DNA Structure and Function PDF | ATAR Biology | 2023

52 3 DNA STRUCTURE CHAPTER 3 CONTENT AND FUNCTION By the end of this chapter, you will have covered the following material. STARTER QUESTIONS 1 Describe the structure of DNA, including the followin...

52 3 DNA STRUCTURE CHAPTER 3 CONTENT AND FUNCTION By the end of this chapter, you will have covered the following material. STARTER QUESTIONS 1 Describe the structure of DNA, including the following: a shape b names of the matching bases c  three main components that make up the molecule. Hint: two form the ‘backbone’ of the molecule. 2 How does the unique structure of DNA enable it to perform its functions? 3 List three roles of DNA in a cell. SCIENCE UNDERSTANDING »» DNA is a helical double-stranded molecule that occurs bound to proteins in chromosomes in the nucleus, and as unbound circular DNA in the cytosol of prokaryotes, and in the mitochondria and chloroplasts of eukaryotic cells »» the structural properties of the DNA molecule, including nucleotide composition and pairing and the hydrogen bonds between strands of DNA, allow for replication »» the genetic code is a base triplet code; genes include ‘coding’ and ‘non-coding’ DNA, and many genes contain information for protein production »» protein synthesis involves transcription of a gene into messenger RNA in the nucleus, and translation into an amino acid sequence at the ribosome »» proteins, including enzymes and structural proteins, are essential to cell structure and functioning SCIENCE INQUIRY SKILLS »» select, construct and use appropriate representations, including models of DNA replication, transcription and translation, Punnett squares and allele frequencies in gene pools, to communicate conceptual understanding, solve problems and make predictions ATAR Biology Syllabus, Government of Western Australia, School Curriculum and Standards Authority Getty Images/Max shen 9780170452922  CHAPTER 3 | DNA structure and function 53 3.1 THE DISCOVERY OF DNA Many secrets of DNA (deoxyribonucleic acid) were unlocked in 1952 at King’s College, when Rosalind Franklin took the first clear X-ray diffraction image of DNA. Franklin’s photograph helped confirm the spiral nature of DNA. Without her consent, her colleague Maurice Wilkins took her photographs to James Watson and Francis Crick. The photographs gave them Alamy Stock Photo/Science History Images evidence for the 3D structure they had previously theorised for DNA. Using these results and other accumulated evidence, Watson and Crick suggested that DNA consists of the now familiar two strands, resembling the uprights of a ladder, linked by ‘rungs’ (made of the four types of nucleotides), twisted to form a double helix. Rosalind Franklin had already died when Wilkins, Watson and Crick received a Nobel prize for their work in discovering the structure of DNA. It is now known that nucleic acids such as DNA do form a FIGURE 3.1 Rosalind Franklin, double helix: two linear strands, each containing a sequence of an expert X-ray crystallographer, worked on the structure of DNA nucleotide subunits, twisted together into a spiral. in the early 1950s. Her work was Some years earlier, an Austro-Hungarian biochemist pivotal in enabling Watson and Crick working in the USA, Erwin Chargaff, used a technique called to propose their hypothesis for the chromatography to work out the ratios of the four types of structure of DNA. nitrogenous bases [adenine (A), cytosine (C), guanine (G) and thymine (T)] present in the nucleotide subunits. He found that the amount of guanine was equal to the amount of cytosine, and the amount of adenine was equal to the amount of thymine. Alamy Stock Photo/Science History Images FIGURE 3.2 An X-ray diffraction photograph of DNA. The DNA molecule was too small to see using conventional methods, so X-rays were used. The image produced an accurate 3D shape. The pairs of nitrogenous bases are known as complementary base pairs. Complementary pairing is the phenomenon whereby guanine always hydrogen bonds with cytosine, and adenine always hydrogen bonds with thymine. Guanine and cytosine share three hydrogen bonds, and adenine and thymine share two hydrogen bonds. The complementary pairing helps produce the 3D helical structure of DNA. Nucleotides are the base units of DNA. 9780170452922 54 UNIT 3 | BIOLOGY WA ATAR UNITS 3 & 4 Where is DNA found in eukaryotes Science Photo Library/A. BARRINGTON BROWN, © GONVILLE & CAIUS and prokaryotes? DNA occurs bound to proteins in chromosomes within the nucleus of eukaryotic cells. The nucleus is enclosed in a nuclear membrane to protect its interior. DNA is also found in prokaryotes, but as unbound circular DNA in the nucleoid region of the cytosol. The nucleoid region is not bound by a nuclear membrane, and therefore the DNA is not contained like it is in a eukaryotic cell. Unbound, circular DNA is also COLLEGE found in the mitochondria and chloroplasts of eukaryotic cells. FIGURE 3.3 Watson (left) and Crick in 1953 with their model of part of a DNA molecule Prokaryote Eukaryote Linear chromosomes Circular in membrane-bound chromosome nucleus in nucleoid region of cell Cell membrane Genetic information encoded by DNA Cytosol Ribosomes Cell wall FIGURE 3.4 Structural similarities and differences between a prokaryotic cell and a eukaryotic cell 9780170452922  CHAPTER 3 | DNA structure and function 55 Core of eight histone molecules DNA double helix 2 nm DNA coiled around histones 30 nm 300 nm More coiling 700 nm 1000 nm Mitotic chromosomes FIGURE 3.5 In the nucleus of eukaryotes, linear DNA is found bound to proteins and becomes tightly coiled to form chromosomes. DNA is found in chloroplasts in eukaryotic plant and protist cells: DNA is found in mitochondria in all eukaryotic cells: Outer Thylakoids Chloroplast Ribosome membrane DNA Intermembrane Inner Mitochondrial space membrane DNA Ribosome Inner membrane Matrix Granum Mitochondrial Thylakoid matrix space (lumen) Intermembrane space Stroma Outer Folds of the Thylakoid membrane membrane inner membrane (cristae) FIGURE 3.6 Circular DNA, which is not bound to proteins, is found in chloroplasts and mitochondria. Key concept DNA is a double-stranded helical molecule. In eukaryotic cells, DNA is found in a linear form bound to proteins in the nucleus, and in an unbound circular form in chloroplasts and mitochondria. In prokaryotic cells, DNA is found in an unbound circular form in the nucleoid region of the cytosol. 9780170452922 56 UNIT 3 | BIOLOGY WA ATAR UNITS 3 & 4 3.1 DNA and who it gets passed to DNA contains instructions that make each species unique, and it is passed from parent cells to APPLICATION daughter cells during cell division, and from adults to their offspring during reproduction. In sexual reproduction, organisms inherit half of their nuclear DNA from the male parent and half from the female parent. However, organisms inherit all of their mitochondrial DNA from the female parent. This occurs because only egg cells, and not sperm cells, keep their mitochondria during fertilisation. Question set 3.1 REMEMBERING ANALYSING 1 List the scientists mentioned so far in this 5 Copy the following table and complete chapter who contributed to the discovery it using Chargaff’s ratio of nucleotide of the structure of DNA. subunits. 2 Describe Watson and Crick’s model of DNA. ADENINE CYTOSINE THYMINE GUANINE UNDERSTANDING 7 3 7 3 Explain why DNA is not found in the 21 25 25 nucleus of a prokaryotic cell. 4 Copy and complete the table of 43 44 complementary base pairs in DNA. 6 The average percentage composition of NITROGENOUS COMPLEMENTARY adenine in human DNA is 30%. Predict the BASE BASE PAIR percentage of the other three nucleotides. A C T 3.2 STRUCTURAL PROPERTIES OF THE DNA MOLECULE Nucleotides – the building blocks of DNA DNA is a nucleic acid made up of nucleotides. Each nucleotide consists of three parts: a five-carbon (pentose) sugar known as deoxyribose sugar, a phosphate group and a nitrogenous base (adenine, cytosine, guanine or thymine). A nucleotide is the basic 5' end O structural unit of DNA. O P T Each phosphate group is attached to two sugar molecules O O– CH2 O (5') by ‘ester’ bonds and is then called a phosphodiester bond (see C C (1') (4') H Figure 3.8). The five carbon atoms in each sugar molecule, H C C(2')H (3') Phosphate group H O Phosphodiester O P P C Nitrogenous bond O base O– CH2 (5') O (e.g. cytosine) C S H C (3') OH Deoxyribose sugar 3' end FIGURE 3.7 A nucleotide FIGURE 3.8 A phosphodiester bond 9780170452922  CHAPTER 3 | DNA structure and function 57 which form a ring, are numbered 1' to 5'. One of the ester bonds is formed with the 3' carbon of one sugar ring and the other is formed with the 5' carbon of the next sugar ring. The chain of alternating sugar molecules and phosphate groups is called the sugar–phosphate backbone. RNA (ribonucleic acid) has a similar structure to DNA, except deoxyribose sugar is replaced with ribose sugar. A strand of nucleotides has directionality described using the phrase 5' to 3'. The 5' end starts with a phosphate and the 3' end finishes with a sugar. DNA and RNA synthesis occurs in the 5' to 3' direction. The structure of the DNA molecule The shape of a DNA molecule is a double helix. The term ‘double’ refers to the two strands, which are joined by the weak hydrogen bonds between complementary pairs of nitrogenous bases. The complementary base pairing means that adenine always pairs with thymine, and cytosine always Build DNA pairs with guanine. The term ‘helix’ describes the helical (spiral) molecular shape: the two linear Practise building DNA strands run in opposite directions to each other (i.e. are anti-parallel) and are twisted into a helix. using complementary pairing. Drawing and labelling DNA Drawing and labelling DNA structure is an essential component of the course. This takes practice. Follow the steps below as you draw your own DNA structure. If you can draw it, then you know it! TABLE 3.1 Steps for how to draw and label DNA structure STEP DRAW AND LABEL 1 Draw the anti-parallel 5' 3' deoxyribose sugars (pentagons) for the two strands. On one strand draw them upright, and on the other strand draw them upside down to show they are anti-parallel. Then draw the phosphodiester bonds (circles, each with a ‘red’ and a ‘black’ bond) connecting the sugars in a chain. The ‘red’ covalent bond joins each 3' 5' phosphate to a 3' carbon, and the ‘black’ covalent bond FIGURE 3.9 Two anti-parallel strands of DNA joins it to a 5' carbon. 2 Draw the complementary 5' 3' C G base pairs. Indicate the C = cytosine weak hydrogen bonds with G = guanine dotted lines: 2 hydrogen A = adenine T = thymine A T bonds between adenine and thymine and 3 hydrogen bonds between cytosine and guanine. G C 3 Make a key listing the full names of the nitrogenous bases and their T A complementary pairs. 3' Hydrogen bonds 5' FIGURE 3.10 Complementary bases are paired 9780170452922 58 UNIT 3 | BIOLOGY WA ATAR UNITS 3 & 4 STEP DRAW AND LABEL 4 Circle one nucleotide. Label C 5 cytosine Sugar– Sugar– Base pairs its three parts. G 5 guanine phosphate phosphate backbone backbone A 5 adenine Nucleotides are very T 5 thymine important because they are the building blocks of each Hydrogen bonds 3' 5' strand. P C G S 5 Label the sugar–phosphate S P backbone sections of P A T the molecule, base pairs, S S ‘phosphate’, ‘nucleotide’ and P ‘deoxyribose’. In each sugar P G C symbol write an ‘S’, and in S S each phosphate circle write P a ‘P’. Nucleotide P T A S 6 Next to the chemical S Phosphate P structure you have just 3' 5' drawn, draw a simplified Nitrogenous Deoxyribose base smaller-scale DNA molecule, sugar demonstrating its double- helix structure. FIGURE 3.11 The sugar–phosphate backbones and the double-helix structure Key concept The base unit of DNA is a nucleotide, which consists of one nitrogenous base, one deoxyribose sugar and one phosphate group. The two strands of DNA are held together by weak hydrogen bonding between the complementary nitrogenous bases: adenine and thymine, cytosine and guanine. The structure of DNA enables it to function DNA is the genetic material common to all organisms. It carries information coded in the segments of its molecule known as genes. DNA thus enables certain traits to be passed on to the next generation. A trait is an inheritable characteristic. DNA is chemically the same in all organisms, although different species usually have different proportions of the various nucleotides, and each organism has a unique DNA sequence. The DNA sequences of individuals within a species have a lot more similarity than those of individuals of different species. In addition, DNA molecules in eukaryotes and prokaryotes generally differ in their associated proteins and in overall shape and appearance. We now understand how DNA is transmitted between generations, how genes are controlled and how differences in genes can cause changes in the way organisms develop and behave. This knowledge has allowed us to manipulate genes to achieve desired characteristics. New technologies have enabled us to accurately examine the interrelationships between species and to account for changes that have occurred in species over time. DNA stores the code for making proteins, and the inheritance of particular gene variants causes an individual to have a specific combination of proteins in its makeup. A section of DNA that codes for a specific protein (or polypeptide) is called a gene. It is now known that genes may code for more than one kind of polypeptide, and that genes interact with one another, causing changes in their expression (i.e. in the production of proteins). DNA, therefore, controls the growth and development of an organism. 9780170452922  CHAPTER 3 | DNA structure and function 59 The structural properties of the DNA molecule [its nucleotide composition, pairing and hydrogen bonding (see Figure 3.12b)] are what allow DNA replication to occur. This is because the DNA strands can function as template strands. a b Hydrogen bonds 3' end Base P S T S A P C G P T A S Sugar– phosphate G C Nitrogenous G backbone bases C C G S P 5' end A T c C G C Sugar C G T A Phosphodiester bond T Phosphate G FIGURE 3.12 a The DNA helix is a double-stranded molecule. b The two strands are held together by hydrogen bonding between complementary nitrogenous bases. c As well as nitrogenous bases, nucleotides have a sugar– phosphate backbone, in which the sugar molecules are linked by phosphodiester bonds. DNA structure View this link to Question set 3.2 reinforce your learning. DNA structure and REMEMBERING UNDERSTANDING replication 1 State the three components of a DNA 4 Explain what is meant by the term ‘anti- DNA learning centre nucleotide. parallel’. 2 Draw and label your own diagram of DNA, ANALYSING following the instructions found in Table 5 Relate DNA’s structure to three of its 3.1 (page 57). functions. 3 What is one main difference between the structure of DNA and RNA? 9780170452922 60 UNIT 3 | BIOLOGY WA ATAR UNITS 3 & 4 CASE DNA: further accumulation of knowledge STUDY When the structure of DNA was deduced, it is not only the roughly 1% of our DNA that seemed to be the final piece of a biological contains genes that code for proteins that has puzzle. It taught us how hereditary a function. The non-coding DNA (DNA that information (the passing on of which was does not code for protein) appears to have demonstrated by Mendel) was encoded. very important roles, such as regulatory and Between the 1950s and 1980s, scientists structural functions. ENCODE scientists are studying DNA thought most of our DNA was theorising that 80% of ‘junk DNA’ is active. useless. They knew that small sections of DNA, See Coding and non-coding DNA, page 63. known as genes, seemed to code for proteins, Today, scientists analyse DNA for many but they wondered about the other sections purposes other than the study of heredity. of the DNA. We now know genes can interact Genomics, such as is researched at the with one another and with the environment to Australian Museum’s Australian Centre for result in traits, but scientists still do not know Wildlife Genomics, is the study of the entire what most of an organism’s DNA does. DNA sequence of an organism and of its In the 1950s–1960s, Watson and Crick genes. Scientists know how to sequence the and other collaborators had determined DNA (work out the order of nucleotides). that DNA guides the production of RNA, and They do this for a range of purposes, such RNA guides the production of protein, which as identification of individuals, paternity may then be manifested as an observable testing, sex determination, measure of characteristic. But the biology of DNA is much species relatedness, conservation, population more complex than was initially thought. management and forensics. Researchers working for the Encyclopedia of DNA Elements (ENCODE) project, a Questions public research consortium launched by 1 Define the term ‘knowledge accumulation’ the US National Human Genome Research and apply it to DNA discoveries. Institute, have been mapping the parts of 2 List five uses of DNA analysis. human chromosomes that are transcribed 3 Explain how Franklin’s first clear X-ray (copied). In addition, they have been studying diffraction image of DNA laid some of the how the copying of DNA is regulated and groundwork for the current uses of DNA how the process is affected by the way the analysis. (Hint: structure can indicate the DNA is packaged. In 2012, they found that it mechanisms for function.) 3.3 DNA STRUCTURE ENABLES DNA REPLICATION DNA contains the genetic code that determines the structure and function of all living things. The product of DNA replication is two identical, double-helix DNA molecules, each consisting of one parental strand and one new strand. DNA replication is referred to as semi-conservative replication because one of the two strands is conserved, or retained, from one generation to the next, while the other strand is new. DNA replication occurs during the S phase of interphase during the cell cycle (see ‘The cell cycle’ in Chapter 2, pages 31–32). The purpose of DNA replication is to duplicate the code it carries. The code can then be passed on to daughter cells. In eukaryotic cells, the chromosomes gain a sister chromatid and become double stranded. DNA replication occurs in preparation for mitosis and meiosis. Key concept DNA replication is semi-conservative. Each of the new DNA molecules that are produced have one strand that is conserved (i.e. the parental strand) and one that is new (i.e. the daughter strand). 9780170452922  CHAPTER 3 | DNA structure and function 61 DNA replication begins with an enzyme called DNA helicase ‘unzipping’ the long molecule of double-stranded DNA by breaking the weak hydrogen bonds between the nucleotides and thus exposing the nucleotide Parental DNA bases. This separation of the parental DNA Replicated DNA molecules strands happens along a small section at a time. Replication fork The hydrogen bonds that hold the two strands of the DNA molecule together are weak, and the FIGURE 3.13 Movement of the replication fork along enzyme is easily able to separate them. parental DNA causes unwinding of the original DNA The junction between the unwound single double strands and rewinding of the two newly strands of DNA and the intact double helix is replicated double strands. called the replication fork. The replication fork moves along the parental DNA strand so that there is a continuous unwinding of the parental strands (Figure 3.13). Within the nucleus, stockpiles of free nucleotides attach to the exposed bases, according to the base-pairing rule (Figure 3.14), with the help of the enzyme DNA polymerase. Another enzyme, DNA ligase, seals the new short stretches of nucleotides into a continuous double strand that rewinds. Ligase catalyses the formation of phosphodiester bonds. The nucleotides link together in what is called a 5' to 3' direction, forming long molecules. Parental double helix Parental strand Daughter strands FIGURE 3.14 Replication of DNA. The specific relationships between A and T and between C and G ensure that the sequence of bases in the daughter DNA is exactly the same as that in the parent DNA. As DNA strands are antiparallel, DNA polymerase moves in opposite directions on the two strands during synthesis. On the leading strand, DNA polymerase is moving towards the replication fork and synthesises continuously. On the lagging strand, DNA polymerase is moving away from the replication fork and synthesises in pieces called Okazaki fragments. The process of DNA replication is summarised in Table 3.2 (page 62). 9780170452922 62 UNIT 3 | BIOLOGY WA ATAR UNITS 3 & 4 TABLE 3.2 DNA replication STEPS VISUAL AID 1 DNA helicase unwinds and separates the double Helicase strand by breaking the weak hydrogen bonds between complementary base pairs. Each half of the parent molecule is used as a template. 2 The enzyme RNA primase attaches a short Primer sequence of RNA, known as a primer, to show DNA polymerase where to start adding nucleotides. RNA primase 3 Free nucleotides are added by DNA polymerase Daughter strand according to complementary base-pairing rules. Synthesis of the new daughter strand is in a 5' to 3' DNA polymerase direction. Adenine pairs with thymine, and cytosine pairs with guanine. 4 DNA ligase removes and replaces the primers. The result is two identical DNA molecules that are each made of one original parent strand and one new DNA daughter strand. DNA replication is described as ligase semi-conservative. 5 In eukaryotic organisms, two identical sister chro- matids are now ready for cell division. In prokary- otes, two identical circular chromosomes are now ready for binary fission. FIGURE 3.15 DNA replication involves enzymes Figure 3.16 illustrates the continuous and discontinuous synthesis of DNA along each of the strands. Synthesis is continuous along the leading strand, with additional nucleotides being added one after the other. It is discontinuous along the lagging strand because it is a 3' to 5' strand and DNA polymerase can only synthesise new DNA in a 5' to 3' direction. Primers are attached at short intervals, starting from the replication fork. DNA polymerase synthesises short strands of new DNA starting at each primer, in a 5' to 3' direction. The short strands are called Okazaki fragments. DNA polymerase moves in opposite directions on the two anti-parallel parent strands. DNA polymerase removes the RNA primers and replaces them with DNA nucleotides. DNA ligase joins the Okazaki fragments together to create a continuous strand. Ligase catalyses the formation of a phosphodiester bond. 9780170452922  CHAPTER 3 | DNA structure and function 63 Free nucleotides DNA polymerase Adenine Thymine Cytosine Guanine Leading strand 3' 5' Helicase 5' 3' Lagging Original strand DNA molecule Replication fork Okazaki fragment DNA polymerase Original (template) DNA strand Chromosome FIGURE 3.16 DNA replication showing the leading and lagging strands Question set 3.3 REMEMBERING 1 State the role of DNA helicase, polymerase and ligase in DNA replication. DNA replication 2 Draw a flow diagram to summarise the process of DNA replication. The one below has been View the link to included as an example of how to set it out. reinforce your understanding of DNA DNA replication. Helicase Primer separates initiates polymerase DNA structure and attaches replication UNDERSTANDING 3 Explain why the process of DNA replication is described as semi-conservative. ANALYSING 4 Describe the relationship between DNA replication and cell division. 5 Predict the nucleotide sequence for the complementary strand of a fragment of a DNA chain with the nucleotide bases GCCTATTGCA. 3.4 CODING AND NON-CODING DNA DNA is a molecule consisting of a sequence of nucleotides. The entire order of the nucleotides in a human cell’s DNA have been sequenced. The sequence of consecutive DNA ‘letters’ spanning all the chromosomes of a cell from start to finish is known as the genome sequence. Some sections of the sequence code for proteins and are called coding DNA. The coding DNA sections are also called genes. The coding DNA specifies sequences of amino acids, which are the building blocks of proteins. Proteins are responsible for nearly all cell functions. Humans have around 20 000 protein-coding genes. (Corn has around 32 000 genes and Escherichia coli (E. coli) bacteria has around 4400 genes.) Approximately 1–2% of the DNA in a human is comprised of coding DNA. Genes contain information 9780170452922 64 UNIT 3 | BIOLOGY WA ATAR UNITS 3 & 4 for the production of proteins, and proteins are the link between the stored genetic code, the genotype, and observable traits, called the phenotype. The majority of the human genome is comprised of non-coding DNA. The German botanist Hans Winkler invented the term ‘genome’ in 1920 by combining the words GENe and chromosOME. A short definition of genome is ‘all the DNA in a cell’, and this includes the genes and also DNA that is not part of any gene. The sections of DNA that do not code for a protein are classified as non-coding DNA. Some non-coding DNA is transcribed into functional non-coding RNA molecules, such as transfer RNAs and regulatory RNAs. Historically, non-coding DNA was referred to as ‘junk DNA’, but through recent advances in knowledge, scientists have found that some of the non-coding DNA is important and therefore not actually ‘junk’. Nucleus Chromosome Regions of DNA that code for protein (coloured section) Gene 1 Gene 2 Regions of DNA that do not code for proteins (white sections); FIGURE 3.17 Coding versus non-coding DNA in a eukaryotic cell: 75% of non-coding DNA occurs between genes. Introns occur within genes and they make up the remaining 25% of non-coding DNA. The genetic code Watch the animation. The genetic code Gene 1 Double Describe how the code The genetic code is the term used for the way strand is stored. that the four nitrogenous bases of DNA, adenine, of DNA Triplet Triplet Triplet thymine, guanine and cytosine, are ordered. The base order is ‘read’ by cellular machinery and 2 Sense turned into a protein via a process called protein strand synthesis. Cellular machinery consist of ‘biological of DNA machines’ that work to manufacture a biological Transcription molecule. The transcription machinery includes RNA polymerase and binding factors/proteins. The translation machine is the ribosome. 3 mRNA In the genetic code, each set of three DNA nucleotides in a row counts as a triplet and codes Codon Codon Codon for an mRNA (messenger RNA) triplet called a codon. The mRNA codon (three nucleotides) is FIGURE 3.18 The genetic code is a base triplet code. 9780170452922  CHAPTER 3 | DNA structure and function 65 again read by cellular machinery and is translated into a single amino acid. Each sequence of three nucleotides codes for an amino acid. Given that some proteins are made up of hundreds of amino acids, the code that would make one protein could have hundreds, sometimes even thousands, of triplets contained in it. Key concept The genome sequence consists of coding DNA (genes) and non-coding DNA. Three coding DNA nucleotides make a triplet, which matches an mRNA codon. Question set 3.4 REMEMBERING UNDERSTANDING 1 Define: 3 Differentiate between a DNA triplet and a a genome sequence codon. b coding DNA ANALYSING c non-coding DNA. 2 State the link between an organism’s 4 Why are nucleotide sequences read in genotype and phenotype. threes? 3.5 PROTEIN SYNTHESIS Proteins are essential to the structure and function of cells, and thus also to the structure and function of organisms. Protein synthesis is the process of making new proteins from the genetic information encoded in DNA. There are two main processes that facilitate the flow of information from gene to protein: transcription and translation. Transcription is the synthesis of mRNA using the stored DNA code. The synthesised mRNA is a chain of RNA nucleotides complementary to the DNA strand, except uracil (rather than thymine) is the base pair of adenine in RNA. Translation is the synthesis of a polypeptide using the information in the mRNA. The RNA nucleotide code is translated into an amino acid sequence. Codons A series of three nucleotides found in mRNA. They act as a code for an amino acid e.g. Enzymes UAU codes for the amino acid tyrosine. Help break or form new bonds A START codon (AUG) initiates translation, and e.g. RNA polymerase a STOP codon (UAG) brings the process to an end. Essential materials needed for the process of protein synthesis Nucleic acids Amino acids DNA stores the code. Twenty amino acids are the building blocks mRNA transports the code from the of the polypeptides and proteins. nucleus into the cytoplasm and to the ribosome. The sequence of amino acids in a protein tRNA is found in the cytoplasm. is a type of code that specifies the structure For each codon, a tRNA carries a specific amino acid to the and function of the protein, making it ribosome for incorporation into the growing polypeptide. different from other proteins. FIGURE 3.19 Major materials required in protein synthesis and their roles Genes are found in chromosomes in cells. They are sequences of DNA that code for a protein. It is only during cell division that the DNA can leave the nucleus of a eukaryotic cell. Otherwise, it remains there, ready for future cell division (mitosis or meiosis). Thus, the DNA code (genes) must be transcribed into messenger RNA (mRNA) while still inside the nucleus. The mRNA can fit through the nuclear pores because it is a short, single-stranded molecule. Therefore, the mRNA can carry the code of instructions 9780170452922 66 UNIT 3 | BIOLOGY WA ATAR UNITS 3 & 4 to the ribosome, where translation can take place. The ribosome binds to the mRNA. Each codon attracts the corresponding anticodon that forms part of a tRNA (transfer RNA) molecule. The tRNA molecule carries the amino acid that is specific to the codon. As one codon at a time moves into and is read by the ribosome, successive tRNAs transport amino acids to it, translating the code by dropping off amino acids in a sequence that matches the sequence of codons. Gradually, a polypeptide is produced (a string of amino acids, joined by peptide bonds). The polypeptide can detach and fold to form a protein by itself, or attach to other polypeptides and then fold to form a protein. Transcription and translation in prokaryotes In a prokaryotic cell (a cell that lacks membrane-bound organelles, including nuclei), the chromosome is generally in the form of a closed circle that is not wrapped around histone proteins. It is found in the region of the cell known as the nucleoid. In addition to the single chromosome, bacteria may contain plasmids, which are small rings of DNA. Plasmids code for traits but are not essential to the survival of the cell (although they may aid in its survival). Transcription begins when a section of the double-stranded chromosome is separated and enzymes synthesise an mRNA product complementary to the template strand. In prokaryotes, transcription and translation are simultaneous; that is, translation begins while the mRNA is still being synthesised, during transcription. Numerous ribosomes concurrently translate the mRNA transcripts into polypeptides. In contrast, a eukaryotic cell performs transcription in the nucleus, and translation in the cytoplasm. Question set 3.5a REMEMBERING function of each in the synthesis of a 1 Describe why protein synthesis is needed. protein. 2 State the two main processes involved in ANALYSING protein synthesis. 4 In eukaryotes, translation follows transcription. Differentiate between UNDERSTANDING transcription and translation in 3 Describe four essential materials required prokaryotes and eukaryotes, and state for protein synthesis and include the why it may be simpler in prokaryotes. Transcription in eukaryotes Transcription, a process that produces mRNA from DNA, occurs in the nucleus in eukaryotes. During transcription, one section of DNA, called a gene, is unwound and separated ready for copying. RNA polymerase moves step by step along the DNA molecule, separating the two strands. Only the template strand is copied. The template strand is also known as the antisense or non-coding strand. The other strand is known as the non-template, sense or coding strand. The coding strand has the same code as the mRNA, except in RNA uracil replaces thymine. The sequence of the DNA nucleotides determines the sequence of the RNA nucleotides, because RNA polymerase attaches the RNA nucleotide that is complementary to each DNA base. The complementary pairs are added according to the base-pair rules, shown in Table 3.3. TABLE 3.3 The complementary base pairs attach during transcription according to base pair rules. DNA NITROGENOUS BASE COMPLEMENTARY RNA NITROGENOUS BASE IN THE RNA NUCLEOTIDE Adenine Uracil Thymine Adenine Cytosine Guanine Guanine Cytosine 9780170452922  CHAPTER 3 | DNA structure and function 67 A promoter attaches to help the DNA template strand to locally separate from the non-template strand, initiating transcription. RNA polymerase binds to the DNA to get ready to start synthesis. RNA polymerase synthesises the mRNA in a 5' to 3' direction, anti-parallel to the template strand. The mRNA nucleotide triplets are called codons. The codons are complementary to the template strand but almost identical to the non-template/coding strand, except for uracil replacing thymine. After the RNA polymerase enables elongation of the strand, the mRNA molecule detaches as pre-mRNA. Pre-mRNA requires processing before it exits the nucleus via the nuclear pore. Stretches of non-coding DNA (known as introns) are removed and the remaining stretches of DNA (known as exons) join to form mature mRNA. Introns and exons SCIENTIFIC LITERACY As part of the normal process of generating proteins from genes stored in DNA, the code for constructing a particular protein is passed from stored DNA to a form that is transportable known as messenger RNA (mRNA). The strip of mRNA that is first formed when the DNA code is copied has excess baggage. In most eukaryotes, the mRNA initially carries the instructions for making a protein, but also carries extra nucleotides that are not needed. The unrefined mRNA is called pre-mRNA. Before the mRNA can leave the cell nucleus, non-coding regions called introns are cut out in a process called (pre-)mRNA splicing. The remaining exons join together as the final set of refined instructions, ready to move out of the nucleus via a nuclear pore. The refined mRNA is called mature mRNA. It performs the function of carrying the code to the translation site, where proteins are built one amino acid at a time according to the code. The discovery of mRNA splicing in the late 1970s was simultaneous with the revelation mRNA splicing that a single species of pre-mRNA could be spliced in different ways, creating multiple, distinct Watch this animation to help you understand mature mRNAs. This is now known as ‘alternative splicing’. The various mature mRNAs pre-mRNA splicing: contain different combinations of exons. The different combinations give rise to different proteins. Scientists at the American Society for Microbiology have studied alternative splicing in Apicomplexan parasites. Apicomplexan parasites are pathogens (organisms that cause an infectious disease) found in humans and domestic animals. Science Photo Library/ALAIN POL/ISM These parasites have also been reported by CSIRO as being parasitic on Australian reptiles, mammals such as the echidna, and the green tree frog. Funding for research into parasites of Australian wildlife is sparse, and anti-parasitic drugs have been hard to develop. Perturbation (interruption) of alternative splicing has been found to be FIGURE 3.20 Toxoplasma gondii, a species of detrimental to these parasites, making it a Apicomplexan parasite worthy drug target to pursue if we wish to reduce disease in humans and domestic animals. Questions 1 Describe pre-mRNA splicing. 2 Explain the rationale for spending money on research into parasites of Australian wildlife. 3 Describe the relationship between studying alternative splicing and anti-parasitic drugs. 9780170452922 68 UNIT 3 | BIOLOGY WA ATAR UNITS 3 & 4 TABLE 3.4 Summary of transcription STEP VISUAL AID Preliminary information Only one of the two strands of DNA is used for Non-template strand transcription: the template DNA DNA rewinds unwinds strand (also known as the non-coding strand or the RNA polymerase antisense strand). 1 RNA polymerase binds Template to a promoter region. strand It breaks the weak hydrogen bonds joining the complementary nucleotides and unzips mRNA transcript (unwinds) a portion of the double helix. FIGURE 3.21 RNA polymerase binds to a promoter region and an area of one gene on the DNA molecule becomes unzipped, beginning transcription. 2 Moving along the 5' Non-template strand 3' template DNA strand T A C T GCC T AG T CGGCG T T CGCC T T A A CCGC T G T A T T in a 3' to 5' direction, the RNA polymerase 5' RNA transcript 3' adds free-floating U A CUGCCU AGUCGGCGUU RNA polymerase nucleotides to the A T GA CGGA T C AGCCGC A AGCGGA A T T GGCGA C A T A A 3' 5' growing mRNA Template strand sequence according to the complementary FIGURE 3.22 mRNA is synthesised in a 5' to 3' direction by RNA polymerase. base-pair rules, but in RNA uracil pairs with adenine. The new strand of mRNA is synthesised in a 5' to 3' direction. 3 The DNA bases are Single coding Double strand in triplets, and the strand of DNA of DNA Gene ­complementary mRNA triplets produced are called codons. The process continues until there is a t­ ermination signal and the Triplet G A C A C T G A C T C T C G T T A C T C T G A C C A T ­pre-­mRNA is released. G A Transcription U C U G U G A C U G A G A G C A A U G A G A C U G G U A C Codon Strand of mRNA FIGURE 3.23 The mRNA produced contains codons that are complementary to the DNA triplets. 9780170452922  CHAPTER 3 | DNA structure and function 69 STEP VISUAL AID 4 Only the coding region Gene (gene) of DNA is transcribed. Pre-mRNA consists of introns and exons. The introns are removed and the exons Promoter region Coding region are joined to create mature mRNA. Mature mRNA then exits the nucleus via a nuclear Transcription pore. Pre-mRNA 5' Intron Exon Exon Intron Exon 3' Untranslated Untranslated region region Mature mRNA 5' UTR Exon Exon Exon UTR 3' FIGURE 3.24 The coding region of DNA is transcribed into pre-mRNA, which is then processed to make mature mRNA. Key concept Protein synthesis in eukaryotes includes the process of transcription. Transcription occurs in the nucleus and is the process of transcribing the code from DNA into a smaller molecule called mRNA, which can then leave the nucleus. Question set 3.5b REMEMBERING APPLYING 1 Explain the purpose of transcription. 4 Apply what you have learned about coding 2 State the site for eukaryotic transcription. and non-coding DNA to differentiate between introns and exons. UNDERSTANDING 3 Differentiate between a DNA triplet and an mRNA codon. 9780170452922 70 UNIT 3 | BIOLOGY WA ATAR UNITS 3 & 4 Translation in eukaryotes Growing polypeptide Protein synthesis involves both transcription 3D animation of DNA to RNA to protein and translation. These processes enable genetic information to flow from DNA to Amino acid mRNA to protein. Transcription is the process tRNA of synthesising a copy of the DNA code in the form of mRNA. Translation is the RNA-directed synthesis of a polypeptide. C CC Ribosomes facilitate the interaction of mRNA A A A UCG and tRNA to position and connect a specific GGG UUU AGC sequence of amino acids. Ribosomes are mostly mRNA composed of ribosomal RNA (rRNA), which is non-coding. Process of translation Movement of ribosome After mRNA moves out from the nucleus through FIGURE 3.25 Ribosomes read codons to construct a nuclear pore, it enters the cytoplasm and polypeptides. travels to a ribosome, where it will be read and translated. The translation process can be divided into three main stages: initiation, elongation and termination. Initiation A ribosome binds to a molecule of mRNA. It ‘reads’ the mRNA nucleotides in threes. A group of three consecutive nucleotides is called a codon. A special codon, AUG, is the start codon and codes for the amino acid methionine. It signals the start of translation and the beginning of a polypeptide chain. The tRNA molecule that contains the anticodon UAC is attracted to the start codon and pairs with it. This tRNA molecule brings with it the amino acid methionine. At initiation, two codons enter and are bound to the ribosome, but after that only one codon enters and is translated at a time. Elongation A tRNA molecule, which includes in its sequence an anticodon, is attracted to the corresponding codon on the mRNA due to complementary base pairing. Each tRNA molecule carries an amino acid specified by the codon that it pairs with. As one codon is read and exits the ribosome, another one slides in to be read. tRNAs transfer the amino acids to the mRNA–ribosomal complex in the order specified by the codons of the mRNA. The ribosomes catalyse the formation of covalent peptide bonds between adjacent amino acids. The mRNA is moved through the ribosome in one direction only. Once a tRNA molecule has dropped off its amino acid, it will return to the cytoplasm to reload with the same type of amino acid. Note that the tRNA is not used up during translation, and some amino acids are coded for by more than one codon. Termination Elongation continues until a stop codon in the mRNA enters the ribosome. The nucleotide base triplets UAG, UAA and UGA do not code for an amino acid. Instead, any one of them acts as a signal From DNA to protein Watch this short video to stop translation. The polypeptide is then released and the mRNA leaves the ribosome. Once on protein synthesis. removed, the polypeptide may fold (or join with another polypeptide to fold) to become a structural or functional protein. The protein will either be used in the cell it was formed in or be transported out of the cell for use elsewhere. Note that the tRNA is not used up during the translation process, and some amino acids are coded for by more than one codon. 9780170452922  CHAPTER 3 | DNA structure and function 71 TABLE 3.5 Summary of translation STEP VISUAL AID 1 The short, single- stranded mRNA leaves Nucleus DNA the nucleus via a Nuclear pore nuclear pore to bind with a ribosome in the cytoplasm. A subunit of the ribosome binds Synthesis of mRNA to mRNA to begin (transcription) mRNA protein synthesis. Movement of mRNA Cytoplasm mRNA Ribosome Synthesis of protein (translation) Polypeptide FIGURE 3.26 mRNA leaves the nucleus and binds with a ribosome in the cytoplasm. 2 A start codon (AUG) in the mRNA molecule Amino acid Met signals for a tRNA molecule with the complementary anticodon (UAC) to arrive for base pairing. Ribosome tRNA The tRNA molecule carries the first amino mRNA acid, methionine. Anticodon UAC A U G G G U G U A C C C (etc.) Start codon FIGURE 3.27 The start codon in mRNA is usually AUG. 9780170452922 72 UNIT 3 | BIOLOGY WA ATAR UNITS 3 & 4 STEP VISUAL AID 3 The next codon attracts a new tRNA molecule with the Met Gly corresponding anticodon and amino acid attached. One tRNA molecule at a time, with its specific amino acid attached, moves into the ribosome while the other, now minus its amino acid, leaves the ribosome. After each codon partners U A C C C A with an anticodon, the A U G G G U G U A C C C (etc.) amino acid is removed from the tRNA and is joined to a growing amino acid chain by a covalent peptide FIGURE 3.28 Amino acids are attached one at a time and according to the order of the codons. bond. The ribosome ‘reads’ one codon at a time. 4 When a stop codon appears, Free elongation ceases polypeptide Release factor and the polypeptide is released from the ribosome. The ribosome separates from the mRNA. tRNA 3' 3' 5' 5' mRNA Ribosome Stop codon (UAG, UAA or UGA) FIGURE 3.29 The polypeptide is formed and released. 9780170452922  CHAPTER 3 | DNA structure and function 73 STEP VISUAL AID 5 The polypeptide Primary structure either folds to form a Amino acid sequence protein or joins with Gly Phe Asn Glu another polypeptide Gln to fold to become a Ala 3D protein ready to Arg perform a function. Pro Tyr Ser Trp Primary, secondary, Asp Ile Met Cys Leu Lys tertiary and quaternary describe different Val His levels of protein structure, from simple to complex. Secondary structure Regular substructures Interactions between side chains create the 3D structure Quaternary structure Tertiary structure Complex of protein molecules 3D structure FIGURE 3.30 Protein synthesis is complete. Key concept Protein synthesis in eukaryotes includes translation. Translation occurs at a ribosome in the cytoplasm, and uses the code in mRNA to produce a sequence of amino acids called a polypeptide. Question set 3.5c REMEMBERING CREATING 1 Define polypeptide. 4 Create a flow diagram showing the 2 State the enzyme(s) involved in separating summarised steps of transcription and the two strands of DNA and synthesising translation. an mRNA strand. UNDERSTANDING 3 Describe the relationship of tRNAs to mRNA and amino acids. 9780170452922 74 UNIT 3 | BIOLOGY WA ATAR UNITS 3 & 4 3.6 PROTEINS Proteins are built of their basic units or monomers (known as amino acids) and are essential to cell structure and functioning. Some proteins are quite rigid, such as collagen (which plays a structural role in the connective tissue of mammals). Other types of protein, such as enzymes, perform functional tasks. Enzymes (e.g. lipase and trypsin) are catalysts that increase the rate of virtually all of the chemical reactions within cells. The protein shape at the active site of an enzyme determines the specificity of the enzyme: only specific enzymes can fit with specific substrates. In addition to providing mechanical support and functioning as catalysts, proteins transport and store other molecules (such as oxygen), provide immune protection, generate movement, transmit nerve impulses, and control growth and differentiation. A protein’s structure is vital to its function. A slight change in structure can alter the function of a protein to the extent that cell death may be triggered. Programmed cell death (apoptosis) is an important strategy for disposing of damaged or infected cells and those no longer needed in a multicellular organism. Proteins are built from a selection of 20 different amino acids. The amino acids are linked together by peptide bonds to form polypeptide chains, which fold and/or are modified to form the protein. The sequence of amino acids in a polypeptide is determined by the sequence of mRNA codons in a strand of mRNA. If the sequence of codons is known, the sequence of amino acids can be determined from an amino acid table (also known as a codon table, Figure 3.31. A codon table is a translation table that identifies the amino acids that correspond to the mRNA codons. To find the amino acid coded for by an mRNA codon, look for the three nitrogenous base letters in the table. There are 64 possible base triplets (4 × 4 × 4), and three of these are stop codons that signal for translation to stop. Second base U C A G Ala = alanine Arg = arginine UUU UCU UAU UGU Phe Tyr Cys U Asn = asparagine UUC UCC UAC UGC C U Ser Asp = aspartic acid UUA UCA UAA Stop UGA Stop A Leu Cys = cysteine UUG UCG UAG Stop UGG Trp G Gln = glutamine CUU CCU CAU CGU Glu = glutamic acid U His CUC CCC CAC CGC C Gly = glycine C Leu Pro Arg CUA CCA CAA CGA A His = histidine Third base First base Gln CUG CCG CAG CGG G Ile = isoleucine Leu = leucine AUU ACU AAU AGU U Lys = lysine Asn Ser A AUC Ile ACC AAC AGC C Met = methionine Thr AUA ACA AAA A Phe = phenylalanine AGA AUG Met/ ACG Lys Arg G AAG AGG Pro = proline Start Ser = serine GUU GCU GAU GGU U Thr = threonine Asp GUC GCC GAC GGC C Trp = tryptophan G Val Ala Gly GUA GCA GAA GGA A Tyr = tyrosine GCG Glu GUG GAG GGG G Val = valine FIGURE 3.31 The genetic code is shown by a codon table. The mRNA codons correspond to the 20 amino acids used to build polypeptides during translation on the ribosomes. Three codons act as stop codons, and (under certain conditions) the codon AUG initiates protein synthesis. 9780170452922  CHAPTER 3 | DNA structure and function 75 The genetic code for keratin, a protein building block for hair, is transcribed and translated by specialist cells underneath growing hair. The following sequence is part of the mRNA molecule that is transcribed from the gene for keratin: AUGUCUCGUGAAUUUUCC. To determine the sequence of amino acids, divide the nucleotides from the gene into sets of three. AUG UCU CGU GAA UUU UCC Then use the codon table (Figure 3.31) to translate the code. The first codon (AUG) is a start codon that codes for an amino acid called methionine. Continuing along the gene, the entire sequence for this section of the code is: methionine–serine–arginine–glutamic acid–phenylalanine–serine Question set 3.6 REMEMBERING ANALYSING 1 Define protein. 5 Use the codon table (Figure 3.31) to 2 State two of the main types of proteins determine the chain of amino acids for the needed in an organism. following DNA code: 3 Give an example of each of the two types UACAGAGCACUUAAAAGG. of protein in your answer to the previous question, and describe their functions. UNDERSTANDING 4 Compare and contrast a codon and anticodon. The koala’s DNA has been sequenced CASE STUDY Koala populations in NSW and Queensland dropped by 42% between 1990 and 2010, according to the federal Threatened Species Scientific Committee. Reasons for the decline included chlamydia (an infectious disease), bushfires (destroying their habitat), wild dogs and climate change. Koalas are particularly vulnerable to climate change because they are heavily reliant on trees for their homes and food. The severe population declines experienced by koalas have prompted WWF- Australia and other groups to nominate the koala for to have its listing changed from vulnerable to endangered. A team of Australian and international scientists, led by Adjunct Professor and Australian Museum Research Institute Fairfax Photos/Jessica Hromas director Rebecca Johnson (pictured) and Professor Katherine Belov at the University of Sydney, completed the world-first full sequencing of the koala genome in 2018. The entire sequence of nucleotides found in a koala’s DNA was recorded. FIGURE 3.32 Researchers such as Rebecca The Australian-led Koala Genome Johnson aim to help koala conservation using Consortium of 54 scientists from 29 genomic studies. 9780170452922 76 UNIT 3 | BIOLOGY WA ATAR UNITS 3 & 4 institutions across seven countries sequenced Some of the 26 000 koala genes the more than 3.4 billion base pairs and more identified in the koala genome project help than 26 000 genes in the koala genome, explain the koala’s extraordinary ability to which makes it slightly larger than the human survive almost exclusively on eucalypt leaves, genome. a diet high in toxins. Researchers found an You may be wondering how this helps abundance of genes for bitter taste receptors the plight of the Australian koala? Koala joeys (which would allow koalas to identify the are born after 35 days of gestation, when they least toxic leaves), as well as genes that code are the size of a jelly bean. While they are in for proteins that help detoxify the poisonous their mother’s pouch, they are protected by substances. antimicrobial peptides in her milk. However, when they are weaned they no longer have Questions this protection. They are then susceptible 1 Propose a logical reason why it took to a bacterial infection called chlamydia. so long for the nucleotides in a koala’s Sequencing the genome has allowed scientists DNA to be sequenced (i.e. the order of to characterise the architecture of their nucleotides to be determined)? immune system and to identify genes that 2 List four changes in the koala’s environment play a role in resistance and susceptibility to that makes them vulnerable today. chlamydia. The DNA data can be used to help 3 Explain how our knowledge of the develop vaccines, manage koala populations sequence of nucleotides (the genome) and ultimately help with their long-term may be useful for the koala’s long-term survival. survival. 9780170452922  CHAPTER 3 | DNA structure and function 77 CHAPTER 3 ACTIVITY AND INVESTIGATION The genetic code 3.1 The nucleus of eukaryotic cells is packed ACTIVITY with DNA, the molecule that is the template Ser Amino acid for all the proteins produced by the cell. Ribosomes, the sites of protein synthesis, are found in the cytoplasm, outside the membrane of the nucleus. DNA is unable to move through the nuclear membrane, so in order to produce a protein, a message must be sent from the nuclear DNA to the ribosome. tRNA To do this, two processes take place: 1 transcription of the message from the DNA into an mRNA molecule 2 translation of the message in the mRNA into a specific amino acid sequence at the ribosome. U C G The molecule of mRNA is transcribed from the template DNA strand using the Anticodon complementary sequences. However, the FIGURE 3.33 The amino acid serine being carried by a thymine present in DNA is replaced with tRNA molecule uracil in RNA. The complementary pairs in RNA are A?