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CBSE Questions for Class 12 Medical Biology Molecular Basis Of Inheritance Quiz 16 - MCQExams.com
CBSE
Class 12 Medical Biology
Molecular Basis Of Inheritance
Quiz 16
The relative percentages of nitrogenous bases of the genetic material isolated from various species (P, Q, R, S and T) are listed in the following table.
Species
Adenine
Guanine
Thymine
Cytosine
Uracil
P
$$21$$
$$29$$
$$21$$
$$29$$
$$0$$
Q
$$29$$
$$21$$
$$29$$
$$21$$
$$0$$
R
$$21$$
$$21$$
$$29$$
$$29$$
$$0$$
S
$$21$$
$$29$$
$$0$$
$$29$$
$$21$$
T
$$21$$
$$29$$
$$0$$
$$21$$
$$29$$
Choose the correct statement from the following.
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0%
Species P, Q and R contain single-stranded DNA; S and T cannot contain double-stranded RNA
0%
Species P, Q and R contain double-stranded RNA; species S and T contain single-stranded RNA
0%
Species P, Q and R cannot contain single-stranded DNA; species S contains single-stranded RNA; species T contains double-stranded RNA
0%
Species P and Q contain double-stranded DNA; species R contains single-stranded DNA; species S contains double-stranded RNA; species T contains single-stranded RNA
Explanation
DNA has base pairs adenine, thymine, guanine and cytosine. RNA has base pairs adenine, guanine, cytosine and uracil. In a double stranded DNA, number of adenine will always be equal to number of thymine and number of cytosine will always be equal to number of guanine as per Chargaff's rule. In species P, number of adenine are equal to number of thymine. Also, it has same number of guanine and cytosine. It does not have uracil in its genome. Hence, species P is double-stranded DNA. Similarly, species Q is double-stranded DNA. Species S has uracil instead of thymine in equal number as adenine. Also, number of guanine are equal to cytosine. Hence, species S is double-stranded RNA. T has uracil in its genome. Also, number of adenine are not equal to uracil and number of guanine are not equal to cytosine. Hence, species T is single-stranded RNA.
Thus, the correct answer is option D.
Transcription unit
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0%
starts with TATA box
0%
starts-with pallendrous regions and ends with rho factor
0%
starts with promoter region and ends in terminator region
0%
starts with CAAT region
Explanation
Transcription unit is the distance between sites of initiation and termination by RNA polymerase. It may include more than one gene. RNA polymerase produces transcription unit that extends from the promoter to the termination sequences.
Match column I with column II and select the correct option from the-given codes.
Column I
(Codons)
Column II
(Translated amino acid)
A.
UUU
(i)
Serine
B.
GGG
(ii)
Methionine
C.
UCU
(iii)
Phenylalanine
D.
CCC
(iv)
Glycine
E.
AUG
(v)
Proline
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0%
A-(iii), B-(iv),C-(i), D-(v), E-(ii)
0%
A-(iii), B-(i),C-(iv), D-(v), E-(ii)
0%
A-(iii), B-(iv),C-(v), D-(i), E-(ii)
0%
A-(ii), B-(iv),C-(i), D-(v), E-(iii)
Explanation
The genetic codes is a degenerate code that arises from the sequence of mRNA and three bases give rise to a single type of amino acid. As there can be about 64 such combinations of the bases A,U,C and G and there are 20 major amino acids more than one code can represent a single amino acid. According to the universal genetic code UUU codes for Phenylalanine,
GGG codes for Glycine,
UCU codes for Serine, CCC codes for Proline and AUG which is also the start codon of translation codes for methionine.
So, the correct option is '
A-(iii), B-(iv),C-(i), D-(v), E-(ii)'.
Choose the correct answer from the alternatives given :
Watson and Crick (1953) proposed DNA double helix model and won the Nobel Prize; their model of DNA was based on
(i) X-ray diffraction studies of DNA done by Wilkins and Franklin
(ii) Chargaffs base equivalence rule
(iii) Griffiths transformation experiment
(iv) Meselson and Stahls experiment.
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0%
(i), (ii) and (iv)
0%
(i) and (ii)
0%
(iii) and (iv)
0%
(i), (ii), (iii) and (iv)
Explanation
In 1953, James Watson and Francis Crick based on the X-ray diffraction data produced by Maurice Wilkins and Rosalind Franklin, proposed double helix model for the structure of DNA. One of the hallmarks of their proposition was base pairing between two strands of polynucleotide chains. However, this proportion was also based on the observation of Erwin Chargaff (1950) that for a double stranded DNA, the ratios between Adenine and Thymine and Guanine and Cytosine are constant and equals one.
So, the correct answer is '(i) and (ii)'.
Choose the correct answer from the alternatives given :
Other than DNA polymerase, which are the enzymes involved in DNA synthesis?
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0%
Topoisomerase
0%
Helicase
0%
RNA primase
0%
All of these
Explanation
Process of DNA synthesis whereby a parent DNA molecule is faithfully copied, giving rise to two identical daughter molecules is called DNA replication. In DNA synthesis, DNA polymerase plays an important role having the capability to elongate an existing DNA strand but cannot initiate the synthesis. So, the synthesis is initiated with the help of RNA primer formed by RNA primase RNA primase synthesises the short RNA primer of about 10 nucleotides that is elongated by DNA polymerase to form an Okazaki fragment of DNA during DNA replication. Helicase unzips the two strands of DNA and topoisomerase reduces the coiling tension developed due to the unwinding of the two strands.
So the correct answer is 'All of these'.
Choose the correct answer from the alternatives given :
Match column I with column II and select the correct option from the given codes.
Column I
(Scientists)
Column II
(Discoveries)
A.
Alec Jeffreys
(i)
Lac operon
B.
F. Sanger
(ii)
Annotated DNA sequences
C.
Jacob and Monod
(iii)
DNA fingerprinting
D.
Avery, MacLeod and McCarty
(iv)
Transforming principle
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A - (ii), B - (iii), C - (iv), D - (i)
0%
A - (iii), B - (ii), C - (i), D - (iv)
0%
A - (iii), B - (ii), C - (iv), D - (i)
0%
A - (i), B - (ii), C - (iii), D - (iv)
Explanation
Alec Jeffreys is a British geneticist, who developed techniques for genetic fingerprinting and DNA profiling which are now used worldwide in forensic science to assist police detective work and to resolve paternity and immigration disputes. F.Sanger developed the Sanger dideoxy method of sequencing DNA and also is credited for the sequencing of fragments of automated DNA sequences called annotated DNA sequences. This was very helpful for the human genome project that could sequence the entire genetic makeup of the human being. In 1961, Francois Jacob and Jacques Monod proposed the operon model of gene regulation in bacteria. The model was based on their study of the genes in E. coli that code for enzymes that affect the breakdown of lactose also called the Lac Operon. Avery, MacLeod and McCarty experimentally proved that DNA is the substance or the genetic material that causes transformation in bacteria.
So, the correct option is '
A - (iii), B - (ii), C - (i), D - (iv)'.
Choose the correct answer from the alternatives given :
Read the following statements.
(i) One codon codes for only one amino acid.
(ii) Some amino acids are coded by more than one codon.
(iii) The, sequence of triplet nitrogenous bases in DNA or mRNA corresponds to the amino acid sequence in the polypeptide chain.
Give suitable terms for the characteristics of genetic code as per the above statements.
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Degeneracy - (i), Colinearity - (ii), Unambiguous - (iii)
0%
Degeneracy - (iii) , Colinearity - (ii) , Unambiguous - (i)
0%
Degeneracy - (ii) , Colinearity - (iii) , Unambiguous - (i)
0%
Degeneracy - (i) , Colinearity (ii) , Unambiguous - (iii)
Explanation
As there are 64 combination of the base pairs A,G,U and C possible that are required to code for the 20 major amino acids more than one code can give rise to a single type of amino acid. This is called degeneracy and is
exhibited as the multiplicity of three-base pair codon combinations that specify an amino acid.
The
genetic code
is a three-letter (triplet)
code
defining the transfer of the information from nucleic acids to proteins. Codon is a successive string of three nucleotide and this corresponds to the amino acid sequence in the polypeptide chain. This property of the genetic code is called Colinearity. Even though one amino acid can be coded by different codes, a particular code can give rise to only to a specific amino acid and this is the unambiguous nature of the genetic code.
Each codon specifies one amino acid only.
So, the correct option is '
Degeneracy - (ii) , Colinearity - (iii) , Unambiguous - (i)'.
Select the correct option that correctly fill the blanks i - iv
I. Less than
(i)
of genome represents structural genes that code for proteins.
II. Chemical substance that binds with repressor and convert it into a non-DNA binding State is
(ii)
.
III. In prokaryotes, during replication RNA primer is removed by
(iii)
whereas in eukaryotes it is removed by
(iv)
.
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(i) - 5%, (ii) regulator, (iii) - DNA polymerase II, (iv) - DNA polymerase $$\beta$$
0%
(i) - 10%, (ii) - regulator, (iii) - DNA polymerase I, (iv) - DNA polymerase $$\alpha$$
0%
(i) - 2%, (ii) inducer, (iii) - DNA polymerase I, (iv) - DNA polymerase $$\beta$$
0%
(i) - 50%, (ii) inducer, (iii) - DNA polymerase III, (iv) - DNA polymerase $$\alpha$$
Explanation
The human genome is made of 3 billion DNA base pairs and out of them 2% only are responsible for coding for proteins and the rest may have regulatory functions. In an inducible operon such as the Lac Operon the inducer binds to the repressor to inactivate it such that the repressor can no longer bind to the operator and the gene expression can take place.
DNA polymerase
I (or Pol I) is an enzyme that participates in the process of prokaryotic DNA replication. This is responsible for proof reading and excising or removing off the RNA primers required for initiation of replication.
DNA polymerase
, beta, also known as
POLB
, is an enzyme present in eukaryotes
involved in base excision and repair, also called gap-filling DNA synthesis.
So, the correct option is '
(i) - 2%, (ii) inducer, (iii) - DNA polymerase I, (iv) - DNA polymerase
β '.
Choose the correct answer from the alternatives given :
If Meselson and Stahls experiment is continued for four generations in bacteria, the ratio of $$N^{15}$$- $$N^{15}$$, $$N^{14}$$ - $$N^{15}$$, $$N^{14}$$ - $$N^{14}$$ containing DNA in the fourth generation would be
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0%
1:1:0
0%
1:4:0
0%
0:1:3
0%
0:1:7
Explanation
As per Messelson and Stahl's experiment, parent DNA was first isolated from E.coli grown in heavy $$^{15}\texterm{N}$$ medium. It was then put in light $$^{14}\texterm{N}$$ medium. DNA shows semi conservative nature, hence, in fourth generation, the ratio of $$^{15}\texterm{N}/^{15}\texterm{N} \, : \, ^{15}\texterm{N}/^{14}\texterm{N} \, : \, ^{14}\texterm{N}/^{14}\texterm{N}$$ will be 0 : 1 : 7.
DNA is a polymer of nucleotides which are linked to each other by 3-5 phosphodiester bond. To prevent polymerisation of nucleotides, which of the following modifications would you choose?
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0%
Replace purines with pyrimidines.
0%
Remove/Replace 3 OH group in deoxyribose.
0%
Remove/Replace 2 OH group with some other group in deoxyribose.
0%
Both (b) and (c).
Explanation
If 3 OH group is removed/replaced in deoxyribose, there will be no formation of pliosphodiester bonds and hence polymerisation of nucleotides will be prevented.
The promoter site and the terminator site for transcription are located at
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0%
3 (downstream) end and 5 (upstream) end, respectively of the transcription unit
0%
5 (upstream) end and 3 (downstream) end, respectively of the transcription unit
0%
the 5 (upstream) end
0%
the 3 (downstream) end.
Explanation
The promoter sequence needs to lie in front of the start site or upstream to it in the 5' end and the terminator should lie in the 3' end downstream as transcription proceeds in the 5' to 3' direction.
So, the correct answer is, '
5 (upstream) end and 3 (downstream) end, respectively of the transcription unit'.
Which of the following statements are correct?
(i) RNA polymerase I transcribes rRNAs.
(ii) RNA polymerase II transcibes snRNAs.
(iii) RNA polymerase III transcibes hnRNA,
(iv) RNA polymerase II transcribes hnRNA.
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0%
(i) and (ii)
0%
(i) and (iii)
0%
(i), (ii) and (iv)
0%
(i) and (iv)
Explanation
RNA polymerase 1 (also known as Pol I) is, in higher eukaryotes, the polymerase that only transcribes ribosomal RNA (but not 5S rRNA, which is synthesized by RNA polymerase III), a type of RNA that accounts for over 50% of the total RNA synthesized in a cell. Pol I is a 590 kDa enzyme that consists of 14 protein subunits (polypeptides). Twelve of its subunits have identical or related counterparts in RNA polymerase II (Pol II) and RNA polymerase III (Pol III). The other two subunits are related to Pol II initiation factors and have structural homologues in Pol III. Ribosomal DNA transcription is confined to the nucleolus, where about 400 copies of the 42.9-kb rDNA gene are present, arranged as tandem repeats in nucleolus organizer regions. Each copy contains a 13.3 kb sequence encoding the 18S, the 5.8S, and the 28S RNA molecules, interlaced with two internal transcribed spacers, ITS1 and ITS2, and flanked upstream by a 5' external transcribed spacer and a downstream 3' external transcribed spacer. Because of the simplicity of Pol I transcription, it is the fastest-acting polymerase and contributes up to 60% of cellular transcription levels in exponentially growing cells.
RNA polymerase II (RNAP II and Pol II) is a multiprotein complex. It is one of the three RNAP enzymes found in the nucleus of eukaryotic cells. It catalyzes the transcription of DNA to synthesize precursors of mRNA and most snRNA and microRNA. A 550 kDa complex of 12 subunits, RNAP II is the most studied type of RNA polymerase. A wide range of transcription factors are required for it to bind to upstream gene promoters and begin transcription. It catalyzes the transcription of DNA to synthesize precursors of mRNA and most snRNA and microRNA. A 550 kDa complex of 12 subunits, RNAP II is the most studied type of RNA polymerase. A wide range of transcription factors are required for it to bind to upstream gene promoters and begin transcription.
So the correct option is ' (i) and (ii)'.
Which of the following statements regarding human genome is incorrect?
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Human genome consists of $$3 \, \times \, 10^9$$ bp and about 20,500 genes.
0%
The average gene size is 3000 bop and dystrophin is the largest known human gene.
0%
Chromosome 1 contains maximum (2968) number of genes and Y-chrbmosome has the least (231) number of genes.
0%
Repeated (or repetitive) sequences are not present in human genome.
Explanation
Repeated or repetitive sequences make up large portion of human genome. Repetitive sequences are nucleotide sequences that are repeated many times, sometimes hundred to thousand times. They have no direct coding function but provide information as to chromosome structure, dynamics and evolution. Approximately 1 million copies of short 5-8 base pair repeated sequences are clustered around centromeres and near the ends of chromosomes. They represent junk DNA.
So, the correct answer is 'Repeated (or repetitive) sequences are not present in human genome'.
Choose the correct answer from the alternatives given :
A DNA template plus primer with the structure
3
P
- TG C GAATTAG C GACAT -
P
5
5
P
- ATCGGTACGACGCTTAAC OH 3
(where
P
= a phosphate group) is placed in an in vitro DNA synthesis system containing $$Mg^{2+}$$, an excess of the four deoxyribonucleoside triphosphates, etc. and a mutant form of E. Coli DNA polymerase I that lacks 5 $$\rightarrow$$ 3 exonuclease activity. The 5 $$\rightarrow$$ 3 polymerase and. 3 $$\rightarrow$$ 5 exonuclease activities of this aberrant enzyme are identical to those of normal E. Coli DNA polymerase I. It simply has no 5 $$\rightarrow$$ 3 exonuclease activity. What will be the structure of the final product?
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0%
3P-TGCGAATTAGCGACAT- P 5
5 P -ATCG GTAC GACG CTTAATCG CTGTA-OH 3
0%
3 P -T6CGAATTGGCGACAT- P 5
5 P -ATCG GTACGAC GCTTAAC C G CTGTA-OH 3
0%
3HOTGCGAATTAGCGACAT P 5
5-ATCG GTAC GACG CTTAATCG CTGTA - P 3
0%
3 P TGCGAATTAGCGACAT P 5
5 P -ACGCTTAATCGCTGTA-OH3
Explanation
The DNA synthesis will not occur on the left end because all DNA polymerases require a free 3 OH terminus, but here the 3 terminus of potential primer strand is blocked with a phosphate group.
How many amino acids will be coded by the mRNA sequence - 5 CCCUCAUAGUCAUAC3 if a adenosine residue is inserted after 12th nucleotide?
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0%
Five amino acids
0%
Six amino acids
0%
Two amino acids
0%
Three amino acids
Explanation
In the given mRNA sequence there are 15 nucleotide . One amino acid codes
triplet
three nucleotide. So, after addition of adenosine residue the total number of nucleotide is 16. So five amino acids will be coded by the mRNA sequence.
The enzyme polynucleotide phosphorylase randomly assembles nucleotides into a polynucleotide phosphorylase to a solution of adenosine triphosphate and guanosine triphosphate, how many types artificial $$mRNA$$ 3 nucleotide codons would be possible?
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0%
$$3$$
0%
$$4$$
0%
$$8$$
0%
$$16$$
Explanation
In the question, it is mentioned only adenosine and guanosine nucleotides are provided and from them 3 nucleotide codons have to be synthesized. So, the three sites of the codon can be occupied by either A or G. So, the possible combinations can give rise to 2 X 2 X 2 = 8 types of codons.
So, the correct option is '8'.
During autocatalysis of DNA, which one of the following provide free OH group and is also used to anchor for addition of new DNA nucleotides?
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0%
DNA polymerase
0%
Ligase
0%
RNA primer
0%
Topoisomerase
Explanation
A primer is a short single strand of RNA or DNA (generally about 18-22 bases) that serves as a starting point for DNA synthesis. It is required for DNA replication because the enzymes that catalyze this process, DNA polymerases, can only add new nucleotides to an existing strand of nucleotides. Since primase produces RNA molecules, the enzyme is a type of RNA polymerase. Primase functions by synthesizing short RNA sequences that are complementary to a single-stranded piece of DNA, which serves as its template. It is critical that primers are synthesized by primase before DNA replication can occur.
So, the correct option is 'RNA primer'.
If the sequence if the nitrogenous bases on template strand is ATCTGGCGT, what would be the sequence of mRNA transcribed from it?
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0%
AUCACCGCU
0%
UUCUGGCGU
0%
UAGACCGCA
0%
AUGUCCGCU
Explanation
The base complementarity occurs between adenine and thymine/uracil or guanine and cytosine. As from the template strand, RNA has to be transcribed so the growing RNA strand will incorporate uracil in it complementary to adenine. The growing strand can only incorporate thymine complementary to adenine when it is a DNA.
So, the correct option is 'UAGACCGCA'.
How many amino acids will be coded by the sequence if the 14$$^{th}$$ base of given mRNA, converts to G?
5' AUG UUU CUC UAG CCG 3'
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0%
Three
0%
Five
0%
Four
0%
Two
Explanation
In the given sequence of mRNA,
First codon is AUG which is the start codon and also codes for methionine.
Second codon is UUU which codes for phenylalanine.
Third codon is CUC which codes for leucine.
Fourth codon is UAG which is the stop codon and does not code for any amino acid.
Now, the translation will terminate at this stop codon and the codons after this stop codon will not be translated as well. So, if the 14$$^{th}$$ base of given mRNA converts to G, the codon CCG will become CGG but this will not have any impact on the translated amino acids because translation will terminate at UAG. Hence, only three amino acids will be produced.
So, the correct answer is 'Three'.
Which one of the incorrect statement regarding DNA polymerase III?
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0%
It is used to ligate any DNA fragment
0%
Can directly bind to ori site
0%
It is isolated from a virus
0%
All of (A), (B) and (C)
Explanation
A. Ligation refers to the joining of two DNA fragments through the formation of a phosphodiester bond. An enzyme known as ligase catalyzes the ligation reaction.
B. During initiation, proteins bind to the origin of replication while helicase unwinds the DNA helix and two replication forks are formed at the origin of replication.
C. It is not isolated from a virus.
DNA polymerase III is essential for the replication of the leading and lagging strands.
So, all statements are incorrect regarding DNA polymerase III.
Which of the following enzyme creates a homonucleotide overhang at the $$3'$$-OH end of the DNA? It can also make blunt ends sticky by adding poly AA and poly TT.
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0%
Terminal Transferase
0%
Methylase
0%
Alkaline phosphatase
0%
DNA Pol III
Explanation
Blunt ends of the end are the simple ends of DNA which are non cohesive, the blunt ends terminate in a base pair so they can't be further extended.
While Terminal transferase are non template enzymes which catalyze the reaction of addition of nucleotide at the 3' prime end of the DNA polymer.
They are also used in the PCR technique to add Adenine and Thymine to the overhang of the DNA strand to create a sticky end so that a circular DNA can be formed.
Adenine and thymine form a base pair and that is why they are used to create sticky ends by the enzyme
Therefore the answer option A is correct.
Analysis of DNA of an organism revealed the following characters. Mark the option that is not likely to be true if the organism is a cat.
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0%
The double helix is antiparallel & a linear molecule
0%
20 bases comprising of a mixture of deoxyribonucleotide phosphates (dNTPs) were found in a single $$360^o$$ turn of helix
0%
17 % of the nitrogenous bases were shown to be cysteines
0%
Uracil was lacking but guanine was 17 %
Explanation
The DNA double helix polymer of nucleic acid held together by nucleotides which base pair together. In B-DNA, the most common double helical structure found in nature, the double helix is right-handed with about 10–10.5 base pairs per turn. The double helix is antiparallel & a linear molecule and uracil are not present in DNA.
So the correct option is '
17 % of the nitrogenous bases were shown to be cysteines'.
If DNA was made of $$6$$ nucleotides instead of $$4$$, what are the total number of triplet codons possible?
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0%
$$24$$
0%
$$18$$
0%
$$64$$
0%
$$216$$
Explanation
If DNA made of 4 nucleotide, then due triplet nature of codon .
No. of triplet codon possible =
4
3
$$\rightarrow$$ 4x4x4 $$\rightarrow$$ 64
Out of 64 codon, 61 code for 20 amino acids and 3 codon are stop codon.
If DNA made of 6 nucleotide -
No. of triplet codon possible =
6
3
$$\rightarrow$$ 6x6x6 $$\rightarrow$$ 216
So, the correct answer is ' 216 '
In
E.coli,
complete DNA replication requires ____ mins.
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0%
38 hours
0%
20 minutes
0%
40 minutes
0%
3 minutes
Explanation
DNA replication is the process by which DNA makes a copy of itself during cell division. It takes
E. coli
about 40 minutes to duplicate its genome of 4.6 × 10$$^6$$ nucleotide pairs.
So, the correct answer is option C.
DNA strands are anti parallel because of
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0%
H-bonds
0%
Phosphodiester bonds
0%
Disulphide bonds
0%
Peptide bonds
Explanation
DNA is double stranded, and the strands are antiparallel because they run in opposite directions. Each DNA molecule has two strands of nucleotides. Each strand has a sugar-phosphate backbone, but the orientation of the sugar molecule is opposite in the two strands. Thus the main reason of this antiparallel nature of the DNA is the presence of a phosphodiester bond that links the phosphate group to the hydroxyl group of the sugar molecule (present in alternate directions on either strand).
So, the correct option is 'phosphodiester bonds'.
Choose the incorrect statement regarding DNA structure.
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0%
Base stacking destabilise DNA double helical structure.
0%
Each chain in DNA constitute sugar-phosphate backbone
0%
Two chains are coiled in right handed fashion
0%
The pitch of the helix has roughly 10 bp.
Explanation
Each chain in DNA constitute sugar-phosphate backbone,
Two chains are coiled in right-handed fashion in case of B-DNA and the
pitch of the helix has roughly 10 bp in it. However, apart from it the base stacking interactions between the nitrogenous bases of the DNA help to stabilize the double helix structure.
So, the correct option is '
Base stacking destabilize DNA double helical structure'.
A couple claimed in court that a child belonged to them. Their claim can be true if the DNA fingerprint pattern of the child shows:
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0%
$$50\%$$ bands similar to father and $$50\%$$ similar to another DNA fingerprint pattern
0%
$$100\%$$ similarity to both the parents DNA fingerprint as both contribute equally to zygote formation
0%
$$100\%$$ similarity to mother's DNA fingerprint because of material inhentance
0%
$$100\%$$ similarity to father's DNA fingerprint due to large number of mitochondria in sperm
DNA fingerprinting can't be performed without:
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0%
Restriction endonucleases
0%
Agarose
0%
Monoclonal antibodies
0%
Both (a) and (b)
Which of the following is not associated with DNA fingerprinting?
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0%
Southern blotting
0%
Gel electrophoresis
0%
Restriction digestion
0%
DNA sequencing
Find the odd one out
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0%
Transcription
0%
Transduction
0%
Translocation
0%
Mutation
Explanation
Correct Option:
A
Explanation:
Transcription
is defined as the process of formation of
mRNA
from DNA template. It is the
copying
of genetic message coded in DNA into mRNA molecule. The transcribed RNA moves out of the nucleus to the
ribosomes
in the
cytoplasm
.
Transduction
is the
transfer
of genetic material from a
virus
into a
bacterium
so as to infect it.
Translocation
is the process in which enzyme
translocase
shifts ribosome on
mRNA by one
codon
, so that the
tRNA
at A-site carrying the dipeptide, moves to P-site.
Mutation
is a sudden
change
in the DNA sequence that
alters
the gene.
Hence,
transcription
is the
odd one out
as it is not a part of protein synthesis.
Join properly Column-I and Column-II.
Column I
Column II
(A) UUU
$$(1)$$ Lysin
(B) GGC
$$(2)$$ Stop codon
(C) AAA
$$(3)$$ Phenyl Alanine
(D) UAA
$$(4)$$ Glycine
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0%
a-$$2$$, b-$$3$$, c-$$4$$, d-$$1$$
0%
a-$$4$$, b-$$3$$, c-$$1$$, d-$$2$$
0%
a-$$1$$, b-$$4$$, c-$$3$$, d-$$2$$
0%
a-$$3$$, b-$$4$$, c-$$1$$, d-$$2$$
Explanation
There are 20 amino acids, which have the triplet codons are code for the process translation UUU codes for phenylalanine, GGC codes for glycine. AAA codes for lysin and UAA is known as stop codon.
Hence, the correct answer is option D.
If the length of DNA strand is $$ { 68A }^{ \circ }$$ then, what will be the number of nucleotides present in DNA molecule
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0%
20
0%
40
0%
60
0%
80
Explanation
The distance between consecutive nitrogenous bases is 3.4 A°.
68 A° means the DNA strand has 68 ÷ 3.4 = 20 base pairs.
Each base pair has 2 bases, hence there are 40 bases.
As we know, 4 nucleotides form 2 base pairs then 40 nucleotides can form 20 base pairs.
So, the correct option is '40'
Which of the following is not include in transcription unit?
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0%
Promoter region
0%
Terminator region
0%
Structural gene
0%
Replicon
The whole genome of Escherichia coli having $$4.6\times 10^6$$ bp will be replaced in?
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0%
$$2$$ minutes
0%
One hour
0%
$$40$$ minutes
0%
$$38$$ minutes
A codon has a sequence of A, and specifies a particular B that is to be incorporated into a C. What are A, B, and C?
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0%
3 bases, amino acid, and carbohydrate
0%
3 acids, carbohydrate, and protein
0%
3 bases, protein, and amino acid
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3 bases, amino acid, and protein
The technique of DNA fingerprint is superior to conventional fingerprint because it can:
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Be generated more rapidly and is inexpensive
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Generate unique fingerprint for each finger
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Compare the whole DNA sequence of two individuas
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Differences between polymorphic DNA sequences among individuals
Which of the following statement is incorrect about DNA fingerprinting?
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It is used to identify racial groups.
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Show high degree of polymorphism.
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It is used to detect sex during fetal development.
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It is used in medico-legal suits.
Which of the following is codons codes for proline.
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CCC, CCU, CCG
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UCC, UGU, CCU
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CUG, CUU, CUA
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CGC, CGG, CCA
Explanation
The genetic code for Proline is CCU, CCG, CCA and CCG.
So, the correct answer is '
CCC, CCU, CCG'
The nucleic acid synthesis takes place in
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3-5 direction
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5’-3’ direction
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4’-3’ direction
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6’-3’ direction
Explanation
A nucleic acid can be synthesized only in the direction of 5-3. The main feature of nucleic acids is that they have two distinctive ends: the 5' (5-prime) and 3' (3-prime) ends. This terminology refers to the 5' and 3' carbons on the sugar. The process involves forming phosphodiester bonds between the 3' carbon of one nucleotide and the 5' carbon of another nucleotide.
Thus the correct answer is option B.
Which of the following statements is correct
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Adenine pairs with thymine through one H-bond
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Adenine pairs with thymine through three H-bonds
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Adenine does not pair with thymine
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Adenine pairs with thymine through two H-bonds
Explanation
Adenine pairs with thymine through two
H-bonds i.e., A == T
Hence, the correct answer is (D).
If the distance between two consecutive base pairs is 0.34 nm and the total number of base pairs of a DNAdouble helix in a typical mammalian cell is $$6.6 \times 10^9$$ bp, then the length of the DNA is approximately:
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2.5 meters
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2.2 meters
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2.7 meters
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2.0 meters
Explanation
Distance between 2 base pair in DNA helix
= $$0.34$$nm = $$0.34 × 10^{–9}$$m
Total number of base pair = $$6.6 × 10^9$$ bp
Length of DNA = $$[0.34 × 10^{–9}]$$m $$× 6.6 × 109$$ bp
= $$2.2$$m
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Practice Class 12 Medical Biology Quiz Questions and Answers
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