Chapter 5: Molecular Basis of Inheritance Application Questions | biology Class 12 CBSE

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Chapter 5: Molecular Basis of Inheritance Application Questions | biology Class 12 CBSE | ByteNotes.in

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A forensic scientist recovers DNA from a hair follicle found at a crime scene. The suspect claims he was never at the scene. With reference to DNA fingerprinting, explain how the scientist would use this DNA sample to support or refute the suspect's claim.

The scientist would extract DNA from the hair follicle and from a blood sample of the suspect, then digest both with restriction endonucleases to generate fragments of varying sizes based on VNTR copy numbers. The fragments would be separated by gel electrophoresis, transferred to a nylon membrane (Southern blotting), and hybridised with a radiolabelled VNTR probe, followed by autoradiography to produce banding patterns.
If the banding pattern from the crime scene hair matches the suspect's pattern, it would indicate the hair belongs to the suspect, placing them at the scene; a mismatch would exonerate the suspect. The uniqueness of VNTR patterns (differing between all individuals except monozygotic twins) makes this comparison highly reliable as a forensic tool.
Since DNA from every tissue — including hair follicles — of an individual shows the same degree of VNTR polymorphism (same genetic profile), even trace amounts of biological material can be used; PCR amplification further increases sensitivity, allowing the technique to work on DNA from a single cell.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 2

Applied Concept

A child of blood group O is born to a father of blood group AB. The mother claims that a certain man (blood group A) is not the biological father. Using principles of molecular genetics, explain how DNA fingerprinting would be used to determine the biological father, and why blood group analysis alone may be insufficient.

Blood group analysis has limitations: blood group O (genotype ii) cannot be a biological child of an AB father (genotype IAIB) under standard ABO inheritance, making paternity by this individual impossible on blood group grounds alone; however, blood typing only examines a few loci and cannot definitively identify one individual from a population.
DNA fingerprinting analyses VNTR polymorphisms across multiple loci: the child's banding pattern consists of bands inherited from both biological parents. By comparing the child's bands with those of the alleged father, scientists determine whether all of the child's bands that are not present in the mother's profile match bands in the alleged father's profile.
Since VNTR patterns are inherited in a Mendelian fashion (half from each biological parent), a true biological father will share specific bands with the child that a non-father would not. This multi-locus analysis provides high statistical certainty of paternity or non-paternity, far exceeding the discriminating power of ABO blood typing.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 3

Applied Concept

A bacterium is growing in a medium containing only glucose. Lactose is then added to the medium. With reference to the lac operon, explain the sequence of molecular events that occur to enable the bacterium to metabolise lactose.

In glucose-only conditions, the lac repressor (produced constitutively by the i gene) is active and bound to the operator, blocking RNA polymerase from transcribing the structural genes z, y, and a. A basal low level of permease exists in the cell membrane, allowing a small amount of lactose to enter.
When lactose is added and transported into the cell, basal β-galactosidase converts some lactose to allolactose; allolactose (the inducer) binds to the lac repressor and causes a conformational change that prevents the repressor from binding the operator.
With the operator free, RNA polymerase binds the promoter and transcribes the polycistronic mRNA encoding β-galactosidase, permease, and transacetylase; β-galactosidase hydrolyses lactose to galactose and glucose (for energy), permease increases lactose uptake, and the operon remains on until all lactose is consumed — after which the active repressor re-binds the operator and turns off the operon.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 4

Applied Concept

During a replication experiment, a scientist inhibits the enzyme DNA ligase in E. coli. Predict the consequence of this inhibition on DNA replication and explain your reasoning with reference to the replication process.

DNA ligase is required to join the discontinuously synthesised Okazaki fragments on the lagging strand; inhibiting DNA ligase would result in the accumulation of many short unjoined Okazaki fragments on the lagging strand, leaving nicks (gaps) in the newly synthesised DNA.
The leading strand synthesis would be unaffected since it is synthesised continuously in the 5'→3' direction without requiring ligase; however, the lagging strand would remain as a series of disconnected fragments and would be incomplete.
The consequence would be failure to complete replication of the lagging strand, resulting in single-stranded DNA breaks, genomic instability, and ultimately inhibition of cell division; if unrepaired, this would trigger DNA damage checkpoints and potentially cell death, highlighting the essential role of DNA ligase in maintaining genome integrity.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 5

Applied Concept

A student reads a sequence on one strand of a DNA double helix as 5'-ATGCATGCATGC-3'. Write the sequence of the complementary strand (in 5'→3' direction), the mRNA transcribed from the template strand, and predict the kind of protein mutation if the 4th nucleotide (C) is deleted.

The complementary (antiparallel) strand, written 5'→3', is: 5'-GCATGCATGCAT-3'. The template strand would be the complement of the coding strand (5'→3': ATGCATGCATGC), read 3'→5' by RNA polymerase; the mRNA transcribed would have the same sequence as the coding strand but with U replacing T: 5'-AUGCAUGCAUGC-3'.
Deletion of the 4th nucleotide (C at position 4 of the original coding strand) is a single nucleotide deletion, causing a frameshift mutation; from the point of deletion onwards, the reading frame shifts by one base, altering every codon downstream and producing a completely different amino acid sequence.
The resulting polypeptide would almost certainly be non-functional because the frameshifted sequence generates different codons throughout the rest of the sequence; it may also introduce a premature stop codon, truncating the protein — this is a severe mutation analogous to a frameshift insertion/deletion described in the chapter.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 6

Applied Concept

E. coli is grown in a medium containing 15N for many generations, then transferred to a 14N medium. With reference to Meselson and Stahl's experiment, predict the DNA density pattern observed after one and two generations using CsCl density gradient centrifugation. Justify your prediction.

After one generation in 14N medium, each replicated DNA molecule has one original 15N-strand (parental) and one newly synthesised 14N-strand; since both strands remain together in the double helix, all DNA molecules have intermediate (hybrid) density — only one band appears at an intermediate position in the CsCl gradient.
After two generations, each hybrid DNA molecule separates its strands during replication: the 15N strand serves as a template for a new 14N strand (giving another hybrid molecule), and the 14N strand serves as a template for a new 14N strand (giving a light molecule); the result is equal amounts of hybrid DNA and light DNA — two bands appear.
These results can only be explained by semiconservative replication; conservative replication would predict heavy and light bands (no hybrid) after one generation, and dispersive replication would predict all-intermediate DNA at all generations — neither matches the actual observations, confirming Watson and Crick's semiconservative model.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 7

Applied Concept

A researcher wishes to propagate a small piece of foreign DNA in bacterial cells. She finds that the DNA is not replicated in bacteria without any modification. With reference to the origin of replication, explain why this occurs and how it can be overcome.

DNA replication in E. coli does not initiate randomly; replication begins only at a specific sequence called the origin of replication (ori) present in the bacterial chromosome. The foreign DNA fragment lacks this ori sequence, so DNA-dependent DNA polymerase cannot initiate replication on it, and it is not duplicated during bacterial cell division.
To overcome this, the foreign DNA must be inserted into a vector (e.g., a plasmid) that contains a functional origin of replication compatible with the bacterial host; the vector provides the ori, allowing the entire recombinant DNA molecule (vector + insert) to be replicated whenever the host cell divides.
This is the molecular basis of recombinant DNA technology and cloning; specialised vectors such as BAC (bacterial artificial chromosomes) and YAC (yeast artificial chromosomes) were also used in the Human Genome Project to propagate large genomic DNA inserts in bacterial and yeast hosts respectively.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 8

Applied Concept

In an experiment, the hnRNA isolated from a eukaryotic cell is larger than the mature mRNA obtained from the same cell for the same gene. Explain the molecular events responsible for this size difference.

The hnRNA (heterogeneous nuclear RNA) is the primary transcript produced by RNA Pol II; it contains both exon sequences (coding) and intron sequences (non-coding intervening sequences), making it larger than the final mRNA.
Post-transcriptional processing removes this size difference through splicing: introns are precisely excised from hnRNA and the exons are joined in a defined order; this significantly reduces the size of the RNA molecule.
Additionally, hnRNA undergoes 5'-capping (addition of 7-methylguanosine triphosphate) and 3'-poly-A tailing (addition of 200–300 adenylate residues at the 3'-end in a template-independent manner), and these modifications are retained in the mature mRNA; the net size reduction comes primarily from intron removal, producing the mature, smaller, functional mRNA that is exported to the cytoplasm.

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Application Question

Easy difficulty • Concept in a practical situation

Easy

Question 9

Applied Concept

A double-stranded DNA sample contains 20% cytosine. Using Chargaff's rules, calculate the percentage of each of the other three bases (adenine, guanine, thymine) in this DNA. Show your reasoning.

By Chargaff's rule, in double-stranded DNA: %C = %G and %A = %T; if cytosine is 20%, then guanine is also 20% (since G pairs with C).
The total percentage of all four bases must equal 100%; therefore: %A + %T = 100% − (%C + %G) = 100% − (20% + 20%) = 60%; since %A = %T, each is 30%.
Final answer: Adenine = 30%, Thymine = 30%, Guanine = 20%, Cytosine = 20%. This calculation confirms that Chargaff's rules are a direct consequence of the complementary base pairing (A-T and G-C) in the Watson-Crick double helix model.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 10

Applied Concept

A student is studying a RNA virus (like Tobacco Mosaic Virus). She wonders whether the Central Dogma applies to this organism. With reference to the properties of RNA as a genetic material and the Central Dogma, explain what modifications to the standard flow of information would be necessary in this virus.

In Tobacco Mosaic Virus (TMV), RNA is the genetic material; the Central Dogma as proposed for most organisms (DNA → RNA → Protein) does not directly apply because there is no DNA intermediate.
Instead, the viral RNA serves a dual role: it acts as both the genetic material (storing and transmitting genetic information) and as a direct template for protein synthesis (translated directly as mRNA), bypassing the need for transcription from DNA.
For replication of the viral RNA genome, the virus uses an RNA-dependent RNA polymerase (RNA replicase) to copy the RNA genome directly (RNA → RNA); this represents a modification of the Central Dogma, as the genetic information flows RNA → RNA (replication) and RNA → Protein (translation), exemplifying that RNA can be sufficient as a genetic material in certain organisms.

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Chapter 5: Molecular Basis of Inheritance | Application Questions

A forensic scientist recovers DNA from a hair follicle found at a crime scene. The suspect claims he was never at the scene. With reference to DNA fingerprinting, explain how the scientist would use this DNA sample to support or refute the suspect's claim.

A bacterium is growing in a medium containing only glucose. Lactose is then added to the medium. With reference to the lac operon, explain the sequence of molecular events that occur to enable the bacterium to metabolise lactose.

During a replication experiment, a scientist inhibits the enzyme DNA ligase in E. coli. Predict the consequence of this inhibition on DNA replication and explain your reasoning with reference to the replication process.

A student reads a sequence on one strand of a DNA double helix as 5'-ATGCATGCATGC-3'. Write the sequence of the complementary strand (in 5'→3' direction), the mRNA transcribed from the template strand, and predict the kind of protein mutation if the 4th nucleotide (C) is deleted.

E. coli is grown in a medium containing 15N for many generations, then transferred to a 14N medium. With reference to Meselson and Stahl's experiment, predict the DNA density pattern observed after one and two generations using CsCl density gradient centrifugation. Justify your prediction.

A researcher wishes to propagate a small piece of foreign DNA in bacterial cells. She finds that the DNA is not replicated in bacteria without any modification. With reference to the origin of replication, explain why this occurs and how it can be overcome.

In an experiment, the hnRNA isolated from a eukaryotic cell is larger than the mature mRNA obtained from the same cell for the same gene. Explain the molecular events responsible for this size difference.

A double-stranded DNA sample contains 20% cytosine. Using Chargaff's rules, calculate the percentage of each of the other three bases (adenine, guanine, thymine) in this DNA. Show your reasoning.

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