Chapter 4: Molecular Basis of Inheritance
In the early 1950s, in a laboratory at King’s College London, a young scientist named Rosalind Franklin was pointing beams of X-rays at fibres of DNA. DNA cannot simply be placed under a microscope and looked at — it is far too small for that. So Franklin used a clever indirect method called X-ray diffraction: she fired X-rays through purified DNA and photographed the pattern of shadows the rays made as they scattered off the molecule’s atoms. One of her images, known simply as “Photo 51,” showed a striking dark cross of spots. To a trained eye, that cross was a signature. It could only be produced by a molecule twisted into a spiral — a helix.
Around the same time, James Watson and Francis Crick were working in Cambridge, trying to build a physical model of DNA out of metal plates and rods. When they saw the clue hidden in Franklin’s photograph, the last piece fell into place. In 1953 they proposed that DNA is a double helix — two strands winding around each other like a twisted ladder. That simple, elegant shape suddenly explained how a single molecule could store the instructions for building a whole organism, and copy itself faithfully every time a cell divides.
This chapter is the story of that molecule: how it was discovered, how it is packed into a tiny nucleus, how it is copied, how it is read to make proteins, and how it can even identify a single person.
Diagram
Fig. 4.0 – Opening illustration: the DNA double helix drawn as a twisted ladder, with one side showing the paired bases as rungs, beside a simplified sketch of the X-shaped “Photo 51” diffraction pattern that revealed the helix.
By the end of this chapter, you will be able to answer each of these questions.
4.1 The Discovery of DNA
Our present understanding of DNA did not arrive in a single flash of insight. It grew slowly, one discovery at a time — beginning with the isolation of a mysterious substance from cells in the nineteenth century, and eventually leading to the double-helix model you met at the start of this chapter.
The story begins inside the nucleus. In a eukaryotic cell, the nucleus is the membrane-bound control centre that houses the cell’s hereditary material. It was from this compartment that the very first sample of what we now call nucleic acid was pulled out.
In 1869, a scientist named Friedrich Miescher was studying white blood cells, which are the main component of the pus that forms at sites of infection. To collect large numbers of these cells, he gathered used bandages from a nearby hospital and washed the pus off them with a salt solution. When he then treated the cells with a weak alkaline (basic) solution, the cells burst open — they lysed — and their nuclei separated out and settled from the liquid. From these nuclei he extracted an unfamiliar chemical substance, which he called nuclein.
Miescher noticed two important things about nuclein. It was unusually rich in phosphorus, and it behaved chemically like an acid. Because it came from the nucleus and showed acidic properties, the substance was later renamed nucleic acid.
By the early 1900s, scientists had realised that Miescher’s “nuclein” was not a single pure substance but a mixture of proteins and nucleic acids. They also discovered that nucleic acids exist in two forms: DNA (deoxyribonucleic acid) and RNA (ribonucleic acid). These two molecules — and DNA in particular — are the central characters of everything that follows in this chapter.
4.2 The Genetic Material is DNA
By the early 1900s, geneticists had already worked out three important facts. They knew that genes control the inheritance of traits, that genes are located on chromosomes, and that chromosomes are made mainly of two kinds of molecules — DNA and proteins. What they did not know was which of these two molecules actually carried the genetic information.
Why proteins were the first suspects
At first, almost every geneticist backed proteins. Their reasoning was not unreasonable. Proteins are large, complex and enormously varied molecules, and they were already known to store the information needed to run the cell’s metabolism. Since proteins differ so widely from one another, it seemed natural to assume that proteins were the source of the variations seen within a species.
DNA, on the other hand, was dismissed. It was thought to be a small, simple molecule whose composition hardly varied from one species to another — hardly the sort of molecule that could hold a whole organism’s instructions.
Over time, both of these ideas about DNA turned out to be wrong. DNA molecules are in fact very large, and they vary tremendously both within a species and between species. The catch is that the variation in DNA is of a different kind from the variation in proteins. Proteins differ visibly in shape, electrical charge and function, while DNA molecules differ only in the order of their building blocks — a difference that is invisible to the methods of that time. Given this, it is not surprising that most researchers initially favoured proteins as the genetic material.
Over a period of roughly 25 years (1928–1952), the tide of opinion turned completely, and geneticists became convinced that DNA — not protein — was the genetic material. Three important contributions caused this shift of opinion, and we shall look at each one in turn.
Flowchart
FC 4.1 – The 25-year shift in opinion (1928–1952): Griffith (1928): a “transforming principle” exists → Avery, MacLeod and McCarty (1944): the transforming principle is DNA → Hershey and Chase (1952): DNA, not protein, enters the bacterial cell → conclusion: DNA is the genetic material
Griffith’s experiments
In 1928, a British medical officer named Frederick Griffith performed an experiment on the bacterium Streptococcus pneumoniae, which causes pneumonia in humans and other mammals. Griffith was not hunting for the genetic material at all. He was searching for a cure for pneumonia, which was a common cause of death at that time.
Griffith worked with two strains — that is, two genetic varieties — of Streptococcus:
- S type — virulent, smooth, pathogenic and encapsulated. The capsule is a slimy protective coat around the bacterium that shields it from the host’s defences, which is why this strain is deadly and why its colonies look smooth.
- R type — non-virulent, rough, non-pathogenic and non-capsulated. Without a capsule, the host’s immune system destroys these bacteria easily, and their colonies look rough.
Griffith conducted four experiments on mice using these bacteria, and the sequence in which he did them matters, because each one rules out one possible explanation. The four experiments are shown in Fig. 4.1.
Diagram
Fig. 4.1 – Griffith’s experiment: four mice, each injected respectively with (a) live rough strain R → mouse alive, (b) live smooth strain S → mouse dead, (c) heat-killed smooth strain S → mouse alive, (d) heat-killed S mixed with live R → mouse dead, with live S bacteria recovered from its blood.
In the first experiment, he injected bacteria of strain R into mice. The mice survived, because they did not develop pneumonia — R cells are harmless.
In the second experiment, he injected bacteria of strain S into mice. The mice developed pneumonia and died — S cells are deadly.
In the third experiment, he injected heat-killed strain S bacteria into mice. Once again the mice survived, showing that dead S cells on their own cannot cause disease.
In the fourth experiment, he mixed heat-killed S bacteria with live bacteria of strain R and injected the mixture into mice. This time the mice died. And when Griffith examined the blood of the dead mice, he recovered large numbers of live strain S bacteria.
This last result is the surprising one. Neither of the two ingredients in the mixture could kill on its own — the S cells were dead, and the R cells were harmless. Yet together they were lethal, and living S bacteria appeared where none had been injected.
Something in the mixture had caused the harmless R strain bacterium to change into the deadly S strain bacterium. Griffith showed that this change was genetic — the transformed bacteria bred true and continued to produce capsules. He suggested that genetic material from the heat-killed S bacterium had somehow changed the living R bacterium into an S bacterium.
Griffith concluded that the R-strain bacterium must have taken up something from the heat-killed S bacterium, which he called a “transforming principle”. This principle allowed the R strain to be transformed into a smooth-coated bacterium and to become virulent.
Notice carefully what Griffith did and did not prove. He proved that a chemical substance can carry a hereditary trait from one cell to another. He did not identify that substance. Heat kills the S bacteria, but heat does not destroy every molecule inside them, so the transforming principle could still have been protein, RNA or DNA. Identifying it was the next task.
Avery, MacLeod and McCarty’s experiment
In 1944, after some ten years of research and experimentation, the U.S. microbiologists Oswald T. Avery, Colin M. MacLeod and Maclyn McCarty (all at Rockefeller University in New York) provided the first direct evidence that DNA is the genetic material — that DNA is Griffith’s transforming principle.
Their strategy was one of elimination. They purified DNA, RNA, proteins (enzymes) and other materials separately from a cell-free extract of S cells, and mixed each purified substance with heat-killed S strain and R cells separately, to confirm which one could transform living R cells into S cells. Only DNA was able to transform harmless strain R into deadly strain S.
They then attacked the problem from the opposite direction, using enzymes that destroy specific molecules. They found that protein-digesting enzymes (proteases) and RNA-digesting enzymes (RNAses) did not affect transformation at all — so the transforming substance was neither a protein nor RNA. But DNA digested with DNAse did inhibit the transformation, showing that once the DNA was destroyed, the transforming ability vanished with it. This suggested that DNA caused the transformation. The result is summarised in Fig. 4.2.
Diagram
Fig. 4.2 – DNA transforms bacteria: rough non-virulent strain R mixed with heat-killed smooth virulent strain S; when protease is added → mouse dead; when DNase is added → mouse alive.
This is powerful reasoning, and it is worth pausing on. Destroying the proteins made no difference — so proteins were not needed. Destroying the RNA made no difference — so RNA was not needed. Destroying the DNA abolished transformation completely — so DNA was the essential ingredient. These experiments proved that the transforming principle is DNA. Even so, not all biologists were convinced; many still suspected that traces of protein contaminating the DNA preparation might be doing the real work.
Hershey–Chase Experiment
Finally, Alfred Hershey and Martha Chase (1952) proved that DNA, and not protein, is the genetic material, by using bacteriophages.
Hershey and Chase worked with viruses that infect bacteria, that is, bacteriophages. A bacteriophage is beautifully suited to this question because it is made of just two things — DNA on the inside and protein on the outside. It infects a bacterium and forces it to make new phages. So whichever of the two molecules actually enters the bacterium must be the one carrying the instructions.
The problem was how to tell DNA and protein apart inside a living cell. Hershey and Chase solved it by using radioactive labels. They used radioactive phosphorus \(^{32}P\) in the medium for some viruses and radioactive sulphur \(^{35}S\) for some others.
The choice of these two elements is the heart of the experiment. DNA contains phosphorus but no sulphur, because of the phosphate groups in its backbone. Protein contains sulphur (in certain amino acids) but no phosphorus. So each label goes to one molecule and only that molecule.
They grew some viruses on a medium that contained radioactive phosphorus and some others on a medium that contained radioactive sulphur. Viruses grown in the presence of radioactive phosphorus contained radioactive DNA (labelled DNA) but not radioactive proteins, because DNA contains phosphorus but proteins do not. Similarly, viruses grown on radioactive sulphur contained radioactive protein but not radioactive DNA, because DNA does not contain sulphur.
These radioactive phages were then allowed to infect E. coli bacteria grown on a medium containing normal ‘P’ and ‘S’. Then, as the infection proceeded, the viral coats were removed from the bacterial surface with the help of a centrifuge. Spinning the mixture at high speed throws the heavy bacterial cells to the bottom as a pellet, while the much lighter empty phage coats stay in the liquid above. The steps are shown in Fig. 4.3.
Diagram
Fig. 4.3 – Hershey–Chase experiment: two parallel tracks. Left: phages with sulphur-labelled protein capsule → infection → blending → centrifugation → no sulphur detected in cells, sulphur detected in supernatant. Right: phages with phosphorus-labelled DNA core → infection → blending → centrifugation → phosphorus detected in cells, no phosphorus detected in supernatant.
Now came the reading of the result. Bacteria which had been infected by viruses carrying radioactive DNA were themselves radioactive, indicating that DNA was the material that passed from the viruses into the bacteria. Bacteria which had been infected by viruses carrying radioactive sulphur (that is, radioactive protein) were not radioactive. This indicates that the proteins from the viruses did not enter the bacteria at all — the protein coat stayed outside, like an empty syringe left on the surface.
In other words, sometime after infection, radioactivity for ‘P’ and for ‘S’ was tested. Only radioactive ‘P’ was found inside the bacterial cell. DNA is therefore the genetic material that is passed from virus to bacteria.
How this section is tested
In NEET and board papers, this section usually appears as assertion–reason questions (“Assertion: Hershey and Chase used \(^{35}S\) to label DNA. Reason: DNA contains sulphur.”) or as sequence-ordering questions asking you to arrange Griffith’s four mouse experiments, or the three landmark experiments, in the correct order. The two facts most often tested are that DNA contains phosphorus but not sulphur, and that protein contains sulphur but not phosphorus — get these the right way round and most questions on this section become straightforward.
4.3 DNA Packaging
Now that we know DNA is the genetic material, a very practical problem arises. The length of the DNA double helix molecule in a typical mammalian cell is approximately 2.2 metres. You can work this out quite simply by multiplying the total number of base pairs by the distance between two consecutive base pairs. Yet the approximate size of a typical nucleus is only \(10^{-6}\) m.
Think about what that means. More than two metres of thread has to be fitted into a container about a millionth of a metre across — and the thread must not tangle, break, or become unreadable, because the cell has to keep using it. The DNA must therefore be condensed, coiled and supercoiled to fit inside such a small nucleus.
This is what we mean by DNA packaging: the orderly folding of a very long DNA molecule into a very small space, in a way that still allows the cell to open up any region it needs to read.
Prokaryotes and eukaryotes solve this problem in different ways, because they have very different cells, so we take them one at a time.
Packaging in Prokaryotes
In prokaryotes like E. coli, the cell size is almost 2–3 μ long. Prokaryotes do not have a well-organised nucleus — there is no nuclear membrane and no nucleolus. Instead, the DNA sits in a region of the cytoplasm called the nucleoid.
The nucleoid is a small, circular, highly folded, naked ring of DNA. “Naked” here means that it is not wrapped around histone proteins the way eukaryotic DNA is. This ring is 1100 μ long in perimeter and contains about 4.6 million base pairs.
Here is the packing problem in miniature. The 1100 μ long nucleoid (approximately 1.1 mm, if cut and stretched out) has to be fitted or packaged into a cell that is hardly 2–3 μ long. The DNA is therefore compacted in a series of steps, as shown in Fig. 4.4.
Diagram
Fig. 4.4 – DNA packaging in prokaryotes: (a) circular, unfolded chromosome, 350 μ → (b) folded chromosome with 40 to 50 loops, 30 μ → (c) supercoiled, folded chromosome, 2 μ, each loop independently supercoiled; also show (d) partially uncoiled chromosome after partial DNase digestion and (e) partially unfolded chromosome after partial RNase digestion, with RNA cleaved and nicked DNA labelled.
First, the negatively charged DNA becomes circular, which reduces the size to 350 μm in diameter. Next, this is further reduced to 30 μm in diameter because of folding and looping — 40 to 50 domains (loops) are formed, and the formation of these loops is assisted by RNA connectors. Finally, each domain is further coiled and supercoiled, thereby reducing the size down to 2 μ in diameter.
This coiling (packaging) is assisted by positively charged HU proteins (Histone-like DNA binding proteins) and by enzymes such as DNA gyrase and DNA topoisomerase I, which maintain the supercoiled state. Notice the recurring principle here: DNA carries a negative charge because of its phosphate backbone, so the proteins that bind and fold it are positively charged. Opposite charges attract, and this simple electrostatic pull is what allows proteins to hold DNA tightly against itself.
Flowchart
FC 4.2 – Levels of prokaryotic DNA packaging: naked circular DNA, 1100 μ perimeter → circular form, 350 μm diameter → folding/looping into 40–50 domains with RNA connectors, 30 μm diameter → supercoiling of each domain (HU proteins, DNA gyrase, topoisomerase I), 2 μ diameter
Packaging in Eukaryotes
Eukaryotes show a well-organised nucleus with a nuclear membrane, a nucleolus, and thread-like material in the form of chromosomes. In these chromosomes, DNA is associated with histone and non-histone proteins, as was reported by R. Kornberg in 1974. The organisation of DNA in eukaryotes is therefore much more complex than in prokaryotes.
To understand how this works, we first need to understand why histones can grip DNA. Depending upon the abundance of amino acid residues with charged side chains, a protein acquires its charge. Histones are proteins that are rich in lysine and arginine residues. Both of these amino acid residues are basic amino acids and carry positive charges. So histones are a set of positively charged, basic proteins (histones + protamine).
These histones organise themselves into a unit of 8 molecules known as the histone octamer. The negatively charged helical DNA is wrapped around the positively charged histone octamer, forming a structure known as the nucleosome, which you can see in Fig. 4.5.
Diagram
Fig. 4.5 – Nucleosome: an octamer of eight histone molecules (two each of H₂A, H₂B, H₃ and H₄) with DNA wound around it, and H1 histone shown binding the DNA where it enters and leaves the octamer.
The nucleosome core is an octamer made up of two molecules of each of four types of histone proteins, namely \(H_2A\), \(H_2B\), \(H_3\) and \(H_4\). A fifth histone, the \(H_1\) protein, is not part of the core — it binds the DNA thread at the point where the thread enters (arrives at) and leaves the nucleosome, rather like a clip holding the two ends of a wrapped bundle.
One nucleosome approximately contains 200 base pair long DNA helix, of which about 146 bp long segment is wound around each octamer, and the remaining bp contribute as linker DNA — the short stretch of DNA that runs between one bead and the next.
Nucleosomes are the repeating units of chromatin, which are thread-like, stained (coloured) bodies present in the nucleus. When chromatin is observed under an electron microscope, these repeating nucleosomes look like ‘beads-on-string’, as seen in Fig. 4.6 — the beads being the nucleosomes and the string being the linker DNA between them.
Diagram
Fig. 4.6 – Chromatin showing beads-on-string: electron-microscope-style view of nucleosomes strung along a DNA thread, with nucleosome and linker DNA labelled.
The DNA helix of 200 bps wraps around the histone octamer by 1¾ turns. Six such nucleosomes then get coiled repeatedly and form a solenoid, which looks like a coiled telephone wire. The stages of this coiling are shown together in Fig. 4.7.
Diagram
Fig. 4.7 – DNA packaging in eukaryotes: (A) beads-on-string magnified, showing nucleosome core of 146 nucleotide pairs wrapped 1¾ turns around the histone octamer and linker DNA of 8 to 114 nucleotide pairs; (B) chain of nucleosomes forming a 10 to 11 nm thick fibre; (C) solenoid forming a 30 nm (300 Å) thick fibre.
The chromatin is a 10 nm thick fibre, and this is packed to form a solenoid structure of 30 nm diameter (300 A⁰). Further supercoiling of the solenoid tends to form a looped structure, which coils still more and finally condenses at the metaphase stage of cell division to form the chromosomes — the compact, visible bodies you have already seen in your study of cell division.
The packaging of chromatin at these higher levels needs an additional set of proteins that are called Non-Histone Chromosomal proteins (NHC).
Flowchart
FC 4.3 – Levels of eukaryotic DNA packaging: DNA double helix (2 nm) → DNA wrapped 1¾ turns around histone octamer → nucleosome → chain of nucleosomes = “beads-on-string” chromatin fibre (10 nm) → six nucleosomes coil into a solenoid (30 nm / 300 Å) → supercoiled looped structure (with NHC proteins) → condensed metaphase chromosome
Non-Histone Chromosomal Proteins (NHC)
These are the additional sets of proteins that contribute to the packaging of chromatin at a higher level — that is, beyond the nucleosome and solenoid stages that the histones themselves manage.
Heterochromatin and Euchromatin
Not all of the chromatin in a nucleus is packed to the same degree, and this difference has a direct consequence for which genes a cell can actually use.
1. Heterochromatin: In eukaryotic cells, some segments of the chromonema (chromosome) remain in a condensed state during interphase and early prophase. These regions constitute heterochromatin. This term was proposed by Heitz. These regions are localised near the centromere and the telomeres, and are also intercalated (scattered in between). Heterochromatin is genetically mostly inactive. It stains strongly and appears dark. Heterochromatin is 2 to 3 times more rich in DNA than the euchromatin.
2. Euchromatin: The regions of the chromonema which are in a non-condensed state constitute euchromatin. Euchromatic regions stain light. Euchromatin is genetically very much active and fast replicating.
Putting these two together gives the key functional statement: euchromatin is transcriptionally active, while heterochromatin is transcriptionally almost inactive. The reason is exactly the one we met a moment ago — loosely packed DNA can be reached by the enzymes that read it, tightly packed DNA cannot. Packing is not just about storage; it is also a way of switching whole regions of the genome on or off.
How this section is tested
Diagram-labelling questions on the nucleosome are extremely common — you may be asked to identify the octamer, the linker DNA, or the \(H_1\) histone. Sequence-ordering questions ask you to arrange the levels of packaging from DNA double helix up to the metaphase chromosome. The numbers most often tested are 146 bp wound around the octamer, 200 bp per nucleosome, 1¾ turns, the 10 nm fibre and the 30 nm solenoid — learn these as a set rather than in isolation.
4.4 DNA Replication
The DNA molecule regulates and controls all the activities of the cell. Because of its unique structure, it is able to control the synthesis of other molecules of the cell. At the same time, when the cell reproduces, the DNA must also duplicate itself so that it can be distributed equally to the daughter cells.
This gives us a useful way of splitting DNA’s job in two. As a carrier of genetic information, DNA has to perform two important functions:
a. Heterocatalytic function: When DNA directs the synthesis of chemical molecules other than itself, such functions of DNA are called heterocatalytic functions. Examples are the synthesis of RNA (transcription) and the synthesis of protein (translation).
b. Autocatalytic function: When DNA directs the synthesis of DNA itself, such a function of DNA is called an autocatalytic function. The example is replication.
The Greek prefixes tell you the difference at a glance: hetero- means “other,” auto- means “self.” This section is about the autocatalytic function; the heterocatalytic one is the subject of the next section.
The process by which DNA duplicates itself is called replication. Through replication, it forms two copies that are identical to each other and also to the parent DNA.
In eukaryotic organisms, replication of DNA takes place only once in the cell cycle. It occurs in the S-phase of interphase in the cell cycle. This timing matters: the cell must finish copying every chromosome before it starts dividing, or the daughter cells would receive incomplete sets of genes.
DNA replicates through the semiconservative mode of replication. The model for semiconservative replication was proposed by Watson and Crick, on the basis of the antiparallel and complementary nature of DNA strands. Because the two strands are complementary — A always pairs with T, and G always pairs with C — each strand already contains all the information needed to rebuild its partner. Separate the two strands, and each one can act as a set of instructions for making a new partner strand.
The steps of semiconservative replication
The process of semiconservative replication takes place through the following steps. The overall picture is shown in Fig. 4.8, and you should keep looking back to it as we go along.
Diagram
Fig. 4.8 – Semiconservative replication of DNA: a Y-shaped replication fork with the leading template (3′ free end) and lagging template (5′ free end), the continuous leading strand growing towards the fork, the discontinuous lagging strand made of Okazaki fragments growing away from the fork, RNA primers, and the parental and new strands labelled with base pairs (A=T, T=A, C=G, G=C).
1. Activation of Nucleotides
The four types of nucleotides of DNA — dAMP, dGMP, dCMP and dTMP — are present in the nucleoplasm. They are activated by ATP in the presence of an enzyme called phosphorylase. This results in the formation of deoxyribonucleotide triphosphates, that is, dATP, dGTP, dCTP and dTTP. The process is known as phosphorylation.
Why is this step needed at all? Because joining nucleotides into a chain costs energy. By converting each monophosphate into a triphosphate, the cell loads every building block with the energy it will later release when it is added to the growing strand. The nucleotides arrive at the site of synthesis already “charged.”
2. Point of Origin or Initiation point
Replication does not begin anywhere at random. It begins at a specific point ‘O’ — the origin — and terminates at point ‘T’. The origin is flanked by ‘T’ sites. The unit of DNA in which replication occurs is called a replicon.
In prokaryotes, there is only one replicon. In eukaryotes, there are several replicons in tandem. This difference is a matter of practicality: a eukaryotic chromosome is enormously longer than a bacterial one, so copying it from a single origin would take far too long. Starting at many origins at once, and letting the copied stretches meet up, finishes the job in a workable time.
At the point ‘O’, the enzyme endonuclease nicks one of the strands of DNA, temporarily. The nick occurs in the sugar-phosphate backbone, that is, in the phosphodiester bond. This small cut gives the unwinding machinery a place to begin.
3. Unwinding of DNA molecule
Now the enzyme DNA helicase operates by breaking the weak hydrogen bonds in the vicinity of ‘O’. The strands of DNA separate and unwind. This unwinding is bidirectional — it proceeds in both directions away from the origin — and continues as a ‘Y’-shaped replication fork. Each separated strand acts as a template.
Notice the beauty of the chemistry here. The hydrogen bonds holding the two strands together are individually weak, so they can be broken without damaging the strands themselves; but there are so many of them along the length of the molecule that the double helix as a whole is very stable. This is exactly what a molecule needs if it is to be both durable and openable.
The two separated strands are prevented from recoiling (rejoining) by SSBP (Single Strand Binding Proteins). SSB proteins remain attached to both the separated strands so as to facilitate the synthesis of new polynucleotide strands. Without them, the two complementary strands would simply snap back together.
4. Replicating fork
The point formed due to the unwinding and separation of the two strands appears like a Y-shaped fork, called the replicating or replication fork. The unwinding of the strands imposes strain on the DNA ahead of the fork — imagine untwisting one end of a rope while the other end is fixed; the twist has to go somewhere. This strain is relieved by a super-helix relaxing enzyme.
5. Synthesis of new strands
Each separated strand acts as a mould or template for the synthesis of a new complementary strand. Synthesis begins with the help of a small RNA molecule called the RNA primer. The RNA primer gets associated with the 3’ end of the template strand and attracts complementary nucleotides from the surrounding nucleoplasm.
These nucleotide molecules bind to the complementary nucleotides on the template strand by forming hydrogen bonds — that is, A=T or T=A; G=C or C=G. The newly bound nucleotides then get interconnected by phosphodiester bonds, forming a polynucleotide strand.
The synthesis of the new complementary strand is catalysed by the enzyme DNA polymerase. There is one absolutely fixed rule here, and everything that follows depends on it: the new complementary strand is always formed in the 5’→3’ direction.
6. Leading and Lagging strand
The two template strands are not identical in orientation, because the two strands of DNA are antiparallel — they run in opposite directions. This creates a genuine problem, and the cell’s solution to it is one of the most elegant details in all of molecular biology.
The template strand with a free 3’ end is called the leading template, and the one with a free 5’ end is called the lagging template. The process of replication always starts at the C-3 end of the template strand and proceeds towards the C-5 end. As both strands of the parental DNA are antiparallel, and new strands are always formed in the 5’→3’ direction, the two new strands cannot both be made in the same way.
One of the newly synthesised strands develops continuously towards the replicating fork. This is called the leading strand. The other new strand develops discontinuously, away from the replicating fork. This is called the lagging strand.
Flowchart
FC 4.4 – Why the lagging strand is discontinuous: DNA polymerase can build only in 5’→3’ direction → the two parental strands are antiparallel → on the 3’-ended (leading) template, 5’→3’ synthesis happens to point towards the fork → continuous leading strand; on the 5’-ended (lagging) template, 5’→3’ synthesis points away from the fork → as the fork opens further, synthesis must restart each time → short Okazaki fragments → discontinuous lagging strand
Maturation of Okazaki fragments
DNA synthesis on the lagging template takes place in the form of small fragments, called Okazaki fragments (named after the scientist Okazaki). These fragments are separate pieces, so the cell must now stitch them into one continuous strand. Okazaki fragments are joined by the enzyme DNA ligase.
There is one more loose end to tidy up. Every fragment began with an RNA primer, and RNA has no place in a finished DNA molecule. The RNA primers are removed by DNA polymerase and replaced by DNA sequence, with the help of DNA polymerase-I in prokaryotes and DNA polymerase-α in eukaryotes.
Finally, the DNA gyrase (topoisomerase) enzyme forms the double helix, giving the daughter DNA molecules their proper twisted shape.
Flowchart
FC 4.5 – Maturation of Okazaki fragments: lagging template → RNA primer laid down → short Okazaki fragment synthesised by DNA polymerase (5’→3’) → RNA primer removed and replaced by DNA (DNA polymerase-I in prokaryotes; DNA polymerase-α in eukaryotes) → adjacent fragments sealed by DNA ligase → continuous lagging strand → DNA gyrase (topoisomerase) restores the double helix
7. Formation of daughter DNA molecules
At the end of replication, two daughter DNA molecules are formed. In each daughter DNA, one strand is parental and the other is completely newly synthesised. Thus 50% of each daughter molecule is contributed by the mother DNA. This is precisely why the process is described as semiconservative replication — “semi” meaning half, and “conservative” meaning preserved. Half of each new molecule is preserved from the old one.
Experimental confirmation
The semiconservative model is a claim about what happens inside a cell — so it had to be tested, not merely assumed. In a newly formed DNA molecule, one strand is old (that is, conserved) and the other strand is newly synthesised. This was experimentally proved by Matthew Meselson and Franklin Stahl (1958) using the equilibrium-density-gradient-centrifugation technique.
Their idea was to make the old strands physically heavier than the new ones, so that old and new DNA could be told apart by weight. They did this using two isotopes of nitrogen — nitrogen being an element found in all four nitrogen bases of DNA. The heavy isotope \(^{15}N\) makes DNA denser; the ordinary isotope \(^{14}N\) leaves it light.
The experiment proceeded as follows.
Meselson and Stahl in 1958 performed an experiment to verify the semiconservative nature (mode) of replication.
E. coli cells growing in \(^{14}N\) were transferred to a \(^{15}N\) medium (heavy isotopic nitrogen) and allowed to replicate for several generations. At the equilibrium point, a density gradient band was obtained by using 6M \(CsCl_2\). The position of this band was recorded.
The heavy DNA (\(^{15}N\)) molecule can be distinguished from normal DNA (\(^{14}N\)) by centrifugation in a 6M caesium chloride (\(CsCl_2\)) density gradient. The density gradient value of 6M \(CsCl_2\) and \(^{15}N\) DNA is almost the same. Therefore, at the equilibrium point, \(^{15}N\) DNA will form a band. In this DNA, both the strands are labelled with \(^{15}N\).
Such E. coli cells were then transferred to another medium containing \(^{14}N\), that is, normal (light) nitrogen. After the first generation, the density gradient band for \(^{14}N\)–\(^{15}N\) (hybrid) was obtained and its position was recorded. After the second generation, two density gradient bands were obtained — one at the \(^{14}N\)–\(^{15}N\) position and the other at the \(^{14}N\) position.
The position of the bands after two generations clearly proved that DNA replication is semiconservative. The results are interpreted in Fig. 4.9.
Diagram
Fig. 4.9 – Interpretation of the results of Meselson’s experiment on the separation of DNA by equilibrium density gradient centrifugation: starting \(^{15}N\)-DNA (heavy band); after 20 min (Generation I) all molecules are \(^{14}N\)–\(^{15}N\) hybrid (single hybrid band); after 40 min (Generation II) half the molecules are \(^{14}N\)-DNA and half are \(^{14}N\)–\(^{15}N\) hybrid (two bands — light and hybrid), with the corresponding centrifuge tubes drawn beside each generation.
How this section is tested
Expect sequence-ordering questions asking you to arrange the enzymes of replication in the order in which they act (endonuclease → helicase → SSBP → DNA polymerase → DNA ligase → DNA gyrase). Assertion–reason questions frequently pair the lagging strand with the antiparallel nature of DNA, and diagram-labelling questions ask you to mark the leading template, lagging template, RNA primer and Okazaki fragments on a replication fork. For Meselson and Stahl, the classic question is what the band pattern looks like after a given number of generations — reason it out from the model rather than memorising it.
4.5 Protein Synthesis
Proteins are very important biomolecules. They serve as structural components, as enzymes, and as hormones. A cell is constantly using up its proteins and constantly needs to synthesise new protein molecules.
But the instructions for building every one of those proteins are stored in the DNA, and DNA never leaves the nucleus in a eukaryotic cell, while proteins are built in the cytoplasm. There must therefore be a messenger — a working copy of the instructions that can travel out and be read where the proteins are made. That messenger is RNA.
The process of protein synthesis accordingly includes two stages: transcription and translation.
The process of copying genetic information from one (template) strand of DNA into a single-stranded RNA transcript is termed transcription. During this process, synthesis of a complementary strand of RNA takes place — except that the adenine nitrogen base pairs with the uracil base instead of thymine, because RNA contains uracil in place of thymine.
Central Dogma
The double-stranded DNA molecule gives rise to mRNA, which acts as a messenger to programme the synthesis of a polypeptide chain (protein). This type of unidirectional flow of information from DNA to RNA to protein is referred to as the central dogma of molecular biology. It was postulated by F. H. C. Crick in 1958.
\[\text{DNA} \xrightarrow{\text{Transcription}} \text{mRNA} \xrightarrow{\text{Translation}} \text{Polypeptide}\]
The word “unidirectional” is important. The classical central dogma says information travels one way: from the stored master copy (DNA), to the working copy (mRNA), to the finished product (protein). Information does not flow backwards from protein to DNA.
But there is one important exception, and it comes from viruses. The present concept of the central dogma in retroviruses (also called riboviruses) is given by Temin (1970) and Baltimore (1970):
\[\text{DNA} \xrightarrow{\text{Transcription}} \text{mRNA} \xrightarrow{\text{Translation}} \text{Polypeptide}\] \[\text{RNA} \xrightarrow{\text{Reverse Transcription}} \text{DNA}\]
Accordingly, the enzyme RNA-dependent DNA polymerase synthesises DNA from RNA. This reverses the usual direction of the first arrow — hence the name reverse transcription.
Flowchart
FC 4.6 – Central dogma and its exception: DNA → (transcription) → mRNA → (translation) → polypeptide; in retroviruses, additionally: RNA → (reverse transcription, by RNA-dependent DNA polymerase) → DNA → back into the normal pathway
A. Transcription
During transcription, the information of only one strand of DNA is copied into RNA. This strand of DNA acts as the template. The enzyme RNA polymerase catalyses the formation of the RNA transcript.
DNA is located in the nucleoid of prokaryotes and in the nucleus of eukaryotes. In eukaryotes, DNA transcription takes place in the nucleus, whereas translation occurs in the cytoplasm. DNA transfers its information to mRNA, which then moves to the ribosomes.
Transcription occurs in the nucleus during the G1 and G2 phases of the cell cycle. DNA has promoter and terminator sites: transcription starts at the promoter site and stops at the terminator site. In both prokaryotes and eukaryotes, the process of transcription involves three stages — initiation, elongation and termination.
Transcription Unit
Each transcribed segment of DNA is called a transcription unit. It consists of three parts: i. a promoter, ii. the structural gene, and iii. a terminator. The arrangement is shown in Fig. 4.10.
Diagram
Fig. 4.10 – Transcription unit: a stretch of double-stranded DNA showing the promoter at the left, the structural gene in the middle, and the terminator at the right; the upper strand labelled template strand (3′ → 5′) and the lower strand labelled coding strand (5′ → 3′).
The two strands of DNA in the structural gene show the following features:
i. The promoter is located towards the 5’ end of the structural gene, that is, upstream. It is a DNA sequence that provides the binding site for the enzyme RNA polymerase. RNA polymerase binds to a specific promoter. In prokaryotes, the enzyme recognises the promoter by its sigma factor subunit.
ii. In the structural gene, the two strands of DNA have opposite polarity. DNA-dependent RNA polymerase catalyses polymerisation only in the 5’→3’ direction. Therefore the DNA strand having 3’→5’ polarity acts as the template strand. The other strand of DNA, having 5’→3’ polarity, is complementary to the template strand. The sequence of bases in this strand is the same as in the RNA (where thymine is replaced by uracil). It is the actual coding strand. The information on this strand of DNA is copied onto the mRNA, and it is therefore called the sense strand.
iii. The terminator is located at the 3’ end of the coding strand, that is, downstream. It defines the end of the transcription process.
Take a moment to see why the template strand must be the 3’→5’ one. RNA polymerase can only build the new RNA chain in the 5’→3’ direction, and the RNA must be antiparallel to the strand it is being read from. So the enzyme has to travel along its template from the template’s 3’ end towards its 5’ end. Only the 3’→5’ strand offers this. The rule is not arbitrary — it follows directly from how the enzyme works.
After binding to the promoter, RNA polymerase moves along the DNA and causes local unwinding of the DNA duplex into two chains in the region of the gene. The exposed A, T, C and G bases project into the nucleoplasm. Only one strand functions as the template (the antisense strand), and the other strand is complementary to it and is actually the coding strand (the sense strand).
The ribonucleoside triphosphates then join to the bases of the DNA template chain. As transcription proceeds, the hybrid DNA–RNA molecule dissociates and makes the mRNA molecule free. The relationship between the template strand, the coding strand and the mRNA is shown in Fig. 4.11.
Diagram
Fig. 4.11 – Formation of template and coding strand of DNA: the mRNA (5′ A G U C A G U A C A U G 3′), the template strand of DNA (antisense strand, 3′ T C A G T C A T G T A C 5′), and the coding strand of DNA (sense strand, 5′ A G T C A G T A C A T G 3′) shown side by side so their correspondence is visible.
When RNA polymerase reaches the terminator signal on the DNA, it leaves the DNA and the fully formed mRNA (the primary transcript) is released. As the mRNA grows, the transcribed region of the DNA molecule becomes spirally coiled again and regains its double helical form — the DNA is left exactly as it was, undamaged and ready for the next round.
RNA polymerases
In bacteria, mRNA does not require any processing, because it has no introns. Prokaryotes possess only one type of RNA polymerase, which makes all their RNA.
In eukaryotes, there are three types of RNA polymerases, each with a specialised job:
- RNA polymerase-I transcribes rRNA.
- RNA polymerase-II transcribes mRNA (the primary transcript) and heterogeneous nuclear RNA (or hnRNA).
- RNA polymerase-III is responsible for the transcription of tRNA and small nuclear RNA (snRNA).
Transcription unit and the gene
The DNA sequence coding for mRNA, tRNA or rRNA is defined as a gene. A cistron is a segment of DNA coding for a polypeptide.
A single structural gene in a transcription unit is said to be monocistronic, whereas a long segment of DNA having a set of various structural genes in one transcription unit is referred to as polycistronic. Bacteria commonly use the polycistronic arrangement, which allows a whole group of related enzymes to be switched on together — a point that will matter greatly when we come to the lac operon.
Structural genes in eukaryotes have interrupted non-coding sequences called introns. The coding or express-sequences are defined as exons. Only exons appear in the processed mRNA in eukaryotes.
Processing of hnRNA
In eukaryotes, the forms of RNA transcribed directly from DNA are called primary transcripts. Such transcripts undergo changes called processing or maturation before becoming functional.
The primary transcript is non-functional, and it contains both exons and introns. If it were sent out to the ribosome as it is, the ribosome would read the intron sequences too and produce a nonsense protein. So the introns must be cut out first.
During processing, only the introns are removed by a process called splicing. The exons are then joined in a definite sequence (order) by the DNA ligase enzyme.
Heterogeneous nuclear RNA also undergoes the processes of capping and tailing. In capping, a methylated guanosine triphosphate is added to the 5’ end of the hnRNA. In tailing, polyadenylation takes place at the 3’ end — a long tail of adenine nucleotides (a poly-A tail) is attached there.
It is this fully processed hnRNA that is now called mRNA. For translation, the mRNA is transported out of the nucleus through a nuclear pore. The whole sequence is shown in Fig. 4.12.
Diagram
Fig. 4.12 – Transcription and processing of hnRNA to mRNA in eukaryotes: DNA transcribed into hnRNA containing exons and introns → capping (methylated guanosine triphosphate, ᵐG_PPP, added at 5′ end) → polyadenylation (poly-A tail added at 3′ end) → RNA splicing removes introns and joins exons → mature messenger RNA (mRNA).
Flowchart
FC 4.7 – Processing of hnRNA in eukaryotes: DNA → transcription by RNA polymerase-II → primary transcript (hnRNA, contains exons + introns, non-functional) → capping at 5′ end with methylated guanosine triphosphate → tailing (polyadenylation) at 3′ end → splicing removes introns; exons joined in definite order → mature mRNA → transported out of nucleus through nuclear pore → ribosome
Genetic Code
It is already known that DNA is the master molecule of the cell — it initiates, guides, regulates and controls the process of protein synthesis. To perform this complicated function, it must carry the requisite information for the synthesis of proteins. Obviously, this information has to be located in the DNA itself. The site for storing this information lies in the sequence of nucleotides (that is, the nitrogen bases), as evidenced by Yanofski and Sarabhai (1964).
Now here is the puzzle that faced biologists. About 20 different types of amino acids are involved in the synthesis of proteins. The DNA molecule has only 4 types of nitrogen bases available to identify these 20 different amino acids. How is it possible that 20 types of amino acids are encoded by only 4 types of nitrogen bases?
According to F. H. C. Crick, this information is stored in the form of a coded language (a cryptogram) called the genetic code, which contains code words (codons), each one specifying a particular amino acid. The genetic code, therefore, is a collection of base sequences that correspond to each amino acid.
Why the code has to be a triplet
The reasoning here is simple counting, and it is worth doing yourself. Ask how many amino acids each possible codon length could specify.
A single nitrogen base in a codon (a singlet codon) will encode for only four different types of amino acids — one for each base. That is nowhere near 20.
A combination of two nitrogen bases (a doublet codon) will specify only 16 different types of amino acids, since \(4 \times 4 = 16\). Still not enough for 20.
A combination of three nitrogen bases (a triplet codon) will specify 64 different types of amino acids, since \(4 \times 4 \times 4 = 64\). This is more than enough.
Hence G. Gamov (1954) suggested that a codon must be a combination of three consecutive nitrogen bases, which is sufficient to specify 20 different types of amino acids.
Thus there would be 64 different codons (code words) in the dictionary of the genetic code, and each code word has to be a triplet codon. Every three consecutive nucleotides in DNA constitute a triplet codon.
\[\text{Number of possible codons} = 4^3 = 64\]
Cracking the code
That the genetic code is a triplet code was first evidenced by Crick (1961) using “frame-shift mutation.” Knowing the code was a triplet, however, is not the same as knowing which triplet means which amino acid. That dictionary had to be built experimentally, codon by codon.
M. Nirenberg and Matthaei were able to synthesise an artificial mRNA which contained only one type of nitrogenous base, namely uracil (a homopolymer). This synthetic poly-U sequence was transferred to protein-synthesising enzymes. A small polypeptide molecule was formed by the linking of phenylalanine molecules. This explained that UUU codes for phenylalanine — the first word of the dictionary. Later, different homopolymer codons were deciphered, and codons formed by two or more bases were also tried.
Dr. Har Gobind Khorana devised a technique for artificially synthesising mRNA with repeated sequences of known nucleotides. Using synthetic DNA, Dr. Khorana prepared chains of polyribonucleotides with known repeated sequences of two or three nucleotides — for example, CUC UCU CUC UCU.
This resulted in the formation of a polypeptide chain having two different amino acids placed alternately (leucine and serine). Similarly, a polynucleotide chain with three-nitrogen-base repeats gave a polypeptide chain with only one amino acid — for example, CUA CUA CUA CUA gave leucine.
The logic is worth appreciating. A repeating two-base sequence, read three at a time, produces two codons that alternate — hence two amino acids alternating. A repeating three-base sequence, read three at a time, produces the same codon over and over — hence a chain of just one amino acid. The pattern of the product reveals the reading frame.
Later, Severo Ochoa established that the enzyme polynucleotide phosphorylase was also helpful in polymerising RNA with defined sequences in a template-independent manner (that is, in the enzymatic synthesis of RNA).
Finally, Nirenberg, Matthaei and Ochoa deciphered all the 64 codons in the dictionary of the genetic code, which is shown in Fig. 4.13.
Diagram
Fig. 4.13 – Dictionary of the genetic code: the standard 4 × 4 × 4 checkerboard table, with the first letter down the left side (U, C, A, G), the second letter across the top (U, C, A, G) and the third letter down the right side, showing all 64 codons and the amino acid each specifies, with UAA, UAG and UGA marked as Stop and AUG marked as Met.
Complementarity and the direction of information
During replication and transcription, a nucleic acid is copied to form another nucleic acid. These two processes are both based on the complementarity principle — base pairs with base. During translation, however, genetic information is transferred from a polymer of nucleotides to a polymer of amino acids. Here, the complementarity principle does not exist: a codon does not chemically “pair” with an amino acid, and this is exactly why an adapter molecule is needed, as we shall shortly see.
It is evident that a change in the nucleic acid (the genetic material) results in a change in the amino acids of proteins. This clearly explains that the genetic code directs the sequence of amino acids during the synthesis of proteins.
Characteristics of Genetic Code
The genetic code of DNA has certain fundamental characteristics.
i. Genetic code is a triplet code: A sequence of three consecutive bases constitutes a codon, which specifies one particular amino acid. The base sequence in a codon is always in the 5’→3’ direction. In every living organism, the genetic code is a triplet code.
ii. Genetic code has distinct polarity: The genetic code shows a definite polarity, that is, a direction. It is therefore always read in the 5’→3’ direction and not in the 3’→5’ direction. Otherwise the message will change — for example, 5’ AUG 3’. Read backwards, the very same three letters would spell a different codon and hence a different amino acid, so direction is part of the meaning.
iii. Genetic code is non-overlapping: The code is non-overlapping, that is, each single base is a part of only one codon. Adjacent codons do not overlap. Since it is non-overlapping, six consecutive bases will place only two amino acid molecules in the chain. Had it been of the overlapping type, six bases would have produced four amino acid molecules in a chain. Experimental evidence is in favour of the non-overlapping nature.
iv. Genetic code is commaless: There is no gap or punctuation mark between successive (consecutive) codons. The message runs straight on, which is exactly why the reading frame matters so much.
v. Genetic code has degeneracy: Usually a single amino acid is encoded by a single codon. However, some amino acids are encoded by more than one codon. For example, cysteine has two codons, while isoleucine has three codons. This is called degeneracy of the code. Degeneracy of the code is explained by the Wobble hypothesis. Here, the first two bases in different codons are identical, but the third one varies.
vi. Genetic code is universal: By and large, in all living organisms the specific codon specifies the same amino acid. For example, the codon AUG always specifies the amino acid methionine, in all organisms from bacteria up to humans.
vii. Genetic code is non-ambiguous: A specific amino acid is encoded by a particular codon. Alternatively, two different amino acids will never be encoded by the same codon.
viii. Initiation codon and termination codon: AUG is always the initiation codon in any and every mRNA. AUG codes for the amino acid methionine. Out of the 64 codons, three codons — namely UAA, UAG and UGA — are termination codons, which terminate (stop) the process of elongation of the polypeptide chain, as they do not code for any amino acid.
ix. Codon and anticodon: A codon is a part of DNA — for example, AUG is a codon. It is always represented as 5’ AUG 3’. An anticodon is a part of tRNA. It is always represented as 3’ UAC 5’.
Mutations and Genetic Code
Mutation is a phenomenon in which a sudden change in the DNA sequence takes place. It results in a change of genotype, which is expressed in terms of the phenotype (that is, the character). Along with recombination, mutation is the raw material for evolution, as it also results in variations.
During mutation, the possibility of loss (deletion) or gain (insertion or duplication) of a segment of DNA results in an alteration of the chromosome.
Mutation can also occur due to a change in a single base pair of DNA. This is known as point mutation. An example is sickle cell anaemia, which you have already studied in an earlier chapter — a single base change in the DNA produces a single amino acid change in the haemoglobin protein, and that one change is enough to distort the shape of the red blood cell.
Deletion or insertion of base pairs of DNA causes frame-shift mutations or deletion mutations. The insertion or deletion of one or two bases changes the reading frame from the point of insertion or deletion onwards. This happens precisely because the code is commaless and non-overlapping — with no punctuation marks to re-anchor the reading, every codon after the change is misread.
However, the insertion or deletion of three bases, or of multiples of three bases, results only in the insertion or deletion of whole amino acids, and the reading frame remains unaltered from that point onwards. Three bases make one complete codon, so removing or adding three leaves everything else in step.
t-RNA — the adapter molecule
We noted earlier that during translation the complementarity principle breaks down: a codon on the mRNA cannot pair directly with an amino acid, because an amino acid has no bases to pair with. Scientists therefore reasoned that there has to be a mechanism in which some molecule reads the codon and simultaneously binds the amino acid, since the amino acid itself does not have any special capacity to read the codon.
That molecule is tRNA. So tRNA is considered an adapter molecule — it plugs into the language of nucleotides at one end and into the language of amino acids at the other. This role of tRNA was understood much later than the code itself.
The cloverleaf structure (the two-dimensional representation) of tRNA possesses an anticodon loop that has bases complementary to the codon. This is called the anticodon. The structure is shown in Fig. 4.14.
Diagram
Fig. 4.14 – tRNA, the adapter molecule: cloverleaf structure showing the amino acid attachment site at the 3′ end (with the amino acid attached), the anticodon loop with its anticodon at the opposite end, intramolecular base pairing in the arms, and the anticodon pairing with the codon on the mRNA below; alongside, the inverted-L three-dimensional form.
tRNA also shows an amino acid acceptor end (the 3’ end), having unpaired CCA bases — this is the amino acid binding site, to which the amino acid binds. For every amino acid, there is a specific tRNA. The initiator tRNA is specific for methionine. There are no tRNAs for stop codons, which is exactly why translation halts when a stop codon appears — no adapter arrives to continue the chain.
In its actual structure, the tRNA molecule looks like an inverted L (its three-dimensional structure).
B. Translation — protein synthesis
Translation is the mechanism in which the codons of mRNA are translated and specific amino acids, in a sequence, form a polypeptide on the ribosomes. All types of proteins are synthesised by the cell within itself, that is, intracellularly.
What translation requires
The process of translation requires amino acids, mRNA, tRNA, ribosomes, ATP, \(Mg^{++}\) ions, enzymes, and elongation, translocation and release factors.
i. Amino acids form the raw material for protein synthesis. About 20 different types of amino acids are known to form proteins. These are available in the cytoplasm.
ii. DNA controls the synthesis of proteins having amino acids in a specific sequence. This control is possible through the transcription of mRNA. The genetic code is specific for a particular amino acid.
iii. RNAs serve as intermediate molecules between DNA and protein.
iv. Ribosomes serve as the site for protein synthesis. Each ribosome consists of a large and a small subunit. These subunits occur separately in the cytoplasm. Only during protein synthesis do these two subunits get associated together, and this association requires \(Mg^{++}\) ions.
A ribosome has one binding site for mRNA and 3 binding sites for tRNA. They are the P site (peptidyl-tRNA site), the A site (aminoacyl-tRNA site) and the E site (exit site). Only the first tRNA–amino acid complex directly enters the P site of the ribosome; every later one arrives at the A site.
In eukaryotes, a groove is present between the two subunits of the ribosome. It protects the polypeptide chain from the action of cellular enzymes, and also protects the mRNA from the action of nucleases.
Mechanism of translation (synthesis of the polypeptide chain)
It involves three steps: i. Initiation, ii. Elongation, iii. Termination. All three are shown in Fig. 4.15.
Diagram
Fig. 4.15 – Translation (protein synthesis): (a) Initiation — smaller subunit of ribosome bound to mRNA, initiator tRNA at the start codon, larger subunit joining; (b) Elongation — E site, P site and A site labelled, charged tRNA entering the A site, peptide link forming between amino acids; (c) Termination — polypeptide chain released, last tRNA leaving, ribosomal subunits separating from the released mRNA.
1. Initiation of the polypeptide chain
Activation of amino acids is essential before translation initiates, and ATP is essential for this. The small subunit of the ribosome binds (attaches) to the mRNA at its 5’ end. The initiator codon, AUG, is present on the mRNA, and it initiates the process of protein synthesis (translation). The initiator tRNA binds with the initiation codon (AUG) by its anticodon (UAC) through hydrogen bonds. It carries the activated amino acid methionine (in eukaryotes) or formyl methionine (in prokaryotes).
Now the large subunit of the ribosome joins with the smaller subunit — a step that requires \(Mg^{++}\) ions.
The initiator charged tRNA (carrying activated methionine) occupies the P site of the ribosome, and the A site is vacant, ready to receive the next amino acid.
2. Elongation of the polypeptide chain
During this process, activated amino acids are added one by one to the first amino acid (methionine). An amino acid is activated by utilising energy from an ATP molecule. This amino acid binds with the amino acid binding site of a tRNA, and this results in the formation of a tRNA–amino acid complex — a “charged” tRNA.
The addition of an amino acid occurs in a 3-step cycle, which then repeats:
Codon recognition — the aminoacyl-tRNA molecule enters the ribosome at the A site. Its anticodon binds with the codon by hydrogen bonds.
Peptide bond formation — the amino acid on the first initiator tRNA at the P site and the amino acid on the tRNA at the A site join by a peptide bond. Here the enzyme ribozyme acts as the catalyst. Note what this means: the catalyst is not a protein but an RNA molecule. At this time, the first tRNA at the P site is kicked off.
Translocation — the tRNA at the A site, now carrying a dipeptide, moves to the P site. This process is called translocation. In translocation, both the subunits of the ribosome move along in relation to the tRNA and the mRNA.
Hence, the tRNA carrying the dipeptide now gets positioned at the P site of the ribosome, making the A site vacant. At this site, the next charged tRNA carrying an amino acid will then be received. During this process, the first, now uncharged, tRNA is discharged from the E site.
This process is repeated as amino acids are added to the polypeptide. It takes less than 0.1 second for the formation of a peptide bond — which is why a cell can build even a large protein in a matter of seconds.
The third charged tRNA, with its amino acid, arrives at the A site of the ribosome. The anticodon and codon bind by hydrogen bonds. A polypeptide bond is formed. The second tRNA is discharged from the P site to the E site and leaves the ribosome.
So the events — arrival of the tRNA–amino acid complex, formation of the peptide bond, ribosomal translocation, and removal of the previous tRNA — are repeated over and over. As the ribosome moves over the mRNA, all the codons on the mRNA are exposed one by one for translation.
Flowchart
FC 4.8 – The elongation cycle of translation: charged tRNA enters A site → anticodon pairs with codon by hydrogen bonds → peptide bond formed between the amino acid at the P site and the amino acid at the A site (catalysed by ribozyme) → tRNA at A site (now carrying the peptide) translocates to the P site → uncharged tRNA leaves via the E site → A site vacant again → back to the start of the cycle with the next codon
3. Termination and release of the polypeptide
At the end of the mRNA, there is a stop codon (UAA, UAG or UGA). It is exposed at the A site. It is not read and joined by the anticodon of any tRNA, because no tRNA exists for a stop codon. Instead, a release factor binds to the stop codon, thereby terminating the translation process. The polypeptide is now released into the cytoplasm.
The two subunits of the ribosome dissociate, and the last tRNA is set free in the cytoplasm.
mRNA also has some additional sequences that are not translated, and these are referred to as untranslated regions (UTR). The UTRs are present at both the 5’ end (before the start codon) and the 3’ end (after the stop codon). They are required for the efficient translation process.
Finally, the mRNA is also released into the cytoplasm. It gets denatured by nucleases immediately. Hence mRNA is short-lived. This is not a flaw but a control mechanism: because the messenger is destroyed soon after use, the cell must keep transcribing a gene in order to keep making its protein — and can therefore stop making that protein simply by stopping transcription.
How this section is tested
This is the richest section of the chapter for application questions. Expect sequence-ordering questions on the steps of translation (binding of small subunit → initiator tRNA at start codon → joining of subunits → arrival of second tRNA → peptide bond), match-the-following on the three eukaryotic RNA polymerases and their products, and diagram-labelling on the transcription unit, the tRNA cloverleaf and the ribosome’s three sites. Assertion–reason questions very often hinge on two facts: that the template strand runs 3’→5’, and that the genetic code is degenerate but non-ambiguous.
4.6 Regulation of Gene Expression
We have now followed a gene all the way from DNA to a finished polypeptide. But a cell does not switch on every gene it possesses, all the time. If it did, it would waste enormous quantities of energy and raw material building proteins it has no use for.
Gene expression is the multistep process by which a gene is regulated and its product is synthesised. Thus, gene expression results in the formation of a polypeptide. And because it is a multistep process, it can be controlled at more than one point — the gene expression process is regulated at different levels.
Levels of regulation in eukaryotes
In eukaryotes, the regulation can be at different levels, such as:
- Transcriptional level (formation of the primary transcript).
- Processing level (regulation of splicing).
- Transport of mRNA from the nucleus to the cytoplasm.
- Translational level.
Notice that these four points are simply the four stages you met in the previous section, taken in order. Anywhere along the assembly line from gene to protein, the cell can slow the line down or stop it. Blocking the very first step — transcription — is the most economical, because nothing is built and then thrown away.
Flowchart
FC 4.9 – Levels at which eukaryotic gene expression can be regulated: gene → [1. transcriptional level] → primary transcript → [2. processing level: regulation of splicing] → mature mRNA → [3. transport of mRNA from nucleus to cytoplasm] → mRNA in cytoplasm → [4. translational level] → polypeptide
Why a cell needs to regulate its genes
Genes of a cell are expressed to perform different functions. Consider a concrete example. The enzyme beta-galactosidase is synthesised by E. coli. It is used for the hydrolysis of lactose into galactose and glucose:
\[\text{Lactose} \xrightarrow[+\,H_2O]{\beta\text{-galactosidase}} \text{Galactose} + \text{Glucose}\]
Now here is the key observation. If E. coli bacteria do not have lactose in the surrounding medium as a source of energy, then the enzyme β-galactosidase is not synthesised at all. The gene is there in the DNA the whole time — but it is only read when there is lactose to be digested.
So it is the metabolic, physiological or environmental conditions that regulate the expression of genes. The bacterium is, in effect, listening to its surroundings and adjusting which proteins it builds.
This is not only a bacterial trick. The development and differentiation of an embryo into an adult organism is also a result of the coordinated regulation, or expression, of several sets of genes. Every cell in your body carries the same DNA, yet a nerve cell and a liver cell look and behave nothing alike — because different sets of genes are switched on in each.
The natural question that follows is this: by what mechanism do organisms regulate gene expression in response to changes in the environment? And is there a single mechanism that regulates the expression of all different genes or sets of genes, or are different genes regulated by different mechanisms?
Inducible enzymes and induction
Certain bacteria like E. coli adapt to their chemical environment by synthesising certain enzymes depending upon the substrate present. Such an adaptive enzyme is called an inducible enzyme.
A set of genes will be switched on when there is a necessity to metabolise a new substrate. This phenomenon is called induction, and the small molecule responsible for it is known as an inducer. Because the presence of the small molecule turns genes on, this is called positive control.
How this section is tested
Questions here usually turn on one distinction: induction versus repression, and which of the two is positive control and which is negative. Assertion–reason questions often pair “β-galactosidase is not made when lactose is absent” with a reason about environmental control of gene expression. Be ready also to list, in order, the four levels at which eukaryotic gene expression can be regulated — this is a common short-answer question, and the order follows the path from gene to protein.
4.7 Operon Concept
In the previous section we saw that bacteria switch genes on and off in response to their surroundings. Now we look at how. The operon is the mechanism — it is a transcriptional control mechanism of gene regulation, meaning that it acts at the very first level of control, deciding whether transcription happens at all.
Francois Jacob and Jacques Monod (1961) explained that metabolic pathways are regulated as a unit. This is the central insight of the operon concept. Genes that work together in one pathway are not switched on one by one; they are grouped along the DNA and switched on together, by a single switch.
Why regulate three genes together?
Consider the example that Jacob and Monod themselves used. In E. coli, when lactose sugar is provided in the culture medium, the cell induces the production of three enzymes necessary for the digestion of lactose. The enzymes are:
i. β-galactosidase: Digests lactose into galactose and glucose.
ii. β-galactoside permease: Permits lactose molecules to enter into the cell.
iii. Transacetylase (β-galactoside acetyltransferase): Transfers an acetyl group from acetyl Co-A to galactoside.
Now think about why all three must appear together. There is no point producing β-galactosidase to split lactose if permease has not first let the lactose into the cell. And there is no point letting lactose in if there is no enzyme to break it down. The three proteins are useless individually and only work as a team — so it makes biological sense for the cell to switch all three on with one command.
The synthesis of these three enzymes is controlled by a long segment of DNA known as an Operon. It consists of an operator site O and three structural genes z, y and a. The action of the structural genes is regulated by the operator site with the help of a repressor protein. The repressor protein is produced by the action of the gene i (inhibitor), which is known as the regulator gene.
Gene expression depends on whether the operator is switched on or switched off.
If the operator is switched on, the three genes z, y and a are transcribed by RNA polymerase into a single mRNA. Each structural gene is generally known as a cistron, and the transcribed long mRNA covering various cistrons is known as polycistronic — a term you met in the previous section, and here you can see exactly why bacteria need it.
Switching the operator on or off is accomplished by a protein called the repressor. When this protein is attached to the operator and blocks it, the switch is turned off and the structural genes are not expressed.
Lac operon
The lactose, or lac, operon of E. coli is an inducible operon. The operon is switched on when a chemical inducer — lactose — is present in the medium.
The lac operon consists of the following components:
- Regulator gene (repressor gene)
- Promoter gene
- Operator gene
- Structural genes
- Inducer — which is not a component of the operon
That last point is worth pausing on. The inducer is listed here because you cannot understand the working of the operon without it, but it is a small molecule from the environment, not a stretch of DNA. It is the signal, not the switch.
1. Regulator gene: This gene controls the operator gene in cooperation with an inducer present in the cytoplasm. The regulator gene precedes the promoter gene. It may not be present immediately adjacent to the operator gene. The regulator gene produces a protein called the repressor protein. The repressor binds with the operator gene and represses (stops) its action. It is therefore also called the regulator protein.
2. Promoter gene: This gene precedes the operator gene and is present adjacent to it. The promoter gene marks the site at which the RNA polymerase enzyme binds. When the operator gene is turned on, the enzyme moves over the operator gene and transcription of the structural genes starts. The promoter gene’s base sequence also determines which strand of DNA acts as the template.
3. Operator gene: This precedes the structural genes and lies adjacent to them. It controls the functioning of the structural genes. When the operator gene is turned on by an inducer, the structural genes produce mRNA. The operator gene is turned off by the product of the repressor gene.
4. Structural gene: When lactose is added to the E. coli culture, the structural genes catalyse (produce) mRNA, which in turn produces polypeptides on the ribosomes. The polypeptides formed act as enzymes to catalyse lactose in the cell. There are 3 structural genes, in the sequence lac-z (1), lac-y (2) and lac-a (3). The enzymes produced are β-galactosidase, β-galactoside permease and transacetylase respectively.
5. Inducer: It is a chemical in the cytoplasm (allolactose) which inactivates the repressor. When the lac operon is switched on, the inducer joins with the repressor protein, preventing the binding of the repressor to the operator gene. So the operator gene is free, and now the enzyme RNA polymerase can move from the promoter to the structural genes via the operator gene.
Working of the lac operon
Everything now comes together. The two states of the operon — inducer absent and inducer present — are shown in Fig. 4.16.
Diagram
Fig. 4.16 – Working of the lac operon, in two panels. (a) Inducer absent: regulatory gene transcribed and translated into repressor protein → repressor binds the operator → operator blocked → RNA polymerase movement blocked at the promoter → structural genes 1, 2, 3 inactive → no synthesis of mRNA. (b) Inducer present: inducer (allolactose) binds the repressor → inactivated repressor cannot bind the operator → operator free → RNA polymerase moves from promoter through operator → transcription of structural genes proceeds → polycistronic mRNA → translation → proteins (enzymes to metabolise lactose).
Flowchart
FC 4.10 – Lac operon, inducer absent (operon OFF): regulator gene i → repressor mRNA → repressor protein → repressor binds operator → operator blocked → RNA polymerase cannot move past operator → structural genes z, y, a not transcribed → no lactose-digesting enzymes made
Flowchart
FC 4.11 – Lac operon, inducer present (operon ON): lactose enters cell (via the few permease molecules already present) → converted to allolactose (inducer) → inducer binds repressor protein → repressor–inducer complex is inactivated and cannot bind operator → operator free → RNA polymerase moves from promoter → structural genes z, y, a transcribed as one polycistronic mRNA → translation on ribosomes → β-galactosidase + permease + transacetylase → lactose digested → back to start when lactose runs out
Role of lactose
There is one loose end in the story above, and it is a genuinely clever one. If permease is needed for lactose to enter the cell, but permease is only made once lactose is inside the cell — how does the very first lactose molecule ever get in?
The answer is this. A few molecules of lactose enter into the cell by the enzyme permease. A small amount of this enzyme is present even when the operon is switched off — the operon leaks a little, and that leak is exactly what makes the system able to start.
These few molecules of lactose act as the inducer and bind to the repressor. This repressor–inducer complex fails to join with the operator gene, which is then turned on. The structural genes now produce all the enzymes.
Thus, lactose acts as an inducer of its own breakdown. This is a striking piece of biological logic: the substance to be destroyed is itself the signal that summons the machinery to destroy it.
When the inducer level falls — that is, when the lactose has all been digested — the operator is blocked again by the repressor. So the structural genes are repressed (inactivated) once more. This is negative feedback: the product of the process shuts the process down.
How this section is tested
The lac operon is one of the most heavily tested topics in the whole chapter. Expect diagram-labelling questions on Fig. 4.16 (identify the regulator gene, promoter, operator and the three structural genes), match-the-following linking z, y and a to their enzymes, and assertion–reason questions on the role of allolactose. The single most common trap is asking whether the inducer binds the operator — it does not; it binds the repressor. Get that right and most operon questions become easy.
4.8 Genomics
So far we have followed single genes — one gene, one mRNA, one protein, or in the case of the operon, a small cluster of genes working as a unit. But an organism is not a collection of isolated genes. It is a complete set of them, working together. Studying that complete set is what genomics is about.
The term Genome (introduced by H. Winkler in 1920) refers to the total genetic constitution of an organism. Alternatively, it is a complete copy of the genetic information (DNA), or one complete set of chromosomes (monoploid or haploid) of an organism.
The term Genomics (coined by T. H. Roderick in 1986) is the study of genomes through the analysis, sequencing and mapping of genes, along with the study of their functions.
Sequencing other organisms
The sequencing of the yeast, Drosophila and mouse genomes was done in order to facilitate comparative studies between humans and other organisms commonly used for genetic studies in the laboratory.
Why sequence a fly or a mouse when the goal is to understand humans? Because a gene means very little in isolation. It is when you find the same gene, doing a similar job, in a fly and a mouse and a human, that you begin to understand what that gene is for — and you can then study it experimentally in the fly, which is far easier and quicker than studying it in a human.
Several additional genomes are now either actively being sequenced or strongly considered for sequencing. These include several microbes, the bee, the tomato and other crops.
Types of genomics
The study of genomics may be classified into two types:
a. Structural genomics: It involves the mapping, sequencing and analysis of the genome. In short, it answers the question — what is written there?
b. Functional genomics: It deals with the study of the functions of all gene sequences and their expression in organisms. In short, it answers the question — what does what is written actually do?
Flowchart
FC 4.12 – From genome to application: DNA of an organism → structural genomics (mapping → sequencing → analysis) → complete genome sequence → functional genomics (study of gene function and gene expression) → applications in medicine, agriculture, forensics and biotechnology
Applications of genomics
Structural and functional genomics are used for different purposes in the improvement of crop plants, human health and livestock. The knowledge and understanding acquired from genomics research can be applied in a number of different sectors, including medicine, biotechnology and the social sciences. It helps in the treatment of genetic disorders through gene therapy.
- Genomics is used in agriculture to develop transgenic crops having more desirable characters.
- Genetic markers developed in genomics have applications in forensic analysis.
- Genomics can lead to the introduction of new genes into microbes, so that they produce enzymes, therapeutic proteins and even biofuels.
How this section is tested
The details most often asked are the two names and dates — H. Winkler (1920) for the term “genome” and T. H. Roderick (1986) for “genomics” — and the distinction between structural and functional genomics, which frequently appears as a match-the-following or a short-answer question. Do not confuse the definition of genome (a haploid set) with the diploid chromosome number of the organism.
4.9 Human Genome Project
Genomics gave scientists the tools. The Human Genome Project was the moment those tools were turned, at enormous scale, on our own species.
The Human Genome Project was initiated in 1990 under the international administration of the Human Genome Organization (HUGO). It was a multinational research project to determine the genomic structure of humans.
The project was co-ordinated by the US Department of Energy and the National Institute of Health. Additional contributors included universities across the United States. International partners included the United Kingdom, France, Germany, Japan, India and China. The project formally began in 1990 and was completed in 2003.
Notice how long that is — thirteen years. Sequencing three billion base pairs was, at the time, one of the largest scientific undertakings ever attempted, comparable in ambition to putting a human on the Moon. No single laboratory or country could have done it alone, which is why it was organised as an international collaboration from the start.
The aims of the project
The main aims of the project were:
I. Mapping the entire human genome at the level of nucleotide sequences.
II. To store the information collected from the project in databases.
III. To develop tools and techniques for the analysis of the data.
IV. Transfer of the related technologies to the private sectors, such as industries.
V. Taking care of the legal, ethical and social issues which may arise from the project.
Read those five aims again and notice something important: only the first is about the DNA itself. The other four are about what to do with three billion letters of information once you have them — how to store it, how to analyse it, how to apply it, and how to prevent it being misused. A sequence that nobody can read, search or interpret is of no use to anyone.
Flowchart
FC 4.13 – The five aims of the Human Genome Project, in logical order: map the entire human genome at nucleotide level → store the data in databases → develop tools and techniques to analyse the data → transfer the technologies to private industry → address the legal, ethical and social issues arising from the project
Bioinformatics
The HGP (Human Genome Project) was closely associated with the rapid development of a new area in biology, called bioinformatics. This is the discipline that uses computing to store, search and analyse biological data — and it grew up alongside the HGP precisely because three billion base pairs cannot be handled by any other means. You cannot read a genome; you must search it.
The work of the Human Genome Project has allowed researchers to begin to understand the blueprint for building and constructing the human genome. As researchers learn more about the functions of genes and proteins, this knowledge will have a major impact on fields such as medicine, biotechnology and the life sciences. This is why the HGP is so important.
What the project set out to find
The Human Genome Project aimed to provide a complete and accurate sequence of the 3 billion DNA base pairs that make up the human genome, and to find out the estimated number of human genes. About 33,000 genes have now been estimated to be present in humans.
The project was also aimed at sequencing the genomes of several other organisms, such as the bacterium Escherichia coli, Caenorhabditis elegans (a free-living, non-pathogenic nematode), Saccharomyces cerevisiae (yeast), Drosophila (the fruit fly), plants (rice and Arabidopsis), Mus musculus (the mouse), and others. Complete genome sequences of these model organisms are useful for comparative studies, which allow researchers to study gene functions in these organisms and then apply that understanding to humans.
The comparative sizes of some of these genomes are given in Table 4.17.
Table 4.17: Comparative genome sizes of humans and other model organisms
| Organism | Chromosome number | Estimated gene number | Estimated size (base pairs) |
|---|---|---|---|
| Human (Homo sapiens) | 46 | 33,000 | 3 billion |
| Mouse (Mus musculus) | 40 | 25,000 | 2.9 billion |
| Fruit fly (Drosophila melanogaster) | 8 | 13,000 | 165 million |
| Plant (Arabidopsis thaliana) | 10 | 25,000 | 157 million |
| Roundworm (Caenorhabditis elegans) | 12 | 19,000 | 97 million |
| Yeast (Saccharomyces cerevisiae) | 32 | 6000 | 12 million |
| Bacteria (Escherichia coli) | 1* | 4400 | 4.6 million |
The secret of our complexity may lie not in the number of our genes but in how we use them. Sequencing will lead to an understanding of gene structure and function in other species too. Since we possess many of the same genes as flies, roundworms and mice, such studies will lead to a greater understanding of human evolution.
How this section is tested
The facts most frequently asked are the start and completion years (1990 and 2003), the coordinating bodies, the figure of 3 billion base pairs, and the estimate of about 33,000 human genes. Table 4.17 is a favourite source for match-the-following and data-based questions — you may be asked which organism has the largest genome, or to compare gene numbers between two organisms. Read such questions carefully: the organism with the most chromosomes is not necessarily the one with the largest genome.
4.10 DNA Fingerprinting
Genes present on chromosomes are responsible for determining the characters of an organism as well as for the inheritance of characters. Due to the recombination of paternal and maternal genes, we differ from our parents. Differences also arise due to the infrequent mutations that occur during gamete formation (cell division).
Because of all these factors, every individual has a unique genetic make-up, which may be called that person’s Fingerprint. The word is borrowed deliberately from the ridges on your thumb — like them, your DNA pattern is yours alone, and unlike them, it is present in every cell of your body.
The technique developed to identify a person with the help of DNA restriction analysis is known as DNA profiling or DNA fingerprinting. The technique of fingerprinting was first given by the British geneticist Dr. Alec Jeffreys in 1984.
What makes fingerprinting possible: VNTRs
Here is the difficulty the technique had to overcome. About 99.9% of the nucleotide sequence in all persons is the same. If you compared two people’s DNA at random, you would almost always find identical sequences — which is useless for telling them apart. So DNA fingerprinting cannot look everywhere. It must look precisely at the small fraction of the genome that does differ.
The DNA fingerprinting technique is therefore based on the identification of particular nucleotide sequences present in this wonder molecule. Only some short sequences of nucleotides differ from person to person. In the population, every person shows unusual sequences of 20–100 base pairs, which are repeated several times. These are termed Variable Number of Tandem Repeats (VNTRs).
The length of the regions having VNTRs is different in each individual, and this is therefore the key factor in DNA profiling. Two people may both carry the same repeated sequence, but one may carry it repeated fifteen times and the other forty times — and that difference in length is what the technique detects.
Steps involved in DNA fingerprinting
The steps involved in DNA fingerprinting are as follows. The whole procedure is shown in Fig. 4.18.
Diagram
Fig. 4.18 – DNA fingerprinting: sample → DNA extraction → isolated purified DNA → restriction digestion with restriction enzyme (scissors) → DNA fragments → gel electrophoresis → transfer of DNA to membrane (Southern blotting) → hybridisation with selected single-stranded DNA probes → X-ray film showing the band pattern.
1. Isolation of DNA: The DNA must be recovered from the cells or tissues of the body (host). Only a small amount of tissue — like blood, hair roots, skin, and so on — is required. This is what makes the technique so powerful in practice: a trace left behind is enough.
2. Restriction digestion: The isolated DNA is treated with restriction enzymes. These enzymes cut the DNA into small fragments having variable lengths. This phenomenon is called Restriction Fragment Length Polymorphism (RFLP). This is the step where the VNTR differences are converted into something measurable: because different people have different numbers of repeats, the fragments cut out from their DNA are of different lengths.
3. Gel electrophoresis: The DNA samples are loaded on an agarose gel for electrophoresis under an electric influence. The DNA fragments, which are negatively charged, move to the positive pole. The movement of these fragments depends on the length of the fragments — shorter fragments slip through the gel more easily and travel further, while longer ones lag behind. This results in the formation of bands. Double-stranded DNA (dsDNA) is split into single-stranded DNA (ssDNA) by alkali treatment.
4. Southern blotting: This is a technique used for detecting a specific DNA sequence, developed by E. Southern. The separated DNA fragments are transferred to a nylon membrane or a nitrocellulose filter paper by placing it over the gel and soaking them with filter paper overnight. The gel itself is fragile and awkward to work with; the membrane holds the same pattern of bands in a stable, usable form.
5. Selection of DNA probe: A known sequence of single-stranded DNA is prepared. It is called a DNA probe. The DNA probe is obtained from organisms or prepared by the cDNA preparation method. The DNA probe is labelled with radioactive isotopes.
6. Hybridisation: The probe DNA is added to the nitrocellulose filter paper containing the host DNA. The single-stranded DNA probe pairs with the complementary base sequence of the host DNA strand. As a result, DNA–DNA hybrids are formed on the nitrocellulose filter paper. The remaining single-stranded DNA probe fragments are washed off. Only the bands that the probe has stuck to now carry a radioactive label.
7. Photography: The nitrocellulose filter paper is photographed on an X-ray film by autoradiography. The film is analysed to determine the presence of hybrid DNA. Wherever the radioactive probe has bound, the film darkens — producing the pattern of bands that we recognise as the DNA fingerprint.
Flowchart
FC 4.14 – Steps in DNA fingerprinting: isolation of DNA from blood/hair root/skin → restriction digestion with restriction enzymes (RFLP) → gel electrophoresis separates fragments by length; alkali treatment splits dsDNA into ssDNA → Southern blotting transfers fragments to nylon/nitrocellulose membrane → radioactively labelled single-stranded DNA probe added → hybridisation: probe pairs with complementary host DNA; unbound probe washed off → autoradiography on X-ray film → band pattern = DNA fingerprint
Applications of DNA fingerprinting
- In forensic science, DNA fingerprinting is used to solve problems of rape and some complicated murder cases.
- DNA fingerprinting is used to find out the biological father or mother, or both, of a child, in cases of disputed parentage.
- DNA fingerprinting is used in pedigree analysis in cats, dogs, horses and humans.
Notice the common thread. In every one of these applications, the question being asked is the same: does this DNA come from this individual, or from someone related to this individual? Because a child inherits half its VNTRs from each parent, a child’s band pattern is never identical to either parent’s — but every band in it must be present in one parent or the other. That single fact is what makes parentage testing possible.
How this section is tested
Sequence-ordering questions on the seven steps of DNA fingerprinting are extremely common — the order isolation → restriction digestion → electrophoresis → Southern blotting → probe → hybridisation → autoradiography must be learnt as a chain, each step feeding the next. Match-the-following questions frequently pair techniques with what they do (Southern blotting → transfer to nitrocellulose sheet; electrophoresis → separation of fragments on a gel slab; alkali treatment → splitting DNA into single strands; autoradiography → X-ray photography). Also be ready for the figures: 99.9% identity between individuals, and VNTRs of 20–100 base pairs.
4.11 Summary
The discovery of DNA
- In 1869, Friedrich Miescher isolated a phosphorus-rich, acidic substance from the nuclei of white blood cells and called it nuclein; it was later renamed nucleic acid.
- By the early 1900s it was known that nuclein is a mixture of proteins and nucleic acids, and that there are two kinds of nucleic acid — DNA (deoxyribonucleic acid) and RNA (ribonucleic acid).
DNA is the genetic material
- Geneticists first favoured proteins as the genetic material, believing DNA to be small and simple. Over roughly 25 years (1928–1952), three experiments overturned this.
- Griffith (1928), working with the S (virulent, smooth, encapsulated) and R (non-virulent, rough, non-capsulated) strains of Streptococcus pneumoniae, showed that heat-killed S bacteria mixed with live R bacteria killed mice, and live S bacteria were recovered from them. He concluded that a “transforming principle” had passed from the dead S cells to the living R cells.
- Avery, MacLeod and McCarty (1944) purified the components of S cells and found that only DNA could transform R into S. Proteases and RNAses did not stop transformation, but DNAse did — so the transforming principle is DNA.
- Hershey and Chase (1952) labelled bacteriophage DNA with \(^{32}P\) and phage protein with \(^{35}S\). Only the phosphorus label entered the bacteria, proving that DNA, and not protein, is the genetic material.
DNA packaging
- A typical mammalian cell contains about 2.2 metres of DNA, while the nucleus is only about \(10^{-6}\) m across, so DNA must be condensed, coiled and supercoiled.
- In prokaryotes such as E. coli, the naked, circular nucleoid (1100 μ in perimeter, about 4.6 million base pairs) is compacted from 350 μm → 30 μm (by forming 40–50 looped domains with RNA connectors) → 2 μ (by supercoiling), assisted by positively charged HU proteins and by DNA gyrase and DNA topoisomerase I.
- In eukaryotes, DNA associates with histone and non-histone proteins (reported by R. Kornberg, 1974). Histones are rich in the basic amino acids lysine and arginine, so they are positively charged and attract the negatively charged DNA.
- Eight histone molecules — two each of \(H_2A\), \(H_2B\), \(H_3\) and \(H_4\) — form the histone octamer. DNA wraps 1¾ turns around it to form a nucleosome. \(H_1\) binds the DNA where it enters and leaves the nucleosome.
- One nucleosome contains about 200 bp of DNA, of which about 146 bp is wound around the octamer; the rest is linker DNA. Nucleosomes are the repeating units of chromatin and look like ‘beads-on-string’ under the electron microscope.
- The 10 nm chromatin fibre coils (six nucleosomes at a time) into a 30 nm (300 Å) solenoid, which supercoils into loops and condenses at metaphase into chromosomes. Higher-level packaging needs Non-Histone Chromosomal proteins (NHC).
- Heterochromatin is condensed, dark-staining, DNA-rich (2–3 times more than euchromatin), located near the centromere and telomeres, and transcriptionally almost inactive. The term was proposed by Heitz. Euchromatin is non-condensed, light-staining, fast-replicating and transcriptionally active.
DNA replication
- DNA has two functions: heterocatalytic (directing the synthesis of molecules other than itself — transcription and translation) and autocatalytic (directing the synthesis of DNA itself — replication).
- Replication occurs once per cell cycle, in the S-phase of interphase. It is semiconservative — proposed by Watson and Crick on the basis of the antiparallel and complementary nature of DNA strands.
- The steps are: activation of nucleotides (dAMP, dGMP, dCMP, dTMP activated by ATP with phosphorylase into dATP, dGTP, dCTP, dTTP); initiation at origin ‘O’ (terminating at ‘T’; the unit is a replicon — one in prokaryotes, several in eukaryotes; endonuclease nicks a strand); unwinding by DNA helicase, held apart by single strand binding proteins (SSBP); formation of the Y-shaped replication fork, with strain relieved by the super-helix relaxing enzyme; synthesis of new strands from an RNA primer, catalysed by DNA polymerase, always in the 5’→3’ direction.
- The leading strand grows continuously towards the fork; the lagging strand grows discontinuously away from it as Okazaki fragments. These are joined by DNA ligase. RNA primers are replaced with DNA by DNA polymerase-I in prokaryotes and DNA polymerase-α in eukaryotes. DNA gyrase (topoisomerase) restores the double helix.
- Each daughter DNA has one parental and one newly synthesised strand — hence “semiconservative.”
- Meselson and Stahl (1958) proved this using \(^{15}N\) and \(^{14}N\) and equilibrium density gradient centrifugation in 6M \(CsCl_2\): after one generation in \(^{14}N\), a single hybrid band appeared; after two generations, a hybrid band and a light band.
Protein synthesis
- Protein synthesis has two stages, transcription and translation. The unidirectional flow DNA → RNA → protein is the central dogma, postulated by F. H. C. Crick in 1958. In retroviruses, Temin (1970) and Baltimore (1970) added reverse transcription (RNA → DNA), carried out by RNA-dependent DNA polymerase.
- Transcription copies only one strand of DNA into RNA, catalysed by RNA polymerase, and has three stages — initiation, elongation and termination. It occurs in the nucleus during the G1 and G2 phases.
- A transcription unit has a promoter (upstream, at the 5’ end; recognised in prokaryotes by the sigma factor), the structural gene, and a terminator (downstream, at the 3’ end). The strand with 3’→5’ polarity is the template (antisense) strand; the strand with 5’→3’ polarity is the coding (sense) strand, whose sequence matches the RNA with uracil in place of thymine.
- Prokaryotes have one RNA polymerase. Eukaryotes have three: RNA polymerase-I (rRNA), RNA polymerase-II (mRNA and hnRNA), and RNA polymerase-III (tRNA and snRNA).
- A gene is a DNA sequence coding for mRNA, tRNA or rRNA; a cistron codes for one polypeptide. A single structural gene in a transcription unit is monocistronic; several structural genes in one unit are polycistronic. Eukaryotic structural genes contain non-coding introns and coding exons; only exons appear in processed mRNA.
- Eukaryotic primary transcripts (hnRNA) undergo processing: splicing removes introns and joins the exons in a definite order; capping adds methylated guanosine triphosphate at the 5’ end; tailing adds a poly-A tail by polyadenylation at the 3’ end. The mature mRNA leaves through a nuclear pore.
- Genetic code: information is stored in the sequence of nitrogen bases (Yanofski and Sarabhai, 1964). Since 4 bases must specify 20 amino acids, a singlet code gives 4 and a doublet 16 — both too few — while a triplet gives \(4^3 = 64\). G. Gamov (1954) suggested the triplet codon; Crick (1961) confirmed it using frame-shift mutation. Nirenberg and Matthaei showed that poly-U gives phenylalanine (UUU); Har Gobind Khorana used repeated synthetic sequences (CUC UCU… → leucine and serine alternately; CUA CUA… → leucine); Severo Ochoa used polynucleotide phosphorylase. Together, Nirenberg, Matthaei and Ochoa deciphered all 64 codons.
- The code is a triplet code; has distinct polarity (read 5’→3’); is non-overlapping; is commaless; is degenerate (some amino acids have more than one codon, explained by the Wobble hypothesis, in which the first two bases are identical and the third varies); is universal; and is non-ambiguous. AUG is always the initiation codon and codes for methionine. UAA, UAG and UGA are termination codons and code for no amino acid. A codon (e.g. 5’ AUG 3’) is on the mRNA; an anticodon (e.g. 3’ UAC 5’) is on the tRNA.
- Complementarity applies in replication and transcription, but not in translation.
- Mutation is a sudden change in the DNA sequence, altering the genotype and hence the phenotype; with recombination it is the raw material of evolution. A point mutation is a change in a single base pair (e.g. sickle cell anaemia). Insertion or deletion of one or two bases causes a frame-shift mutation; insertion or deletion of three or a multiple of three bases adds or removes whole amino acids and leaves the reading frame unaltered.
- tRNA is the adapter molecule. In its cloverleaf (2-D) form it has an anticodon loop complementary to the codon, and an amino acid acceptor end at the 3’ end with unpaired CCA bases. There is a specific tRNA for every amino acid; the initiator tRNA is specific for methionine; there are no tRNAs for stop codons. In three dimensions it resembles an inverted L.
- Translation takes place on ribosomes and requires amino acids, mRNA, tRNA, ribosomes, ATP, \(Mg^{++}\) ions, enzymes, and elongation, translocation and release factors. A ribosome has one binding site for mRNA and three for tRNA — the P site (peptidyl), A site (aminoacyl) and E site (exit). Only the first tRNA–amino acid complex enters the P site directly.
- Initiation: the small subunit binds the mRNA at the 5’ end; the initiator tRNA (anticodon UAC) pairs with the initiation codon AUG, carrying methionine (or formyl methionine in prokaryotes); the large subunit joins, requiring \(Mg^{++}\) ions.
- Elongation is a three-step cycle: codon recognition at the A site; peptide bond formation, catalysed by ribozyme; and translocation of the tRNA from the A site to the P site, with the uncharged tRNA discharged from the E site. A peptide bond forms in less than 0.1 second.
- Termination: a stop codon (UAA/UAG/UGA) at the A site is read by a release factor, not a tRNA; the polypeptide is released; the ribosomal subunits dissociate. mRNA has untranslated regions (UTRs) at both the 5’ end (before the start codon) and the 3’ end (after the stop codon), needed for efficient translation. mRNA is quickly denatured by nucleases and is therefore short-lived.
Flowchart
FC 4.15 – The chapter in one chain: DNA (packaged as nucleosomes → chromatin → chromosome) → replication (semiconservative, S-phase) → transcription (template strand → hnRNA → capping, tailing, splicing → mRNA) → translation (mRNA read as triplet codons; tRNA adapters deliver amino acids; ribosome forms peptide bonds) → polypeptide → protein → trait
Regulation of gene expression
- Gene expression is the multistep process by which a gene is regulated and its polypeptide product is synthesised. In eukaryotes it can be regulated at the transcriptional level, the processing (splicing) level, the level of mRNA transport out of the nucleus, and the translational level.
- E. coli makes β-galactosidase only when lactose is present: \(\text{Lactose} \xrightarrow[+\,H_2O]{\beta\text{-galactosidase}} \text{Galactose} + \text{Glucose}\). Metabolic, physiological and environmental conditions therefore regulate gene expression. The development of an embryo into an adult is likewise the result of coordinated gene expression.
- An inducible enzyme is made only when its substrate is present; the switching on of such genes is induction, the small molecule responsible is the inducer, and this is positive control. Repressible regulation, in which an end product switches off the genes that make it (feedback repression), is negative control.
Operon concept
- Jacob and Monod (1961) showed that metabolic pathways are regulated as a unit. An operon is a segment of DNA consisting of an operator site O and the structural genes z, y and a, regulated by a repressor protein produced by the regulator gene i.
- When lactose is supplied to E. coli, three enzymes are induced: β-galactosidase (digests lactose into galactose and glucose), β-galactoside permease (lets lactose into the cell) and transacetylase (transfers an acetyl group from acetyl Co-A to galactoside).
- The lac operon of E. coli is an inducible operon. Its components are the regulator gene, promoter gene, operator gene and structural genes; the inducer (allolactose) is not a component of the operon.
- With the inducer absent, the repressor binds the operator, blocking RNA polymerase — the structural genes are not expressed. With the inducer present, allolactose binds the repressor and inactivates it, freeing the operator, so RNA polymerase transcribes z, y and a as a single polycistronic mRNA.
- A few lactose molecules enter through the small amount of permease present even when the operon is off. Lactose thus acts as an inducer of its own breakdown. When the inducer level falls, the repressor blocks the operator again — this is negative feedback.
Flowchart
FC 4.16 – Lac operon summary: lactose absent → repressor (from gene i) binds operator → RNA polymerase blocked → no enzymes; lactose present → allolactose binds repressor → repressor inactivated → operator free → z, y, a transcribed as polycistronic mRNA → β-galactosidase + permease + transacetylase → lactose digested → inducer level falls → repressor re-binds operator → operon off
Genomics
- The genome (term introduced by H. Winkler, 1920) is the total genetic constitution of an organism — one complete (monoploid or haploid) set of chromosomes. Genomics (term coined by T. H. Roderick, 1986) is the study of genomes through analysis, sequencing and mapping of genes and the study of their functions.
- Structural genomics involves mapping, sequencing and analysis of the genome. Functional genomics deals with the functions of gene sequences and their expression.
- Genomics is applied in crop improvement, human health, livestock, gene therapy, transgenic crops, forensic analysis using genetic markers, and the engineering of microbes to produce enzymes, therapeutic proteins and biofuels.
Human Genome Project
- Initiated in 1990 under HUGO, coordinated by the US Department of Energy and the National Institute of Health, with international partners including the UK, France, Germany, Japan, India and China. It began in 1990 and was completed in 2003.
- Its aims were: to map the entire human genome at the nucleotide level; to store the information in databases; to develop tools and techniques for analysing the data; to transfer the technologies to the private sector; and to address the legal, ethical and social issues arising from the project.
- It sequenced the 3 billion base pairs of the human genome; about 33,000 genes are estimated to be present in humans. It also sequenced model organisms such as E. coli, Caenorhabditis elegans, Saccharomyces cerevisiae, Drosophila, rice, Arabidopsis and Mus musculus, for comparative study.
- The HGP drove the growth of bioinformatics. The secret of human complexity may lie not in the number of our genes but in how we use them.
DNA fingerprinting
- Every individual has a unique genetic make-up, arising from recombination of paternal and maternal genes and from mutations during gamete formation.
- About 99.9% of the nucleotide sequence is the same in all persons. Short sequences of 20–100 base pairs, repeated several times and varying in repeat number between individuals, are called Variable Number of Tandem Repeats (VNTRs). Their differing lengths are the key to DNA profiling.
- The technique of DNA fingerprinting (DNA profiling) was first given by Dr. Alec Jeffreys in 1984.
- The steps are: isolation of DNA; restriction digestion with restriction enzymes, giving Restriction Fragment Length Polymorphism (RFLP); gel electrophoresis on agarose, in which negatively charged fragments move to the positive pole and separate by length into bands (dsDNA is split into ssDNA by alkali treatment); Southern blotting (developed by E. Southern) onto a nylon or nitrocellulose membrane; selection of a radioactively labelled single-stranded DNA probe; hybridisation of the probe with complementary host DNA; and photography by autoradiography on X-ray film.
- Applications include forensic science, resolving disputed parentage, and pedigree analysis in cats, dogs, horses and humans.
- Dr. Lalji Singh (1947–2017), the Father of DNA Fingerprinting in India, obtained the BKM-DNA probe from the banded krait, established the CDFD and founded LaCONES.
4.12 Points to Ponder
Most geneticists of the early twentieth century backed proteins as the genetic material, and their reasoning was careful rather than careless. What does this tell you about how scientific consensus can be both reasonable and wrong at the same time? What kind of evidence is needed to overturn such a consensus?
Griffith never identified his “transforming principle,” yet his experiment is still considered a landmark. Is an experiment that raises the right question as valuable as one that supplies the answer?
The DNA in one of your cells is about 2.2 metres long, and it is packed, unpacked and repacked continuously without tangling or breaking. What would go wrong in a cell if the packaging were slightly too tight, or slightly too loose?
DNA polymerase can add nucleotides only in the 5’→3’ direction, and this single restriction forces the cell into the awkward business of Okazaki fragments, primers and ligase. Why might evolution have kept such a seemingly inconvenient rule rather than producing an enzyme that works in both directions?
The genetic code has 64 codons for only 20 amino acids. Consider what would change — for the organism, and for evolution — if the code had exactly 20 codons and no degeneracy at all.
The code is described as “universal,” the same codon meaning the same amino acid from bacteria to humans. What does this universality suggest about the origin of life on Earth? And what would it imply if an organism were found that used a different code?
Every cell in your body carries the same DNA, yet a nerve cell and a bone cell could hardly be more different. If the information is identical, where does the difference actually come from?
In the lac operon, lactose acts as the inducer of its own breakdown. Can you think of other biological systems in which a substance triggers the very machinery that destroys or uses it? What advantage does such an arrangement give an organism living in a changing environment?
Humans have about 33,000 genes; the small plant Arabidopsis thaliana has about 25,000. If gene number does not explain our complexity, what does?
The Human Genome Project deliberately included an aim about the legal, ethical and social issues that its own results might raise. Should every large scientific project be required to do the same? Who should decide what counts as misuse of genetic information?
DNA fingerprinting can identify a person from a trace of blood or a single hair root. Consider the value of this in solving crimes, and set it against the question of who should be allowed to hold such information, and for how long. Where would you draw the line?
Mutation damages individuals and yet, together with recombination, supplies the raw material of evolution. Can a process be both harmful and necessary at the same time? What does this suggest about the difference between what is good for an individual and what is good for a species?
4.13 Objective Questions
- Griffith worked on ………….
- Bacteriophage
- Drosophila
- Frog eggs
- Streptococci
- The molecular knives of DNA are …………..
- Ligases
- Polymerases
- Endonucleases
- Transcriptase
- Translation occurs in the ……………
- Nucleus
- Cytoplasm
- Nucleolus
- Lysosomes
- The enzyme required for transcription is ………………
- DNA polymerase
- RNA polymerase
- Restriction enzyme
- RNAase
- Transcription is the transfer of genetic information from …………..
- DNA to RNA
- tRNA to mRNA
- DNA to mRNA
- mRNA to tRNA
- Which of the following is NOT part of protein synthesis?
- Replication
- Translation
- Transcription
- All of these
- In the RNA molecule, which nitrogen base is found in place of thymine?
- Guanine
- Cytosine
- Thymine
- Uracil
- How many codons are needed to specify three amino acids?
- 3
- 6
- 9
- 12
- Which out of the following is NOT an example of inducible operon?
- Lactose operon
- Histidine operon
- Arabinose operon
- Tryptophan operon
- Place the following events of translation in the correct sequence.
- Binding of met-tRNA to the start codon.
- Covalent bonding between two amino acids.
- Binding of second tRNA.
- Joining of small and large ribosome subunits.
- iii, iv, i, ii
- i, iv, iii, ii
- iv, iii, ii, i
- ii, iii, iv, i
- Match the entries in Column I with those in Column II and choose the correct answer.
| Column I | Column II |
|---|---|
| a. Alkali treatment | i. Separation of DNA fragments on gel slab |
| b. Southern blotting | ii. Split DNA fragments into single strands |
| c. Electrophoresis | iii. DNA transferred to nitrocellulose sheet |
| d. PCR | iv. X-ray photography |
| e. Autoradiography | v. Produce fragments of different sizes |
| f. DNA treated with REN | vi. DNA amplification |
- Justify the statements. If the answer is false, change the underlined word(s) to make the statement true. (150–200 words)
- The DNA molecule is double stranded and the RNA molecule is single stranded.
- The process of translation occurs at the ribosome.
- The job of mRNA is to pick up amino acids and transport them to the ribosomes.
- Transcription must occur before translation may occur.
4.14 Check Your Concepts
Assertion: Hershey and Chase used radioactive sulphur (\(^{35}S\)) to label the DNA of bacteriophages. Reason: DNA contains sulphur but not phosphorus. State whether the assertion and the reason are true or false, and whether the reason correctly explains the assertion.
Griffith’s fourth experiment (heat-killed S + live R) killed the mice, but his third experiment (heat-killed S alone) did not. Explain why this pair of results, taken together, was essential to his conclusion — and state clearly what Griffith did NOT prove.
In the Avery–MacLeod–McCarty experiment, transformation continued after treatment with proteases and RNAses but stopped after treatment with DNAse. Explain the logic by which this identifies the transforming principle.
Why must histones be rich in lysine and arginine in order to do their job? Answer with reference to the charge on DNA.
Arrange in the correct sequence, from least condensed to most condensed: solenoid; metaphase chromosome; DNA double helix; nucleosome; looped supercoiled structure; beads-on-string chromatin fibre.
What would happen to a region of a chromosome if it remained permanently in the heterochromatic state? Answer in terms of transcription.
Assertion: The lagging strand is synthesised discontinuously. Reason: DNA polymerase can add nucleotides only in the 5’→3’ direction, and the two parental strands are antiparallel. State whether both are true and whether the reason explains the assertion.
What would be the consequence for a replicating DNA molecule if single strand binding proteins (SSBP) were absent? Explain.
Why can DNA polymerase not begin a new strand on its own? Name the molecule that solves this problem and explain what happens to it later.
Arrange the following enzymes of DNA replication in the order in which they act: DNA ligase; DNA helicase; endonuclease; DNA gyrase; DNA polymerase; phosphorylase.
In the Meselson–Stahl experiment, why was only ONE band obtained after the first generation in \(^{14}N\) medium, and TWO bands after the second? Explain what result the conservative model would have predicted instead.
In a transcription unit, the strand with 3’→5’ polarity acts as the template. Explain WHY it must be this strand and not the other.
Match the following:
| Column I | Column II |
|---|---|
| a. RNA polymerase-I | i. tRNA and snRNA |
| b. RNA polymerase-II | ii. rRNA |
| c. RNA polymerase-III | iii. mRNA and hnRNA |
A eukaryotic primary transcript is described as “non-functional.” What two kinds of processing must it undergo before it can be translated, and why does a bacterial mRNA need neither?
Explain, using simple counting, why a doublet code would be inadequate for the 20 amino acids but a triplet code is sufficient. What is the number of possible codons in each case?
Distinguish between a code that is degenerate and a code that is ambiguous. Which one is the genetic code, and why does this matter for protein synthesis?
Khorana synthesised the repeating RNA sequence CUC UCU CUC UCU and obtained a polypeptide of two amino acids placed alternately, while CUA CUA CUA CUA gave a polypeptide of only one amino acid. Explain why the two synthetic RNAs gave such different products.
Why is the insertion of three bases into a gene usually far less damaging than the insertion of a single base? Answer with reference to the reading frame.
Explain why tRNA is called an adapter molecule. Which two ends of the tRNA molecule are involved, and what does each do?
There are no tRNAs with anticodons for UAA, UAG or UGA. What would happen during translation if such a tRNA did exist?
Label the three tRNA binding sites of a ribosome and state, in order, what happens to a tRNA as it passes through them.
Assertion: The catalyst of peptide bond formation in the ribosome is a protein enzyme. Reason: All biological catalysts are proteins. State whether the assertion and reason are true or false, and give the correct catalyst.
Why is mRNA described as short-lived? Explain how this short life is actually useful to the cell.
List, in order, the four levels at which gene expression can be regulated in a eukaryote. Which of these is the most economical for the cell, and why?
Distinguish between induction and repression. Which is positive control and which is negative control?
Name the three structural genes of the lac operon in sequence, and name the enzyme that each one produces. Explain why the cell benefits from switching all three on together rather than one at a time.
In the lac operon, the inducer does not bind to the operator. What does it bind to, and how does this switch the operon on?
If permease is needed for lactose to enter the cell, and permease is only made after lactose enters, how does the lac operon ever get started? Explain.
What would happen to the lac operon if a mutation destroyed the regulator gene i so that no repressor protein could be made? Would the structural genes be expressed in the absence of lactose?
Distinguish between structural genomics and functional genomics. Which must be done first, and why?
Look at the comparative genome table in the chapter. Humans have about 33,000 genes and Arabidopsis thaliana about 25,000. Yeast has 32 chromosomes and humans 46. What do these two comparisons tell you about the relationship between gene number, chromosome number and biological complexity?
Only one of the five aims of the Human Genome Project concerns the DNA sequence itself. Explain why the other four aims were considered necessary.
About 99.9% of human nucleotide sequence is identical from person to person. Explain how DNA fingerprinting nevertheless identifies a single individual.
Arrange the steps of DNA fingerprinting in the correct sequence: Southern blotting; autoradiography; restriction digestion; hybridisation; isolation of DNA; selection of DNA probe; gel electrophoresis.
During gel electrophoresis, DNA fragments move towards the positive pole and shorter fragments travel further. Explain both of these observations.
Why must double-stranded DNA be converted to single-stranded DNA by alkali treatment before the DNA probe is added? What would happen if this step were skipped?
A child’s DNA fingerprint is never identical to either parent’s, yet every band in the child’s pattern is present in one parent or the other. Explain why, and state how this makes parentage testing possible.
Compare heterochromatin and euchromatin under the following heads: state of packing, staining, DNA richness, and transcriptional activity.
Compare transcription and translation under the following heads: template used, site in a eukaryotic cell, key catalyst, and whether the complementarity principle applies.
State whether each of the following is true or false, and correct the false ones. (i) The coding strand of DNA is the strand read by RNA polymerase. (ii) The initiation codon AUG codes for methionine in eukaryotes and formyl methionine in prokaryotes. (iii) Introns appear in the processed mRNA of eukaryotes. (iv) The inducer is a component of the lac operon. (v) A cistron is a segment of DNA coding for one polypeptide.
What is the function of an RNA primer during protein synthesis? (30–40 words)
Why is the genetic code considered as commaless? (30–40 words)
What is genome? (30–40 words)
Which enzyme removes supercoils from replicating DNA? (30–40 words)
Why are Okazaki fragments formed on the lagging strand only? (30–40 words)
When does DNA replication take place? (30–40 words)
Define the terms codon and codogen. (30–40 words)
What is degeneracy of the genetic code? (30–40 words)
Which are the nucleosomal ‘core’ histones? (30–40 words)
Write a short note on DNA packaging in a eukaryotic cell. (80–100 words)
Enlist the characteristics of the genetic code. (80–100 words)
Write a note on the applications of DNA fingerprinting. (80–100 words)
Explain the role of lactose in the ‘Lac Operon’. (80–100 words)
Write a note on the Human Genome Project (HGP). (80–100 words)
Describe the structure of the ‘Operon’. (80–100 words)
Explain the process of DNA replication. (150–200 words)
Describe the process of transcription in protein synthesis. (150–200 words)
Describe the process of translation in protein synthesis. (150–200 words)
Describe the ‘Lac-operon’. (150–200 words)
4.15 Case-Based Questions
In the figure below, A, B and C are three types of ____________________. Figure: a strand containing uracil labelled A; a ribosome labelled B; a clover-leaf shaped molecule carrying an amino acid labelled C.
Identify the labelled structures on the following diagram of translation.
- Figure: an mRNA reading CCG GCC ACU CCC, with structures labelled A, B and C — A carries the anticodon UGA, C carries the anticodon GGG.
- Part A is the ________________________.
- Part B is the ________________________.
- Part C is the ________________________.
Guess (i) the possible locations of DNA on the collected evidence from a crime scene, and (ii) the possible sources of DNA. Complete the blanks in the table.
| Evidence | Possible location of DNA on the evidence | Sources of DNA |
|---|---|---|
| e.g. Eyeglasses | e.g. Ear pieces | e.g. Sweat, Skin |
| Bottle, Can, Glass | Sides, mouthpiece | ———— |
| —————- | Handle | Sweat, skin, blood |
| Used cigarette | Cigarette butt | ————- |
| Bite mark | ————- | Saliva |
| ————- | Surface area | Hair, semen, sweat, urine |