NEET Biology Molecular Basis of Inheritance MCQs Set 25

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Multiple Choice Questions

Question 1: Diploid chromosome number in humans is
(a) 46
(b) 44
(c) 48
(d) 42
Show Answer & Explanation

Answer: (a) 46

Explanation:
1. The diploid chromosome number of humans is 46 (23 pairs).

Teacher's Note:
1. The haploid number is 23.
2. 44 are autosomes and 2 are sex chromosomes.

Question 2: DNA replication is
(a) conservative and discontinuous
(b) semi-conservative and semi-discontinuous
(c) semi-conservative and discontinuous
(d) conservative
Show Answer & Explanation

Answer: (b) semi-conservative and semi-discontinuous

Explanation:
1. DNA replication is semi-conservative (each daughter has one old strand) and semi-discontinuous (leading strand continuous, lagging strand in fragments).

Teacher's Note:
1. Meselson and Stahl proved semi-conservative replication.
2. Both words describe replication.

Question 3: The process of transfer of genetic information from DNA to RNA/formation of RNA from DNA is
(a) trnsversion
(b) transcription
(c) translation
(d) translocation
Show Answer & Explanation

Answer: (b) transcription

Explanation:
1. Formation of RNA from a DNA template is transcription.

Teacher's Note:
1. Translation is protein synthesis from mRNA.
2. Transcription copies DNA into RNA.

Question 4: Experimental material in the study of DNA replication has been
(a) Escherichia coli
(b) Neurospora crassa
(c) Pneumococcus
(d) Drosophila melanogaster
Show Answer & Explanation

Answer: (a) Escherichia coli

Explanation:
1. E. coli was used by Meselson and Stahl for the study of DNA replication, because it divides every 20 minutes.

Teacher's Note:
1. Pneumococcus was used by Griffith.
2. Remember the experimental organism.

Question 5: The transforming principle of Pneumococcus as found out by Avery, MacLeod and McCarty was
(a) mRNA
(b) DNA
(c) protein
(d) polysaccharide
Show Answer & Explanation

Answer: (b) DNA

Explanation:
1. Avery, MacLeod and McCarty showed that DNA is the transforming principle of Pneumococcus.

Teacher's Note:
1. Griffith had observed the transformation.
2. DNA is the transforming principle.

Question 6: Because most of the amino acids are represented by more than one codon, the genetic code is
(a) overlapping
(b) Wobbling
(c) degenerate
(d) generate
Show Answer & Explanation

Answer: (c) degenerate

Explanation:
1. Because most amino acids are coded by more than one codon, the genetic code is degenerate.

Teacher's Note:
1. Wobbling refers to the third base pairing.
2. Degenerate means redundant.

Question 7: Who proved that DNA is basic genetic material?
(a) Griffith
(b) Watson
(c) Boveri and Sutton
(d) Hershey and Chase
Show Answer & Explanation

Answer: (d) Hershey and Chase

Explanation:
1. Hershey and Chase proved with radioactive phosphorus (32P) and sulphur (35S) that DNA is the genetic material.

Teacher's Note:
1. Griffith showed transformation only.
2. Bacteriophage experiment.

Question 8: During DNA replication, the strands separate by
(a) DNA polymerase
(b) topoisomerase
(c) unwindase/helicase
(d) gyrase
Show Answer & Explanation

Answer: (c) unwindase/helicase

Explanation:
1. Helicase (unwindase) separates the two strands of DNA at the replication fork.

Teacher's Note:
1. Topoisomerase relieves supercoiling and polymerase builds new strands.
2. Helicase unwinds the DNA.

Question 9: The process of translation is
(a) ribosome synthesis
(b) protein synthesis
(c) DNA synthesis
(d) RNA synthesis
Show Answer & Explanation

Answer: (b) protein synthesis

Explanation:
1. Translation is the process of protein synthesis on ribosomes from mRNA.

Teacher's Note:
1. Transcription makes RNA.
2. Translation means mRNA to protein.

Question 10: A DNA with unequal nitrogen bases would most probably be
(a) single stranded
(b) double stranded
(c) triple stranded
(d) four stranded
Show Answer & Explanation

Answer: (a) single stranded

Explanation:
1. In double-stranded DNA A = T and G = C. If the bases are unequal, the DNA must be single stranded.

Teacher's Note:
1. Chargaff's rules apply to double strands.
2. Check the base ratios.

Question 11: Nucleosome core is made of
(a) H1, H2A, H2B and H3
(b) H1, H2A, H2B, H4
(c) H1, H2A, H2B, H3 and H4
(d) H2A, H2B, H3 and H4
Show Answer & Explanation

Answer: (d) H2A, H2B, H3 and H4

Explanation:
1. The histone octamer core of a nucleosome is made up of two molecules each of H2A, H2B, H3 and H4. H1 is the linker histone.

Teacher's Note:
1. It is wrapped by 200 bp of DNA.
2. H1 is outside the core.

Question 12: Initiation codon of protein synthesis (in eukaryotes) is
(a) GUA
(b) GCA
(c) CCA
(d) AUG
Show Answer & Explanation

Answer: (d) AUG

Explanation:
1. AUG is the initiation codon in eukaryotes, which codes for methionine.

Teacher's Note:
1. It is formylmethionine in prokaryotes.
2. AUG starts translation.

Question 13: Reverse transcriptase is
(a) RNA dependent RNA polymerase
(b) DNA dependent RNA polymerase
(c) DNA dependent DNA polymerase
(d) RNA dependent DNA polymerase
Show Answer & Explanation

Answer: (d) RNA dependent DNA polymerase

Explanation:
1. Reverse transcriptase is an RNA-dependent DNA polymerase, which makes DNA from RNA.

Teacher's Note:
1. It is found in retroviruses like HIV.
2. RNA to DNA.

Question 14: In Escherichia coli lac Operon is induced by
(a) lactose
(b) promoter gene
(c) -galactosidase
(d) I-gene
Show Answer & Explanation

Answer: (a) lactose

Explanation:
1. The lac operon of E. coli is induced by lactose (allolactose), which binds to the repressor.

Teacher's Note:
1. The i gene makes the repressor.
2. Lactose is the inducer.

Question 15: DNA template sequence of CTGATAGC is transcribed over mRNA as
(a) GUCTUTCG
(b) GACUAUCG
(c) GAUTATUG
(d) UACTATCU
Show Answer & Explanation

Answer: (b) GACUAUCG

Explanation:
1. The mRNA is complementary to the template strand. For CTGATAGC the mRNA is GACUAUCG, with U in place of T.

Teacher's Note:
1. The template is read 3' to 5'.
2. Pair G with C, A with U.

Question 16: Out of A=T, G  C pairing, bases of DNA may exist in alternate valency state owing to arrangement called
(a) analogue substitution
(b) tautomerizational mutation
(c) frameshift mutation
(d) point mutation
Show Answer & Explanation

Answer: (b) tautomerizational mutation

Explanation:
1. The bases of DNA can exist in rare alternate (tautomeric) forms, which cause wrong pairing. This is a tautomerisation mutation.

Teacher's Note:
1. Normally A pairs with T and G with C.
2. Tautomers cause mismatches.

Question 17: The wild type E. coli cells are growing in normal medium with glucose. They are transferred to a medium containing only lactose as sugar. Which of the following changes takes place?
(a) The lac operon is repressed
(b) All operons are induced
(c) The lac operon is induced
(d) E. coli cells stop dividing
Show Answer & Explanation

Answer: (c) The lac operon is induced

Explanation:
1. In a lactose-only medium the repressor is inactivated by lactose, so the lac operon is induced.

Teacher's Note:
1. With glucose it is repressed.
2. Lactose induces the operon.

Question 18: If the sequence of bases in DNA is ATTCGATG, then the sequence of bases in its transcript will be
(a) CAUCGAAU
(b) UAAGCUAC
(c) GUAGCUUA
(d) AUUCGAUG
Show Answer & Explanation

Answer: (b) UAAGCUAC

Explanation:
1. The transcript is complementary to the template: ATTCGATG gives UAAGCUAC.

Teacher's Note:
1. U replaces T in RNA.
2. Pair A with U, T with A, C with G.

Question 19: In split genes, the coding sequence are called
(a) introns
(b) operons
(c) exons
(d) cistrons
Show Answer & Explanation

Answer: (c) exons

Explanation:
1. In split genes the coding sequences are the exons and the non-coding sequences are the introns.

Teacher's Note:
1. Introns are removed by splicing.
2. Exons are expressed.

Question 20: The polytene chromosomes were discovered for the first time in
(a) Drosophila
(b) Chironomus
(c) Musca nebulo
(d) Musca domestica
Show Answer & Explanation

Answer: (b) Chironomus

Explanation:
1. Polytene chromosomes were first discovered by Balbiani in the salivary glands of Chironomus larvae.

Teacher's Note:
1. They are also found in Drosophila.
2. Chironomus is the first discovery.

Question 21: The most striking example of point mutation is found in a disease called
(a) thalassemia
(b) night blindness
(c) Down’s syndrome
(d) sickle cell anaemia
Show Answer & Explanation

Answer: (d) sickle cell anaemia

Explanation:
1. Sickle cell anaemia is the classic example of point mutation, where a single base change replaces glutamic acid by valine.

Teacher's Note:
1. Thalassemia is due to reduced globin synthesis.
2. One base, one amino acid.

Question 22: The translation termination triplet is
(a) UAU
(b) UAA
(c) UAC
(d) UGC
Show Answer & Explanation

Answer: (b) UAA

Explanation:
1. UAA is a termination (stop) codon, together with UAG and UGA.

Teacher's Note:
1. They do not code for any amino acid.
2. Stop codons end translation.

Question 23: Okazaki fragments are seen during
(a) transcription
(b) translation
(c) replication
(d) transduction
Show Answer & Explanation

Answer: (c) replication

Explanation:
1. Okazaki fragments are short DNA pieces formed on the lagging strand during replication.

Teacher's Note:
1. DNA ligase joins them.
2. Okazaki means replication.

Question 24: An enzyme that joins the ends of two strands of nucleic acid is a
(a) polymerase
(b) synthetase
(c) helicase
(d) ligase
Show Answer & Explanation

Answer: (d) ligase

Explanation:
1. DNA ligase joins the ends of two DNA strands by making phosphodiester bonds.

Teacher's Note:
1. Polymerase adds nucleotides and helicase unwinds.
2. Ligase seals nicks.

Question 25: H. J. Muller was awarded Nobel Prize for his
(a) discovery that chemicals can induce gene mutations
(b) discovery that ionizing radiations can induce gene mutations
(c) work on gene mapping in Drosophila
(d) efforts to prevent the use of nuclear weapons
Show Answer & Explanation

Answer: (b) discovery that ionizing radiations can induce gene mutations

Explanation:
1. H. J. Muller was awarded the Nobel Prize for discovering that ionising radiation (X-rays) can induce mutations.

Teacher's Note:
1. It was awarded in 1946.
2. Muller used X-rays on Drosophila.

Question 26: Different mutations referrable to the same locus of chromosome give rise to
(a) pseudoalleles
(b) polygenes
(c) oncogenes
(d) multiple alleles
Show Answer & Explanation

Answer: (d) multiple alleles

Explanation:
1. Different mutations at the same locus give rise to multiple alleles.

Teacher's Note:
1. Pseudoalleles are closely linked loci.
2. Same locus, several alleles.

Question 27: The hereditary material present in the bacterium E. coli is
(a) single stranded RNA
(b) double stranded RNA
(c) single stranded DNA
(d) double stranded DNA
Show Answer & Explanation

Answer: (d) double stranded DNA

Explanation:
1. The hereditary material of E. coli is a double-stranded circular DNA.

Teacher's Note:
1. It is present in the nucleoid.
2. Bacteria have double-stranded DNA.

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