If a geneticist uses the blind approach for sequencing the whole genome of an organism, followed by assignment of function to different segments, the methodology adopted by him is called as :
A.
Sequence annotation
B.
Gene mapping
C.
Expressed sequence tags
D.
Bioinformatics
Correct Answer: A
Explanation:
Sequencing the whole set of genome that contained all the coding and non-coding sequences and later assigning different regions in the sequence with functions is called sequence annotation.
In an E. Coli strain i gene gets mutated and its product can not bind the inducer molecule. If growth medium is provided with lactose, what will be the outcome?
A.
Only z gene will get transcribed
B.
z, y, a genes will be transcribed
C.
z, y, a genes will not be translated
D.
RNA polymerase will bind the promoter region
Correct Answer: C
Explanation:
As the product of 'i' gene binds with the operator region and blocks the transcription and translation of z, y and a genes.
It's product is prevented from binding to the operator by attaching it with the inducer. As the inducer can not no more capable of binding with the repressor, thus, in all the cases, operator always gets attached with the repressor thereby preventing the transcription and transmission of z, y and a.
Even in the presence of lactose, transcription and translation of z, y and a would not occur.
Ten E.coli cells with 15N - dsDNA are incubated in medium containing 14N nucleotide. After 60 minutes, how many E.coli cells will have DNA totally free from 15N?
A.
20 cells
B.
40 cells
C.
60 cells
D.
80 cells
Correct Answer: C
Explanation:
From 10 parent E.coli cells
Therefore, after 60 minutes, 60 E.coli cells will have DNA totally free from 15N.
DNA strands on a gel stained with ethidium bromide when viewed under UV radiation, appear as
A.
Bright blue bands
B.
Yellow bands
C.
Bright orange bands
D.
Dark red bands
Correct Answer: C
Explanation:
To make the DNA visible in the gel,
ethidium bromide is added to the gel
solution and the buffer. This positively
charged polycyclic aromatic compound
binds to DNA by inserting itself between
the basepairs (intercalation). The DNA
fragments when exposed to ultraviolet
light appear as orange colour bands, due
to the large increase in fluorescence of
the ethidium bromide upon binding to
the DNA.
Split gene arrangement is characteristic of prokaryotes
B.
In capping, methyl guanosine triphosphate is added to the 3' end of hnRNA
C.
RNA polymerase binds with Rho factor to terminate the process of transcription in bacteria
D.
The coding strand in a transcription unit is copied to an mRNA
Correct Answer: C
Explanation:
$\bullet$ Split gene arrangement is characteristic of eukaryotes.
$\bullet$ In capping 5-methyl guanosine triphosphate is added at 5' end of hnRNA.
$\bullet$ At 3' end poly-A tail is added.
$\bullet$ The non coding or template strand is copied to an mRNA. RNA polymerase associate with $\rho$ factor (Rho factor) and it alters the specificity of the RNA polymerase to terminate the processes.
Which of the following RNAs is not required for the synthesis of protein?
A.
siRNA
B.
mRNA
C.
tRNA
D.
rRNA
Correct Answer: A
Explanation:
siRNA mainly protect the cell from
exogenous mRNA attacks. It degrades
the growing mRNA and stop gene
expression. It is highly specific and
reduces the synthesis of particular
proteins by reducing the translation of
specific messenger RNAs. Hence, siRNA
is not required for protein synthesis but
is used to reduce its synthesis.
mRNA is messenger RNA that carries genetic information provided by DNA.
tRNA carries amino acids to the mRNA during translation.
rRNA is structural RNA that forms ribosomes which are involved in translation.
Which is the "Only enzyme" that has "Capability" to catalyse Initiation, Elongation and Termination in the process of transcription in prokaryotes?
A.
DNase
B.
DNA dependent DNA polymerase
C.
DNA dependent RNA polymerase
D.
DNA Ligase
Correct Answer: C
Explanation:
Prokaryotes utilize one RNA
polymerase for transcription of all
types of RNA. The enzyme RNA
polymerase is needed for RNA
formation from DNA, i.e. DNA
dependent RNA polymerase. It occurs
in the cytoplasm of prokaryotic cells.
RNA polymerase is the only enzyme
which, has the capability to catalyse
all initiation, elongation and
termination in prokaryotes.
Which one of the following statements about histones is wrong?
A.
Histones carry positive charge in the side chain
B.
Histones are organized to form a unit of 8 molecules
C.
The pH of histones is slightly acidic
D.
Histones are rich in amino acids - Lysine and Arginine
Correct Answer: C
Explanation:
$\bullet$ Histones are rich in basic amino acids residue lysine and arginine with charged side chain.
$\bullet$ There are five types of histone proteins i.e. H1, H2A, H2B, H3 and H4. Four of them occur in pairs to produce a unit of 8 molecules (histone octamer).
If the distance between two consecutive base
pairs is 0.34 nm and the total number of base
pairs of a DNA double helix in a typical
mammalian cell is 6.6 × 109 bp, then the length
of the DNA is approximately
A.
2.7 meters
B.
2.0 meters
C.
2.5 meters
D.
2.2 meters
Correct Answer: D
Explanation:
Length of DNA = [0.34 × 10–9] m × 6.6 ×
109 bp = 2.2 m
Distance between 2 base pair in DNA helix =
0.34 nm = 0.34 × 10–9 m
Based on the observation of Erwin
Chargaff that for a double stranded DNA,
the ratios between Adenine and Thymine and
Guanine and Cytosine are constant and equals
one. Adenine pairs with thymine through two
H-bonds i.e., A = T and guanine pairs with
cytosine with three H-bonds.
The first phase of translation involves
activation of amino acid in the presence of ATP
and linked to their cognate tRNA - a process
commonly called as charging of tRNA or
aminoacylation of tRNA.
Name the enzyme that facilitates opening of
DNA helix during transcription.
A.
DNA polymerase
B.
RNA polymerase
C.
DNA helicase
D.
DNA ligase
Correct Answer: B
Explanation:
The enzyme that facilitates opening of the DNA helix during transcription is RNA polymerase.
RNA polymerase is responsible for the synthesis of RNA from a DNA template during transcription. It binds to the DNA at a specific site called the promoter region and then unwinds the DNA helix to expose the template strand. This opening of the DNA helix is facilitated by the activity of RNA polymerase, which breaks the hydrogen bonds between the base pairs of the DNA.
Once the DNA is unwound, RNA polymerase can begin the synthesis of RNA by adding complementary RNA nucleotides to the template strand. The RNA polymerase moves along the DNA strand, opening the helix as it goes, and synthesizing a new RNA molecule.
Therefore, option B, "RNA polymerase," is the correct answer.
Which of the following features of genetic code does allow bacteria to produce human insulin by
recombinant DNA technology?
A.
Genetic code is nearly universal
B.
Genetic code is specific
C.
Genetic code is not ambiguous
D.
Genetic code is redundant
Correct Answer: A
Explanation:
In recombinant DNA technology, a bacterium is able to produce human insulin because genetic code is nearly universal. Human insulin is used to treat diabetes.
Key idea Genetic code is a sequence of nucleotides in nucleic acid (DNA or RNA) that determine the amino acid sequence during protein synthesis.
Genetic codes for arginine are CGU, CGC, CGA and CGG.
CAU and CAC code for histidine. CAA and CAG code for glutamine. AGU and AGC code for serine. AAC codes for asparagine. GAU, GAC code for Aspartic acid. GAA code for glutamic acid.
The human genome contains 3164.7 million nucleotide bases
B.
Less than $10 \%$ of the genome codes for protein
C.
Repeated sequences make up very large portion of the human genome
D.
Chromosome 1 has most genes (2968) and $Y$ has the fewest (231)
Correct Answer: B
Explanation:
Key idea The content of the human genome is commonly divided into coding and non-coding DNA sequences. Coding DNA are those sequences that can be transcribed into mRNA and translated into protein, they occupy only a small fraction of genome. Non-coding DNA sequences do not encode proteins.
Statement in option (b) is wrong. The correct form of statement is "less than $2 \%$ of the genome codes for protein".
Note In most of organisms more than $98 \%$ of human genome is composed of non-coding DNA (or Junk DNA).
Gene library or DNA library has collection of
cDNA only. It is a cloned/copy DNA of
cells/tissues, organs of an organism in a
preserved form for future use.
Key idea The synthesis of RNA (transcription) in prokaryotes utilises a DNA dependent RNA polymerase to catalyse the bacterial RNA's. It complete in three steps - 1. Initiation 2. Elongation 3. Termination.
In prokaryotes, transcription of all types of bacterial RNA is catalysed by a single DNA dependent RNA polymerase in three steps.
Initiation It involves binding of RNA polymerase with the promoter site by associating with a initiation or sigma factor.
Elongation This step involves elongation of RNA chain as the RNA polymerase travel in $5^{\prime} \rightarrow 3^{\prime}$ direction. This step is characterised by dissociation of '$\sigma$' initiation factor from RNA polymerase.
Termination In this step, RNA polymerase on reaching termination region separates and RNA falls off from transcription unit. This step is characterised by binding of termination or rho ($\rho$) factor with RNA polymerase, which marks termination of transcription.
Assertion : In eukaryotes, transcription
occurs in nucleus.
Reason : In bacteria, transcription and
translation occurs in cytoplasm.
A.
If both assertion and reason are true and reason is the correct explanation of assertion.
B.
If both assertion and reason are true, but reason is not the correct explanation of assertion.
C.
If assertion is true, but reason is false.
D.
If both assertion and reason are false.
Correct Answer: B
Explanation:
Both the assertion and reason are true, but
reason is not the correct explanation of assertion.
In eukaryotes, transcription, i.e. synthesis of RNA
from DNA, occurs inside nucleus, because the
genetic material DNA is enclosed inside the
nuclear membrane in eukaryotes whereas, in the
prokaryotes like bacteria, their genetic
material-DNA remains suspended within the
cytoplasm, thus in prokaryotes the process of
transcription and translation occurs in cytoplasm.
Francois Jacob and Jacque Monod
proposed model of gene regulation known
as operon model/lac operon. Alec Jeffreys
gave DNA fingerprinting technique.
Matthew Meselson and F. Stahl gave semiconservative DNA replication in E.coli.
Alfred Hershey and Martha Chase proved
DNA as genetic material not protein.
Many ribosomes may associate with a single
mRNA to form multiple copies of a polypeptide
simultaneously. Such strings of ribosomes are
termed as
A.
polysome
B.
polyhedral bodies
C.
plastidome
D.
nucleosome
Correct Answer: A
Explanation:
Ribosomes may occur singly a monosomes
or in rosettes and helical groups called
polysomes. The different ribosomes of a
polysome are connected with a strand of
m-RNA.
Nucleosome is a basic unit of DNA packaging
in eukaryotes.
Plastidome are the plastids of a cell when they
are referred to a functional unit.
Polyhedral bodies are involved in carbon
fixation are present in autotrophic bacteria.
The codons of glycine are GGU, GGC, GGA,
GGG. A genetic code is the smallest unit to code
one amino acid. This genetic code is defined as
nucleotide sequence of nitrogenous bases, which
specifies the amino acid sequence in a
polypeptide molecule.
During DNA replication, Okazaki fragments are used to elongate
A.
The leading strand away from replication fork
B.
The lagging strand towards replication fork
C.
The lagging strand away from the replication fork
D.
The leading strand towards replication fork
Correct Answer: C
Explanation:
Two DNA polymerase molecules
simultaneously work at the DNA fork, one
on the leading strand and the other on the
lagging strand.
DNA polymerase synthesizes each Okazaki
fragment at lagging strand in 5′-3′ direction.
As the replication fork opens further, new
Pkazaki fragments appear. The first Okazaki
fragment appears away from the replication
fork and thus the direction of elongation
would be away from replication fork.
Spliceosomes helps in removal of introns.
They will not occur in prokaryotes because
prokaryotes do not have introns and thus, processing
does not require splicing of mRNA.
The association of histone H1 with a nucleosome indicates :
A.
DNA replication is occurring
B.
Transcription is occurring
C.
The DNA is condensed into a Chromatin Fibre
D.
The DNA double helix is exposed
Correct Answer: C
Explanation:
Histones help in packaging of DNA. In
eukaryotes, DNA packaging is carried out with the
help of positively charged basic proteins called
histones. Histones are of five types – H1
, H2A, H2B,
H3
and H4
. H1
is attached over the linker DNA.
Histone contains a large proportion of the positively
charged (basic) amino acids, lysine and arginine in
their structure. DNA is negatively charged due to
the phosphate groups on its backbone. The result
of these opposite charges is strong attraction and
therefore, high binding affinity between histones
and DNA.
If there are 999 bases in an RNA that codes for a protein with 333 amino acids, and the base at position 901 is
deleted such that the length of the RNA becomes 998 bases, how many codons will be altered ?
A.
1
B.
11
C.
333
D.
33
Correct Answer: D
Explanation:
If deletion happen at 901st position than
the remaining 98 bases specifying for 33
codons of amino acids will be altered.
Cistron (or gene) is a length of DNA that
contains the information for coding a specific
polypeptide chain or a functional RNA molecule (i.e.,
transfer RNA or ribosomal RNA). Hence, cistron is a
unit of function. Currently such a gene is called
structural gene.
Taylor conducted the experiments to prove semiconservative mode of chromosome replication on
A.
Vicia faba
B.
Drosophila melanogaster
C.
Vinca rosea
D.
E. coli
Correct Answer: A
Explanation:
Taylor et al. (1957) conducted experiment
on Vicia faba (broad bean) to prove semi-conservative
replication of DNA. He fed dividing cells of root tips
of Vicia faba with radioactive 3H containing thymine
instead of normal thymine and found that all the
chromosomes became radioactive. Labelled thymine
was then replaced with normal one. Next generation
came to have radioactivity in one of the two
chromatids of each chromosome while in subsequent
generation radioactivity was present in 50% of the
chromosomes. This is possible only if out of the two
strands of a chromosome, one is formed afresh while
the other is conserved at each replication.
A molecule that can act as a genetic material must fulfill the traits given below, except
A.
it should be able to express itself in the form of ‘Mendelian characters’
B.
it should be unstable structurally and chemically
C.
it should provide the scope for slow changes that are required for evolution
D.
it should be able to generate its replica
Correct Answer: B
Explanation:
Genetic material should be structurally and
chemically stable otherwise its expression will change
and lead to loss of several metabolic functions, etc.
DNA-dependent RNA polymerase catalyzes transcription on one strand of the DNA which is called the
A.
coding strand
B.
antistrand
C.
template strand
D.
alpha strand
Correct Answer: C
Explanation:
The strand of DNA on which RNA
polymerase binds to catalyse transcription is called
template strand. It is also known as master or antisense
strand. It has the polarity of 3' $ \to $ 5'.
The start codon is the first codon of
a messenger RNA (mRNA) transcript
translated by a ribosome. The start codon
always codes for methionine in eukaryotes
and a modified Met (fMet) in prokaryotes.
The most common start codon is AUG.
Which of the following is required as inducer(s) for the expression of Lac operon?
A.
lactose
B.
lactose and galactose
C.
glucose
D.
galactose
Correct Answer: A
Explanation:
In Lac operon, lactose is an inducer. It binds
with suppressor and inactivates it. It allows RNA
polymers access to the promoter and transcription
proceeds.