Chemistry 301 Introduction to Biochemistry

Study Guide :: UNIT 5

Flow of Genetic Information

Overview

Of all areas of biochemistry, deoxyribonucleic acids (DNA) and ribonucleic acids (RNA) generate the most research excitement. The study of DNA and RNA, commonly called “molecular genetics,” forms the basis of genetics and explains the production of enzymes. Enzymes control all biological processes, so the study of DNA and RNA brings us as close to understanding life as chemistry allows.

In Unit 3, we introduced the structure of nucleic acids. In this unit we describe how DNA and RNA work, emphasizing the three‑dimensional structure of these molecules, and discussing the importance of the three‑dimensional structures in DNA and RNA function. We also describe how DNA and RNA synthesis is controlled, and how DNA, in particular, is protected from chemical reaction. Finally, we explain the differences between DNA and RNA, and the biochemical consequences of these differences. We then go on to discuss protein synthesis, which is under the direction of RNA. In translation, the information in DNA is finally converted to a protein.

Unit 5 is divided into six lessons:

Learning Objectives

After completing this unit, you should be able to:

• Explain the chemical differences between DNA and RNA.
• Describe DNA replication.
• Explain the major differences and similarities between the synthesis of DNA and that of RNA.
• Use the genetic code to translate mRNA sequences into (protein) amino acid sequences.
• Explain the difference between a “codon” and an “anti‑codon.”
• Describe the role of the ribosome and tRNA in protein synthesis.
• Explain how eukaryotic protein synthesis is controlled.
• Outline the present biochemical knowledge of the molecular basis of cancer.

Glossary

Lesson 1: DNA Replication

Overview/Objectives

After completing this lesson, you should be able to:

1. Explain the steps involved in DNA replication.
2. Describe the differences and similarities in replication of the two opposing DNA strands in the double helix.
3.  Name the major proteins involved in DNA replication, and describe their functions.

Readings and Activities

1. Read “DNA Replication” (pages 102–111 of the textbook).
2. You can also watch four video lectures on DNA Replication and Repair:
   • #41 Biochemistry DNA Replication I Lecture for Kevin Ahern’s BB 451/551
   • #42 Biochemistry DNA Replication II Lecture for Kevin Ahern’s BB 451/551
   • #43 Biochemistry DNA Replication III Lecture for Kevin Ahern’s BB 451/551

   • #45 Biochemistry DNA Repair / Transcription I Lecture for Kevin Ahern’s BB 451/551

     (Links to these are also provided on pages 105, 108, and 110 of the textbook.)

Commentary

DNA exists as a double stranded species. The strands are held together by hydrogen‑bonding between opposing nitrogenous bases and by van der Waals stacking forces among the bases. One of the advantages of double stranding is protection of the individual bases from chemical attack. This advantage is preserved during replication of DNA. A DNA duplex acts as the template to make a new duplex—each parent strand is replicated at the same time. Therefore, DNA remains double stranded at all times. At the precise point of replication, the replication enzymes themselves protect the (momentarily) single stranded DNA.

When DNA is replicated, each of the daughter strands of DNA contains one of the original strands and a newly synthesized strand. The process is called “semiconservative replication.”

Replication of DNA proceeds in one direction only:

5′ → 3′

Recall that primed numbers refers to the carbons of ribose. As DNA is normally drawn, the 5′ position is above the plane of ribose and the 3′ position is below. Refer to the picture of adenylic acid in Unit 3: C5 is phosphorylated; C3 contains only a hydroxyl group. An incoming deoxynucleotide will add to C3 end of the existing chain. Therefore the chain is said to grow 5′ → 3′.

The two strands of a DNA double helix are anti‑parallel. That is, the 5′ end of one chain is opposite the 3′ end of the other.

The arrows in the diagram refer to the direction of replication; that is, the new partner strands are synthesized in the 5′ to 3′ direction to complement the existing strands.

There seems to be a problem here: replication proceeds in one direction, but at no time during the replication process is there a free single strand of DNA. This problem is solved by continuous synthesis of the new partner to one parent strand, and discontinuous production of the new partner to the other strand.

Short pieces of the new partner to strand b (Okazaki fragments) above, will be synthesized in a 5′ → 3′ direction, and later joined together. Therefore, in any given area of the DNA duplex, replication is occurring for both strands. (Synthesis and joining of the Okazaki fragments will be slower than synthesis of the continuous strand, so the former is referred to as the “lagging strand.”)

Synthesis of DNA begins with a primer strand of RNA that is complementary to the beginning of DNA. This strand is later excised and replaced by DNA. This mechanism decreases the chance of incorrect bases in the new DNA strand.

Examine the figure, on page 107 in the textbook, of the replication fork and the associated proteins involved in DNA replication.

The proteins involved in DNA synthesis are:

• gyrase, which unwinds DNA supercoils in an ATP‑dependent process.
• rep protein and helicase II, which unwind the double helix at the replication site.

They bind, respectively, to the leading and lagging strands of DNA. These proteins travel along the parent DNA as replication proceeds.

Unwinding the duplex and moving along it are processes that require ATP, including:

• single‑stranded binding proteins, which bind to and stabilize the parent strands of DNA as replication occurs.
• DNA polymerase III (Pol III), which catalyses the reaction of a complementary 5′‑triphosphate and the growing DNA chain.

The figure on page 108 of the textbook demonstrates the activity of DNA Pol III. DNA Pol II catalyzes the following reaction:

deoxynucleoside triphosphate + DNA_n ⇌ diphosphate + DNA_n+1

The reaction is driven in a forward direction by the highly exothermic hydrolysis of pyrophosphate to two phosphate groups:

• Pol I, which replaces the RNA primer strands with DNA. See the outline of different types and properties of DNA polymerases on the Wikipedia page DNA polymerase.
• DNA ligase, which seals the Okazaki fragments after the RNA primer on each has been replaced by DNA.

Study Questions

1. The synthesis of DNA starts with a primer strand of RNA, which is later excised and replaced with DNA. Why does this mechanism decrease the chance of error?
2. What is a DNA supercoil? What function or functions does supercoiling perform?
3. Before DNA polymerase can add nucleotides to replicate DNA, a number of steps must occur that involve a number of proteins. List these steps and the proteins involved.
4. Distinguish between the “leading” and the “lagging” strand in DNA replication.

If you wish to discuss any of these questions or need assistance with the material, please contact your Academic Expert (AE) by emailing the Student Success Centre at fst_success@athabascau.ca.

Lesson 2: DNA Repair

Overview/Objectives

After completing this lesson, you should be able to:

1. Describe how the chemical strategy of DNA replication reduces errors in its synthesis, and explain the features of DNA polymerase activity that increase replication activity.
2. Describe the major mechanisms of DNA repair in vivo.
3. Explain why repair of DNA is much more important than repair of RNA.
4. Explain why uracil is not used in DNA.

Readings and Activities

1. Read “Maintaining the Integrity of the Cell’s Information: DNA Repair,” “Post‑Replicative Mismatch Repair,” “Systems to Repair Damage to DNA,” “Nucleotide Excision Repair (NER),” and “Base Excision Repair (BER),” (pages 111–117 of the textbook).
2. You can also watch the four video lectures on DNA Replication and Repair:
   • #41 Biochemistry DNA Replication I Lecture for Kevin Ahern’s BB 451/551
   • #42 Biochemistry DNA Replication II Lecture for Kevin Ahern’s BB 451/551
   • #43 Biochemistry DNA Replication III Lecture for Kevin Ahern’s BB 451/551

   • #45 Biochemistry DNA Repair / Transcription I Lecture for Kevin Ahern’s BB 451/551

     (Links to these were provided in Lesson 1, are also provided on pages 102, 105, 108, and 110 of the textbook.)

Commentary

Pol I, II, and III all share an important 5′ → 3′ exonuclease function that edits out mistakes, meaning that these polymerases can also cleave a base after it has been added. If an incorrect base is added, inefficient hydrogen‑bonding and stacking will occur, and the base can be easily cleaved by the polymerases. On the other hand, if the newly added base is complementary to the parent strand of DNA, this base will hydrogen‑bond and stack efficiently. Therefore, it will be difficult to cleave.

Repair processes exist for DNA because mature DNA can be modified, mutations can be induced after replication, or both. The repair enzymes are:

• alkyltransferases, which remove alkyl groups incorrectly esterified to reactive hydroxyl groups of the nitrogenous base
• photoylase, which cleaves pyrimidine dimers. (Pyrimidine dimers are caused by UV light.)
• excision repair enzymes, which recognize and remove incorrect bases from an intact daughter strand. (Of particular importance is uracil‑N‑glycosylase.)
• recombination repair processes, which recognize and repair certain gaps in DNA. These gaps may occur when a chemically modified daughter strand replicates before the three processes listed above can occur. (That is, the gaps are in the new, second‑generation DNA.)
• SOS response, which is called up when there is a great deal of DNA damage. This is a complex response by the cell, in which several proteins are synthesized to help repair the damage. Under normal conditions, these SOS proteins are not synthesized; their expression is repressed by a protein called LexA. LexA is cleaved when DNA damage is extensive, thereby allowing the synthesis of the SOS proteins. This repair process, as might be expected, is error prone.

Study Questions

1. Suppose an incorrect base was inserted into a daughter strand of DNA and this error was not corrected. Now suppose the daughter strand replicates. Can the error be corrected at this stage?
2. Why is error control less critical for RNA than for DNA?
3. Does proofreading by DNA polymerase eliminate all replication errors?
4. How are errors recognized by the mut genes in mismatch repair?
5. What are some causes of DNA damage? What repair system is used to fix this damage and what are the two types of repair?

If you wish to discuss any of these questions or need assistance with the material, please contact your Academic Expert (AE) by emailing the Student Success Centre at fst_success@athabascau.ca.

Lesson 3: Transcription

Overview/Objectives

After completing this lesson, you should be able to:

1. Name the steps involved in RNA synthesis.
2. Describe the major differences between, and similarities in DNA and RNA synthesis.
3. Explain how RNA synthesis is initiated and terminated in vivo.

Readings and Activities

1. Read “Transcription” (pages 117–124 of the textbook).
2. You can also watch three video lectures on Transcription:
   • #45 Biochemistry DNA Repair / Transcription I Lecture for Kevin Ahern’s BB 451/551
   • #46 Biochemistry Transcription II Lecture for Kevin Ahern’s BB 451/551

   • #47 Biochemistry Transcription III / Translation I Lecture for Kevin Ahern’s BB 451/551

     (Links to these are also provided on pages 117, 121, and 124 of the textbook.)

Commentary

The actual synthesis of RNA from a DNA template is simpler than the synthesis of DNA, because there are fewer quality controls and virtually no repair processes in RNA synthesis.

NTP + RNA_n ⇌ diphosphate + RNA_n+1

NTP is a ribonucleotide (ATP, GTP, UTP, or CTP); “n” refers to the number of ribonucleotides.

RNA polymerase catalyses this reaction, which is driven in a forward direction by the hydrolysis of pyrophosphate. The RNA strand is synthesized in a 5′ → 3′ direction. RNA polymerase transcribes only one side of a DNA duplex in any given area of the chromosome. However, the codes for different proteins may not all be on the same strand.

Note the figures on pages 121 and 123 of the text and on the Wikipedia page Transcription (genetics).

RNA polymerase from E. coli contains five subunits: α₂ββ′σ. This protein is called the “holoenzyme.”

The four major steps in RNA synthesis are:

1. Recognition: RNA polymerase binds to a specific site on DNA in search of an initiation site. This site on DNA is called the “promoter region.” All genes contain a promoter region “upstream” of the base sequences, which are actually transcribed into RNA.

   The promoter regions of various genes are shown at the top of page 120 of the textbook. Promoter sequences elements (upstream region, TATA box, and transcription start site) are shown near the bottom of page 120.

   Note that two short nucleotide sequences within the promoter region on DNA are conserved: the Pribnow box (or TATA box) and another 9–12 nucleotide stretch farther away from the initiation site. Both of these nucleotide stretches are on the same side of the double helix. The more tightly RNA polymerase binds to a specific promoter region, the greater the RNA transcription of that gene.

2. Initiation: Two nucleoside triphosphates, complementary to the first two DNA bases in the gene itself, are coupled by RNA polymerase. At this point, the s protein dissociates to give “core” RNA polymerase: α2ββ′σ.
3. Elongation: The core enzyme moves rapidly along DNA synthesizing RNA exactly complementary to the DNA. Because RNA polymerase does not have an exonuclease (editing) function, the error rate in RNA synthesis is considerably higher than it is for DNA synthesis.
4. Termination: When the core enzyme reaches a stretch of DNA rich in adenine residues, it dissociates from DNA, thereby terminating RNA synthesis. Frequently another protein, the Rho factor, aids in dissociating RNA polymerase from DNA.

Study Questions

1. Is any one σ‑factor unique to one RNA polymerase holoenzyme?
2. Outline the steps involved in transcription (mRNA synthesis).

If you wish to discuss any of these questions or need assistance with the material, please contact your Academic Expert (AE) by emailing the Student Success Centre at fst_success@athabascau.ca.

Lesson 4: Regulation of Transcription

Overview/Objectives

After completing this lesson, you should be able to:

1. Describe how transcription is controlled in vivo.
2. Define the terms “enhancer”, “silencer”, and “repressor.”
3. Demonstrate an understanding of how transcription is regulated in the lac operon.

Readings and Activities

1. Read “Regulation of Transcription” (pages 124–129 of the textbook).
2. You can also watch the video lectures on Transcription Regulation:
   • #49 Biochemistry Translation III/Gene Expression I Lecture for Kevin Ahern’s BB 451/551
   • #50 Biochemistry Gene Expression II Lecture for Kevin Ahern’s BB 451/551

   • #51 Biochemistry Gene Expression III Lecture for Kevin Ahern’s BB 451/551

     (Links to these are also provided on pages 126 and 128 of the textbook.)

Commentary

Control of RNA transcription depends on the degree of binding of the RNA polymerase holoenzyme to a promoter region of DNA, but other factors, outlined below, also help to control RNA transcription.

• Repressors: A repressor protein is sometimes synthesized by an organism. The repressor binds to a DNA promoter region to prevent RNA polymerase from binding. Addition of an exogenous inducer dislodges the repressor protein, and so the gene can be transcribed. This is the control mechanism of the lac operon—the gene specifying the enzymes for lactose metabolism.
• Gene activation: Just as repressors can decrease gene expression, so there are positive regulators which enhance gene transcription. A positive regulator may interact directly with RNA polymerase (enhancing its activity); or it may bind directly to DNA, enhancing the ability of RNA polymerase to bind.
• Stringent response: There may be several genes that code for the same RNA. This is the case for rRNA: there are seven rRNA operons. When massive synthesis of rRNA is required, all the rRNA operons can be activated.

Study Questions

1. Why is it necessary to have repressor proteins to prevent RNA polymerase binding to DNA?
2. How does the presence of lactose affect the lac operon?
3. What are enhancers and silencers? Are these typically eukaryotic or prokaryotic?

If you wish to discuss any of these questions or need assistance with the material, please contact your Academic Expert (AE) by emailing the Student Success Centre at fst_success@athabascau.ca.

Lesson 5: RNA Processing

Overview/Objectives

After completing this lesson, you should be able to:

1. Describe the post‑transcriptional processing of mRNA, tRNA, and rRNA in eukaryotes.
2. Explain why mRNA in prokaryotes is not processed.
3. Define the term “ribozyme,” and explain the role of ribozymes in RNA processing.

Readings and Activities

1. Read “RNA Processing” (pages 129–132 of the textbook).

Commentary

RNA molecules may be chemically modified before they become functional. This modification occurs after the RNA has been synthesized. Common modifications are cleavage of nucleotide sequences at one or both ends of the RNA molecule or in the middle of the RNA sequence, splicing of RNA from which a middle sequence has been removed, addition of nucleotide sequences to the 3′ or 5′ end, and methylation of selected adenylate residues.

Post‑transcriptional modification of eukaryotic RNA is more common than modification of prokaryotic RNA. Because eukaryotes have a nuclear membrane, synthesis of RNA (which occurs in the nucleus) is complete before the RNA molecule migrates to the cytosol. Modification takes place in the nucleus. In prokaryotes, which lack a nuclear membrane, protein synthesis can occur at the same time as RNA is being transcribed. Thus, post‑transcriptional modification of mRNA is unlikely.

Until very recently it was thought that only proteins could be enzymes. However, RNase P, an RNA‑protein complex, has been shown to be catalytically active even when its protein segment is missing. This intriguing enzyme cleaves RNA to generate the 5′ ends of all E. coli tRNAs. In addition, a precursor rRNA in Tetrahymena self‑splices. The liberated piece of RNA then catalyzes the transformation of other RNA molecules. These are two examples of true enzymatic activity by RNA.

It is generally believed that RNA preceded DNA in evolutionary time. The existence of ribozymes may have allowed RNA to catalyze its own replication in this early world. In addition, the primitive organism may have developed various catalytic functions within its store of RNA to give the organism an edge over competing organisms. Therefore, the ribozymes may be holdovers from the DNA‑lacking days. Alternatively, ribozymes may be a whole new class of functional molecules that have evolved to meet specific challenges. Questions like this make biochemical research a challenging field.

Study Questions

1. Why do eukaryotic cells process mRNA?
2. What are the three processing steps for mRNA?
3. What is the spliceosome? What are the two main steps in splicing?
4. What is the purpose of alternative splicing?

If you wish to discuss any of these questions or need assistance with the material, please contact your Academic Expert (AE) by emailing the Student Success Centre at fst_success@athabascau.ca.

Lesson 6: Translation

Overview/Objectives

Translation occurs in the cytosol of all cells. Because proteins are constantly being degraded, translation is a never‑ending process.

The three forms of RNA (mRNA, tRNA, and rRNA) participate in translation. The code for the protein is found in mRNA; rRNA in ribosomes provides the protein synthesis platform; tRNA molecules bring amino acids to the ribosomes. Translation is more complex than DNA replication or RNA transcription. This complexity occurs because, while replication and transcription use the common language of nitrogenous bases, translation involves two languages: nitrogenous bases and amino acids.

The genetic code is a set of ribonucleotide bases (on mRNA) which are recognized by tRNA molecules. That is, the genetic code is the link between a sequence of bases and a sequence of amino acids. The binding of appropriate tRNA molecules to mRNA brings amino acids close enough to be combined to make proteins.

After completing this lesson, you should be able to:

1. Explain the difference between a “codon” and an “anti‑codon.”
2. Use the genetic code to translate mRNA sequences into (protein) amino acid sequences.
3. Explain why the genetic code is said to be “degenerate.”
4. Define the term “reading frame.”
5. Identify the initation and stop codons.
6. Describe the role of the ribosome and tRNA in protein synthesis.
7. Explain how eukaryotic protein synthesis is controlled.
8. Outline the present biochemical knowledge of the molecular basis of cancer.
9. Describe a tRNA structure.
10. Illustrate the spatial relationship between the anti‑codon on tRNA and the amino acid which the tRNA molecule carries.
11. Explain the processes that ensure the correct amino acid is carried by the appropriate tRNA molecule.
12. Define the “wobble hypothesis,” and explain how it accounts for codon degeneracy.
13. Describe the basic structure of a ribosome.
14. List the major features of protein synthesis.
15. Describe how a ribosome, mRNA, and tRNA molecules interact during initiation, elongation, and termination of protein synthesis.
16. Outline three methods by which eukaryotic translation can be controlled.
17. List the major mechanisms involved in eukaryotic gene expression.
18. Describe the types of post‑translational modification of eukaryotic proteins.
19. Describe how cancer cells differ from normal cells.
20. Outline the differences between benign and malignant tumours.
21. Define an “oncogenes,” and explain how viral oncogenes can lead to cancerous growth of eukaryotic cells.
22. List the genetic alterations that may lead to cancer.

Readings and Activities

1. Read “Translation” (pages 132–139 of the textbook).
2. You can also watch three video lectures on Translation:
   • #47 Biochemistry Transcription III / Translation I Lecture for Kevin Ahern’s BB 451/551
   • #48 Biochemistry Translation II Lecture for Kevin Ahern’s BB 451/551

   • #49 Biochemistry Translation III/Gene Expression I Lecture for Kevin Ahern’s BB 451/551

     (Links to these are also provided on page 133 of the textbook.)

Commentary

The Genetic Code

A sequence of three ribonucleotides is the “unit” of the genetic code called a codon. There are four common ribonucleotides (U, C, A, G), so there are 64 possible codons (4 × 4 × 4). Indeed, all 64 codons are used in vivo: there are no nonsense codons. Since there are 20 amino acids and 64 codons, some amino acids must be specified by more than one codon. Thus, the genetic code is said to be “degenerate.”

For example, there is only one codon for trp: UGG; there are four codons for pro: CCU, CCC, CCA, and CCG. There are also three “stop‑making‑this‑protein” codons: UAA, UGA, UAG.

Note the pattern of ribonucleotides when more than one codon specifies an amino acid. The first two ribonucleotides are usually the same (e.g., ile = AUU, AUC, AUA). It is the third nucleotide that varies. This fact indicates that the primordial code may have been a 2‑nucleotide sequence rather than a 3‑nucleotide sequence. If this were the case, only 16 amino acids are possible (4 × 4), or 15 amino acids plus a STOP codon:

Indeed, in one of the oldest species of bacteria, Clostridia, there are only 13 different amino acids in the ferredoxin proteins.

Codons for the amino acids are three bases on mRNA, read in a 5′ → 3′ direction. Anti‑codons are three bases on tRNA, read in a 3′ → 5′ direction, which are complementary to the codons.

For example, to specify tyr (codon = UAC) in the protein:

The genetic code described in this lesson is a standard code but not a universal one. Certain mitochondria and ciliated protozoa possess genetic codes that are variations of the standard code. (Mitochondria contain genes and can synthesize some, but not all, of their own proteins. It is believed that mitochondria, in early evolutionary time, were free‑living organisms.) We noted above that the standard three‑base code may have evolved from a two‑base code. Mitochondria and ciliated protozoa branched off very early in eukaryotic evolution, and it is interesting to note that their code variations are usually in the direction of a simpler code. Genetically speaking, things may have been quite chaotic before the supremacy of the standard three‑base code was established.

Transfer RNA (tRNA)

There is at least one unique tRNA molecule for each common amino acid. All tRNA molecules have similar primary (and therefore secondary) structures. Each tRNA molecule is distinguished by:

• at least three separate areas of internal base‑pairing
• a distinctive three‑dimensional “L”‑shape
• a narrow (20–25Å) profile
• an appropriate amino acid at its 3′ terminus
• a 5′ phosphate group
• a triplet anti‑codon in a nucleotide loop distant from the attached amino acid
• modification of approximately 20% of its bases

Three important points about the three‑dimensional shape of a tRNA molecule should be highlighted. First, the anti‑codon and the amino acid are as distant from each other as possible. Second, the narrow shape of the tRNA‑amino acid complex allows two such complexes to bind to adjacent codons on mRNA. Third, the structural similarity of tRNA molecules allows them to fit efficiently when two are H$\ce{-}$bonded, side by side, to an mRNA molecule. In this configuration, tRNA molecules are sandwiched between mRNA and the developing protein.

An amino acid is attached to the 3′ terminus of a tRNA molecule by a specific aminoacyl‑tRNA synthetase. This is a critical step if the amino acid coded by the DNA is to end up in the protein. (If we consider how similar the three‑dimensional shapes of the tRNA molecules are, it is not just a critical step, but also a difficult one.)

The amino acid + tRNA reaction occurs in two steps, both catalyzed by the aminoacyl‑tRNA synthetase:

aa₁ + ATP → aminoacyl‑AMP + P_i

aminoacyl‑AMP + tRNA 1 → aminoacyl‑tRNA1 + AMP

The overall reaction, then, is:

aa₁ + tRNA₁ + ATP → aa₁‑tRNA₁ + AMP + PP_i

There are similarities among amino acids, just as there are among the tRNA molecules, although they are not as prominent. For example, phe and tyr differ by only one hydroxyl group, and leu and ile differ only in geometry. Three factors, however, ensure accurate attachment of an amino acid to the appropriate tRNA:

1. A two‑step process (described above) ensures greater accuracy than a one‑step process; correct geometries and charge distribution within the enzyme active site must be achieved twice for the same amino acid to get a product.
2. tRNA molecules are almost completely surrounded by the aminoacyl‑tRNA synthetase enzyme during enzymatic linking. This milieu provides many contact points between enzyme, amino acid, and tRNA for precise chemical identification.
3. Several aminoacyl‑tRNA synthetases possess an editing catalytic site. An inappropriate aminoacyl‑AMP (first step above) will be cleaved before the actual attachment → aa‑tRNA (second step above).

Ribosomes and Protein Synthesis

A ribosome is a very large rRNA‑protein complex, in which rRNA makes up ~66% of the ribosome mass. Composed of two unequal subunits, the ribosome is the platform for protein synthesis.

The anti‑codon of a tRNA molecule H$\ce{-}$bonds to a triplet code on an mRNA molecule. Because of the three‑dimensional structure of tRNA, a perfect fit between the anti‑codon and the codon is not possible. The third position (i.e., 3′ end of the codon and 5′ end of the anti‑codon) binds less strongly than the other two. This third position is called the “wobble position.” The consequence is that dissociation is easier, and thus protein synthesis can move more quickly.

Initiation  The initiation of protein synthesis involves the association of mRNA, the ribosome, a unique aa‑tRNA (fMet‑tRNA_f^Met), and three soluble proteins, called “initiation factors.” The three steps of the process are outlined below.

1. mRNA fits in a cleft between the two ribosome subunits. There is a purine‑rich region (the Shine‑Dalgarno sequence) on mRNA, near the 5′ end. This Shine‑Dalgarno sequence is complementary to one of the rRNA molecules in the ribosome, so mRNA is initially held on the ribosome by H$\ce{-}$bonding of RNA base pairs.
2. Translation begins approximately 25 nucleotides from the 5′ end of mRNA, at an AUG or GUG “start signal.” The first amino acid of a newly synthesized protein is formylmethionine (fMet). The initiation factors make possible the binding of the two ribosome subunits, mRNA, and fMet‑tRNA_f^Met.
3. There are two tRNA side‑by‑side binding sites on the ribosome: the P site and the A site. fMet‑tRNA_f^Met occupies the P site, leaving the A site vacant. The A site waits for the tRNA that holds the second amino acid for the new protein.

Elongation  The elongation cycle of protein synthesis begins when a tRNA molecule occupies the A site adjacent to fMet‑tRNA_f^Met. Which tRNA occupies the A site depends, of course, on the second codon on mRNA. Elongation can also be described as a three step process.

1. Protein elongation factors deliver the second (and subsequent) aa‑tRNA molecules to a ribosome. The elongation factors must dissociate before fMet binds to aa₂. During the several milliseconds this takes, an incorrect tRNA will leave mRNA before a peptide bond can form.

2. A peptidyl transferase catalyzes the formation of the bond between fMet and aa₂ by transferring fMet from its tRNA. The energy necessary to form the peptide bond is that stored earlier when the aa‑tRNA bond was formed.

   

3. Three things now happen so that the new dipeptide is in position to accept the next amino acid. First, the uncharged tRNA leaves the ribosome P site. Second, the dipeptide (attached to tRNA₂) moves from the A to the P site on the ribosome. Finally, mRNA moves three nucleotides along the ribosome groove.

   

Termination  Termination of protein synthesis occurs when mRNA has moved along the ribosome groove so that the mRNA STOP code or codes are at the A site. Protein release factors, rather than tRNAs, recognize the STOP codons. The three steps of termination are described below.

1. Peptidyl transferase catalyzes the release of the new protein from the final tRNA.
2. The uncharged, last used tRNA drops away, then mRNA is released from the ribosome.
3. Finally, the ribosome dissociates into its two subunits.

Control of Eukaryotic Transcription and Translation and Post‑translational Protein Modification

Prokaryotes are single‑celled organisms. Therefore, all the cells of a particular species will look pretty much alike, and will have the same relative concentrations of proteins. Eukaryotes, on the other hand, are multicellular organisms whose cells are not all the same. This lesson asks how a liver cell, for example, knows it is a liver cell and should produce liver‑specific proteins, while a brain cell produces brain‑specific proteins and not liver proteins.

This question does not yet have a complete answer. This lesson will start you thinking about the subject, rather than learning “the answer.” Two processes are certainly involved: control of transcription (the “messages” produced from the DNA), and control of translation (the product of proteins from mRNA). In eukaryotes, both processes are very complex, but control of transcription is more fundamentally involved in determining which proteins a cell will produce.

Prokaryote protein production is related almost completely to the rate of mRNA transcription: the more mRNA produced, the more protein is synthesized. Eukaryotes depend far more on control of translation to vary protein production (and therefore vary cell identity).

As eukaryotes are more difficult to study, the mechanisms controlling gene expression in eukaryotes are not as well understood as those in prokaryotes. The division time is slower, the growth conditions more stringent, and genetic manipulation is more demanding for eukaryotes.

Three methods of translational control in eukaryotes are:

• phosphorylation of one of the initiation factors, eIF‑2
• “masking” of mRNA by associated proteins
• activation of a pre‑existing RNase L to destroy mRNA.

Remember that the initiation factors are proteins that assist the assembly of ribosome, mRNA, and Met‑tRNA_f^Met. When eIF‑2 is phosphorylated, it is not as effective in initiating protein synthesis. Masked mRNA is bound by cytosol proteins, which prevent it from binding to ribosomes.

Eukaryotic control of gene expression (transcription) is an extremely complex process. Ultimately, it is the turning on or off of certain genes that dictates which proteins will be produced in a cell, and hence determines its identity.

The proteins coded by mRNA are not mature, functional proteins. Proteolytic cleavage, modification of amino acid R groups, and formation of precise disulphide bonds are necessary to convert the so‑called pro‑proteins to fully functional proteins. Proteolytic cleavage may be minor (e.g., removal of the initial fMet) or extensive (e.g., production of mature insulin). Chemical modification usually involves the more reactive R‑groups (the polar and ionic ones) and includes acetylation, hydroxylation, glycosylation, methylation, and phosphorylation.

Glycosylation is a special modification that increases the solubility and establishes the identity of membrane bound and extracellular proteins. It is the sugars attached to proteins that mark them as “self” or “other” for the immune system.

The Molecular Basis of Cancer

In a noncancerous or normal condition, all the cells of a multicellular organism are coordinated; development and senescence of these cells is controlled. In the adult human, most cells are in the quiescent phase of the cell cycle. The malignant state is characterized by:

• uncontrolled growth of cells
• loss of normal functioning by the malignant cell
• cellular immortality.

It is easier to culture malignant cells than it is to grow normal eukaryotic cells in culture. Normal cells are inhibited from further growth by contact with adjoining cells. Malignant cells lack contact inhibition. The first mammalian cells to be successfully grown in culture were the HeLa cells. These were taken from a malignant tumour from a woman whose name was Henrietta Lacks (not Helen Lane, as many authors have incorrectly stated).

The loss of normal function and the invasiveness of malignant cells are the features that make cancer a major health problem. A “malignant” tumour grows by invading healthy tissue, and by shedding daughter cells that can migrate and proliferate elsewhere in the body. A “benign” tumour grows by expansion and usually remains encapsulated. It does not invade healthy tissue. Not all tumours grow at the same rate. Those with the fastest doubling times (such as ovarian cancer) create the greatest health risk.

Damage to DNA or disruption of repair or replication processes are the basis of most cancers. Such damage or disruption may be caused by:

• free radicals (e.g., those found in cigarette smoke)
• chemicals (e.g., polycyclic hydrocarbons)
• radiation (e.g., sunlight or X‑rays)
• specific viruses (RNA viruses carrying oncogenes).

The viruses that cause cancer are called “retroviruses.” Relatively few human cancers are caused by retroviruses, yet much of what we know about normal stimuli to cell growth and about malignancy comes from the study of animal retroviruses. A retrovirus contains the sequence for an oncogene. The oncogene protein products seem to be modified or inappropriately expressed components of the normal system for growth regulation. To date ~40 viral and cellular oncogenes have been characterized.

The various DNA modifications that can give rise to malignancies are:

• point mutations yielding altered proteins
• modification of DNA control sequences enhancing protein production from oncogenes
• rearrangement of chromosomal segments (and consequent alteration of control sequences)
• gene amplification (many copies of an oncogene)
• insertion of a viral genome into a cellular chromosome
• loss or inactivation of oncogene control proteins (anti‑oncogenes).

Study Questions

1. The following fragment of an RNA transcript from a bacterium has been isolated. What polypeptide or polypeptides are encoded?

   5′‑AAAUGAGGACCG‑3′

2. Suppose you were told that your colleague had discovered a new mutant cytochrome c in which aspartate replaced lysine. How would you react?
3. Why would gly‑tRNA synthetase have less need of an editing function than the other amino acid‑tRNA synthetases?
4. A nonsense mutation is one that yields a STOP codon. Such a mutation could be overcome by an altered tRNA. What does this mean?
5. In the synthesis of a 100 residue protein, how many high‑energy phosphate bonds are needed? (Assume that not all the amino acids are available; none need be synthesized de novo.)
6. One set of bacteria is grown in heavy medium (which contains ¹³C and ¹⁵N) and another in normal medium (¹²C and ¹⁴N). Ribosomes from the two sets of bacteria are mixed in a medium in which cell‑free protein synthesis is occurring. Several hours later, the ribosomes from the mixed tube are separated by density gradient centrifugation. How many ribosome bands will be observed?
7. Which high‑energy bond provides the energy to drive DNA, RNA, and protein synthesis?
8. What biological advantage is given by post‑translational modification of proteins?
9. Malignant tumour cells have a higher dependence on glycolysis than normal cells. Why?
10. Why does a mutation in the p53 gene lead to cancer?

If you wish to discuss any of these questions or need assistance with the material, please contact your Academic Expert (AE) by emailing the Student Success Centre at fst_success@athabascau.ca.

Answers to Unit 5 Study Questions

Lesson 1: DNA Replication

1. The first few bases to be added during replication have much more steric mobility than bases added to a strand already in place. Therefore, incorrectly paired bases will “fit” better at the beginning of a strand than they will farther along the strand.
2. Circular DNA (the type found in bacteria) is twisted in one of two ways: the duplex axis is wound as if around a cylinder or the duplex axis is twisted around itself. This is called supercoiling. It gives DNA a more compact shape and provides additional protection to the nitrogenous bases.

3. Before DNA polymerase can add nucleotides to replicate DNA, a number of steps must occur first which involve a number of proteins. List these steps and the proteins involved.

   Special recognition proteins bind the origin of replication, a special sequence where DNA replication begins and help open up a region of single‑stranded DNA (ssDNA). Helicase unwinds the DNA helix to open up the DNA further. Topoisomerase (gyrase in E. coli) prevent topological stress due to helicase activity by cutting the DNA and allowing the strands to swivel and then join at the ends. Single‑stranded DNA binding proteins (SSB proteins) bind the ssDNA strands at the replication site to prevent them from annealing to each other. Primase adds short RNA primers to the 3′ to 5′ strand because DNA polymerase can only replicate from 5′ to 3′ directly. The clamp loader protein and the sliding clamp protein help in loading the DNA polymerase onto the DNA at the site of replication and help keep it attached to the DNA. DNA polymerase can then perform its function of adding nucleotides in a 5′ to 3′ direction that are complementary to the existing strand of DNA.

4. The leading strand is the 5′ to 3′ strand that is directly replicated in one long piece (continuously). The lagging strand is the 3′ to 5′ strand that primase adds nucleotides to. These are first composed of RNA. DNA polymerase must then come in and replace the RNA nucleotides with DNA nucleotides. These short pieces are called Okazaki fragments. These pieces are joined by DNA ligase.

   5′AAGTCCGTCATTAAGCTATGTACACCATGACC3′

   3′TTCAGGCAGTAATTCGATACATGTGGTACTGG5′

   ↓  Strands separate

   5′AAGTCCGTCATTAAGCTATGTACACCATGACC3′ parental 5′ to 3′

   3′UUCAGGCAGUAA UUCGAUACAUG UGGUACUGG3′ discontinuous synthesis primase

   5′AAGTCCGTCATTAAGCTATGTACACCATGACC3′ continuous synthesis DNA pol

   3′TTCAGGCAGTAATTCGATACATGTGGTACTGG5′ parental 3′ to 5′

   ↓  DNA polymerase replaces RNA
    nucleotides and DNA ligase joins
    Okazaki fragments

   5′AAGTCCGTCATTAAGCTATGTACACCATGACC3′ parental 5′ to 3′

   3′TTCAGGCAGTAATTCGATACATGTGGTACTGG5′ new DNA strand

   5′AAGTCCGTCATTAAGCTATGTACACCATGACC3′ new DNA strand

   3′TTCAGGCAGTAATTCGATACATGTGGTACTGG5′ parental 3′ to 5′

Lesson 2: DNA Repair

1. The error cannot be corrected when the daughter strand replicates. The second generation DNA duplex (i.e., the “incorrect” daughter strand plus the partner strand synthesized to complement it) appear perfectly normal to the repair processes, because the base pairing is correct.
2. RNA is a throwaway molecule. It is synthesized on demand. Errors in one mRNA molecule will create defective proteins (or no proteins at all). However, subsequent error‑free RNA molecules will correct the problem. Permanent errors in replication (i.e., mutation) are passed on to subsequent generations of cells and organism.
3. No, the proofreading mechanism of DNA polymerase does not catch all errors in replication. These errors are fixed post‑replication by mismatch repair.
4. E. coli methylates DNA strands. New DNA strands are not yet methylated, so the mismatch repair system can distinguish between the old and new strands of DNA. The mut genes detect the new strand and excise the error. DNA polymerase and DNA ligase fill in the correct bases and join the strand.
5. Some causes of DNA damage are: radiation, exposure to dangerous chemicals, and chemical reactions in the cell. The general repair system to fix these errors is called excision repair and there are two types: nucleotide excision repair (NER), for UV damage and chemicals; and base excision repair (BER), for the removal of uracil bases resulting from the deamination of cytosine to uracil.

Lesson 3: Transcription

1. No σ factor is unique to one RNA polymerase holoenzyme. The σ‑factor dissociates from the holoenzyme after binding to DNA. The factor can then be picked up by any other core RNA polymerase.
2. The steps in transcription are:
   1. Recognition: RNA polymerase binds to a specific site on DNA in search of at the promoter region. The more tightly RNA polymerase binds to a specific promoter region, the greater the RNA transcription of that gene.
   2. Initiation: Two nucleoside triphosphates, complementary to the first two DNA bases in the gene itself, are coupled by RNA polymerase. The bases used in mRNA synthesis are A, U, G, and C.
   3. Elongation: The core enzyme moves rapidly along DNA synthesizing RNA exactly complementary to the DNA. Because RNA polymerase does not have an exonuclease (editing) function, the error rate in RNA synthesis is considerably higher than it is for DNA synthesis.
   4. Termination: When the core enzyme reaches a stretch of DNA rich in adenine residues, it dissociates from DNA, thereby terminating RNA synthesis. The rho factor aids in dissociating RNA polymerase from DNA.

Lesson 4: Regulation of Transcription

1. Some proteins (e.g., the enzymes for synthesizing lactose) are only required at infrequent intervals. (Lactose is milk sugar, it is only synthesized when a female is lactating.) Repressor proteins keep a gene turned off until a positive factor, which indicates a need for the gene’s protein product, is present.
2. When lactose is present, it binds to the repressor, which is normally bound to the operator site on the DNA. Binding of lactose to the repressor makes the repressor unable to bind the operator region which overlaps the promoter region for the lac genes. This means that RNA polymerase can now bind the promoter region and transcribe the lac genes, which are needed to utilize the lactose.
3. Enhancers and silencers are regulatory DNA sequences. Enhancers activate transcription by binding transcriptional activators and silencers inhibit transcription of genes by binding to repressors. These are found in eukaryotic organisms.

Lesson 5: RNA Processing

1. Eukaryotic cells process mRNA because transcription occurs in the nucleus and translation (protein synthesis) occurs in the cytoplasm. Processing is needed for the mRNA to leave the nucleus and be protected from degradation. Processing is also needed so that DNA sequences that are not needed in the final protein are removed. Bacteria do not have nuclei, and so the mRNA is not processed because it is translated into proteins directly.
2. The three steps for processing are:
   1. addition of a 7‑methyl guanosine cap to the 5′ end
   2. addition of about 200 A nucleotides (PolyA) tail to the 3′ end
   3. removal of introns by splicing.
3. The spliceosome is a complex of proteins and small RNAs that form small nuclear ribonucleoproteins (snRNPs). These are protein‑RNA enzymes that recognize intron sequences, cut them out, and join the exon sequences together to form a mature mRNA for protein translation. The two main steps in splicing are:
   1. The pre‑mRNA is cut at the 5′ splice site (junction of the 5′ exon and intron) and joined to the branch point sequence in the intron which forms a loop structure.
   2. The 3′ splice site is cut, the two exons are joined together, and the intron is released.
4. The purpose of alternative splicing is to provide diversity and minimize the number of genes needed to encode proteins. If different introns can be spliced and a variety of exons joined together, a cell can produce many different proteins using fewer genes.

Lesson 6: Translation

1. There are three possibilities, since we do not know if the first base in the fragment is the beginning, middle, or end of a codon:
   1. ‑lys (end of one polypeptide); gly‑pro‑ (beginning of second polypeptide)
   2. ‑asn‑glu‑asp‑arg‑ (Arg is coded by CGX, so it doesn’t matter which base is lost from the 3′ end of the fragment.)
   3. ‑met‑arg‑thr‑
2. One would react to such an announcement with scepticism. Two bases differ between the two codons. A double mutation in mRNA is not unreasonable, but not in the same three base area.
3. Gly is the smallest amino acid. If the active site of gly‑tRNA synthetase is small enough, no other amino acid can fit in the amino acid cleft to be attached the t‑RNA. Conversely, gly would fit spatially, if not by charge, in the other amino acid‑tRNA synthetases. Thus, other synthetases would need a means of deleting gly.
4. A mutation in the tRNA anticodon can occur so that it is complementary to the STOP nonsense mutation.
5. Approximately 399 high‑energy bonds are hydrolyzed: 200 are hydrolyzed to form the 100 aa‑tRNA complexes; initiation of protein synthesis uses one high‑energy phosphate bond; 198 high‑energy phosphate bonds are needed to form the peptide bonds.
6. Six ribosome bands will be observed. A ribosome is composed of two unequal subunits, which associate for protein synthesis and dissociate after protein synthesis. After several hours, the two sets of ribosomes will have split and recombined many times, so the subunits are randomly arranged: H, L, H‑H, L‑L, H‑L, and L‑H.
7. DNA and RNA synthesis is driven by hydrolysis of phosphate bonds. Protein synthesis is driven by hydrolysis of amino‑acyl‑tRNA bonds and of GTP (by elongation factors).
8. Digestion enzymes, such as lysozyme, and hormones, such as insulin, are powerful biochemical agents. To ensure that they are active only when needed, they are synthesized in an inactive pro‑form. Proteolytic cleavage to the active form occurs when these agents are required to perform their biological function, not before. This is only one answer; you may have thought of others.
9. Malignant tumour cells grow very fast. The capillary networks that supply oxygenated hemoglobin do not develop as rapidly. Therefore, a malignant tumour will be more successful if its need for oxygen is not great.
10. The p53 gene is a tumour suppressor gene, which means it functions to control cell proliferation that is characteristic of cancer. If p53 is mutated, then there is no mechanism for controlling cell proliferation and tumour growth. The p53 gene activates the transcription of cyclin‑dependant kinase inhibitor (CKI), which binds to Cdk‑cyclin complexes to control the cell cycle. It may also activate the proteins involved in apoptosis as a means to control cell proliferation.

If you wish to discuss any of these questions or need assistance with the material, please contact your Academic Expert (AE) by emailing the Student Success Centre at fst_success@athabascau.ca.