1. Group the following as
nitrogenous bases and nucleosides: Adenine, Cytidine, Thymine, Guanosine, Uracil
and Cytosine.
Ans: Nitrogenous Bases –
Adenine, Uracil and Cytosine, Thymine; Nucleosides – Cytidine, guanosine.
2. If a double stranded DNA
has 20 per cent of cytosine, calculate the per cent of adenine in the
DNA.
Ans: In a DNA molecule,
the number of cytosine molecule is equal to guanine molecules & the number
of adenine molecules are equal to thymine molecules. As a result, if a double
stranded DNA has 20% of cytosine, it has 20% of guanine. The remaining 60%
includes both adenine & thymine which are in equal amounts. So, the
percentage of adenine is 30%.
3. 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.
Ans:
If the sequence of one strand of DNA is written as follows:
5′ – ATGCATGCATGCATGCATGCATGCATGC – 3′
The sequence of the complementary
strand in 5′ —> 3′ direction will be:
5′ – GCATGCATGCATGCATGCATGCATGCAT –
3′
4. If the sequence of the
coding strand in a transcription unit is written as follows:
5-ATGCATGCATGCATGCATGCA TGCATGC-3′
Write down the sequence of mRNA.
Ans: mRNA:
5′ -A U G CAU G CAU G C AU G CA UGCAUGCAUGC-3′.
5. Which property of DNA
double helix led Watson and Crick to hypothesise semi-conservative mode of DNA
replication? Explain
Ans: The antiparallel,
double-stranded nature of the DNA molecule led Watson and Crick to hypothesise
semi-conservative mode of DNA replication. They suggested that the two strands
of DNA molecule uncoil and separate, and each strand serves as a template for
the synthesis of a new (complementary) strand alongside it. The template and its
complement, then form a new DNA double strand, identical to the original DNA
molecule. The sequence of bases which should be present in the new strands can
be easily predicted because these would be complementary to the bases present in
the old strands. A will pair with T, T with A, C with G, and G with C. Thus, two
daughter DNA molecules identical to the parent molecule are formed and each
daughter DNA molecule consists of one old (parent) strand and one new strand.
Since only one parent strand is conserved in each daughter molecule, this mode
of replication is said to be semiconservative. Meselson and Stahl and Joseph
Taylor, later proved it by experiments.
6. Depending upon the
chemical nature of the template (DNA or RNA) and the nature of nucleic acids
synthesized from it (DNA or RNA), list the types of nucleic acid
polymerases.
Ans: (i) DNA dependent
DNA polymerase – synthesis.
(ii) DNA dependent RNA polymerase –
synthesis.
(iii) RNA dependent DNA polymerase – Retroviral nucleic acid.
(iv) RNA dependent RNA polymerase – cDNA synthesis.
7. How did Hershey and Chase
differentiate between DNA and protein in their experiment white proving that DNA
is the genetic material?
Ans: Alfred Hershey and
Martha Chase (1952) worked with viruses that infect bacteria called
bacteriophages. In 1952, they chose a bacteriophage known as T2 for their
experimental material.
They grew some viruses on a medium that contained
radioactive phosphorus (p32) and some others on medium that contained
radioactive sulphur (s35). Viruses grown in the presence of radioactive
phosphorus contained radioactive DNA but not radioactive protein because DNA
contains phosphorus but protem does not. Similarly, viruses grown on radioactive
sulphur contained radioactive protein but not radio’active DNA because DNA does
not contain sulphur.
Radioactive phages were allowed to attach to E. coli
bacteria. Then, as the infection proceeded, the viral coats were removed from
the bacteria by agitating them in a blender. The virus particles were separated
from the bacteria by spinning them in a centrifuge. ,
Bacteria which was
infected with viruses that had radioactive DNA were radioactive, indicating that
DNA was the material that passed from the virus to the bacteria. Bacteria that
were infected with viruses that had radioactive proteins were not radioactive.
This indicates that proteins did not enter the bacteria from the viruses. DNA is
therefore the genetic material that is passed from virus to bacteria.
8. Differentiate between the
followings:
(a) Repetitive DNA and Satellite DNA
(b) mRNAand tRNA
(c) Template strand and Coding
strand
Ans: (a) The main
differences between repetitive DNA and satellite DNA are as following:
(b)
The main difference between mRNA and tRNA are as following:
(c) The main
difference between template strand and coding strand are as follows :
9. List two essential roles
of ribosome during translation.
Ans: Two
essential roles of ribiosomes during translation are ;o
(i) they provide
surface for binding of mRNA in the groove of smaller sub unit of ribosome.
(ii) As larger sub unit of ribosome has peptidy transferase on its ‘P’ site,
therefore, it helps in joining amino acids by forming peptide bonds. .
10. 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?
Ans:
Lac operon is switched on, on adding lactose in medium, as
lactose acts as inducer and makes repressor inactive by binding with it. When
the lac operon system is switched on, β-galactosidase is formed, which converts
lactose into glucose and galactose. As soon as all the lactose is consumed,
repressor again becomes active and causes the system to switch off (shut
down).
11. Explain (in one or two
lines) the function of the followings:
(a) Promoter
(b) tRNA
(c) Exons
Ans: Promoter: It is one
of the three components of a transcription unit that takes part in
transcription. It is located at the start 5′ end and provides site for
attachment of transcription factors (TATA Box) and RNA polymerase. tRNA: It
takes part in the transfer of activated amino acids from cellular pool to
ribosome for their taking part in protein formation.
Exons: In eukarytoes,
DNA is mosaic of exons and introns. Exons are coding sequences of DNA which are
transcribed and translated both.
12. Why is the Human genome
project called a mega project?
Ans: Human
genome project is called a mega project because
(i) it required
bioinformatics data basing and other high speed computational devices for
analysis, storage and retrieval of information.
(ii) it generated lot of
information in the form of sequence annotation.
(iii) it was carried out in
number of labs and coordinated on extensive scale.
13. What is DNA
fingerprinting? Mention its application.
Ans: DNA
fingerprinting or DNA typing is a technique of determining nucleotide sequences
of certain areas (VNTRs) of DNA which are unique to each individual. Each person
has a unique DNA fingerprint. Unlike a conventional fingerprint that occurs only
on the fingertips and can be altered by surgery, a DNA fingerprint is the same
for every cell, tissue and organ of a person. It cannot be changed by any known
treatment. Applications of DNA fingerprinting are as follows:
14. Briefly describe the
following:
(a) Transcription
(b)
Polymorphism
(c) Translation
(d) Bioinformatics
Ans: Transcription
: It is DNA directed synthesis of RNA in which the RNA is transcribed on
3*—>5’ template strand of DNA in 5’—>3’ direction. Polymorphism: Variation
at genetic level arisen due,to mutation, is called polymorphism. Such variations
are unique at particular site of DNA, forming satellite DNA. The polymorphism in
DNA sequences is the basis of genetic mapping and DNA finger printing.
Translation : Protein synthesis from mRNA, tRNA, rRNA.
Bioinformatics :
Computational method of handling and analyzing biological databases.