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Error Notes β€” Botany Double Paper

Aamirah Fathima  Β·  Molecular Basis of Inheritance (80Q) and Biotechnology and its Applications (80Q)  Β·  marked +4 / βˆ’1 / 0  Β·  13 October 2026

597out of 640
93.3%percentage
151correct
7wrong
2left blank
95.6%accuracy
PaperCorrectWrongBlankMarks%
Molecular Basis of Inheritance7352287/32089.7%
Biotechnology and its Applications7820310/32096.9%
Both papers15172597/64093.3%
Read these two scores against Sunday's analysis, not on their own. 310 and 287 out of 320 are excellent numbers on papers that were shown two days ago to be pitched below her real level. Three pieces of evidence from this very attempt confirm it, and they matter more than the totals: The marks below are real and the chapters are genuinely well learnt. But treat 97% and 90% as the ceiling of what these papers can measure, not as her NEET standing. The Thermal Properties paper sent yesterday is the first one built to the corrected specification; that score will be the informative one.

Molecular Basis of Inheritance

Transcription Β· The Genetic Code Β· Translation  Β·  287/320 (89.7%)  Β·  73 correct, 5 wrong, 2 blank

SectionQβœ“βœ—β€”Marks
Transcription28243193/112
The Genetic Code24231091/96
Translation282611103/112

The transcription unit and its strands

Q4Moderatewrong  βˆ’1
A template strand reads 3'-ATGCATGCATGC-5'. The mRNA transcribed from it is
  1. (A) 3'-UACGUACGUACG-5'
  2. (B) 5'-AUGCAUGCAUGC-3'  βœ— her answer
  3. (C) 5'-UACGUACGUACG-3'  βœ“
  4. (D) 5'-TACGTACGTACG-3'
Why C: Pair each base of the template and reverse the polarity: 3'-ATGCATGCATGC-5' gives 5'-UACGUACGUACG-3'.
Q5Moderateleft blank  0
The coding strand of a gene reads 5'-ATGGGCTAA-3'. The corresponding mRNA is
  1. (A) 5'-UACCCGAUU-3'
  2. (B) 3'-AUGGGCUAA-5'
  3. (C) 5'-UUAGCCCAU-3'
  4. (D) 5'-AUGGGCUAA-3'  βœ“
Why D: The transcript matches the coding strand with U for T, so 5'-ATGGGCTAA-3' gives 5'-AUGGGCUAA-3'.
Q6Moderatewrong  βˆ’1
With reference to the coding strand, the promoter of a transcription unit is situated
  1. (A) upstream, towards the 5' end  βœ“
  2. (B) within the structural gene
  3. (C) immediately after the terminator
  4. (D) downstream, towards the 3' end  βœ— her answer
Why A: The promoter is upstream, towards the 5' end of the coding strand.

Processing of hnRNA and split genes

Q24Moderatewrong  βˆ’1
Tailing of hnRNA involves the addition of
  1. (A) about 200 to 300 adenylate residues at the 5' end, using the DNA as template  βœ— her answer
  2. (B) about 20 guanylate residues at the 5' end, using the DNA as template
  3. (C) a single methylated base at the 3' end
  4. (D) about 200 to 300 adenylate residues at the 3' end, independently of any template  βœ“
Why D: Tailing adds 200–300 adenylate residues at the 3' end, and no template is used.

Features of the genetic code

Q32Moderatewrong  βˆ’1
The statement that the code is read continuously, with no base being skipped and no base being shared between two codons, describes it as
  1. (A) ambiguous and overlapping
  2. (B) comma-less and non-overlapping  βœ“
  3. (C) degenerate and universal
  4. (D) unambiguous and degenerate  βœ— her answer
Why B: Comma-less means no base is skipped; non-overlapping means no base is shared between two codons.

The ribosome and its sites

Q66Easywrong  βˆ’1
A ribosome that is not engaged in translation exists as
  1. (A) a single compact particle
  2. (B) a complex with the nuclear membrane  βœ— her answer
  3. (C) two separate subunits  βœ“
  4. (D) a ring of several particles
Why C: An inactive ribosome exists as two separate subunits; they come together on the mRNA.

Initiation, elongation and termination

Q80Hardleft blank  0
A bacterial mRNA is 1500 nucleotides long, of which 90 nucleotides make up the two untranslated regions and the stop codon together. The largest polypeptide it can specify contains about
  1. (A) 500 amino acids
  2. (B) 1410 amino acids
  3. (C) 300 amino acids
  4. (D) 470 amino acids  βœ“
Why D: 1500 − 90 = 1410 coding nucleotides; 1410 ÷ 3 = 470 codons, so about 470 amino acids.

Biotechnology and its Applications

Agriculture Β· Medicine Β· Transgenic Animals and Ethics  Β·  310/320 (96.9%)  Β·  78 correct, 2 wrong, 0 blank

SectionQβœ“βœ—β€”Marks
Applications in Agriculture26251099/104
Applications in Medicine282710107/112
Transgenic Animals and Ethical Issues262600104/104

Bt crops and the cry genes

Q18Hardwrong  βˆ’1
How many of the following statements are correct?
A. Bt cotton carries a gene taken from a bacterium.
B. The Bt toxin is sprayed on the crop as a chemical insecticide.
C. Bt varieties of corn, rice, tomato, potato and soyabean have been produced.
D. The activated toxin acts on the midgut of the insect.
  1. (A) Only one
  2. (B) Only two  βœ— her answer
  3. (C) Only three  βœ“
  4. (D) All four
Why C: A, C and D are correct. The whole point of a Bt crop is that the toxin is made by the plant and not sprayed on it.

Gene therapy

Q41Moderatewrong  βˆ’1
celllymphocytesfrom patientfunctional ADAcDNA in a retroviral vectorgrown in culture,returned to the patientlymphocytes are not immortal β†’ the infusion must be repeateda permanent cure needs the gene put into bone-marrow cellsat an early embryonic stageGene therapy for ADA deficiencyIn the procedure shown, the cells taken from the patient, grown in culture and returned after receiving the functional gene are
  1. (A) bone-marrow stem cells  βœ— her answer
  2. (B) epithelial cells of the gut
  3. (C) lymphocytes  βœ“
  4. (D) red blood cells
Why C: Lymphocytes from the patient's blood are grown in culture, given the gene and returned.

The complete papers, question by question

All 160 questions are reproduced below in the order she sat them, with the option she ticked and the correct one marked on each, and the working underneath. The seven she got wrong carry a red edge and the two she left blank a purple one.

Key:correctwrongleft blank

Molecular Basis of Inheritance

TranscriptionQ1–Q28  Β·  24 of 28 correct  Β·  93/112
Q1Easycorrect  +4
5'3'3'5'coding strandtemplate strandpromoterterminatorstructural genedirection of transcriptionA transcription unitIn the transcription unit shown, the region that defines the beginning of transcription is
  1. (A) the terminator
  2. (B) the coding strand
  3. (C) the promoter  βœ“ her answer
  4. (D) the structural gene
Why C: The promoter lies at the 5' end of the coding strand and is where RNA polymerase binds; it defines the start of transcription.
Q2Easycorrect  +4
5'3'3'5'coding strandtemplate strandpromoterterminatorstructural genedirection of transcriptionA transcription unitIn the same figure, the strand that is actually copied by RNA polymerase is the one with polarity
  1. (A) 5'β†’3' called the coding strand
  2. (B) 3'β†’5' called the template strand  βœ“ her answer
  3. (C) 5'β†’3' called the template strand
  4. (D) 3'β†’5' called the coding strand
Why B: Only the 3'→5' strand serves as template, because RNA can be built only 5'→3'.
Q3Easycorrect  +4
The strand of a transcription unit that has the same base sequence as the RNA transcript, except that thymine replaces uracil, is called the
  1. (A) lagging strand
  2. (B) coding strand  βœ“ her answer
  3. (C) antisense strand
  4. (D) template strand
Why B: The coding strand is not copied, yet it has the same sequence as the transcript with T in place of U.
Q4Moderatewrong  βˆ’1
A template strand reads 3'-ATGCATGCATGC-5'. The mRNA transcribed from it is
  1. (A) 3'-UACGUACGUACG-5'
  2. (B) 5'-AUGCAUGCAUGC-3'  βœ— her answer
  3. (C) 5'-UACGUACGUACG-3'  βœ“
  4. (D) 5'-TACGTACGTACG-3'
Why C: Pair each base of the template and reverse the polarity: 3'-ATGCATGCATGC-5' gives 5'-UACGUACGUACG-3'.
Q5Moderateleft blank  0
The coding strand of a gene reads 5'-ATGGGCTAA-3'. The corresponding mRNA is
  1. (A) 5'-UACCCGAUU-3'
  2. (B) 3'-AUGGGCUAA-5'
  3. (C) 5'-UUAGCCCAU-3'
  4. (D) 5'-AUGGGCUAA-3'  βœ“
Why D: The transcript matches the coding strand with U for T, so 5'-ATGGGCTAA-3' gives 5'-AUGGGCUAA-3'.
Q6Moderatewrong  βˆ’1
With reference to the coding strand, the promoter of a transcription unit is situated
  1. (A) upstream, towards the 5' end  βœ“
  2. (B) within the structural gene
  3. (C) immediately after the terminator
  4. (D) downstream, towards the 3' end  βœ— her answer
Why A: The promoter is upstream, towards the 5' end of the coding strand.
Q7Easycorrect  +4
The terminator of a transcription unit
  1. (A) is located towards the 3' end of the coding strand and defines the end of transcription  βœ“ her answer
  2. (B) is the site where RNA polymerase first binds
  3. (C) is transcribed into the 5' cap of the mRNA
  4. (D) is located towards the 5' end of the coding strand
Why A: The terminator lies towards the 3' end of the coding strand and defines where transcription stops.
Q8Easycorrect  +4
The number of types of DNA-dependent RNA polymerase present in a bacterium is
  1. (A) three, as in eukaryotes
  2. (B) one, which makes all three kinds of RNA  βœ“ her answer
  3. (C) four, one for each kind of RNA
  4. (D) two, one for mRNA and one for rRNA
Why B: A bacterium has a single DNA-dependent RNA polymerase that makes mRNA, tRNA and rRNA alike.
Q9Easycorrect  +4
The sigma (Οƒ) factor of bacterial RNA polymerase is required for
  1. (A) termination of transcription
  2. (B) initiation of transcription  βœ“ her answer
  3. (C) elongation of the RNA chain
  4. (D) removal of introns
Why B: σ is the initiation factor; it is released once the chain starts to grow.
Q10Easycorrect  +4
The rho (ρ) factor acts as a
  1. (A) termination factor  βœ“ her answer
  2. (B) elongation factor
  3. (C) splicing factor
  4. (D) initiation factor
Why A: ρ is the termination factor.
Q11Moderatecorrect  +4
Bacterial RNA polymerase by itself is capable of catalysing only
  1. (A) capping
  2. (B) initiation
  3. (C) elongation  βœ“ her answer
  4. (D) termination
Why C: The core enzyme catalyses elongation only; it needs σ to initiate and ρ to terminate.
Q12Moderatecorrect  +4
In a bacterium, transcription and translation can occur simultaneously because
  1. (A) bacterial mRNA needs no processing
  2. (B) the ribosomes enter the nucleus
  3. (C) the mRNA is polycistronic
  4. (D) there is no nuclear membrane separating the two processes  βœ“ her answer
Why D: A bacterium has no nuclear membrane, so ribosomes can begin translating an mRNA while it is still being transcribed.
Q13Moderatecorrect  +4
During transcription, RNA polymerase moves along the template strand in the direction
  1. (A) 5'β†’3', synthesising RNA 3'β†’5'
  2. (B) 3'β†’5', synthesising RNA 5'β†’3'  βœ“ her answer
  3. (C) 5'β†’3', synthesising RNA 5'β†’3'
  4. (D) 3'β†’5', synthesising RNA 3'β†’5'
Why B: The enzyme reads the template 3'→5' and therefore builds the RNA 5'→3'.
Q14Moderatecorrect  +4
RNA polymerasenascent RNA3'5'5'3'templateTranscription in a bacteriumIn the figure, the RNA chain being formed
  1. (A) grows at its 3' end, its 5' end having been made first  βœ“ her answer
  2. (B) grows from its 5' end, which stays attached to the enzyme
  3. (C) is a copy of the coding strand base for base, including thymine
  4. (D) is synthesised in both directions from the middle
Why A: Nucleotides are added to the free 3'-OH end, so the 5' end of the transcript is the oldest part.
Q15Hardcorrect  +4
Assertion (A): In bacteria a single RNA polymerase transcribes all classes of RNA.
Reason (R): The enzyme associates with different factors at the start and at the end of transcription.
  1. (A) Both A and R are true and R is the correct explanation of A
  2. (B) Both A and R are true but R is NOT the correct explanation of A  βœ“ her answer
  3. (C) A is true but R is false
  4. (D) A is false but R is true
Why B: Both statements are true, but the association with σ and ρ explains how one enzyme manages all three steps, not why it makes all classes of RNA.
Q16Moderatecorrect  +4
How many of the following statements are correct?
A. Transcription needs a promoter.
B. RNA polymerase binds to the terminator to begin transcription.
C. Only one of the two DNA strands of a gene is transcribed.
D. The DNA duplex unwinds only over a short stretch at a time.
  1. (A) Only one
  2. (B) Only two
  3. (C) Only three  βœ“ her answer
  4. (D) All four
Why C: A, C and D are correct. RNA polymerase binds the promoter, not the terminator.
Q17Easycorrect  +4
In a eukaryotic nucleus, the 28S, 18S and 5.8S rRNAs are transcribed by
  1. (A) RNA polymerase II
  2. (B) RNA polymerase I  βœ“ her answer
  3. (C) the same enzyme that makes tRNA
  4. (D) RNA polymerase III
Why B: RNA polymerase I transcribes the 28S, 18S and 5.8S rRNAs.
Q18Easycorrect  +4
The precursor of mRNA in a eukaryote is transcribed by RNA polymerase II and is called
  1. (A) snRNA
  2. (B) hnRNA  βœ“ her answer
  3. (C) tRNA
  4. (D) 5S rRNA
Why B: RNA polymerase II transcribes the precursor of mRNA, heterogeneous nuclear RNA.
Q19Moderatecorrect  +4
tRNA, 5S rRNA and snRNAs are transcribed by
  1. (A) RNA polymerase I
  2. (B) mitochondrial RNA polymerase
  3. (C) RNA polymerase II
  4. (D) RNA polymerase III  βœ“ her answer
Why D: RNA polymerase III transcribes tRNA, 5S rRNA and the snRNAs.
Q20Moderatecorrect  +4
Match Column I with Column II.
Column IColumn II
ARNA polymerase IIhnRNA
BRNA polymerase IIIItermination in bacteria
CRNA polymerase IIIIIIrRNAs (28S, 18S, 5.8S)
Dρ factorIVtRNA and snRNAs
  1. (A) A-IV, B-I, C-III, D-II
  2. (B) A-III, B-I, C-IV, D-II  βœ“ her answer
  3. (C) A-I, B-III, C-IV, D-II
  4. (D) A-III, B-IV, C-I, D-II
Why B: Pol I → rRNAs, Pol II → hnRNA, Pol III → tRNA and snRNAs, ρ → termination in bacteria.
Q21Hardcorrect  +4
Assertion (A): A eukaryotic cell needs three different nuclear RNA polymerases.
Reason (R): Each of the three enzymes transcribes a different set of genes in the nucleus.
  1. (A) Both A and R are true and R is the correct explanation of A  βœ“ her answer
  2. (B) Both A and R are true but R is NOT the correct explanation of A
  3. (C) A is true but R is false
  4. (D) A is false but R is true
Why A: The division of labour among three enzymes, each for its own set of genes, is exactly the reason three are needed.
Q22Easycorrect  +4
ccaptail123ABhnRNA (exons 1-3, introns A-B)splicingc123mature mRNAProcessing of hnRNA into mRNAIn the figure, the step that removes the regions marked A and B is called
  1. (A) splicing  βœ“ her answer
  2. (B) capping
  3. (C) termination
  4. (D) tailing
Why A: Removal of the intervening sequences and joining of the exons is splicing.
Q23Easycorrect  +4
In capping, the unusual nucleotide added to the 5' end of hnRNA is
  1. (A) thymidine triphosphate
  2. (B) methyl guanosine triphosphate  βœ“ her answer
  3. (C) adenylate
  4. (D) uridine monophosphate
Why B: Capping adds methyl guanosine triphosphate, an unusual nucleotide, to the 5' end.
Q24Moderatewrong  βˆ’1
Tailing of hnRNA involves the addition of
  1. (A) about 200 to 300 adenylate residues at the 5' end, using the DNA as template  βœ— her answer
  2. (B) about 20 guanylate residues at the 5' end, using the DNA as template
  3. (C) a single methylated base at the 3' end
  4. (D) about 200 to 300 adenylate residues at the 3' end, independently of any template  βœ“
Why D: Tailing adds 200–300 adenylate residues at the 3' end, and no template is used.
Q25Moderatecorrect  +4
E1I1E2I2E3I3E4E = exon (coding, retained) I = intron (intervening, removed)A split gene of a eukaryoteA eukaryotic gene is arranged as shown. The number of intervening sequences that must be removed from its primary transcript is
  1. (A) three  βœ“ her answer
  2. (B) four
  3. (C) two
  4. (D) seven
Why A: Four exons are separated by three introns, so three intervening sequences must be removed.
Q26Moderatecorrect  +4
A monocistronic eukaryotic structural gene is described as a split gene because
  1. (A) each of its two strands codes for a different protein
  2. (B) it is split between the nucleus and the cytoplasm
  3. (C) its coding sequences are interrupted by non-coding intervening sequences  βœ“ her answer
  4. (D) it is transcribed by two different polymerases
Why C: The coding exons of a eukaryotic structural gene are interrupted by non-coding introns.
Q27Hardcorrect  +4
Assertion (A): hnRNA cannot be translated as it is.
Reason (R): hnRNA contains intervening sequences and lacks the cap and the tail needed for efficient translation.
  1. (A) Both A and R are true and R is the correct explanation of A  βœ“ her answer
  2. (B) Both A and R are true but R is NOT the correct explanation of A
  3. (C) A is true but R is false
  4. (D) A is false but R is true
Why A: hnRNA still carries its introns and lacks the cap and tail, which is precisely why it cannot be translated as it is.
Q28Hardcorrect  +4
Which of the following statements are correct?
A. Introns are removed and exons are joined in a defined order.
B. Fully processed hnRNA is called mRNA and is exported from the nucleus.
C. Bacterial mRNA is extensively spliced before translation.
D. Capping adds a nucleotide to the 3' end of the transcript.
  1. (A) 1, 2 and 4 only
  2. (B) 2 and 3 only
  3. (C) 1, 3 and 4 only
  4. (D) 1 and 2 only  βœ“ her answer
Why D: Statements 1 and 2 are right. Bacterial mRNA is not spliced, and capping is at the 5' end, not the 3'.
The Genetic CodeQ29–Q52  Β·  23 of 24 correct  Β·  91/96
Q29Easycorrect  +4
Of the 64 codons of the genetic code, the number that actually specify amino acids is
  1. (A) 60
  2. (B) 63
  3. (C) 64
  4. (D) 61  βœ“ her answer
Why D: Three of the 64 codons are stop codons, leaving 61 that specify amino acids.
Q30Easycorrect  +4
The genetic code is said to be degenerate because
  1. (A) some codons specify no amino acid at all
  2. (B) some amino acids are specified by more than one codon  βœ“ her answer
  3. (C) one codon can specify more than one amino acid
  4. (D) the code differs slightly between organisms
Why B: Degeneracy means several codons may stand for the same amino acid.
Q31Easycorrect  +4
That a given codon always specifies one and only one amino acid makes the code
  1. (A) overlapping
  2. (B) comma-less
  3. (C) unambiguous  βœ“ her answer
  4. (D) degenerate
Why C: Unambiguous means one codon never stands for two different amino acids.
Q32Moderatewrong  βˆ’1
The statement that the code is read continuously, with no base being skipped and no base being shared between two codons, describes it as
  1. (A) ambiguous and overlapping
  2. (B) comma-less and non-overlapping  βœ“
  3. (C) degenerate and universal
  4. (D) unambiguous and degenerate  βœ— her answer
Why B: Comma-less means no base is skipped; non-overlapping means no base is shared between two codons.
Q33Moderatecorrect  +4
The genetic code is described as nearly universal rather than universal because
  1. (A) the stop codons differ in every kingdom
  2. (B) it does not apply to viruses
  3. (C) plants and animals use entirely different codons
  4. (D) a few exceptions occur, for instance in mitochondrial codons and in some protozoans  βœ“ her answer
Why D: A few exceptions exist, notably among mitochondrial codons and in some protozoans.
Q34Moderatecorrect  +4
A code of two bases per amino acid was rejected on the simple ground that it would give only
  1. (A) 8 combinations, too few for 20 amino acids
  2. (B) 16 combinations, too few for 20 amino acids  βœ“ her answer
  3. (C) 32 combinations, too many for 20 amino acids
  4. (D) 64 combinations, too many for 20 amino acids
Why B: 42 = 16, which is fewer than the 20 amino acids that must be specified; 43 = 64 is enough.
Q35Hardcorrect  +4
Assertion (A): The genetic code is degenerate but it is not ambiguous.
Reason (R): Several codons may stand for the same amino acid, yet no single codon stands for two different amino acids.
  1. (A) Both A and R are true and R is the correct explanation of A  βœ“ her answer
  2. (B) Both A and R are true but R is NOT the correct explanation of A
  3. (C) A is true but R is false
  4. (D) A is false but R is true
Why A: Degeneracy is about several codons per amino acid; ambiguity would be several amino acids per codon. The second never happens, so R explains A exactly.
Q36Moderatecorrect  +4
How many of the following statements are correct?
A. The code is a triplet code.
B. Some amino acids are coded by more than one codon.
C. A codon may be read in more than one reading frame at the same time.
D. UUU codes for phenylalanine in bacteria as well as in humans.
  1. (A) Only one
  2. (B) Only two
  3. (C) Only three  βœ“ her answer
  4. (D) All four
Why C: A, B and D are correct. The code is non-overlapping, so a base is never read in two frames at once.
Q37Easycorrect  +4
The three codons that do not code for any amino acid and act as chain terminators are
  1. (A) UAA, UAG and AUG
  2. (B) UGA, UGG and UAG
  3. (C) AUG, UGG and UAA
  4. (D) UAA, UAG and UGA  βœ“ her answer
Why D: UAA, UAG and UGA are the chain-terminating or stop codons.
Q38Easycorrect  +4
AUG is unusual among codons because it
  1. (A) acts as the initiator codon and also codes for methionine  βœ“ her answer
  2. (B) terminates as well as initiates translation
  3. (C) codes for two different amino acids
  4. (D) is read in two different reading frames
Why A: AUG serves both as the initiator codon and as the codon for methionine.
Q39Moderatecorrect  +4
The two amino acids that are each specified by one codon only are
  1. (A) leucine and serine
  2. (B) phenylalanine and valine
  3. (C) methionine and tryptophan  βœ“ her answer
  4. (D) glutamic acid and valine
Why C: Methionine (AUG) and tryptophan (UGG) each have a single codon.
Q40Moderatecorrect  +4
Tryptophan is coded by
  1. (A) AUG
  2. (B) UAG
  3. (C) UGG  βœ“ her answer
  4. (D) UGA
Why C: UGG is the only codon for tryptophan. UGA, which differs by one base, is a stop codon.
Q41Moderatecorrect  +4
There is no tRNA whose anticodon can pair with UAA, UAG or UGA. The consequence is that
  1. (A) a random amino acid is inserted at that point
  2. (B) translation stops, the polypeptide being released with the help of a release factor  βœ“ her answer
  3. (C) the mRNA is at once degraded by the ribosome
  4. (D) these codons are skipped by the ribosome
Why B: With no tRNA to read them, the stop codons are recognised by a release factor, which ends translation.
Q42Moderatecorrect  +4
Match Column I with Column II.
Column IColumn II
AAUGIphenylalanine
BUGGIIstop
CUAGIIImethionine
DUUUIVtryptophan
  1. (A) A-III, B-II, C-IV, D-I
  2. (B) A-I, B-IV, C-II, D-III
  3. (C) A-III, B-IV, C-II, D-I  βœ“ her answer
  4. (D) A-II, B-IV, C-III, D-I
Why C: AUG → methionine, UGG → tryptophan, UAG → stop, UUU → phenylalanine.
Q43Easycorrect  +4
The suggestion that the genetic code should be a combination of three nucleotides came from
  1. (A) Marshall Nirenberg
  2. (B) Francis Crick
  3. (C) Har Gobind Khorana
  4. (D) George Gamow  βœ“ her answer
Why D: George Gamow argued on mathematical grounds that the code had to be a triplet.
Q44Moderatecorrect  +4
The cell-free system that first allowed a codon to be assigned to an amino acid was developed by
  1. (A) Khorana and Gamow
  2. (B) Beadle and Tatum
  3. (C) Severo Ochoa
  4. (D) Nirenberg and Matthaei  βœ“ her answer
Why D: Nirenberg and Matthaei developed the cell-free system that cracked the first codon.
Q45Moderatecorrect  +4
The enzyme polynucleotide phosphorylase, used to make RNA of a known composition in the laboratory, was discovered by
  1. (A) Marshall Nirenberg
  2. (B) Har Gobind Khorana
  3. (C) Severo Ochoa  βœ“ her answer
  4. (D) Francois Jacob
Why C: Severo Ochoa's polynucleotide phosphorylase allowed RNA of known composition to be made.
Q46Moderatecorrect  +4
Match Column I with Column II.
Column IColumn II
AHar Gobind KhoranaIcell-free system for protein synthesis
BSevero OchoaIIthe code must be a triplet
CNirenberg and MatthaeiIIIchemical synthesis of RNA with defined base combinations
DGeorge GamowIVpolynucleotide phosphorylase
  1. (A) A-IV, B-III, C-I, D-II
  2. (B) A-III, B-IV, C-I, D-II  βœ“ her answer
  3. (C) A-I, B-IV, C-III, D-II
  4. (D) A-III, B-I, C-IV, D-II
Why B: Khorana → chemical synthesis of defined RNA; Ochoa → polynucleotide phosphorylase; Nirenberg and Matthaei → cell-free system; Gamow → the triplet proposal.
Q47Easycorrect  +4
Sickle-cell anaemia arises from a single base change in which the codon
  1. (A) GAG becomes UAG, ending the chain early
  2. (B) GUG becomes GAG, putting glutamic acid in place of valine
  3. (C) GAA becomes GAG, with no change in the amino acid
  4. (D) GAG becomes GUG, putting valine in place of glutamic acid  βœ“ her answer
Why D: A single base change turns GAG into GUG, so valine replaces glutamic acid.
Q48Moderatecorrect  +4
The substitution responsible for sickle-cell anaemia occurs at the
  1. (A) hundred and forty-sixth position of the beta globin chain
  2. (B) sixth position of the beta globin chain  βœ“ her answer
  3. (C) sixth position of the alpha globin chain
  4. (D) first position of the beta globin chain
Why B: The substitution is at the sixth position of the beta globin chain of haemoglobin.
Q49Easycorrect  +4
normalAUGCCUGGAUAAMet - Pro - Gly - stopinsert one basemutantAUGACCUGGAUAevery codon after the insertion is misreadFrameshift caused by a single insertionThe kind of mutation illustrated is called
  1. (A) a point substitution
  2. (B) a silent mutation
  3. (C) a frameshift mutation  βœ“ her answer
  4. (D) a chromosomal deletion
Why C: One extra base shifts the reading frame, so every codon beyond the insertion is misread.
Q50Hardcorrect  +4
Insertion of exactly three bases into a coding sequence usually
  1. (A) adds one amino acid while leaving the rest of the reading frame unaltered  βœ“ her answer
  2. (B) always creates a stop codon at the site of insertion
  3. (C) shifts the reading frame from that point onwards
  4. (D) has no effect of any kind on the protein
Why A: Three bases make one codon, so inserting three adds a single amino acid and leaves the frame intact. This observation is what proved the code is a triplet.
Q51Hardcorrect  +4
Assertion (A): The insertion or deletion of a single base has a far greater effect on a protein than the substitution of one base.
Reason (R): An insertion or a deletion shifts the reading frame, so that every codon downstream of it is misread.
  1. (A) Both A and R are true and R is the correct explanation of A  βœ“ her answer
  2. (B) Both A and R are true but R is NOT the correct explanation of A
  3. (C) A is true but R is false
  4. (D) A is false but R is true
Why A: A substitution changes at most one amino acid; an insertion or deletion of one base misreads everything downstream, which is why its effect is far greater.
Q52Hardcorrect  +4
Which of the following statements are correct?
A. A single base substitution may change one amino acid in the protein.
B. The insertion of two bases shifts the reading frame.
C. Deletion of three consecutive bases restores the original reading frame beyond the site.
D. Every point mutation must produce a non-functional protein.
  1. (A) 1, 2 and 4 only
  2. (B) 1, 2 and 3 only  βœ“ her answer
  3. (C) 1 and 4 only
  4. (D) 2, 3 and 4 only
Why B: Statements 1, 2 and 3 are right. Some point mutations are silent or harmless, so 4 is too strong.
TranslationQ53–Q80  Β·  26 of 28 correct  Β·  103/112
Q53Easycorrect  +4
The existence of an adaptor molecule that could read the code on one side and carry an amino acid on the other was postulated by
  1. (A) James Watson
  2. (B) George Gamow
  3. (C) Marshall Nirenberg
  4. (D) Francis Crick  βœ“ her answer
Why D: Francis Crick postulated an adaptor that reads the codon at one end and holds the amino acid at the other.
Q54Easycorrect  +4
aaamino acid3'5'DHUTψCanticodonClover-leaf model of tRNAIn the clover-leaf model shown, the loop marked at the bottom of the molecule
  1. (A) holds the amino acid
  2. (B) is the site of peptide bond formation
  3. (C) carries the anticodon that reads the mRNA codon  βœ“ her answer
  4. (D) binds to the large subunit of the ribosome
Why C: The loop at the bottom of the clover leaf carries the anticodon.
Q55Moderatecorrect  +4
AUGUUCGGACAUACGUGGCCUUAA5'3'mRNAUACMetAAGPhetRNA = adaptoranticodon pairs withthe mRNA codontRNA reading codons on mRNAIn the figure, the anticodon of the tRNA that reads the initiator codon AUG is
  1. (A) UAC  βœ“ her answer
  2. (B) AUG
  3. (C) UUC
  4. (D) AAG
Why A: The anticodon complementary to AUG is UAC.
Q56Moderatecorrect  +4
The clover-leaf is only the secondary structure of tRNA. In three dimensions the molecule is
  1. (A) a closed circle
  2. (B) a double helix throughout
  3. (C) a compact, inverted L shape  βœ“ her answer
  4. (D) a straight rod
Why C: In three dimensions tRNA folds into a compact, inverted L; the clover leaf is only its secondary structure.
Q57Moderatecorrect  +4
Which of the following tRNAs does not exist in a cell?
  1. (A) The tRNA for methionine
  2. (B) The tRNA for tryptophan
  3. (C) A tRNA whose anticodon reads UAG  βœ“ her answer
  4. (D) The initiator tRNA
Why C: No tRNA exists for a stop codon, so no anticodon reads UAG.
Q58Easycorrect  +4
The end of a tRNA molecule to which the amino acid becomes attached is
  1. (A) the TψC loop
  2. (B) the DHU loop
  3. (C) the anticodon loop
  4. (D) the amino acid acceptor end  βœ“ her answer
Why D: The amino acid is attached at the amino acid acceptor end, the other end of the molecule from the anticodon.
Q59Hardcorrect  +4
Assertion (A): Each amino acid has at least one tRNA of its own.
Reason (R): The anticodon of a tRNA must be complementary to the codon that specifies its amino acid.
  1. (A) Both A and R are true and R is the correct explanation of A  βœ“ her answer
  2. (B) Both A and R are true but R is NOT the correct explanation of A
  3. (C) A is true but R is false
  4. (D) A is false but R is true
Why A: The anticodon must match the codon for that amino acid, so every amino acid needs a tRNA of its own.
Q60Moderatecorrect  +4
How many of the following statements are correct?
A. tRNA has an anticodon loop and an amino acid acceptor end.
B. There is a separate initiator tRNA for the start codon.
C. tRNAs exist for each of the three stop codons.
D. tRNA was called soluble RNA before its role was known.
  1. (A) Only one
  2. (B) Only two
  3. (C) Only three  βœ“ her answer
  4. (D) All four
Why C: A, B and D are correct. There are no tRNAs for the stop codons.
Q61Easycorrect  +4
The process in which an amino acid is activated and linked to its own tRNA is called
  1. (A) aminoacylation or charging of tRNA  βœ“ her answer
  2. (B) capping
  3. (C) translocation
  4. (D) splicing
Why A: Activation of the amino acid and its attachment to its own tRNA is aminoacylation, or charging.
Q62Moderatecorrect  +4
The activation of an amino acid before it is attached to its tRNA requires
  1. (A) NADPH
  2. (B) acetyl CoA
  3. (C) ATP  βœ“ her answer
  4. (D) GTP
Why C: Activation uses ATP.
Q63Moderatecorrect  +4
A peptide bond forms when
  1. (A) two charged tRNAs are brought close together on the ribosome  βœ“ her answer
  2. (B) two uncharged tRNAs lie side by side on the mRNA
  3. (C) the release factor binds to the stop codon
  4. (D) the mRNA is capped at its 5' end
Why A: A peptide bond forms only between two charged tRNAs held side by side on the ribosome.
Q64Hardcorrect  +4
Assertion (A): Amino acids are charged on to tRNA before translation begins, not afterwards.
Reason (R): Charging supplies the energy that makes the subsequent formation of the peptide bond possible.
  1. (A) Both A and R are true and R is the correct explanation of A  βœ“ her answer
  2. (B) Both A and R are true but R is NOT the correct explanation of A
  3. (C) A is true but R is false
  4. (D) A is false but R is true
Why A: The energy for the peptide bond comes from the earlier activation step, so charging must precede translation.
Q65Moderatecorrect  +4
The sequence of amino acids in a polypeptide is decided by
  1. (A) the number of ribosomes on the mRNA
  2. (B) the proteins of the large subunit
  3. (C) the order of tRNAs in the cytoplasm
  4. (D) the base sequence of the mRNA  βœ“ her answer
Why D: The order of codons in the mRNA fixes the order of amino acids in the protein.
Q66Easywrong  βˆ’1
A ribosome that is not engaged in translation exists as
  1. (A) a single compact particle
  2. (B) a complex with the nuclear membrane  βœ— her answer
  3. (C) two separate subunits  βœ“
  4. (D) a ring of several particles
Why C: An inactive ribosome exists as two separate subunits; they come together on the mRNA.
Q67Easycorrect  +4
5'3'mRNAPAlarge subunitsmallgrowing polypeptideA ribosome translating mRNAThe two sites marked P and A on the ribosome are the places where
  1. (A) the release factor and the initiation factor bind
  2. (B) two charged tRNAs are held close enough for a peptide bond to form  βœ“ her answer
  3. (C) the mRNA is capped and tailed
  4. (D) the two ribosomal subunits join together
Why B: The two sites hold successive charged tRNAs close enough for the peptide bond to be made.
Q68Moderatecorrect  +4
In a bacterium the peptide bond is catalysed by
  1. (A) a protein enzyme of the small subunit
  2. (B) the release factor
  3. (C) the initiator tRNA
  4. (D) the 23S rRNA of the large subunit, acting as a ribozyme  βœ“ her answer
Why D: In bacteria the 23S rRNA of the large subunit is the catalyst, that is, a ribozyme.
Q69Moderatecorrect  +4
Translation of an mRNA begins when
  1. (A) the large subunit binds to the terminator
  2. (B) the small subunit encounters the mRNA  βœ“ her answer
  3. (C) the poly-A tail is removed
  4. (D) the release factor binds to the start codon
Why B: Translation begins when the small subunit encounters the mRNA.
Q70Easycorrect  +4
5'3'one mRNA translated by several ribosomes at onceA polysomeThe structure shown is called
  1. (A) a ribozyme
  2. (B) a spliceosome
  3. (C) a nucleosome
  4. (D) a polysome  βœ“ her answer
Why D: Several ribosomes on one mRNA make a polysome.
Q71Moderatecorrect  +4
Several ribosomes translating the same mRNA at the same time allow the cell to
  1. (A) splice the mRNA while it is being read
  2. (B) make several different proteins from one mRNA
  3. (C) proof-read the mRNA before translation
  4. (D) make many copies of the same polypeptide quickly  βœ“ her answer
Why D: A polysome lets the cell turn out many copies of the same polypeptide at once.
Q72Hardcorrect  +4
Assertion (A): The ribosome is described as both a factory and an enzyme.
Reason (R): It holds the mRNA and the charged tRNAs in place and its rRNA itself catalyses the joining of amino acids.
  1. (A) Both A and R are true and R is the correct explanation of A  βœ“ her answer
  2. (B) Both A and R are true but R is NOT the correct explanation of A
  3. (C) A is true but R is false
  4. (D) A is false but R is true
Why A: The ribosome both holds the components in place and, through its rRNA, catalyses the joining, so both descriptions are justified.
Q73Easycorrect  +4
c5' UTRAUGcoding sequenceUAA3' UTRstartstopUTRs are transcribed but not translatedA monocistronic mRNAThe shaded regions marked 5' UTR and 3' UTR in the mRNA shown are
  1. (A) neither transcribed nor translated
  2. (B) removed along with the introns
  3. (C) translated into a signal peptide
  4. (D) transcribed but not translated  βœ“ her answer
Why D: The UTRs are part of the transcript but lie outside the start and stop codons, so they are never translated.
Q74Moderatecorrect  +4
The untranslated regions of an mRNA are required
  1. (A) for efficient translation of the message  βœ“ her answer
  2. (B) to protect the mRNA from the ribosome
  3. (C) to code for the first few amino acids
  4. (D) to join the two ribosomal subunits
Why A: The UTRs are needed for translation to proceed efficiently.
Q75Easycorrect  +4
Translation is terminated when
  1. (A) the two subunits of the ribosome fuse permanently
  2. (B) the initiator tRNA leaves the P site
  3. (C) the ribosome reaches the poly-A tail
  4. (D) a release factor binds to the stop codon and the polypeptide is set free  βœ“ her answer
Why D: A release factor binds the stop codon, translation ends and the polypeptide is released.
Q76Moderatecorrect  +4
During elongation, charged tRNAs bind to the mRNA
  1. (A) only at the 3' untranslated region
  2. (B) at random, the ribosome rejecting the wrong ones later
  3. (C) in the order dictated by the codons, each anticodon pairing with its codon  βœ“ her answer
  4. (D) two codons at a time
Why C: Each charged tRNA binds where its anticodon matches the codon, so the order is dictated by the mRNA.
Q77Moderatecorrect  +4
Match Column I with Column II.
Column IColumn II
AInitiator tRNAIcatalyses peptide bond formation
BRelease factorIIneeded for efficient translation
C23S rRNAIIIrecognises AUG at the start of the message
DUntranslated regionIVbinds the stop codon
  1. (A) A-III, B-I, C-IV, D-II
  2. (B) A-III, B-IV, C-I, D-II  βœ“ her answer
  3. (C) A-IV, B-III, C-I, D-II
  4. (D) A-I, B-IV, C-III, D-II
Why B: Initiator tRNA → recognises AUG; release factor → binds the stop codon; 23S rRNA → catalyses the peptide bond; UTR → needed for efficient translation.
Q78Hardcorrect  +4
Which of the following statements are correct?
A. Amino acids are polymerised in a template-dependent manner.
B. A peptide bond is formed between two amino acids held on adjacent sites of the ribosome.
C. Translation proceeds from the 3' end of the mRNA towards its 5' end.
D. The polypeptide is released only when a stop codon is reached.
  1. (A) 1, 2 and 4 only  βœ“ her answer
  2. (B) 2, 3 and 4 only
  3. (C) 1, 2 and 3 only
  4. (D) 1 and 3 only
Why A: Statements 1, 2 and 4 are right. Translation reads the mRNA 5'→3', not the other way round.
Q79Hardcorrect  +4
How many of the following statements are correct?
A. Translation needs a start codon in the mRNA.
B. An mRNA without a 5' untranslated region is translated more efficiently.
C. The growing polypeptide stays attached to a tRNA until the next peptide bond forms.
D. No tRNA enters the A site when a stop codon reaches it.
  1. (A) Only one
  2. (B) Only two
  3. (C) Only three  βœ“ her answer
  4. (D) All four
Why C: A, C and D are correct. An mRNA lacking a 5' UTR is translated less efficiently, not more.
Q80Hardleft blank  0
A bacterial mRNA is 1500 nucleotides long, of which 90 nucleotides make up the two untranslated regions and the stop codon together. The largest polypeptide it can specify contains about
  1. (A) 500 amino acids
  2. (B) 1410 amino acids
  3. (C) 300 amino acids
  4. (D) 470 amino acids  βœ“
Why D: 1500 − 90 = 1410 coding nucleotides; 1410 ÷ 3 = 470 codons, so about 470 amino acids.

Biotechnology and its Applications

Applications in AgricultureQ1–Q26  Β·  25 of 26 correct  Β·  99/104
Q1Easycorrect  +4
The three options before man for increasing food production are agrochemical-based farming, organic farming and
  1. (A) mixed cropping
  2. (B) genetically engineered crop-based farming  βœ“ her answer
  3. (C) hydroponics
  4. (D) shifting cultivation
Why B: The three options are agrochemical-based, organic and genetically engineered crop-based farming.
Q2Moderatecorrect  +4
The Green Revolution succeeded in tripling the food supply, yet it was not enough to feed the growing population mainly because
  1. (A) the new varieties were not accepted by farmers
  2. (B) the irrigation needed was not available anywhere in India
  3. (C) further increase depended on agrochemicals that most small farmers cannot afford  βœ“ her answer
  4. (D) the new varieties could not be stored
Why C: The Green Revolution tripled the supply, but going further meant more agrochemicals, which most small farmers cannot afford.
Q3Easycorrect  +4
Plants, bacteria, fungi and animals whose genes have been altered by manipulation are called
  1. (A) clones
  2. (B) genetically modified organisms  βœ“ her answer
  3. (C) mutants
  4. (D) hybrid organisms
Why B: Organisms whose genes have been altered by manipulation are genetically modified organisms.
Q4Moderatecorrect  +4
Which of the following statements are correct?
A. Genetic modification has made crops more tolerant of cold, drought, salt and heat.
B. It has reduced the dependence of crops on chemical pesticides.
C. It has helped to reduce post-harvest losses.
D. It has increased the amount of fertiliser a crop needs each season.
  1. (A) 1, 2 and 4 only
  2. (B) 1, 2 and 3 only  βœ“ her answer
  3. (C) 2, 3 and 4 only
  4. (D) 1 and 4 only
Why B: Statements 1, 2 and 3 are right. Genetic modification raises the efficiency of mineral use; it does not increase the fertiliser needed.
Q5Moderatecorrect  +4
Golden rice is an example of the use of genetic modification to
  1. (A) enhance the nutritional value of a food  βœ“ her answer
  2. (B) reduce the time a crop takes to mature
  3. (C) make a crop resistant to insects
  4. (D) make a crop tolerant of salt
Why A: Golden rice is vitamin A enriched rice, an example of enhanced nutritional value.
Q6Moderatecorrect  +4
Increasing the efficiency with which a crop uses minerals is valuable because it
  1. (A) makes the crop resistant to all pests
  2. (B) raises the market price of the crop
  3. (C) removes the need for irrigation
  4. (D) prevents early exhaustion of the fertility of the soil  βœ“ her answer
Why D: A crop that uses minerals efficiently does not drain the soil, so fertility is not exhausted early.
Q7Easycorrect  +4
A crop plant engineered so that it makes its own insecticidal protein is best described as
  1. (A) a biofertiliser
  2. (B) a biosensor
  3. (C) a bio-pesticide producing plant  βœ“ her answer
  4. (D) a bioreactor
Why C: A plant that makes its own insecticidal protein is in effect producing a bio-pesticide.
Q8Hardcorrect  +4
Assertion (A): Genetically modified crops can reduce the quantity of chemical pesticide sprayed on a field.
Reason (R): The crop itself expresses a protein that is toxic to the insect that attacks it.
  1. (A) Both A and R are true and R is the correct explanation of A  βœ“ her answer
  2. (B) Both A and R are true but R is NOT the correct explanation of A
  3. (C) A is true but R is false
  4. (D) A is false but R is true
Why A: The crop expresses the toxin itself, which is exactly why less pesticide has to be sprayed.
Q9Easycorrect  +4
The bacterium from which the Bt toxin gene is obtained is
  1. (A) Thermus aquaticus
  2. (B) Agrobacterium tumefaciens
  3. (C) Escherichia coli
  4. (D) Bacillus thuringiensis  βœ“ her answer
Why D: The toxin gene comes from the soil bacterium Bacillus thuringiensis.
Q10Easycorrect  +4
Bacillusthuringiensisinactiveprotoxin crystalinsect eats it;alkaline midgutpH > 9.5soluble activetoxin (Cry protein)midgut epitheliumporescell swells, lyses,insect diesMode of action of the Bt toxinThe figure shows why the Bt protein does not harm the bacterium that makes it. The protein exists in the bacterium as
  1. (A) an active toxin kept in a vacuole
  2. (B) a toxin that is secreted at once into the soil
  3. (C) an inactive protoxin in the form of a crystal  βœ“ her answer
  4. (D) a toxin bound to a specific inhibitor protein
Why C: Inside the bacterium the toxin exists as an inactive crystalline protoxin, so the bacterium is unharmed.
Q11Moderatecorrect  +4
Bacillusthuringiensisinactiveprotoxin crystalinsect eats it;alkaline midgutpH > 9.5soluble activetoxin (Cry protein)midgut epitheliumporescell swells, lyses,insect diesMode of action of the Bt toxinIn the same figure, the crystal is solubilised and the toxin activated because the insect's midgut is
  1. (A) strongly acidic
  2. (B) neutral
  3. (C) alkaline  βœ“ her answer
  4. (D) rich in bile salts
Why C: The alkaline pH of the insect midgut solubilises the crystal and converts the protoxin into active toxin.
Q12Moderatecorrect  +4
Once activated, the Bt toxin kills the insect by
  1. (A) blocking the nerve impulses at the synapse
  2. (B) binding to the surface of midgut epithelial cells and creating pores, so that the cells swell and lyse  βœ“ her answer
  3. (C) digesting the cuticle of the insect
  4. (D) preventing the insect from moulting
Why B: The active toxin binds the midgut epithelial cells and makes pores, so the cells swell and burst.
Q13Easycorrect  +4
The gene that codes for the Bt toxin is named
  1. (A) nif
  2. (B) tox
  3. (C) bt
  4. (D) cry  βœ“ her answer
Why D: The toxin is coded by the cry gene, of which there are several forms.
Q14Moderatecorrect  +4
cry geneinsect controlledcryIAccotton bollwormcryIIAbcotton bollwormcryIAbcorn borerthe gene is inserted into the plant, which then makes its own toxincry genes used in Bt cropsFrom the figure, the genes that protect cotton against the bollworm are
  1. (A) cryIAb alone
  2. (B) cryIAb and cryIAc
  3. (C) cryIAc and cryIIAb  βœ“ her answer
  4. (D) cryIIAb and cryIAb
Why C: cryIAc and cryIIAb both control the cotton bollworm.
Q15Moderatecorrect  +4
The corn borer is controlled by the protein encoded by
  1. (A) cryIAb  βœ“ her answer
  2. (B) cryIAc
  3. (C) cryIIIA
  4. (D) cryIIAb
Why A: cryIAb controls the corn borer.
Q16Moderatecorrect  +4
The choice of which cry gene to put into a particular crop depends on the pest to be controlled, because
  1. (A) cry genes are destroyed by the sap of some plants
  2. (B) each cry gene works only at a particular temperature
  3. (C) each crop can accept only one kind of gene
  4. (D) most Bt toxins are specific to a particular group of insects  βœ“ her answer
Why D: Most Bt toxins act only on a particular group of insects, so the gene must be matched to the pest.
Q17Moderatecorrect  +4
Match Column I with Column II.
Column IColumn II
AcryIAcIsource of the toxin gene
BcryIAbIIsolubilises the crystal
CBacillus thuringiensisIIIcotton bollworm
DAlkaline gut pHIVcorn borer
  1. (A) A-III, B-I, C-IV, D-II
  2. (B) A-IV, B-III, C-I, D-II
  3. (C) A-I, B-IV, C-III, D-II
  4. (D) A-III, B-IV, C-I, D-II  βœ“ her answer
Why D: cryIAc → cotton bollworm; cryIAb → corn borer; Bacillus thuringiensis → source of the gene; alkaline gut pH → solubilises the crystal.
Q18Hardwrong  βˆ’1
How many of the following statements are correct?
A. Bt cotton carries a gene taken from a bacterium.
B. The Bt toxin is sprayed on the crop as a chemical insecticide.
C. Bt varieties of corn, rice, tomato, potato and soyabean have been produced.
D. The activated toxin acts on the midgut of the insect.
  1. (A) Only one
  2. (B) Only two  βœ— her answer
  3. (C) Only three  βœ“
  4. (D) All four
Why C: A, C and D are correct. The whole point of a Bt crop is that the toxin is made by the plant and not sprayed on it.
Q19Easycorrect  +4
The nematode that infects the roots of tobacco and greatly reduces its yield is
  1. (A) Meloidogyne incognita  βœ“ her answer
  2. (B) Ascaris lumbricoides
  3. (C) Ancylostoma duodenale
  4. (D) Wuchereria bancrofti
Why A: Meloidogyne incognita infects tobacco roots and reduces the yield greatly.
Q20Easycorrect  +4
RNA interference is a method of
  1. (A) cellular defence found in all eukaryotic organisms  βœ“ her answer
  2. (B) DNA repair found only in bacteria
  3. (C) gene transfer peculiar to plants
  4. (D) protein folding in the endoplasmic reticulum
Why A: RNA interference is a defence mechanism present in all eukaryotic organisms.
Q21Moderatecorrect  +4
host plant celltransgene makes dsRNAnematodeMeloidogyne incognitaparasite feedsdsRNA enters the parasitebinds the complementary mRNA→ mRNA silencedparasite cannot surviveRNA interference against a nematodeIn the strategy shown, the molecule that silences the nematode's mRNA is
  1. (A) a toxin protein made by the plant
  2. (B) a double-stranded RNA complementary to that mRNA  βœ“ her answer
  3. (C) a ribozyme made by the plant
  4. (D) a single-stranded antisense DNA
Why B: A double-stranded RNA complementary to the nematode mRNA binds it and prevents its translation.
Q22Moderatecorrect  +4
The vector used to introduce the nematode-specific genes into the tobacco plant was
  1. (A) a bacteriophage
  2. (B) a retrovirus
  3. (C) Agrobacterium  βœ“ her answer
  4. (D) a liposome
Why C: Agrobacterium vectors were used to introduce the nematode-specific genes into the plant.
Q23Hardcorrect  +4
The introduced DNA was designed so that the host cell would produce
  1. (A) a double-stranded DNA that enters the nematode
  2. (B) only antisense RNA
  3. (C) only sense RNA
  4. (D) both sense and antisense RNA, which pair to form double-stranded RNA  βœ“ her answer
Why D: The construct makes both sense and antisense RNA; being complementary, these pair to form the dsRNA that triggers RNAi.
Q24Moderatecorrect  +4
The natural sources of the complementary RNA that triggers RNA interference in a cell include
  1. (A) RNA viruses and transposons that replicate through an RNA intermediate  βœ“ her answer
  2. (B) introns released during splicing
  3. (C) the poly-A tail of every mRNA
  4. (D) ribosomal RNA and transfer RNA
Why A: The natural triggers are RNA viruses and transposons that replicate through an RNA intermediate.
Q25Hardcorrect  +4
Assertion (A): The transgenic tobacco plant survives an attack by Meloidogyne incognita.
Reason (R): The parasite takes up double-stranded RNA from the host, which silences a specific nematode mRNA.
  1. (A) Both A and R are true and R is the correct explanation of A  βœ“ her answer
  2. (B) Both A and R are true but R is NOT the correct explanation of A
  3. (C) A is true but R is false
  4. (D) A is false but R is true
Why A: The dsRNA taken up from the host silences a nematode mRNA, so the parasite cannot survive and the plant is protected.
Q26Hardcorrect  +4
Which of the following statements are correct?
A. RNA interference silences a specific mRNA.
B. The silencing RNA must be complementary to the target mRNA.
C. In the nematode strategy the plant itself is killed by the dsRNA.
D. The transgenic plant becomes protected from the parasite.
  1. (A) 1, 2 and 3 only
  2. (B) 2, 3 and 4 only
  3. (C) 1 and 3 only
  4. (D) 1, 2 and 4 only  βœ“ her answer
Why D: Statements 1, 2 and 4 are right. The dsRNA silences the parasite, not the plant.
Applications in MedicineQ27–Q54  Β·  27 of 28 correct  Β·  107/112
Q27Easycorrect  +4
Before recombinant DNA technology, the insulin given to diabetics was extracted from
  1. (A) the milk of transgenic cows
  2. (B) the pancreas of slaughtered cattle and pigs  βœ“ her answer
  3. (C) cultures of yeast
  4. (D) human cadavers
Why B: Insulin was extracted from the pancreas of slaughtered cattle and pigs.
Q28Moderatecorrect  +4
The chief disadvantage of insulin obtained from an animal source was that it
  1. (A) caused allergy or other reactions in some patients because it was a foreign protein  βœ“ her answer
  2. (B) could not be injected
  3. (C) lost its activity within a day
  4. (D) raised the blood sugar instead of lowering it
Why A: Being a foreign protein, animal insulin caused allergy or other reactions in some patients.
Q29Easycorrect  +4
Mature insulin consists of
  1. (A) two identical chains joined end to end
  2. (B) a single polypeptide chain
  3. (C) three chains, A, B and C, held by hydrogen bonds
  4. (D) two chains, A and B, held together by disulphide bridges  βœ“ her answer
Why D: Mature insulin has two chains, A and B, linked by disulphide bridges.
Q30Easycorrect  +4
B chainC peptideremoved during maturationA chainpro-insulin (made as one chain)mature insulin = A chain + B chain onlyPro-insulin and the extra C peptideThe stretch marked in the figure, which is present in pro-insulin but absent from mature insulin, is the
  1. (A) A chain
  2. (B) B chain
  3. (C) C peptide  βœ“ her answer
  4. (D) signal peptide
Why C: Pro-insulin carries an extra stretch, the C peptide, which is removed as the hormone matures.
Q31Moderatecorrect  +4
Insulin is synthesised in mammals as a pro-hormone. This means that it
  1. (A) must be processed before it becomes a fully mature and functional hormone  βœ“ her answer
  2. (B) is active as soon as it is made
  3. (C) is made only in the liver
  4. (D) is made of RNA rather than protein
Why A: A pro-hormone must be processed before it becomes a mature, functional hormone.
Q32Moderatecorrect  +4
E. coli withA-chain geneE. coli withB-chain geneA chain (21 aa)B chain (30 aa)combineddisulphide bonds β†’ insulinMaking human insulin in bacteriaAccording to the route shown, the main difficulty in making human insulin by recombinant DNA technology was
  1. (A) finding a promoter that works in bacteria
  2. (B) purifying the mRNA of the insulin gene
  3. (C) assembling the separately produced chains into a mature, functional hormone  βœ“ her answer
  4. (D) getting E. coli to take up the plasmid
Why C: The chains could be made in bacteria readily enough; the difficulty was assembling them into a mature, functional hormone.
Q33Moderatecorrect  +4
The company that prepared the two DNA sequences corresponding to the A and B chains of human insulin and introduced them into plasmids of E. coli, in 1983, was
  1. (A) Monsanto
  2. (B) Novartis
  3. (C) Genentech
  4. (D) Eli Lilly  βœ“ her answer
Why D: Eli Lilly, an American company, did this in 1983.
Q34Moderatecorrect  +4
In that process the A and B chains were
  1. (A) extracted from pig pancreas and modified chemically
  2. (B) produced together as one chain and then cut
  3. (C) synthesised chemically without using any living cell
  4. (D) produced separately, extracted and then combined by creating disulphide bonds  βœ“ her answer
Why D: The A and B chains were produced separately, extracted, and then joined by creating disulphide bonds.
Q35Hardcorrect  +4
Assertion (A): Human insulin made in bacteria does not provoke the reactions that animal insulin sometimes did.
Reason (R): The bacterially produced hormone has the same amino acid sequence as human insulin.
  1. (A) Both A and R are true and R is the correct explanation of A  βœ“ her answer
  2. (B) Both A and R are true but R is NOT the correct explanation of A
  3. (C) A is true but R is false
  4. (D) A is false but R is true
Why A: Because the sequence is identical to the human hormone, the body does not treat it as foreign.
Q36Hardcorrect  +4
How many of the following statements are correct?
A. Insulin from cattle and pigs was once the only source for diabetics.
B. The C peptide is removed as pro-insulin matures.
C. Recombinant human insulin is produced in E. coli.
D. The A and B chains of insulin are linked by peptide bonds between their ends.
  1. (A) Only one
  2. (B) Only two
  3. (C) Only three  βœ“ her answer
  4. (D) All four
Why C: A, B and C are correct. The two chains are held by disulphide bridges, not by peptide bonds between their ends.
Q37Easycorrect  +4
Gene therapy may be defined as
  1. (A) the use of drugs that block a faulty enzyme
  2. (B) a collection of methods that allows the correction of a diagnosed gene defect  βœ“ her answer
  3. (C) the injection of antibodies against a defective protein
  4. (D) the removal of a defective organ and its replacement
Why B: Gene therapy is a set of methods for correcting a gene defect that has been diagnosed.
Q38Easycorrect  +4
The first clinical gene therapy, given in 1990, was for a four-year-old girl suffering from
  1. (A) cystic fibrosis
  2. (B) sickle-cell anaemia
  3. (C) adenosine deaminase deficiency  βœ“ her answer
  4. (D) phenylketonuria
Why C: The first clinical gene therapy, in 1990, was for a four-year-old girl with adenosine deaminase deficiency.
Q39Moderatecorrect  +4
Adenosine deaminase deficiency is caused by
  1. (A) the deletion of the gene for adenosine deaminase  βœ“ her answer
  2. (B) an extra copy of chromosome 21
  3. (C) a frameshift in the gene for haemoglobin
  4. (D) the loss of the thymus gland
Why A: The disorder follows the deletion of the gene for adenosine deaminase.
Q40Moderatecorrect  +4
ADA deficiency is serious because the enzyme is crucial for the working of the
  1. (A) endocrine system
  2. (B) immune system  βœ“ her answer
  3. (C) digestive system
  4. (D) nervous system
Why B: ADA is crucial for the immune system, so its absence leaves the child without immune defence.
Q41Moderatewrong  βˆ’1
celllymphocytesfrom patientfunctional ADAcDNA in a retroviral vectorgrown in culture,returned to the patientlymphocytes are not immortal β†’ the infusion must be repeateda permanent cure needs the gene put into bone-marrow cellsat an early embryonic stageGene therapy for ADA deficiencyIn the procedure shown, the cells taken from the patient, grown in culture and returned after receiving the functional gene are
  1. (A) bone-marrow stem cells  βœ— her answer
  2. (B) epithelial cells of the gut
  3. (C) lymphocytes  βœ“
  4. (D) red blood cells
Why C: Lymphocytes from the patient's blood are grown in culture, given the gene and returned.
Q42Moderatecorrect  +4
The functional ADA cDNA is introduced into those cells using
  1. (A) a plasmid of Agrobacterium
  2. (B) a retroviral vector  βœ“ her answer
  3. (C) a liposome of cholesterol
  4. (D) a gene gun
Why B: A functional ADA cDNA is delivered using a retroviral vector.
Q43Hardcorrect  +4
The patient must receive periodic infusions of the engineered cells because
  1. (A) the retroviral vector loses the gene after one division
  2. (B) the enzyme produced is unstable at body temperature
  3. (C) the introduced gene is destroyed by the immune system
  4. (D) the cells used are not immortal  βœ“ her answer
Why D: The lymphocytes used are not immortal, so the infusion has to be repeated periodically.
Q44Hardcorrect  +4
A permanent cure for ADA deficiency would be possible if the gene producing ADA were introduced into
  1. (A) the bacteria of the patient's gut
  2. (B) the liver cells of the patient
  3. (C) the lymphocytes of an adult patient
  4. (D) cells at an early embryonic stage  βœ“ her answer
Why D: Introducing the gene into cells at an early embryonic stage could give a permanent cure.
Q45Hardcorrect  +4
Assertion (A): Bone-marrow transplantation and enzyme replacement therapy are not completely curative for ADA deficiency.
Reason (R): Neither approach supplies the patient's own cells with a working copy of the ADA gene.
  1. (A) Both A and R are true and R is the correct explanation of A  βœ“ her answer
  2. (B) Both A and R are true but R is NOT the correct explanation of A
  3. (C) A is true but R is false
  4. (D) A is false but R is true
Why A: Neither bone-marrow transplantation nor enzyme replacement gives the patient's own cells a working ADA gene, which is why neither is completely curative.
Q46Easycorrect  +4
The limitation of conventional methods of diagnosis such as serum analysis, urine analysis and biopsy is that they
  1. (A) cannot detect a disease at an early stage  βœ“ her answer
  2. (B) are too expensive for most patients
  3. (C) cannot be repeated on the same patient
  4. (D) require a hospital stay
Why A: Conventional methods cannot pick up a disease early, which is the gap molecular methods fill.
Q47Easycorrect  +4
a very low number of copies cannot be seen1 copyPCR248amplified until it can be detected β€” used for HIV and cancerPCR in molecular diagnosisThe technique shown, which allows a pathogen to be detected while its number is still very low, is
  1. (A) autoradiography
  2. (B) the polymerase chain reaction  βœ“ her answer
  3. (C) gel electrophoresis
  4. (D) ELISA
Why B: PCR amplifies the pathogen's nucleic acid, so a very small number of copies can be detected.
Q48Moderatecorrect  +4
PCR is routinely used in medicine to
  1. (A) separate proteins of the blood serum
  2. (B) measure the blood sugar of diabetics
  3. (C) count the red blood cells of anaemic patients
  4. (D) detect HIV in suspected AIDS patients and mutations in suspected cancer patients  βœ“ her answer
Why D: PCR is used to detect HIV in suspected AIDS patients and mutations in suspected cancer patients.
Q49Moderatecorrect  +4
By the time a pathogen produces a visible symptom of disease, its concentration in the body is
  1. (A) still too low to be detected by any method
  2. (B) falling steadily
  3. (C) already very high  βœ“ her answer
  4. (D) exactly at the threshold of detection
Why C: By the time symptoms appear the pathogen is already present in high concentration.
Q50Moderatecorrect  +4
single-stranded DNA on a membrane*labelled probehybridisationa dark spot on the film means the gene is presentDetecting a gene by autoradiographyIn the method shown, the single-stranded DNA or RNA tagged with a radioactive molecule is called
  1. (A) a primer
  2. (B) a probe  βœ“ her answer
  3. (C) a ligase
  4. (D) a vector
Why B: A single-stranded DNA or RNA carrying a radioactive label is a probe.
Q51Hardcorrect  +4
A probe for a normal gene is hybridised with DNA from a clone of cells and the result is examined by autoradiography. A clone carrying the mutated gene will
  1. (A) give two spots instead of one
  2. (B) not appear on the photographic film, because the probe is not complementary to the mutated gene  βœ“ her answer
  3. (C) give a darker spot than the normal clone
  4. (D) appear on the film only after a second round of hybridisation
Why B: The probe is complementary to the normal gene, so it will not hybridise with the mutated gene and that clone leaves no spot on the film.
Q52Easycorrect  +4
well coated with antigenpatient's antibody bindsEenzyme-linkedsecond antibodyEsubstrate β†’ colourcolour = infection detected (antigen or antibody)Principle of ELISAThe technique shown works on the principle of
  1. (A) complementary base pairing
  2. (B) enzyme-substrate specificity alone
  3. (C) differential centrifugation
  4. (D) antigen-antibody interaction  βœ“ her answer
Why D: ELISA rests on antigen-antibody interaction.
Q53Moderatecorrect  +4
An infection can be detected by ELISA through the presence of
  1. (A) either the antigens of the pathogen or the antibodies made against it  βœ“ her answer
  2. (B) the enzymes of the host liver
  3. (C) only the nucleic acid of the pathogen
  4. (D) only the antigens of the pathogen
Why A: Infection shows up either as the antigens of the pathogen or as the antibodies made against it.
Q54Hardcorrect  +4
Match Column I with Column II.
Column IColumn II
APCRIcarries a normal gene into a lymphocyte
BELISAIIantigen-antibody interaction
CRadioactive probeIIIamplifies a very small amount of nucleic acid
DRetroviral vectorIVlocates a gene by autoradiography
  1. (A) A-III, B-IV, C-II, D-I
  2. (B) A-III, B-II, C-IV, D-I  βœ“ her answer
  3. (C) A-I, B-II, C-IV, D-III
  4. (D) A-II, B-III, C-IV, D-I
Why B: PCR → amplifies a small amount of nucleic acid; ELISA → antigen-antibody interaction; radioactive probe → locates a gene by autoradiography; retroviral vector → carries a normal gene into a lymphocyte.
Transgenic Animals and Ethical IssuesQ55–Q80  Β·  26 of 26 correct  Β·  104/104
Q55Easycorrect  +4
An animal whose DNA has been manipulated so that it possesses and expresses an extra foreign gene is called
  1. (A) a mutant
  2. (B) a clone
  3. (C) a hybrid
  4. (D) a transgenic animal  βœ“ her answer
Why D: An animal carrying and expressing an extra foreign gene is a transgenic animal.
Q56Easycorrect  +4
Of all the transgenic animals produced so far, more than ninety-five per cent are
  1. (A) mice  βœ“ her answer
  2. (B) sheep
  3. (C) rabbits
  4. (D) rats
Why A: Over 95 per cent of all existing transgenic animals are mice.
Q57Moderatecorrect  +4
Transgenic animals are used in the study of normal physiology and development mainly to find out
  1. (A) how long an animal can live
  2. (B) how animals respond to changes in climate
  3. (C) how genes are regulated and how they affect normal body function and development  βœ“ her answer
  4. (D) which animals breed fastest
Why C: They are used to study how genes are regulated and how they affect normal function and development.
Q58Moderatecorrect  +4
The complex factors involved in growth that have been studied with the help of transgenic animals include
  1. (A) adrenaline
  2. (B) insulin-like growth factor  βœ“ her answer
  3. (C) thyroxine
  4. (D) melatonin
Why B: Insulin-like growth factor is the example given for the study of complex growth factors.
Q59Moderatecorrect  +4
Transgenic models have been developed for all the following human diseases except
  1. (A) rheumatoid arthritis
  2. (B) scurvy  βœ“ her answer
  3. (C) cystic fibrosis
  4. (D) cancer
Why B: Models exist for cancer, cystic fibrosis, rheumatoid arthritis and Alzheimer's. Scurvy is a vitamin C deficiency, not a genetic disease modelled this way.
Q60Easycorrect  +4
The human protein used to treat emphysema, and obtained from transgenic animals, is
  1. (A) interferon
  2. (B) adenosine deaminase
  3. (C) alpha-lactalbumin
  4. (D) alpha-1-antitrypsin  βœ“ her answer
Why D: Alpha-1-antitrypsin, obtained from transgenic animals, is used to treat emphysema.
Q61Moderatecorrect  +4
Attempts to obtain biological products from transgenic animals are also being made for the treatment of
  1. (A) phenylketonuria and cystic fibrosis  βœ“ her answer
  2. (B) malaria and dengue
  3. (C) anaemia and rickets
  4. (D) tuberculosis and leprosy
Why A: Similar attempts are being made for phenylketonuria and cystic fibrosis.
Q62Easycorrect  +4
egghuman geneembryotransgenic cowRosie β€” milk with human alpha-lactalbumin, 2.4 g per litremore balanced for human babies than ordinary cow milkA transgenic animal as a bioreactorThe transgenic cow referred to in the figure, produced in 1997, was named
  1. (A) Molly
  2. (B) Dolly
  3. (C) Polly
  4. (D) Rosie  βœ“ her answer
Why D: Rosie, the first transgenic cow, was produced in 1997.
Q63Moderatecorrect  +4
egghuman geneembryotransgenic cowRosie β€” milk with human alpha-lactalbumin, 2.4 g per litremore balanced for human babies than ordinary cow milkA transgenic animal as a bioreactorThe human protein present in the milk of that cow, at 2.4 grams per litre, was
  1. (A) alpha-1-antitrypsin
  2. (B) human alpha-lactalbumin  βœ“ her answer
  3. (C) human growth hormone
  4. (D) human insulin
Why B: Her milk contained human alpha-lactalbumin at 2.4 grams per litre.
Q64Moderatecorrect  +4
That milk was considered nutritionally better than ordinary cow milk for
  1. (A) patients with emphysema
  2. (B) patients with diabetes
  3. (C) human babies  βœ“ her answer
  4. (D) elderly patients
Why C: That milk was nutritionally more balanced for human babies than natural cow milk.
Q65Moderatecorrect  +4
Transgenic mice are being used to test the safety of
  1. (A) the polio vaccine before it is used on humans  βœ“ her answer
  2. (B) antibiotics before they are given to cattle
  3. (C) pesticides before they are sprayed on crops
  4. (D) food colours before they are sold
Why A: Transgenic mice are used to test the safety of the polio vaccine before it is given to humans.
Q66Hardcorrect  +4
In chemical safety testing, the transgenic animals used are made to carry genes that
  1. (A) allow them to break the toxic substance down
  2. (B) make them more sensitive to the toxic substance than non-transgenic animals  βœ“ her answer
  3. (C) prevent them from absorbing the toxic substance
  4. (D) make them resistant to the toxic substance
Why B: The animals are engineered to be more sensitive to the toxic substance than ordinary animals.
Q67Hardcorrect  +4
The advantage of toxicity testing in such transgenic animals is that
  1. (A) the results apply to plants as well
  2. (B) results are obtained in less time  βœ“ her answer
  3. (C) no control group is required
  4. (D) fewer animals are needed in total
Why B: Greater sensitivity means the effect shows up sooner, so results are obtained in less time.
Q68Hardcorrect  +4
How many of the following statements are correct?
A. Transgenic animals help us to understand how genes contribute to disease.
B. Transgenic animals can be used as a source of expensive biological products.
C. Transgenic animals are used to test the safety of vaccines.
D. Transgenic rats make up the majority of transgenic animals produced.
  1. (A) Only one
  2. (B) Only two
  3. (C) Only three  βœ“ her answer
  4. (D) All four
Why C: A, B and C are correct. It is mice, not rats, that make up the great majority of transgenic animals.
Q69Easycorrect  +4
The Indian body set up to decide on the validity of GM research and the safety of introducing GM organisms for public services is
  1. (A) ICMR
  2. (B) CSIR
  3. (C) NBPGR
  4. (D) GEAC  βœ“ her answer
Why D: The Genetic Engineering Approval Committee was set up for this purpose.
Q70Moderatecorrect  +4
GEAC stands for
  1. (A) Genetically Engineered Animal Council
  2. (B) Genetic Engineering Approval Committee  βœ“ her answer
  3. (C) Government Ethics and Agriculture Commission
  4. (D) General Environment Assessment Council
Why B: GEAC stands for Genetic Engineering Approval Committee.
Q71Easycorrect  +4
The use of bio-resources by multinational companies and other organisations without proper authorisation from the countries and people concerned, and without compensatory payment, is called
  1. (A) biopiracy  βœ“ her answer
  2. (B) bio-patenting
  3. (C) bioremediation
  4. (D) bio-fortification
Why A: Using bio-resources without authorisation or compensation is biopiracy.
Q72Moderatecorrect  +4
An American company obtained patent rights on Basmati rice. The variety it claimed as new had in fact been derived from
  1. (A) an African variety of rice
  2. (B) a laboratory mutant of Japanese rice
  3. (C) Indian farmers' varieties  βœ“ her answer
  4. (D) a wild rice of South America
Why C: The variety patented had been derived from Indian farmers' varieties.
Q73Moderatecorrect  +4
The number of documented varieties of Basmati grown in India is
  1. (A) 37
  2. (B) 7
  3. (C) 27  βœ“ her answer
  4. (D) 17
Why C: India grows 27 documented varieties of Basmati.
Q74Moderatecorrect  +4
The main concern behind the patenting of biological resources by foreign firms is that
  1. (A) the resources will become extinct
  2. (B) the patented varieties cannot be grown in India
  3. (C) the resources will lose their medicinal value
  4. (D) the country and the people who conserved and developed them get no share of the benefit  βœ“ her answer
Why D: The people and the country that conserved and developed the resource receive no share of the benefit.
Q75Moderatecorrect  +4
The main ethical objection raised against genetic modification is that it
  1. (A) always produces toxic food
  2. (B) involves the unpredictable manipulation of living organisms and their use for human purposes  βœ“ her answer
  3. (C) is technically impossible to control
  4. (D) reduces the yield of every crop
Why B: The objection is to the unpredictable manipulation of living organisms and their use for human purposes.
Q76Hardcorrect  +4
Assertion (A): Nations have been developing laws to prevent the unauthorised exploitation of their bio-resources.
Reason (R): Traditional knowledge and genetic resources of a country have commercial value that others may exploit without compensation.
  1. (A) Both A and R are true and R is the correct explanation of A  βœ“ her answer
  2. (B) Both A and R are true but R is NOT the correct explanation of A
  3. (C) A is true but R is false
  4. (D) A is false but R is true
Why A: Because traditional knowledge and genetic resources carry commercial value that others may exploit, nations have framed laws to prevent it.
Q77Hardcorrect  +4
Which of the following statements are correct?
A. Biopiracy involves the use of bio-resources without authorisation.
B. GEAC examines the safety of introducing GM organisms for public services.
C. Genetic modification of organisms raises no ethical questions at all.
D. Developing countries are rich in bio-resources and in traditional knowledge about them.
  1. (A) 1, 2 and 4 only  βœ“ her answer
  2. (B) 2, 3 and 4 only
  3. (C) 1 and 3 only
  4. (D) 1, 2 and 3 only
Why A: Statements 1, 2 and 4 are right. Genetic modification does raise ethical questions, so 3 is false.
Q78Hardcorrect  +4
Match Column I with Column II.
Column IColumn II
AGEACIcotton bollworm
BBiopiracyIItransgenic cow
CRosieIIIsafety of GM organisms for public services
DcryIAcIVunauthorised use of bio-resources
  1. (A) A-III, B-II, C-IV, D-I
  2. (B) A-IV, B-III, C-II, D-I
  3. (C) A-III, B-IV, C-II, D-I  βœ“ her answer
  4. (D) A-I, B-IV, C-II, D-III
Why C: GEAC → safety of GM organisms for public services; biopiracy → unauthorised use of bio-resources; Rosie → transgenic cow; cryIAc → cotton bollworm.
Q79Moderatecorrect  +4
Biotechnology may be defined as the use of
  1. (A) only micro-organisms to produce useful substances
  2. (B) chemical catalysts to replace living cells in industry
  3. (C) radiation to create useful mutations in crops
  4. (D) live organisms or enzymes from organisms to produce products and processes useful to humans  βœ“ her answer
Why D: Biotechnology is the use of live organisms, or enzymes from them, to make products and processes useful to humans.
Q80Hardcorrect  +4
How many of the following statements are correct?
A. Developing countries hold a large share of the world's bio-resources.
B. A patent gives its holder exclusive rights over the use of what is patented.
C. GEAC is concerned only with the import of foreign seeds.
D. Traditional knowledge about a plant may be used to obtain a patent elsewhere.
  1. (A) Only one
  2. (B) Only two
  3. (C) Only three  βœ“ her answer
  4. (D) All four
Why C: A, B and D are correct. GEAC's remit is the validity of GM research and the safety of GM organisms, not merely the import of seeds.

What the nine lost marks actually say

One real cluster

Three of the four Transcription losses sit in the same block β€” Q4, Q5 and Q6 β€” and all three are about polarity. Q4 she copied the template strand straight across with U for T instead of taking the complement. Q5, the mirror question, she left blank. Q6 she put the promoter downstream at the 3' end instead of upstream at the 5' end. That is one idea failing three times, not three separate slips, and it is the single most worthwhile thing to drill this week.

The fix is a thirty-second drill, not a chapter re-read. Write any template strand, mark its 3' end, and build the RNA by complementing base by base while running from that 3' end: the RNA always comes out antiparallel, so its 5' end pairs with the template's 3' end. Then check it against the coding strand β€” the transcript must match the coding strand letter for letter with U in place of T. If it does not, the complement was taken the wrong way.

Four isolated slips

The one that counts most

MBI Q80 was the only question across 160 that needed more than one step, and it is the one she left blank. The work is two lines: 1500 βˆ’ 90 = 1410 coding nucleotides, and 1410 Γ· 3 = 470 codons. Nothing in it is hard. What stopped her was almost certainly that it did not look like the other 159 questions. That is precisely the gap the revised papers are built to close, and it is worth watching on the Thermal paper, which carries twenty such questions.

This week

  1. The polarity drill. Ten template strands, ten transcripts, written out by hand, checked against the coding strand each time. Fifteen minutes, once.
  2. Re-do MBI Q80 and then the twenty chained questions on the Thermal paper β€” the point is the habit of starting a question whose shape is unfamiliar, not the arithmetic.
  3. Leave the assertion–reason format alone until the next paper. Drilling it against these two papers will only reinforce the reflex, because every answer in them is the same. The new papers vary it across all four options.
  4. Do not re-read either chapter. At 97% and 90% the content is in place. Nine marks across 160 questions is a proof-reading problem and one polarity hole, not a knowledge gap.
Marks read directly from the two scanned scripts and checked against the published answer keys. Every wrong and blank answer in this report was re-read from the scan individually before being written down.