NCERT Solutions for Class 12 Biology Chapter 5 Molecular Basis of Inheritance

NCERT
SOLUTIONS
CLASS - 12th

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Class : 12th
Subject : Biology
Chapter : 6
Chapter Name : Molecular Basis of Inheritance

Q1 Group the following as nitrogenous bases and nucleosides: Adenine, Cytidine,
Thymine, Guanosine, Uracil and Cytosine.

Answer. Nitrogenous bases present in the list are adenine, thymine, uracil, and cytosine.
Nucleosides present in the list are cytidine and guanosine.

Q2 If a double stranded DNA has 20 percent of cytosine, calculate the percent of
adenine in the DNA.

Answer. According to Chargaff's rule, the DNA molecule should have an equal ratio of pyrimidine
(cytosine and thymine) and purine (adenine and guanine). It means that the number of adenine
molecules is equal to thymine molecules and the number of guanine molecules is equal to
cytosine molecules. % A = % T and % G = % C
If dsDNA has 20% of cytosine, then according to the law, it would have 20% of guanine. Thus,
percentage of G + C content = 40% The remaining 60% represents both A + T molecule. Since
adenine and guanine are always present in equal numbers, the percentage of adenine molecule .

Q3 If the sequence of one strand of DNA is written as follows:
5'-ATGCATGCATGCATGCATGCATGCATGC-3'
Write down the sequence of complementary strand in 5'3' direction.

Answer. The DNA strands are complementary to each other with respect to base sequence. Hence,
if the sequence of one strand of DNA is 5'- ATGCATGCATGCATGCATGCATGCATGC - 3' Then, the
sequence of complementary strand in 3'- TACGTACGTACGTACGTACGTACGTACG - 5' direction will
be Therefore, the sequence of nucleotides on DNA polypeptide in direction is 5'-
GCATGCATGCATGCATGCATGCATGCAT- 3'

Q4 If the sequence of the coding strand in a transcription unit is written as follows: 5'-
ATGCATGCATGCATGCATGCATGCATGC-3' Write down the sequence of mRNA

Answer. If the coding strand in a transcription unit is 5'- ATGCATGCATGCATGCATGCATGCATGC-3'

Then, the template strand in 3' to 5' direction would be 3' -
TACGTACGTACGTACGTACGTACGTACG-S'
It is known that the sequence of mRNA is same as the coding strand of DNA. However, in RNA,
thymine is replaced by uracil. Hence, the sequence of mRNA will be 5' -
AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3'

Q5 Which property of DNA double helix led Watson and Crick to hypothesise semi- conservative
mode of DNA replication? Explain.

Answer. Watson and Crick observed that the two strands of DNA are antiparallel and
complementary to each other with respect to their base sequences. This type of arrangement in
DNA molecule led to the hypothesis that DNA replication is semiconservative. It means that the
double stranded DNA molecule separates and then, each of the separated strand acts as a template
for the synthesis of a new complementary strand. As a result, each DNA molecule would have one
parental strand and a newly synthesized daughter strand. Since only one parental strand is
conserved in each daughter molecule, it is known as semi-conservative mode of replication.

Q6 Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic
acids synthesised from it (DNA or RNA), list the types of nucleic acid polymerases.

Answer. There are two different types of nucleic acid polymerases.
(1) DNA-dependent DNA polymerases
(2) DNA-dependent RNA polymerases
The DNA-dependent DNA polymerases use a DNA template for synthesizing a new strand of DNA,
whereas DNA-dependent RNA polymerases use a DNA template strand for synthesizing RNA.

Q7 How did Hershey and Chase differentiate between DNA and protein in their experiment while
proving that DNA is the genetic material?

Answer. Hershey and Chase worked with bacteriophage and E.coli to prove that DNA is the genetic
material. They used different radioactive isotopes to label DNA and protein coat of the
bacteriophage. They grew some bacteriophages on a medium containing radioactive phosphorus
(32P) to identify DNA and some on a medium containing radioactive sulphur (35S) to identify
protein. Then, these radioactive labelled phages were allowed to infect E.coli bacteria. After
infecting, the protein coat of the bacteriophage was separated from the bacterial cell by blending

and then subjected to the process of centrifugation. Since the protein coat was lighter, it was
found in the supernatant while the infected bacteria got settled at the bottom of the centrifuge
tube. Hence, it was proved that DNA is the genetic material as it was transferred from virus to
bacteria.

Q8 Differentiate between the followings:
(a) Repetitive DNA and Satellite DNA
(b) mRNA and tRNA
(c) Template strand and Coding strand

Answer.
(a) Repetitive DNA and Satellite DNA

(b) mRNA and tRNA

(c) Template strand and Coding strand

Q9 List two essential roles of ribosome during translation.

Answer. The important functions of ribosome during translation are as follows
(a) Ribosome acts as the site where protein synthesis takes place from individual amino acids. It is
made up of two subunits. The smaller subunit comes in contact with mRNA and forms a protein
synthesizing complex whereas the larger subunit acts as an amino acid binding site.
(b) Ribosome acts as a catalyst for forming peptide bond. For example, 23s rRNA in bacteria acts as
a ribozyme.

Q10 In the medium where E. coli was growing, lactose was added, which induced the lac operon.
Then, why does lac operon shut down some time after addition of lactose in the medium?

Answer. Lac operon is a segment of DNA that is made up of three adjacent structural genes,
namely, an operator gene, a promoter gene, and a regulator gene. It works in a coordinated
manner to metabolize lactose into glucose and galactose. In lac operon, lactose acts as an inducer.
It binds to the repressor and inactivates it.
Once the lactose binds to the repressor, RNA polymerase binds to the promoter region. Hence,
three structural genes express their product and respective enzymes are produced. These enzymes
act on lactose so that lactose is metabolized into glucose and galactose. After sometime, when the
level of inducer decreases as it is completely metabolized by enzymes, it causes synthesis of the
repressor from regulator gene. The repressor binds to the operator gene and prevents RNA

polymerase from transcribing the operon. Hence, the transcription is stopped. This type of
regulation is known as negative regulation.

Q11 Explain (in one or two lines) the function of the followings:
(a) Promoter
(b) tRNA
(c) Exons

Answer.
(a) Promoter :Promoter is a region of DNA that helps in initiating the process of transcription. It
serves as the binding site for RNA polymerase.
(b) tRNA :tRNA or transfer RNA is a small RNA that reads the genetic code present on mRNA. It
carries speci c amino acid to mRNA on ribosome during translation of proteins.
(c) Exons :Exons are coding sequences of DNA in eukaryotes that transcribe for proteins.

Q12 Why is the Human Genome project called a mega project?

Answer. Human genome project was considered to be a mega project because it had a speci c goal
to sequence every base pair present in the human genome. It took around 13 years for its
completion and got accomplished in year 2006. It was a large scale project, which aimed at
developing new technology and generating new information in the eld of genomic studies. As a
result of it, several new areas and avenues have opened up in the eld of genetics, biotechnology,
and medical sciences. It provided clues regarding the understanding of human biology.

Q13 What is DNA ngerprinting? Mention its application.

Answer. DNA ngerprinting is a technique used to identify and analyze the variations in various

individuals at the level of DNA. It is based on variability and polymorphism in DNA sequences.
Application
(1) It is used in forensic science to identify potential crime suspects.
(2) It is used to establish paternity and family relationships.
(3) It is used to identify and protect the commercial varieties of crops and livestock.
(4) It is used to nd out the evolutionary history of an organism and trace out the linkages
between groups of various organisms.

Q14 Brie y describe the following:
(a) Transcription
(b) Polymorphism
(c) Translation
(d) Bioinformatics

Answer. (a) Transcription :
Transcription is the process of synthesis of RNA from DNA template. A segment of DNA gets
copied into mRNA during the process. The process of transcription starts at the promoter region of
the template DNA and terminates at the terminator region. The segment of DNA between these
two regions is known as transcription unit. The transcription requires RNA polymerase enzyme, a
DNA template, four types of ribonucleotides, and certain cofactors such as M g . The three
2+

important events that occur during the process of transcription are as follows.
(i) Initiation
(ii) Elongation
(iii) Termination
The DNA-dependent RNA polymerase and certain initiation factors (o) bind at the double stranded
DNA at the promoter region of the template strand and initiate the process of transcription. RNA
polymerase moves along the DNA and leads to the unwinding of DNA duplex into two separate
strands. Then, one of the strands, called sense strand, acts as template for mRNA synthesis. The
enzyme, RNA polymerase, utilizes nucleoside triphosphates (dNTPs) as raw material and
polymerizes them to form mRNA according to the complementary bases present on the template
DNA. This process of opening of helix and elongation of polynucleotide chain continues until the
enzyme reaches the terminator region. As RNA polymerase reaches the terminator region, the
newly synthesized mRNA transcripted along with enzyme is released. Another factor called
terminator factor (p) is required for the termination of the transcription.

(b)Polymorphism :Polymorphism is a form of genetic variation in which distinct nucleotide
sequence can exist at a particular site in a DNA molecule. This heritable mutation is observed at a
high frequency in a population. It arises due to mutation either in somatic cell or in the germ cells.
The germ cell mutation can be transmitted from parents to their offsprings. This results in
accumulation of various mutations in a population, leading to variation and polymorphism in the
population. This plays a very important role in the process of evolution and speciation.

(c) Translation
Translation is the process of polymerizing amino acid to form a polypeptide chain. The triplet
sequence of base pairs in mRNA de nes the order and sequence of amino acids in a polypeptide
chain. The process of translation involves three steps.
(i) Initiation
(ii) Elongation
(iii) Termination
During the initiation of the translation, tRNA gets charged when the amino acid binds to it using
ATR The start (initiation) codon (AUG) present on mRNA is recognized only by the charged tRNA.
The ribosome acts as an actual site for the process of translation and contains two separate sites in
a large subunit for the attachment of subsequent amino acids. The small subunit of ribosome
binds to mRNA at the initiation codon (AUG) followed by the large subunit. Then, it initiates the
process of translation. During the elongation process, the ribosome moves one codon downstream
along with mRNA so as to leave the space for binding of another charged tRNA. The amino acid
brought by tRNA gets linked with the previous amino acid through a peptide bond and this process
continues resulting in the formation of a polypeptide chain. When the ribosome reaches one or
more STOP codon (VAA, I-JAG, and UGA), the process of translation ets terminated. The 01 e tide
chain is released and the ribosomes get detached from mRNA

(d) Bioinformatics
Bioinformatics is the application of computational and statistical techniques to the eld of
molecular biology. It solves the practical problems arising from the management and analysis of
biological data. The eld of bioinformatics developed after the completion of human genome
project (HGP). This is because enormous amount of data has been generated during the process of
HGP that has to be managed and stored for easy access and interpretation for future use by various
scientists. Hence, bioinformatics involves the creation of biological databases that store the vast
information of biology. It develops certain tools for easy and ef cient access to the information
and its utilization. Bioinformatics has developed new algorithms and statistical methods to nd
out the relationship between the data, to predict protein structure and their functions, and to
cluster the protein sequences into their related families.