Molecular Basis of Inheritance (last 10 yeas) Flashcards
- If the length of a DNA molecule is 1.1 metres, what will be the approximate number of base pairs? (2022)
a. 6.6 × 106 bp
b. 3.3 × 109 bp
c. 6.6 × 109 bp
d. 3.3 × 106 bp
Grade 12 / Chapter 5 molecular basis of inheritance (page 80)
b. 3.3 × 109 bp
NCERT
DNA is a long polymer of deoxyribonucleotides.
The length of DNA is usually defined as number of nucleotides
(or a pair of nucleotide referred to as base pairs) present in it.
This also is the characteristic of an organism.
For example, a bacteriophage known as φ ×174 has 5386 nucleotides,
Bacteriophage lambda has 48502 base pairs (bp),
Escherichia coli has 4.6 × 106 bp, and
haploid content of human DNA is 3.3 × 109 bp.
.
- Read the following statements and choose the set of correct statements (2022)
A. Euchromatin is loosely packed chromatin
B. Heterochromatin is transcriptionally active
C. Histone octomer is wrapped by negatively charged DNA in nucleosome
D. Histones are rich in lysine and arginine
E. A typical nucleosome contains 400 bp of DNA helix
Choose the correct answer from the options given below.
a. A, C and E only
b. B, D and E only
c. A, C and D only
d. B and E only
Grade 12 / Chapter 5 molecular basis of inheritance (page 83-84)
c. A, C and D only
NCERT
A. Euchromatin is loosely packed chromatin
C. Histone octomer is wrapped by negatively charged DNA in nucleosome
D. Histones are rich in lysine and arginine
NCERT
A. Non-histone Chromosomal (NHC) proteins. In a typical nucleus, some region of chromatin are loosely packed (and stains light) and are referred to as euchromatin (page 84)
B. Euchromatin is said to be transcriptionally active chromatin, whereas heterochromatin is inactive (page 84)
C. The negatively charged DNA is wrapped around the positively charged histone octamer to form a structure called nucleosome (page 83)
D. Histones are rich in the basic amino acid residues lysine and arginine. (page 83)
E. A typical nucleosome contains 200 bp of DNA helix (page 83)
- Complete the flow chart on central dogma. (2021)
Insert the image
a. (A)-Translation;(B)-Replication; (C)-Transcription;(D)- Transduction
b. (A)-Replication;(B)-Transcription; (C)-Translation; (D)-Protein
c. (A)-Transduction;(B)-Translation; (C)-Replication; (D)-Protein
d. (A)-Replication;(B)-Transcription (C)-Transduction;(D)-Protein
Grade 12 / Chapter 5 molecular basis of inheritance (page 82)
b. (A)-Replication;(B)-Transcription; (C)-Translation; (D)-Protein
NCERT
Very soon, Francis Crick proposed the Central dogma in molecular biology, which states that the genetic information flows from DNA arrow RNA arrow Protein.
Add the image
.
- If Adenine makes 30% of the DNA molecule, what will be the percentage of Thymine, Guanine and Cytosine in it? (2021)
a. T : 20; G : 20; C : 30
b. T : 30; G : 20; C : 20
c. T : 20; G : 25; C : 25
d. T : 20; G : 30; C : 20
Grade 12 / Chapter 5 molecular basis of inheritance (page 81)
b. T : 30; G : 20; C : 20
NCERT
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.
- Which one of the following statement about histones is wrong? (2021)
a. The pH of histones is slightly acidic.
b. Histones are rich in amino acids - Lysine and Arginine.
c. Histones carry positive charge in the side chain.
d. Histones are organized to form a unit of 8 molecules.
Grade 12 / Chapter 5 molecular basis of inheritance (page 83)
a. The pH of histones is slightly acidic.
NCERT
a. In eukaryotes, this organization is much more complex. There is a set of positively charged, basic proteins called histones.
b. Histones are rich in the basic amino acid residues lysine and arginine.
c. A protein acquires charge depending upon the abundance of amino acids residues with charged side chains. Both the amino acid residues carry positive charges in their side chains.
d. Histones are organized to form a unit of eight molecules called histone octamer
- Which of the following statements is correct? (2020)
a. Adenine pairs with thymine through one H-bond
b. Adenine pairs with thymine through three H-bonds.
c. Adenine does not pair with thymine.
d. Adenine pairs with thymine through two H-bonds.
Grade 12 / Chapter 5 molecular basis of inheritance
(page 81 - summary)
d. Adenine pairs with thymine through two H-bonds.
NCERT
Adenine forms two hydrogen bonds with Thymine from opposite strand and vice-versa.
Similarly, Guanine is bonded with Cytosine with three H-bonds.
(page 81)
The rule is that Adenine pairs with Thymine through two H-bonds, and
Guanine with Cytosine through three H-bonds.
This makes one strand complementary to the other (summary)
- 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: (2020)
a. 2.5 meters
b. 2.2 meters
c. 2.7 meters
d. 2.0 meters
Grade 12 / Chapter 5 molecular basis of inheritance (page 83)
b. 2.2 meters
NCERT
Taken the distance between two consecutive base pairs as 0.34 nm (0.34×10–9 m), if the length of DNA double helix in a typical mammalian cell is calculated
(simply by multiplying the total number of bp with distance between two consecutive bp, that is,
6.6 × 109 bp × 0.34 × 10-9m/bp), it comes out to be approximately 2.2 metres
- In the polynucleotide chain of DNA, a nitrogenous base is linked to the –OH of: (2020-Covid)
a. 3’C pentose sugar
b. 5’C pentose sugar
c. 1’C pentose sugar
d. 2’C pentose sugar
Grade 12 / Chapter 5 molecular basis of inheritance (page 80)
c. 1’C pentose sugar
NCERT
A nitrogenous base is linked to the OH of 1’ C pentose sugar through a N-glycosidic linkage to form a nucleoside, such as
adenosine or deoxyadenosine,
guanosine or deoxyguanosine,
cytidine or deoxycytidine and
uridine or deoxythymidine.
UNDERSTAND THIS TOPIC AND TERMINOLOGY
- E. Coli has only 4.6 × 106 base pairs and completes the process of replication within 18 minutes; then the average rate of polymerization is approximately- (2020-Covid)
a. 3000 base pairs/second
b. 4000 base pairs/second
c. 1000 base pairs/second
d. 2000 base pairs/second
Grade 12 / Chapter 5 molecular basis of inheritance (page 90)
d. 2000 base pairs/second
NCERT
E. coli that has only 4.6 ×106 bp
(compare it with human whose diploid content is 6.6 × 109 bp), completes the process of replication within 18 minutes; that means the average rate of polymerization has to be approximately 2000 bp per second.
- Purines found both in DNA and RNA are (2019)
a. Adenine and thymine
b. Adenine and guanine
c. Guanine and cytosine
d. Cytosine and thymine
Grade 12 / Chapter 5 molecular basis of inheritance (page 80)
b. Adenine and guanine
NCERT
There are two types of nitrogenous bases –
Purines (Adenine and Guanine), and
Pyrimidines (Cytosine, Uracil and Thymine).
Cytosine is common for both DNA and RNA and Thymine is present in DNA.
- The association of histone H1 with a nucleosome indicates: (2017-Delhi)
a. Transcription is occurring
b. DNA replication is occurring
c. The DNA is condensed into a chromatin fibre
d. The DNA double helix is exposed
Grade 12 / Chapter 5 molecular basis of inheritance (page 83)
c. The DNA is condensed into a chromatin fibre
NCERT (add the image)
The negatively charged DNA is wrapped around the positively charged histone octamer to form a structure called nucleosome (Figure 5.4 a).
A typical nucleosome contains 200 bp of DNA helix.
Nucleosomes constitute the repeating unit of a structure in nucleus called chromatin, threadlike stained (coloured) bodies seen in nucleus.
The nucleosomes in chromatin are seen as ‘beads-on-string’ structure when viewed under
electron microscope (EM) (Figure 5.4 b).
Theoretically, how many such beads (nucleosomes) do you imagine are present in a mammalian cell?
The beads-on-string structure in chromatin is packaged to form chromatin fibers that are further coiled and condensed at metaphase stage of cell division to form chromosomes
- DNA fragments are: (2017-Delhi)
a. Positively charged
b. Negatively charged
c. Neutral
d. Either positively or negatively charged depending on their size
Grade 12 / Chapter 5 molecular basis of inheritance (page 83)
b. Negatively charged
NCERT
DNA (being negatively charged) is held with some proteins
(that have positive charges) in a region termed as ‘nucleoid’.
The DNA in nucleoid is organised in large loops held by
proteins.
The negatively charged DNA is wrapped around the positively charged histone octamer to form a structure called nucleosome
- Identify the correct order of organization of genetic material from largest to smallest: (2015 Re)
a. Genome, chromosome, nucleotide, gene
b. Genome, chromosome, gene, nucleotide
c. Chromosome, genome, nucleotide, gene
d. Chromosome, gene, genome, nucleotide
b. Genome, chromosome, gene, nucleotide
(see this question)
DIDN’T GET THE QUESTION / COULDN’T FIND IT IN NCERT
DIDNT WRITE IT IN NCERT
- The diagram shows an important concept in the genetic implication of DNA Fill in the blanks A to C: (2013)
insert an image
a. A-translation, B-extension, C-Rosalind Franklin
b. A-transcription, B-replication, C-James Watson
c. A-translation, B-transcription, C-Erwin Chargaff
d. A-transcription, B-translation, C-Francis Crick
Grade 12 / Chapter 5 molecular basis of inheritance (page 82)
d. A-transcription, B-translation, C-Francis Crick
NCERT
Very soon, Francis Crick proposed the Central dogma in molecular biology, which states that the genetic information flows from
DNA arrow RNA arrow Protein.
ADD THE IMAGE
- Ten E.coli 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? (2022)
a. 80 cells
b. 20 cells
c. 40 cells
d. 60 cells
d. 60 cells
(see this question)
DIDN’T GET THE QUESTION / COULDN’T FIND IT IN NCERT
DIDNT WRITE IT IN NCERT
- The term ‘Nuclein’ for the genetic material was used by:
(2020-Covid)
a. Meischer
b. Chargaff
c. Mendel
d. Franklin
Grade 12 / Chapter 5 molecular basis of inheritance (page 81)
a. Meischer
NCERT
DNA as an acidic substance present in nucleus was first identified by
Friedrich Meischer in 1869. He named it as ‘Nuclein’.
Even though the discovery of nuclein by Meischer and the proposition for principles of inheritance by Mendel were almost at the same time, but that the DNA acts as a genetic material took long to be discovered and proven.
- The experimental proof for semi-conservative replication of DNA was first shown in a: (2018)
a. Fungus
b. Bacterium
c. Plant
d. Virus
Grade 12 / Chapter 5 molecular basis of inheritance (page 88)
b. Bacterium
NCERT
It is now proven that DNA replicates semiconservatively. It was shown first in Escherichia coli (bacteria) and subsequently in higher organisms, such as plants and human cells.
- Select the correct match (2018)
a. Ribozyme Nucleic acid
b. F2 × Recessive parent Dihybrid cross
c. T.H. Morgan Transduction
d. G. Mendel Transformation
Grade 12 / Chapter 5 molecular basis of inheritance
(page 99 - summary)
a. Ribozyme Nucleic acid
NCERT
A. The ribosome also acts as a catalyst (23S rRNA in bacteria
is the enzyme- ribozyme) for the formation of peptide bond.
(page 99)
One of the rRNA acts as a catalyst for peptide bond formation, which is an example of RNA enzyme (ribozyme). (Summary)
(couldn’t find the other options)
- The final proof for DNA as the genetic material came from the experiments of (2017-Delhi)
a. Griffith
b. Hershey and Chase
c. Avery, Mcleod and McCarty
d. Hargobind Khorana
Grade 12 / Chapter 5 molecular basis of inheritance (page 85)
b. Hershey and Chase
NCERT
The unequivocal proof that DNA is the genetic material came from the experiments of Alfred Hershey and Martha Chase (1952).
They worked with viruses that infect bacteria called bacteriophages (page 85)
- A molecule that can act as a genetic material must fulfill the traits given below, except: (2016 - II)
a. It should be unstable structurally and chemically
b. It should provide the scope for slow changes that are required for evolution
c. It should be able to express itself in the form of ‘Mendelian characters’
d. It should be able to generate its replica
a. It should be unstable structurally and chemically
(see this question)
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DIDNT WRITE IT IN NCERT
- Taylor conducted the experiment to prove semi- conservative mode of chromosome replication on: (2016 - II)
a. Drosophila melanogaster
b. E. coli
c. Vinca rosea
d. Vicia faba
Grade 12 / Chapter 5 molecular basis of inheritance (page 90)
d. Vicia faba
NCERT
Very similar experiments involving use of radioactive thymidine to
detect distribution of newly synthesised DNA in the chromosomes was performed on Vicia faba (faba beans) by Taylor and colleagues in 1958.
The experiments proved that the DNA in chromosomes also replicate
semiconservatively.
- Which of the following rRNA acts as structural RNA as well as ribozyme in bacteria? (2016 - II)
a. 23 S rRNA
b. 5.8 S rRNA
c. 5 S rRNA
d. 18 S rRNA
Grade 12 / Chapter 5 molecular basis of inheritance (page 99)
a. 23 S rRNA
There are two sites in the large subunit, for subsequent amino acids
to bind to and thus, be close enough to each other for the formation of a peptide bond.
The ribosome also acts as a catalyst
(23S rRNA in bacteria is the enzyme- ribozyme) for the formation of peptide bond.
- In sea urchin DNA, which is double stranded, 17% of the bases were shown to be cytosine. The percentages of the other three bases expected to be present in this DNA are: (2015)
a. G = 17%, A = 33%, T = 33%
b. G = 8.5 %, A = 50 %, T = 24.5 %
c. G = 34%, A = 24.5%, T = 24.5%
d. G = 17%, A = 16.5%, T = 32.5%
Grade 12 / Chapter 5 molecular basis of inheritance (page 81)
a. G = 17%, A = 33%, T = 33%
NCERT
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.
- Which one of the following is not applicable to RNA? (2015 Re)
a. 5′ phosphoryl and 3′ hydroxyl ends
b. Heterocyclic nitrogenous bases
c. Chargaff’s rule
d. Complementary base pairing
Grade 12 / Chapter 5 molecular basis of inheritance (page 81)
c. Chargaff’s rule
NCERT
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
LOOK IN DEPTH STILL
- Transformation was discovered by: (2014)
a. Watson and Crick
b. Messelson and Stahl
c. Hershey and Chase
d. Griffith
Grade 12 / Chapter 5 molecular basis of inheritance (page 84)
d. Griffith
NCERT
In 1928, Frederick Griffith, in a series of experiments with Streptococcus pneumoniae (bacterium responsible for pneumonia), witnessed a miraculous transformation in the bacteria. During the course of his experiment, a living organism (bacteria) had changed in physical form.
- During DNA replication, Okazaki fragments are used to elongate (2017-Delhi)
a. The leading strand towards replication fork
b. The lagging strand towards replication fork
c. The leading strand away from replication fork
d. The lagging strand away from the replication fork
d. The lagging strand away from the replication fork
COULDN’T FIND IT IN NCERT
DIDN’T WRITE IT IN NCERT BOOK
- Select the correct option: (2014)
Direction of RNA synthesis Direction of reading of the
template DNA strand
a. 3′ → 5′ 3′ → 5′
b. 5′ → 3′ 3′ → 5′
c. 3′ → 5′ 5′ → 3′
d. 5′ → 3′ 5′ → 3′
Grade 12 / Chapter 5 molecular basis of inheritance (page 90)
b. 5′ → 3′ 3′ → 5′
NCERT
The DNA-dependent DNA polymerases catalyse polymerisation only in one direction, that is 5’à3’.
This creates some additional complications at the replicating fork.
Consequently, on one strand (the template with polarity 3’à5’), the
replication is continuous, while on the other
(the template withnpolarity 5’à3’), it is discontinuous.
The discontinuously synthesised fragments are later joined by the enzyme DNA ligase (Figure 5.8).
ADD THE IMAGE - understand the topic and question
- What is the role of RNA ploymerase III in the process of transcription in eukaryotes? (2021)
a. Transcribes tRNA, 5s rRNA and sn RNA
b. Transcribes precursor of mRNA
c. Transcribes only snRNAs
d. Transcribes rRNAs (28S, 18S and 5.8S)
Grade 12 / Chapter 5 molecular basis of inheritance (page 95)
a. Transcribes tRNA, 5s rRNA and sn RNA
NCERT
RNA polymerase III is responsible for transcription of
tRNA, 5srRNA, and snRNAs (small nuclear RNAs)
- Identify the correct statement. (2021)
a. RNA polymerase binds with Rho factor to terminate the process of transcription in bacteria.
b. The coding strand in transcription unit is copied to an mRNA.
c. Split gene arrangement is characteristic of prokaryotes.
d. In capping, methyl guanosine triphosphate is added to the 3′ end of hnRNA of transcription in bacteria.
Grade 12 / Chapter 5 molecular basis of inheritance (page 93)
a. RNA polymerase binds with Rho factor to terminate the process of transcription in bacteria.
ADD THE IMAGE
- Which is the “Only enzyme” that has “Capability” to catalyse Initiation, Elongation and Termination in the process of transcription in prokaryotes? (2021)
a. DNA dependent RNA polymerase
b. DNA Ligase
c. DNase
d. DNA dependent DNA polymerase
Grade 12 / Chapter 5 molecular basis of inheritance (page 93)
a. DNA dependent RNA polymerase
NCERT
There is single DNA-dependent RNA polymerase that catalyses transcription of all types of RNA in bacteria.
- Name the enzyme that facilitates opening of DNA helix during transcription. (2020)
a. DNA helicase
b. DNA polymerase
c. RNA polymerase
d. DNA ligase
c. RNA polymerase
- AGGTATCGCAT is a sequence from the coding strand of a gene. What will be the corresponding sequence of the transcribed mRNA? (2018)
a. AGGUAUCGCAU
b. UGGTUTCGCAT
c. ACCUAUGCGAU
d. UCCAUAGCGUA
a. AGGUAUCGCAU
- Spliceosomes are not found in cells of: (2017-Delhi)
a. Plants
b. Fungi
c. Animals
d. Bacteria
d. Bacteria
- Which of the following RNAs should be most abundant in animal cell? (2017-Delhi)
a. r-RNA
b. t-RNA
c. m-RNA
d. mi-RNA
a. r-RNA
- DNA-dependent RNA polymerase catalyses transcription on one strand of the DNA which is called the: (2016 - II)
a. Alpha strand
b. Antistrand
c. Template strand
d. Coding strand
c. Template strand
- Statement I: The codon ‘AUG’ codes for methionine and
phenylalanine.
Statement II: ‘AAA’ and ‘AAG’ both codons code for the
amino acid lysine.
In the light of the above statements, choose the correct answer from the options given below. (2021)
a. Both statement I and statement II are false
b. Statement I is correct but statement II is false
c. Statement I is incorrect but statement II is true
d. Both statement I and statement II are true
c. Statement I is incorrect but statement II is true
- Under which of the following conditions will there be no change in the reading frame of following mRNA? (2019)
5′AACAGCGGUGCUAUU3′
a. Insertion of G at 5th position
b. Deletion of G from 5th position
c. Insertion of A and G at 4th and 5th positions respectively
d. Deletion of GGU from 7th, 8th and 9th positions
d. Deletion of GGU from 7th, 8th and 9th positions
- Which of the following features of genetic code does allow bacteria to produce human insulin by recombinant DNA technology? (2019)
a. Genetic code is not ambiguous
b. Genetic code is redundant
c. Genetic code is nearly universal
d. Genetic code is specific
c. Genetic code is nearly universal
- 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? (2017-Delhi)
a. 1
b. 11
c. 33
d. 333
c. 33
- Which one of the following is the starter codon? (2016 - I)
a. AUG
b. UGA
c. UAA
d. UAG
a. AUG
- The process of translation of mRNA to proteins begins as soon as: (2022)
a. The tRNA is activated and the larger subunit of ribosome encounters mRNA
b. The small subunit of ribosome encounters mRNA
c. The larger subunit of ribosome encounters mRNA
d. Both the subunits join together to bind with mRNA
b. The small subunit of ribosome encounters mRNA
- Which of the following RNAs is not required for the synthesis of protein? (2021)
a. tRNA
b. rRNA
c. siRNA
d. mRNA
c. siRNA
- The first phase of translation is: (2020)
a. Recognition of DNA molecule
b. Aminoacylation of tRNA
c. Recognition of an anti-codon
d. Binding of mRNA to ribosome
b. Aminoacylation of tRNA
- Which one of the following is wrongly matched? (2014)
a. Operon-Structural genes, operator and promoter
b. Transcription-Writing information from DNA to tRNA
c. Translation-Using information in mRNA to make protein
d. Repressor protein-Binds to operator to stop enzyme synthesis
b. Transcription-Writing information from DNA to tRNA
- In an E.coil 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? (2022)
a. RNA polymerase will bind the promoter region
b. Only z gene will get transcribed
c. z, y, a genes will be transcribed
d. z, y, a genes will not be translated
d. z, y, a genes will not be translated
- Match the following genes of the Lac operon with their respective products : (2019)
A. i gene i. β-galactosidase
B. z gene ii. Permease
C. a gene iii. Repressor
D. y gene iv. Transacetylase
Select the correct option.
(A) (B) (C) (D)
a. (i) (iii) (ii) (iv)
b. (iii) (i) (ii) (iv)
c. (iii) (i) (iv) (ii)
d. (iii) (iv) (i) (ii)
c. (iii) (i) (iv) (ii)
- Select the correct match: (2018)
a. Alec Jeffreys – Streptococcus pneumoniae
b. Alfred Hershey and Martha Chase – TMV
c. Matthew Meselson and F. Stahl – Pisum sativum
d. Francois Jacob and Jacques Monod – Lac operon
d. Francois Jacob and Jacques Monod – Lac operon
- All of the following are part of an operon except: (2018)
a. An operator
b. Structural genes
c. An enhancer
d. A promoter
c. An enhancer
- The equivalent of a structural gene is: (2016 - II)
a. Operon
b. Recon
c. Muton
d. Cistron
d. Cistron
- Gene regulation governing lactose operon of E. coli that involves the lac I gene product is: (2015)
a. Negative and repressible because repressor protein prevents transcription
b. Feedback inhibition because excess of β-galactosidase can switch off transcription
c. Positive and inducible because it can be induced lactose
d. Negative and inducible because repressor protein prevents transcription
d. Negative and inducible because repressor protein prevents transcription
- Which enzyme/s will be produced in a cell in which there is a nonsense mutation in the lac Y gene? (2013)
a. Lactose permease and transacetylase
b. β-galactosidase
c. Lactose permease
d. Transacetylase
b. β-galactosidase
- DNA polymorphism forms the basis of: (2022)
a. Translation
b. Genetic mapping
c. DNA finger printing
d. Both genetic mapping and DNA finger printing
d. Both genetic mapping and DNA finger printing
- 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: (2022)
a. Bioinformatics
b. Sequence annotation
c. Gene mapping
d. Expressed sequence tags
b. Sequence annotation
- DNA fingerprinting involves identifying differences in some specific regions in DNA sequence, called as: (2021)
a. Repetitive DNA
b. Single nucleotides
c. Polymorphic DNA
d. Satellite DNA
a. Repetitive DNA
- Which is the basis of genetic mapping of human genome as well as DNA finger printing?
(2020-Covid)
a. Single nucleotide polymorphism
b. Polymorphism in hnRNA sequence
c. Polymorphism in RNA sequence
d. Polymorphism in DNA sequence
d. Polymorphism in DNA sequence
- Expressed Sequence Tags (ESTs) refers to : (2019)
a. Genes expressed as RNA
b. Polypeptide expression
c. DNA polymorphism
d. Novel DNA sequences
a. Genes expressed as RNA
- Which of the following is not required for any of the techniques of DNA fingerprinting available at present? (2016 - I)
a. Polymerase chain reaction
b. Zinc finger analysis
c. Restriction enzymes
d. DNA-DNA hybridisation
b. Zinc finger analysis
- Which of the following is required as inducer(s) for the expression of Lac operon? (2016 - I)
a. Glucose
b. Galactose
c. Lactose
d. Lactose and Galactose
c. Lactose
- Satellite DNA is important because it: (2015 Re)
a. Shows high degree of polymorphism in population and also the same degree of polymorphism in an individual, which are heritable form parents to children.
b. Does not code for proteins and is same in all members of the population
c. Codes for enzymes needed for DNA replication
d. Codes for proteins needed in cell cycle.
a. Shows high degree of polymorphism in population and also the same degree of polymorphism in an individual, which are heritable form parents to children.
- Commonly used vectors for human genome sequencing are: (2014)
a. T/A Cloning Vectors
b. T-DNA
c. BAC and YAC
d. Expression Vectors
c. BAC and YAC
- An analysis of chromosomal DNA using the southern hybridization technique does not use: (2014)
a. PCR
b. Electrophoresis
c. Blotting
d. Autoradiography
a. PCR
