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Maharashtra HSC Biology study materials
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Maharashtra HSC Biology practice questions

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Question 1Genetic Basis of Inheritance and Molecular Basis of Inheritance

In a dihybrid cross between two plants heterozygous for both seed shape (Rr) and seed colour (Yy), what phenotypic ratio is expected in the F2 generation under independent assortment?

  1. 9:3:3:1 Correct answer.
  2. 1:1:1:1
  3. 3:1
  4. 1:2:1

Explanation

When two gene pairs assort independently, each heterozygote gives four gamete types in equal frequency, and random fertilisation produces the classic 9:3:3:1 phenotypic ratio for the two dominant and two recessive trait combinations. This ratio arises from multiplying the two independent 3:1 monohybrid ratios.

Question 2Genetic Basis of Inheritance and Molecular Basis of Inheritance

A cross between a heterozygous tall pea plant (Tt) and a homozygous dwarf plant (tt) produces offspring. What proportion of the progeny is expected to be dwarf?

  1. Three quarters
  2. One quarter
  3. All of them
  4. One half Correct answer.

Explanation

The heterozygote Tt produces T and t gametes in equal numbers, while tt produces only t gametes. Combining these gives Tt and tt in a 1:1 ratio, so half the offspring are dwarf.

Question 3Genetic Basis of Inheritance and Molecular Basis of Inheritance

Which nitrogenous base is found in RNA but not in DNA?

  1. Adenine
  2. Guanine
  3. Uracil Correct answer.
  4. Thymine

Explanation

RNA contains uracil in place of the thymine found in DNA, while adenine, guanine and cytosine occur in both nucleic acids. Uracil pairs with adenine during RNA synthesis and translation.

Question 4Genetic Basis of Inheritance and Molecular Basis of Inheritance

According to the Watson-Crick model, the two strands of a DNA molecule are held together mainly by which type of bond between complementary bases?

  1. Hydrogen bonds Correct answer.
  2. Glycosidic bonds
  3. Phosphodiester bonds
  4. Peptide bonds

Explanation

Complementary bases on opposite strands pair through hydrogen bonds: two between adenine and thymine, three between guanine and cytosine. Phosphodiester bonds instead link nucleotides along each strand, forming the sugar-phosphate backbone.

Question 5Genetic Basis of Inheritance and Molecular Basis of Inheritance

In a DNA sample the adenine content is 30 percent of the total bases. What percentage of the bases is guanine?

  1. 40 percent
  2. 20 percent Correct answer.
  3. 70 percent
  4. 30 percent

Explanation

By Chargaff's rule, adenine equals thymine, so A + T = 60 percent, leaving 40 percent for G + C. Since guanine equals cytosine, guanine is half of 40 percent, that is 20 percent.

Question 6Genetic Basis of Inheritance and Molecular Basis of Inheritance

A woman who is a carrier of an X-linked recessive disorder marries an unaffected man. What is the probability that a son born to them will be affected?

  1. All sons
  2. One quarter
  3. One half Correct answer.
  4. Zero

Explanation

A son inherits his X chromosome from his mother and a Y from his father. Since the carrier mother has one normal and one affected X, each son has a 50 percent chance of receiving the affected allele and being affected.

Question 7Genetic Basis of Inheritance and Molecular Basis of Inheritance

In the lac operon of Escherichia coli, which molecule acts as the inducer that inactivates the repressor and allows transcription of the structural genes?

  1. Cyclic AMP
  2. Lactose permease
  3. Allolactose Correct answer.
  4. Glucose

Explanation

When lactose enters the cell, beta-galactosidase converts some of it to allolactose, which binds the repressor and prevents it from blocking the operator. This frees RNA polymerase to transcribe the lac genes.

Question 8Genetic Basis of Inheritance and Molecular Basis of Inheritance

During transcription in eukaryotes, which enzyme is responsible for synthesising the mRNA precursor from a DNA template?

  1. DNA polymerase III
  2. Reverse transcriptase
  3. RNA polymerase I
  4. RNA polymerase II Correct answer.

Explanation

RNA polymerase II transcribes protein-coding genes to produce the mRNA precursor, which is later processed by capping, polyadenylation and splicing. RNA polymerase I makes rRNA, and DNA polymerase synthesises DNA.

Question 9Genetic Basis of Inheritance and Molecular Basis of Inheritance

A template DNA strand reads 3'-TAC GGA-5'. Following base-pairing rules, what is the sequence of the mRNA synthesised from it?

  1. 5'-AUG CCU-3' Correct answer.
  2. 5'-TAC GGA-3'
  3. 5'-UAC GGA-3'
  4. 5'-ATG CCT-3'

Explanation

RNA is synthesised antiparallel to the template, so adenine pairs with thymine, uracil pairs with adenine, and guanine pairs with cytosine. Reading the template 3' to 5' gives the mRNA 5'-AUG CCU-3'.

Question 10Genetic Basis of Inheritance and Molecular Basis of Inheritance

Which feature distinguishes the genetic code as 'degenerate'?

  1. Most amino acids are specified by more than one codon Correct answer.
  2. Codons overlap along the mRNA
  3. The code differs between bacteria and humans
  4. Each codon specifies more than one amino acid

Explanation

Degeneracy means that several synonymous codons can encode the same amino acid, which buffers the effect of some point mutations. The code is nevertheless unambiguous, non-overlapping and nearly universal.

Question 11Genetic Basis of Inheritance and Molecular Basis of Inheritance

In a pedigree, a trait appears in every generation, affects both males and females equally, and an affected father never transmits it to his sons but transmits it to all his daughters. Which mode of inheritance best fits?

  1. X-linked dominant Correct answer.
  2. Y-linked
  3. Autosomal recessive
  4. X-linked recessive

Explanation

An affected father passes his only X chromosome to all daughters but his Y to sons, so X-linked dominant traits appear in all daughters and no sons. Appearance in every generation with equal sex distribution among affected individuals is consistent with dominance.

Question 12Genetic Basis of Inheritance and Molecular Basis of Inheritance

A test cross is performed on a plant showing the dominant phenotype for flower colour. All the progeny show the dominant phenotype. What is the most likely genotype of the tested plant?

  1. Homozygous dominant Correct answer.
  2. Homozygous recessive
  3. Heterozygous
  4. Hemizygous

Explanation

In a test cross the unknown individual is crossed with a homozygous recessive. If all offspring show the dominant phenotype, the tested parent must have contributed only dominant alleles, so it is homozygous dominant.

Question 13Genetic Basis of Inheritance and Molecular Basis of Inheritance

Which observation from Griffith's experiments with Streptococcus pneumoniae first suggested the existence of a transforming principle?

  1. Live avirulent bacteria did not kill mice
  2. DNA was shown to be the genetic material by chemical analysis
  3. Heat-killed virulent bacteria mixed with live avirulent bacteria produced live virulent bacteria Correct answer.
  4. Live virulent bacteria killed mice

Explanation

Griffith found that a mixture of heat-killed virulent and live avirulent cells killed mice and yielded live virulent cells, showing that some component of the dead cells transformed the avirulent strain. This transforming principle was later identified as DNA.

Question 14Genetic Basis of Inheritance and Molecular Basis of Inheritance

In the Hershey-Chase experiment, radioactive phosphorus (32P) was used to label which component of bacteriophages, and where was the radioactivity found after infection?

  1. Protein, which entered the bacterial cells
  2. Protein, which remained outside the cells
  3. DNA, which remained outside the cells
  4. DNA, which entered the bacterial cells Correct answer.

Explanation

Phosphorus occurs in DNA but not in protein, so 32P labelled the phage DNA. After infection the radioactivity was recovered inside the bacteria, supporting DNA as the genetic material.

Question 15Genetic Basis of Inheritance and Molecular Basis of Inheritance

During DNA replication, why is one newly synthesised strand called the lagging strand?

  1. It contains only RNA nucleotides
  2. It is synthesised discontinuously as short Okazaki fragments Correct answer.
  3. It is synthesised continuously towards the replication fork
  4. It is synthesised before the leading strand

Explanation

DNA polymerase can add nucleotides only in the 5' to 3' direction. On the template oriented 5' to 3' towards the fork, synthesis must proceed away from the fork in short Okazaki fragments that are later joined by ligase, making this the lagging strand.

Question 16Genetic Basis of Inheritance and Molecular Basis of Inheritance

A point mutation changes a codon from GAA to GAG, but the amino acid incorporated remains glutamic acid. What type of mutation is this?

  1. Silent mutation Correct answer.
  2. Nonsense mutation
  3. Frameshift mutation
  4. Missense mutation

Explanation

A silent mutation changes the nucleotide sequence but not the amino acid, because of the degeneracy of the genetic code. Both GAA and GAG code for glutamic acid.

Question 17Genetic Basis of Inheritance and Molecular Basis of Inheritance

Which of the following best describes the function of a promoter in a gene?

  1. It is the region of mRNA that is translated into protein
  2. It is the sequence that signals termination of translation
  3. It is a DNA sequence where RNA polymerase binds to initiate transcription Correct answer.
  4. It is the site where introns are removed after transcription

Explanation

A promoter is a specific DNA sequence upstream of a gene that binds RNA polymerase and associated factors, positioning the enzyme to begin transcription at the correct site. It is not transcribed into protein-coding sequence.

Question 18Genetic Basis of Inheritance and Molecular Basis of Inheritance

In a cross between two individuals of genotype AaBb, what proportion of offspring is expected to be homozygous recessive for both genes (aabb), assuming independent assortment?

  1. One quarter
  2. One half
  3. One sixteenth Correct answer.
  4. One eighth

Explanation

For each gene, Aa x Aa gives a one-quarter chance of aa. Multiplying the independent probabilities gives one-quarter times one-quarter, which is one sixteenth.

Question 19Genetic Basis of Inheritance and Molecular Basis of Inheritance

Colour blindness is an X-linked recessive trait. A colour-blind man marries a homozygous normal woman. What is the expected phenotype of their daughters?

  1. All daughters are colour blind
  2. All daughters are homozygous normal
  3. All daughters are carriers with normal vision Correct answer.
  4. Half the daughters are colour blind

Explanation

Daughters receive the father's affected X and a normal X from the mother, so they are heterozygous carriers with normal vision. They can pass the allele to their sons in the next generation.

Question 20Genetic Basis of Inheritance and Molecular Basis of Inheritance

Which statement correctly describes the relationship between a codon and an anticodon?

  1. An anticodon codes for an amino acid while a codon does not
  2. Both codon and anticodon are found on the same tRNA molecule
  3. A codon on tRNA pairs with an anticodon on mRNA
  4. A codon on mRNA pairs with a complementary anticodon on tRNA Correct answer.

Explanation

During translation the mRNA codon is read by the tRNA anticodon through complementary base pairing, ensuring the correct amino acid is delivered. The anticodon is part of the tRNA, and the codon is part of the mRNA.

Question 21Genetic Basis of Inheritance and Molecular Basis of Inheritance

A man with genotype IAi and a woman with genotype IBi have children. What blood group phenotypes are possible in their offspring?

  1. Only A and B
  2. A, B, AB and O Correct answer.
  3. Only O
  4. Only AB

Explanation

Each parent can contribute either an I allele or an i allele, so the offspring can be IAIB (AB), IAi (A), IBi (B) or ii (O). Multiple alleles and codominance make all four blood groups possible.

Question 22Genetic Basis of Inheritance and Molecular Basis of Inheritance

In the lac operon, what is the consequence of a mutation that inactivates the operator sequence so the repressor cannot bind?

  1. The structural genes are never transcribed
  2. The structural genes are transcribed constitutively Correct answer.
  3. The repressor protein is not synthesised
  4. Lactose can no longer enter the cell

Explanation

The operator is the binding site for the repressor. If the repressor cannot bind, RNA polymerase is not blocked, so the lac genes are expressed continuously regardless of lactose availability, a constitutive phenotype.

Question 23Genetic Basis of Inheritance and Molecular Basis of Inheritance

A segment of mRNA has the sequence 5'-AUG UUU GCU UAA-3'. How many amino acids are incorporated into the polypeptide before termination?

  1. Three Correct answer.
  2. One
  3. Four
  4. Two

Explanation

AUG, UUU and GCU each specify an amino acid, while UAA is a stop codon that terminates translation without adding an amino acid. Three amino acids are therefore incorporated.

Question 24Genetic Basis of Inheritance and Molecular Basis of Inheritance

Which feature of the DNA double helix directly explains semiconservative replication?

  1. Complementary base pairing between the two strands Correct answer.
  2. The occurrence of thymine in place of uracil
  3. The presence of phosphodiester bonds in the backbone
  4. The antiparallel orientation of the strands

Explanation

Because each base on one strand is paired with a specific complementary base on the other, either strand can act as a template to build a new partner strand. This produces two daughter molecules each containing one parental and one new strand.

Question 25Genetic Basis of Inheritance and Molecular Basis of Inheritance

A person is heterozygous for a dominant allele that causes a genetic disorder, yet shows no symptoms. Which concept best accounts for this observation?

  1. Incomplete penetrance Correct answer.
  2. Pleiotropy
  3. Codominance
  4. Polygenic inheritance

Explanation

Incomplete penetrance means that an individual carrying a dominant disease allele does not always express the phenotype, often because of modifying genes or environment. The genotype is present but the expected trait is absent.

Question 26Genetic Basis of Inheritance and Molecular Basis of Inheritance

In a dihybrid cross between two pea plants heterozygous for both seed shape (Rr) and seed colour (Yy), what is the expected phenotypic ratio of the offspring?

  1. 3:1
  2. 1:2:1
  3. 9:3:3:1 Correct answer.
  4. 1:1:1:1

Explanation

When two individuals heterozygous for two independently assorting gene pairs (RrYy × RrYy) are crossed, each gene pair shows a 3:1 phenotypic ratio. The combined ratio is the product (3:1) × (3:1) = 9:3:3:1, corresponding to the four phenotypic classes in the order round yellow : round green : wrinkled yellow : wrinkled green.

Question 27Genetic Basis of Inheritance and Molecular Basis of Inheritance

Which statement best explains why the genetic code is described as non-overlapping?

  1. The same amino acid can be encoded by several different codons.
  2. Each nucleotide is part of only one codon, so reading frames do not share bases. Correct answer.
  3. A single codon can specify more than one amino acid depending on the context.
  4. Some codons do not code for any amino acid and act as stop signals.

Explanation

In a non-overlapping code, the three bases of a codon are read as a discrete unit and are not shared with the adjacent codon. This means the reading frame advances by three nucleotides at a time, so each base belongs to exactly one codon. Overlapping codes would allow a single base to contribute to more than one amino acid specification.

Question 28Genetic Basis of Inheritance and Molecular Basis of Inheritance

During elongation of translation, which event directly precedes the formation of a peptide bond between the growing polypeptide and the new amino acid?

  1. The addition of the poly(A) tail to the 3' end of the mRNA.
  2. The codon–anticodon pairing of the incoming aminoacyl-tRNA at the A site. Correct answer.
  3. The release of the completed polypeptide from the tRNA in the P site.
  4. The dissociation of the small ribosomal subunit from the mRNA.

Explanation

Elongation begins with the delivery of an aminoacyl-tRNA to the A site, where its anticodon must base-pair with the mRNA codon. Only after this recognition does the ribosome catalyse peptide bond formation between the amino group of the A-site amino acid and the carboxyl group of the polypeptide in the P site. Thus codon–anticodon pairing at the A site is the immediate prerequisite for peptide bond formation.

Question 29Genetic Basis of Inheritance and Molecular Basis of Inheritance

A mutation changes a single base pair in the template strand of DNA from 3'-TAC-5' to 3'-TAG-5'. What is the consequence for the encoded amino acid?

  1. The mRNA codon becomes UAG, which is a stop codon, so translation terminates.
  2. The mRNA codon remains AUG because the third base is degenerate.
  3. The mRNA codon changes from AUG to AUC, so a different amino acid is incorporated. Correct answer.
  4. The mRNA codon becomes AUG, so the amino acid remains methionine.

Explanation

The template strand is read antiparallel to the mRNA, so 3'-TAC-5' gives 5'-AUG-3' and 3'-TAG-5' gives 5'-AUC-3'. AUG codes for methionine, while AUC codes for isoleucine; therefore the amino acid changes. This is a missense mutation because a single base substitution alters the amino acid specified.

Question 30Genetic Basis of Inheritance and Molecular Basis of Inheritance

Which of the following best describes the role of the TATA box in eukaryotic gene expression?

  1. It is a core promoter element that helps position RNA polymerase II at the transcription start site. Correct answer.
  2. It is a splice site that marks the boundary between an exon and an intron.
  3. It is an enhancer sequence that binds tissue-specific transcription factors far from the gene.
  4. It is a terminator sequence that signals RNA polymerase to stop transcription.

Explanation

The TATA box is a conserved AT-rich sequence found about 25–35 base pairs upstream of the transcription start site in many eukaryotic promoters. It is recognised by TATA-binding protein (TBP), a subunit of TFIID, which helps assemble the preinitiation complex and correctly position RNA polymerase II. Therefore it functions as a core promoter element rather than a distant enhancer or terminator.

Question 31Genetic Basis of Inheritance and Molecular Basis of Inheritance

In a cross between a heterozygous tall pea plant (Tt) and a homozygous dwarf pea plant (tt), what proportion of the offspring is expected to be dwarf?

  1. 1/4
  2. All
  3. 3/4
  4. 1/2 Correct answer.

Explanation

The cross Tt × tt produces gametes T and t from the heterozygote and only t from the dwarf parent. The possible offspring genotypes are Tt and tt in a 1:1 ratio, so half the progeny receive two recessive alleles and show the dwarf phenotype. This is a test cross, and the 1:1 phenotypic ratio is characteristic of a heterozygote crossed with a homozygous recessive.

Question 32Genetic Basis of Inheritance and Molecular Basis of Inheritance

In the Meselson–Stahl experiment, E. coli cells were grown in 15N medium for many generations and then transferred to 14N medium. After one round of DNA replication, what banding pattern was observed in the CsCl density gradient?

  1. Two bands, one heavy and one light.
  2. A single band of intermediate density. Correct answer.
  3. Two bands, one intermediate and one light.
  4. A single band of heavy density.

Explanation

After growth in 15N, both DNA strands are heavy. Upon transfer to 14N and one round of replication, each daughter molecule contains one 15N strand and one newly synthesised 14N strand, giving hybrid DNA of intermediate density. This result ruled out conservative replication and supported semiconservative replication, because a single intermediate band was seen after the first generation.

Question 33Genetic Engineering and Genomics

Which enzyme is used to join two DNA fragments by forming phosphodiester bonds between them during the construction of a recombinant DNA molecule?

  1. DNA polymerase III
  2. Reverse transcriptase
  3. DNA ligase Correct answer.
  4. Restriction endonuclease

Explanation

DNA ligase catalyses the formation of phosphodiester bonds between adjacent nucleotides, sealing the sugar-phosphate backbone of a vector and insert. This sealing step is essential after a restriction enzyme has cut both DNA molecules and they have been annealed.

Question 34Genetic Engineering and Genomics

A plasmid vector used in bacterial transformation carries an ampicillin-resistance gene. What is the main purpose of this gene in the selection of transformed cells?

  1. It acts as a promoter that drives transcription of the inserted foreign gene
  2. It codes for the restriction enzyme that cuts the plasmid before insertion of the gene
  3. It provides the origin of replication needed for the plasmid to multiply inside the bacterium
  4. It allows transformed bacteria to grow on medium containing ampicillin while non-transformed cells die Correct answer.

Explanation

A selectable marker such as an ampicillin-resistance gene lets transformed cells survive on antibiotic-containing medium, whereas cells that failed to take up the plasmid are killed. This allows only transformants to form colonies that can be picked and cultured.

Question 35Genetic Engineering and Genomics

Which feature makes a bacterial artificial chromosome (BAC) more suitable than a standard plasmid for cloning very large DNA inserts?

  1. It can stably carry inserts of several hundred kilobase pairs Correct answer.
  2. It carries its own ribosome-binding sites that increase protein yield
  3. It is maintained in bacteria with high copy number, giving a stronger signal on gels
  4. It replicates independently of the host chromosome in eukaryotic nuclei

Explanation

BACs are derived from the F plasmid and can stably maintain inserts of roughly 100 to 300 kilobase pairs, far larger than the few kilobases a typical plasmid accepts. This capacity makes them useful for genomic libraries and physical mapping.

Question 36Genetic Engineering and Genomics

In Sanger's dideoxy chain-termination sequencing, why does incorporation of a dideoxynucleotide stop further chain elongation?

  1. It pairs with two different template bases, distorting the double helix
  2. It carries an extra phosphate that blocks the polymerase active site permanently
  3. It lacks a 3'-hydroxyl group, so the next phosphodiester bond cannot form Correct answer.
  4. It is removed from the reaction by the proofreading activity of the polymerase

Explanation

A dideoxynucleotide has no 3'-hydroxyl group, so DNA polymerase cannot add the next nucleotide to it. Each incorporation therefore terminates a growing chain and generates a fragment whose length identifies the base at that position.

Question 37Genetic Engineering and Genomics

Bt cotton is produced by introducing genes from Bacillus thuringiensis into the cotton genome. What is the direct effect of the expressed Bt protein on the targeted insect larvae?

  1. It disrupts the insect midgut epithelium after ingestion, leading to death Correct answer.
  2. It prevents the insect from locating the plant by masking its volatile odours
  3. It inhibits the germination of insect eggs laid on the leaf surface
  4. It competitively inhibits digestive enzymes without damaging any tissue

Explanation

The Bt gene encodes a crystal protein that becomes toxic after ingestion and activation in the alkaline insect midgut, where it binds receptors and creates pores in the epithelial cells. The damaged gut stops feeding and the larva dies.

Question 38Genetic Engineering and Genomics

A researcher amplifies a gene by PCR using a primer pair, but the reaction produces no product. Which change is most likely to correct the failure if the primers were designed with mismatched sequences?

  1. Redesign the primers so they are complementary to the flanking regions Correct answer.
  2. Add more dNTPs to the reaction mixture
  3. Lengthen the final extension step to ten minutes
  4. Increase the annealing temperature in the thermal cycle

Explanation

Primers must be complementary to the sequences flanking the target region so that they anneal and allow polymerase extension. If the primer sequences are mismatched, no specific product can form, and redesigning them is the direct remedy.

Question 39Genetic Engineering and Genomics

A gene of interest is inserted into a plasmid at the site within the tetracycline-resistance gene, while the plasmid retains an intact ampicillin-resistance gene. After transformation, which procedure identifies bacteria carrying the recombinant plasmid?

  1. Grow on medium lacking both antibiotics and select the largest colonies
  2. Grow on ampicillin medium and select colonies that also grow on tetracycline
  3. Grow on ampicillin medium, then replica-plate onto tetracycline medium and select colonies that fail to grow there Correct answer.
  4. Grow on tetracycline medium and select colonies that grow, then test them for ampicillin sensitivity

Explanation

Insertional inactivation destroys the tetracycline-resistance gene, so recombinants are ampicillin resistant but tetracycline sensitive. Selection on ampicillin followed by replica plating onto tetracycline reveals recombinants as colonies absent from the tetracycline plate.

Question 40Genetic Engineering and Genomics

Reverse transcriptase is used to prepare a cDNA library from eukaryotic cells. Which property of the resulting library makes it useful for expressing a protein in bacteria?

  1. It contains introns that bacteria splice efficiently
  2. It contains the entire genome including promoter regions
  3. It contains only mitochondrial DNA sequences
  4. It lacks introns because it is copied from mature mRNA Correct answer.

Explanation

cDNA is synthesised from processed mRNA, so introns have already been removed. Bacteria cannot splice eukaryotic introns, so a cDNA copy allows correct expression of the protein in a bacterial host.

Question 41Genetic Engineering and Genomics

In agarose gel electrophoresis of DNA fragments, why do smaller fragments migrate farther than larger fragments toward the anode?

  1. Smaller fragments are attracted more strongly by the positive electrode
  2. The gel matrix retards larger molecules more, while charge-to-mass ratio is similar Correct answer.
  3. Larger fragments bind ethidium bromide more tightly and become immobile
  4. Smaller fragments carry a greater negative charge per molecule

Explanation

DNA has a nearly uniform charge-to-mass ratio because each nucleotide contributes one phosphate, so all fragments experience a similar driving force. The agarose mesh impedes larger molecules more, so smaller fragments travel farther through the gel.

Question 42Genetic Engineering and Genomics

A restriction enzyme recognises the palindromic sequence GAATTC and cuts between G and A on both strands. What type of ends does this produce?

  1. Covalently sealed ends that cannot be ligated
  2. 5' overhanging sticky ends with the sequence AATT Correct answer.
  3. Blunt ends with no overhanging nucleotides
  4. 3' overhanging sticky ends with the sequence TTAA

Explanation

Staggered cuts on the two strands of a palindromic site leave short single-stranded projections. Cutting between G and A on both strands of GAATTC leaves an AATT projection with a free 5' phosphate, which can base-pair with a compatible end.

Question 43Genetic Engineering and Genomics

In the human insulin production workflow, why is the synthetic A and B chain DNA inserted separately into plasmids in bacteria rather than as one continuous gene?

  1. Bacteria cannot translate any mRNA longer than one hundred codons
  2. The two chains are produced separately and later joined by disulfide bonds to form active insulin Correct answer.
  3. A single plasmid cannot carry two antibiotic-resistance genes
  4. The two chains must be transcribed by different RNA polymerases

Explanation

Proinsulin processing in human cells removes the connecting C peptide, so the mature hormone consists of separate A and B chains linked by disulfide bridges. Producing each chain in bacteria and combining them chemically yields functional insulin.

Question 44Genetic Engineering and Genomics

A forensic sample contains a very small amount of DNA. Which combination of techniques allows comparison of its short tandem repeat profile with a reference sample?

  1. Southern blotting of genomic DNA followed by autoradiography without amplification
  2. Conversion of the sample to cDNA followed by library screening
  3. PCR amplification of STR loci followed by capillary or gel electrophoresis Correct answer.
  4. Restriction digestion followed by ligation into a plasmid vector

Explanation

PCR amplifies the STR loci from minute quantities of DNA, and electrophoresis then resolves the products so that allele sizes can be compared. The repeat number varies between individuals, giving a profile suitable for identification.

Question 45Genetic Engineering and Genomics

A patient with a suspected inherited disorder is screened using a DNA microarray. What does a strong fluorescent signal at a particular probe spot indicate?

  1. The target sequence is absent from the patient's genome
  2. The DNA sample was degraded and needs to be re-extracted
  3. The probe was not synthesised correctly and should be discarded
  4. The patient's labelled DNA hybridised to the probe, showing that the complementary sequence is present Correct answer.

Explanation

In a microarray, labelled target DNA hybridises to immobilised probes whose sequences are known. A strong signal at a spot means the complementary sequence was present in the sample, allowing many sequences to be screened simultaneously.

Question 46Genetic Engineering and Genomics

The human genome contains roughly 20,000 protein-coding genes, yet the proteome is far more diverse. Which explanation best accounts for this discrepancy?

  1. Non-coding DNA is translated into additional proteins in the cytoplasm
  2. Each gene is translated repeatedly to produce chemically distinct proteins
  3. Alternative splicing and post-translational modifications generate multiple protein forms from one gene Correct answer.
  4. The number of chromosomes varies between different tissues of one person

Explanation

A single gene can yield several mRNA variants through alternative splicing, and the resulting polypeptides can be further modified after translation. Together these processes greatly expand the variety of proteins beyond the gene count.

Question 47Genetic Engineering and Genomics

A gene-editing experiment uses CRISPR-Cas9 with a guide RNA complementary to a target locus. Which outcome is expected if the guide RNA is designed to match a sequence that occurs at several sites in the genome?

  1. The Cas9 enzyme will lose its nuclease activity and only bind DNA
  2. Off-target cleavage may occur at the additional matching sites, reducing precision Correct answer.
  3. Editing efficiency at the intended locus will increase because more sites are available
  4. The guide RNA will be degraded before it can associate with Cas9

Explanation

Specificity of CRISPR-Cas9 depends on base pairing between the guide RNA and the target, so a guide matching several genomic sites can direct Cas9 to cleave at unintended loci. Such off-target cleavage is a major concern in editing experiments.

Question 48Genetic Engineering and Genomics

A scientist wants to determine whether a cloned DNA fragment is actually transcribed in a transformed cell line. Which approach provides the most direct evidence of transcription?

  1. Detect the specific mRNA in the cells by RT-PCR or northern hybridisation Correct answer.
  2. Sequence the cloned fragment to verify that the insert is intact
  3. Measure the antibiotic resistance of the transformed cells
  4. Confirm the presence of the plasmid by agarose gel electrophoresis of undigested DNA

Explanation

Transcription produces mRNA, so detecting the specific transcript by RT-PCR or northern hybridisation directly demonstrates that the cloned gene is being expressed. Presence of the plasmid or an intact sequence shows only that the DNA is there.

Question 49Cell Cycle, Cell Division and Reproduction in Organisms

During which phase of the cell cycle does DNA replication occur in a typical eukaryotic cell?

  1. Prophase of mitosis
  2. G1 phase of interphase
  3. S phase of interphase Correct answer.
  4. G2 phase of interphase

Explanation

Interphase is divided into G1, S and G2. DNA synthesis happens only in S phase, where each chromosome is duplicated into two sister chromatids. G1 and G2 are growth phases, and mitosis is the separation phase, so replication cannot occur there.

Question 50Cell Cycle, Cell Division and Reproduction in Organisms

In which stage of mitosis do chromosomes align at the equatorial plate with their centromeres on the metaphase plate?

  1. Metaphase Correct answer.
  2. Telophase
  3. Anaphase
  4. Prophase

Explanation

Metaphase is defined by the arrangement of all chromosomes at the cell equator, with kinetochores attached to spindle fibres from opposite poles. This alignment is checked by the spindle assembly checkpoint before separation begins. Anaphase and telophase follow this alignment, and prophase precedes it.