SCHOLAR Study Guide Higher Biology Unit 1 PDF

SCHOLAR Study Guide Higher Biology Unit 1: DNA and the Genome Authored by: Bryony Clutton (North Berwick High School) Reviewed by: Fiona Stewart (Perth Grammar School) Previously authored by: Jaquie Burt Eileen Humphrey Lorraine Knight Nadine Randle Fergus Forsyth Patrick Hartie Heriot-Watt University Edinburgh EH14 4AS, United Kingdom. First published 2018 by Heriot-Watt University. This edition published in 2018 by Heriot-Watt University SCHOLAR. Copyright © 2018 SCHOLAR Forum. Members of the SCHOLAR Forum may reproduce this publication in whole or in part for educational purposes within their establishment providing that no profit accrues at any stage, Any other use of the materials is governed by the general copyright statement that follows. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, without written permission from the publisher. Heriot-Watt University accepts no responsibility or liability whatsoever with regard to the information contained in this study guide. Distributed by the SCHOLAR Forum. SCHOLAR Study Guide Higher Biology: Unit 1 Higher Biology Course Code: C807 76 ISBN 978-1-911057-36-9 Print Production and Fulfilment in UK by Print Trail www.printtrail.com Acknowledgements Thanks are due to the members of Heriot-Watt University's SCHOLAR team who planned and created these materials, and to the many colleagues who reviewed the content. We would like to acknowledge the assistance of the education authorities, colleges, teachers and students who contributed to the SCHOLAR programme and who evaluated these materials. Grateful acknowledgement is made for permission to use the following material in the SCHOLAR programme: The Scottish Qualifications Authority for permission to use Past Papers assessments. The Scottish Government for financial support. The content of this Study Guide is aligned to the Scottish Qualifications Authority (SQA) curriculum. All brand names, product names, logos and related devices are used for identification purposes only and are trademarks, registered trademarks or service marks of their respective holders. v Contents 1 Structure and organisation of DNA 1 1.1 The structure of DNA.................................. 3 1.2 The organisation of DNA in prokaryotes and eukaryotes.............. 7 1.3 Learning points..................................... 9 1.4 Extension materials................................... 10 1.5 End of topic test..................................... 11 2 Replication of DNA 13 2.1 DNA replication..................................... 15 2.2 The polymerase chain reaction (PCR)......................... 17 2.3 Learning points..................................... 20 2.4 Extended response question.............................. 21 2.5 Extension materials................................... 21 2.6 End of topic test..................................... 23 3 Gene expression 25 3.1 Introduction....................................... 27 3.2 The structure and functions of RNA.......................... 27 3.3 Transcription....................................... 30 3.4 Translation........................................ 34 3.5 One gene, many proteins................................ 39 3.6 Protein structure and function............................. 40 3.7 Learning points..................................... 41 3.8 Extended response question.............................. 43 3.9 End of topic test..................................... 43 4 Differentiation in multicellular organisms 47 4.1 Introduction....................................... 49 4.2 Meristems........................................ 49 4.3 Stem cells........................................ 54 4.4 Embryonic stem cells.................................. 55 4.5 Tissue (adult) stem cells................................ 56 4.6 Differentiation...................................... 57 4.7 Therapeutic use of stem cells............................. 57 4.8 Research involving stem cells............................. 58 4.9 Ethical issues regarding stem cells.......................... 58 4.10 Learning points..................................... 60 4.11 Extended response question.............................. 61 4.12 Extension materials................................... 61 4.13 End of topic test..................................... 63 vi CONTENTS 5 Structure of the genome 65 5.1 Introduction to the Genome.............................. 66 5.2 The genome....................................... 66 5.3 Learning points..................................... 68 5.4 Extension materials................................... 69 5.5 End of topic test..................................... 70 6 Mutations 71 6.1 Single gene mutations................................. 73 6.2 Chromosome structure mutations........................... 84 6.3 The importance of mutations and gene duplication.................. 87 6.4 Learning points..................................... 87 6.5 Extended response question.............................. 88 6.6 Extension materials................................... 88 6.7 End of topic test..................................... 89 7 Evolution 91 7.1 Evolution......................................... 93 7.2 Gene transfer...................................... 94 7.3 Selection......................................... 94 7.4 Speciation........................................ 99 7.5 Learning points..................................... 103 7.6 End of topic test..................................... 104 8 Genomics 107 8.1 Genomic sequencing.................................. 109 8.2 Phylogenetics...................................... 110 8.3 Comparative genomics................................. 112 8.4 Personal genomics................................... 113 8.5 Learning points..................................... 114 8.6 Extension materials................................... 115 8.7 End of topic test..................................... 118 9 End of unit test 119 Glossary 129 Answers to questions and activities 133 © H ERIOT-WATT U NIVERSITY 1 Topic 1 Structure and organisation of DNA Contents 1.1 The structure of DNA......................................... 3 1.2 The organisation of DNA in prokaryotes and eukaryotes..................... 7 1.3 Learning points............................................ 9 1.4 Extension materials.......................................... 10 1.5 End of topic test............................................ 11 Prerequisites You should already know that: DNA takes the form of a double-stranded helix; the two strands of DNA are held together by complementary base pairs; DNA contains the four bases A, T, G and C which make up the genetic code. 2 TOPIC 1. STRUCTURE AND ORGANISATION OF DNA Learning objective By the end of this topic, you should be able to: appreciate that DNA is found in all living organisms and is the chemical which carries hereditary information; understand that the sequence of bases within DNA is the genetic code; appreciate that the sum total of all the DNA bases is the genome; describe the structure of a nucleotide; describe the structure of DNA; describe how nucleotides combine to form a backbone and that base pairing holds the two strands together, forming a double helix; describe the base pairing rule; name the bonds which hold bases together; explain that the two strands of DNA lie anti-parallel to each other and are read in different directions; understand that DNA can be organised in a number of ways, either circular or linear; state that circular forms of DNA can be found in prokaryotes, mitochondria and chloroplasts; state that bacteria and yeast also contain smaller rings of DNA, called plasmids; state that the linear form of DNA found in eukaryotes is tightly packaged with associated proteins called histones. © H ERIOT-WATT U NIVERSITY TOPIC 1. STRUCTURE AND ORGANISATION OF DNA 3 1.1 The structure of DNA In 1953, the structure of the DNA molecule was explained for the first time by two scientists, James Watson and Francis Crick, at the Cavendish Laboratory in Cambridge. Double helix structure of DNA They found that a molecule of DNA consists of two strands of repeating units called nucleotides. Nucleotides are composed of phosphate, deoxyribose sugar and a base. A DNA nucleotide There are four bases in DNA: (A) adenine, (T) thymine, (G) guanine and (C) cytosine. Each nucleotide has a different base. Nucleotide bases © H ERIOT-WATT U NIVERSITY 4 TOPIC 1. STRUCTURE AND ORGANISATION OF DNA These bases follow the base pairing rules: A always pairs with T, and G always pairs with C. The bases are held together by hydrogen bonds. DNA nucleotide DNA nucleotide A pairs with T T pairs with A G pairs with C C pairs with G Base pairing rules The phosphate of one nucleotide is attached by a strong chemical bond to the deoxyribose sugar of the next. This forms the sugar-phosphate backbone of the DNA. DNA takes the form of a double helix. DNA is made up of two antiparallel strands. This means the strands run in opposite directions to each other. The carbon atoms on the deoxyribose sugar are numbered as shown below. Deoxyribose sugar with carbon atoms numbered Using this system, each end of the DNA can be labelled to show the antiparallel strands. The structure of DNA is shown as follows. © H ERIOT-WATT U NIVERSITY TOPIC 1. STRUCTURE AND ORGANISATION OF DNA 5 DNA structure The structure of DNA: Interactive 3D model Go online An activity that shows the double helix structure of DNA as an interactive 3D model is available in the online materials at this point. The following illustration gives an idea of what to expect. Double helix structure of DNA - molecular model © H ERIOT-WATT U NIVERSITY 6 TOPIC 1. STRUCTURE AND ORGANISATION OF DNA The structure of DNA: Questions Go online Q1: Complete the diagram using the words from the list. Word list: base, deoxyribose sugar, phosphate................................................ Q2: Complete the blanks using the base pairing rules. The nucleotide guanine pairs with. The nucleotide thymine pairs with. The nucleotide cytosine pairs with. The nucleotide adenine pairs with. Q3: What shape is the DNA molecule?............................................... Q4: What type of bonding holds two DNA strands together?............................................... Q5: Which three components make up a DNA nucleotide? © H ERIOT-WATT U NIVERSITY TOPIC 1. STRUCTURE AND ORGANISATION OF DNA 7 1.2 The organisation of DNA in prokaryotes and eukaryotes Prokaryotes are organisms which lack a true membrane-bound nucleus. Bacteria are an example of a prokaryote. Their DNA is found in the cytoplasm of the cell. Eukaryotes are organisms which have a membrane bound nucleus that stores their genetic material. Animals, plants and fungi are examples of eukaryotes. DNA is a double-stranded molecule that can either be circular or linear. Circular and linear DNA Prokaryotes have a large circular chromosome. They may also have smaller rings of DNA called plasmids. A bacterial (prokaryotic) cell Circular plasmids may also be found in yeast (a fungus), which is classified as a eukaryote. In eukaryotes, DNA is found tightly coiled into linear chromosomes. DNA is also found within mitochondria (mtDNA) where it forms circular chromosomes. It is sometimes described as the smallest chromosome and is inherited from the mother in humans. A eukaryotic cell © H ERIOT-WATT U NIVERSITY 8 TOPIC 1. STRUCTURE AND ORGANISATION OF DNA DNA can also be found within the chloroplasts of plant cells. It is usually larger than mitochondrial DNA and takes the form of circular chromosomes containing the genes involved in the photosynthetic process. Where circular DNA is found in eukaryotes, it is thought that it has been incorporated from early bacteria or prokaryotes. Typically, the DNA content of a single human cell, if completely unravelled, would measure around two metres in length. This DNA must be packaged so it can fit inside the nucleus. DNA found in the linear chromosomes of the nucleus of eukaryotes is tightly coiled and packaged with associated proteins called histones. This is shown in the diagram below. DNA packaging The organisation of DNA in prokaryotes and eukaryotes: Questions Go online Q6: In which type of organism is circular DNA mainly found?............................................... Q7: In which structures of eukaryotic cells might circular DNA be found?............................................... Q8: What is the term for sections of extra-chromosomal DNA sometimes found in organisms? © H ERIOT-WATT U NIVERSITY TOPIC 1. STRUCTURE AND ORGANISATION OF DNA 9 1.3 Learning points Summary DNA encodes hereditary information in a chemical language. All cells store their genetic information in the base sequence of DNA. The structure of a DNA nucleotide is composed of deoxyribose sugar, a phosphate and a base. Nucleotides bond to form a sugar-phosphate backbone. Base pairs (adenine, thymine, guanine and cytosine) hold the two strands together by hydrogen bonds, forming a double helix. Adenine always pairs with thymine, and guanine always pairs with cytosine. DNA is a double-stranded, antiparallel (one strand goes from 3 to 5 , the other from 5 to 3 ) structure with a deoxyribose and phosphate backbone held together by internal base pairs. The DNA molecule can be circular or linear. Circular chromosomal DNA and plasmids are found in prokaryotes. Circular plasmids are found in yeast. Circular chromosomes are in the mitochondria and chloroplasts of eukaryotes. The DNA found in the linear chromosomes of the nucleus of eukaryotes is tightly coiled and packaged with associated proteins called histones. The sum total of all genetic material is the genome. Chromosomes in eukaryotes are contained within a membrane-bound nucleus. © H ERIOT-WATT U NIVERSITY 10 TOPIC 1. STRUCTURE AND ORGANISATION OF DNA 1.4 Extension materials The material in this section is not examinable. It includes information which will widen your appreciation of this section of work. Extension materials: The Discovery of DNA The significance of DNA and its role in hereditary can be traced from the work of Griffiths, who in 1928 demonstrated the "transforming principle" in bacteria. He, and later others (Avery, McCartney & McLeod, 1944), would show this "transforming principle" to be DNA. Later, in the 1950s, Hershey & Chase, working with bacteriophage and radioactive forms of phosphorus and sulfur, would confirm DNA as the genetic material and eliminated protein as the carrier of genetic information. Edwin Chargaff, using paper chromatography and ultraviolet spectroscopy techniques, demonstrated two findings, now known as Chargaff's rules: firstly, that adenine and thymine always occur together, and similarly that cytosine and guanine pair up - this is called base pairing; secondly, that DNA sequences vary between species. In the early 1950s, work by Maurice Wilkins and Rosalind Franklin uncovered some characteristic features of the DNA molecule. Using a method called X-ray crystallography, it was shown that the molecule had a helical structure. Using this, and other evidence, Francis Crick and James Watson were able to construct the model of DNA that we recognise today. The main features of the model are that it not only shows the configuration of the molecule, but it allows for the explanation of two processes: the first is the mechanism for DNA replication (semi-conservative) and the second is how it codes for proteins. The story of the discovery of the structure of DNA, and an excellent book about how science works, is: The Double Helix by James D. Watson, published by Penguin Books. © H ERIOT-WATT U NIVERSITY TOPIC 1. STRUCTURE AND ORGANISATION OF DNA 11 1.5 End of topic test End of Topic 1 test Go online Q9: What is the term for the sum total of DNA in an organism?............................................... Q10: What is the complementary base sequence for the following DNA sequence: ATGGACTTTAGGT?............................................... Q11: A DNA molecule contains 1500 bases, of which 450 are adenine. What percentage of the bases in the DNA molecule are guanine?............................................... Q12: The following diagram shows part of a DNA molecule. Which of the following correctly identifies base, phosphate and sugar, in that order? 1. A, B, C 2. A, C, B 3. B, A, C 4. C, A, B © H ERIOT-WATT U NIVERSITY 12 TOPIC 1. STRUCTURE AND ORGANISATION OF DNA............................................... Q13: What is the name of the sugar that is found in DNA?............................................... Q14: Base pairs in a DNA molecule are linked by weak bonds. What is the name for these?............................................... Q15: What name is given to the backbone of DNA?............................................... Q16: The diagram below illustrates the structure of a DNA molecule. Fill in the missing labels to complete the diagram................................................ Q17: Name two sources of circular DNA in eukaryotes................................................ Q18: Name the substance which DNA is packaged with in the nucleus of eukaryotes. © H ERIOT-WATT U NIVERSITY 13 Topic 2 Replication of DNA Contents 2.1 DNA replication............................................ 15 2.2 The polymerase chain reaction (PCR)............................... 17 2.3 Learning points............................................ 20 2.4 Extended response question..................................... 21 2.5 Extension materials.......................................... 21 2.6 End of topic test............................................ 23 Prerequisites You should already know that: chromosomes (and therefore DNA) are replicated during mitosis; the two strands of DNA are held together by complementary base pairs; DNA contains the four bases A, T, G and C which make up the genetic code. 14 TOPIC 2. REPLICATION OF DNA Learning objective By the end of this topic, you should be able to: explain the function of DNA polymerase, when it acts, and what conditions are necessary for it to function; describe how DNA replicates in terms of DNA unwinding and DNA polymerase adding complementary nucleotides; explain the importance and significance of adding new nucleotides to the 3 end of an existing DNA chain; state that replication occurs at various points on a DNA molecule; outline the process where DNA polymerase can act continuously on the leading strand, but discontinuously on the lagging strand; describe the action and significance of the enzyme ligase; state the purpose of the polymerase chain reaction (PCR); describe the action and purpose of primers in PCR; describe the sequence of PCR in terms of heating DNA, adding primers, and cooling DNA; explain why heat tolerant DNA polymerase is required; explain the significance and outcome of 'cycling' sequences; describe the practical applications of PCR. © H ERIOT-WATT U NIVERSITY TOPIC 2. REPLICATION OF DNA 15 2.1 DNA replication DNA replication takes place prior to cell division. The replication of DNA is semi-conservative. Each strand acts as a template for the synthesis of a new DNA molecule by the addition of complementary base pairs, thereby generating a new DNA strand that is the complementary sequence to the parental DNA. Each daughter DNA molecule ends up with one of the original strands and one newly synthesised strand. DNA replication DNA replication is an enzyme controlled process which relies on the activities of DNA polymerase and DNA ligase. DNA polymerase: adds DNA nucleotides, using complementary base pairing, to the deoxyribose (3 ) end of the new DNA strand which is forming. DNA ligase: joins fragments of DNA together. Before replication begins, there must be a pool of free nucleotides present; however, DNA polymerase cannot start adding nucleotides on its own. Short sections of RNA nucleotides called primers are added to the DNA and the enzyme extends from them. A primer is a short strand of nucleotides which binds to the 3 end of the template DNA strand allowing polymerase to add DNA nucleotides. Due to the action of the enzyme DNA polymerase, the two strands of DNA are copied differently. The leading strand is made continuously while the lagging strand is made in fragments, which are then joined together. © H ERIOT-WATT U NIVERSITY 16 TOPIC 2. REPLICATION OF DNA DNA replication: Steps Go online The following provides a summary of the steps involved in DNA replication. DNA is unwound and hydrogen bonds between bases are broken to form two template strands. two replication forks form and open the double-strand in opposite directions, exposing the bases. on the leading strand, a primer binds to the DNA and DNA polymerase adds nucleotides to the 3 end. DNA polymerase catalyses the formation of a chemical bond between nucleotides and continues to add nucleotides to the 3 end of the growing strand. on the lagging strand, a primer binds to the DNA once it is exposed and DNA polymerase adds nucleotides to the 3 end. As more DNA is exposed, a new primer is added. DNA polymerase extends the new strand from this primer until it meets the previous fragment. The old primer is replaced by DNA and the enzyme DNA ligase joins the fragments together. As the DNA unzips further, another fragment will be made and connected to the previous one. © H ERIOT-WATT U NIVERSITY TOPIC 2. REPLICATION OF DNA 17 2.2 The polymerase chain reaction (PCR) PCR can be used to amplify a desired DNA sequence of any origin (virus, bacteria, plant or animal) millions of times in a matter of hours. It is especially useful because: it is highly specific; it is easily automated; it is capable of amplifying minute amounts of sample. To amplify a target DNA sequence, several components are required: buffer; nucleotides; primers; Taq polymerase; template DNA. The section of DNA which is to be amplified must be added to the reaction mixture. It acts as a template to copy from. The buffer keeps the reaction mixture at the correct pH to ensure the reaction will proceed. Polymerase enzyme is found in all animals. It has an optimum temperature of 37 ◦ C. PCR requires polymerase to operate at high temperatures. Thermus aquaticus is a bacterium that lives in hot springs and hydrothermal vents. Taq polymerase, an enzyme which adds nucleotides to DNA, was first isolated from this bacteria. It is special type of polymerase which is stable at high temperatures, having an optimum temperature of 70 ◦ C. Polymerase can only add nucleotides to an existing strand of DNA, therefore, PCR requires primers. In PCR, primers are short strands of nucleotides which are complementary to specific target sequences at the two ends of the region of DNA to be amplified. The PCR process involves repeated cycles of the following steps: the DNA molecule which is to be amplified is denatured by heating to between 92 and 98 ◦ C, breaking the hydrogen bonds between base pairs to separate the strands; the solution is cooled to between 50 and 65 ◦ C to allow the primers to bind to target sequences; the solution is heated to between 70 and 80 ◦ C for heat-tolerant DNA polymerase to replicate the region of DNA. This cycle is usually repeated at least 30 times. Polymerase chain reaction allows DNA to be amplified in vitro; this means it is performed outwith a living organism. The opposite, in vivo, means carried out within an organism. © H ERIOT-WATT U NIVERSITY 18 TOPIC 2. REPLICATION OF DNA The polymerase chain reaction (PCR): Stages Go online The following provides a summary of the stages of PCR. © H ERIOT-WATT U NIVERSITY TOPIC 2. REPLICATION OF DNA 19 PCR has revolutionised many areas of research science. PCR is involved in the process of DNA sequencing and has allowed amplification of DNA from ancient sources, such as Neanderthal bones, enabling in-depth DNA analysis. PCR also has forensic applications, allowing minute quantities of DNA from a crime scene to be amplified, sequenced, and compared to DNA sequences from suspects. PCR has medical applications, for example the diagnosis of human immunodeficiency virus (HIV). Finally, PCR can be used to settle paternity disputes by amplifying and comparing a child's DNA to their potential father. © H ERIOT-WATT U NIVERSITY 20 TOPIC 2. REPLICATION OF DNA 2.3 Learning points Summary Prior to cell division, DNA polymerase replicates a DNA strand precisely using DNA nucleotides. DNA polymerase needs a primer to start replication. A primer is a short strand of nucleotides which binds to the 3 end of the template DNA strand allowing polymerase to add DNA nucleotides. DNA unwinds to form two template strands. DNA polymerase adds DNA nucleotides, using complementary base pairing, to the deoxyribose (3 ) end of the new DNA strand which is forming. This process occurs at several locations on a DNA molecule. DNA polymerase can only add DNA nucleotides in one direction resulting in the leading strand being replicated continuously and the lagging strand replicated in fragments. Fragments of DNA are joined together by ligase. The polymerase chain reaction (PCR) is a technique for the amplification of DNA in vitro. In PCR, primers are short strands of nucleotides which are complementary to specific target sequences at the two ends of the region of DNA to be amplified. PCR is a three step process: ◦ heating (92-98 ◦ C) separates the DNA strands; ◦ annealing (50-65 ◦ C) is the binding of the primers to target sequences; ◦ extension (70-80◦ C) of the primers to complete the complementary strand is carried out by heat stable DNA polymerase. Repeated cycles of heating and cooling amplify this region of DNA. PCR can amplify DNA to help solve crimes, settle paternity suits and diagnose genetic disorders. © H ERIOT-WATT U NIVERSITY TOPIC 2. REPLICATION OF DNA 21 2.4 Extended response question The activity which follows presents an extended response question similar to the style that you will encounter in the examination. You should have a good understanding of DNA structure and replication before attempting the question. You should give your completed answer to your teacher or tutor for marking, or try to mark it yourself using the suggested marking scheme. Extended response question: DNA structure and replication Give an account of DNA structure and replication. (8 marks) 2.5 Extension materials The material in this section is not examinable. It includes information which will widen your appreciation of this section of work. Extension materials: Meselson and Stahl's experiment The mechanism of DNA replication is said to be semi-conservative. That is, after replication, each of the two resulting DNA molecules is composed of one original (or conserved) strand and one new strand. This hypothesis, put forward by Watson and Crick, was proved experimentally by Meselson and Stahl in the late 1950s. Their experiment involved growing a culture of the bacterium Escherichia coli in a growth medium containing heavy nitrogen ( 15 N). As the bacteria grew, they incorporated the heavy nitrogen into their nitrogenous bases. The bacteria were then inoculated into growth media containing light nitrogen ( 14 N) and three classes of DNA were subsequently extracted after the change to the light nitrogen medium. The three classes of DNA were: parental DNA; first generation DNA; second generation DNA. The results of the experiment showed that parental DNA grown in heavy medium was 'heavier' than when grown in light medium. First generation growth showed that the DNA was all of medium density. Lastly, the second generation showed DNA of both medium and light intensities. Second generation growth supported the semi-conservative model of DNA replication since there were two bands of growth (one with both conserved and new DNA, and a band of light DNA). © H ERIOT-WATT U NIVERSITY 22 TOPIC 2. REPLICATION OF DNA Meselson and Stahl's experiment © H ERIOT-WATT U NIVERSITY TOPIC 2. REPLICATION OF DNA 23 2.6 End of topic test End of Topic 2 test Go online Q1: Name the enzyme that is required to add nucleotides to a growing strand of DNA................................................ Q2: What name is given to the short strand of nucleotides added to DNA at the start of the replication process?............................................... Q3: Name the enzyme that is required to join fragments of DNA together................................................ Q4: The following four stages occur during DNA replication: a) Base pairing occurs between free nucleotides and each of the DNA strands. b) The hydrogen bonds between DNA strands break. c) The DNA molecules coil up to form double helices. d) Nucleotides are bonded together by DNA polymerase. Which of the following gives these stages in the correct order? 1. a, b, d, c 2. b, a, d, c 3. c, b, a, d 4. d, a, b, c............................................... Q5: The DNA polymerase used in the polymerase chain reaction possesses a particular characteristic that makes it ideally suited to the purpose. What is it? a) The enzyme is relatively stable at high temperatures. b) The enzyme synthesises DNA at a very rapid rate. c) The enzyme is very accurate at copying DNA from a template. d) The enzyme can seal together fragments of DNA................................................ © H ERIOT-WATT U NIVERSITY 24 TOPIC 2. REPLICATION OF DNA Q6: The steps involved in the polymerase chain reaction are given below. a) Temperature of the reaction adjusted to 70-80 ◦ C. b) Temperature of the reaction adjusted to 50-65 ◦ C. c) The DNA strands separate. d) Temperature of the reaction adjusted to 92-98 ◦ C. e) Synthesis of DNA by the enzyme DNA polymerase. f) Annealing of the primers to the single-stranded DNA. Which of the following describes the correct order in which the steps would occur? 1. d, c, b, e, a, f 2. d, c, a, f, b, e 3. d, c, b, f, a, e 4. d, c, a, e, b, f............................................... Q7: occurs at 70-80◦ C. Word list: extension, annealing, heating, exponential................................................ Q8: PCR leads to an amplification of desired DNA sequences. Word list: extension, annealing, heating, exponential................................................ Q9: Starting with a single molecule of DNA, the polymerase chain reaction was allowed to go through three complete cycles. How many molecules of DNA would be produced? a) 4 b) 8 c) 16 d) 32 © H ERIOT-WATT U NIVERSITY 25 Topic 3 Gene expression Contents 3.1 Introduction.............................................. 27 3.2 The structure and functions of RNA................................. 27 3.3 Transcription.............................................. 30 3.4 Translation............................................... 34 3.5 One gene, many proteins...................................... 39 3.6 Protein structure and function.................................... 40 3.7 Learning points............................................ 41 3.8 Extended response question..................................... 43 3.9 End of topic test............................................ 43 Prerequisites You should already know that: DNA carries the genetic information for making proteins; the base sequence of DNA determines the amino acid sequence in proteins; messenger RNA (mRNA) is a molecule which carries a copy of the code from the DNA in the nucleus to a ribosome in the cytoplasm; at the ribosome, proteins are assembled from amino acids. 26 TOPIC 3. GENE EXPRESSION Learning objective By the end of this topic, you should be able to: explain that the genetic code is universal to all forms of life; explain that the phenotype is a combination of the genotype and environmental factors; describe the structure of RNA; describe the function of mRNA, tRNA and rRNA; describe the process of transcription including the role of RNA polymerase and complementary base pairing; describe the process of RNA splicing; describe the process of translation, including the role of tRNA and ribosomes; state that codons are found on mRNA and anticodons are found on tRNA; name the bonds which hold amino acids together in a protein; explain how different proteins can be expressed from one gene; state that polypeptide chains fold to give the final structure of the protein; describe the role of hydrogen bonds and the interactions between amino acids in the 3D shape of a protein. © H ERIOT-WATT U NIVERSITY TOPIC 3. GENE EXPRESSION 27 3.1 Introduction The many thousands of proteins that our cells use are synthesised inside the cells during a complex process involving DNA and RNA (the nucleic acids), as well as ribosomes. You will remember from previous studies that a protein is composed of amino acids joined together in a specific sequence. The information to determine the sequence of amino acids in a protein is contained in the DNA in the nucleus of our cells. In this topic, we will study how the instructions for making a protein are transferred into the cytoplasm using RNA, and how a protein is actually constructed on the ribosomes. Gene expression involves two major stages. The first process is transcription, during which the DNA is used to produce an RNA molecule that is called a primary transcript. This RNA has the same sequence as the gene. Human genes can be divided into exons and introns, but it is only the exons that carry the information needed for protein synthesis. The next stage of gene expression is known as translation, which allows amino acids to come together in a certain order at the ribosome, where they form a polypeptide chain. DNA makes mRNA makes protein Process: transcription translation Occurs in: the nucleus cytoplasm Protein synthesis summary 3.2 The structure and functions of RNA Ribonucleic acid (RNA) provides a bridge between DNA and protein synthesis. RNA consists of nucleotides that are composed of phosphate, ribose sugar and a base. An RNA nucleotide © H ERIOT-WATT U NIVERSITY 28 TOPIC 3. GENE EXPRESSION Important points to remember about RNA structure are that: RNA nucleotides contain the sugar ribose; RNA has the base (U) uracil rather than (T) thymine (as found in DNA); RNA molecules are usually single-stranded. There are three main types of RNA involved in protein synthesis: 1. mRNA (messenger RNA), which carries a copy of the DNA code from the nucleus to the ribosome; 2. tRNA (transfer RNA), which are molecules found in the cytoplasm that become attached to specific amino acids, bringing them to the ribosomes where they are joined together; 3. rRNA (ribosomal RNA), which forms a complex with protein molecules to make the ribosome. For the synthesis of a protein, the particular sequence of bases on the DNA is first transcribed into the complementary sequence of mRNA. This messenger RNA can then carry the Messenger RNA (mRNA) information for a protein through the nuclear envelope to the sites of protein synthesis (the ribosomes). Each triplet of bases on the mRNA molecule is called a codon and codes for a specific amino acid. This type of RNA is responsible for the transport and transfer of individual amino acids during protein synthesis. Amino acids are transported by specific tRNA molecules, which Transfer RNA (tRNA) recognise the genetic code presented by the mRNA. The three bases exposed at the bottom form the anticodon. This is the complementary base sequence to the base sequence on mRNA coding for a particular amino acid. This type of RNA is bound to structural proteins to form a Ribosomal RNA (rRNA) ribosome. The ribosome is used in the synthesis of proteins. The three forms of RNA © H ERIOT-WATT U NIVERSITY TOPIC 3. GENE EXPRESSION 29 The structure and functions of RNA: Questions Go online Q1: Complete the diagram using the labels from the list. Labels: base, phosphate, ribose sugar................................................ Q2: Complete the table using the properties of RNA and DNA from the list. RNA DNA Structure Preferred form Number of types Present in Bases Properties: >1; 1; adenine, cytosine, guanine and thymine; adenine, cytosine, guanine and uracil; double-stranded; double helix; not a double helix; single-stranded; the cytoplasm and the nucleus; the nucleus. Q3: What are the components of an RNA nucleotide?............................................... Q4: List the four types of bases found in an RNA molecule................................................ Q5: Name the three types of RNA found in the cell................................................ © H ERIOT-WATT U NIVERSITY 30 TOPIC 3. GENE EXPRESSION Q6: What are main functional differences between mRNA and tRNA?............................................... Q7: Nucleotides are the building blocks of: a) DNA only b) RNA only c) both DNA and RNA d) neither DNA nor RNA 3.3 Transcription Transcription is the first step in protein synthesis. Information from DNA is copied into an RNA molecule, a process which takes place in the nucleus. The RNA polymerase enzyme moves along the DNA, unwinding the double helix and breaking the hydrogen bonds that hold the base pairs together. Free RNA nucleotides bond with the complementary base pairs on the DNA. The base pairing rules are summarised in the table below. The RNA nucleotides are held in place by hydrogen bonds while strong bonds form between the phosphate of one nucleotide and the ribose sugar of the adjacent nucleotide. When transcription is complete, the RNA polymerase enzyme and the mRNA strand that has been constructed are released. The mRNA that has been produced at this stage is known as the primary transcript. DNA nucleotide RNA nucleotide A pairs with U T pairs with A G pairs with C C pairs with G RNA base pairing rules © H ERIOT-WATT U NIVERSITY TOPIC 3. GENE EXPRESSION 31 An overview of the transcription process © H ERIOT-WATT U NIVERSITY 32 TOPIC 3. GENE EXPRESSION The steps involved in RNA transcription © H ERIOT-WATT U NIVERSITY TOPIC 3. GENE EXPRESSION 33 After a eukaryotic cell transcribes a protein coding gene, the RNA transcript, called the primary transcript, is processed. One type of processing is called RNA splicing. This takes place in the nucleus, after which the mature mRNA is released into the cytoplasm where ribosomes translate the mRNA transcript into protein. RNA splicing forms a mature mRNA transcript. The introns of the primary transcript are non-coding regions and are removed. The exons are coding regions and are joined together to form the mature transcript. The order of the exons is unchanged during splicing. RNA splicing: Steps Go online The following provides a summary of the steps involved in RNA splicing. © H ERIOT-WATT U NIVERSITY 34 TOPIC 3. GENE EXPRESSION Transcription: Questions Go online Q8: Name the enzyme which is responsible for producing mRNA................................................ Q9: Give the location of transcription................................................ Q10: Name the bonds between DNA bases which must be broken during transcription................................................ Q11: What name is given to the mRNA molecule which has just been released from the DNA?............................................... Q12: Name the process which removes introns from the primary transcript. 3.4 Translation Once the DNA in a gene has been transcribed into mRNA, translation can take place. First, the mRNA molecules pass through the nuclear pores. Translation of mRNA into protein takes place on ribosomes in the cytoplasm and requires a second type of RNA, transfer RNA (tRNA). tRNA Go online The following provides structural information about a tRNA molecule, in this case with the amino acid serine attached to it. Amino acids are attached to tRNA at the amino acid attachment site at the top of the molecule. Unlike mRNA, there are some regions of base pairing in a tRNA molecule. The three bases exposed at the base of the tRNA molecule is called the anticodon. (Each group of three bases on the mRNA which codes for an amino acid is called a codon). © H ERIOT-WATT U NIVERSITY TOPIC 3. GENE EXPRESSION 35 As an mRNA molecule passes through a ribosome, each codon is translated into an amino acid. The genetic code table indicates which amino acid corresponds to each mRNA codon. The tRNA molecule carrying the complementary anticodon binds briefly to the mRNA codon. The amino acid attached to the tRNA is then added to the polypeptide chain being synthesised. Amino acids are joined together by strong peptide bonds. After the amino acid has been added to a polypeptide chain during translation, the tRNA is free to pick up another amino acid in the cytoplasm. Genetic code table When the polypeptide chain is completed, it is released from the ribosome. Further processing, such as folding and binding to other polypeptide chains, results in the formation of a mature protein. The mRNA molecule is usually reused to produce more identical polypeptide chains. There are three codons that do not code for amino acids: UGA, UAA and UAG. These codons are known as stop codons and signal where translation ends. The genetic code also includes start codons where translation begins. In eukaryotes this is almost always AUG, which also codes for the amino acid methionine. © H ERIOT-WATT U NIVERSITY 36 TOPIC 3. GENE EXPRESSION Translation: Steps Go online © H ERIOT-WATT U NIVERSITY TOPIC 3. GENE EXPRESSION 37 The steps involved in translation Translation: Questions Go online The following steps describe the role of messenger RNA in the cell, providing a summary of protein synthesis transcription to translation. Q13: The following steps describe the role of messenger RNA in the cell, providing a summary of protein synthesis from transcription to translation. List the steps in the correct order. Hydrogen bonds are formed between the first codon of the mRNA and the complementary anticodon on a tRNA Hydrogen bonds form between the bases on the RNA and the DNA nucleotides The second tRNA binds to the mRNA The first tRNA leaves the ribosome, and another tRNA enters and base-pairs with the mRNA The double-stranded DNA unwinds, hydrogen bonds in the DNA break and the DNA strands separate A second peptide bond is then formed. The process continues, with the ribosome moving along the mRNA The mRNA leaves the nucleus and enters the cytoplasm A peptide bond forms between the amino acids carried by the tRNA molecules A ribosome attaches to the mRNA. Two transfer RNA (tRNA) molecules are also contained within the ribosome As each mRNA codon is exposed, incoming tRNA pairs with it and polypeptide synthesis continues until completed When synthesis of the mRNA is completed, the mRNA separates from the DNA An RNA nucleotide binds to a complementary nucleotide on one of the DNA strands The RNA nucleotides are linked together to form messenger RNA (mRNA)............................................... © H ERIOT-WATT U NIVERSITY 38 TOPIC 3. GENE EXPRESSION Q14: What name is given to the three bases exposed at the bottom of a tRNA molecule?............................................... Q15: Where does translation take place?............................................... Q16: Which cellular organelle is required for translation?............................................... Q17: Name the bond which holds amino acids together in a protein................................................ Q18: Three mRNA codons do not code for amino acids. What is their role in translation? Q19: Complete the sequence of bases encoded in the mRNA and then determine the sequence of amino acids in the protein. Use the genetic code table below to help. DNA C A C A G T G T T T G T C C G mRNA protein © H ERIOT-WATT U NIVERSITY TOPIC 3. GENE EXPRESSION 39 3.5 One gene, many proteins Until quite recently, there was a theory that stated "one gene, one protein". This, however, has been superseded. With the completion of the Human Genome Project it is now accepted that the human genome contains between 20,000 and 25,000 genes, and yet it is also accepted that there are in excess of one million proteins in humans. Clearly there must be some mechanism that allows the genes to be expressed in a variety of ways. One gene, many proteins The mRNA (sometimes called pre-mRNA) can be edited in different ways by assembling a different sequence of exons for translation. As a result, many different mature transcripts of mRNA can be derived from one section of DNA. This process is known as alternative splicing. As a result of alternative splicing, different mature mRNA transcripts are produced from the same primary transcript depending on which exons are retained. Alternative splicing © H ERIOT-WATT U NIVERSITY 40 TOPIC 3. GENE EXPRESSION 3.6 Protein structure and function Proteins do not simply exist and function as strings of amino acids. After translation, the protein is folded to produce its final 3D shape. The folded polypeptide chains of a protein are held in place by hydrogen bonds and other interactions between individual amino acids. Description Diagram Chain of amino acids linked by strong peptide bonds Polypeptide structure determined by weak hydrogen bonds Strong bonds form between special groups of amino acids More than one polypeptide makes up the final structure Protein structure Different types of proteins have a variety of different functions within a living organism, examples of which are given in the following table. Type of protein Example Structural collagen, elastin Contractile actin and myosin in muscle cells Hormones insulin insulin receptor in liver cells (forming part of the structure of the Receptors plasma membrane) Transport proteins transporter of glucose into the cell Defence proteins immunoglobulins Enzymes lipase, pepsin, maltase Protein functions © H ERIOT-WATT U NIVERSITY TOPIC 3. GENE EXPRESSION 41 Protein structure and function: Questions Go online Q20: Complete the table by selecting from the listed images. Description Diagram Chain of amino acids linked by strong peptide bonds Polypeptide structure determined by weak hydrogen bonds Strong bonds form between special groups of amino acids More than one polypeptide makes up the final structure a) b) c) d) 3.7 Learning points Summary Gene expression involves the transcription and translation of DNA sequences. Only a fraction of the genes in a cell are expressed. RNA is single stranded and is composed of nucleotides containing ribose sugar, phosphate and one of four bases: cytosine, guanine, adenine and uracil. Transcription and translation involves three types of RNA (mRNA, tRNA and rRNA). Messenger RNA (mRNA) carries a copy of the DNA code from the nucleus to the ribosome. mRNA is transcribed from DNA in the nucleus and translated into proteins by ribosomes in the cytoplasm. Each triplet of bases on the mRNA molecule is called a codon and codes for a specific amino acid. © H ERIOT-WATT U NIVERSITY 42 TOPIC 3. GENE EXPRESSION Summary continued Transfer RNA (tRNA) folds due to complementary base pairing. Each tRNA molecule carries its specific amino acid to the ribosome. A tRNA molecule has an anticodon (an exposed triplet of bases) at one end and an attachment site for a specific amino acid at the other end. Ribosomal RNA (rRNA) and proteins form the ribosome. RNA polymerase moves along DNA unwinding the double helix and breaking the hydrogen bonds between the bases. RNA polymerase synthesises a primary transcript of mRNA from RNA nucleotides by complementary base pairing. Uracil in RNA is complementary to adenine. RNA splicing forms a mature mRNA transcript. The introns of the primary transcript are non-coding regions and are removed. The exons are coding regions and are joined together to form the mature transcript. The order of the exons is unchanged during splicing. tRNA is involved in the translation of mRNA into a polypeptide at a ribosome. Translation begins at a start codon and ends at a stop codon. Anticodons bond to codons by complementary base pairing, translating the genetic code into a sequence of amino acids. Peptide bonds join the amino acids together. Each tRNA then leaves the ribosome as the polypeptide is formed. Different proteins can be expressed from one gene, as a result of alternative RNA splicing. Different mature mRNA transcripts are produced from the same primary transcript depending on which exons are retained. Amino acids are linked by peptide bonds to form polypeptides. Polypeptide chains fold to form the three-dimensional shape of a protein, held together by hydrogen bonds and other interactions between individual amino acids. Proteins have a large variety of shapes which determines their functions. Phenotype is determined by the proteins produced as the result of gene expression. Environmental factors also influence phenotype. © H ERIOT-WATT U NIVERSITY TOPIC 3. GENE EXPRESSION 43 3.8 Extended response question The activity which follows presents an extended response question similar to the style that you will encounter in the examination. You should have a good understanding of protein synthesis before attempting the question. You should give your completed answer to your teacher or tutor for marking, or try to mark it yourself using the suggested marking scheme. Extended response question: Protein synthesis Describe the role of RNA in protein synthesis. (7 marks) 3.9 End of topic test End of Topic 3 test Go online Q21: The diagram shows two nucleotides that form part of a messenger RNA molecule. Which of the following correctly identifies phosphate, sugar and base, in order? 1. A, B and C 2. B, A and C 3. A, C and B 4. B, C and A............................................... Q22: Which of the following correctly describes the structure of an RNA molecule? a) It is single-stranded; it has deoxyribose in its backbone; it contains the base thymine. b) It is double-stranded; it has ribose in its backbone; it contains the base uracil. c) It is single-stranded; it has ribose in its backbone; it contains the base uracil. d) It is double-stranded; it has deoxyribose in its backbone; it contains the base thymine. © H ERIOT-WATT U NIVERSITY 44 TOPIC 3. GENE EXPRESSION............................................... Q23: a) What is the messenger RNA (mRNA) sequence encoded by the following DNA sequence: GCATTCATTGCA? b) What name is given to the above process? c) Name the enzyme required to carry out the above process................................................ Q24: Complete the following sentence by selecting the correct options from the word list. The mRNA produced after transcription is called the ; the are removed, leaving only the in the final. Word list: exons, introns, mature transcript, primary transcript................................................ Q25: What name is given to the process which removes introns from mRNA?............................................... Q26: What name is given to the process by which ribosomes use messenger RNA to produce a polypeptide chain?............................................... Q27: The diagram below shows six transfer RNA (tRNA) molecules, each of which carries a different amino acid (indicated by the letters G, A, S, L, C and V). Part of a messenger RNA (mRNA) molecule is also shown. What is the sequence of amino acids encoded by the mRNA molecule? © H ERIOT-WATT U NIVERSITY TOPIC 3. GENE EXPRESSION 45............................................... Q28: What name is given to the site indicated by letter X in the diagram below? x............................................... Q29: Name the process which allows different mature mRNA transcripts to be produced from the same primary transcript................................................ Q30: The following diagram shows part of a peptide chain from a protein. What type of bond is indicated by the arrow? © H ERIOT-WATT U NIVERSITY 47 Topic 4 Differentiation in multicellular organisms Contents 4.1 Introduction.............................................. 49 4.2 Meristems.............................................. 49 4.3 Stem cells.............................................. 54 4.4 Embryonic stem cells........................................ 55 4.5 Tissue (adult) stem cells...................................... 56 4.6 Differentiation............................................ 57 4.7 Therapeutic use of stem cells.................................... 57 4.8 Research involving stem cells................................... 58 4.9 Ethical issues regarding stem cells................................. 58 4.10 Learning points........................................... 60 4.11 Extended response question.................................... 61 4.12 Extension materials......................................... 61 4.13 End of topic test........................................... 63 Prerequisites You should already know that: stem cells have the potential to become different types of cell; stem cells are involved in growth and repair. 48 TOPIC 4. DIFFERENTIATION IN MULTICELLULAR ORGANISMS Learning objective By the end of this topic, you should be able to: understand and explain the term differentiation; describe the role of plant meristems and how they generate new tissue during plant growth; describe the role of animal stem cells; describe some therapeutic uses of stem cells; describe the importance of stem cell research and take into account the ethical issues. © H ERIOT-WATT U NIVERSITY TOPIC 4. DIFFERENTIATION IN MULTICELLULAR ORGANISMS 49 4.1 Introduction All living things are characterised by levels of organisation that are hierarchical. The cell is the lowest level of organisation that can exist independently. Multicellular organisms have cells organised into groups of cells called tissues, the next level of organisation. Tissues are formed from specialised cells that carry out a particular function. The columnar cells in the lining of the intestine, for example, are specialised for absorption of nutrients. Tissues can themselves become grouped together to form an organ. Most organs, such as the heart, lungs and liver, are also specialised for a certain function. The final level of organisation is the organ system where a group of organs work together at a particular function. Examples are the nervous system, the endocrine system and the vascular system. Although the vast majority of cells contain identical genomes, cells within the same organism differ from one another because they express different genes and make different proteins. Differentiation is the process by which cells or tissues undergo a change towards a more specialised function. 4.2 Meristems Growth is restricted to specific regions (the meristems) of a plant, but it can occur throughout the plant's lifetime. In animals, growth can occur throughout the animal's body, but it stops when the animal reaches maturity. Animals do not have meristems; they are exclusive to plants. Meristems are regions of unspecialised cells in plants that can divide (self-renew) and/or differentiate. These cells divide rapidly by mitosis to differentiate and form new plant tissues. Apical meristems are located at the tips of the roots and shoots of a plant. The name is derived from the position at the tip, which is also known as the apex. They contain a cluster of actively dividing cells that increase the length of the plant. Therefore, in order for a plant to increase in length, it has to produce new cells at the apical meristems. © H ERIOT-WATT U NIVERSITY 50 TOPIC 4. DIFFERENTIATION IN MULTICELLULAR ORGANISMS Differentiation: Phloem and xylem vessels Go online The following illustrates how cells differentiate to form phloem and xylem vessels. Differentiation within phloem vessels © H ERIOT-WATT U NIVERSITY TOPIC 4. DIFFERENTIATION IN MULTICELLULAR ORGANISMS 51 Differentiation within xylem vessels © H ERIOT-WATT U NIVERSITY 52 TOPIC 4. DIFFERENTIATION IN MULTICELLULAR ORGANISMS Shoot growth: Zones Go online The following provides a summary of the zones involved in shoot growth. © H ERIOT-WATT U NIVERSITY TOPIC 4. DIFFERENTIATION IN MULTICELLULAR ORGANISMS 53 Root growth: Zones Go online The following provides a summary of the zones involved in root growth. © H ERIOT-WATT U NIVERSITY 54 TOPIC 4. DIFFERENTIATION IN MULTICELLULAR ORGANISMS 4.3 Stem cells Stem cells are unspecialised cells. When a stem cell divides, each new cell has the potential to remain a stem cell. This process is called self-renewal. In addition to self-renewal, stem cells can differentiate to become another type of cell with a more specialised function, such as a muscle cell, a red blood cell or a brain cell. Stem cells Stem cells are different from other body cells as they have the following characteristics: 1. Undifferentiated (unspecialised cell type), allowing them to divide and maintain a supply of stem cells for the body. 2. Found in all multicellular organisms. 3. Self-renewing and can differentiate; in some organs like the gut, stem cells regularly divide to repair and replace worn out or damaged tissues. The two types of stem cells found in humans are: 1. Embryonic stem cells. 2. Tissue (adult) stem cells. Stem cells: Questions Go online Q1: What do you understand by self-renewal in stem cells?............................................... Q2: What is the unique property of stem cells which makes them different from a specialised cell?............................................... Q3: Can you give examples of differentiated cells and their functions? © H ERIOT-WATT U NIVERSITY TOPIC 4. DIFFERENTIATION IN MULTICELLULAR ORGANISMS 55 4.4 Embryonic stem cells Most embryonic stem cells are derived from embryos that develop from eggs that have been fertilised, but before preimplantation. Stages in the early development of the human embryo After the process of fertilisation, the zygote undergoes rapid cell division (stage A) and produces a multicellular ball called a blastula. The blastula contains a fluid-filled cavity called the blastocoel (stage B). In humans, the blastula becomes implanted in the uterus, and the cells of the inner cell mass begin to differentiate (stage C). Embryonic stem cells are derived from embryos at the blastocyst stage. All the genes in embryonic stem cells can be switched on so these cells can differentiate into any type of cell. These cells are described as pluripotent as they can differentiate into all the cell types that make up the organism. Human embryonic stem cells can be formed by transferring cells from a preimplantation-stage embryo into a plastic laboratory culture dish that contains a nutrient broth, known as culture medium. In more recent research into embryonic stem cells, scientists have reliably directed the differentiation of embryonic stem cells into specific cell types by using different culture conditions. They are able to use the resulting, differentiated cells to treat certain diseases. © H ERIOT-WATT U NIVERSITY 56 TOPIC 4. DIFFERENTIATION IN MULTICELLULAR ORGANISMS Embryonic stem cells: Question Go online Q4: Put the following stages of the process of using human embryonic stem cells (hESCs) to form specialised cells into the correct order. Formation of mass of undifferentiated stem cells Stem cell cultured in the laboratory Embryo stem cell removed Undifferentiated stem cells cultured in different culture conditions Formation of specialised cells: nerve cell, muscle cell, gut cells Early human embryo Blastocyst 4.5 Tissue (adult) stem cells Early work on tissue stem cells started in the 1950s. Researchers discovered that bone marrow contains at least two kinds of stem cells. These were haematopoietic stem cells, which form all of the types of blood cells in the body, and bone marrow stem cells, which can generate bone, cartilage and fat cells that support the formation of blood and fibrous connective tissue. Tissue stem cells are involved in the growth, repair and renewal of the cells found in that tissue. They are multipotent. Tissue stem cells are multipotent as they can differentiate into all of the types of cell found in a particular tissue type. For example, blood stem cells located in bone marrow can give rise to all types of blood cell. Tissue stem cells have been identified in many organs and tissues, including brain, bone marrow, peripheral blood, blood vessels, skeletal muscle and skin. It is also worth knowing that these stem cells are also found in foetuses and babies. Although found in many types of tissues, only a very small number of stem cells actually occur in these tissues. © H ERIOT-WATT U NIVERSITY TOPIC 4. DIFFERENTIATION IN MULTICELLULAR ORGANISMS 57 4.6 Differentiation Cellular differentiation is the process by which a cell develops more specialised functions by expressing the genes characteristic for that type of cell. For example, a white blood cell only expresses genes which relate to its function, such as those which produce antibodies. Cellular differentiation is the result of gene expression, which is under the influence of many factors. Not all cells complete the process of differentiation; some cells pause at the stage where they can still undergo duplication. This allows them to generate replacement cells which may die or be damaged - it allows for growth and repair. These cells are known as tissue (or adult) stem cells. Differentiation The original cell has the power to differentiate into several different varieties of cell. Cells differentiate gradually over several rounds of division until the final differentiated cells can no longer reproduce themselves. 4.7 Therapeutic use of stem cells Some types of stem cells have been used in medicine for a number of years to repair damaged or diseased organs. Some examples are listed below: Skin: a rich source of tissue stem cells. Patients with serious burns can be treated using a technique which grows new skin in the lab from skin stem cells. Blood: a type of stem cell found in the bone marrow is capable of giving rise to all of the different types of blood cells. Bone marrow transplants have been carried out for many years as a treatment for diseases such as leukaemia and other blood disorders. Cornea: corneal stem cells are removed and cultured in a laboratory. They are then transplanted onto the diseased cornea to repair it. © H ERIOT-WATT U NIVERSITY 58 TOPIC 4. DIFFERENTIATION IN MULTICELLULAR ORGANISMS 4.8 Research involving stem cells Almost all animals contain a very small population of cells that retain the ability to reproduce daughter cells which will, in turn, replace differentiated tissues that have become worn, diseased or damaged. However, the power of the stem cell to regenerate could be potentially dangerous and, if poorly regulated, give rise to many forms of cancer. An understanding of how stem cells regulate their own growth and development is therefore extremely important. To begin studying stem cells, it is necessary to develop a stem cell line. A stem cell line is a group of constantly dividing cells from a single parent group of stem cells. Stem cell lines are grown in culture dishes, allowing them to divide and grow as undifferentiated cells for many years. Stem cell research allows scientists to discover more information about key cell processes such as growth, differentiation and gene regulation. Stem cells can also be used to study how diseases develop. Stem cells may be a viable alternative to animals for testing new drugs in the future. The potential benefits of using stem cells in medicine seem endless. In the future, scientists hope that stem cells will be used to cure conditions such as Alzheimer's disease, Parkinson's disease, diabetes, traumatic spinal cord injury, vision and hearing loss, Duchenne's muscular dystrophy, stroke and heart disease. 4.9 Ethical issues regarding stem cells Stem cell research and therapy are regulated in this country by the Human Fertilisation and Tissue Authority, and the Human Tissue Authority. Researchers and clinicians have to work within strict guidelines outlined in the Human Fertilisation and Embryology Act 1990, which was revised in 2000 and updated in 2008. In some other countries, all work in this area is outlawed. Stem cell research and the sourcing of stem cells have produced a great deal of argument and discussion. Many, from principally religious and moral stand-points, have argued against this work. This is because they believe that life begins at the point of fertilisation and that the zygote or blastocyst should be thought of, and treated, as if it was the same as a living being. Currently, embryos up to 14 days may be used. Other groups, such as patients awaiting treatment, are supportive of stem cell research. Other areas of research are attempting to grow stem cells from adult and differentiated tissue using a technique known as induced pluripotentency. In this process, adult differentiated cells are taken and re-engineered back to embryonic-like cells. This could be a way of expanding research in stem cells by avoiding the ethical issues associated with the use of embryonic stem cells. © H ERIOT-WATT U NIVERSITY TOPIC 4. DIFFERENTIATION IN MULTICELLULAR ORGANISMS 59 A study of ethics The ethical matrix (designed by Professor Ben Mepham, Centre for Applied Bioethics at the University of Nottingham) is a tool to help people analyse an ethical issue and make an informed choice. It is based on three key ethical principles (for further information on these see Mepham: Bioethics: an introduction for the Biosciences Oxford University Press (2008), now in 2nd edition): 1. Wellbeing: the safety, welfare and health of an individual or group. 2. Autonomy: an individual's right to be free to choose and make their own decisions. 3. Justice: to what extent a situation is just or fair for an individual or group. You are asked to consider the following questions with reference to the information in the ethical matrix to help explore your own opinions and feelings using, as far as possible, the evidence here and any other source you feel appropriate. a) What do you think might be the priority of each of the interest groups? b) In what way do you think that the three principals apply to each interest group? c) To what extent do you think others might agree or disagree with you? d) Might your decision be influenced by the thoughts and beliefs of others? e) Can you suggest any way round some of the ethical issues that are raised by others? Wellbeing Autonomy Justice Interest Groups (safety, welfare (freedom and (fairness) and health) choice) Patients - people who are hoping that stem cell therapies will treat an illness, disease or injury. Scientists - people working in stem cell research, developing stem cell therapies to treat patients. Embryo - the source of embryonic stem cells for research. Society - issues for wider society, such as social priorities, research and medical priorities, and how money should be allocated. © H ERIOT-WATT U NIVERSITY 60 TOPIC 4. DIFFERENTIATION IN MULTICELLULAR ORGANISMS 4.10 Learning points Summary Cellular differentiation is the process by which a cell expresses certain genes to produce proteins characteristic for that type of cell. This allows a cell to carry out specialised functions. Meristems are regions of unspecialised cells in plants that can divide (self-renew) and/or differentiate. Stem cells are unspecialised cells in animals that can divide (self-renew) and/or differentiate. In the very early embryo, a group of cells, the inner cell mass, can differentiate into almost all body tissues and so are pluripotent. Tissue stem cells are involved in the growth, repair and renewal of the cells found in that tissue. They are multipotent. Tissue stem cells are multipotent as they can differentiate into all of the types of cell found in a particular tissue type. For example, blood stem cells located in bone marrow can give rise to all types of blood cell. Stem cell research provides information about how cell processes, such as cell growth, gene expression and gene regulation, occur. Stem cells are used in therapies to repair and replace damaged organs and tissues; more therapies are being developed to treat diseases such as diabetes. Research uses involve stem cells being used as model cells to study how diseases develop or being used for drug testing. There are major ethical issues surrounding stem cell use and research; for example, the use of embryonic stem cells involves the destruction of embryos which some people believe is akin to murder. © H ERIOT-WATT U NIVERSITY TOPIC 4. DIFFERENTIATION IN MULTICELLULAR ORGANISMS 61 4.11 Extended response question The activity which follows presents an extended response question similar to the style that you will encounter in the examination. You should have a good understanding of stem cells before attempting the question. You should give your completed answer to your teacher or tutor for marking, or try to mark it yourself using the suggested marking scheme. Extended response question: Stem cells Describe the differences between and similarities of embryonic stem cells and tissue stem cells. (6 marks) 4.12 Extension materials The material in this section is not examinable. It includes information which will widen your appreciation of this section of work. Extension materials: The Jacob-Monod hypothesis Although bacterial gene control is not strictly needed for the exam, the simpler mechanisms in prokaryotes lead to better understanding of control in complex cells. There are many different ways in which gene expression is controlled. The mechanisms of gene expression are very complicated and are not fully understood in higher organisms. However, a lot is known about gene expression in bacteria thanks to work carried out by two French scientists. In the 1950s, Francois Jacob and Jacques Monod, two scientists at the Pasteur Institute in Paris, performed a series of experiments and determined how the production of the enzyme β-galactosidase was controlled in the bacterium Escherichia coli. They later won the Nobel prize for their work. E. coli uses glucose during respiration to release energy. It will always use glucose if it is present in the environment. If no glucose is available, then it will utilise the glucose found in other energy sources, such as lactose. However, it is only able to use the glucose in lactose once it has been separated from galactose. The enzyme β-galactosidase breaks down lactose into glucose and galactose. Jacob and Monod found that E. coli only produces β-galactosidase when lactose is present in the nutrient medium in which the bacteria are growing. If lactose is absent, then no β-galactosidase is produced. The gene that controls β-galactosidase production is 'switched on' (expressed) in the presence of lactose and 'switched off' (not expressed) when lactose is absent. Enzyme induction occurs, meaning that the gene is only switched on when the enzyme it codes for is required. In other words, E. coli is able to regulate the expression of the genes needed for lactose metabolism. © H ERIOT-WATT U NIVERSITY 62 TOPIC 4. DIFFERENTIATION IN MULTICELLULAR ORGANISMS The Jacob-Monod hypothesis suggests that the production of β-galactosidase is controlled by an operon. An operon is a region of DNA that contains an operator gene that controls the expression of a structural gene which make the proteins or enzyme. A regulator gene that is found further along the DNA strand codes for a repressor molecule that interacts with the operator gene, preventing the expression of the structural gene. When lactose is absent, the regulator gene produces a repressor molecule. This repressor molecule binds to the operator gene and this means that the structural gene is 'switched off'. When lactose is present, it binds to the repressor molecule which means the operator gene is free. The operator gene then 'switches on' the structural gene which produces the enzyme β-galactosidase. This enzyme breaks down lactose into glucose and galactose. When the lactose is used up, the repressor molecule binds to the operator gene and the structural gene is once again 'switched off'. The lac operon © H ERIOT-WATT U NIVERSITY TOPIC 4. DIFFERENTIATION IN MULTICELLULAR ORGANISMS 63 4.13 End of topic test End of Topic 4 test Go online Q5: Which of the following statements is not true of a plant meristem? a) Growth can only occur at a meristem in a plant. b) It contains unspecialised cells that differentiate. c) It contains specialised cells that differentiate. d) The roots and shoots have meristems................................................ Q6: The following diagram shows a cross section through a root. Which letter indicates the meristem?............................................... Q7: What is a stem cell?............................................... Q8: State two properties of stem cells................................................ Q9: Name two sources of tissue (adult) stem cells................................................ Q10: Why do our bodies need stem cells?............................................... Q11: Describe a current medical use of stem cells mentioned in this topic................................................ Q12: What is the main ethical consideration regarding the use of embryonic stem cells? © H ERIOT-WATT U NIVERSITY 65 Topic 5 Structure of the genome Contents 5.1 Introduction to the Genome..................................... 66 5.2 The genome.............................................. 66 5.3 Learning points............................................ 68 5.4 Extension materials.......................................... 69 5.5 End of topic test............................................ 70 Prerequisites You should already know that: DNA carries the genetic information for making proteins. Learning objective By the end of this topic, you should be able to: understand the meaning of the term genome; describe that a section of DNA which produces a polypeptide is regarded as a gene; recognise that some DNA sequences code for proteins, while other sections do not; describe how some of the non-coding DNA has a role in controlling and regulating transcription; recognise that not all of the functions of non-coding DNA are as yet known or understood; describe the functions of RNA, including tRNA and rRNA. 66 TOPIC 5. STRUCTURE OF THE GENOME 5.1 Introduction to the Genome The genome is the sum total of all the hereditary material within an organism. It is usually taken to be the complete complement of DNA, although in some viruses this could be DNA or RNA. The majority of the genetic information is carried within the nuclear DNA (linear chromosomes), but other sources exist. In bacteria, the DNA is found as a circular strand, sometimes called a chromosome, but lacking the associated packaging proteins. Bacteria often contain plasmids, which are small circular sections of DNA. In eukaryotes, organelles such as chloroplasts and mitochondria also contain circular sections of DNA. Mitochondrial DNA is of great importance in hereditary studies as it is only passed down the maternal line. More often, the term genome is used to refer to nuclear DNA only. The mitochondrial DNA may be referred to as the mitochondrial genome and the DNA of the chloroplast as the plastome. The human genome is recognised as consisting of 3 × 10 9 nucleotides. These are found as approximately 20 - 25,000 genes, arranged on 22 autosomal chromosomes, and a pair of sex chromosomes, either two X chromosomes or an X and a Y chromosome. The study of the properties of genomes is referred to as genomics, compared to the study of single genes or groups of genes, which is genetics. 5.2 The genome Processing from DNA to protein An organism's genome is its genetic information encoded in its DNA. The genome contains many genes which carry instructions for making all of the proteins found in an organism. These regions of DNA are known as coding regions. The genome contains both coding and non-coding regions. © H ERIOT-WATT U NIVERSITY TOPIC 5. STRUCTURE OF THE GENOME 67 In fact, most of the genome is made up of non-coding sequences. These non-coding regions can have several functions: regulating transcription, transcription of RNA, no known function. Some non-coding sections of DNA are used to regulate transcription. This means they can bind proteins which promote or prevent transcription of a gene. The diagram below illustrates how a sequence of DNA can regulate transcription of a gene. Regulation of transcription Some sections of DNA are transcribed into RNA but are not translated, for example tRNA, which carries specific amino acids to the ribosome during translation, and rRNA, which together with protein forms the ribosome. Another type of RNA which is not translated is RNA fragments. These are small sections of RNA which are involved in splicing, and other processes such as post- transcriptional regulation of genes. © H ERIOT-WATT U NIVERSITY 68 TOPIC 5. STRUCTURE OF THE GENOME The function of large sections of the genome are still unknown. It was once referred to as junk DNA, but it is now widely acknowledged that it serves a purpose. The genome: Question Go online Q1: Complete the table of genome terms by matching the descriptions from the list with the processes. Process Description Transcription Splicing Translating Descriptions: DNA copied to RNA Exons pass to ribosome where polypeptides are assembled Introns removed from pre-mRNA 5.3 Learning points Summary The genome of an organism is its hereditary information encoded in DNA. Coding regions are sections of DNA which contain a gene. Much of the genome is non-coding in that it does not contain genes. Some regions of the genome contain regulatory sequences which control the transcription of genes. Other non-coding regions contain sequences for producing non-translated forms of RNA such as tRNA, rRNA and RNA fragments. While some sections of non-coding DNA assist in the control and regulation of gene expression, there are sections whose function remains unknown. © H ERIOT-WATT U NIVERSITY TOPIC 5. STRUCTURE OF THE GENOME 69 5.4 Extension materials The material in this section is not examinable. It includes information which will widen your appreciation of this section of work. Extension materials: The Human Genome Project The Human Genome Project began in October 1990 and was completed in 2003. The project involved the discovery of all the estimated 20,000 - 25,000 human genes, making them available for further studies. The project also led to the discovery of the complete sequence of the 3 billion DNA sub-units (the bases in the human genome). In April 2003, the completion of the human DNA sequence coincided with the 50 th anniversary of Crick and Watson's description of the DNA structure in 1953. Only about 3 percent of the human genome is actually used as the set of instructions. These regions are called coding regions. At present, little is known functionally for most of the remaining 97 percent of the genome; these regions are called non-coding regions. Single nucleotide polymorphisms, or SNPs (pronounced "snips"), are DNA sequence variations where a single nucleotide (A,T,C,or G) in the genome sequence is altered. For example, a SNP might change the DNA sequence ACGGCTCA to ATGGCTCA. Remember that there are 20 different amino acids. DNA is made up of four nucleotides, so if each were to specify (or code for) a single amino acid, only four amino acids could be coded for. A two letter code would give 16 (4 2 ) possible arrangements - still not enough to code for all 20 amino acids. The shortest unit that can code for all amino acids is a triplet code or codon. A triplet code produces 64 (4 3 ) possibilities - more than enough! A variation is considered a SNP when it occurs in at least 1% of the population. SNPs make up about 90% of all human genetic variation and can occur every 100 to 300 bases along the 3 billion base human genome. About 66% of SNPs involve the replacement of cytosine (C) with thymine (T). SNPs can occur in coding and non-coding regions of the genome. They can act as biological markers, helping scientists in locating genes that are associated with certain diseases. SNPs have no effect on cell function; scientists believe SNP maps will help them identify the multiple genes associated with complex ailments such as cancer, diabetes, vascular disease, and some forms of mental illness. Craig Venter's genome was published in 2007. His genome contains 4.1 million variations; 3.2 million were SNPs. The following year, James Watson's genome was published, costing about £8 million and taking only 4 months. In 2010, the first personal genome machine came onto the market. This machine can sequence an individual's genome in about 12 days at a cost of £6,000! Although SNPs do not cause diseases, they can help determine the likelihood that someone will develop a particular illness. One of the genes associated with Alzheimer's disease, apolipoprotein E (or ApoE), is a good example of how SNPs affect disease development. Scientists believe that SNPs may help them to discover and catalogue the unique sets of changes involved in different types of cancers. They are confident that SNPs can play an important role in the different methods used in the treatment of cancer. © H ERIOT-WATT U NIVERSITY 70 TOPIC 5. STRUCTURE OF THE GENOME Scientists are trying to identify all of the different SNPs in the human genome. They are sequencing the genomes of a large number of people and then comparing the base sequences to discover SNPs. The sequence data is being stored in computers that can generate a single map of the human genome, containing all possible SNPs. 5.5 End of topic test End of Topic 5 test Go online Q2: Complete the sentence using the words from list. The of an organism is its information encoded in. Word list: DNA; genome; hereditary............................................... Q3: DNA sequences that code for proteins are: a) cistrons b) exons c) genes d) introns............................................... Q4: All DNA codes for proteins. True or false?............................................... Q5: Name two types of RNA................................................ Q6: The function of all DNA was worked out with the completion of the Human Genome Project. True or false?............................................... Q7: By working out the nucleotide sequence of a genome, it is possible to describe all the organism's genes. True or false? © H ERIOT-WATT U NIVERSITY 71 Topic 6 Mutations Contents 6.1 Single gene mutations...............................

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