Molecular Basis of Inheritance | CBSE Class 12 Biology Notes

This note covers the structure of DNA and the polynucleotide chain, the double helix and the central dogma, DNA packaging, the experiments that identified DNA as the genetic material, DNA versus RNA and the RNA world, replication, transcription and RNA processing, the genetic code and mutations, tRNA and translation, the lac operon, the Human Genome Project, and DNA fingerprinting.

What is DNA made of, and how is a polynucleotide chain built?

DNA (deoxyribonucleic acid) and RNA (ribonucleic acid) are the two nucleic acids of living systems; both are polymers of nucleotides. DNA is the genetic material in most organisms. RNA is the genetic material in some viruses but mostly works as a messenger, and also as an adapter, a structural molecule and sometimes a catalyst.

The length of DNA is usually given as its number of nucleotides or base pairs (bp), and it is characteristic of an organism.

How are nucleotides joined into a chain?

1. A nucleotide has a nitrogenous base, a pentose sugar (ribose in RNA, deoxyribose in DNA) and a phosphate group.
2. The base is linked to the OH of the 1′ carbon of the sugar by an N-glycosidic linkage, forming a nucleoside such as adenosine or deoxyadenosine.
3. A phosphate linked to the OH of the 5′ carbon by a phosphoester linkage makes a nucleotide (or deoxynucleotide).
4. Two nucleotides joined by a 3′-5′ phosphodiester linkage form a dinucleotide; more joined the same way form a polynucleotide chain.

The purines are adenine and guanine; the pyrimidines are cytosine, uracil and thymine. Cytosine occurs in both nucleic acids, thymine (5-methyl uracil) in DNA, and uracil in RNA in place of thymine.

A chain has a free phosphate on the 5′ carbon of the sugar at its 5′-end and a free 3′ OH at its 3′-end. Sugars and phosphates form the backbone, and the bases project from it. In RNA every nucleotide also has an -OH group at the 2′ position of ribose.

What are the features of the double helix model of DNA?

Friedrich Meischer first identified DNA as an acidic substance in the nucleus in 1869 and named it 'Nuclein'. In 1953, James Watson and Francis Crick proposed the Double Helix model, using X-ray diffraction data produced by Maurice Wilkins and Rosalind Franklin.

The model also used Chargaff's rule: Erwin Chargaff observed that in double-stranded DNA the ratios of adenine to thymine and of guanine to cytosine are constant and equal one. Base pairing makes the two strands complementary, so one sequence predicts the other.

1. Two polynucleotide chains form the helix, with a sugar-phosphate backbone and the bases projecting inside.
2. The chains have anti-parallel polarity: if one runs 5′→3′, the other runs 3′→5′.
3. Adenine forms two hydrogen bonds with thymine, and guanine forms three with cytosine. A purine always comes opposite a pyrimidine, keeping the distance between the strands approximately uniform.
4. The chains coil in a right-handed fashion. The pitch is 3.4 nm, with roughly 10 bp per turn, so base pairs are about 0.34 nm apart.
5. The plane of one base pair stacks over the next; this stacking, with the hydrogen bonds, stabilises the helix.

The upper strand runs 5′ to 3′ with bases A, T, G, C; the lower runs 3′ to 5′ with T, A, C, G. Two dashed hydrogen bonds join each A and T, and three join each G and C.

See Fig. 5.2 in your NCERT textbook

What is the central dogma?

Francis Crick then proposed the central dogma: genetic information flows from DNA → RNA → protein. DNA copies itself by replication, makes mRNA by transcription, and mRNA makes protein by translation. In some viruses the flow is reversed, from RNA to DNA, a process called reverse transcription.

How is the long DNA molecule packaged in a cell?

With 0.34 × 10⁻⁹ m between base pairs, the DNA of a typical mammalian cell is 6.6 × 10⁹ bp × 0.34 × 10⁻⁹ m/bp, about 2.2 metres long. A typical nucleus is only about 10⁻⁶ m across.

Worked example 1. If E. coli DNA is 1.36 mm long, how many base pairs does it have?

Answer: Number of base pairs = (1.36 × 10⁻³ m) ÷ (0.34 × 10⁻⁹ m per bp) = 4 × 10⁶ bp, of the same order as the 4.6 × 10⁶ bp given for E. coli in the table.

Prokaryotes such as E. coli lack a defined nucleus, but their negatively charged DNA is held with positively charged proteins in a region called the nucleoid, organised in large loops held by proteins.

Eukaryotes use histones, positively charged basic proteins rich in lysine and arginine, whose side chains carry positive charges. Packaging proceeds as follows.

1. Eight histone molecules form a histone octamer.
2. Negatively charged DNA wraps around the octamer to form a nucleosome, typically containing 200 bp of DNA.
3. Nucleosomes are the repeating unit of chromatin and appear as 'beads-on-string' under the electron microscope.
4. This structure is packaged into chromatin fibres, which coil and condense at metaphase to form chromosomes, using non-histone chromosomal (NHC) proteins.

DNA is wrapped around a large sphere labelled histone octamer, the core of histone molecules. A separate rod labelled H1 histone lies against the DNA on one side.

See Fig. 5.4a in your NCERT textbook

How did experiments prove that DNA is the genetic material?

By 1926, work by Gregor Mendel, Walter Sutton, Thomas Hunt Morgan and others had narrowed the search to the chromosomes, but the molecule itself was unknown.

What was Griffith's transforming principle?

In 1928, Frederick Griffith worked with Streptococcus pneumoniae, which causes pneumonia. The S strain forms smooth, shiny colonies because it has a mucous (polysaccharide) coat; the R strain forms rough colonies and has none. He killed bacteria by heating them.

A 'transforming principle' from the heat-killed S cells had enabled R cells to make the polysaccharide coat and become virulent. This had to be genetic material, but its biochemical nature was not defined.

What did Avery, MacLeod and McCarty show?

Until their work (1933 to 1944), the genetic material was thought to be a protein. Oswald Avery, Colin MacLeod and Maclyn McCarty purified proteins, DNA and RNA from heat-killed S cells and found that DNA alone transformed R cells.

Proteases and RNases did not affect transformation, but DNase inhibited it. They concluded that DNA is the hereditary material, though not all biologists were convinced.

How did Hershey and Chase settle the question?

The unequivocal proof came from Alfred Hershey and Martha Chase in 1952, using bacteriophages, viruses that infect bacteria.

1. Phages grown with radioactive phosphorus got radioactive DNA, since DNA contains phosphorus and protein does not. Phages grown with radioactive sulfur got radioactive protein, since DNA contains no sulfur.
2. The radioactive phages were allowed to infect E. coli.
3. The viral coats were removed from the bacteria by agitating them in a blender.
4. The virus particles were separated from the bacteria by centrifugation.

Only DNA passed from virus to bacterium, so DNA is the genetic material.

Two columns start with phages labelled ³⁵S (protein capsule) and ³²P (DNA), and pass through 1. Infection, 2. Blending and 3. Centrifugation. ³⁵S ends in the supernatant; ³²P ends in the cells.

See Fig. 5.5 in your NCERT textbook

Why is DNA a better genetic material than RNA, and what is the RNA world?

In some viruses, such as Tobacco Mosaic Virus and QB bacteriophage, RNA is the genetic material. A genetic material must meet four criteria.

1. It should generate its replica (replication).
2. It should be chemically and structurally stable.
3. It should allow slow changes (mutation) needed for evolution.
4. It should express itself as 'Mendelian characters'.

Proteins fail the first criterion. Stability showed in Griffith's work: heat killed the bacteria but did not destroy some properties of the genetic material, and separated DNA strands come together again under suitable conditions.

What is the RNA world?

RNA was the first genetic material. Essential processes such as metabolism, translation and splicing evolved around RNA, which acted as both genetic material and catalyst. Some important biochemical reactions in living systems are catalysed by RNA catalysts, not by protein enzymes.

Being reactive, RNA was unstable, so DNA evolved from it with chemical modifications for stability. Double-stranded DNA, with its complementary strand, further resists change by evolving repair.

How does DNA replicate?

Watson and Crick noted that their base pairing immediately suggested a copying mechanism. The two strands separate and each templates a new complementary strand, so each daughter molecule has one parental and one new strand: semiconservative replication.

How did Meselson and Stahl prove it?

1. In 1958, Matthew Meselson and Franklin Stahl grew E. coli for many generations on ¹⁵NH₄Cl as the only nitrogen source, so ¹⁵N entered its DNA.
2. Heavy DNA could be told from normal DNA by centrifugation in a caesium chloride (CsCl) density gradient. ¹⁵N is not radioactive.
3. The cells were moved to normal ¹⁴NH₄Cl, and DNA samples were taken at definite intervals and run on CsCl gradients.
4. After one generation (20 minutes, as E. coli divides in 20 minutes) the DNA had a hybrid density.
5. After 40 minutes (generation II) there were equal amounts of hybrid and light DNA.

One ¹⁵N helix gives two hybrid helices at 20 minutes and four helices at 40 minutes, two hybrid and two light. Centrifuge tubes show a heavy band, then a hybrid band, then light and hybrid bands.

See Fig. 5.7 in your NCERT textbook

Worked example 2. What proportion of DNA is hybrid after 80 minutes?

Answer: 80 minutes is four generations, giving 16 molecules. The two heavy parental strands sit in two hybrid molecules, so 2 are hybrid and 14 light: 12.5 per cent hybrid.

In 1958, Taylor and colleagues used radioactive thymidine on Vicia faba (faba beans) to show that chromosomal DNA also replicates semiconservatively.

Which enzymes and conditions does replication need?

The main enzyme is DNA-dependent DNA polymerase. E. coli (4.6 × 10⁶ bp) replicates within 18 minutes, about 2000 bp per second, and errors cause mutations. Deoxyribonucleoside triphosphates act as substrates and supply energy through their two terminal high-energy phosphates.

1. Replication starts at a definite origin of replication, since polymerases cannot initiate on their own.
2. Separating the whole molecule needs too much energy, so replication occurs in a small opening, the replication fork.
3. The polymerase works only 5′→3′, so on the 3′→5′ template synthesis is continuous.
4. On the 5′→3′ template it is discontinuous, and the fragments are joined by DNA ligase.

Two parental template strands separate in a Y. One arm shows a long arrow of continuous synthesis; the other shows short arrows of discontinuous synthesis.

See Fig. 5.8 in your NCERT textbook

Because an origin is needed, DNA propagated in recombinant DNA work needs a vector. In eukaryotes replication occurs in the S-phase; if cell division fails after it, polyploidy results.

What is transcription, and how is a transcription unit organised?

Transcription copies genetic information from one DNA strand into RNA, with adenine pairing with uracil instead of thymine. Unlike replication, only a segment of DNA and only one strand is copied.

Copying both strands would give two RNAs of different sequence, so one segment would code for two proteins. The two RNAs would also be complementary, forming double-stranded RNA that could not be translated.

A transcription unit has a promoter, a structural gene and a terminator. RNA polymerase also works only 5′→3′, so the 3′→5′ strand is the template strand. The other, the coding strand, codes for nothing but matches the RNA sequence.

Worked example 3. The coding strand is 5′-TACGTACGTACGTACGTACGTACG-3′. Write the RNA.

Answer: 5′-UACGUACGUACGUACGUACGUACG-3′, the coding sequence with U for T.

The promoter lies upstream, towards the 5′-end of the coding strand, and binds RNA polymerase; its position defines which strand is template. The terminator lies downstream, towards the 3′-end, and usually ends transcription.

The template strand runs 3′ to 5′ above the coding strand. A promoter box and transcription start site lie at the left, the structural gene in the middle, and a terminator box at the right.

See Fig. 5.9 in your NCERT textbook

How does a transcription unit relate to a gene?

A gene is the functional unit of inheritance. A cistron is a DNA segment coding for a polypeptide; structural genes are monocistronic mostly in eukaryotes and polycistronic mostly in prokaryotes.

Eukaryotic genes are split. Exons appear in mature RNA; introns interrupt them and do not. Regulatory sequences are loosely called regulatory genes, though they code for no RNA or protein.

How does transcription take place in bacteria and in eukaryotes?

Bacteria have three major RNAs: mRNA provides the template, tRNA brings amino acids and reads the code, and rRNA plays structural and catalytic roles. One RNA polymerase transcribes them all.

1. Initiation: RNA polymerase, with the initiation factor σ, binds the promoter.
2. Using nucleoside triphosphates, it polymerises RNA in a template-dependent way by complementarity.
3. Elongation: it opens the helix and continues; only a short stretch of RNA stays bound.
4. Termination: at the terminator, with the termination factor ρ, the RNA and polymerase fall off.

The polymerase itself catalyses only elongation; σ and ρ alter its specificity. Bacterial mRNA needs no processing, and cytosol and nucleus are not separated, so translation can often begin before the mRNA is fully transcribed; the two are coupled.

What extra complexities do eukaryotes have?

Eukaryotes have at least three RNA polymerases in the nucleus, in addition to the one found in organelles, with a clear division of labour.

The primary transcript contains exons and introns and is non-functional, so it is processed.

1. Splicing: introns are removed and exons joined in a defined order.
2. Capping: methyl guanosine triphosphate is added to the 5′-end of hnRNA.
3. Tailing: 200 to 300 adenylate residues are added at the 3′-end without a template.
4. The processed hnRNA, now mRNA, leaves the nucleus for translation.

A transcript of exons and introns leaves the DNA. It is shown being capped at the 5′-end, spliced to remove introns, and polyadenylated with a poly A tail at the 3′-end, giving messenger RNA.

See Fig. 5.11 in your NCERT textbook

Split genes are probably an ancient feature, and splicing reflects the dominance of the RNA world.

What is the genetic code, and how do mutations affect it?

No complementarity exists between nucleotides and amino acids, so a genetic code must direct the amino acid sequence. Physicists, organic chemists, biochemists and geneticists all helped decipher it.

• The codon is a triplet: 61 codons code for amino acids and 3 are stop codons (UAA, UAG, UGA).
• The code is degenerate: some amino acids have more than one codon.
• It is read contiguously, without punctuation.
• It is nearly universal; UUU codes for phenylalanine from bacteria to humans, with exceptions in mitochondria and some protozoans.
• AUG codes for methionine and is the initiator codon.

Worked example 4. Translate -AUG UUU UUC UUC UUU UUU UUC-.

Answer: Met-Phe-Phe-Phe-Phe-Phe-Phe. The reverse cannot be predicted uniquely, because Phe has two codons, UUU and UUC.

How do point and frameshift mutations change the message?

A point mutation, a single base pair change in the beta globin gene, replaces glutamate with valine and causes sickle cell anaemia. Inserting or deleting bases is shown with a sentence read in triplets.

Inserting or deleting one or two bases changes the reading frame from that point, causing frameshift mutations. Three bases, or multiples of three, add or remove whole codons, leaving the frame unaltered.

How does tRNA read the code, and how is a protein made?

Francis Crick postulated an adapter molecule to read the code and bind amino acids. tRNA, once called sRNA (soluble RNA), was later given this role.

tRNA has an anticodon loop complementary to the code and an amino acid acceptor end. Each amino acid has specific tRNAs, an initiator tRNA starts translation, and no tRNAs exist for stop codons. Drawn as a clover-leaf, the real molecule is an inverted L.

Two tRNAs, each drawn with an anticodon loop at the bottom and an amino acid at the 3′ end, sit on an mRNA. One carries Ser, its anticodon UCA pairing with codon AGU; the other carries Tyr, its anticodon AUG pairing with codon UAC.

See Fig. 5.12 in your NCERT textbook

How does translation take place?

Translation joins amino acids by peptide bonds in the order set by mRNA. Amino acids are first activated with ATP and linked to their tRNA: aminoacylation, or charging of tRNA.

The ribosome has structural RNAs and about 80 proteins, in a large and a small subunit. The large subunit has two sites for adjacent amino acids, and in bacteria 23S rRNA catalyses peptide bonds as a ribozyme.

The translational unit runs from the start codon (AUG) to the stop codon. Untranslated regions (UTRs) at the 5′-end (before the start codon) and the 3′-end (after the stop codon) are required for efficient translation.

1. Translation begins when the small subunit encounters an mRNA.
2. Initiation: the ribosome binds the start codon, recognised only by the initiator tRNA.
3. Elongation: charged tRNAs pair with codons by their anticodons, and the ribosome moves codon to codon, adding amino acids.
4. Termination: a release factor binds the stop codon and the polypeptide is released.

A ribosome on mRNA holds two tRNAs carrying Ala and Val, paired with codons GCA and GUU. A growing polypeptide chain hangs from Ala; an empty tRNA leaves on the left and a tRNA carrying Asn arrives on the right.

See Fig. 5.13 in your NCERT textbook

How is gene expression regulated, and how does the lac operon work?

In eukaryotes, regulation can act at the transcriptional level, the processing level (splicing), the transport of mRNA to the cytoplasm, and the translational level. Metabolic, physiological or environmental conditions decide which genes are expressed.

In prokaryotes, the rate of transcriptional initiation is the main control point. Accessory proteins act as activators or repressors. In most operons an operator lies next to the promoter and, in most cases, binds a repressor protein. The lac operator occurs only in the lac operon and binds only the lac repressor.

What is the lac operon?

Francois Jacob, a geneticist, and Jacque Monod, a biochemist, first elucidated this transcriptionally regulated system. An operon is a polycistronic structural gene regulated by a common promoter and regulatory genes; examples include the lac, trp, ara, his and val operons.

Lactose, the substrate of beta-galactosidase, switches the operon on and off, so it is the inducer.

1. When a preferred carbon source such as glucose is absent and lactose is provided, permease carries lactose into the cell. A very low level of lac operon expression is present all the time, so some permease is available.
2. The repressor is made constitutively from the i gene.
3. Without inducer, the repressor binds the operator and blocks RNA polymerase.
4. Lactose or allolactose binds and inactivates the repressor.
5. RNA polymerase reaches the promoter, and the z, y and a genes are transcribed.

This is regulation of enzyme synthesis by its substrate. Glucose and galactose cannot induce the operon. Control by the repressor is negative regulation; the lac operon is also under positive regulation.

Upper panel, no inducer: p, i, p, o, z, y, a in a row; the repressor binds o and blocks transcription. Lower panel: inducer inactivates the repressor, and lac mRNA gives beta-galactosidase, permease and transacetylase.

See Fig. 5.14 in your NCERT textbook

What was the Human Genome Project, and what did it find?

The Human Genome Project (HGP) was launched in 1990. It was a mega project: about 3 × 10⁹ bp at an estimated US $3 per bp would cost about 9 billion US dollars, and the sequence of one cell (taking the haploid content of 3.3 × 10⁹ bp) would fill 3300 books of 1000 pages with 1000 letters each. It was closely associated with the rapid development of bioinformatics. Its important goals were as follows.

1. Identify all the approximately 20,000 to 25,000 human genes.
2. Sequence the 3 billion chemical base pairs of human DNA.
3. Store the information in databases and improve tools for data analysis.
4. Transfer related technologies to other sectors, such as industries.
5. Address the ethical, legal and social issues (ELSI).

The 13-year project was coordinated by the U.S. Department of Energy and the National Institute of Health, with the Wellcome Trust (U.K.) as a major partner and contributions from Japan, France, Germany, China and others. It was completed in 2003.

Methods included Expressed Sequence Tags (ESTs) for expressed genes and whole-genome sequencing followed by Sequence Annotation. Random DNA fragments were cloned in bacteria and yeast using BAC and YAC vectors, sequenced by Frederick Sanger's method, and aligned by computer programs. Chromosome 1, the last of the 24 human chromosomes (22 autosomes, X and Y), was completed in May 2006.

Repeated sequences make up a very large portion of the genome. SNPs (single nucleotide polymorphisms) promise to help locate disease-associated sequences and trace human history. Model organisms sequenced include bacteria, yeast, Caenorhabditis elegans, Drosophila, rice and Arabidopsis.

How does DNA fingerprinting work?

DNA fingerprinting is a quick way to compare the DNA of individuals without sequencing it. It targets repetitive DNA, where a short stretch repeats many times. In density gradient centrifugation, bulk DNA forms a major peak and repetitive DNA forms small peaks called satellite DNA.

By base composition, length of segment and number of repetitive units, satellite DNA is classed as micro-satellites, mini-satellites and so on. These sequences normally do not code for proteins and are highly polymorphic.

DNA from every tissue of an individual, such as blood, hair follicle, skin, bone, saliva and sperm, shows the same polymorphism, which makes it useful in forensics. Because polymorphisms are inherited, it is also the basis of paternity testing.

Polymorphism, variation at the genetic level, arises from mutation. Allelic variation is called a DNA polymorphism if more than one variant (allele) at a locus occurs in the human population with a frequency greater than 0.01.

Such variation is more likely in non-coding DNA, where mutations may have no immediate effect on reproductive ability, and it accumulates generation after generation.

Alec Jeffreys developed the technique using Variable Number of Tandem Repeats (VNTR) as a probe in Southern blot hybridisation.

1. Isolation of DNA.
2. Digestion by restriction endonucleases.
3. Separation of fragments by electrophoresis.
4. Blotting onto nitrocellulose or nylon membranes.
5. Hybridisation with a labelled VNTR probe.
6. Detection by autoradiography.

VNTR is a mini-satellite whose copy number varies between chromosomes, so it ranges from 0.1 to 20 kb. The band pattern differs from individual to individual, except in monozygotic (identical) twins. PCR has increased sensitivity, so DNA from a single cell is enough. The technique is also used to study population and genetic diversities.

Chromosomes 7, 2 and 16 of individuals A and B, and crime-scene DNA C, carry different repeat numbers. On the gel, C's bands match B, not A.

See Fig. 5.16 in your NCERT textbook

Glossary

• Nucleoside — A nitrogenous base linked to the 1′ carbon of a pentose sugar by an N-glycosidic linkage, such as adenosine.
• Nucleosome — Typically 200 bp of DNA wrapped around a histone octamer; the repeating unit of chromatin.
• Heterochromatin — Densely packed chromatin that stains dark and is transcriptionally inactive.
• Transforming principle — The substance from heat-killed S pneumococci that made R bacteria virulent; later shown to be DNA.
• Semiconservative replication — Replication in which each daughter DNA molecule has one parental and one new strand.
• Replication fork — The small opening of the DNA helix within which replication occurs.
• Template strand — The 3′→5′ DNA strand of a transcription unit that is copied into RNA.
• Cistron — A segment of DNA that codes for a single polypeptide.
• Exon — A coding or expressed sequence of a eukaryotic gene that appears in mature or processed RNA.
• Intron — An intervening sequence in a eukaryotic gene that is removed and does not appear in mature RNA.
• Codon — A triplet of mRNA bases coding for an amino acid or a stop signal.
• Ribozyme — An RNA acting as an enzyme, such as bacterial 23S rRNA forming peptide bonds.
• Operon — A polycistronic structural gene regulated by a common promoter and regulatory genes, common in bacteria.
• SNP — Single nucleotide polymorphism; a location of single-base DNA difference, of which about 1.4 million were found in humans.
• VNTR — Variable Number of Tandem Repeats; a highly polymorphic mini-satellite used as a DNA fingerprinting probe.

Common errors and misconceptions

• Misconception: The coding strand is copied into RNA. Correct: The template strand is copied; the coding strand matches the RNA (T for U) and is displaced.
• Misconception: Meselson and Stahl used radioactive nitrogen. Correct: ¹⁵N is heavy but not radioactive; it is separated by density.
• Misconception: Radioactive sulfur labelled phage DNA. Correct: Sulfur labelled protein and phosphorus labelled DNA.
• Misconception: The i gene is named after the inducer. Correct: i comes from inhibitor; it codes for the repressor.
• Misconception: Glucose can induce the lac operon. Correct: Lactose or allolactose induce it; glucose and galactose cannot.
• Misconception: RNA polymerase alone initiates and terminates transcription. Correct: It only catalyses elongation; it needs σ to initiate and ρ to terminate.
• Misconception: Every person has a unique DNA fingerprint. Correct: Monozygotic twins share the same pattern.

Exam-style questions with model answers

Q1. A double-stranded DNA has 20 per cent cytosine. Calculate the percentage of adenine. [1 mark]

1. G pairs with C, so G = C = 20 per cent, totalling 40 per cent. A and T share the remaining 60 per cent, and since A = T, adenine is 30 per cent.

Q2. The coding strand is 5′-ATGCATGCATGCATGCATGCATGCATGC-3′. Write the mRNA. [2 marks]

1. The mRNA is complementary to the template strand, so it has the same sequence and the same 5′→3′ polarity as the coding strand.
2. In RNA, uracil replaces thymine, so every T becomes U: mRNA 5′-AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3′.

Q3. How did Hershey and Chase distinguish DNA from protein? [3 marks]

1. They grew phages on radioactive phosphorus, labelling DNA (protein has no phosphorus), and others on radioactive sulfur, labelling protein (DNA has no sulfur).
2. After infecting E. coli, blending and centrifuging, only bacteria infected by phages with radioactive DNA were radioactive.
3. So DNA, not protein, entered the bacteria and is the genetic material.

Q4. Lactose was added to an E. coli culture and induced the lac operon. Why does the operon shut down later? [3 marks]

1. Lactose acts as the inducer. It binds and inactivates the repressor, so RNA polymerase can transcribe the z, y and a genes.
2. The beta-galactosidase produced hydrolyses lactose into glucose and galactose, which cannot act as inducers. As the lactose is used up, the inducer disappears.
3. The repressor, which is made constitutively from the i gene, is no longer inactivated. It binds the operator again, blocks RNA polymerase and the operon shuts down. This is regulation of enzyme synthesis by its substrate.

Q5. State four features of the genetic code. [4 marks]

1. It is a triplet code: 61 codons code for amino acids and 3 are stop codons.
2. It is degenerate, as some amino acids have several codons.
3. It is read contiguously, without punctuation.
4. It is nearly universal, with exceptions in mitochondria and some protozoans; AUG codes for methionine and also starts translation.

Q6. Describe the Meselson and Stahl experiment and its conclusion. [5 marks]

1. In 1958 they grew E. coli for many generations on ¹⁵NH₄Cl, the only nitrogen source, so its DNA became heavy; ¹⁵N is not radioactive.
2. Heavy DNA was distinguished from normal DNA by CsCl density gradient centrifugation.
3. The cells were moved to ¹⁴NH₄Cl, and DNA samples were taken at definite intervals.
4. After one generation (20 minutes) all DNA was hybrid; after two (40 minutes) there were equal amounts of hybrid and light DNA.
5. A hybrid band after one generation means each daughter molecule had one old heavy strand and one new light strand. After the second generation, the hybrid molecules give one hybrid and one light molecule each. This proves semiconservative replication, as proposed by Watson and Crick.

Q7. Explain the lac operon with a labelled diagram. [5 marks]

1. Draw the genes p, i, p, o, z, y, a in two panels, with and without inducer. The lac operon has one regulatory gene (i) and three structural genes (z, y and a), worked out by Jacob and Monod.
2. The i gene codes for the repressor. The z gene codes for beta-galactosidase, which hydrolyses lactose into galactose and glucose; y codes for permease, which increases permeability to β-galactosides; a codes for transacetylase.
3. Without inducer, the constitutively made repressor binds the operator and prevents RNA polymerase from transcribing the operon.
4. A very low level of expression lets permease bring lactose in. Lactose or allolactose, the inducer, binds and inactivates the repressor.
5. RNA polymerase then reaches the promoter and transcribes lac mRNA, which is translated into the three enzymes. The substrate thus regulates its own enzyme synthesis; this repressor control is negative regulation.

Q8. What is DNA fingerprinting? List its steps and two applications. [5 marks]

1. DNA fingerprinting is a quick way to compare the DNA of individuals by identifying differences in repetitive DNA, which is highly polymorphic. Alec Jeffreys developed it using a VNTR probe.
2. Steps: isolate the DNA; digest it with restriction endonucleases; separate the fragments by electrophoresis; blot them onto nitrocellulose or nylon membranes; hybridise with a labelled VNTR probe; detect the hybridised fragments by autoradiography.
3. VNTR copy numbers vary between chromosomes, so the autoradiogram shows bands of differing sizes. The pattern is characteristic of each person, except monozygotic twins.
4. Applications: forensic science, since DNA from blood, hair follicle, skin, bone, saliva or sperm shows the same polymorphism; and paternity testing, since polymorphisms are inherited.

Key takeaways

• DNA is a right-handed double helix of two anti-parallel chains, with A pairing with T by two hydrogen bonds and G with C by three.
• About 2.2 metres of mammalian DNA is packed around histone octamers into nucleosomes, chromatin fibres and chromosomes.
• Griffith, then Avery, MacLeod and McCarty, and finally Hershey and Chase showed that DNA, not protein, is the genetic material.
• DNA is more stable and stores information; RNA, the first genetic material, is better at transmitting it.
• Meselson and Stahl proved semiconservative replication; DNA polymerase works only 5′→3′, giving continuous and discontinuous synthesis.
• Eukaryotic hnRNA is spliced, capped and tailed into mRNA, while bacterial transcription and translation can be coupled.
• The genetic code is triplet, degenerate, contiguous and nearly universal, with AUG as start and UAA, UAG and UGA as stop codons.
• In the lac operon, lactose inactivates the repressor so the z, y and a genes are transcribed.
• The Human Genome Project found 3164.7 million bp and about 30,000 genes, and DNA fingerprinting uses VNTR polymorphism to identify individuals.

Test yourself

Who first identified DNA, and what did he call it?

Friedrich Meischer identified it in the nucleus in 1869 and named it 'Nuclein'.

Why are histones positively charged?

They are rich in lysine and arginine, which carry positive charges on their side chains.

Which enzyme stopped transformation in Avery's work?

DNase stopped it, while proteases and RNases did not.

Which RNA polymerase transcribes hnRNA?

RNA polymerase II transcribes hnRNA, the precursor of mRNA, in eukaryotes.

What is the real shape of tRNA?

Drawn as a clover-leaf, tRNA is actually a compact molecule shaped like an inverted L.

Which factors help bacterial RNA polymerase to start and stop transcription?

The initiation factor sigma (σ) helps it initiate, and the termination factor rho (ρ) helps it terminate.

Why does a mammalian cell need to package its DNA?

Its DNA is about 2.2 metres long, far greater than a typical nucleus of about 10⁻⁶ m.

How many genes does the human genome have?

The total number of genes is estimated at about 30,000.