Unit C3 Biology 30 - Past Paper PDF

Ch 20_Bio_Alberta30 1/8/07 12:40 PM Page 660 chapter 20 In this chapter Exploration: Similarities Molecular Genetics By the mid-1950s, scientists had determined that...

Ch 20_Bio_Alberta30 1/8/07 12:40 PM Page 660 chapter 20 In this chapter Exploration: Similarities Molecular Genetics By the mid-1950s, scientists had determined that chromosomes contained DNA and that DNA was the genetic material (Figure 1). Building on the work of other scientists, and Differences Watson and Crick deduced the structure of this complex molecule. This knowledge laid Mini Investigation: the basis for the field of molecular biology, which aims to understand the inheritance of Building a DNA Model traits at the level of interactions between molecules in the cell. Web Activity: DNA A primary goal of molecular genetics is to understand how DNA determines the phe- Replication notype of an organism. What happens to DNA during duplication of chromosomes in mitosis? How does the structure of DNA relate to its function? How does one molecule, Lab Exercise 20.A: Synthesis of a Protein identical in every somatic cell of an organism, determine the characteristics of the many different types of cells that are found in that organism? Investigation 20.1: Today, questions such as these continue to drive research in the fields of biology, Protein Synthesis and Inactivation of Antibiotics biotechnology, biochemistry, and medicine. We now know the sequence of all the nucleotides that make up the genome of many organisms, including that of our own Web Activity: species, Homo sapiens. This information has given scientists new ways to study the rela- Electrophoresis tionships between species and the mechanisms of evolution. It also allows law enforce- Web Activity: Researchers ment agencies to identify individuals with incredible accuracy from minute quantities in Human Genetic of DNA. Disorders Using genetic technologies, scientists can move genes from one species to another. In Mini Investigation: fields such as agriculture, corporations have patented the genomes of these organisms Examining the Human in order to profit from the advantages they offer over conventional organisms. Similar Genome manipulation of human cells may one day lead to treatments for previously untreat- Investigation 20.2: able debilitating diseases. The research and application of these technologies raises many Restriction Enzyme social, ethical, and legal issues that society has yet to fully resolve. Digestion of Bacteriophage DNA Web Activity: STARTING Points Transformation of Eukaryotes Answer these questions as best you can with your current knowledge. Then, using the concepts and skills you have learned, you will revise your answers at the end of Case Study: Gene the chapter. Mutations and Cancer Lab Exercise 20.B: 1. Differentiate between DNA and proteins. What cellular roles do they play? Looking for SINEs of 2. Describe the physical and chemical characteristics of DNA. Evolution 3. What is the significance of DNA replication in your body? 4. Write a short overview, in paragraph form, of the process of DNA replication. Career Connections: Biological Technician; Biotechnologist 660 Chapter 20 NEL Ch 20_Bio_Alberta30 1/8/07 12:40 PM Page 661 Figure 1 DNA sequences are represented by the letters A, T, C, and G. Exploration The Size of the Genome All organisms, no matter how simple they may seem to us, (a) Why does it take two lengths of thread to represent the require DNA in each cell to encode the instructions necessary chromosome? to live and reproduce. The total DNA of an organism is referred (b) Is the thread that you tried to place in the capsule too to as its genome. In bacteria, the genomic DNA is circular, thick to represent the actual thickness of the DNA? accounts for 2 % to 3 % of the cell’s mass, and occupies about (What percentage of bacterial cell volume does your 10 % of its volume. In this activity, you will make a model of an thread fill, and what is the actual volume that the DNA Escherichia coli cell that will be 10 000 times bigger than actual occupies in the bacteria?) size. You will also gain an appreciation for how compactly DNA (c) If the human genome is 1000 times bigger than the E. coli is packed within a cell. genome, how many metres of thread would it take to Materials: 2 cm gelatin capsule, 10 m of white thread, 10 m of represent the human genome? coloured thread (d) What size container would you need to hold the thread representing the human genome? Try to construct the bacterium by placing the long lengths of thread inside the gelatin capsule. Good luck! It’s not easy! NEL Molecular Genetics 661  20.1 DNA Structure and Replication According to the model proposed by Watson and Crick, DNA consists of two strands of nucleotides. Each nucleotide contains a deoxyribose sugar, a phosphate group, and a CAREER CONNECTION nitrogenous base, all covalently bonded to each other. Each strand of DNA has a back- Biological Technician bone of sugar and phosphate groups (Figure 1). The nitrogenous bases stick out from the Biological technicians may work in backbone of each DNA strand. the field, in the laboratory, or both. Watson and Crick’s model also indicates that the two strands of DNA form a structure They perform routine analysis and technical duties to support the that resembles a twisted ladder. The base pairs are the rungs of the ladder and the sugar– work of scientists and engineers phosphate backbones are the struts. This structure is called a double helix (see Figure 1). working in fields that include Each DNA strand in the double helix twists in a clockwise direction. molecular biology. What educational background is a simplified required to enter this field? 5’ end P deoxyribose sugar molecule version of the nitrogenous bases phosphate molecule DNA molecule www.science.nelson.com GO S showing the S T A double coil P P or helix S S G C P P S S C G P P S S A T Figure 1 3’ end P nitrogenous bases A simplified diagram of the one nucleotide sugar–phosphate backbone structure of DNA In the DNA molecule, the bases of one DNA strand are paired with bases in the other strand. A purine is always paired with a pyrimidine. Adenine (a purine) is always paired with thymine (a pyrimidine), and guanine (a purine) is always paired with cytosine (a complementary base pairing pyrimidine). This type of pairing is termed complementary base pairing. Hydrogen pairing of the nitrogenous base of bonds, between the complementary bases (A-T and G-C) on opposite strands, hold the one strand of DNA with the double helix together (Figure 2). Although a single hydrogen bond is very weak, large nitrogenous base of another strand numbers of hydrogen bonds are collectively strong, so the DNA molecule is very stable. CH3 H H H O H N H H H T C N N N N N O sugar H N sugar H N Figure 2 O N O N Adenine forms two hydrogen bonds A H H G H with thymine, while guanine forms H N N N N N three hydrogen bonds with sugar sugar cytosine. H 662 Chapter 20 NEL  Section 20.1 The sequence of bases on any one strand of DNA can vary greatly between species, but antiparallel parallel but running in its opposite strand will always have the complementary sequence of bases. For example, the opposite directions; the 5
end of one strand of DNA aligns with the sequences of the strands below are complementary: 3
end of the other strand in a 5
–ATGCCGTTA–3
double helix 3
–TACGGCAAT–5
The two strands of nucleotides are antiparallel. They run parallel but in opposite directions to one another. One strand will have a 5
carbon and phosphate group at one end and a 3
carbon and the hydroxyl group of a deoxyribose sugar at its other end. Its antiparallel strand will have a 3
carbon and the hydroxyl group of a deoxyribose sugar at the first end and a 5
carbon and phosphate group at its other end (Figure 1, pre- vious page). Learning Tip The direction of the strand is important when enzymes interact with DNA, either to copy the DNA prior to cell division or to “read” genes in order to make proteins. Enzymes The rules of complementary can read or copy DNA in only one direction. The sequence of only one DNA strand is DNA base pairing are A to T given when sequences are written out since the complementary strand is easily deduced G to C according to the rules of complementary base pairing. When you know the sequence on one strand, you also know Practice the sequence on the complementary strand. 1. Define the following terms: nucleotide, complementary base pairing, and antiparallel. 2. In a DNA molecule, a purine pairs with a pyrimidine. If this is the case, then why can’t A–C and G–T pairs form? (Hint: Look closely at the bonds between the base pairs in Figure 2 on the previous page.) 3. The following is a segment taken from a strand of DNA: 5
–ATGCCTTA–3
. Write out the complementary strand for this segment. Be sure to show directionality. mini Investigation Building a DNA Model What would a section of a DNA molecule look like if you could see one close up? You can find out by building your own model the bonds between complementary base pairs that hold the two strands together of the double helix. For this activity, you need to select materials that will allow you to model the following features: Your model should show a minimum of 12 base pairs. It should be free-standing and approximately 15 cm tall by 6 cm wide. the sugar–phosphate backbone Include a legend with your model that clearly identifies each the anti-parallel strands part of the DNA strand. the four different nitrogenous bases DNA Replication In Chapter 17, you saw that mitosis involves the duplication of chromosomes. For mitosis DNA replication the process to occur, DNA must copy itself and be equally divided between the daughter cells. To have whereby DNA makes exact copies all the correct genetic information, the DNA in each daughter cell must be an exact copy of itself of the DNA in the parent cell. DNA replication is the process by which a cell makes an semiconservative replication exact copy of its DNA. The main stages of DNA replication are the same in both prokary- process of replication in which each otic cells (without a membrane-bound nucleus) and eukaryotic cells (with a membrane- DNA molecule is composed of one bound nucleus). parent strand and one newly DNA replication is semiconservative. Semiconservative replication involves sepa- synthesized strand rating the two parent strands and using them to synthesize two new strands (Figure 3, next template a single-stranded DNA page). The hydrogen bonds between complementary bases break, allowing the DNA helix sequence that acts as the guiding to unzip. Each single DNA strand acts as a template to build the complementary strand. pattern for producing a Finally, any errors are repaired, resulting in two identical DNA molecules, one for each complementary DNA strand daughter cell. NEL Molecular Genetics 663  Key Figure 3 DNA replicates semiconservatively. semiconservative Each daughter molecule receives old strand replication one strand from the parent molecule plus one newly new strand synthesized strand. Separating the DNA Strands The two strands of the DNA helix cannot simply pull apart because they are tightly held together by the hydrogen bonds between bases and by the twists of the helix. The enzyme DNA helicase the enzyme that DNA helicase unwinds the helix by breaking the hydrogen bonds between the com- unwinds double-helical DNA by plementary bases. As this happens, the bonds between bases tend to reform. To prevent disrupting hydrogen bonds this, proteins bind to the separated DNA strands, helping to hold them apart. The two strands are now separated along part of the DNA molecule and are the template strands for the next step in replication. The point at which the two template strands are separating is called the replication fork. One template strand runs in the 3
to 5
direction in rela- tion to the replication fork, while the other runs in the 5
to 3
direction (Figure 4). 3
DID YOU KNOW ? 5
DNA Polymerases There are several DNA polymerases 2. proteins nd in a cell, all with their own role. g s tra Each has a unique name, created 1. DNA helicase din by adding a roman numeral after lea direction of replication "DNA polymerase." The main DNA 5
3
4. DNA polymerase III polymerase involved in DNA 3
5
replication is DNA polymerase III. It 3
5
lag g adds the 5
phosphate group of a stra ing free nucleotide to the 3
carbon of nd 3. primer the sugar in the last nucleotide. replication fork 3
5
Figure 4 1. DNA helicase opens the double helix. 2. Proteins bind to the DNA to keep the two strands separate. 3. RNA primers are attached to the template strands. 4. DNA polymerase synthesizes the new DNA strands. The leading strand is synthesized continuously, and the lagging strand is synthesized in short fragments. DNA polymerase III adds complementary nucleotides in the 5
to 3
direction, using single-stranded primers as starting points. One nucleotide is attached to the next by bonding the phosphate on the 5
end of the new nucleotide to the hydroxyl group on the 3
end of the last nucleotide. Building the Complementary Strands The next stage of DNA replication synthesizes two new DNA strands on the template strands through complementary base pairing. The new strands are synthesized by an DNA polymerase III the enzyme enzyme called DNA polymerase III. This DNA polymerase builds a new strand by that synthesizes complementary linking together free nucleotides that have bases complementary to the bases in the tem- strands of DNA during DNA plate. A short piece of single-stranded ribonucleic acid, called a primer, is attached to the replication template strand. This gives DNA polymerase III a starting point to begin synthesizing the 664 Chapter 20 NEL  Section 20.1 new DNA strand. DNA polymerase III adds nucleotides to a growing strand in only one leading strand the new strand direction—the 5
to 3
direction. The phosphate group at the 5
end of a free nucleotide of DNA that is synthesized towards the replication fork and is connected to the hydroxyl group on the 3
carbon of the sugar on the last nucleotide continuously during DNA in the strand. As a result, one of the new strands will be synthesized continuously as replication DNA polymerase III moves in the 5
to 3
direction toward the replication fork. This strand is called the leading strand. lagging strand the new strand The other new strand, the lagging strand, is synthesized in short fragments. This of DNA that is synthesized away allows the lagging strand to be synthesized in the 5
to 3
direction. RNA primers are from the replication fork and in short fragments, which are later required. To complete the replication of the DNA, the primers are cut out from the lag- joined together ging strand and are replaced by DNA nucleotides by a different enzyme called DNA polymerase I. DNA polymerase I an enzyme that Another enzyme, DNA ligase, links the sugar–phosphate backbone of the DNA frag- removes RNA primers and replaces ments together (Figure 5). them with the appropriate nucleotides during DNA replication RNA primer DNA fragment DNA ligase an enzyme that joins DNA polymerase III adds DNA fragments together nucleotides to the 3
RNA primer 5
primers to form short fragments of DNA. 5
3
direction of synthesis DNA polymerase III DNA polymerase I 3
5
removes the RNA primers and replaces them with DNA nucleotides. A nick 5
3
is left between fragments. DNA polymerase I DNA ligase 3
5
DNA ligase joins the fragments together. Figure 5 5
3
Building the lagging strand DNA Repair As complementary strands of DNA are synthesized, both DNA polymerase I and III act as quality control checkers by proofreading the newly synthesized strands. When a mis- DID YOU KNOW ? take occurs, the DNA polymerases backtrack to the incorrect nucleotide, cut it out, and Okazaki Fragments The short fragments that are then continue adding nucleotides to the complementary strand. The repair must be synthesized to form the lagging made immediately to avoid the mistake from being copied in later replications. Other DNA strand during DNA replication are repair mechanisms can correct any errors that were missed during proofreading. called Okazaki fragments. They were named after Reiji Okazaki, who first described them in the 1960s. NEL Molecular Genetics 665 Ch 20_Bio_Alberta30 1/8/07 12:40 PM Page 666 WEB Activity Simulation—DNA Replication The Escherichia coli genome consists of 4.7 million nucleotide pairs. This entire genome is replicated in 40 min. Proofreading by DNA polymerase I and polymerase III maintains the error rate at roughly one error per 1000 cells duplicated! View a complete animation of DNA replication by accessing the Nelson Web site. www.science.nelson.com GO SUMMARY DNA Structure and Replication Separating the Strands DNA helicase unzips the double helix by breaking the hydrogen bonds between the complementary bases in the two strands of the parent DNA molecule. Proteins attach to the newly exposed DNA strands, preventing the hydrogen bonds from re-forming and keeping the strands apart. Building the Complementary Strands DNA polymerase III adds complementary nucleotides to the growing strands, using the exposed strands of the parent DNA molecule as a template. The leading strand is formed continuously. The lagging strand is formed in short fragments, starting from an RNA primer. DNA polymerase I cuts out the RNA primers and replaces them with the appropriate DNA nucleotides. DNA ligase joins the fragments together to form a complete DNA strand. DNA Repair DNA polymerase enzymes cut out incorrectly paired nucleotides and add the correct nucleotides in a process called proofreading. Section 20.1 Questions 1. Summarize the key physical and chemical properties of 5. Define a replication fork. DNA. 6. In a double helix, there is a complete turn every 3.4 nm, or 2. Differentiate between a purine and a pyrimidine. 10 nucleotides. Assume that the DNA molecule in a 3. Copy Table 1 into your notebook, fill in the missing particular chromosome is 75 mm long. Calculate the information, and supply an appropriate title. number of nucleotide pairs in this molecule. 7. Copy Table 2 into your notebook and complete the Table 1 missing information. Explain how you determined the Enzyme Function missing values. DNA helicase Table 2 DNA polymerase I Nucleotide Sample A Sample B Sample C DNA polymerase III adenine 10 % 20 % DNA ligase guanine 40 % 15 % thymine 35 % 20 % 4. A molecule of DNA was analyzed and found to contain 20 % thymine. Calculate the percentage of adenine, cytosine guanine, and cytosine in this molecule. 666 Chapter 20 NEL  Gene Expression 20.2 As you learned in previous chapters, specific segments of DNA on a chromosome are called genes. Genes determine the inherited characteristics, or traits, of an organism. Every somatic (body) cell in an organism contains identical copies of DNA, and each of these DNA copies is a genetic blueprint for the organism. Once scientists knew the structure of DNA and how it replicated, they used this knowledge to further investigate another question: How do the genes in DNA determine an inherited trait? The way the information in a gene is converted into a specific characteristic or trait through the production of a polypeptide is called gene expression. Recall that a polypep- gene expression conversion of a tide is a chain of amino acids and that proteins are made up of polypeptides. Proteins gene into a specific trait through form many structures in an organism, such as skin and muscle, and they also form all the production of a particular polypeptide of the enzymes in a cell. The products of all genes are polypeptides. A second type of nucleic acid is involved in converting the instructions in a gene into a polypeptide chain. Ribonucleic acid (RNA) is a polymer of nucleotides similar to ribonucleic acid (RNA) a nucleic DNA. There are three main structural differences between RNA and DNA. First, the acid consisting of nucleotides sugar in RNA has an extra hydroxyl group and is called ribose rather than deoxyribose comprised of the sugar ribose and nitrogenous bases (Figure 1). Second, instead of the base thymine found in DNA, RNA contains the base uracil. Like thymine, uracil can form complementary base pairs with adenine (Figure 2). Third, RNA is single-stranded and not double-stranded like DNA. There are three types of RNA that are needed to convert genes into proteins: messenger RNA (mRNA), transfer RNA (tRNA), and ribosomal RNA (rRNA). 5
5
HO CH2 O OH HO CH2 O OH 4
H H 1
4
H H 1
3
2
3
2
DNA A T G C A A H H H H OH OH OH H RNA U A C G U U ribose sugar deoxyribose sugar Figure 2 Figure 1 Base pairing of RNA with DNA A ribose sugar possesses an OH group (hydroxyl) on the 2
carbon. The during transcription. Notice that deoxyribose sugar is missing the OH group on the 2
carbon. The deoxy part thymine does not exist in RNA but is of the name deoxyribose indicates a “loss of oxygen” at position 2. substituted with uracil. The Central Dogma There are two main stages of gene expression, transcription and translation. In transcription, the genetic information is converted from a DNA sequence into transcription the process of converting DNA into messenger messenger RNA (mRNA). In all cells, the mRNA carries the genetic information from RNA the chromosome to the site of protein synthesis. In eukaryotic cells, which contain a nucleus, the mRNA carries the genetic information from the nucleus to the cytoplasm messenger RNA (mRNA) the as it passes through the pores in the nuclear envelope. product of transcription of a gene; The second stage of gene expression is translation. During translation, the genetic mRNA is translated by ribosomes into protein information carried by the mRNA is used to synthesize a polypeptide chain. The two-step process of transferring genetic information from DNA to RNA and then translation the process of from RNA to protein is known as the central dogma of molecular genetics (Figure 3, next synthesizing a specific polypeptide page). We will explore transcription and translation in more detail in this section. You as coded for by messenger RNA NEL Molecular Genetics 667  nucleus Transcription nuclear Translation cytoplasm envelope protein DNA gene mRNA polypeptide Figure 3 The central dogma of molecular genetics will see that the sequence of nucleotides in a gene determines the sequence of amino acids in a polypeptide. Transcription During transcription, the DNA sequence of a gene is copied (transcribed) into the sequence of a single-stranded mRNA molecule. Transcription is divided into three processes: initiation, elongation, and termination. RNA polymerase enzyme that During initiation, an enzyme called RNA polymerase binds to the DNA at a specific transcribes DNA site near the beginning of the gene. During elongation, RNA polymerase uses the DNA as a template to build the mRNA molecule. During termination, the RNA polymerase passes the end of the gene and comes to a stop. The mRNA is then released from the template strand of DNA. Initiation Transcription starts when the RNA polymerase enzyme binds to the segment of DNA to be transcribed and opens the double helix. Figure 4 shows an electron micrograph of this process. The RNA polymerase binds to the DNA molecule in front of the gene to be transcribed in a region called the promoter. In most genes, the promoter sequence contains a string of adenine and thymine bases that serves as the recognition site for RNA polymerase. The promoter indicates which DNA strand should be transcribed and where the RNA polymerase should start transcribing the DNA. Since the binding site Figure 4 of RNA polymerase only recognizes the promoter region, it can only bind in front of The RNA polymerase (dark circles) a gene. binds to the DNA strand and initiates transcription. Transcription Elongation occurs simultaneously at numerous locations along the DNA. Once the RNA polymerase binds to the promoter and opens the double helix, it starts building the single-stranded mRNA in the 5
to 3
direction. The promoter is not tran- scribed. The process of elongation of the mRNA molecule is similar to DNA replica- tion. However, RNA polymerase does not require a primer and it copies only one of the promoter sequence of DNA that binds RNA polymerase in front of a DNA strands. The transcribed DNA strand is called the template strand. The mRNA gene sequence is complementary to the DNA template strand except that it contains the base uracil in place of thymine. template strand the strand of DNA that the RNA polymerase uses as a Termination guide to build complementary mRNA Synthesis of the mRNA continues until RNA polymerase reaches the end of the gene. RNA polymerase recognizes the end of the gene when it comes to a stop signal called a termination sequence sequence termination sequence. Transcription stops and the newly synthesized mRNA discon- of bases at the end of a gene that nects from the DNA template strand. RNA polymerase is then free to bind to another pro- signals the RNA polymerase to stop moter region and transcribe another gene. Figure 5, on the next page, summarizes the transcribing steps in transcription. 668 Chapter 20 NEL  Section 20.2 (a) Initiation RNA polymerase RNA polymerase binds to DNA at a promoter. DNA promoter (b) DNA double helix is unwound, exposing the template strand. template strand (c) Elongation mRNA is synthesized using one strand of DNA as a template. mRNA is synthesized in the 5
to 3
direction. mRNA (d) As elongation proceeds, RNA direction of polymerase moves along DNA, DNA has rewound transcription synthesizing mRNA. DNA that has already been transcribed rewinds into a double helix. 5
mRNA (e) RNA polymerase reaches the termination sequence at end of gene. 3
5
(f) Termination Transcription stops. mRNA and RNA polymerase RNA polymerase are released. 5
3
mRNA Figure 5 A summary of the process of transcription Practice + EXTENSION 1. A short fragment of a particular gene includes the following sequence of Regulation of Transcription nucleotides: This Audio Clip discusses the TACTACGGT regulatory factors that control when and how much mRNA is Write out the corresponding mRNA transcript. transcribed from a given gene. 2. A short fragment of another gene includes the following sequence of nucleotides: ACCATAATATTACCGACCT TCG www.science.nelson.com GO (a) Explain the purpose of the promoter region in transcription. (b) Copy the sequence into your notebook and circle the promoter region. Explain the rationale for your selection. NEL Molecular Genetics 669  Translation The second part of the central dogma of molecular biology (Figure 3, page 668) is the translation of the genetic information carried by mRNA into a chain of amino acids to form a polypeptide. Therefore, the process of translation involves protein synthesis, and it depends on the remarkable nature of the genetic code. Only 20 amino acids are found in proteins. The DNA in a gene codes for these 20 amino acids by combinations of the four nitrogenous bases. During translation, the codon sequence of three bases in DNA code is read in groups of three nucleotides, called a codon. Each codon calls for a DNA or complementary mRNA that specific amino acid to be placed in the growing polypeptide chain. Codons can consist serves as a code for a particular of any combination of the four nitrogenous bases, so there are 64 (43 = 64) possible dif- amino acid ferent codons for the 20 different amino acids. The groups of three bases in both DNA and mRNA are both called codons, so it is important to clarify which code is being pre- sented when writing out a genetic sequence. The remainder of this description will use mRNA codons. Table 1 shows the mRNA codons. One of these codons (AUG) is the start codon specific codon (AUG) start codon, where translation begins. It also codes for the insertion of the amino acid that signals the start of translation methionine, so all polypeptide chains initially start with the methionine, but it may later be edited out. Three other codons (UAA, UAG, and UGA) do not code for amino acids stop codon specific codon that and are called the stop codons because they cause protein synthesis to stop. The other signals the end of translation 60 codons code for one of the 20 amino acids. Some amino acids have more than one codon; for example, both serine and leucine each have 6 different codons. Table 2, on the next page, lists the abbreviations for the amino acids to help you look them up in Table 1. Like transcription, translation can be divided into the same three stages: initiation, elon- gation, and termination. Table 1 Codons and Their Amino Acids 2nd (middle) Base of a Codon 1st Base U C A G 3rd Base U UUU Phe UCU Ser UAU Tyr UGU Cys U UUC Phe UCC Ser UAC Tyr UGC Cys C UUA Leu UCA Ser UAA STOP UGA STOP A UUG Leu UCG Ser UAG STOP UGG Trp S + EXTENSION C CUU CUC Leu Leu CCU CCC Pro Pro CAU CAC His His CGU CGC Arg Arg U C Why Three Nucleotides CUA Leu CCA Pro CAA Gln CGA Arg A per Codon? CUG Leu CCG Pro CAG Gln CGG Arg S Why are there always three nucleotides in a codon? Why not A AUU Ile ACU Thr AAU Asn AGU Ser U two or four? Listen to this Audio AUC Ile ACC Thr AAC Asn AGC Ser C Clip to find out the reason behind AUA Ile ACA Thr AAA Lys AGA Arg A the triplet code found in DNA and AUG Met ACG Thr AAG Lys AGG Arg S mRNA sequences. G GUU Val GCU Ala GAU Asp GGU Gly U GUC Val GCC Ala GAC Asp GGC Gly C www.science.nelson.com GO GUA Val GCA Ala GAA Glu GGA Gly A GUG Val GCG Ala GAG Glu GGG Gly S Initiation ribosome an organelle composed Initiation of translation occurs when a ribosome recognizes a specific sequence on the of RNA and protein and located in mRNA and binds to that site. In eukaryotes, the ribosome consists of two subunits, a the cytoplasm that carries out large subunit and a small subunit (Figure 6, next page). The two subunits bind to the protein synthesis mRNA, clamping it between them. The ribosome then moves along the mRNA in the 5
to 3
direction, adding a new amino acid to the growing polypeptide chain each time it 670 Chapter 20 NEL  Section 20.2 polypeptide Table 2 Amino Acids and Their 3’ chain Abbreviations mRNA Amino Three-letter direction of acid abbreviation translation alanine Ala large small subunit subunit arginine Arg asparagine Asn aspartic acid Asp cysteine Cys large subunit ribosome small subunit glutamic acid Glu 5’ glutamine Gln intact ribosome glycine Gly Figure 6 Figure 7 histidine His Ribosomes consist of a large subunit and a The large and small subunit of a ribosome isoleucine Ile small subunit. work together to translate a strand of mRNA into a polypeptide. The polypeptide leucine Leu grows as the ribosome moves farther along lysine Lys the mRNA strand. methionine Met phenylalanine Phe reads a codon (Figure 7). Ribosomes synthesize different proteins by associating with dif- proline Pro ferent mRNAs and reading their coding sequences. serine Ser A ribosome must begin reading the coding sequence at the correct place in the mRNA, threonine Thr or it will misread all the codons. The first codon that it recognizes is the start codon AUG. Binding to the start codon ensures that the ribosome translates the genetic code tryptophan Trp using the reading frame of the mRNA molecule. It is critical that the mRNA be positioned tyrosine Tyr in the ribosome in its reading frame so that the genetic code is translated into the cor- valine Val rect sequence of amino acids. Once the ribosome has bound the mRNA, how does it get the amino acids that cor- respond to the codon? This job falls to a second type of RNA molecule known as transfer transfer RNA (tRNA) the form of RNA (tRNA). At one end of the tRNA there is a sequence of three bases, the anticodon, RNA that delivers amino acids to a that is complementary to the codon of the mRNA. The opposite end carries the corre- ribosome during translation sponding amino acid (Figure 8, next page). For example, if the mRNA has the codon UAU, anticodon group of three the complementary base sequence of the anticodon is AUA, and the tRNA would carry complementary bases on tRNA that the amino acid tyrosine. Check Table 1 to find the mRNA codon and prove to yourself recognizes and pairs with a codon that it calls for tyrosine. Every tRNA carries only one specific amino acid, which means on the mRNA that at least 20 different tRNAs are required. Recall that there are 64 possible codons. In reality, anywhere from 20 to 64 types of tRNA molecules are available, depending on the organism. Practice 3. Transcribe the following sequence of DNA into mRNA. TACGGATTTCTCCGCAAATTAGGG 4. Translate the following mRNA sequence into an amino acid sequence. 5
-AUGCCCUCUAUUCCGGGAAGAUAG-3
5. How many nucleotides are necessary in the DNA to code for the following sequence of amino acids? Leu-Tyr-Arg-Trp-Ser NEL Molecular Genetics 671 Ch 20_Bio_Alberta30 1/8/07 12:40 PM Page 672 DID YOU KNOW ? aminoacyl site for attachment 3⬘ end of amino acid tyrosine RNA Polymerase I, II, III A OH There are three forms of the RNA C polymerase in eukaryotes: RNA C polymerase I transcribes ribosomal 5⬘ end A RNA; RNA polymerase II transcribes P G C mRNA; and RNA polymerase III transcribes tRNA and other short C G genes that are about 100 base pairs C G in length. G C A U A U A U U U G G G C C C A C G A G C A C U C G C C C G G U C G C G U G A G C C G G G A G U A G U A A U G C A U C A U G anticodon arm A A U anticodon Figure 8 The tRNA molecule has a cloverleaf structure. The molecule folds to form this structure because of hydrogen bonding. The anticodon is located on the anticodon arm and the amino acid is covalently bound to the adenine nucleotide at the 3
end (aminoacyl). In this case, the amino acid that would be added is tyrosine because the anticodon is AUA. Elongation The first codon that is recognized by the ribosome is the start codon AUG. The AUG codon also codes for methionine, so every protein initially starts with the amino acid methionine. The ribosome has two sites for tRNA to attach: the A (aminoacyl) site and the P (peptidyl) site. The tRNA with the anticodon complementary to the start codon DID YOU KNOW ? enters the P site, as shown in Figure 9 (a). The next tRNA carrying the required amino From DNA to Protein acid enters the A site, as shown in Figure 9 (b). In Figure 9 (c), a peptide bond has The discovery of the relationship formed between the methionine and the second amino acid, alanine. The ribosome has between DNA, mRNA, ribosomes, shifted over one codon so that the second tRNA is now in the P site. This action has tRNA, and protein was the result of numerous scientists working on released the methionine-carrying tRNA from the ribosome and allowed a third tRNA to separate pieces of the puzzle. enter the empty A site. The process is similar to a tickertape running through a ticker- Watch an online animation of their tape machine, except that the ribosome “machine” moves along the mRNA “tickertape.” studies. The tRNAs that have been released are recycled in the cell cytoplasm by adding new www.science.nelson.com GO amino acids to them. The process continues until the entire code of the mRNA has been translated and the ribosome reaches a stop codon, as shown in Figures 9 (d) and (e). 672 Chapter 20 NEL Ch 20_Bio_Alberta30 1/8/07 12:40 PM Page 673 Section 20.2 (a) (b) tRNA Ile tRNA Met P site A site Ala UAU Met 3
UAC 5
5
AGG AGG AUG GCA AUA 3
mRNA P site A site tRNA 3
UAC CGU 5
5
AGG AGG AUG GCA AUA 3
mRNA ribosome shifts (c) UAC (d) Ile Arg Met Ala Met Ala Ile peptide bond P site A site P site A site 3
CGU UAU 5
3
UAU UUG 5
5
AGG AGG AUG GCA AUA 3
mRNA 5
AGG AGG AUG GCA AUA AAC CUA 3
mRNA ribosome shifts ribosome shifts (e) (f) tRNA polypeptide chain polypeptide chain 3
UGU 5
Val Thr Ser Asn P A site site large subunit of ribosome P site A site GCU small subunit of ribosome 5
GUU ACU AGU CGA UAG 3
mRNA stop codon Figure 9 5
AGU CGA UAG 3
mRNA Protein synthesis (a) The first tRNA that is brought into the P site carries methionine because the start codon is AUG. (b) The second tRNA enters the A site. (c) A peptide bond forms between methionine and alanine. The ribosome shifts one codon over and the next tRNA brings in the appropriate amino acid into the A site. (d) The ribosome moves the mRNA and another amino acid is added to the chain. (e) The process is repeated until the ribosome reaches a stop codon for which no tRNA exists. (f) A release-factor protein helps break apart the ribosome–mRNA complex, releasing the polypeptide chain. NEL Molecular Genetics 673 Ch 20_Bio_Alberta30 1/8/07 12:40 PM Page 674 Termination Eventually, the ribosome reaches one of the three stop codons: UGA, UAG, or UAA. Since these three codons do not code for an amino acid, there are no corresponding tRNAs. A protein known as a release factor recognizes that the ribosome has stalled and helps release the polypeptide chain from the ribosome. As shown in Figure 9 (f), on the previous page, the two subunits of the ribosome now fall off the mRNA and transla- tion stops. LAB EXERCISE 20.A Report Checklist Purpose Design Analysis Synthesis of a Protein Problem Materials Evaluation Hypothesis Procedure Synthesis In this activity, you are provided with a DNA nucleotide Prediction Evidence sequence that codes for a hypothetical protein. The code is given in three fragments. This DNA code is from a eukary- 4. Identify the middle, end, and beginning sequence. otic cell so in the mRNA transcript there are extra codons Use your knowledge of start and stop codons to help called introns. Eukaryotic cells cut these sequences out of the you figure it out. (Hint: You will need to examine the mRNA before it leaves the nucleus, so the codons are tran- codons that start and end a fragment.) scribed but are not translated. 5. Remove codons 24 to 51, including codon 51. These In this exercise, you will transcribe the three pieces of DNA codons are the intron, or extra codons, found in this code into mRNA and identify the beginning fragment, the DNA segment. middle fragment, and the end fragment. In addition, you will remove the intron segment and translate the mRNA into the 6. Translate the mRNA into protein using the genetic protein. code. Procedure Analysis 1. Copy each of the following sequences onto a separate (a) Which fragment was the beginning fragment? piece of paper. (Hint: Turn your paper so you can How do you know? write the sequence out along the horizontal length of (b) Which fragment was the end fragment? How do you the paper. Leave room below each sequence to write know? your mRNA sequence directly below.) (c) Codons 24 to 51 represent an intron. If the introns were not cut out of the mRNA before it leaves the Sequence A nucleus and attaches to a ribosome, what would CTCGCGCCGAAACTTCCCTCCTAAACGTTCAAC happen to the protein structure? Is it likely that this CGGTTCTTAATCCGCCGCCAGGGCCCC protein would still perform the same function? Explain your answer. Sequence B (d) How many amino acids does this protein contain? CGTAACAACTTGTTACAACATGGTCATAAACGTCA GATGGTCAATCTCTTAATGACT (e) Is this genetic sequence eukaryotic or prokaryotic? How do you know? Sequence C (f) If you worked backward, starting with the amino TACAAACATGTAAACACACCCTCAGTGGACCAA acid sequence of the protein, would you obtain the CTCCGCAACATAAACCAAACACCG same DNA nucleotide sequence? Why or why not? 2. Divide the sequences into triplets (codons) by putting (g) Provide the anticodon sequence that would build this a slash between each group of three bases. protein. 3. Transcribe the DNA into mRNA. 674 Chapter 20 NEL Ch 20_Bio_Alberta30 1/8/07 12:40 PM Page 675 Section 20.2 INVESTIGATION 20.1 Introduction Report Checklist Protein Synthesis and Inactivation of Purpose Design Analysis Problem Materials Evaluation Antibiotics Hypothesis Procedure Synthesis Each protein has a specific function. Its presence or absence in a Prediction Evidence cell may make the difference between life and death. Bacteria that carry an ampicillin-resistance gene produce a protein that grown on ampicillin-rich media? This investigation allows you to inactivates the antibiotic ampicillin. What happens when they are observe the effects of the presence and function of a specific gene. To perform this investigation, turn to page 695. SUMMARY Gene Expression Table 3 Summary of Transcription Initiation Initiation of transcription starts when the RNA polymerase binds to the promoter region of the gene to be transcribed. The DNA is unwound and the double helix is disrupted. Elongation A complementary messenger RNA (mRNA) molecule is synthesized in the 5
to 3
direction, using one strand of DNA as a template. Adenine (A) bases in the DNA are paired with uracil (U) in the mRNA. Transcription continues until the RNA polymerase reaches a termination sequence. Termination When the RNA polymerase comes to a termination sequence, it falls off the DNA molecule. The mRNA separates from the DNA. DNA template 3
5
strand C A A C G G T T T G G A transcription G U U G C C A A A C C U mRNA 5
3
codon translation polypeptide valine alanine lysine proline Figure 10 amino acid An overview of gene expression NEL Molecular Genetics 675 Ch 20_Bio_Alberta30 1/8/07 12:40 PM Page 676 Table 4 Summary of Translation Initiation Ribosome subunits (large and small) bind to the mRNA transcript, sandwiching the mRNA between them. The ribosome moves along the mRNA, reading the codons. Translation begins when the ribosome reaches the start codon, AUG. Elongation Through the genetic code, each codon specifies a particular one of the 20 amino acids that make up polypeptides. Transfer RNA (tRNA) molecules have an anticodon that is complementary to the codon in the mRNA. The tRNA carries the amino acid specified by the codon. The ribosome contains two sites, the A (aminoacyl) site and the P (peptidyl) site. When the start codon is in the P site, the first tRNA delivers methionine. Since the start codon codes for methionine, all polypeptides initially start with this amino acid. The second codon of the mRNA is exposed at the same time in the A site. When the tRNA delivers the second amino acid, a peptide bond is formed between the two amino acids. The ribosome shifts over one codon. The tRNA that delivered the methionine is released from the P site. When the ribosome shifts, the tRNA containing the growing polypeptide moves to the P site. A third amino acid, specified by the third codon, is brought in to the A site by the next tRNA. A peptide bond is formed between the second and third amino acid. Amino acids continue to be added to the polypeptide until a stop codon is read in the A site. Termination The stop codons are UAG, UGA, and UAA. At this point the ribosome stalls. A protein known as the release factor recognizes that the ribosome has stalled and causes the ribosome subunits to disassemble, releasing the mRNA and newly formed polypeptide. Section 20.2 Questions 1. State the central dogma of molecular genetics. 8. Construct a table to compare the processes of replication 2. Describe the role of the following molecules in gene and transcription. Remember to consider both similarities expression: ribosomes, mRNA, tRNA. and differences. 3. The genetic code is read in groups of three nucleotides 9. Distinguish between the following terms: called codons. Explain why reading the code in pairs of (a) P site and A site nucleotides is not sufficient. (b) codon and anticodon (c) start and stop codon 4. The following is the sequence of a fragment of DNA: (d) DNA and RNA GGATCAGGTCCATAC 10. Identify which of the following selections correctly lists the Transcribe this sequence into mRNA. anticodons for the amino acids threonine, alanine, and 5. Using the genetic code, decipher the following mRNA proline: sequence: A. ACU GCU CCA 5
- AUGGGACAUUAUUUUGCCCGUUGUGGU - 3
B. ACT GCT CCA C. TGA CGA GGT 6. The amino acid sequence for a certain peptide is D. UGA CGA GGU Leu–Tyr–Arg–Trp–Ser. How many nucleotides are necessary in the DNA to code for this peptide? 11. Errors are occasionally made during the process of transcription. Explain why these errors do not always result 7. Identify which step in transcription would be affected and in an incorrect sequence of amino acids. Describe at least predict what would happen in each situation: two examples to illustrate your answer. (a) The termination sequence of a gene is removed. (b) RNA polymerase fails to recognize the promoter. 676 Chapter 20 NEL  DNA and Biotechnology 20.3 Carpenters require tools such as hammers, screwdrivers, and saws, and surgeons require scalpels, forceps, and stitching needles. The tools of the molecular biologist are living biological organisms or biological molecules. Using these tools, scientists can treat spe- cific DNA sequences as modules and move them from one DNA molecule to another, forming recombinant DNA. Research in exploring and using this type of biotechnology recombinant DNA fragment has led to exciting new advances in biological, agricultural, and medical technology. of DNA composed of sequences Biotechnology research has also found ways to introduce specific DNA sequences into originating from at least two different sources a living cell. For example, the gene that encodes insulin has been introduced into bac- terial cells so that they become living factories producing this vital hormone. The intro- duction and expression of foreign DNA in an organism is called genetic transformation. genetic transformation In this section, you will explore some of the key tools used by molecular geneticists in introduction and expression of producing recombinant DNA and genetically transformed organisms. foreign DNA in a living organism DNA Sequencing Before a DNA sequence can be used to make recombinant DNA or to transform an organism, the scientist or technician must first identify and isolate a piece of DNA con- taining that sequence. One of the tools used to do this is DNA sequencing. DNA sequencing determines the exact sequence of base pairs for a particular DNA fragment or molecule. In 1975, the first DNA sequencing techniques were simultaneously devel- G oped by Frederick Sanger and his colleagues and by Alan Maxim and Walter Gilbert. C Sanger’s technique relied on first replicating short segments of DNA that terminate due T to a chain-terminating nucleotide. Four separate reaction tubes are run, each with a A A chain-terminating nucleotide incorporating a different base (i.e., A, T, G, and C). The var- T ious lengths of DNA segments are then separated by loading and running the contents G of the tubes on a sequencing gel (Figure 1). Because the end nucleotide of each segment C A is chain-terminating, its base is already known. Consequently, the sequence can be read T directly from the gel in ascending order (shortest to longest segments). The sequence G of the strand is written along the edge of the gel diagram, starting from the bottom C A where the shortest strands have travelled. This method is comparatively slow and can only T sequence short fragments of DNA. T DNA can also be sequenced in a test tube using isolated segments of DNA. This tech- Figure 1 nique depends on a primer, DNA polymerase, and the four DNA nucleotides, each of A sequencing gel is a matrix which is labelled with a specific dye. The complementary strand is built from these dye- containing many small spaces. The labelled nucleotides. The nucleotides in the synthesized strand can then be identified DNA fragments are charged and by their colours, allowing the original strand sequence to be deduced according to will move towards one pole of an the rules of complementary base pairing. electric field. Smaller DNA fragments move through the spaces more quickly than larger fragments and are found at the bottom of the WEB Activity gel. The larger fragments will remain towards the top of the gel. The resulting ladder of fragments Simulation—Electrophoresis can be read, giving the sequence of Electrophoresis is an important tool in molecular biology. In addition to nucleic acids, it is also used the initial DNA fragment. to separate proteins from a mixture. Electrophoresis of nucleic acids and proteins depend on the similar factors. In this Virtual Biology Lab, you will perform polyacrylamide gel electrophoresis (PAGE) to identify proteins involved in the biochemistry of shell colour in an extinct organism. www.science.nelson.com GO NEL Molecular Genetics 677 Ch 20_Bio_Alberta30 1/8/07 12:40 PM Page 678 WEB Activity Canadian Achievers—Researchers in Human Genetic Disorders Advances in biotechnology have led a greater understanding of many human genetic disorders. These advances have involved many research teams working together, either directly or by publishing their work in peer-reviewed articles. The following list shows some Canadians who are among the researchers making important contributions: Dr. Michael Hayden, University of British Columbia: Huntington disease Dr. Lap-Chee Tsui, Hospital for Sick Children, Toronto: cystic fibrosis Dr. Judith Hall, University of British Columbia: cystic fibrosis Dr. Christine Bear, University of Toronto: cystic fibrosis Dr. Ron Warton, Hospital for Sick Children, Toronto: Duchenne muscular dystrophy Go to the Nelson Web site to find more information on the work of these people. After you Figure 2 have completed reading this material, write a short paragraph that describes your view on the Dr. Judith Hall importance of genetic research. Defend your position. www.science.nelson.com GO DID YOU KNOW ? The Human Genome Project In a series of meetings held in the mid-1980s, plans were developed to begin the process Genome Facts On February 15, 2001, scientists of producing maps of the entire genetic makeup of a human being. The international from the Human Genome Project project began in the United States in October 1990 with James Watson, of Nobel Prize and Celera Genomics confirmed fame, as one of the first directors. The human genome consists of approximately that there were approximately 30 000 genes, with the 23 pairs of chromosomes containing an estimated 3 billion pairs 30 000 genes in the human of nucleotides. Constructing the genome map involved using mapping techniques (sim- genome—a number far less than ilar to those you read about in Chapter 19) and DNA sequencing technology. When the the original estimate of 120 000. This was determined using two project began, only about 4500 genes had been identified and sequenced. The collabo- different DNA sequencing rative efforts of many scientists from numerous countries and rapid improvements in techniques. sequencing techniques helped complete the gene map by June 2000 (Table 1). Other Facts 99.9 % of the nitrogenous base Table 1 Milestones in Genome Mapping sequences is the same in all Milestone Date humans. Only 5 % of the genes contains human chromosome 22 completed (the first chromosome December 1999 the instructions for producing to be mapped) functional proteins; the remaining Drosophila genome completed March 2000 95 % does not have any known function. human chromosomes 5, 16, 19 mapped April 2000 A worm has approximately 18 000 human chromosome 21 completely mapped May 2000 genes; a yeast cell has about human genome completely mapped June 2000 6000. A DNA sequencing technique based on the one developed by Sanger was the most common method used in the project. In this technique, pieces of DNA are replicated and changed so that the fragments, each ending with one of the four nucleotides, can be detected by a laser. Automated equipment can then determine the exact number of nucleotides in the chain. A computer is used to combine the huge amount of data and reconstruct the original DNA sequence. Prior to the Human Genome Project, the genes for hereditary disorders such as cystic fibrosis, muscular dystrophy, and Huntington disease had been identified. The aim of the project is to add to this list so that new drugs and genetic therapies can be developed to 678 Chapter 20 NEL Ch 20_Bio_Alberta30 1/8/07 12:40 PM Page 679 Section 20.3 mini Investigation Examining the Human Genome In this activity, you will go to an online map of the human Go to the Nelson Web site, and follow the link for Mini genome. On the map, you will find diagrams containing Investigation: Examining the Human Genome. On the genome information about every chromosome in the genome. The map, click on each chromosome diagram to discover the traits magenta and green regions on the diagrams reflect the unique and disorders located on that chromosome. For example, patterns of light and dark bands seen on human chromosomes Figure 3 shows traits and disorders that are found on that have been stained for viewing through a light microscope. chromosome 20. The red region represents the centromere or constricted Touch each chromosome pair to find the number of genes portion of the chromosome. On other chromosome diagrams, mapped on that chromosome. you will see yellow regions that mark chromosomal areas that Use the information you find to answer the questions below. vary in staining intensity. The chromatin in these areas is (a) Which chromosome pair contains the greatest number of condensed and sometimes known as heterochromatin, genes? meaning “different colour.” Some diagrams have yellow regions (b) Which chromosome contains the fewest genes? overlaid by thin horizontal magenta lines. This colour pattern indicates variable regions called stalks that connect very small (c) Estimate the size of the human genome. Show how you chromosome arms (satellit