However, some amino acids are m-RNA
encoded by more than one codons. e.g. Consider given m-RNA strand which
Cysteine has two codons, while isoleucin
has three codons. This is called degeneracy has undergone mutation and lost nucleotides
of the code. Degeneracy of the code is
explained by Wobble hypothesis. Here, A, C, and G sequentially. Resultant mRNA
the first two bases in different codons are
identical but the third one, varies. is represented by 1, 2 and 3. With the help of

vi. Genetic code is universal : By and checker board of amino acids, explain the
large in all living organisms the specific
codon specifies same amino acid. e.g. changes in amino acid sequence that will
codon AUG always specifies amino acid
methionine in all organisms from bacteria occur due to such mutation.
up to humans.
5′ 3′
vii. Genetic code is non-ambiguous : Specific 1.
amino acid is encoded by a particular
codon. Alternatively, two different amino A UG U C G C G C C C U A A
acids will never be encoded by the same
codon. 5′ 3′

viii. Initiation codon and termination codon: 2.
AUG is always an initiation codon in any
and every mRNA. AUG codes for amino A UG U C G G C C C UA A
acid methionine. Out of 64 codons, three
codons viz. UAA, UAG and UGA are 5′ 3′
termination codons which terminate/ stop
the process of elongation of polypeptide 3.
chain, as they do not code for any amino
acid. A UG U C G C C C U A A

ix. Universal : Usually in all organisms the Mutations and Genetic Code:
specific codon specifies same amino acid.
Mutation is a phenomenon in which
x. Codon and anticodon : Codon is a part sudden change in the DNA sequence takes
of DNA e.g. AUG is codon. It is always place. It results in the change of genotype
represented as 5’ AUG 3’. Anticodon is a (i.e. character). Along with recombination,
part of tRNA. It is always represented as mutation is raw material for evolution as it
3’UAC 5’. also results in variations. During mutation,
possibility of loss (deletion) or gain (insertion/
• It is possible to predict sequence of codon duplication) of a segment of DNA results in
on mRNA by studying the sequence of alteration in the chromosome. Mutation can
amino acids in a polypeptide chain. also occur due to change in a single base pair
of DNA. This is known as point mutation. Eg.
Sickle cell anaemia (Refer to earlier chapter).
Deletion or insertion of base pairs of DNA
causes frame – shift mutations or deletion
mutation. Insertion or deletion of one or two
bases changes the reading frame from the point
of insertion or deletion. Insertion or deletion
of three or multiples of three bases (insert or
delete) results in insertion or deletion of amino
acids and reading frame remains unaltered

from that point onwards.

t-RNA- the adapter molecule:
Scientists considered that there has to

be a mechanism in which t-RNA will read the

82

codon and also simultaniously binds with the B. Translation - protein synthesis :
amino acid as amino acid does not have any Translation is the mechanism in which
special capacity to read the codon. So t-RNA is
considered as an adapter molecule. This role of codons of mRNA are translated and specific
tRNA was understood much later. amino acids in a sequence form a polypeptide on
ribosomes. All types of proteins are synthesised
Glutamic acid by the cell, within itself (i.e. intracellularly).

3’ Amino acid 5’ Process of translation requires amino
tRNA 5’ attachment site 3’ acids, mRNA, tRNA, ribosomes, ATP, Mg++
ions, enzymes, elongation, translocation and
Anticodon Intramolecular release factors.
loop base pairing
i. Amino acids form raw material for
C U C Anticodon protein synthesis. About 20 different
types of amino acids are known to form
GUC CAG GAG CUA UAG proteins. These are available in the
cytoplasm.
mRNA
ii. DNAcontrols synthesis of proteins having
Fig. 4.14: t-RNA - the adapter molecule amino acids in specific sequence. This
control is possible through transcription
Cloverleaf structure (2 dimentional) of of m-RNA. Genetic code is specific for
t-RNA possess an anticodon loop that has particular amino acid.
bases complementary to the codon. It is called
anticodon. It shows amino acid acceptor end iii. RNAs serve as intermediate molecules
(3’ end) having unpaired CCA bases (i.e. amino between DNA and protein.
acid binding site) to which amino acid binds.
For every amino acid, there is specific t- RNA. iv. Ribosomes serve as site for protein
Initiator t-RNA is specific for methionine. synthesis. Each ribosome consists of large
There are no t-RNA’s for stop codons. In the and small subunits. These subunits occur
actual structure, the t-RNA molecule looks like separately in cytoplasm. Only during
inverted L (3 dimentional structure). protein synthesis, these two subunits get
associated together due to Mg++ ions.
Can you tell?
A ribosome has one binding site for m-RNA
Why t-RNA is called as adapter and 3 binding sites for t-RNA. They are P
molecule? site (peptidy t-RNA site), A site (aminoacyl
– t-RNA site) and E site (exit site). Only first
t- RNA- amino acid complex, directly enters P
site of ribosome.

Can you recall? In Eukaryotes, a groove is present between
two subunits of ribosomes. It protects the
1. Name different types of RNAs. Polypeptide chain from the action of cellular
2. Name the site of protein synthesis. enzymes and also protects mRNA from the
3. Which molecule carries information of action of nucleases.

protein synthesis from gene? Mechanism of translation (i.e. synthesis of
4. Which molecule carries amino acid from polypeptide chain) :
It involves three steps :
cytoplasm to ribosome? i. Initiation, ii. Elongation, iii. Termination

83

1. Initiation of Polypeptide chain : tRNA carrying dipeptide now gets
a. Activation of amino acids is essential positioned at ‘P’ site of ribosome, making
before translation initiates for which ‘A’ site vacant. At this site, then next
ATP is essential. Small subunit of charged tRNA molecule carrying amino
ribosome binds (attaches) to the m-RNA acid will be received. During this process,
at 5’ end. Initiator codon, AUG is present first uncharged tRNA is discharged from
on m-RNA which initiates the process of E-site.
protein synthesis (translation). Initiator This process is repeated as amino acids
t- RNA binds with initiation codon are added to Polypeptide. It takes less than 0.1
(AUG) by its anticodon (UAC) through second for formation of peptide bond.
hydrogen bonds. It carries activated
amino acid methionine (in Eukaryotes) Third charged t-RNA with its amino acid,
or formyl methionine (in prokaryotes). arrives at A-site of ribosome. Anticodon and
b. Now the large subunit of ribosome joins codon bind by hydrogen bond. Polypeptide
with the smaller subunit, that requires bond is formed. Second t-RNA is discharged
Mg++ ions. from P-site to E-site and leaves the ribosome.
c. Initiator charged t-RNA (with activated So the events like arrival of t-RNA- amino acid
amino acid methionine) occupies the P- complex, formation of peptide bond, ribosomal
site of ribosome and A- site is vacant. translocation and removal of previous tRNA,
are repeated. As ribosome move over the m-
2. Elongations of polypeptide chain : RNA, all the codons on mRNA are exposed
During this process, activated amino acids one by one for translation.
are added one by one to first amino acid
(methionine). Amino acid is activated by Initiator tRNA Larger sub unit of
utilising energy form ATP molecule. This ribosome
amino acid binds with amino acid binding
site of t-RNA- This results in formation of mRNA
t-RNA- amino acid complex.
Smaller sub unit a. E site A site
Addition of Amino acid occurs in 3 Step of ribosome P site
cycle -
a. Condon recognition- Amino acyl t- RNA b. Peptide A site
link
molecule enters the ribosome at A-site. Polypeptide
Anticodon binds with the codon by chain E site
hydrogen bonds.
b. Amino acid on the first initiator t-RNA A site
at P-site and amino acid on t-RNA at
A-site join by peptide bond. Here enzyme Last
Ribozyme acts as a catalyst. At this time tRNA
first tRNA at ‘P’ site is kicked off.
c. Translocation- The t- RNA at A-site c.
carrying a dipeptide at A-site moves
to the P-site. This process is called Released mRNA
translocation. In translocation, both
the subunits of ribosome move along in Fig. 4.15 : Translation Protein synthesis
relation to tRNA and mRNA. Hence, a. Initiation, b. Elongation, c. Termination

84

3. Termination and release of polypeptide: Genes of a cell are expressed to perform
At the end of m-RNA, there is a stop
different functions. For eg. An enzyme beta
codon (UAA/ UAG/ UGA). It is exposed at the
A-site. It is not read and joined by anticodon galactosidase is synthesised by E-coli. It is
of any t-RNA. The release factor binds to the
stop codon, thereby terminating the translation used for hydrolysis of lactose into galactose
process. The Polypeptide is now released in the
cytoplasm. and glucose.

Two subunits of Ribosome dissociate and Lactose (b – galactosidase) Galactose + Glucose.
last tRNA is set free in the cytoplasm.
+H2O
m-RNAalso has some additional sequences
that are not translated and are referred as If E.coli bacteria do not have lactose
untranslated regions (UTR). The UTRs are
present at both 5’-end (before start codon) and in the surrounding medium as a source of
at 3’-end (after stop codon). They are required
for efficient translation process. energy, then enzyme b-galactosidase is

Finally mRNA is also released in the not synthesised. So, it is the metabolic or
cytoplasm. It gets denatured by nucleases
immidiately. Hence mRNA is short -lived. physiological or environmental conditions that

Can you tell? regulate expression of genes. The development

1. Enlist different steps of protein synthesis. and differentiation of embryo into an adult
2. Name the initiator codon of protein
organism, is also a result of the coordinated
synthesis.
3. Explain in brief the process of initiation regulation or expression, of several sets of

during protein synthesis. genes.
4. Name three binding sites of ribosome.
5. What is translocation? Now one has to understand and know the

4.6 Regulation of gene expression: mechanism by which the organisms regulate
It is the multistep process by which a gene
gene expression in response to changes in the
is regulated and its product is synthesized.
Thus, gene expression results in the formation environment. If so, whether single mechanism
of a Polypeptide. Gene expression process is
regulated at different levels. exists for regulation of the expression of

In eukaryotes, the regulation can be at different genes/ sets of genes or different genes
different levels like-
are regulated by different mechanisms.
1. Transcriptional level (formation of
primary transcript) Certain bacteria like E.coli adapt to their

2. Processing level ( regulation of splicing) chemical environment by synthesizing certain
3. Transport of m-RNA from nucleus to the
enzymes depending upon the substrate present.
cytoplasm.
4. Translational level. Such adaptive enzyme is called inducible

enzymes. A set of genes will be switched on

when there is necessity to metabolise a new

substrate. This phenomenon is called induction

and small molecule responsible for this, is

known as inducer. It is positive control.

Do you know ?

Repressible regulation of gene is
seen when the end product of a biosynthetic
pathway like amino acid, is provided in the
medium. At this time, internal biosynthesis
of amino acid stops. It is negative control so
the metabolite (amino acid) turns off a set of
genes involved in producing that metabolite.
This is called feedback repression.

85

4.7 Operon concept : Lac operon consists of following
It is a transcriptional control mechanism components :

of gene regulation. Francois Jacob and Jacques 1. Regulator gene (repressor gene)
Monod (1961) explained that metabolic 2. Promoter gene
pathways are regulated as a unit. 3. Operator gene
4. Structural genes
For example in E.coli, when lactose sugar 5. Inducer - It is not a component of
is provided to the culture medium, cell induces
production of three enzymes necessary for operon.
digestion of lactose. The enzymes are :
1. Regulator gene : This gene controls the
i. b-galactosidase : Digests lactose into operator gene in cooperation with an
galactose and glucose. inducer present in the cytoplasm. Regulator
gene preceeds the promoter gene. It may not
ii. b-galactoside permease : Permits be present immidiately adjacent to operator
lactose molecules to enter into the cell. gene. Regulator gene produces a protein
called repressor protein. Repressor binds
iii. Transacetylase (b-Galactoside with operator gene and represses (stops) its
action. It is called regulator protein.
acetyltransferease) : Transfers an
2. Promoter gene : This gene preceeds the
acetyl group from acetyl CO-A to operator gene. It is present adjacent to
operator gene. The promoter gene marks
galactoside. the site at which the RNA Polymerase
enzyme binds. When the operator gene
Synthesis of these three enzymes, is is turned on, the enzyme moves over
controlled by a long segment of DNA known the operator gene and transcription of
as Operon. It consists of an operator site O and structural genes starts. Promoter gene base
three structural genes Z, Y and A .The action sequence determines which strand of DNA
of structural genes is regulated by operator acts a template.
site with the help of a repressor protein.
Repressor protein is produced by the action of 3. Operator gene : It preceeds the structural
gene i (inhibitor) known as regulator gene. genes. This controls the functioning of
structural genes. It lies adjacent to the
The gene expression depends on whether Structural genes. When operator gene is
operator is switched on or switched off. turned on by an inducer, the Structural
genes produce m-RNA. Operator gene is
If the operator is switched on, the three genes turned off by a product of repressor gene.
z, y and a are transcribed by RNA Polymerase
into a single m-RNA. Each structural gene is 4. Structural gene : When lactose is added
generally known as cistron and the transcribed to the E.coli culture, the structural genes
long m-RNA covering various cistrons is catalyse (produce) m-RNA which in turn
known as Polycistronic. produces polypeptides, on the ribosomes.
The polypeptides formed, act as enzymes
Switching on or switching off of the to caltalyse lactose in the cell. There are
operator is achieved (acomplished) by a protein 3 structural genes in the sequence lac-Z,
called repressor. When this protein is attached lac-Y and lac-A. Enzymes produced are
to the operator and blocks it, the switch is turned b-galactosidase, b-galactoside permease
off and structural genes are not expressed. and transacetylase respectively.

Lac operon :
Lactose or lac operon of E.coli is inducible

operon. The operon is switched on when a chemical
inducer- lactose is present in the medium.

86

lac operon

Regulatory gene RNA polymerase Operator blocked
Promoter
3’ 123 5’ Template
DNA strand
Transcription Repressor
(a) 5’ mRNA Structural genes

Inducer absent Translation 3’ in active no synthesis of mRNA
Movement

blocked Transcription

Repressor proceeds

Regulatory gene Promoter Operator 23 Template
3’ 1 DNA strand
5’
Transcription Repressor
5’ mRNA Transcription
3’
5’ mRNA
Translation
Translation

Proteins (enzymes to metabolize lactose)

(b) Repressor Inactivated
Inducer present repressor

Inducer (allolactose)

Fig. 4.16 : Working of Lac Operon

5. Inducer : It is a chemical in the cytoplasm Can you tell?
(allolactose) which inactivates the
repressor. When lac operon is switched on, 1. What is the role of a repressor gene?
then inducer joins with repressor protein 2. Name the different structural genes in
preventing the binding of repressor to the
operator gene. So the Operator gene is free sequence of lac operon.
and now enzyme RNA polymerase can 3. Which molecule does act as inducer
move from promoter to structural genes via
operator gene. molecule in lac operon?
4. In which condition, lac operon is
Role of lactose :
A few molecules of lactose enter into the switched off?

cell by an enzyme permease.A small amount Internet my friend
of this enzyme is present even when operon is
switched off. A few molecules of lactose, act as Find out information about Trp-operon, Ara-
inducer and bind to repressor. This repressor – operon, His-operon, Val-operon.
inducer complex fails to join with the operator
gene, which is then turned on. Structural genes 4.8 Genomics:
produce all enzymes. Thus, lactose acts as The term Genome (introduced by H.Winkler
an inducer of its own break down. When the
inducer level falls, the operator is blocked again in 1920) is the total genetic constitution of an
by repressor. So structural genes are repressed/ organism. Alternatively, it is a complete copy
inactivated again. This is negative feedback. of genetic information (DNA) or one complete
set of chromosomes (monoploid or haploid) of
Use your brain power an organism.

If operator gene is deleted due to The term Genomics (term coined by T.H.
mutation, how will E.coli metabolise lactose? Roderick in 1986) is the study of genomes
through analysis, sequencing and mapping of
genes along with the study of their functions.

87

The sequencing of yeast, Drosophila and Human Genome Project formally began in
mouse genome was done in order to facilitate 1990 and was completed in 2003.The human
comparative studies between humans and other genome project is a multinational research
organisms commonly used for genetic studies, project to determine the genomic structure of
in laboratory. Several additional genomes are humans. The main aims of project are –
now either actively being sequenced or strongly I. Mapping the entire human genome at the
considered for sequencing. These include
several microbes, bee, tomato and other crops. level of nucleotide sequences.
Genomics study may be classified into two II. To store the information collected from the
types:
a. Structural genomics: It involves mapping, project in databases.
III. To develop tools and techniques for analysis
sequencing and analysis of genome.
of the data.
b. Functional genomics: It deals with the IV. Transfer of the related technologies to the
study of functions of all gene sequences
and their expression in organisms. private sectors, such as industries.
V. Taking care of the legal, ethical and social

issues which may arise from project.

Application of genomics: HGP (Human Genome Project) was closely
Structural and functional genomics is used associated with rapid development of a new
area in biology, called Bioinformatics. The
for different purposes in the improvement of work of human genome project has allowed
crop plant, human health and livestock. The researchers to begin to understand the blueprint
knowledge and understanding acquired from in building and constructing the human genome.
genomics research can be applied in a number As researchers learn more about the functions
of different sectors, including medicine, of genes and proteins, this knowledge will have
biotechnology and social sciences. It helps in a major impact in the fields like Medicine,
the treatment of genetic disorders through gene Biotechnology and the Life sciences. Therefore
therapy. HGP is very important.
• Genomics is used in agriculture to develop
Human Genome Project was to provide a
transgenic crops having more desirable complete and accurate sequence of the 3 billion
characters. DNA base pairs that make up the human
• Genetic markers developed in genomics, genome and to find out the estimated number
have applications in forensic analysis. of human genes. Now about 33000 genes have
• Genomics can lead to introduce new gene in been estimated to be present in humans.
microbes to produce enzymes, therapeutic
proteins and even biofuels. The project was also aimed to sequence
the genomes of several other organisms such
4.9 Human Genome Project : as bacteia e.g. E.coli, Caenorhabditis elegans
The human genome project was initiated (a free living non-pathogenic nematode),
Saccharomyces cerevisiae (yeast), Drosophila
in 1990 under the International administration (fruit fly), plants (rice and Arabidopsis), Mus
of the Human Genome Organization (HUGO). musculus (mouse), etc. Complete genome
This project was co-ordinated by the US sequences of these model organisms will be
department of Energy and National institute useful for comparative studies that will allow
of health. Additional contributors included researchers to study gene functions in these
universities across the United States and organisms.
international partners in the United Kingdom,
France, Germany, Japan, India and China. The

88

Table 4.17 : Comparative genome sizes of humans and other models organisms.

Organism Chromosome Estimated gene Estimated size
Human (Homo sapiens) number number
(base pairs)

46 33,000 3 billion

Mouse (Mus musculus) 40 25,000 2.9 billion
Fruit fly (Drosophila melanogaster) 8 13,000 165 million

Plant (Arabidopsis thaliana) 10 25,000 157 million

Roundworm (Caenorhabditis elegans) 12 19,000 97 million
Yeast (Saccharomyces cerevisiae) 32 6000 12 million

Bacteria (Escherichia coli) 1* 4400 4.6 million

The secret of our complexity may lie not DNA fingerprinting technique is based on
in the number of our genes but how we use identification of nucleotide sequence present
them. It will lead to the understanding of gene in this wonder molecule. About 99.9% of
structure and function in other species. Since nucleotide sequence in all persons, is same.
we possess many of the genes same as these Only some short sequences of nucleotides
of flies, round worms and mice, such studies differ from person to person. In the population,
will lead to a greater understanding of human every person shows unusual sequences of 20-
evolution. 100 base pairs, which are repeated several
times. They are termed as Variable Number of
Use your brain power Tandem Repeats (VNTRs).

1. What have we learnt from the Human The length of the regions having VNTRs
Genome Project? is different in each individual and hence is the
key factor in DNA profiling. Steps involved in
2. Why is HGP important? DNA finger printing are as follows:

Can you tell? 1. Isolation of DNA: The DNA must be
recovered from the cells or tissues of the
Do different organisms have the same DNA? body (host). Only small amount of tissue
like blood, hair roots, skin, etc. is required.
4.10 DNA Fingerprinting:
Genes present on chromosomes are 2. Restriction digestion: The isolated DNA
is treated with restriction enzymes. The
responsible for determining characters of restriction enzymes cut the DNA into
the organism as well as for inheritance of small fragments having variable lengths.
characters. Due to recombination of paternal This phenomenon is called Restriction
and maternal genes, we differ from our parents. Fragment Length Polymorphism (RFLP).
Differences also arise due to infrequent
mutations that occur during gamete formation 3. Gel electrophoresis: The DNA samples
(cell division). Due to all these factors, every are loaded for agarose gel electrophoresis
individual has its unique genetic make-up, under an electric influence. The DNA
which may be called its Fingerprint. The fragments, which are negatively charged
technique developed to identify a person with move to the positive pole. The movement
the help of DNA restriction analysis, is known of these fragments depends on length of
as DNA profiling or DNA fingerprinting. The the fragments. This results in formation of
technique of finger printing was first given by bands. dsDNA splits into ssDNA by alkali
British geneticist, Dr. Alec Jeffreys in 1984. treatment.

89

4. Southern blotting: The separated DNA the nitrocellulose filter paper. Remaining
fragments are transferred to a nylon single stranded DNA probe fragments are
membrane or a nitrocellulose filter paper washed off.
by placing it over the gel and soaking them
with filter paper overnight. 7. Photography: The nitrocellulose filter
paper is photographed on an X-ray film by
5. Selection of DNAprobe: Aknown sequence autoradiography. The film is analysed to
of single- stranded DNA is prepared. It is determine the presence of hybrid DNA.
called DNA Probe. DNA Probe is obtained
from organisms or prepared by cDNA Application of DNA fingerprinting
preparation method. The DNA probe is
labelled with radioactive isotopes. 1. In forensic science, DNA finger printing is
used to solve problems of rape and some
6. Hybridization: Probe DNA is added to complicated murder cases.
the nitrocellulose filter paper containing
host DNA. The single-stranded DNA 2. DNA finger printing is used to find out the
probe pairs with the complementary base biological father or mother or both, of the
sequence of the host DNA strand. As a child, in case of disputed parentage.
result DNA-DNA hybrids are formed on
3. DNA finger printing is used in pedigree
Restriction analysis in cats, dogs, horses and humans.
enzyme

Isolated purified
DNA

Sample

DNA extraction Restriction Electrophoresis
X-ray
digestion DNA fragments Transfer of

DNA to Southern blotting

Selected single membrane

stranded DNA

probes

Hybridization of DNA

Fig. 4.18 : DNA Fingerprinting

Know the scientists

Father of DNA Fingerprinting in India. He was instrumental in making DNA
fingerprinting mainstream in India, for research and its forensic applications. He
obtained DNA probe from Y chromosome of female banded krait snake (in this snake
female has XY and male has YY chromosome). The unique segment obtained from this
chromosome is, banded krait minor (BKM - DNA). It was used to developed probe for
the Indigenous DNA fingerprinting technique.
Dr. Lalji Singh Contributions of Dr. Lalji Singh: i. He installed several dedicated laboratories on aspects
( 1947 - 2017) of genetics such as population biology, structural biology and transgenic research. ii.
His work in the field of DNA fingerprinting technology, contributed for, wildlife conservation, forensics,
evolution and phylogeny. iii. Established Centre for DNA Fingerprinting and Diagnostics (CDFD) in
late 1990s- making it nodal centre for DNA fingerprinting and diagnostics for all species and several
diseases. iv. Founded Laboratory for Conservation of Endangered Species (LaCONES).

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Activity :
Prepare physical model of DNA molecule (Watson-Crick model)
Requirements :

Labelled Diagram :

Functions :

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