NEET Biology Molecular Basis of Inheritance MCQs Set 29

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Chapter-wise Objective Questions: Molecular Basis of Inheritance

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Multiple Choice Questions

Question 1: Which is true for repressible operon :–
(a) Operon is off; the inducer turns it on
(b) Inactive repressor + corepressor = active repressor
(c) Active repressor + inducer = inactive repressor
(d) Operon is on; the inducer turns it off
Show Answer & Explanation

Answer: (b) Inactive repressor + corepressor = active repressor

Explanation:
1. In a repressible operon the repressor is inactive alone.
2. It becomes active when a corepressor (end product) binds, and then the operon is switched off.
3. Hence option (b) is correct: Inactive repressor + corepressor = active repressor

Teacher's Note:
1. Repressible = switched off by the end product.
2. Inducible = switched on by the substrate.

Question 2: Which of the following is not produced by E.Coli in the lactose operon –
(a) b galactosidase
(b) Thiogalactoside transacetylase
(c) Lactose dehydrogenase
(d) Lactose permease
Show Answer & Explanation

Answer: (c) Lactose dehydrogenase

Explanation:
1. The lac operon makes beta-galactosidase, permease and transacetylase.
2. Lactose dehydrogenase is not made by it.
3. Hence option (c) is correct: Lactose dehydrogenase

Teacher's Note:
1. The lac operon has three enzymes only.
2. Learn z, y and a.

Question 3: A functional complex comprising a cluster of genes including structural gene, a promoter gene, an operator gene and a regulator gene was discovered by:
(a) Beadle and Tatum (1958)
(b) Watson and crick (1953)
(c) Jacob and Monod (1961)
(d) Britten and Davidson (1961)
Show Answer & Explanation

Answer: (c) Jacob and Monod (1961)

Explanation:
1. Jacob and Monod (1961) proposed the operon model with structural, promoter, operator and regulator genes.
2. Beadle and Tatum gave one gene - one enzyme.
3. Hence option (c) is correct: Jacob and Monod (1961)

Teacher's Note:
1. The lac operon was worked out in E. coli.

Question 4: The accessibility of promoter regions of prokaryotic DNA by RNA polymerase is in many cases regulated by the interaction of some protein with sequences termed as –
(a) Promoter
(b) Operator
(c) Regulator
(d) Cistron
Show Answer & Explanation

Answer: (b) Operator

Explanation:
1. The operator is a DNA sequence to which the repressor binds.
2. This blocks the RNA polymerase from reaching the promoter.
3. Hence option (b) is correct: Operator

Teacher's Note:
1. Repressor + operator = operon off.
2. The inducer removes the repressor.

Question 5: Find out the correct sequence of structural gene in lac operon
(a) y, a, z
(b) a, z, y
(c) z, y, a
(d) z, a y
Show Answer & Explanation

Answer: (c) z, y, a

Explanation:
1. The structural genes of the lac operon are arranged as z, y, a.
2. They are transcribed together as one mRNA.
3. Hence option (c) is correct: z, y, a

Teacher's Note:
1. Z-Y-A is the order.
2. Remember it as the order of enzymes.

Question 6: Mutation is:
(a) An abrupt or discontinuous change which is inherited
(b) A factor for plant growth
(c) A change which affects parents only and is never inherited
(d) A change which affects the offspring of F2 generation
Show Answer & Explanation

Answer: (a) An abrupt or discontinuous change which is inherited

Explanation:
1. A mutation is a sudden change in the genetic material that is inherited.
2. It is not just a change in the body that disappears.
3. Hence option (a) is correct: An abrupt or discontinuous change which is inherited

Teacher's Note:
1. Mutations are the raw material of evolution.
2. Only germ-line mutations are inherited.

Question 7: The change of chromosomal parts between non homologous pairs of chromosome:
(a) Crossing over/Transduction
(b) Translocation
(c) Inversion
(d) Transition
Show Answer & Explanation

Answer: (b) Translocation

Explanation:
1. Exchange of parts between non-homologous chromosomes is translocation.
2. Crossing over occurs between homologous chromosomes.
3. Hence option (b) is correct: Translocation

Teacher's Note:
1. Translocation can cause genetic disorders.
2. Inversion reverses a segment within one chromosome.

Question 8: In hexaploid wheat (2n = 42), what are the haploid number (n) and the basic number (x) of chromosomes?
(a) n = 21, x = 7
(b) n = 28, x = 7
(c) n = 7, x = 28
(d) n = 7, x = 21
Show Answer & Explanation

Answer: (a) n = 21, x = 7

Explanation:
1. Hexaploid wheat has 2n = 42, so n = 21.
2. The basic number x = 7 because 6x = 42.
3. Hence option (a) is correct: n = 21, x = 7

Teacher's Note:
1. Haploid number n is half of 2n, and the basic number x is 2n divided by the ploidy level.
2. In wheat x = 7.

Question 9: Non-ionizing radiations commonly used for inducing mutations in organisms are:
(a) UV-rays
(b) Beta-rays
(c) X-rays
(d) Gamma-rays
Show Answer & Explanation

Answer: (a) UV-rays

Explanation:
1. UV rays are non-ionizing radiation used to induce mutations.
2. Beta rays, X rays and gamma rays are ionizing.
3. Hence option (a) is correct: UV-rays

Teacher's Note:
1. UV is absorbed by purines and pyrimidines.
2. It causes thymine dimers.

Question 10: Type of gene mutation which involves replacement of purine with pyrimidine or vice versa (OR) The substitution of one type of base with another type of base is:
(a) Transduction
(b) Transversion
(c) Translocation
(d) Transcription
Show Answer & Explanation

Answer: (b) Transversion

Explanation:
1. A change of a purine to a pyrimidine or the reverse is a transversion.
2. A change of purine to purine is a transition.
3. Hence option (b) is correct: Transversion

Teacher's Note:
1. Transition = like for like.
2. Transversion = purine and pyrimidine swapped.

Question 11: Sickle cell anaemia is an example of:
(a) Frame shift mutation
(b) Point mutation
(c) Segmental mutation
(d) Gibberish mutation
Show Answer & Explanation

Answer: (b) Point mutation

Explanation:
1. Sickle-cell anaemia is caused by a single base change, GAG to GUG, in the beta chain gene.
2. So glutamic acid becomes valine.
3. Hence option (b) is correct: Point mutation

Teacher's Note:
1. A single base change is a point mutation.
2. Frame shift needs insertion or deletion.

Question 12: The "cri-du-chat" syndrome is caused by change in chromosome structure involving:
(a) Deletion
(b) Duplication
(c) Inversion
(d) Translocation
Show Answer & Explanation

Answer: (a) Deletion

Explanation:
1. Cri-du-chat syndrome is caused by deletion of the short arm of chromosome 5.
2. Babies have a cat-like cry.
3. Hence option (a) is correct: Deletion

Teacher's Note:
1. Deletion = loss of a segment.
2. Learn syndromes with the chromosome number.

Question 13: A class of mutation induced by addition or deletion of a nucleotide is called:
(a) Missense
(b) Non-sense
(c) Substitution
(d) frame shift
Show Answer & Explanation

Answer: (d) frame shift

Explanation:
1. Addition or deletion of one or two nucleotides shifts the reading frame.
2. This is called a frame-shift mutation.
3. Hence option (d) is correct: frame shift

Teacher's Note:
1. Frame shift changes all codons after the change.
2. Substitution does not shift the frame.

Question 14: Chromosomes with genes abcdefg becoming abedcfg is :
(a) duplication
(b) deletion
(c) translocation
(d) inversion
Show Answer & Explanation

Answer: (d) inversion

Explanation:
1. The segment 'cde' has been reversed to 'edc'.
2. A reversed segment within a chromosome is an inversion.
3. Hence option (d) is correct: inversion

Teacher's Note:
1. Inversion = 180 degree turn of a segment.
2. No gene is lost or gained.

Question 15: Chromosome number 2n-1 is an example of
(a) trisomy
(b) euploidy
(c) polyploidy
(d) monosomy
Show Answer & Explanation

Answer: (d) monosomy

Explanation:
1. Loss of one chromosome from the diploid number (2n - 1) is monosomy.
2. 2n + 1 is trisomy.
3. Hence option (d) is correct: monosomy

Teacher's Note:
1. Aneuploidy changes one or a few chromosomes.
2. Euploidy changes whole sets.

Question 16: Which step does not involve in DNA finger printing
(a) Southern blotting
(b) Gel electrophoresis
(c) Restriction enzyme digestion
(d) Northern blotting
Show Answer & Explanation

Answer: (d) Northern blotting

Explanation:
1. DNA fingerprinting uses restriction digestion, gel electrophoresis, Southern blotting and hybridisation with a probe.
2. Northern blotting is for RNA.
3. Hence option (d) is correct: Northern blotting

Teacher's Note:
1. Southern = DNA, Northern = RNA, Western = protein.

Question 17: Which of the following techniques are used in analyzing restriction fragment length polymorphism (RFLP): (a) Electrophoresis (b) Electroporation (c) Methylation (d) Restriction digestion
(a) 'a' and 'c'
(b) 'c' and 'd'
(c) 'a' and 'd'
(d) 'b' and 'd'
Show Answer & Explanation

Answer: (c) 'a' and 'd'

Explanation:
1. RFLP is analysed by digesting DNA with restriction enzymes and separating the fragments by electrophoresis.
2. Electroporation and methylation are not used.
3. Hence option (c) is correct: 'a' and 'd'

Teacher's Note:
1. RFLP = differences in fragment lengths.
2. They are seen as band patterns on a gel.

Question 18: The transfer of protein from electrophoretic gel to nitrocellulose membrane is known as:
(a) transferase
(b) northern blotting
(c) western blotting
(d) southern blotting
Show Answer & Explanation

Answer: (c) western blotting

Explanation:
1. Transfer of proteins from a gel to a nitrocellulose membrane is western blotting.
2. Southern is for DNA and Northern for RNA.
3. Hence option (c) is correct: western blotting

Teacher's Note:
1. Antibodies are used to detect the protein on the membrane.

Question 19: In density gradient centrifugation , the bulk DNA forms_____ while satellite DNA forms________.
(a) Major peak; Minor peak
(b) Minor peak; Major peak
(c) Major peak; Major peak
(d) Minor peak; Minor peak
Show Answer & Explanation

Answer: (a) Major peak; Minor peak

Explanation:
1. On density gradient centrifugation the bulk DNA forms a major peak and the satellite DNA forms a minor peak.
2. Satellite DNA has a different base composition.
3. Hence option (a) is correct: Major peak; Minor peak

Teacher's Note:
1. Satellite DNA is repetitive DNA.
2. It is the basis of VNTR and DNA fingerprinting.

Question 20: Which step is not correct in DNA finger printing–
(a) Isolation of DNA
(b) Digestion of DNA by DNA ligase enzyme
(c) Separation of DNA by electophoresis
(d) Hybridisation using labelled VNTR probe
Show Answer & Explanation

Answer: (b) Digestion of DNA by DNA ligase enzyme

Explanation:
1. DNA is cut with restriction enzymes, not with DNA ligase.
2. Ligase joins DNA fragments.
3. Hence option (b) is correct: Digestion of DNA by DNA ligase enzyme

Teacher's Note:
1. Restriction enzyme cuts and ligase joins.
2. Do not mix them up.

Question 21: DNA fingerprinting method is very useful for -
(a) DNA tests for identity & relation ships
(b) Forensic studies
(c) Polymorphism
(d) All of the above
Show Answer & Explanation

Answer: (d) All of the above

Explanation:
1. DNA fingerprinting is used to determine identity and relationships, in forensic work and to study polymorphism.
2. So all of the above are correct.
3. Hence option (d) is correct: All of the above

Teacher's Note:
1. VNTR patterns are unique to each person except identical twins.

Multiple Choice Questions (MCQs) for NEET Biology Molecular Basis of Inheritance

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