CBSE Class 12 Biology Molecular Basis Of Inheritance Notes Set A

Download CBSE Class 12 Biology Molecular Basis Of Inheritance Notes Set A in PDF format. All Revision notes for Class 12 Biology have been designed as per the latest syllabus and updated chapters given in your textbook for Biology in Standard 12. Our teachers have designed these concept notes for the benefit of Grade 12 students. You should use these chapter wise notes for revision on daily basis. These study notes can also be used for learning each chapter and its important and difficult topics or revision just before your exams to help you get better scores in upcoming examinations, You can also use Printable notes for Class 12 Biology for faster revision of difficult topics and get higher rank. After reading these notes also refer to MCQ questions for Class 12 Biology given our website

Molecular Basis Of Inheritance Class 12 Biology Revision Notes

Class 12 Biology students should refer to the following concepts and notes for Molecular Basis Of Inheritance in standard 12. These exam notes for Grade 12 Biology will be very useful for upcoming class tests and examinations and help you to score good marks

Molecular Basis Of Inheritance Notes Class 12 Biology

Chapter 6. MOLECULAR BASIS OF INHERITANCE

DNA largest macromolecule made of helically twisted, two, antiparallel polydeoxyribonucleotide chains held together by hydrogen bonds.

→X-ray diffraction pattern of DNA by Rosalind Franklin showed DNA a helix.

→Components of DNA are (i) deoxyribose sugar, (ii) a phosphate, and (iii) nitrogen containing organic bases.

→ DNA contains four different bases called adenine (A), guanine (G) cytosine (C), and thymine (T).

→ These are grouped into two classes on the basis of their chemical structure: (i) Purines (with a double ring structure) and (ii) Pyrimidines (with a single ring structure)

→1953.James Watson and Francis Crick proposed three dimensional structure of DNA and won the Nobel prize.

→ DNA double helix with sugar phosphate back bone on outside and paired bases inside.

→Planes of the bases perpendicular to helix axis.

→ Each turn has ten base pairs.( 34 A0)

→ Diameter of helix 20 A0.

→ Two strands of DNA antiparallel.

→ DNA found both in nucleus and cytoplasm.

→ Extranuclear DNA found in mitochondria and chloroplasts.

→Two chains complementary

→ Two chains held together by hydrogen bond.

→ Adenine-Thymine pair has two hydrogen bonds.

→ Guanine-Cytosine pair has three hydrogen bonds.

→ Upon heating at temperature above 80-90 degree two strands uncoil and separate (Denaturation)

→ On cooling two strands join together (renaturation /annealing)

→DNA is mostly right handed and B form.

→ Bacterial nucleoid consists of a single circular DNA molecule .

class_12-biology_concept_235

 

DNA eukaryotes is wrapped around positively charged histone proteins to form nucleosome.

# Nucleosome contains 200 base pairs of DNA helix.

# Histone octamer =2(H2a+H2b+H3+H4)

# Linker DNA bears H1 protein

# Chromatin fibres formed by repeated units of nucleosomes.

# Non histone proteins required for packaging.

# Regions of chromatin, loosely packed and stains lightly called euchromatin.

# Regions of chromatin, densely packed and stains darkly is called heterochromatin.

 

DNA AS THE GENETIC MATERIAL

 

_Transformation experiment or Griffith effect.

• Griffith performed his experiments on Mice using Diplococcus pneumoniae.

• Two strains of bacteria are S-type and R-type cells.

• Experiments

_ Living S-strain Injected into mice →Mice killed

_ Living R-strain Injected into mice → Mice lived

_ Heat Killed S-strain Injected into mice →Mice lived

_ Living R-strain + Heat Killed S-strain Injected into mice→Mice killed

Griffith concluded that R type bacteria is transformed into virulent form.

# Transformation - change in the genetic constitution of an organism by picking up genes present in the remains of its relatives.

 

BIOCHEMICAL CHARACTERISATION OF TRANSFORMING PRINCIPLE

 

# Proved by Oswarld Avery, Colin Macleod, Maclyn Mc Cartyclass_12-biology_concept_236

Question. Number of regulatory and structural genes present in lac operon are
(a) one and two
(b) one and three
(c) one and one
(d) three and one
Answer : B

Question. Gene regulation governing lactose operon of E. coli that involves the lac i gene products is
(a) positive and inducible because it can be induced by lactose
(b) negative and inducible because its repressor protein prevents transcription
(c) negative and repressible because its repressor protein prevents transcription
(d) feedback inhibition because excess of b-galactosidase can switch off transcription
Answer : B

Question. The relationship between genes and DNA are best understood by
(a) mutation
(b) recombination
(c) enzymatic synthesis of amino acid
(d) enzymatic synthesis of codons
Answer : A

Question. Sickle-cell anaemia is a classical example of point mutation in which valine amino acid comes in place of
(a) glutamate
(b) tryptophane
(c) alanine
(d) guanine
Answer : A

Question. Under which of the following conditions will there be no change in the reading frame of following mRNA?
(a) Deletion of G from 5th position
(b) Insertion of A and G at 4th and 5th positions, respectively
(c) Deletion of GGU from 7th, 8th and 9th positions
(d) Insertion of G at 5th position
Answer : C

Question. 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) 33
(d) 333
Answer : C

Question. Which mutation of the genetic bases gives the proof that codon is triplet and reads in a contagious manner?
(a) Frameshift mutation
(b) Point mutation
(c) Both (a) and (b)
(d) Inversion mutation
Answer : A

Question. Choose the incorrect option for tRNA molecule.
(a) It has an anticodon loop that has bases complementary to the code
(b) It has an amino acid acceptor end to which it binds to amino acids
(c) tRNA are not specific for each amino acid
(d) tRNA looks like a clover leaf
Answer : C

Question. tRNA is a compact molecule which looks like
(a) M-shaped
(b) P-shaped
(c) L-shaped
(d) K-shaped
Answer : C

Question. RNA binds to mRNA through 
(a) anticodon loop
(b) T y C loop
(c) amino acid binding loop
(d) D-loop
Answer : A

Question. Removal of RNA polymerase-III from nucleoplasm will affect the synthesis of 
(a) tRNA
(b) hnRNA
(c) mRNA
Answer : A

Question. The process of polymerisation of amino acids to form a polypeptide is
(a) transcription
(b) replication
(c) translation
(d) polymerisation
Answer : C

Question. Many ribosomes may associate with a single mRNA to form multiple copies of a polypeptide simultaneously. Such strings of ribosomes are termed as 
(a) plastidome
(b) polyhedral bodies
(c) polysome
(d) nucleosome
Answer : C

Question. In the protein synthesis, tRNA carrying the amino acid enters from which site of the ribosome?
(a) A-site
(b) P-site
(c) Anticodon site
(d) R-site
Answer : A

Question. The order and sequences of amino acids are defined by the sequences of the bases in
(a) rRNA
(b) mRNA
(c) tRNA
(d) All of these
Answer : B

Question. Which of the following rRNAs act as structural RNA as well as ribozyme in bacteria? 
(a) 5 srRNA
(b) 18 srRNA
(c) 23 srRNA
(d) 5.8 srRNA
Answer : C

Question. UTRs present on mRNA refer to
(a) Untranscribed regions at both 5¢ end and 3¢ end
(b) Untranslated regions at 5¢ end
(c) Untranslated regions at both 5¢ end and 3¢ end
(d) Untranslated regions at 3¢ end
Answer : C

Question. For initiation, the ribosomes binds to the ...A... at the start codon and ...B... is recognised by the ...C... . Choose the correct option for A, B and C.
(a) A–mRNA, B–AUG, C–initiator tRNA
(b) A–mRNA, B–AUG, C–tRNA
(c) A–rRNA, B–AUG, C–tRNA
(d) A–rRNA B–AUG, C–initiator mRNA
Answer : A

Question. Choose the incorrect pair. (Image 137)
Answer : C

Question. For the given DNA sequence 3¢–TACATGGGTCCG–5¢.
Choose the correct sequential code of tRNA anticodon.
I. UAC II. AUG
III. GGC IV. CCA
(a) I, II, III and IV
(b) II, III, IV and I
(c) I, III, II and IV
(d) II, I, IV and III
Answer : D

Question. RNA molecule Anti-codon pairing
I UCA
II AUG
III AUU
IV AUC
Which of the following tRNA molecules does not exist?
(a) I and I
I (b) II and III
(c) III and IV
(d) IV and I
Answer : C

Question. Termination of protein synthesis or translation requires
(a) Both stop signal and starting codon
(b) Both starting codon and release factor
(c) Both release factor and stop codon
(d) GUG and AUG codon
Answer : C

Question. The regulation of gene expression in eukaryotes can be exerted at
I. transcriptional level.
II. processing level.
III. transport ofmRNA(from nucleus to cytoplasm).
IV. translational level.
Choose the correct combination to complete the statement.
(a) I and II
(b) II and III
(c) III and IV
(d) All of these
Answer : D

Question. In prokaryotes, control of the rate of ...A... is the pre-dominant site for the control of gene expression.
In a transcription unit, the activity of ...B... at a given promoter is in turn regulated by interaction with ...C...
proteins, which affects its ability to recognise the start sites.
Complete the statement filling the correct options in given blanks.
(a) A–RNA replication, B–DNA polymerase, C–accessory
(b) A–transcriptional initiation, B–RNA polymerase, C–accessory
(c) A–translational initiation, B–RNA polymerase, C–accessory
(d) A–DNA replication, B–RNA polymerase, C–accessory
Answer : B

Question. Positively regulatory proteins are called
(a) activator
(b) repressors
(c) necessary proteins
(d) accessory proteins
Answer : A

Question. The accessibility of the promoter regions of prokaryotic DNA is (in many cases) regulated by the interaction of proteins with the sequences termed as
(a) regulator
(b) promoter
(c) operator
(d) structural genes
Answer : C

Question. Which of the following enzymes are required in protein synthesis?
I. Ligase II. Permease
III. Endonuclease IV. Ribozyme
V. RNApolymerase VI. Peptidyl transferase
VII. Amino acid activating enzyme
Choose the correct option.
(a) IV, VI and VII
(b) I, II and III
(c) II, III, IV and V
(d) All of these
Answer : A

Question. Which among the following process occur(s) during charging or aminoacylation of tRNA?
(a) Activation of amino acids in the presence of ATP
(b) Linking of amino acids to their cognate tRNA
(c) Both (a) and (b)
(d) None of the above
Answer : C

Question. The cellular factory responsible for the synthesis of proteins is
(a) mitochondria
(b) endoplasmic reticulum
(c) Golgi body
(d) ribosome
Answer : D

Question. Control of gene expression takes place at the level of
(a) DNA-replication
(b) transcription
(c) translation
(d) None of these
Answer : B,C

Question. Who amongst the following scientists had no contribution in the development of the double helix model for the structure of DNA?
(a) Rosalind Franklin
(b) Maurice Wilkins
(c) Erwin Chargaff
(d) Meselson and Stahl
Answer : D

Question. While analysing the DNA of an organism a total number of 5386 nucleotides were found out of which the proportion of different bases were Adenine =
29%, Guanine = 17%, Cytosine = 32%, Thymine =
17%. Considering the 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

Question. In some viruses, DNA is synthesised by using RNA as template. Such a DNA is called
(a) A-DNA
(b) B-DNA
(c) cDNA
(d) rDNA
Answer : C

Question. 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

Question. Which was the last human chromosome to be completely sequenced?
(a) Chromosome 1
(b) Chromosome 11
(c) Chromosome 21
(d) Chromosome-X
Answer : A

Question. The human chromosome with the highest and least number of genes in them are respectively
(a) chromosome 21 and Y
(b) chromosome 1 and X
(c) chromosome 1 and Y
(d) chromosome X and Y
Answer : C

Question. The net electric charge on DNA and histones is
(a) positive
(b) negative
(c) negative and positive, respectively
(d) zero
Answer : C

Question. The first genetic material could be
(a) protein
(b) carbohydrates
(c) DNA
(d) RNA
Answer : D

Question. The fact that a purine always paired base through hydrogen bonds with a pyrimidine base leads to, in the DNA double helix
(a) the antiparallel nature
(b) the semiconservative nature
(c) uniform width throughout DNA
(d) uniform length in all DNA
Answer : C

Question. Which of the following are the functions of RNA?
(a) It is 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

Question. If the base sequence of a codon in mRNA is 5'–AUG–3', the sequence of tRNA pairing with it must be
(a) 5¢ – UAC – 3¢
(b) 5¢ – CAU – 3¢
(c) 5¢ – AUG – 3¢
(d) 5¢ – GUA – 3¢
Answer : B

Question. Both deoxyribose and ribose belong to a class of sugars called
(a) trioses
(b) hexoses
(c) pentoses
(d) polysaccharides
Answer : C

Question. The amino acid attaches to the tRNA at its
(a) 5¢-end
(b) 3¢-end
(c) Anticodon site
(d) DHU loop
Answer : B

Question. If Meselson and Stahl’s experiment is continued for four generations in bacteria, the ratio of 15 15 15 14 14 14 N / N : N / N : N / N containing DNA in the fourth generation would be
(a) 1:1:0
(b) 1:4:0
(c) 0:1:3
(d) 0:1:7
Answer : D

Question. 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?
(a) Replace purine with pyrimidines
(b) Remove/Replace 3¢ OH group in deoxyribose
(c) Remove/Replace 2¢ OH group with some other group in deoxyribose
(d) Both (b) and (c)
Answer : B

Question. Discontinuous synthesis of DNA occurs in one strand, because
(a) DNA molecule being synthesised is very long
(b) DNA-dependent DNA polymerase catalyses polymerisation only in one direction (5¢ → 3¢)
(c) it is a more efficient process
(d) DNA ligase has to have a role
Answer : B

Question. If the sequence of nitrogen bases of the coding strand of DNA in a transcription unit is 5' – A T G A A T G – 3',the sequence of bases in its RNA transcript would be
(a) 5¢ – A U G A A U G – 3¢
(b) 5¢ – U A C U U A C – 3¢
(c) 5¢ – C A U U C A U – 3¢
(d) 5¢ – G U A A G U A – 3¢
Answer : A

Question. 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

Question. The RNA polymerase holoenzyme transcribes
(a) the promoter, structural gene and the terminator region
(b) the promoter and the terminator region
(c) the structural gene and the terminator regions
(d) the structural gene only
Answer : C

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