Molecular Basis of Inheritance Class 12 MCQ Questions With Answers
Molecular Basis Of Inheritance MCQ Class 12 Question 1.
In a DNA strand, the nucleotides are linked together by
(A) glycosidic bonds
(B) phosphodiester bonds
(C) peptide bonds
(D) hydrogen bonds
Answer:
(B) phosphodiester bonds
Explanation :
In a DNA strand; the nucleotides are linked together by 3′-5′ phosphodiester linkage (bonds) to form a dinucleotide. More nucleotides can be joined in such a manner to form a polynucleotide chain.
Class 12 Biology Chapter 6 MCQ Question 2.
A nucleoside differs from a nucleotide. It lacks the
(A) base
(B) sugar
(C) phosphate group
(D) hydroxyl group
Answer:
(C) phosphate group
Explanation :
A nitrogenous base is attached to the pentose sugar by an N-glycosidic linkage to form a nucleoside, that is, Nucleoside = Nitro¬gen base + Pentose sugar When a phosphate group is attached to the 5′-OH of a nucleoside through phosphodiester linkage, a nucleotide is formed, that is, Nucleotide = Nitrogen base + Pentose sugar + Phosphate (P04). So, a nu¬cleoside differs from a nucleotide as it lacks the phosphate group.
Molecular Basis Of Inheritance Class 12 MCQ Question 3.
Both deoxyribose and ribose belong to a class of sugars called
(A) trioses
(B) hexoses
(C) pentoses
(D) polysaccharides
Answer:
(C) pentoses
Explanation :
Both deoxyribose and ribose belongs to the class pentoses as it contains ‘5’ carbon atoms.
Chapter 6 Biology Class 12 MCQs Question 4.
The fact that a purine base always paired through hydrogen bonds with a pyrimidine base leads to, in the DNA double helix
(A) the antiparallel nature
(B) the semi-conservative nature
(C) uniform width throughout DNA
(D) uniform length in all DNA
Answer:
(C) uniform width throughout DNA
Explanation :
The diameter of the strand is al-ways constant due to a pairing of purine (ad-enine and guanine) and pyrimidine (cytosine and thymine). This specific bonding gives uniform width to the DNA.
MCQ On Molecular Basis Of Inheritance Question 5.
The net electric charge on DNA and histones is
(A) both positive
(B) both negative
(C) negative and positive, respectively
(D) zero
Answer:
(C) negative and positive, respectively
Explanation :
DMA consists of a nitrogenous base, pentose .sugar and a phosphate group. DNA has negative charge due to the presence of phosphate group. Histones are rich in the basic amino acids lysine and arginine, which carry positive charges in their side chains. Therefore, histones are positively charged.
MCQ Of Molecular Basis Of Inheritance Class 12 Question 6.
The first genetic material could be
(A) protein
(B) carbohydrates
(C) DNA
(D) RNA
Answer:
(D) RNA
Explanation :
RNA was the first genetic material. There is now enough evidence to suggest hat essential life processes e.g., metabolism, translation and splicing, evolved from RNA. RNA is used to act as a genetic material as well as a catalyst (there are some important s ” biochemical reactions in living systems that are catalysed by RNA catalysis and not by protein enzymes). But, RNA being a catalyst was y reactive and hence unstable. Therefore, DNA has evolved from RNA with chemical modifications that make it more stable. DNA being double-stranded and having complementary strand, further resists changes by evolving a process of repair.
Molecular Basis Of Inheritance MCQ Class 12 Question 7.
With regard to mature mRNA in eukaryotes
(A) exons and introns do not appear in the mature RNA.
(B) exons appear, but introns do not appear in the mature RNA.
(C) introns appear but exons do not appear in the mature RNA.
(D) both exons and introns appear in the mature RNA.
Answer:
(B) exons appear, but introns do not appear in the mature RNA.
Explanation :
In eukaryotes, the mono-cistron- ic structural genes have interrupted coding sequences, that is, the genes in eukaryotes are Split. The coding sequences or expressed se-quences are defined as exons. These sequences (exons) appear in mature or processed RNA, thus exons are interrupted by introns or inter-vening sequences which do not appear in ma-ture or processed RNA.
Biology Class 12 Chapter 6 MCQ Question 8.
Which of the following are the functions of RNA?
(A) It is a carrier of genetic information from DNA to ribosomes synthesising polypeptides.
(B) It carries amino acids to ribosomes.
(C) It is a constituent component of ribosomes.
(D) All of the above
Answer:
(D) All of the above
Explanation :
Ribosomal RNA (rRNA), messenger RNA (mRNA) and transfer RNA “(tRNA) are major classes of RNAs that are ” involved in gene expression. rRNAs bind protein molecules and give rise to ribosomes. RNA carries coded information for translation into polypeptide formation. RNA – is also called soluble or adaptor RNA and carries amino acids to mRNA during protein synthesis.
MCQ Of Chapter 6 Biology Class 12 Question 9.
While analyzing the DNA of an organism a total number of 5,386 nucleotides were found out of which the proportion of different bases were: Adenine = 29%, Guanine = 17%, Cytosine = 32%, and Thymine = 17%. Considering Chargaff’s rule, it can be concluded that
(A) it is a double-stranded circular DNA.
(B) it is single-stranded DNA.
(C) it is a double-stranded linear DNA.
(D) no conclusion can be drawn.
Answer:
(B) it is single-stranded DNA.
Explanation :
According to Chargaff’s rules of base pairing, (i) The amount of adenine is always equal to the amount of thymine and the amount of guanine is always equal to the amount of cytosine, (ii) Adenine is joined to thymine with two hydrogen bonds and guanine is joined to cytosine by three hydrogen bonds, (iii) The ratio of adenine to thymine and that of guanine to cytosine is always equal to one, that is., AG: TC = 1. In the given or¬ganism, the DNA is not following the Chargaff’s rule, hence it can be concluded that it is a single-stranded DNA, not double-stranded.
MCQ Questions For Class 12 Biology Chapter 6 Question 10.
If the sequence of nitrogen bases of the coding strand of DNA in a transcription unit is: 5′-ATGAATG-3′, the sequence of bases in its RNA transcript would be
(A) 5′-AUGAAUG-3′
(B) 5′-UACUUAC-3′
(C) 5′-CAUUCAU-3’
(D) 5′-GUAAGUA-3′
Answer:
(A) 5′-AUGAAUG-3′
Explanation :
5′-ATGAATG-3′ (coding strand) I ! 5′-TAC ITAC-3′ (complementary strand) 5-AU- I l GAAUG-3′ (RNA)
Class 12 Molecular Basis Of Inheritance MCQ Question 11.
One of the following is true with respect to AUG.
(A) It codes for methionine only.
(B) It is also an initiation codon.
(C) It codes for methionine in both prokaryotes and eukaryotes.
(D) All of the above
Answer:
(D) All of the above
Explanation:
Polypeptide synthesis is signaled by two initiation codons – commonly AUG or methionine codon and rarely GUG or valine codon. AUG serves two main functions. It signals the start of translation and codes for the incorporation of the methionine into the growing polypeptide chain. AUG codes for methionine in both prokaryotes and eukaryotes.
Biology Class 12 Chapter 6 MCQs Question 12.
The promoter site and the terminator site for transcription are located at
(A) 3′ (downstream) end and 5′ (upstream) end, respectively of the transcription unit.
(B) 5′ (upstream) end and 3′ (downstream) end, respectively of the transcription unit.
(C) the 5′ (upstream) end.
(D) the 3′ (downstream) end.
Answer:
(B) 5′ (upstream) end and 3′ (downstream) end, respectively of the transcription unit.
Explanation :
The promoter site and the terminator site for transcription are located at 5’ (up-stream) end and 3′ (downstream) end, respectively of the transcription unit. The promoter is the binding site for RNA polymerase for initiation of transcription.
Molecular Basis Of Inheritance Class 12 MCQ Pdf Question 13.
Which of the following steps in transcription is catalyzed by RNA polymerase?
(A) Initiation
(B) Elongation
(C) Termination
(D) All of the above 0
Answer:
(B) Elongation
Explanation :
The DNA-dependent RNA polymerase helps in transcription by catalyzing the polymerization in only one direction (i.e., 5′-3′).
MCQ On Molecular Basis Of Inheritance Class 12 Question 14.
In some viruses, DNA is synthesized by using RNA as a template. Such a DNA is called
(A) A-DNA
(B) B-DNA
(C) cDNA
(D) rDNA
Answer:
(C) cDNA
Explanation :
In some viruses, like retro-virus- (e.g., HIV), an en/yme called reverse Iranscriptase is used to generate complementary DNA (cDNA) from an RNA template. This process is termed reverse transcription.
Chapter 6 Biology Class 12 MCQ Question 15.
To initiate translation, the mRNA first binds to
(A) the smaller ribosomal subunit.
(B) the larger ribosomal subunit.
(C) the whole ribosome.
(D) No such specificity exists.
Answer:
(A) the smaller ribosomal subunit.
Explanation :
The ribosome consists of structural RNAs and about 80 different proteins. In its inactive state, it exists as two subunits, a large subunit, and a small subunit. When the smaller subunit encounters the mRNA, the process of translation of the mRNA to protein begins.
MCQ Chapter 6 Biology Class 12 Question 16.
Control of gene expression takes place at the level of
(A) DNA-replication
(B) transcription
(C) translation
(D) None of the above 0
Answer:
(B) transcription
Explanation :
Gene expression, which results in a in the formation of a polypeptide, can be regulated at several levels. In eukaryotes, the regulation occurs at the transcriptional level (formation of primary transcript), processing level (regulation of splicing), and transport of mRNA from nucleus to the cytoplasm, and translational level. While in prokaryotes, control of the rate of transcriptional initiation is the predominant site for control of gene expression.
MCQ Molecular Basis Of Inheritance Class 12 Question 17.
Regulatory proteins are the accessory proteins that interact with RNA polymerase and affect its role in transcription. Which of the following statements is correct about regulatory protein?
(A) They only increase expression.
(B) They only decrease expression.
(C) They interact with RNA polymerase, but do not affect the expression.
(D) They can act both as activators and as repressors.
Answer:
(D) They can act both as activators and as repressors.
Explanation:
Regulatory proteins, the accessory proteins that interact with RNA polymerase and affect its role in transcription. It controls the functions of structural genes and are called regulatory genes. Promoters, terminators, operators and repressdr are some important reg-ulatory genes. They can act both as activators and as repressors.
Class 12 Biology Ch 6 MCQ Question 18.
The RNA polymerase holoenzyme transcribes
(A) the promoter, structural gene and the terminator region.
(B) the promoter and the terminator gene.
(C) the structural gene and the terminator regions.
(D) the structural gene only.
Answer:
(C) the structural gene and the terminator regions.
Explanation :
The RNA polymerase holoenzyme transcribes the structural gene and the terminator regions. RNA polymerase consists of a number of subunits, including a sigma factor (transcription factor) that catalyses the process of transcription. It recognises the start signals or promoter region on DNA which then along with RNA polymerase binds to promoter to initiate the transcription. In eukaryotes there are three RNA polymerases I, II and III. The process includes a proofreading mechanism.
Class 12 Biology Molecular Basis Of Inheritance MCQ Question 19.
Which one of the following pairs of codons is correctly matched with their function or the signal for the particular amino add?
(A) GUU, GCU – Alanine
(B) UAG, UGA – stop
(C) AUG, ACG – Start/methionine
(D) UUA, UCA – Leudne
Answer:
(B) UAG, UGA – stop
Explanation :
Three codons UAG, UAA, and GGA are the stop or termination codons.
MCQs On Molecular Basis Of Inheritance Class 12 Question 20.
Which one of the following is not a part of the transcription unit in DNA?
(A) The inducer
(B) A terminator
(C) A promoter
(D) The structural gene
Answer:
(A) The inducer
Explanation :
The segment of DNA that takes part in transcription is called transcription unit, It has three components. (i) a promoter (ii) the If structural gene and (iii) a terminator.
Assertion and Reason Based MCQs
Directions: In the following questions a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as :
(A) Both assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of assertion
(B) Both assertion (A) and Reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(C) Assertion (A) is true but reason (R) is false.
(D) Assertion (A) is false but reason (R) is true.
Chapter 6 Class 12 Biology MCQ Question 1.
Assertion (A): In RNA uracil is present at the place of thymine.
Reason (R): 5-methyl uracil is chemical name of thymine.
Answer:
(B) Both assertion (A) and Reason (R) are true but reason (R) is not the correct explanation of assertion (A).
Explanation:
Both assertion (A) and Reason (R) are true. Since the half-life of the RNA molecule shorter uracil would suffice to achieve the function of RNA. On the other hand, DNA remains same until cell dies/divides. The functions of thymine and uracil are the same.
Question 2.
Assertion (A): Chargaff s rule is applicable to RNA.
Reason (R): RNA contains deoxyribose sugar in them.
Answer:
(D) Assertion (A) is false but reason (R) is true.
Explanation :
According to Chargaffs is rule £ the DNA helix contain equal molar ratio of A & T,G & C …… This does not apply to RNA, as uracil is present in RNA instead of DNA. RNA contains ribose sugar in them.
Question 3.
Assertion (A): The enzyme involved in the continuous replication of PNA strand is DNA polymerase.
Reason (R): The polarity of the template strand is 3’→5′.
Answer:
(C) Assertion (A) is true but reason (R) is false.
Explanation :
DNA polymerase en/.vme is responsible for synthesizing DNA, they add nucleotides one by one to the growing DNA chain, adding those which are complementary to the template. The template strand has polarh ity in 3’→5′.
Question 4.
Assertion (A): Primary transcripts in eukaryotes are non-functional.
Reason (R): Methyl guanosine triphosphate is attached to 5’ – end of hnRNA.
Answer:
(B) Both assertion (A) and Reason (R) are true but reason (R) is not the correct explanation of assertion (A).
Explanation:
Primary transcripts contain both introns and exons, in which introns are non-coding parts.
Question 5.
Assertion (A): In Griffith’s experiment, the dead R strain bacteria were capable of causing the transformation of the live S-strain bacteria.
Reason(R): The S-strain is non-virulent strain.
Answer:
(D) Assertion (A) is false but reason (R) is true.
Explanation :
It was concluded by the transforming principle, that bacteria is transferred from heat-killed S-strain to live R-strain which is nonvirulent.
Question 6.
Assertion (A): Termination codons or stop codons are UAA, UAG, and UGA.
Reason (R): Stop codons represent termination of translation.
Answer:
(A) Both assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of assertion
Explanation :
Both assertion and reason are true, slop codons are also known as non-sense code for an amino add. If they are not present, then protein synthesis will continue and result in defective protein.
Question 7.
Assertion (A): Aminoacylation is an essential step for the synthesis of protein.
Reason (R): It is the process of adding an activated amino add to the acceptor arm of a transfer RNA.
Answer:
(A) Both assertion (A) and reason (R) are true and Reason (R) is the correct explanation of assertion
Explanation :
Both assertion and reason true and reason (R) is the correct explanation of the assertion (A). This attachment is an es-sential step in the synthesis of protein. Bril attachment is brought by aminoacvI-t-RNA synthetase.
Question 8.
Assertion (A): Genetic codes are commaless.
Reason (R): Genetic codes are overlapping.
Answer:
(C) Assertion (A) is true but reason (R) is false.
Explanation:
Genetic codes arc commaless, bul ‘t they are not overlapping as no single base lake part in the formation of more than one codon, therefore Lhey are non-overlapping.
Question 9.
Assertion (A): The newly formed mRNA has same sequence as the coding strand of transcriptional unit with uracil present at place of thymine.
Reason (R): The rule of complementarity guides the formation of DNA and RNA.
Answer:
Option (A) is correct
Explanation :
The sequence of mRNA will be identical to the given sequence of coding strands except for the presence of uracil in place of thymine in mRNA.
Question 10.
Assertion (A): DNA fingerprinting is applied in paternity testing in case of disputes.
Reason(R): It employs the principle of polymorphism in DNA sequences as polymorphisms are inheritable from parent to child.
Answer:
(B) Both assertion (A) and Reason (R) are true but reason (R) is not the correct explanation of assertion (A).
Case-based MCQs
Attempt any four sub-parts from each question. Each sub-part carries 1 mark.
I. Read the following and answer the question Question l to Question S. given below:
The lac operon consists of a regulation gene and three structural genes. The lactose acts as inducer. In the presence of an Inducer such as lactose, the repressor is in inactivated during the interaction. This allows RNA polymerase access to the promoter and transcription proceeds. The repressor is synthesized which in turn binds with the operator region of the operon and prevents RNA polymerase from transcribing the operon.
Question 1.
When the process of Lac operon is blocked by a repressor it represents:
(A) Positive regulation
(B) Negative regulation
(C) sometimes positive sometimes negative
(D) both positive and negative regulation
Answer:
(B) Negative regulation
Explanation :
The lac operon regulation can be in both negative and positive ways. It is a negative control system because expression is typically blocked by an active repressor (the lac repressor) that turns off transcription. And when CAP (catabolite gene activating protein) binds upstream of this operator region near the promoter and transcription increases, this is an example of a positive system.
Question 2.
Identify the correct sequence of the structural genes in the lac operon.
(A) lacA-lacZ-lacY
(B) lacZ-lacA-lacY
(C) lacZ-lacY-lacA
(D) lacA-lacY-lacZ
Answer:
(C) lacZ-lacY-lacA
Explanation :
The lac operon consists of structural genes, and a promoter, terminator, regulator, and operator. The three structural genes are lacZ, lacY, and lace.
Question 3.
Which of the following statement is true in reference to the lac operon process in E.coli?
(i) Galactosidase is the only enzyme produced in large quantities when lac operon is turned on
(ii) The messenger RNA in lac operon is a polycistronic mRNA
(A) Only i is correct
(B) Only ii is correct
(C) Both (i) and (ii) are correct
(D) None of them are correct
Answer:
(B) Only ii is correct
Explanation :
The messenger RNA produced by transcription carries information for the synthesis of all three proteins found in all three structural genes. Hence, it is a polycistronic messenger RNA.
Question 4.
What provides a binding site to RNA polymerase?
(A) Exons
(B) Promoter
(C) Inducer
(D) Repressor
Answer:
(B) Promoter
Explanation :
Promoter helps in starting the process transcription and provides a binding site to RNA polymerase.
Question 5.
The lac operon of E. coli contains genes involved in lactose metabolism. It’s expressed only when lactose is (1) and glucose is (2).
(A) 1: Present, 2: Absent
(B) 1: Absent, 2: Present
(C) 1: More, 2: less
(D) 1: repressed, 2: promoted
Answer:
(A) 1: Present, 2: Absent
Explanation:
The lac operon of E. coli contains genes involved in lactose metabolism. It’s ex-pressed only when lactose is present and glucose is absent.
II. Read the following text and answer the following questions on the basis of the same :
DNA, a long polymer of a deoxyribonucleotide. Altmann and these substances to be acidic hence he named nucleic acid. The basic unit of DNA is a nucleotide which has three components nitrogenous base, a pentose sugar (deoxyribose), and a phosphate group. There are two types of nitrogenous bases in DNA, Purine, and Pyrimidine. Watson and F. Crick proposed a double helix model for the structure of DNA. There are four types of DNA i.e., A, B, C, Z.
Question 1.
Which DNA form has maximum number of base pairs per turn?
(A) A-DNA
(B) B-DNA
(C) C-DNA
(D) Z-DNA.
Answer:
(D) Z-DNA.
Explanation :
Z DNA is the zigzag DNA that has maximum number of base pairs per turn. It is left-handed helix. There are 12 base pairs per turn, with a rise of 0.38 nm per base pair.
Question 2.
Which among the following does not confer stability to the helical structure of DNA?
(A) Phosphodiester bond
(B) H-bond
(C) N-glycosidic linakge
(D) All of these.
Answer:
(C) N-glycosidic linkage
Explanation :
N-glycosidic linkage in DNA is the nitrogen carbon linkage between the nitrogen of purine or pyrimidine bases and the carbon of the sugar group. This bond does not provide stability.
Question 3.
Cytidine is a :
(A) Nucleoside
(B) Nitrogen base
(C) Nucleotide
(D) Common dinucleotide in DNA and RNA.
Answer:
(A) Nucleoside
Explanation :
Cytosine (C) is a pyrimidine i.e., a nitrogenous base. A combination of a nitrogenous base (purine or pyrimidine) with a pen-tose sugar is known as a nucleoside. Thus, the combination of cytosine with ribose sugar results in the formation of a nucleoside called as cytidine.
Question 4.
Heaviest molecule of protoplasm is :
(A) Lipids
(B) Proteins
(C) DNA
(D) RNA.
Answer:
(C) DNA
Explanation :
The heaviest molecule in the protoplasm is DNA. DNA is a compound of very high molecular weight (over one million), it has a giant molecule made of smaller molecules linked together, but its molecular weight is variable.
Question 5.
Phosphoric acid is found in :
(A) Nucleic acids
(B) NAD and FAD
(C) Phosphoprotein
(D) All of these.
Answer:
(D) All of these.
Explanation:
Phosphoric acid. H3P04 provides the unit that holds the various segments of the nucleic acid chain to each other. It is present in all the three I;e: Nucleic acids; NAD and FAD; Phosphoprotein.
III. The DNA replication is semi-conservative is proved by an experiment conducted by Meselson and Stahl in 1958. To perform their experiment they use heavy nitrogen (15N) in E. coli. The process of replication in living cells requires a set of enzymes. The main enzyme is DNA-dependent DNA polymerase. The DNA-A-dependent DNA polymerase catalyzes polymerization only in one direction, that is 5→3′. In eukaryotes, the replication of DNA takes place at the S-phase of the cell cycle.
Question 1.
Viruses grown in the presence of radioactive phosphorus contained radioactive but not radioactive.
(A) DNA, protein
(B) Protein, DNA
(C) RNA, Nucleoside
(D) mRNA, Protein
Answer:
(A) DNA, protein
Explanation:
DNA is the genetic material came from the experiments of Hershey and Chase (1952). They worked with viruses that infect bacteria called bacteriophages. They worked to discover whether it was protein or DNA from the virus that entered the bacteria. They grew some viruses on a medium that contained radioactive phosphorus and some others on a medium that contained sulfur. Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein because DNA contains phosphorus but protein does not. Similarly, viruses grown on radioactive sulfur contained radioactive protein but not radioactive DNA because DNA does not contain sulfur.
Question 2.
During DNA replication, the breaking of H-bonds is performed by:
(A) Topoisomerase
(B) Gyrase
(C) Helicases
(D) None.
Answer:
(C) Helicases
Explanation :
DNA replication starts with un-winding of DNA duplexes which are held together by hydrogen bonds. Helicases move along the double-stranded DNA and separate the strands by breaking hydrogen bonds between base pairs.
Question 3.
How many types of DNA polymerases are associated with eukaryotic cell?
(A) Three
(B) Six
(C) Five
(D) One.
Answer:
(C) Five
Explanation :
5 types of DNA polymerases are associated with a eukaryotic cells.
Question 4.
DNA replication is :
(A) Semi-conservative, continuous
(B) Semi-continuous, conservative
(C) Semi-conservative, semi-discontinuous
(D) Conservative.
Answer:
(C) Semi-conservative, semi-discontinuous
Explanation :
DNA replication is said to be semi-conservative because of the process of replication, where the resulting double helix is composed of both an old strand and a new strand. … The two resulting double helices, which each contain one “old” strand and one “new” strand of DNA, are identical to the initial double helix.
Due to this reason, replication occurs continuously on one strand and discontinuously on the other strand. This is known as the semi-discontinuous mode of replication. Every new DNA molecule that is formed has a new and an old strand of the DNA. Thus, during DNA replication, entirely new DNA copies are not generated.
Directions: In the following questions a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as :
(A) Both assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of assertion (A).
(B) Both assertion (A) and Reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(C) Assertion (A) is true but reason (R) is false.
(D) Assertion (A) is false but reason (R) is true.
Question 5.
Assertion (A): Feminism is a bidirectional flow of information.
Reason (R): It requires DNA dependent RNA polymerase enzyme.
Answer:
(C) Assertion (A) is true but reason (R) is false.
Explanation :
Teminism is popularly known as 1 reverse transcription, i.c., DNA can be synthe sized by RNA.