Molecular Basis of Inheritance: Class 12 Biology · Chapter 5
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What you will learn in Molecular Basis of Inheritance
- Describe how nucleotides build a polynucleotide chain and how two antiparallel chains pair into the double helix, and use base pairing to find base percentages and partner strands
- Explain how DNA is packed into the nucleoid of a bacterium and, through nucleosomes and chromatin, into eukaryotic chromosomes
- Trace how the experiments of Griffith, of Avery, MacLeod and McCarty, and of Hershey and Chase identified DNA as the genetic material, and compare DNA with RNA as a store of information
- Explain semiconservative replication, how the Meselson–Stahl experiment proved it, and why one new strand is made continuously and the other in pieces
- Describe a transcription unit, the template and coding strands, split genes and the steps of transcription and RNA processing in bacteria and eukaryotes
- Use the genetic code to translate an mRNA, state its main features, predict the effect of point and frameshift mutations, and describe tRNA and the steps of translation
- Explain negative regulation of the lac operon by its repressor and inducer
- Summarise the aims, methods and main findings of the Human Genome Project, and explain the principle, steps and uses of DNA fingerprinting
Original companion text aligned to NCERT Biology Class 12, Chapter 5, pp. 79–109. Not affiliated with or endorsed by NCERT.
Scrape the inside of your cheek with a clean toothpick and you collect a few hundred cells. Each one carries a full copy of your instructions. Stretched end to end, the DNA in a single one of those cells would run about two metres, yet it sits inside a nucleus only a few micrometres wide, and it is copied, read and switched on and off with remarkable precision.
In the previous chapter you followed Mendel's 'factors' from parents to children without asking what they are made of. Here we open the factor up. You will see how DNA is built and packed, how scientists proved that it, and not protein, carries heredity, and why RNA is the likely ancestor of both. Then we follow the information as it is copied (replication), rewritten into RNA (transcription) and decoded into protein (translation), and see how a bacterium decides when a gene should work. The chapter ends with two big applications: reading the whole human genome, and DNA fingerprinting. In Hindi the chapter is वंशागति का आणविक आधार.
5.1 The DNA
How do you measure a molecule that can be a metre long? By counting its building blocks. DNA, a polymer of deoxyribonucleotides, is sized in nucleotides when it is single-stranded and in base pairs (bp) when it is double-stranded. Each species has a characteristic genome size, and the range is huge: a few thousand units in small viruses, billions in mammals (Table 5.1).
| Organism | Size | Stretched |
|---|---|---|
| Phage φX174 | 5386 nucleotides (one strand) | — |
| Phage lambda | 48 502 bp | 16 µm |
| E. coli | bp | 1.6 mm |
| Baker's yeast | bp | 4 mm |
| Human (haploid set) | bp | 1.1 m |
5.1.1 Structure of Polynucleotide Chain
You met the building block in Class XI. Every nucleotide combines a phosphate group, a five-carbon (pentose) sugar and a nitrogenous base. Two details tell DNA and RNA apart at this level. DNA's sugar is deoxyribose and RNA's is ribose; and while both use adenine (A), guanine (G) and cytosine (C), DNA uses thymine (T) where RNA uses uracil (U). By shape the bases fall into two families: purines (A and G) are built on a double ring, pyrimidines (C, T and U) on a single ring.
The parts join in a fixed way (Fig. 5.1). The base attaches to carbon 1′ of the sugar through an N-glycosidic bond, giving a nucleoside. A phosphate then links to the OH on carbon 5′ through a phosphoester bond, giving a nucleotide. To grow a chain, the phosphate of one nucleotide bonds to the 3′ OH of the sugar in the next, a 3′–5′ phosphodiester bond. Two nucleotides make a dinucleotide; hundreds or millions make a polynucleotide.
4/4Phosphodiester bonds link sugar to sugar; the chain has a free phosphate at the 5′ end and a free OH at the 3′ end.
Step 1 shows the three separate parts: a phosphate circle marked P, a pentagon labelled sugar with carbons 5′, 3′ and 1′ marked, and a box for the base adenine. Step 2 joins the base to carbon 1′ through an N-glycosidic bond, and a bracket names the pair a nucleoside. Step 3 joins the phosphate to carbon 5′ through a phosphoester bond, and the bracket now reads nucleotide. Step 4 replaces the single unit with a chain of three nucleotides: phosphate and sugar alternate along a backbone, bases A, C and G hang below, a free phosphate marks the 5′ end on the left, a free OH marks the 3′ end on the right, and a label points to a 3′–5′ phosphodiester bond.
The repeating unit of DNA and RNA: a nitrogenous base joined to a pentose sugar (the nucleoside), with a phosphate attached at the sugar's 5′ carbon.
ExampleDeoxyadenosine with a 5′ phosphate is a deoxyadenosine nucleotide of DNA.
| Base | In RNA | In DNA |
|---|---|---|
| Adenine | adenosine | deoxyadenosine |
| Guanine | guanosine | deoxyguanosine |
| Cytosine | cytidine | deoxycytidine |
| Uracil | uridine | — |
| Thymine | — | deoxythymidine |
Because every link runs from a 3′ carbon to the next 5′ carbon, a chain has direction. One end carries a free phosphate on a 5′ carbon (the 5′ end); the other ends in a free OH on a 3′ carbon (the 3′ end). Sugars and phosphates form the backbone, and the bases stick out from it. RNA differs in two details: every ribose carries an extra OH at carbon 2′, and uracil stands where DNA has thymine (thymine is simply 5-methyluracil).
Base pairing is what made the model famous. Because A can face only T, and G only C, each strand carries the complete information of its partner: the two strands are complementary. That at once suggested how DNA could be inherited. Separate the strands, rebuild a partner on each by the same rules, and you get two molecules identical to the original (Section 5.4).
- Two chains. Two polynucleotide chains wind around one axis, with the sugar–phosphate backbones outside and the bases inside.
- Antiparallel. The chains run in opposite directions: if one runs 5′→3′, its partner runs 3′→5′.
- Specific pairs. A pairs with T through two hydrogen bonds; G pairs with C through three. A purine always faces a pyrimidine, which keeps the width almost constant (about 2 nm).
- Right-handed twist. One full turn (the pitch) is 3.4 nm long and holds about 10 base pairs, so neighbouring pairs are about 0.34 nm apart.
- Stacking. The flat base pairs lie on top of one another like a pile of coins. Stacking, together with the hydrogen bonds, holds the helix firm.
4/4The strands are antiparallel and twist into a helix: 3.4 nm and about 10 bp per turn.
Step 1 shows two vertical backbone rails with an A–T pair between them: the purine A is a wide box, the pyrimidine T a narrow box, joined by two dashed hydrogen bonds; a bracket below marks the same width of about 2 nm. Step 2 adds a C–G pair, this time with the pyrimidine on the left and the purine on the right, joined by three dashed bonds, and it fills exactly the same width. Step 3 adds two impossible pairs below: two purines together stick out past the right rail and are labelled too wide, and two pyrimidines fall short of it and are labelled too narrow. Step 4 removes the mismatches, labels the left strand 5′ at the top and 3′ at the bottom and the right strand the other way round, with arrows showing opposite directions, and draws a twisted double helix beside it with one full turn of 3.4 nm holding 10 base pairs, 0.34 nm per pair.
Base percentages from one value
A double-stranded DNA sample from a soil bacterium contains 28 % adenine. What percentages of thymine, guanine and cytosine does it contain?
Try it yourself first. Then reveal the solution one step at a time.
Writing the partner strand
One strand of a DNA stretch reads 5′-GCA TTA CGG ATC-3′. Write its complementary strand in the 5′→3′ direction.
Try it yourself first. Then reveal the solution one step at a time.
Soon after the double helix, Crick summed up how genetic information flows in a cell: DNA is copied into RNA, and RNA is used to make protein. He called this the central dogma (Fig. 5.3). DNA also copies itself before a cell divides. Some RNA viruses, such as HIV, run part of the path backwards: an enzyme called reverse transcriptase makes DNA from their RNA, a step aptly called reverse transcription.
Three boxes stand in a row: DNA drawn as a short ladder, mRNA as a single wavy strand, and protein as a chain of beads. An arrow labelled transcription runs from DNA to mRNA and one labelled translation from mRNA to protein. A curved arrow loops back onto the DNA box, labelled replication. A dashed arrow runs from mRNA back to DNA, labelled reverse transcription in some RNA viruses. Under the protein box a note says there is no arrow back from protein.
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Answer (A) Guanine and cytosine
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Answer 16 %
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Answer False
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Answer (C) 3′→5′, the opposite way
5.1.2 Packaging of DNA Helix
How long is the DNA in one of your cells? A diploid human nucleus holds about bp. Multiplying by 0.34 nm per pair gives about 2.2 m of double helix, packed into a nucleus only a few micrometres ( m) across. Cells solve this with tight, orderly folding.
From length to base pairs
The DNA of a rod-shaped bacterium, carefully spread out, measures 1.70 mm. Roughly how many base pairs does it contain?
Try it yourself first. Then reveal the solution one step at a time.
A bacterium manages without a nucleus. Its chromosome is confined to one part of the cell, the nucleoid, where it is folded into large loops held in place by proteins; positive charges on those proteins grip the negatively charged DNA.
In eukaryotes the basic unit of packing is the nucleosome: roughly 200 bp of DNA coiled around a core of eight histone proteins, the histone octamer (Fig. 5.4). Histones are small proteins loaded with lysine and arginine, amino acids whose side chains carry positive charges, so the negatively charged DNA clings to them. Chromatin, the material of the nucleus that takes up stain, is essentially nucleosome after nucleosome along the DNA, and in electron micrographs it looks like beads on a string.
2/5About 200 bp of DNA wound on a positive histone octamer.
A strip at the top names the five levels: helix, nucleosome, beads, fibre and chromosome, and the current one is highlighted. Step 1 shows a bare double helix 2 nm wide with minus signs, because its phosphates are negatively charged. Step 2 shows DNA wrapped almost twice around a disc of eight histones marked plus: a nucleosome holding about 200 bp. Step 3 shows a row of five nucleosomes joined by short stretches of linker DNA, the beads-on-a-string form seen in electron micrographs. Step 4 shows the beads coiled into a thick chromatin fibre, with a note that non-histone proteins help pack it further. Step 5 shows a condensed metaphase chromosome with two chromatids joined at the centromere.
Further folding needs another group of proteins, the non-histone chromosomal (NHC) proteins. With their help the bead string coils into thicker chromatin fibres, and at metaphase these fold so tightly that each chromosome can be seen under a light microscope. Even in a non-dividing nucleus the packing varies from place to place, which gives two kinds of chromatin.
Two kinds of chromatin
Euchromatin
- Loosely packed
- Stains light
- Genes here are being transcribed (active)
Heterochromatin
- Densely packed
- Stains dark
- Largely inactive in transcription
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Answer 50000000 nucleosomes
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Answer (A) DNA is negatively charged and histones carry positive charges
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Answer (C) euchromatin, which is loosely packed
5.2 The Search for Genetic Material
Miescher found nuclein in 1869, only four years after Mendel announced his laws, yet nobody connected the two for decades. By the 1920s the work of Mendel, Walter Sutton, Thomas Hunt Morgan and many others had placed genes on chromosomes inside the nucleus. Chromosomes, however, contain both DNA and protein. Which one carries the instructions? Many biologists bet on protein: with twenty kinds of amino acids it seemed far richer than DNA with its four bases.
From nuclein to genomes
- 1869
Nuclein isolated
Miescher extracts an acidic substance from cell nuclei.
- 1928
Transformation
Griffith finds that dead bacteria can change live ones.
- 1944
DNA transforms
Avery, MacLeod and McCarty show the transforming principle is DNA.
- 1950
Chargaff's rules
In double-stranded DNA, A equals T and G equals C.
- 1952
Phage DNA enters cells
Hershey and Chase follow labelled DNA and protein.
- 1953
Double helix
Watson and Crick propose the structure.
- 1958
Semiconservative copying
Meselson and Stahl separate heavy and light DNA.
- 1961
The operon
Jacob and Monod explain how the lac genes are switched.
- 1961–66
Code cracked
Nirenberg, Khorana, Ochoa and others assign all 64 codons.
- 1984–85
DNA fingerprints
Alec Jeffreys uses repeat DNA to tell people apart.
- 1990–2003
Human Genome Project
The human genome sequence is read.
Transforming principle
The first clue came from a medical microbiologist. In 1928 Frederick Griffith was studying Streptococcus pneumoniae, a bacterium that causes pneumonia. On a culture plate it grows in two forms. The S strain wraps each cell in a slimy polysaccharide capsule and forms smooth, shiny colonies; it kills mice. The R strain has no capsule, forms rough colonies and is harmless.
Griffith injected mice with four different preparations (Fig. 5.5). Live S cells killed the mice and live R cells did not, as expected. S cells killed by heat were also harmless. But a mixture of heat-killed S cells and live R cells killed the mice, and live S cells could be isolated from their blood. Something from the dead S cells had transformed R cells into capsule-making S cells, and their offspring stayed S. He called it the 'transforming principle'. It had to be hereditary material, but its chemistry was unknown.
4/4Dead S plus live R kills the mouse, and live S cells are recovered from it.
A results board with columns for what was injected, the mouse, and the result; rows appear one per step. Step 1: live S cells, drawn with capsules, are injected and the mouse dies. Step 2: live R cells without capsules are injected and the mouse lives. Step 3: heat-killed S cells, drawn with dashed outlines, are injected and the mouse lives. Step 4: heat-killed S cells mixed with live R cells are injected; the mouse dies and a note says live S bacteria were found in it.
A change in the inherited properties of a bacterium brought about by DNA it takes in from its surroundings.
ExampleHarmless R cells becoming capsule-forming S cells in Griffith's mixture.
Biochemical characterisation of transforming principle
Until the 1940s most scientists still assumed genes were proteins. Between 1933 and 1944 Oswald Avery worked with Colin MacLeod and Maclyn McCarty to pin down the chemical identity of Griffith's principle. They separated the components of heat-killed S cells and tested which could transform live R cells. Only the DNA fraction worked. Enzymes then gave the decisive test (Table 5.4).
| Treated with | Destroys | Transforms R? |
|---|---|---|
| nothing | — | yes |
| protease | protein | yes |
| RNase | RNA | yes |
| DNase | DNA | no |
Destroying protein or RNA made no difference, but destroying DNA abolished transformation, so the principle had to be DNA. Not everyone was convinced: some suspected that a trace of protein hidden in the DNA did the work.
5.2.1 The Genetic Material is DNA
In 1952 Alfred Hershey and Martha Chase put a sharp question to a very simple system. A bacteriophage, a virus that attacks bacteria, is little more than DNA packed inside a protein shell. When it infects a cell, something passes into the bacterium and takes over, so that the cell starts turning out new phages. If that something is protein, protein is the genetic material; if it is DNA, the case for DNA is closed.
They exploited a chemical difference. DNA contains phosphorus but no sulfur, while the coat protein contains sulfur but no phosphorus. Phages grown with radioactive phosphorus (P) therefore had labelled DNA, and phages grown with radioactive sulfur (S) had labelled coats. Each batch infected E. coli. A kitchen blender then shook the empty coats off the cells, and a centrifuge spun the heavy bacteria into a pellet, leaving the light coats in the liquid above (Fig. 5.6).
4/4S stays in the liquid with the coats; P goes into the bacterial pellet.
Two rows run side by side: the upper row uses phages whose protein coats are labelled with sulfur-35, the lower row phages whose DNA is labelled with phosphorus-32; a star marks radioactivity. Step 1, label: three phages in each row, with radioactive coats above and radioactive DNA inside the heads below. Step 2, infect: phages sit on a bacterium and their DNA is inside the cell; in the lower row the DNA inside the cell is radioactive. Step 3, blend: the empty coats have been shaken off and float away from the bacterium. Step 4, spin: a centrifuge tube shows liquid with empty coats above a pellet of bacteria; in the upper row the radioactivity is in the liquid, in the lower row it is in the pellet.
Bacteria infected by P phages were radioactive; those infected by S phages were not, and their radioactivity stayed in the liquid. Protein never entered the cells, while DNA did, and the next generation of phages was made from it. DNA, not protein, is the genetic material that passes from virus to bacterium.
5.2.2 Properties of Genetic Material (DNA versus RNA)
After Hershey and Chase, protein was out of the race. Yet DNA is not the only answer: in some viruses, such as tobacco mosaic virus, the bacteriophage Qβ and the influenza virus, the genes are made of RNA. Why, then, did most life settle on DNA, while RNA works as messenger and adapter? The answer lies in what any genetic material has to do.
- Replicate: make accurate copies of itself.
- Stay stable: keep its chemistry and structure through the life of the organism.
- Change slowly: allow occasional mutations, the raw material of evolution.
- Be expressed: show itself as traits that are inherited in Mendelian fashion.
| Test | DNA | RNA |
|---|---|---|
| Replicate | yes, by base pairing | yes, by base pairing |
| Stability | high: two strands, no 2′-OH, T in place of U | low: reactive 2′-OH, often catalytic |
| Mutation | slow | fast |
| Expression | works through RNA | can be translated directly |
Replication. Base pairing lets both nucleic acids direct their own copying. Proteins have no such pairing rule, so they fail the very first test.
Stability. Here the two molecules really differ. Every ribose in RNA carries a reactive OH group on carbon 2′, which makes RNA chemically fragile and quick to break down; that some RNAs can catalyse reactions is itself a sign of reactivity. DNA lacks that group and has a second, complementary strand. Even heat, which pulls the strands apart, does not ruin it, because they pair up again on cooling; that is how the DNA in Griffith's heat-killed S cells could still transform. Using thymine instead of uracil makes DNA steadier still.Cytosine slowly loses an amino group and becomes uracil. Because normal DNA contains no uracil, repair enzymes can spot and fix every U they find; in a U-containing DNA they could not tell damage from normal bases.
Mutation. Change is needed too, but slowly. Being less stable, RNA picks up mutations faster than DNA. Combine that with the short generation times of viruses and RNA viruses evolve at great speed; influenza changes so quickly that its vaccine has to be updated most years.
Expression. Here RNA has the edge. A ribosome translates RNA directly, while DNA's information must first be copied into RNA, and the protein-making apparatus is itself built from RNA. The balance sheet: DNA wins as an archive and RNA as a courier, which is why cells store their genes in DNA and send RNA copies out to work.
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Answer (A) In the pellet, inside the bacteria
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- (A) An OH group on carbon 2′ of every ribose
- (B) It is usually single-stranded, with no partner for repair
- (D) It is reactive enough to act as a catalyst
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Answer False
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Answer (B) R cells picked up material from the dead S cells and began making capsules
5.3 RNA World
Today DNA needs proteins (its polymerases) to be copied, and proteins need DNA's instructions to be made. Which came first? A molecule that could do both jobs would break the circle, and RNA can: it stores information, and some RNAs, the ribozymes, catalyse reactions like enzymes. Tellingly, some of the oldest core processes of the cell, such as translation, splicing and parts of metabolism, still run on RNA; the peptide bond in every ribosome is made by RNA. Hence the idea of an early RNA world, in which RNA was the first genetic material.
That double life had a price: a molecule reactive enough to catalyse is also reactive enough to fall apart. DNA, a chemically modified RNA with no 2′-OH and with thymine, proved far steadier, and its second strand gave a backup copy from which damage could be repaired. So storage passed to DNA and most catalysis to proteins, leaving RNA in the middle (Fig. 5.7). You will meet this idea again in the chapter on evolution.
3/3Proteins do most catalysis; RNA is messenger, adapter and occasional catalyst.
Three columns stand for DNA, RNA and protein. Step 1, early life: only RNA is solid, the other two are faint, and RNA carries two tags, stores genes and catalyses. Step 2: DNA appears; the stores genes tag has moved under DNA with a second tag, more stable, while RNA keeps catalyses. Step 3, today: protein appears with the tag most catalysis, and RNA now carries messenger, adapter and some catalysis; headings read store, carry and help, and do the work.
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Answer (D) Some RNAs catalyse reactions, including peptide bond formation
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Answer ribozyme
5.4 Replication
Watson and Crick saw the copying method in their own model. Near the end of their 1953 paper they remarked, almost in passing, that the pairing rule pointed straight to a way of duplicating the molecule. Unzip the two strands, let each one serve as a template, and add new nucleotides according to the pairing rules. Each finished molecule would then hold one old (parental) strand and one new strand: semiconservative replication.
DNA copying in which the two parental strands separate and each becomes half of a daughter molecule, paired with a newly made strand.
ExampleAfter one round, each of the two daughter helices is one-half old and one-half new.
This was a hypothesis, and two others were possible. In conservative copying the parent helix would stay intact and an entirely new helix would be built beside it. In dispersive copying old and new pieces would be scattered through both strands of every daughter. An experiment had to tell them apart.
5.4.1 The Experimental Proof
In 1958 Matthew Meselson and Franklin Stahl found a way to tell old DNA from new by its weight. First they made all of a bacterium's DNA heavy. E. coli was grown for many generations on a medium whose only nitrogen source was , so nitrogen-15, a heavy but stable isotope (it is not radioactive), ended up in every base. Then the cells were switched to ordinary , and from then on every strand they made was light. Samples were taken as the cells went on dividing, about once every 20 minutes. Each was spun at very high speed in a caesium chloride (CsCl) density gradient, a salt solution that grows denser towards the bottom of the tube, where every DNA molecule, still double-stranded, settles at the level that matches its own density (Fig. 5.8).
- Start: a single heavy band.
- After one generation (20 min): a single band of hybrid density, halfway between heavy and light, and no heavy band at all.
- After two generations (40 min): two bands of equal size, one hybrid and one light.
3/4Two generations: half hybrid, half light.
Columns at the top mark where light, hybrid and heavy DNA settle in a caesium chloride gradient; on the right, solid bars stand for nitrogen-15 strands and pale dashed bars for nitrogen-14 strands. Each row is one sample, drawn as a peak at the density where its DNA settles, with the molecules beside it. Step 1, start: one peak at heavy, one molecule with two solid strands. Step 2, after 20 minutes: one peak at hybrid, two molecules each with one solid and one pale strand. Step 3, after 40 minutes: two equal peaks at hybrid and light, four molecules, two hybrid and two fully pale. Step 4 adds a dashed prediction row for conservative copying at 20 minutes: peaks at heavy and light with no hybrid, which does not match the real result.
The results rule out both rivals. Conservative copying predicts a heavy band and a light band after one generation, but the real sample showed only hybrid DNA. Dispersive copying predicts a single band at every generation, drifting steadily towards light, but at 40 minutes the hybrid and light DNA formed two separate bands. Only semiconservative copying fits every sample. Try all three models yourself in Fig. 5.9.
Bacteria were not a special case. At about the same time, J. Herbert Taylor and colleagues fed radioactive thymidine to dividing root-tip cells of the broad bean Vicia faba and followed where the label went. The chromosomes, too, replicated semiconservatively, and the same was later shown for other plants and for human cells.
5.4.2 The Machinery and the Enzymes
How fast must copying go? E. coli duplicates its bp in about 18 minutes, starting at one point and working both ways round its circular chromosome. Each of the two copying sites must therefore add roughly 2000 nucleotides every second, and add the right one almost every time, because an uncorrected slip becomes a mutation. The enzyme that manages this is DNA-dependent DNA polymerase: it reads a DNA template and strings deoxynucleotides into a matching strand.
The fuel comes with the building blocks. Nucleotides arrive as deoxyribonucleoside triphosphates (dATP, dGTP, dCTP and dTTP). When one is added to a growing strand its two outer phosphates are released; these are high-energy bonds of the same kind as in ATP, and breaking them drives the reaction. Copying DNA is expensive, and the dNTPs pay for it themselves.
DNA polymerase obeys one strict rule: it can attach a nucleotide only to the free 3′ end of a growing strand, so every new strand lengthens 5′→3′. The cell does not unwind the whole helix before copying, which would cost far too much energy; it opens a short stretch at a time, a Y-shaped replication fork that travels along the molecule (Fig. 5.10). The two templates exposed at the fork point in opposite directions, so the rule plays out differently on each. Along the template that runs 3′→5′ in the direction of travel, the new strand grows the same way the fork moves and is made continuously. Along the other template it has to grow away from the fork, so it is made discontinuously, in short pieces started as fresh template is exposed, and DNA ligase later seals the pieces into one strand.
3/4On the other template short pieces are made 5′→3′, away from the fork.
Parental double-stranded DNA lies on the left, its upper strand labelled 5′ and its lower strand 3′ at the left end; at the fork the strands separate into two arms that open to the right, the upper arm ending at 3′ and the lower at 5′. An arrow shows the fork moving left. Step 1 shows only the opening fork. Step 2 adds a long blue new strand along the upper arm, its arrowhead pointing towards the fork, labelled leading: continuous. Step 3 adds three short orange new pieces along the lower arm, each arrowhead pointing away from the fork, labelled lagging: short pieces. Step 4 shows the pieces joined into one strand with small circles at the former gaps, labelled joined by DNA ligase. A note says new strands grow only 5′ to 3′.
Where does copying begin? Not at random, but only at particular sequences, the origins of replication; E. coli has exactly one. This has a practical consequence. A gene cut out of one organism and put into a bacterium carries no origin of its own, so it would never be copied. Genetic engineers therefore join it to a vector, which brings an origin along (Chapter 9). Note also that DNA polymerase can lengthen a strand but cannot begin one from scratch; a short RNA primer provides its starting point.
In eukaryotic cells copying is confined to one part of the cell cycle, the S phase, and it starts at many origins along each chromosome. The cell must then divide exactly once for each round of copying; a cell that replicates its DNA but skips division is left with extra sets of chromosomes, a state called polyploidy. Many details of how eukaryotic origins work are still being worked out.
How long does one round take?
A bacterium has a circular chromosome of bp with one origin. Two forks leave the origin in opposite directions, and each adds 1000 nucleotides per second. How long does one round of replication take?
Try it yourself first. Then reveal the solution one step at a time.
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Answer 25 %
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Answer (B) The one running 5′→3′ in the direction the fork moves
- DNA ligase seals the gap between neighbouring pieces
- The helix is opened at the fork
- The fork moves on and the next piece begins
- A short piece is made 5′→3′, away from the fork
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- The helix is opened at the fork
- A short piece is made 5′→3′, away from the fork
- The fork moves on and the next piece begins
- DNA ligase seals the gap between neighbouring pieces
Show answer
Answer (D) Both the building blocks and the energy for joining them
5.5 Transcription
Replication copies everything; transcription is selective. In any one cell only some genes are copied into RNA, each as often as needed, and from each gene only one of the two strands is used. Transcription copies the information of that strand into RNA by the familiar pairing rules, except that the RNA puts uracil (U) opposite adenine. Because the process is so selective, the DNA must carry signals that mark where a transcribed stretch begins and ends, and which strand is to be read.
Suppose a cell did copy both strands of a gene. It would get two RNAs whose sequences are complementary, not identical. Those RNAs would find each other and pair, forming double-stranded RNA that ribosomes cannot read, so the work would be wasted. Even if they stayed apart, they would spell two unrelated messages, and one stretch of DNA would specify two different proteins, a recipe for confusion. Reading a single strand avoids both problems.
5.5.1 Transcription Unit
A transcription unit is a stretch of DNA with three parts: a promoter, a structural gene and a terminator.
Which strand gets read? The enzyme that makes RNA, DNA-dependent RNA polymerase, can only build 5′→3′, and its product must run antiparallel to the DNA it copies. So within the structural gene it reads the strand oriented 3′→5′, the template strand. The partner strand is not copied, but its sequence matches the RNA's (with T for U), so it is the one written down whenever we quote a gene. It is called the coding strand. Do not let the name mislead you: the coding strand is the copy we read, not the one the enzyme reads (Fig. 5.11).
Two DNA strands run across the figure: the coding strand on top runs 5′ to 3′ and the template strand below runs 3′ to 5′. A promoter box sits at the left end and a terminator box at the right, with the structural gene between them showing the bases ATG GCT TCA AAG on the coding strand and TAC CGA AGT TTC on the template. Below, the RNA reads AUG GCU UCA AAG from 5′ to 3′: the same as the coding strand with U in place of T. Arrows at the bottom mark upstream to the left and downstream to the right.
RNA polymerase finds a gene by docking at its promoter, which sits at the upstream end, towards the 5′ end as measured along the coding strand. The promoter's orientation also fixes the direction of reading, and with it which strand serves as template: a promoter placed at the other end of the gene would make the opposite strand the template. At the downstream end, towards the coding strand's 3′ end, the terminator usually signals where transcription stops. Other regulatory sequences can lie further upstream or downstream; you will meet them in Section 5.8.
From template strand to mRNA
The template strand of a short gene reads 3′-TAC CGA AAG TTT ACT-5′. Write the mRNA made from it and the coding strand, each 5′→3′.
Try it yourself first. Then reveal the solution one step at a time.
5.5.2 Transcription Unit and the Gene
Where does one gene end and the next begin? The classical definition, a functional unit of inheritance, does not translate neatly into a stretch of bases. Think of the DNA that is transcribed into a tRNA or an rRNA: it never codes for protein, yet it is a gene too. For protein-coding DNA biologists use a sharper term, the cistron: the stretch that codes for one polypeptide. A transcription unit carrying one cistron is monocistronic; one carrying several is polycistronic.
One message or several?
Monocistronic
- One polypeptide per transcription unit
- Typical of eukaryotes
- Genes usually split into exons and introns
Polycistronic
- Several polypeptides from one transcription unit
- Typical of bacteria
- Example: the lac operon (5.8.1)
Most eukaryotic genes are also split. Their coding sequence is broken into pieces, the exons, separated by stretches called introns (intervening sequences). Both are copied into the first RNA transcript, but only the exons survive into the finished, mature RNA; the introns are removed. A eukaryotic gene is therefore not one continuous coding stretch, which makes defining it as 'a piece of DNA' even trickier.
A mutation does not have to fall inside a coding sequence to change a trait: damage to a gene's promoter or other control sequences can silence it just as surely. That is why such sequences are sometimes loosely called regulatory genes, although strictly they make neither RNA nor protein.
5.5.3 Types of RNA and the process of Transcription
Making a protein takes teamwork among three kinds of RNA, all transcribed from DNA. mRNA (messenger RNA) is the message itself. tRNAs (transfer RNAs) are the adapters that bring amino acids and read the code. rRNAs (ribosomal RNAs) form the core of the ribosome and carry out part of its chemistry.
A bacterium gets by with a single RNA polymerase for all its RNA. Follow one molecule of it along a gene (Fig. 5.12). It first settles on the promoter: initiation. It then travels along the template, prising the helix open just ahead of itself and joining ribonucleoside triphosphates that pair correctly with the template, while the RNA trails behind with only its newest end still held by the enzyme: elongation. At the terminator, the finished RNA and then the enzyme let go of the DNA: termination.
The enzyme does not manage all this alone. The core polymerase is good only at lengthening RNA. To find a promoter and begin, it briefly carries a partner protein, the initiation factor σ (sigma); to stop at a terminator it is helped by the termination factor ρ (rho). Each partner changes what the polymerase recognises, and each leaves once its job is done.
2/3σ leaves; the enzyme moves along, building RNA 5′→3′.
A double line of DNA runs across, with a promoter box at the left and a terminator box at the right; the current stage is highlighted in a strip at the top. Step 1, initiation: RNA polymerase sits on the promoter with a small sigma factor attached, which helps it find the start. Step 2, elongation: the polymerase has moved to the middle of the gene, the sigma factor has gone, and a growing RNA strand trails out with its 5′ end free while nucleotides are added 5′ to 3′. Step 3, termination: the polymerase has reached the terminator with a rho factor attached, and the finished RNA floats free.
Bacterial mRNA is ready to use the moment it is made, and it is made in the same compartment as the ribosomes. Ribosomes therefore climb onto the 5′ end of an mRNA and start translating while RNA polymerase is still busy further along the gene: in bacteria transcription and translation are coupled.
Eukaryotes add two complications. The first is a division of labour. Their nuclei use three different RNA polymerases, each responsible for its own set of RNAs, while mitochondria and chloroplasts have polymerases of their own (Table 5.7).
| Enzyme | Makes |
|---|---|
| RNA polymerase I | rRNAs: 28S, 18S and 5.8S |
| RNA polymerase II | hnRNA, the precursor of mRNA |
| RNA polymerase III | tRNA, 5S rRNA and snRNAs (small nuclear RNAs) |
The second is that the first copy of a gene, the primary transcript or hnRNA (heterogeneous nuclear RNA), contains introns as well as exons and cannot be used as it is. Three kinds of processing turn it into mRNA (Fig. 5.13). Capping fixes a modified nucleotide, a methylated guanosine joined through a triphosphate link, onto the 5′ end. In tailing, 200–300 adenylate (A) residues are added to the 3′ end without any template. In splicing, the introns are cut out and the exons joined in the right order. The finished mRNA then leaves the nucleus to be translated.
4/4Introns are removed and exons joined; the mRNA leaves the nucleus.
Step 1 shows the primary transcript, hnRNA, as a bar of three blue exon blocks with two dashed grey introns between them. Step 2 adds a round cap at the 5′ end. Step 3 adds a poly-A tail, written AAAA, at the 3′ end. Step 4 shows the two introns dropped below the bar and labelled introns cut out, while the three exons have slid together into one continuous mature mRNA with its cap and tail; an arrow points to the cytoplasm.
Why would cells bother cutting and pasting their own messages? One view is historical. Introns seem to be very old, and the splicing machinery is built around small nuclear RNAs that do much of the cutting and joining themselves, a likely relic of the RNA world (Section 5.3). The more biologists look, the more jobs they find that RNA does in cells.
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Answer (B) 5′-AUG CCA-3′
Tap an item on the left, then its match on the right.
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- RNA polymerase I : 28S, 18S and 5.8S rRNA
- RNA polymerase II : hnRNA (pre-mRNA)
- RNA polymerase III : tRNA, 5S rRNA and snRNA
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Answer (D) Cutting out the exons
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Answer True
5.6 Genetic Code
Copying DNA into DNA, or DNA into RNA, needs no dictionary: each base simply pairs with its partner. Translation is a different kind of problem. A sequence of four kinds of nucleotide has to become a sequence of twenty kinds of amino acid, and nothing in an amino acid's shape fits a base. What biologists did know was that mutations in genes produced altered amino acids in proteins. So the cell must follow a set of rules relating base sequence to amino acid sequence: the genetic code.
The first clue was arithmetic, and it came from a physicist, George Gamow. If one base stood for one amino acid, four bases could name only four amino acids. Pairs of bases give combinations, still short of twenty. Triplets give , comfortably enough even with many left over. So a codon was probably three bases long. Showing that this was really so needed chemists, biochemists and geneticists working together.
The decisive tool was Marshall Nirenberg's cell-free system, a test-tube mix of cell contents that makes protein from whatever RNA is added. In 1961 an artificial RNA containing only U gave a chain containing only phenylalanine, so UUU had to mean phenylalanine. Artificial RNAs for such tests could be made with Severo Ochoa's enzyme, polynucleotide phosphorylase, which strings nucleotides together without a template, while Har Gobind Khorana's chemical methods produced RNAs with precisely repeating sequences (homopolymers and copolymers) that decoded codon after codon. By 1966 all 64 codons were known (Table 5.8).
| Amino acid | Codons |
|---|---|
| Alanine (Ala) | GCU, GCC, GCA, GCG |
| Arginine (Arg) | CGU, CGC, CGA, CGG, AGA, AGG |
| Asparagine (Asn) | AAU, AAC |
| Aspartic acid (Asp) | GAU, GAC |
| Cysteine (Cys) | UGU, UGC |
| Glutamic acid (Glu) | GAA, GAG |
| Glutamine (Gln) | CAA, CAG |
| Glycine (Gly) | GGU, GGC, GGA, GGG |
| Histidine (His) | CAU, CAC |
| Isoleucine (Ile) | AUU, AUC, AUA |
| Leucine (Leu) | UUA, UUG, CUU, CUC, CUA, CUG |
| Lysine (Lys) | AAA, AAG |
| Methionine (Met) | AUG (also the start codon) |
| Phenylalanine (Phe) | UUU, UUC |
| Proline (Pro) | CCU, CCC, CCA, CCG |
| Serine (Ser) | UCU, UCC, UCA, UCG, AGU, AGC |
| Threonine (Thr) | ACU, ACC, ACA, ACG |
| Tryptophan (Trp) | UGG |
| Tyrosine (Tyr) | UAU, UAC |
| Valine (Val) | GUU, GUC, GUA, GUG |
| Stop (no amino acid) | UAA, UAG, UGA |
The table is arranged by amino acid so you can go from protein back to RNA. Reading in the other direction, from a codon to its amino acid, is easiest in Fig. 5.14. Either way, a few features of the code stand out.
- It is a triplet code. Of the 64 codons, 61 stand for amino acids and 3 (UAA, UAG and UGA) code for none; they are stop codons that end a chain.
- It is degenerate. Most amino acids have more than one codon; leucine, serine and arginine have six each. Only methionine and tryptophan have a single codon.
- It is unambiguous. Each codon means one amino acid and nothing else.
- It has no punctuation. Codons follow one another without gaps or commas, so the message is read three bases at a time from a fixed starting point.
- It is nearly universal. UUU means phenylalanine in bacteria, plants and people alike; the few exceptions occur in mitochondria and in some protozoans.
- AUG has two jobs. It codes for methionine and also serves as the start (initiator) codon.
Reading an mRNA, and reading it backwards
An mRNA reads 5′-CU AUG CAC GGA UAC UGG CCC AAG UGA GC-3′. What polypeptide does it code for? How many different mRNA sequences could code for just the first three amino acids?
Try it yourself first. Then reveal the solution one step at a time.
5.6.1 Mutations and Genetic Code
Mutations are the natural experiments that reveal how the code works. A big deletion or rearrangement can remove or duplicate whole genes, with an obvious loss or gain of function, but it says little about codons. Changes of one or a few bases are far more revealing.
A point mutation changes a single base pair. The classic example is sickle-cell anaemia. In the gene for the β-globin chain of haemoglobin, one base change turns the codon GAG into GUG, so glutamic acid (glutamate), the sixth amino acid of the finished chain, is replaced by valine. That single swap makes haemoglobin molecules stick together when oxygen is low, distorting red blood cells into sickles.
Because codons are read without punctuation, inserting or deleting bases has a special effect. Adding or removing one or two bases shifts the reading frame from that point onward, so every codon after it is read wrongly: a frameshift insertion or deletion. Adding or removing three bases (or a multiple of three) instead adds or removes whole codons. One or more amino acids are gained or lost, and the frame after the change stays intact (Table 5.9).
| Change | mRNA | Polypeptide |
|---|---|---|
| None | AUG GCU UCA AAA GGC UGG UAA | Met–Ala–Ser–Lys–Gly–Trp |
| A→G in codon 4 (silent) | AUG GCU UCA AAG GGC UGG UAA | Met–Ala–Ser–Lys–Gly–Trp |
| C→U in codon 2 (missense) | AUG GUU UCA AAA GGC UGG UAA | Met–Val–Ser–Lys–Gly–Trp |
| A→U in codon 4 (nonsense) | AUG GCU UCA UAA … | Met–Ala–Ser, then stop |
| G inserted after AUG (frameshift) | AUG GGC UUC AAA AGG CUG GUA A… | Met–Gly–Phe–Lys–Arg–Leu–Val… |
| Codon 3 deleted (in frame) | AUG GCU AAA GGC UGG UAA | Met–Ala–Lys–Gly–Trp |
5.6.2 tRNA – the Adapter Molecule
Something has to stand between codon and amino acid. Transfer RNA does: it reads a codon at one end and carries the matching amino acid at the other. Francis Crick predicted such an adapter on logical grounds, since amino acids have no shape that could recognise a codon. Curiously, tRNA had already been discovered, under the name soluble RNA (sRNA), before anyone knew this was its job.
Drawn flat, a tRNA resembles a clover leaf; its true folded shape is a compact, upside-down L (Fig. 5.15). Two places matter. The anticodon loop carries three bases that pair, antiparallel, with a codon. The amino acid acceptor end (the 3′ end) is where the amino acid is attached. Each tRNA is charged with one particular amino acid, a special initiator tRNA carries the methionine that starts every chain, and stop codons have no tRNA at all.
On the left a tRNA is drawn as a clover leaf: a paired stem rises to the acceptor end, where an amino acid circle is attached at the 3′ end; two side arms end in loops; a lower arm ends in the anticodon loop. The anticodon bases, written 3′ C C G 5′, pair through dashed hydrogen bonds with the codon G G C on an mRNA strand running 5′ to 3′. On the right an inset shows the real folded shape, an inverted L, with the amino acid end at the tip of the top arm and the anticodon end at the bottom of the long arm.
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Answer 8 sequences
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Answer (B) 16
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Answer (D) is changed from that point onward, often cut short
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Answer CCG
5.7 Translation
Translation strings amino acids together by peptide bonds, in exactly the order of an mRNA's codons. A peptide bond does not form on its own, so the cell pays in advance.
Before translation, an enzyme uses ATP to activate each amino acid and load it onto a tRNA that matches it (its cognate tRNA). This loading is called charging of tRNA, or more precisely aminoacylation. A charged tRNA holds its amino acid by a high-energy link, so when two charged tRNAs sit side by side, joining their amino acids is energetically easy, and a catalyst makes it fast.
The work happens on ribosomes. Each is assembled from a small subunit and a large subunit, which float apart when idle; translation begins when a small subunit binds an mRNA. Together the subunits contain structural RNAs and roughly 80 kinds of protein. The large subunit has two neighbouring sites for charged tRNAs (usually called the P site and the A site), which hold two amino acids next to each other. The bond between them is formed by RNA: in bacteria the catalyst is the 23S rRNA, a ribozyme.
An mRNA is longer than the part that is read. At each end lies an untranslated region (UTR): the 5′ UTR before the start codon and the 3′ UTR after the stop codon. They help the mRNA to be translated efficiently. The stretch between, from AUG to the stop codon, is the translational unit that actually codes for the polypeptide.
- Initiation. The ribosome assembles on the mRNA at the start codon AUG, which only the initiator tRNA (carrying methionine) can recognise.
- Elongation. Charged tRNAs arrive one at a time; each binds only if its anticodon pairs with the codon in the free site. A peptide bond joins the growing chain to the new amino acid, and the ribosome moves along the mRNA by one codon. The cycle repeats.
- Termination. When a stop codon reaches the ribosome, no tRNA fits. A release factor binds instead, the finished polypeptide is freed, and the ribosome comes apart.
3/5The next tRNA pairs with GCU and a peptide bond forms.
An mRNA runs along the bottom with the codons AUG, GCU, UGG and UAG. Step 1, charging: a methionine and a tRNA with anticodon UAC combine, using ATP, into a charged tRNA. Step 2, initiation: a ribosome with a small and a large subunit sits over AUG and GCU, with its two sites marked P and A; the initiator tRNA carrying methionine pairs with AUG in the P site. Step 3: a tRNA with anticodon CGA carrying alanine pairs with GCU in the A site, and a peptide bond now joins methionine to alanine on the second tRNA. Step 4: the ribosome has moved one codon to the right; the empty first tRNA drifts away, the tRNA holding Met–Ala is in the P site, and a tRNA with anticodon ACC carrying tryptophan has arrived at UGG. Step 5: the ribosome sits over UGG and the stop codon UAG; a release factor occupies the A site and the chain Met–Ala–Trp floats free.
How long is the message?
A bacterial polypeptide has 300 amino acids, counting the first methionine. How many nucleotides lie between the first base of its start codon and the last base of its stop codon? Is the whole mRNA exactly this long?
Try it yourself first. Then reveal the solution one step at a time.
- A charged tRNA pairs with the next codon
- A release factor binds the stop codon
- Amino acids are attached to their tRNAs using ATP
- The ribosome assembles at AUG with the initiator tRNA
- The ribosome moves along by one codon
- A peptide bond forms
Show answer
- Amino acids are attached to their tRNAs using ATP
- The ribosome assembles at AUG with the initiator tRNA
- A charged tRNA pairs with the next codon
- A peptide bond forms
- The ribosome moves along by one codon
- A release factor binds the stop codon
Show answer
Answer (C) The 23S rRNA
Tap an item on the left, then its match on the right.
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- mRNA : carries the codons
- tRNA : brings an amino acid and reads a codon
- rRNA : builds the ribosome and makes peptide bonds
- Release factor : recognises a stop codon
5.8 Regulation of Gene Expression
A gene is 'expressed' when its product, usually a polypeptide, is made, and a cell can turn that output up or down at several points along the way. In eukaryotes, regulation can act at four levels:
- the transcriptional level, deciding whether a primary transcript is made;
- the processing level, for example which way an RNA is spliced;
- the transport of mRNA from the nucleus to the cytoplasm;
- the translational level, deciding how much protein each mRNA yields.
Cells do not run every gene all the time; that would waste energy and materials. A liver cell and a nerve cell carry identical DNA but use different sets of genes, and turning an embryo into an adult means switching whole batteries of genes on and off in the right order. Even a single bacterium tunes its genes to its surroundings. E. coli makes the enzyme β-galactosidase, which splits lactose into glucose and galactose for energy, mainly when lactose is actually there. Gene expression, in short, responds to the cell's metabolic, physiological and environmental conditions.
Bacteria do most of their regulating at the very first step: whether RNA polymerase starts transcribing a gene at all. Proteins that bind DNA near the promoter decide this. A repressor blocks the polymerase (negative control); an activator helps it (positive control). Many bacterial repressors work by sitting on an operator, a short DNA sequence next to the promoter, and each operator is recognised by its own repressor: the lac repressor binds the lac operator and no other.
5.8.1 The Lac operon
Bacterial genes that serve the same pathway are often lined up side by side and transcribed together into one polycistronic mRNA, sharing one promoter and one set of control sequences. Such a unit is an operon; the trp, ara, his and val operons are examples. The best known is the lac operon, which handles the sugar lactose. Its control was worked out by two French scientists, the geneticist François Jacob and the biochemist Jacques Monod, and it was the first system of transcriptional regulation to be understood.
The operon has one regulatory gene and three structural genes (Table 5.11, Fig. 5.17). The regulatory gene is called i, short for inhibitor (not inducer). All three structural-gene products serve lactose use, and in most operons, likewise, the grouped genes work in one pathway.
| Gene | Product | Its job |
|---|---|---|
| i (regulatory) | repressor | binds the operator and blocks transcription |
| z | β-galactosidase | splits lactose into galactose and glucose |
| y | permease | lets lactose (a β-galactoside) into the cell |
| a | transacetylase | a third enzyme of lactose metabolism |
2/4The repressor binds the operator and RNA polymerase cannot transcribe.
A gene map runs across: the i gene on the left, then the promoter p and operator o, then the structural genes z, y and a. Step 1 labels the promoter and operator and shows the repressor, always made from the i gene, as a notched block below it. Step 2, no lactose: the repressor sits on the operator and RNA polymerase, waiting on the promoter, is blocked. Step 3, lactose added: inducer molecules appear, one fits into the repressor's notch, and the tilted, inactive repressor has left the operator. Step 4: RNA polymerase moves along z, y and a and one mRNA is made, giving three proteins: beta-galactosidase, permease and transacetylase.
With no lactose around, the operon is off. The i gene is expressed nonstop (constitutively), so repressor is always present, and it sits on the operator where it blocks RNA polymerase. Now give the cells lactose, and no glucose, their preferred food. A trickle of transcription always escapes the repressor, so each cell has a few permease molecules, enough to let some lactose in; without them the sugar could never enter. Inside, lactose (more exactly allolactose, made from it) binds the repressor and changes its shape, so it lets go of the operator. RNA polymerase can now run through z, y and a, and the enzymes for using lactose are made. Because lactose both feeds the enzyme and flips the switch, it is called the inducer.
It is a neat economy: the substrate itself calls for the enzymes that use it. The products, glucose and galactose, cannot switch the operon on. Explore every combination, including two broken switches, in Fig. 5.18.
Because this switch works by removing a block, it is called negative regulation. The lac operon also has a second, positive control, an activator that keeps it low while glucose is plentiful, but that part of the story is beyond this chapter.
Show answer
Answer (A) made and sitting on the operator
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Answer (D) All the time, with or without lactose
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- (A) β-galactosidase
- (B) Permease
- (C) Transacetylase
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Answer False
5.9 Human Genome Project
In 1990 biologists set out to read the complete DNA sequence of human beings: the Human Genome Project (HGP). The goal followed directly from this chapter. If a person's genetic make-up is written in the order of bases, then knowing that order, and how it differs between people, should explain a great deal. Two advances made the attempt realistic: genetic engineering allowed any piece of DNA to be isolated and cloned, and sequencing had become quick and routine.
It was a mega project by any measure. The genome has about bp. Early planners guessed it would cost about US$3 to read each base pair, which comes to about US$9 billion. The data are just as daunting. Stored as one letter per base, one copy of the genome fills a text file of about 3 gigabytes; read aloud at five letters a second without a break, it would take nearly twenty years. Storing, searching and analysing so much information needed fast computers, and the project grew up together with a new field, bioinformatics, which uses computers to handle biological data.
Goals of HGP
- Find all the genes in human DNA, then estimated at roughly 20 000–25 000.
- Work out the order of all the roughly 3 billion base pairs.
- Keep the information in databases open to researchers.
- Build better tools for analysing the data.
- Share the new technologies with industry and other sectors.
- Think through the ethical, legal and social issues (ELSI) the project might raise.
The HGP ran for 13 years under the US Department of Energy and the National Institutes of Health, with the UK's Wellcome Trust soon joining as a major partner and Japan, France, Germany, China and other countries contributing. The sequence was declared complete in 2003. Mapping how DNA differs between people could change how diseases are detected, treated and eventually prevented. The genomes of other species were read too, among them bacteria, yeast, the nematode Caenorhabditis elegans, the fruit fly Drosophila, rice and Arabidopsis; knowing them helps us use such organisms in medicine, farming, fuel production and the clean-up of polluted sites.
Methodologies
The project attacked the problem from two directions. Expressed sequence tags (ESTs) picked out genes by starting from the RNA that cells actually make. The whole-genome route read everything, genes and the long stretches between them, without choosing, and worked out afterwards which parts do what; that labelling is called sequence annotation.
The whole-genome route was a giant jigsaw that the scientists first cut up themselves (Fig. 5.19). Sequencing machines read only short pieces, so DNA from many cells was broken at random into fragments. Each fragment was multiplied by cloning it in bacteria or yeast, carried in vectors called BACs (bacterial artificial chromosomes) and YACs (yeast artificial chromosomes). Automated sequencers read the clones by a method devised by Frederick Sanger, the same Sanger who first worked out the amino acid sequence of a protein. Because the pieces were cut at random they overlapped, and the overlaps showed which pieces lay next to each other. Matching millions of overlaps is impossible by hand, so special computer programs did the assembly.
4/4Line up the reads where their ends overlap to rebuild the sequence.
Step 1 shows three long bars, many copies of the same genome. Step 2 shows each bar broken at different random places into fragments, which are cloned in BACs or YACs. Step 3 shows three short sequence reads, ATGCCTGAA, TGAAGTTCG and TCGATCCAG, scattered at random positions. Step 4 slides the reads into line: the end TGAA of the first matches the start of the second, and TCG at the end of the second matches the start of the third; the overlaps are highlighted and the joined sequence ATGCCTGAAGTTCGATCCAG is written below.
The assembled sequences were annotated and assigned to their chromosomes. The last of the 24 different human chromosomes (22 autosomes plus X and Y) to be finished was chromosome 1, in May 2006. A further task was to build genetic and physical maps of the genome. These used variation (polymorphism) in the sites cut by restriction endonucleases, and short repeated sequences called microsatellites; the next section shows how such repeat polymorphisms are used.
5.9.1 Salient Features of Human Genome
| Feature | Finding |
|---|---|
| Size | 3164.7 million bp |
| Average gene | about 3000 bases; the largest, dystrophin, is 2.4 million bases |
| Number of genes | estimated at about 30 000, far below earlier guesses of 80 000–140 000; later counts put protein-coding genes near 20 000 |
| Shared between people | 99.9 % of bases are the same in everyone |
| Known function | unknown for over half the genes found |
| Protein-coding DNA | less than 2 % of the genome |
| Repeated sequences | a very large share; some repeated hundreds or thousands of times |
| Most and fewest genes | chromosome 1 (2968); Y chromosome (231) |
| SNPs | about 1.4 million places where single bases differ between people |
Two findings deserve a word. Repeats were once dismissed as 'junk', but their arrangement is a record of how chromosomes are organised, how they move and change, and how they evolved. Single nucleotide polymorphisms (SNPs, said 'snips') have become the workhorse markers of human genetics: by checking which SNP variants travel with a disease in families, researchers can home in on the gene responsible, and SNP patterns across populations trace the migrations of our ancestors.
5.9.2 Applications and Future Challenges
The sequence was a beginning, not an end. Knowing the letters is not the same as understanding them, and turning sequence into biology will occupy many thousands of researchers, in laboratories and companies, for decades. The biggest change is in how questions are asked. Instead of studying one gene at a time, biologists can now measure thousands at once: which genes are switched on in a tumour compared with the healthy tissue beside it, which variants make a medicine work well or badly in a patient, or how a crop variety copes with drought. Reading a person's genome now takes days and costs a tiny fraction of the first one, which is bringing genome tests into hospitals, including in India.
Show answer
Answer (B) Bacteria and yeast, with BAC and YAC vectors
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Answer (A) assigning functions to regions of the sequence
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Answer 3200000 bp
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Answer False
5.10 DNA Fingerprinting
Forensic scientists and courts often need to know whether two DNA samples come from the same person, or whether two people are related. Since any two of us differ at only about 0.1 % of our bases, comparing whole genomes for every case would be wasteful. DNA fingerprinting takes a shortcut: it compares just a few regions that differ wildly from person to person.
The chosen regions are repetitive DNA, in which a short sequence is repeated over and over. Some of it first showed up as an oddity of centrifugation. In a density gradient most of a cell's DNA forms one big peak, but small extra peaks appear beside it, made of repeats whose base composition gives them a slightly different density: satellite DNA. Satellites come in many types, depending on how rich they are in A and T or in G and C, how long the repeating unit is and how many times it repeats; microsatellites and minisatellites are two of them. They code for no protein, yet they make up a large part of the genome, and the number of repeats varies enormously between people.
Every tissue of a person, whether blood, hair roots, skin, bone, saliva or semen, carries the same pattern of repeats. That makes it a powerful identification tool in forensic work. And because the repeats are inherited from parents, the same test can settle disputes about parentage.
Since the method rests on DNA polymorphism, we should be clear about the term.
Inherited variation at a DNA locus. A locus is called polymorphic when more than one allele occurs in the population and a variant allele has a frequency above 0.01 (1 %).
ExampleA minisatellite with 8 repeats in some people and 11 in others.
Polymorphisms are inherited, so they begin as mutations in germ cells; a mutation in a body (somatic) cell dies with the person. If a germ-line change does not stop its carriers having children, it can spread through the population over many generations. Most polymorphisms lie in non-coding DNA, where a change rarely affects health or fertility, so it is not weeded out and keeps accumulating. They range from a single changed base to large rearrangements, and they supply the variation on which evolution and the formation of new species depend.
The method was invented by Alec Jeffreys in 1984–85. His probe was a highly variable satellite sequence, a variable number of tandem repeats (VNTR), and the original procedure detected it by Southern blot hybridisation with the probe made radioactive.
DNA fingerprinting by Southern blotting
Isolate the DNA
Extract DNA from a sample such as blood, a hair root or saliva.
Why does this give a personal pattern (Fig. 5.20)? A VNTR is a minisatellite: a short sequence repeated head to tail. People differ in how many repeats they carry at each VNTR site, and even the two copies of a chromosome in one person usually differ, so cutting beside the repeats gives fragments anywhere from about 0.1 to 20 kb long. The probe lights up every one of them, and the film shows a ladder of bands unique to each person. Only identical (monozygotic) twins, who come from one zygote, share it.
5/5Each child band matches the mother or F2; F1 shares none of the father's bands.
A gel on the right has four lanes labelled M for mother, C for child, F1 and F2 for two possible fathers, with wells at the negative end at the top and the positive end at the bottom. Step 1: on the left, the child's two copies of one VNTR locus are drawn as bars, the copy from the mother with 11 repeat units and the copy from the father with 8. Step 2: triangles mark restriction cuts on either side of each repeat block, and the cut pieces are shown below, the longer piece carrying more repeats. Step 3: the child's two pieces appear as two bands in lane C, the smaller one lower down because small pieces run farther. Step 4: a labelled VNTR probe reveals bands for all four people at two loci. Step 5: dashed lines link each of the child's four bands to a matching band in the mother's lane or in F2's lane; F1 shares none of the bands that did not come from the mother, so F2 is the father.
Today the test is far more sensitive. The polymerase chain reaction (PCR, Chapter 9) multiplies chosen stretches of DNA millions of times, so the DNA of a single cell can be enough, and many different probes and markers are in use. Beyond crime and paternity cases, DNA fingerprints help measure genetic diversity, for example within an endangered species or among crop varieties.
- Hybridise with a labelled VNTR probe
- Detect bands by autoradiography
- Separate fragments by gel electrophoresis
- Blot fragments onto a nylon membrane
- Isolate DNA
- Digest it with restriction endonucleases
Show answer
- Isolate DNA
- Digest it with restriction endonucleases
- Separate fragments by gel electrophoresis
- Blot fragments onto a nylon membrane
- Hybridise with a labelled VNTR probe
- Detect bands by autoradiography
Show answer
Answer (C) The number of repeats varies hugely between people
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Answer True
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Answer (A) the biological father
Molecular Basis of Inheritance: key points and summary
- DNA and RNA are chains of nucleotides joined by 3′–5′ phosphodiester bonds, each chain with a 5′ and a 3′ end; RNA has a 2′-OH on every sugar and uracil in place of thymine (Section 5.1.1).
- The double helix has two antiparallel strands paired A with T (two hydrogen bonds) and G with C (three); it is 2 nm wide with about 10 bp in each 3.4 nm turn (Fig. 5.2).
- DNA is packed in nucleosomes of about 200 bp wound on histone octamers, then into chromatin fibres and chromosomes; euchromatin is active, heterochromatin largely inactive (Section 5.1.2).
- Griffith's transformation, the enzyme tests of Avery, MacLeod and McCarty, and the labelled phages of Hershey and Chase showed that DNA is the genetic material (Section 5.2).
- DNA is the more stable store of information and RNA the better messenger; RNA probably came first, as both store and catalyst (Section 5.2.2, Section 5.3).
- Replication is semiconservative, as Meselson and Stahl showed with heavy nitrogen and CsCl gradients (Section 5.4.1).
- DNA polymerase builds only 5′→3′, so at the fork one strand is made continuously and the other in pieces joined by ligase; copying starts at origins of replication (Section 5.4.2).
- A transcription unit has a promoter, a structural gene and a terminator; RNA polymerase copies the 3′→5′ template strand, and the RNA matches the coding strand (Section 5.5.1).
- Bacteria use one RNA polymerase aided by σ and ρ; eukaryotes use three, and turn hnRNA into mRNA by capping, tailing and splicing out introns (Section 5.5.3).
- The genetic code is triplet, degenerate, unambiguous, without punctuation and nearly universal; AUG starts and UAA, UAG, UGA stop. One- or two-base insertions and deletions shift the frame (Section 5.6, Section 5.6.1).
- tRNA is the adapter; on the ribosome, charged tRNAs pair with codons, rRNA forms peptide bonds and a release factor ends the chain at a stop codon (Section 5.7).
- In the lac operon the repressor from the i gene blocks the operator until lactose (allolactose) inactivates it; z, y and a are then transcribed together (Section 5.8.1).
- The Human Genome Project (1990–2003) read about bp: under 2 % codes for protein and people share 99.9 % of bases. DNA fingerprinting compares VNTR band patterns to identify people and settle parentage (Section 5.9, Section 5.10).
Practice questions with answers (NCERT “Exercises”)
Which of these are nucleosides rather than free nitrogenous bases? Pick all that apply.
Select all that applyShow solution
Nitrogenous bases: guanine, thymine and uracil. Nucleosides: uridine (uracil + ribose), adenosine (adenine + ribose) and deoxycytidine (cytosine + deoxyribose).
A sample of double-stranded DNA is 35 % thymine. What percentage of its bases are guanine?
%Show solution
Thymine pairs with adenine, so A = 35 % and A + T = 70 %. The remaining 30 % is shared equally by G and C, so G = 15 %.
One strand of a DNA stretch reads 5′-AGC TTG CAA TCG GAT-3′. Write its complementary strand in the 5′→3′ direction.
Show solution
Answer5′-ATC CGA TTG CAA GCT-3′
Pairing base by base gives the partner written under the strand as 3′-TCG AAC GTT AGC CTA-5′.
Read from its 5′ end, it is 5′-ATC CGA TTG CAA GCT-3′.
The coding strand of a transcription unit reads 5′-ATG TCC GGA CTT AAG TGA-3′. Write the mRNA, and the polypeptide it codes for.
Show solution
AnswermRNA 5′-AUG UCC GGA CUU AAG UGA-3′, coding Met–Ser–Gly–Leu–Lys.
The mRNA has the coding strand's sequence and direction with U for T: 5′-AUG UCC GGA CUU AAG UGA-3′.
Reading from AUG: AUG Met, UCC Ser, GGA Gly, CUU Leu, AAG Lys, then UGA stop. The polypeptide is Met–Ser–Gly–Leu–Lys.
Watson and Crick suggested how DNA is copied using nothing but their model of its structure. Which feature of the double helix made this possible, and what copying scheme did it suggest?
Show solution
AnswerComplementary base pairing; each strand can act as a template, giving semiconservative replication.
Because A always pairs with T and G with C, the two strands are complementary: each fully specifies the other. If the strands separate, each can serve as a template on which a new partner is assembled by the same pairing rules. Each daughter molecule then contains one parental strand and one new strand, which is semiconservative replication.
Nucleic acid polymerases are named after the template they read and the nucleic acid they make. Match each template → product with its enzyme.
Tap an item on the left, then its match on the right.
Show solution
DNA-dependent DNA polymerase copies DNA in replication; DNA-dependent RNA polymerase makes RNA in transcription; RNA-dependent DNA polymerase (reverse transcriptase) makes DNA from the RNA of retroviruses; RNA-dependent RNA polymerase copies the RNA genomes of many RNA viruses.
How did using two different radioactive isotopes let Hershey and Chase follow a phage's DNA and its protein separately?
Show solution
Answer³²P labels only DNA and ³⁵S labels only protein; only ³²P entered the bacteria.
DNA contains phosphorus but no sulfur, while the phage coat protein contains sulfur but no phosphorus. Phages grown with P therefore had radioactive DNA, and those grown with S had radioactive protein. After infection, blending and centrifuging, the bacteria infected by P phages were radioactive, while the S stayed in the liquid with the empty coats. So DNA, not protein, entered the cells.
State one clear difference between the members of each pair.
Repetitive DNA and satellite DNA
Show solution
AnswerSatellite DNA is the part of repetitive DNA that forms separate minor peaks in density gradient centrifugation.
Repetitive DNA is any DNA in which a sequence is repeated many times. Satellite DNA is repetitive DNA whose base composition makes it separate from the bulk DNA as small satellite peaks in a density gradient; it includes micro- and minisatellites.
mRNA and tRNA
Show solution
AnswermRNA carries the codons; tRNA carries an amino acid and reads a codon with its anticodon.
mRNA is a long, linear copy of a gene that carries the codons to the ribosome. tRNA is a small folded (clover-leaf, inverted-L) RNA with an anticodon at one end and an amino acid at its 3′ end; it acts as the adapter.
Template strand and coding strand
Show solution
AnswerThe template (3′→5′) is copied; the coding strand (5′→3′) is not copied but has the RNA's sequence.
RNA polymerase reads the template strand, which runs 3′→5′. The coding strand runs 5′→3′, is displaced rather than copied, and has the same sequence as the RNA except for T in place of U.
A classmate calls the ribosome 'just a workbench where proteins are assembled'. Give two essential jobs it does during translation that show it is more than that.
Show solution
AnswerIt holds the mRNA and positions charged tRNAs at its two sites, and its rRNA catalyses peptide bonds.
(1) The ribosome binds the mRNA and holds two charged tRNAs at neighbouring sites so their anticodons pair with consecutive codons; it then moves one codon at a time, keeping the reading frame. (2) Its rRNA (23S rRNA in bacteria) is a ribozyme that catalyses the peptide bond between the amino acids.
An E. coli culture is given lactose and starts making β-galactosidase. A few hours later, although nobody has changed the medium, production of the enzyme falls almost to nothing. Explain why.
Show solution
AnswerThe enzyme uses up the lactose; with no inducer left, the repressor binds the operator again.
Lactose (as allolactose) is the inducer: it inactivates the repressor so the operon can be transcribed. But β-galactosidase breaks lactose down. Once the lactose is used up, no inducer remains, the repressor (made constitutively from the i gene) binds the operator again, and transcription of z, y and a stops. Existing enzyme is then diluted as the cells divide.
In one or two lines each, say what would go wrong if
the promoter of a gene were deleted;
Show solution
AnswerRNA polymerase could not bind, so the gene would not be transcribed.
The promoter is the binding site where RNA polymerase starts transcription and it fixes which strand is the template. Without it the gene is silent.
a cell lacked the tRNA for one amino acid;
Show solution
AnswerCodons for that amino acid could not be read, so translation would stall there.
tRNA is the adapter that reads a codon with its anticodon and delivers the matching amino acid. Without it that amino acid cannot be added and proteins containing it cannot be completed.
one exon were cut out along with the introns during splicing.
Show solution
AnswerPart of the coding sequence would be missing from the mRNA, so the protein would be altered.
Exons are the expressed sequences that remain in mature mRNA and carry the code for the protein. Losing one removes those codons, and if its length is not a multiple of three it also shifts the reading frame.
Give three different reasons why the Human Genome Project is described as a mega project.
Show solution
AnswerIts size (about bp), its cost (billions of dollars), its data needs (bioinformatics) and its international scale over 13 years.
- It set out to read about bp, the entire human sequence.
- At early estimates of about US$3 per base pair the cost ran to roughly US$9 billion.
- The data, about 3 gigabytes per genome, needed fast computers and a new field, bioinformatics, to store and analyse them.
- It took 13 years and partners in many countries, with goals that also included new technologies and ethical, legal and social questions.
A court must decide which of two men is a child's biological father. Explain what DNA fingerprinting is, how it would settle the case, and name two of its other uses.
Show solution
AnswerIt compares band patterns of highly variable VNTR DNA; every band of the child not found in the mother must be found in the father.
DNA fingerprinting compares people at highly polymorphic repeat regions such as VNTRs. DNA from each person is cut with restriction enzymes, separated by electrophoresis, blotted and probed with a labelled VNTR probe, giving each person a pattern of bands.
A child inherits one allele of each locus from each parent. So every band in the child's pattern must appear either in the mother's lane or in the biological father's lane. The man whose bands account for all the child's non-maternal bands is the father; the other man will lack several of them.
Other uses: identifying suspects or victims in forensic cases from blood, hair or saliva, and studying genetic diversity within and between populations.
Explain each term in two or three lines.
Transcription
Show solution
AnswerMaking RNA from the template strand of a DNA segment.
RNA polymerase binds a promoter and copies one strand of a transcription unit into RNA, 5′→3′, pairing A with U, until it reaches a terminator.
Polymorphism
Show solution
AnswerInherited DNA variation at a locus with a variant allele above 1 % frequency.
DNA polymorphism is variation in a DNA sequence between members of a population, arising from inherited mutations, when more than one allele exists and a variant has a frequency above 0.01. It is common in non-coding DNA.
Translation
Show solution
AnswerMaking a polypeptide on a ribosome by reading mRNA codons.
Charged tRNAs pair with mRNA codons on the ribosome, peptide bonds join their amino acids in order, and a release factor ends the chain at a stop codon.
Bioinformatics
Show solution
AnswerUsing computers to store, search and analyse biological data.
Bioinformatics applies computing to biology: storing huge sequence databases, aligning and assembling sequences, finding genes and comparing genomes. It grew rapidly alongside the Human Genome Project.
E. coli grown on N is moved to N medium and divides every 20 minutes. What percentage of its DNA molecules are fully light after 100 minutes?
%Show answer
Answer 93.75 %
100 min = 5 generations: one molecule becomes 32, of which 2 are hybrid and 30 light. 30/32 = 93.75 %.Which change in a gene's coding sequence is least likely to alter the protein?
Show answer
Answer (C) Changing the third base of a codon
Codons for the same amino acid often differ only in the third base (for example GGU, GGC, GGA and GGG are all glycine), so such a change is often silent. The others cause frameshifts or an early stop.Put these events in order after lactose is added to E. coli that has had none.
- β-galactosidase, permease and transacetylase accumulate
- Allolactose binds the repressor
- The repressor leaves the operator
- A few permease molecules let lactose into the cell
- RNA polymerase transcribes z, y and a
Show answer
Answer- A few permease molecules let lactose into the cell
- Allolactose binds the repressor
- The repressor leaves the operator
- RNA polymerase transcribes z, y and a
- β-galactosidase, permease and transacetylase accumulate
The basal level of permease admits the first lactose; the inducer releases the repressor, transcription follows, and the enzymes build up.In the standard genetic code, UGA codes for tryptophan.
Show answer
Answer False
Tryptophan's only codon is UGG; UGA is a stop codon. (Human mitochondria are one of the rare exceptions, reading UGA as tryptophan.)In eukaryotes the cap is added to the end of hnRNA and the poly-A tail to the end.
Show answer
Answer (1) 5′, (2) 3′
Methyl guanosine triphosphate caps the 5′ end; 200–300 adenylates are added to the 3′ end.A stretch of double helix makes 150 complete turns.
About how many base pairs does it contain?
bpShow answer
Answer 1500 bp
150 turns × 10 bp = 1500 bp.About how long is it, in nanometres?
nmShow answer
Answer 510 nm
150 turns × 3.4 nm = 510 nm (the same as 1500 bp × 0.34 nm).
Find your NCERT exercise question
Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.
| In your NCERT textbook | Practice question here |
|---|---|
| NCERT Exercises Q1 | Practice Q1 |
| NCERT Exercises Q2 | Practice Q2 |
| NCERT Exercises Q3 | Practice Q3 |
| NCERT Exercises Q4 | Practice Q4 |
| NCERT Exercises Q5 | Practice Q5 |
| NCERT Exercises Q6 | Practice Q6 |
| NCERT Exercises Q7 | Practice Q7 |
| NCERT Exercises Q8 | Practice Q8 |
| NCERT Exercises Q9 | Practice Q9 |
| NCERT Exercises Q10 | Practice Q10 |
| NCERT Exercises Q11 | Practice Q11 |
| NCERT Exercises Q12 | Practice Q12 |
| NCERT Exercises Q13 | Practice Q13 |
| NCERT Exercises Q14 | Practice Q14 |
Quick check: MCQs on Molecular Basis of Inheritance
- 5.1.1 Structure of Polynucleotide Chain4 questions
- 5.1.2 Packaging of DNA Helix3 questions
- 5.2.2 Properties of Genetic Material (DNA versus RNA)4 questions
- 5.3 RNA World2 questions
- 5.4.2 The Machinery and the Enzymes4 questions
- 5.5.3 Types of RNA and the process of Transcription4 questions
- 5.6.2 tRNA – the Adapter Molecule4 questions
- 5.7 Translation3 questions
- 5.8.1 The Lac operon4 questions
- 5.9.2 Applications and Future Challenges4 questions
- 5.10 DNA Fingerprinting4 questions
Key terms (English–Hindi)
- codonप्रकूटprakoot
- A group of three mRNA bases that stands for one amino acid or a stop signal; a tRNA reads it with a matching anticodon.First seen in the chapter →
- complementary strands
- Two strands whose bases pair A with T (or U) and G with C, so the sequence of one fixes the other.First seen in the chapter →
- DNA fingerprintingडीएनए अंगुलिछापीDNA angulichhapi
- Comparing people by the band pattern of highly variable repeat DNA, such as VNTRs.First seen in the chapter →
- DNA polymorphismबहुरूपताbahuroopata
- Inherited variation at a DNA locus, counted as a polymorphism when a second allele has a frequency above 0.01 in a population.First seen in the chapter →
- exon
- A part of a split gene that remains in the mature RNA. The intervening parts that are cut out are introns.First seen in the chapter →
- nucleosome
- About 200 bp of DNA wrapped around an octamer of histone proteins; the repeating unit of chromatin.First seen in the chapter →
- nucleotideन्यूक्लियोटाइडnyookliyotaid
- A unit of a nucleic acid: a nitrogenous base and a pentose sugar (together a nucleoside) with a phosphate on the 5′ carbon.First seen in the chapter →
- operon
- A group of bacterial genes transcribed together from one promoter and controlled by one operator, such as the lac operon.First seen in the chapter →
- promoter
- The DNA sequence at the upstream end of a transcription unit where RNA polymerase binds to start transcription.First seen in the chapter →
- semiconservative replication
- Copying in which each daughter DNA keeps one parental strand and gains one newly made strand.First seen in the chapter →
- splicing
- Removing the introns from a primary transcript and joining its exons in order.First seen in the chapter →
- template strand
- The DNA strand (3′→5′) that RNA polymerase reads. Its partner, the coding strand, has the RNA's sequence with T for U.First seen in the chapter →
- transcriptionअनुलेखनanulekhan
- Making RNA from one strand of a DNA segment, by base pairing, with U in place of T.First seen in the chapter →
- transformation
- A heritable change in a bacterium caused by DNA taken up from outside, as in Griffith's R to S change.First seen in the chapter →
- translation
- Building a polypeptide on a ribosome in the order set by the codons of an mRNA.First seen in the chapter →