📌 Snapshot
- Understand DNA structure, replication, transcription, the genetic code, the lac operon and DNA fingerprinting.
- The classic experiments of Griffith, Hershey and Chase, and Meselson and Stahl, explained step by step.
- Replication, transcription and translation compared side by side, with every number NTA likes to test.
- The Human Genome Project findings and DNA fingerprinting steps, plus the traps that cost marks.
📖 Detailed Notes
Written and reviewed by Anjali Sharma, Biology faculty at UniDrill (M.Sc. Botany, Banaras Hindu University · 8 years teaching NCERT Biology). Based on NCERT Biology Class XII, Chapter 5.
How do two metres of DNA fit inside one cell?
Multiply the number of base pairs in a diploid human cell (6.6 × 10⁹) by the distance between two base pairs (0.34 nm) and you get about 2.2 metres of DNA. It sits inside a nucleus only a few micrometres across. This chapter explains how that molecule is built, packed, copied, read and switched on and off, and how scientists proved it is the genetic material in the first place.
DNA and RNA: the building blocks
A nucleotide has three parts: a nitrogenous base, a pentose sugar and a phosphate group. A base joined to a sugar is a nucleoside, and a nucleoside with a phosphate (at the 5′ carbon) is a nucleotide. Nucleotides link through 3′–5′ phosphodiester bonds to make a polynucleotide chain.
The bases are of two kinds. Purines (adenine and guanine) have two rings. Pyrimidines (cytosine, thymine and uracil) have one ring. Thymine is found in DNA, and uracil replaces it in RNA.
| Feature | DNA | RNA |
|---|---|---|
| Sugar | Deoxyribose | Ribose (has a reactive 2′ hydroxyl) |
| Pyrimidines | Cytosine, thymine | Cytosine, uracil |
| Strands | Two, coiled as a double helix | Usually one |
| Stability | High | Lower, mutates faster |
| Main role | Genetic material | Messenger, adapter, structural and catalytic roles |
Sizes worth knowing. Bacteriophage φX174 has 5386 nucleotides, bacteriophage lambda has 48,502 base pairs, E. coli has 4.6 × 10⁶ base pairs, and the haploid human genome has about 3.3 × 10⁹ base pairs.
The double helix
Friedrich Miescher first isolated DNA from the nucleus in 1869 and called it nuclein. In 1953 James Watson and Francis Crick proposed the double helix, using X-ray diffraction data from Maurice Wilkins and Rosalind Franklin, and Chargaff's rule: in double stranded DNA, the amount of A equals T, and G equals C, so the ratio A+G to T+C is 1.
| Feature of B-DNA | Value or fact |
|---|---|
| Strands | Two, antiparallel (one runs 5′ to 3′, the other 3′ to 5′) |
| Backbone | Sugar and phosphate, on the outside |
| Pairing | A with T by 2 hydrogen bonds; G with C by 3 hydrogen bonds |
| Handedness | Right handed |
| Pitch (one full turn) | 3.4 nm |
| Base pairs per turn | 10 |
| Distance between adjacent base pairs | 0.34 nm |
A purine always pairs with a pyrimidine, so the width of the helix stays constant. The stacking of base pairs also stabilises the structure.
Packaging DNA
A bacterium has no nucleus. Its negatively charged DNA is held by some positively charged proteins in a region called the nucleoid, organised in large loops.
Eukaryotes use histones, positively charged basic proteins rich in lysine and arginine. Eight histone molecules form a histone octamer, and about 200 base pairs of negatively charged DNA wrap around it to form a nucleosome. Under the electron microscope, nucleosomes look like "beads on a string", and this is chromatin. Chromatin is then folded into fibres and into the metaphase chromosome, and the higher levels of packing need non-histone chromosomal (NHC) proteins.
| Euchromatin | Heterochromatin | |
|---|---|---|
| Packing | Loose | Dense |
| Staining | Light | Dark |
| Transcription | Active | Inactive |
Beyond the textbook. For the exam: a nucleosome has about 200 bp of DNA on a histone octamer. In reality: the DNA tightly wound on the octamer core is about 147 base pairs, and the remaining stretch of linker DNA between nucleosomes brings the repeat to about 200. If a question gives 147 or 146 as an option, the stem is asking about the core, not the repeat.
Which molecule carries the genes? Three experiments
Until the 1940s many scientists believed proteins were the genetic material. Three experiments settled the question.
Griffith, 1928. Griffith worked with Streptococcus pneumoniae and mice. Smooth (S) strain bacteria have a capsule and cause pneumonia. Rough (R) strain bacteria have none and are harmless.
| Injection | Result |
|---|---|
| Live S | Mouse dies |
| Live R | Mouse lives |
| Heat killed S | Mouse lives |
| Heat killed S plus live R | Mouse dies, and live S bacteria are recovered |
Something from the dead S cells had transformed the living R cells into virulent S cells. Griffith called it the transforming principle but could not say what it was.
Avery, MacLeod and McCarty (1933 to 1944). They purified the biochemicals in the heat killed S cells and tested each. Only DNA caused transformation. Protein digesting enzymes (proteases) and RNA digesting enzymes (RNases) did not stop it, but a DNA digesting enzyme (DNase) did. They concluded DNA was the transforming material, though many scientists were still unconvinced.
Hershey and Chase, 1952. The experiment was designed around a chemical difference: DNA contains phosphorus but no sulphur, and proteins contain sulphur but no phosphorus. They grew T2 bacteriophages in two media, one with radioactive 32P (labelling DNA) and the other with radioactive 35S (labelling protein). They let the phages infect E. coli, then blended the mixture to shake the empty phage coats off the bacteria and centrifuged it so the heavy bacteria settled to the bottom.
The bacteria that received 32P phages carried radioactivity inside. In the 35S experiment, the radioactivity stayed in the supernatant with the coats. So DNA entered the cell, and the protein did not. DNA is the genetic material.
Pause and predict: in the Hershey and Chase experiment, which labelled molecule would you find in the pellet of infected bacteria if the phage's protein had been the genetic material?
Check your answer: 35S, the protein label. The pellet carries only what entered the bacteria, so the label that is found there shows which molecule carries the genetic message. Because 32P was found in the pellet, it is DNA.
Why DNA and not RNA?
A genetic material must be able to replicate, be chemically and structurally stable, allow slow changes (mutation) for evolution, and be able to express itself as characters. DNA scores well on stability. RNA's 2′ hydroxyl group makes it reactive, and DNA's thymine (rather than uracil) and double strands add further stability. RNA, however, can both carry information and act as a catalyst, so it mutates faster and is the genetic material of many viruses.
Beyond the textbook. For the exam: the RNA world idea says RNA was the first genetic material, and DNA evolved from it later because it is more stable. In reality: this is a leading hypothesis, not an observed fact. It rests on the discovery that some RNAs, called ribozymes, are catalysts. The bacterial 23S rRNA forms the peptide bonds during translation, for instance.
Replication: copying the code
Watson and Crick predicted that DNA replicates by a semi-conservative mechanism: the two strands separate, and each acts as a template for a new complementary strand. Each daughter DNA therefore has one old strand and one new strand.
Meselson and Stahl, 1958. They grew E. coli for many generations in a medium containing the heavy isotope 15N (as 15NH4Cl), so all its DNA became heavy. They then shifted the bacteria to a medium with ordinary 14N, and took samples every 20 minutes (one generation). The DNA was separated by density gradient centrifugation in caesium chloride.
| Time after shift | DNA density found | Meaning |
|---|---|---|
| 0 minutes | All heavy | Starting DNA, both strands 15N |
| 20 minutes (1 generation) | All hybrid | Each molecule has one old, one new strand |
| 40 minutes (2 generations) | Equal hybrid and light | Half the molecules have two new strands |
Had replication been conservative, the first generation would have shown separate heavy and light bands, not a single hybrid band. Taylor and colleagues (1958) showed the same semi-conservative pattern in the chromosomes of Vicia faba (broad bean) using radioactive thymidine.
Pause and predict: after three generations in 14N, what will be the ratio of hybrid to light DNA molecules?
Check your answer: 1 hybrid to 3 light. Eight molecules result from one original, and the two original heavy strands are always found in two hybrid molecules. The other six are fully light.
E. coli copies 4.6 × 10⁶ base pairs in about 18 minutes, which works out to roughly 2000 base pairs per second. This speed means that the copying and the cell division cycle need to be tightly coordinated, because a failure leads to unequal chromosome distribution.
How the cell does it.
- Replication begins at a specific site, the origin of replication, where the strands open into a Y shaped replication fork.
- The main enzyme is a DNA dependent DNA polymerase. It uses deoxynucleoside triphosphates, which are both the raw material and the energy source.
- DNA polymerase works only in the 5′ to 3′ direction. On the template read 3′ to 5′, the new strand is built continuously. On the other template, the new strand is built in short pieces (Okazaki fragments), discontinuously, and the pieces are joined by DNA ligase.
- In eukaryotes, replication takes place in the S phase of the cell cycle.
Transcription: DNA to RNA
Transcription copies one strand of DNA into RNA, following the same base pairing rules except that adenine pairs with uracil.
A transcription unit has three regions: a promoter, a structural gene and a terminator. The promoter lies at the 5′ end of the coding strand, upstream of the gene, and is where RNA polymerase binds. The terminator lies at the 3′ end, downstream.
| Template strand | Coding strand | |
|---|---|---|
| Polarity | 3′ to 5′ | 5′ to 3′ |
| Role | Read by RNA polymerase | Not transcribed |
| Sequence | Complementary to RNA | Same as RNA, with T in place of U |
Why only one strand? If both strands were transcribed, two complementary RNAs would form that could pair with each other into double stranded RNA, and the message would not be translated. The cell would also make two different proteins from one gene.
Pause and predict: a template strand reads 3′ TACGGA 5′. What is the mRNA?
Check your answer: 5′ AUGCCU 3′. The RNA is complementary and antiparallel to the template, with U in place of T. Notice it matches the coding strand, 5′ ATGCCT 3′, except for U.
A cistron is a segment of DNA that codes for one polypeptide.
| Prokaryotes | Eukaryotes | |
|---|---|---|
| Structural genes | Polycistronic (many cistrons on one mRNA), continuous | Monocistronic, split into exons and introns |
| RNA polymerase | One enzyme makes all RNAs | Three nuclear polymerases |
| Initiation and termination | Sigma factor and rho factor | Other protein factors |
| mRNA processing | None | Splicing, capping, tailing |
| Place | Transcription and translation in the same compartment | Transcription in nucleus, translation in cytoplasm |
Because bacterial mRNA needs no processing, translation can begin before transcription has finished.
| Eukaryotic polymerase | Makes |
|---|---|
| RNA polymerase I | rRNAs (28S, 18S, 5.8S) |
| RNA polymerase II | hnRNA, the precursor of mRNA |
| RNA polymerase III | tRNA, 5S rRNA and snRNAs |
Processing in eukaryotes. The primary transcript is heterogeneous nuclear RNA (hnRNA). It contains exons (the sequences that appear in mature RNA) and introns (intervening sequences that do not). Three steps turn it into mRNA. Splicing removes the introns and joins the exons. Capping adds methyl guanosine triphosphate to the 5′ end. Tailing adds 200 to 300 adenylate residues to the 3′ end, without a template.
The genetic code
Gamow argued that 4 bases must be read in groups of 3, because 4³ gives 64 codons, enough for 20 amino acids. Har Gobind Khorana synthesised RNAs of known sequence chemically. Marshall Nirenberg developed a cell free protein synthesis system, and Severo Ochoa's enzyme polynucleotide phosphorylase helped make RNA polymers. Together these work broke the code.
| Property | Meaning |
|---|---|
| Triplet | Three bases per codon, 64 codons |
| Degenerate | Most amino acids have more than one codon (61 codons code for 20 amino acids) |
| Unambiguous | One codon codes for only one amino acid |
| Contiguous | Read continuously with no punctuation |
| Nearly universal | The same across organisms, with exceptions in some mitochondria and protozoa |
| Start and stop | AUG is the start codon and codes for methionine; UAA, UAG and UGA are stop codons and code for no amino acid |
Mutations and the code. A point mutation changes one base, as in sickle cell anaemia, where one base change replaces glutamic acid with valine in the beta globin chain. If bases are inserted or deleted in numbers that are not multiples of three, the reading frame shifts and every codon downstream is changed. This is a frameshift mutation. Insertions or deletions of exactly three bases (or multiples) leave the frame intact and add or remove just one amino acid.
Translation: RNA to protein
tRNA, the adapter. Crick predicted an adapter molecule that reads the code. A tRNA has an anticodon loop that pairs with the mRNA codon and an amino acid acceptor end that carries the amino acid. There is a specific tRNA for every amino acid, plus a separate initiator tRNA, and no tRNA for stop codons. On paper the tRNA is drawn as a clover leaf, but in three dimensions it is an inverted L.
The mRNA also has untranslated regions (UTRs) at both ends, which are needed for efficient translation. A ribosome is made of rRNA and about 80 proteins, and it exists as separate large and small subunits when inactive.
Steps of translation.
- Charging. Each amino acid is activated using ATP and attached to its own tRNA (aminoacylation).
- Initiation. The small ribosomal subunit binds the mRNA at the start codon AUG, and then the large subunit joins.
- Elongation. The ribosome reads one codon at a time. Each incoming charged tRNA pairs through its anticodon, and the ribosome forms a peptide bond between the amino acids. This reaction is catalysed by 23S rRNA in bacteria, which acts as a ribozyme. The ribosome then moves along the mRNA.
- Termination. At a stop codon no tRNA can bind. A release factor binds instead, and the completed polypeptide is released.
Regulation of gene expression: the lac operon
A cell does not need every protein all the time. Eukaryotes control gene expression at several levels: transcription, processing, transport of mRNA to the cytoplasm, and translation. In bacteria, the main control point is the start of transcription. A cluster of genes under common control is an operon.
The lac operon, worked out by François Jacob and Jacques Monod, controls the use of lactose in E. coli. Lactose is a disaccharide, and the cell only needs the enzymes to digest it when lactose is present.
| Gene or element | Function |
|---|---|
| i gene (regulator) | Makes the repressor; i stands for inhibitor |
| Promoter (p) | RNA polymerase binds here |
| Operator (o) | The repressor binds here |
| z gene | Beta galactosidase, which splits lactose into glucose and galactose |
| y gene | Permease, which lets lactose into the cell |
| a gene | Transacetylase |
Without lactose, the repressor binds the operator and blocks RNA polymerase, so z, y and a are silent. With lactose, the lactose (or its isomer allolactose) acts as the inducer. It binds the repressor and inactivates it, the operator is free, and RNA polymerase transcribes all three genes as one polycistronic mRNA. This is negative regulation, because the regulator acts by turning genes off. Glucose and galactose are not inducers.
Why it makes sense. The cell saves energy by making the lactose enzymes only when their substrate is present. The substrate itself throws the switch.
Beyond the textbook. For the exam: lactose is the inducer, and the repressor is always made. In reality: if glucose is also available, E. coli still prefers glucose, and the operon stays low even when lactose is present. This second layer of control (catabolite repression) is positive regulation.
The Human Genome Project (HGP)
The HGP was a 13 year project, launched in 1990 and completed in 2003. It was coordinated by the US Department of Energy and the National Institutes of Health, with major contributions from the Wellcome Trust (UK) and partners in Japan, France, Germany and China. It created a need for storing and analysing huge amounts of data, and so helped to build bioinformatics. At the start, its cost was estimated at around 9 billion US dollars, at 3 US dollars per base pair for 3 × 10⁹ base pairs.
Goals included identifying all human genes, sequencing the 3 × 10⁹ bases, storing the data in databases, improving data analysis tools, transferring the technology and addressing the ethical, legal and social issues (ELSI).
Methods. One approach labelled the genes that are expressed as RNA, using Expressed Sequence Tags (ESTs). The other, sequence annotation, sequenced the whole genome and assigned functions later. The DNA was broken into fragments, cloned in bacteria or yeast using BAC and YAC (bacterial and yeast artificial chromosomes), and sequenced by automated machines based on Frederick Sanger's method. Chromosome 1 was the last to be completed, in May 2006.
| Finding | Value |
|---|---|
| Total bases | 3164.7 million base pairs |
| Genes | About 30,000 (earlier estimates were far higher) |
| Average gene | About 3000 bases |
| Largest known gene | Dystrophin, 2.4 million bases |
| Genome shared by any two people | 99.9 percent |
| Protein coding sequence | Less than 2 percent of the genome |
| Genes with unknown function | More than 50 percent |
| Most genes | Chromosome 1 (2968) |
| Fewest genes | Y chromosome (231) |
| SNPs identified | About 1.4 million (single nucleotide polymorphisms, "snips") |
Beyond the textbook. For the exam: about 30,000 genes. In reality: current counts put protein coding genes at roughly 20,000, and the "complete" 2003 sequence actually left gaps in repetitive regions, which were filled in only in 2022.
DNA fingerprinting
Everyone's DNA is 99.9 percent the same, so fingerprinting looks at the part that differs. About 0.1 percent of base differences make each person unique, and the technique finds those differences in repetitive DNA. Alec Jeffreys developed the method.
Most of the genome's bulk DNA forms one main peak on a density gradient. Smaller peaks of repetitive, non coding satellite DNA appear beside it. Satellite DNA is divided by base composition, length of segment and number of repeats into micro satellites, mini satellites and so on. It does not code for protein, but the number of repeats varies greatly between individuals. The probe used is a VNTR (Variable Number of Tandem Repeats), a mini satellite whose size varies from 0.1 to 20 kb. Because you inherit one allele from each parent, the repeat pattern is passed on following Mendel's rules.
The steps: isolate DNA → digest it with restriction enzymes → separate fragments by gel electrophoresis → transfer them to a nitrocellulose or nylon membrane (Southern blotting) → hybridise with a labelled VNTR probe → detect by autoradiography. The bands then form a pattern. PCR can be used to amplify small DNA samples.
The pattern is the same in every tissue of one person (blood, hair root, semen) and different between any two individuals, except identical twins. Its main uses are in forensic science, paternity disputes and studying genetic diversity of populations.
Exam traps
| The trap | Why it's tempting | The truth |
|---|---|---|
| "A=T has 3 hydrogen bonds." | Both pairs sound similar | A with T has 2; G with C has 3 |
| "The distance between base pairs is 3.4 nm." | You remember 3.4 | 3.4 nm is the pitch; adjacent base pairs are 0.34 nm apart |
| "The coding strand is the one transcribed." | The word "coding" sounds active | Template strand is read (3′ to 5′); the coding strand matches RNA but is not transcribed |
| "The i gene is the inducer." | It starts with i | i is the inhibitor (regulator) gene; the inducer is lactose or allolactose |
| "Glucose and galactose induce the lac operon." | They are the products of lactose digestion | Only lactose (allolactose) is the inducer |
| "RNA polymerase I makes mRNA." | I comes first | Pol II makes hnRNA. Pol I makes rRNA, Pol III makes tRNA |
| "Hershey and Chase used 14N and 15N." | Both involve isotopes | They used 32P and 35S. 15N is for Meselson and Stahl |
| "Chromosome X has the fewest genes." | X is the other sex chromosome | Y has the fewest (231) and chromosome 1 the most (2968) |
60-second revision
| Topic | Lock this in |
|---|---|
| Nucleotide | Base + sugar + phosphate · nucleoside = base + sugar |
| Helix | Antiparallel · A=T (2 bonds), G≡C (3) · pitch 3.4 nm · 10 bp/turn · 0.34 nm/bp |
| Nucleosome | Histone octamer + about 200 bp DNA |
| Griffith 1928 | Transforming principle (heat killed S + live R → mouse dies) |
| Avery et al. | DNase stops transformation → DNA |
| Hershey and Chase 1952 | 32P (DNA) enters · 35S (protein) stays outside |
| Meselson and Stahl 1958 | 15N → 14N · hybrid after 20 min · hybrid + light after 40 min |
| Replication | 5′ to 3′ only · leading continuous · lagging in fragments + ligase · about 2000 bp/s in E. coli |
| Transcription | Promoter + gene + terminator · template 3′ to 5′ · Pol I rRNA, II hnRNA, III tRNA |
| hnRNA processing | Splicing · 5′ capping · 3′ poly A tail (200 to 300 A) |
| Genetic code | Triplet · degenerate · unambiguous · AUG start · UAA UAG UGA stop |
| lac operon | i repressor · z beta galactosidase · y permease · a transacetylase · lactose inducer |
| HGP | 1990 to 2003 · 3164.7 million bp · about 30,000 genes · chr 1 most (2968), Y fewest (231) |
| DNA fingerprinting | Jeffreys · VNTR probe · Southern blot · autoradiography |
Where this chapter connects
- Previous: Principles of Inheritance and Variation gave Mendel's factors and the chromosome theory. This chapter shows what those factors are made of.
- Next: Evolution uses mutation and gene frequencies to explain how populations change.
- Class XII link: Biotechnology: Principles and Processes uses restriction digestion, gel electrophoresis and PCR, the same tools as DNA fingerprinting.
🎯 Practice MCQs
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Q1. In the B form of the DNA double helix, the distance between two adjacent base pairs is approximately:
▸ Show answer & explanation
Answer: B
One full turn (pitch) is 3.4 nm and contains about 10 base pairs, so each base pair adds 0.34 nm. Option A is the pitch, not the spacing between neighbouring base pairs.
Q2. A diploid human cell has about 6.6 × 10⁹ base pairs, and adjacent base pairs are 0.34 nm apart. The approximate length of its DNA is:
▸ Show answer & explanation
Answer: A
6.6 × 10⁹ multiplied by 0.34 × 10⁻⁹ metres gives about 2.2 metres. The other options are off by factors of 10 or 1000.
Q3. Which statement about a eukaryotic nucleosome is correct?
▸ Show answer & explanation
Answer: B
DNA is negatively charged because of its phosphates, and histones are rich in basic lysine and arginine, so they bind. Option A reverses the charges, C is wrong because bacteria use a nucleoid, and D is wrong because euchromatin is loosely packed and active.
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Q4. In the Hershey and Chase experiment, which observation showed that DNA, not protein, is the genetic material?
▸ Show answer & explanation
Answer: B
DNA contains phosphorus and protein contains sulphur. Only the phosphorus label entered the bacteria, so DNA carried the genetic message. Option A is reversed, and C and D do not match the result.
Q5. In Griffith's experiment with Streptococcus pneumoniae, mice died when injected with:
▸ Show answer & explanation
Answer: C
Live S bacteria were recovered from the dead mice, showing that the R cells had been transformed by something from the dead S cells. Heat killed S alone, live R alone and heat killed R alone did not kill the mice.
Q6. Match List I with List II. List I: (i) Griffith (ii) Avery, MacLeod and McCarty (iii) Hershey and Chase (iv) Meselson and Stahl. List II: (p) Semi conservative replication using 15N (q) Transforming principle (r) DNase stopped transformation, so DNA is the transforming principle (s) 32P and 35S labelled bacteriophages
▸ Show answer & explanation
Answer: A
Griffith described the transforming principle, Avery and colleagues identified it as DNA using enzymes, Hershey and Chase used labelled phages, and Meselson and Stahl used 15N to show semi conservative replication. The other options swap the experiments.
Q7. E. coli grown for many generations in 15N medium is shifted to 14N medium. After 40 minutes (two generations), the DNA contains:
▸ Show answer & explanation
Answer: C
After one generation all the DNA is hybrid. After the second generation half of the molecules are hybrid and half are fully light. Only heavy or only hybrid DNA would be seen at 0 and 20 minutes respectively, and the DNA never becomes only light so early.
Q8. E. coli with 4.6 × 10⁶ base pairs completes replication in about 18 minutes. The approximate rate of polymerisation is:
▸ Show answer & explanation
Answer: B
The average rate is about 2000 base pairs per second. The other values are off by factors of 10 or more.
Q9. Which enzyme joins the fragments made discontinuously on one template strand during DNA replication?
▸ Show answer & explanation
Answer: C
The strand built in short pieces is sealed together by DNA ligase. DNA polymerase makes the fragments, RNA polymerase makes RNA in transcription, and reverse transcriptase copies RNA into DNA in some viruses.
Q10. In eukaryotic cells, the precursor of mRNA, called hnRNA, is transcribed by:
▸ Show answer & explanation
Answer: B
RNA polymerase II transcribes hnRNA. Polymerase I makes the larger rRNAs (28S, 18S, 5.8S) and polymerase III makes tRNA, 5S rRNA and snRNAs. DNA polymerase copies DNA.
Q11. Assertion (A): In transcription, only one strand of the DNA acts as a template. Reason (R): If both strands were transcribed, the two RNA molecules would be complementary and could form double stranded RNA, which would block translation.
▸ Show answer & explanation
Answer: A
Transcription of both strands would produce complementary RNAs that pair and prevent the message from being read, which is why the cell transcribes one strand only. So R explains A.
Q12. In the processing of hnRNA in eukaryotes, capping refers to the addition of:
▸ Show answer & explanation
Answer: A
Capping adds methyl guanosine triphosphate to the 5′ end. Option B describes tailing, which is done at the 3′ end without a template, and C and D are not steps in hnRNA processing.
Q13. Which of the following codons does NOT act as a stop codon?
▸ Show answer & explanation
Answer: D
UAA, UAG and UGA are the three stop codons. UGG codes for the amino acid tryptophan.
Q14. Which statement about the genetic code is correct?
▸ Show answer & explanation
Answer: B
The code is read in triplets without overlap, several codons can code for one amino acid, one codon never codes for two amino acids, and it is nearly the same in all organisms. UAA is a stop codon, not tryptophan.
Q15. The codon AUG has a dual function. It:
▸ Show answer & explanation
Answer: B
AUG codes for methionine and also starts translation. The stop codons are UAA, UAG and UGA, so options A and D are wrong, and phenylalanine is coded by UUU and UUC, not AUG.
Q16. Deletion of exactly three consecutive bases within the coding region of a gene would most likely result in:
▸ Show answer & explanation
Answer: B
Three bases make one codon, so removing them shortens the protein by one amino acid but keeps the later codons in frame. A frameshift needs a change of one or two bases (not a multiple of three), and a stop codon is not guaranteed.
Q17. Which statement about tRNA is correct?
▸ Show answer & explanation
Answer: A
The anticodon loop reads the mRNA codon and the acceptor end binds the amino acid. There is no tRNA for stop codons. The clover leaf is only the two dimensional drawing and the real shape is an inverted L, and mRNA, not tRNA, carries the code.
Q18. In the lac operon, in the presence of lactose, the inducer:
▸ Show answer & explanation
Answer: C
The inducer (lactose or allolactose) binds the repressor, which can then no longer bind the operator, so RNA polymerase can transcribe the structural genes. The inducer does not bind RNA polymerase, the operator or the promoter directly.
Q19. Match the elements of the lac operon with their roles. List I: (i) i gene (ii) z gene (iii) y gene (iv) a gene. List II: (p) Permease (q) Beta galactosidase (r) Transacetylase (s) Repressor
▸ Show answer & explanation
Answer: A
The i gene codes for the repressor, z for beta galactosidase, y for permease and a for transacetylase. Options B, C and D interchange the products of these genes.
Q20. According to the Human Genome Project, which chromosome has the most genes and which has the fewest?
▸ Show answer & explanation
Answer: A
Chromosome 1 has 2968 genes and the Y chromosome has 231 genes. The other options reverse them or choose the wrong chromosomes.
Q21. Which of the following is NOT a finding of the Human Genome Project?
▸ Show answer & explanation
Answer: C
Less than 2 percent of the human genome codes for proteins. The other three statements are findings of the project.
Q22. Which sequence correctly lists the steps of DNA fingerprinting?
▸ Show answer & explanation
Answer: A
DNA must be isolated and cut before the fragments are separated, transferred to a membrane, probed with the labelled VNTR and finally detected by autoradiography. In the other sequences, a step comes before the material it needs.
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