NEET Biology Molecular Basis of Inheritance MCQs Set 11

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

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Practice Molecular Basis of Inheritance MCQs for NEET Biology

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Multiple Choice Questions

Question 1: The untranslated regions (UTRs) are present at the ___ end before the start codon and at the ___ end after the stop codon.
(a) 5'; 3'
(b) 3'; 3'
(c) 3'; 5'
(d) 5'; 5'
Show Answer & Explanation

Answer: (a) 5'; 3'

Explanation:
1. The 5' UTR lies before the start codon.
2. The 3' UTR lies after the stop codon.

Teacher's Note:
1. UTRs are required for efficient translation.
2. They are part of mRNA but not translated.

Question 2: At the end of translation, a ___ binds to the stop codon, terminating translation and releasing the complete polypeptide.
(a) sigma factor
(b) release factor
(c) rho factor
(d) tRNA
Show Answer & Explanation

Answer: (b) release factor

Explanation:
1. No tRNA recognises the stop codon.
2. A release factor binds to it and ends translation.

Teacher's Note:
1. Rho is used in transcription termination.
2. Release factors act in translation.

Question 3: In prokaryotes, the predominant site for control of gene expression is the rate of
(a) mRNA transport
(b) splicing
(c) translation
(d) transcriptional initiation
Show Answer & Explanation

Answer: (d) transcriptional initiation

Explanation:
1. The rate of transcription initiation is the main control point in prokaryotes.
2. Eukaryotes have more levels.

Teacher's Note:
1. Regulation saves energy.
2. Prokaryotes lack splicing and nuclear transport.

Question 4: The activity of RNA polymerase at a given promoter is regulated by interaction with
(a) histones only
(b) accessory (regulatory) proteins
(c) DNA ligase
(d) ribosomes
Show Answer & Explanation

Answer: (b) accessory (regulatory) proteins

Explanation:
1. Accessory proteins affect the ability of RNA polymerase to recognise the start site.
2. They can act as activators or repressors.

Teacher's Note:
1. Regulatory proteins can act positively or negatively.
2. Sigma factor is one example.

Question 5: The z gene of the lac operon codes for ___, which is primarily responsible for the hydrolysis of ___.
(a) beta-galactosidase; maltose
(b) beta-galactosidase; lactose
(c) permease; lactose
(d) transacetylase; sucrose
Show Answer & Explanation

Answer: (b) beta-galactosidase; lactose

Explanation:
1. Beta-galactosidase hydrolyses lactose into glucose and galactose.
2. It is coded by the z gene.

Teacher's Note:
1. The z gene is the first structural gene.
2. The y gene makes permease.

Question 6: The human genome contains about ___ million nucleotide base pairs.
(a) 3164.7
(b) 164
(c) 316
(d) 31647
Show Answer & Explanation

Answer: (a) 3164.7

Explanation:
1. The human genome has about 3164.7 million base pairs.
2. This is about 3 x 10^9 bp.

Teacher's Note:
1. The number of genes is about 30,000.
2. Only a small part codes for proteins.

Question 7: Less than ___ of the human genome codes for proteins.
(a) 50 percent
(b) 20 percent
(c) 98 percent
(d) 2 percent
Show Answer & Explanation

Answer: (d) 2 percent

Explanation:
1. Only about 2 percent of the genome codes for proteins.
2. The rest is repetitive and non-coding.

Teacher's Note:
1. Repetitive sequences make up a large part.
2. Their function is not fully known.

Question 8: Chromosome 1 has the most genes and the Y chromosome has the fewest. What are the numbers?
(a) 2968 on chromosome 1 and 231 on Y
(b) 231 on chromosome 1 and 2968 on Y
(c) 1968 on chromosome 1 and 131 on Y
(d) 3968 on chromosome 1 and 331 on Y
Show Answer & Explanation

Answer: (a) 2968 on chromosome 1 and 231 on Y

Explanation:
1. Chromosome 1 has 2968 genes.
2. The Y chromosome has 231 genes.

Teacher's Note:
1. Remember the pair 2968 and 231.
2. Chromosome 1 is the biggest.

Question 9: Under the electron microscope, nucleosomes in chromatin look like a
(a) cloverleaf
(b) beads-on-string structure
(c) double helix
(d) solenoid fibre only
Show Answer & Explanation

Answer: (b) beads-on-string structure

Explanation:
1. Nucleosomes appear as beads connected by thin DNA threads.
2. This is the beads-on-string structure.

Teacher's Note:
1. It is the first level of packaging.
2. Higher levels give the 30 nm fibre.

Question 10: Transcriptionally active chromatin is called
(a) euchromatin
(b) centromere
(c) nucleolus
(d) heterochromatin
Show Answer & Explanation

Answer: (a) euchromatin

Explanation:
1. Euchromatin is loosely packed and stains light.
2. Genes in it can be transcribed.

Teacher's Note:
1. Heterochromatin is densely packed and inactive.
2. Euchromatin is active.

Question 11: The biochemical nature of the transforming principle in Griffith's experiment was determined by
(a) Hershey and Chase
(b) Meselson and Stahl
(c) Avery, MacLeod and McCarty
(d) Watson and Crick
Show Answer & Explanation

Answer: (c) Avery, MacLeod and McCarty

Explanation:
1. Avery, MacLeod and McCarty purified the transforming principle.
2. They showed it was DNA.

Teacher's Note:
1. They used enzymes that digest proteins, RNA and DNA.
2. Only DNase stopped transformation.

Question 12: The unequivocal proof that DNA is the genetic material came from the experiments of
(a) Alfred Hershey and Martha Chase
(b) Mendel
(c) Avery
(d) Griffith
Show Answer & Explanation

Answer: (a) Alfred Hershey and Martha Chase

Explanation:
1. Hershey and Chase showed that DNA, not protein, entered the bacteria.
2. It was done on bacteriophage T2.

Teacher's Note:
1. Their blender experiment was done in 1952.
2. It is the final proof.

Question 13: The first genetic material was most probably
(a) RNA
(b) DNA
(c) protein
(d) lipid
Show Answer & Explanation

Answer: (a) RNA

Explanation:
1. RNA could act as both genetic material and catalyst.
2. So the first genetic material was RNA.

Teacher's Note:
1. DNA evolved later.
2. It is more stable and chemically less reactive.

Question 14: The semi-conservative replication of DNA in chromosomes was shown on Vicia faba by Taylor using radioactive
(a) thymidine
(b) phosphorus
(c) uridine
(d) sulphur
Show Answer & Explanation

Answer: (a) thymidine

Explanation:
1. Taylor used tritiated thymidine in Vicia faba root tips.
2. The radioactivity pattern in chromatids showed semi-conservative replication.

Teacher's Note:
1. Thymidine is a DNA-specific precursor.
2. Uridine goes into RNA.

Question 15: In E. coli the average rate of polymerisation during DNA replication is approximately
(a) 2000 bp per second
(b) 20 bp per second
(c) 20,000 bp per second
(d) 200 bp per second
Show Answer & Explanation

Answer: (a) 2000 bp per second

Explanation:
1. The E. coli genome of 4.6 million bp replicates in about 40 minutes.
2. This gives about 2000 bp per second.

Teacher's Note:
1. Replication is very fast.
2. It is also accurate.

Question 16: In a transcription unit, the promoter is located towards the ___ of the structural gene.
(a) terminator end
(b) 3' end (downstream)
(c) 5' end (upstream)
(d) middle
Show Answer & Explanation

Answer: (c) 5' end (upstream)

Explanation:
1. The promoter lies at the 5' end of the coding strand, upstream.
2. The terminator lies at the 3' end.

Teacher's Note:
1. Upstream means toward the 5' end.
2. RNA polymerase binds at the promoter.

Question 17: A piece of DNA to be propagated in recombinant DNA work requires a vector because the vector provides
(a) the origin of replication
(b) the stop codon
(c) restriction enzymes
(d) the promoter of the host only
Show Answer & Explanation

Answer: (a) the origin of replication

Explanation:
1. The vector has an origin of replication (ori).
2. The foreign DNA multiplies with it.

Teacher's Note:
1. Without ori the DNA will not replicate.
2. Vectors also carry selectable markers.

Question 18: Which of the following statements is correct?
(a) Amino acids act as templates for mRNA
(b) tRNA is not needed in translation
(c) No complementarity exists between nucleotides and amino acids, so an adapter molecule is needed
(d) Nucleotides pair directly with amino acids
Show Answer & Explanation

Answer: (c) No complementarity exists between nucleotides and amino acids, so an adapter molecule is needed

Explanation:
1. Amino acids have no chemical affinity for nucleotides.
2. Crick therefore proposed an adapter, the tRNA.

Teacher's Note:
1. tRNA reads the codon and carries the amino acid.
2. It is the adapter.

Question 19: Which statement about the enzyme polynucleotide phosphorylase of Severo Ochoa is correct?
(a) It polymerised DNA from RNA
(b) It polymerised RNA with no need of a template
(c) It polymerised RNA with a defined sequence in a template dependent manner
(d) It cut RNA at specific sites
Show Answer & Explanation

Answer: (b) It polymerised RNA with no need of a template

Explanation:
1. The enzyme made RNA without a template, so the sequence was not defined.
2. It was still helpful for the code work.

Teacher's Note:
1. Khorana's method gave defined sequences.
2. Ochoa's enzyme gave random sequences.

Question 20: For initiation of translation, the ribosome binds to the mRNA at the start codon, which is recognised only by the
(a) release factor
(b) any tRNA
(c) rRNA
(d) initiator tRNA
Show Answer & Explanation

Answer: (d) initiator tRNA

Explanation:
1. The initiator tRNA carries methionine and reads AUG.
2. No other tRNA can start translation.

Teacher's Note:
1. The small subunit binds first.
2. Then the large subunit joins.

Question 21: Regulatory proteins in gene expression
(a) can act only negatively
(b) can act only positively
(c) do not affect transcription
(d) can act positively or negatively
Show Answer & Explanation

Answer: (d) can act positively or negatively

Explanation:
1. Activators act positively and repressors act negatively.
2. So regulatory proteins can act in both ways.

Teacher's Note:
1. The lac repressor is a negative regulator.
2. CAP is a positive regulator.

Question 22: In the lac operon, the three structural genes are regulated by
(a) three operators
(b) the promoter of the i gene only
(c) a common promoter and regulator gene
(d) separate promoters
Show Answer & Explanation

Answer: (c) a common promoter and regulator gene

Explanation:
1. The three structural genes z, y and a are transcribed from one promoter.
2. The regulator gene controls them.

Teacher's Note:
1. They give one polycistronic mRNA.
2. They are controlled together.

Question 23: A very low level of lac operon expression is present in the cell all the time because
(a) the operator is absent
(b) the repressor never binds
(c) lactose is always present
(d) this ensures permease is present to bring the inducer in when lactose appears
Show Answer & Explanation

Answer: (d) this ensures permease is present to bring the inducer in when lactose appears

Explanation:
1. Basal expression gives a few permease and beta-galactosidase molecules.
2. These allow lactose to enter and induce the operon.

Teacher's Note:
1. Without basal expression induction would not start.
2. Regulation is therefore leaky.

Question 24: The development and differentiation of an embryo into an adult organism are the result of
(a) replication errors
(b) coordinated regulation of the expression of several sets of genes
(c) mutation of all genes
(d) loss of DNA
Show Answer & Explanation

Answer: (b) coordinated regulation of the expression of several sets of genes

Explanation:
1. Different genes switch on at different times and places.
2. This gives differentiation.

Teacher's Note:
1. All cells have the same genes.
2. Only some are expressed.

Question 25: In bacteria, transcription and translation can be coupled because
(a) mRNA is capped
(b) the ribosome enters the nucleus
(c) mRNA is spliced first
(d) there is no nuclear membrane separating them
Show Answer & Explanation

Answer: (d) there is no nuclear membrane separating them

Explanation:
1. Bacteria lack a nucleus.
2. So ribosomes can begin translating mRNA while it is still being transcribed.

Teacher's Note:
1. In eukaryotes, the nuclear membrane separates the two.
2. mRNA must be processed first.

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