Can a single strand of mRNA form different polypeptide chains?

My book has a statement:

A single strand of mRNA is capable of forming a number of different polypeptide chains.

In my opinion this statement is wrong because a single strand of mRNA will have same sequence of codons. So, every time it gets translated, it produces same polypeptide chain. But, according to my book this statement is correct.

Am I wrong or is my book wrong?

I believe, your book refers to polycistronic mRNA. This is mRNA where multiple genes are encoded together in one mRNA and are often (but not necessarily) translated one after another. This is mainly found in prokaryotes where the proteins encoded on the same mRNA often form a metabolic pathway together.

You can come at mRNA from two directions. Remember that human genes are largely monocistronic, in other words they often code for a single protein. If we look at prokaryotes, however, polycistronic genes such as those in operons are very much possible. From a single lac mRNA you can get multiple proteins.

For humans with monocistronic genes, you can code for multiple isoforms, however. A good example is protein kinase C: You have PKC-α, PKC-δ, and so forth. So how can this be? In part, you can look at splicing. What gets initially transcribed from your human gene is a pre-mRNA, in that the entire gene contains introns and exons. The spliceosome knows where to cut out introns, and you technically get a mature mRNA that is made of only exons and codes for protein.

In special cases, however, RNA-binding proteins and-at times in concert with, at times to the detriment of-the spliceosome machinery alternatively splice some other combination of introns and exons (sometimes, introns are kept!). And so from the PKC gene, you still get PKC, but a different isoform of it. The different proteins that different types of cells or tissues express that can interact with the splicing machinery are capable of taking the same genetic information, the same pre-mRNA, and make different polypeptides out of it.

Apart from the polycistronic case there is another possibility of one mRNA yielding multiple proteins. By one mRNA, I mean that the RNA is not in any way altered because of RNA editing or other mechanisms.

Translation can be initiated at alternative sites leading to production of different protein products (Touriol et al., 2003). VEGF is one of the genes that exhibits this phenomenon. However, the precise factors that regulate the choice of the translation start site are not very well known. The basic mechanism, however, seems to be masking of the alternative start sites by different factors such as secondary structures and/or trans regulatory factors.

Similarly a stop codon can be skipped. This is frequently seen in the case of theUAG(Amber) stop codon. This phenomenon is called amber suppression (see this post). Again, the factors that regulate this phenomenon are not well known.

Having said all this, I am 100% sure that your textbook refers to polycistronic mRNAs only.

6.4: Protein Synthesis

  • Contributed by Suzanne Wakim & Mandeep Grewal
  • Professors (Cell Molecular Biology & Plant Science) at Butte College

The Central Dogma of Biology

Your DNA, or deoxyribonucleic acid, contains the genes that determine who you are. How can this organic molecule control your characteristics? DNA contains instructions for all the proteins your body makes. Proteins, in turn, determine the structure and function of all your cells. What determines a protein&rsquos structure? It begins with the sequence of amino acids that make up the protein. Instructions for making proteins with the correct sequence of amino acids are encoded in DNA.

Figure (PageIndex<1>): Transcription and translation (Protein synthesis) in a cell.

DNA is found in chromosomes. In eukaryotic cells, chromosomes always remain in the nucleus, but proteins are made at ribosomes in the cytoplasm or on the rough endoplasmic reticulum (RER). How do the instructions in DNA get to the site of protein synthesis outside the nucleus? Another type of nucleic acid is responsible. This nucleic acid is RNA or ribonucleic acid. RNA is a small molecule that can squeeze through pores in the nuclear membrane. It carries the information from DNA in the nucleus to a ribosome in the cytoplasm and then helps assemble the protein. In short:

DNA &rarr RNA &rarr Protein

Discovering this sequence of events was a major milestone in molecular biology. It is called the central dogma of biology. The two processes involved in the central dogma are transcription and translation.

Figure (PageIndex<2>): An overview of transcription and translation. The top panel shows a gene. A gene is composed of the open reading frame (aka coding sequence) that is flanked by regulatory sequences. At the beginning of the gene, the regulatory sequence contains a promoter where RNA polymerase attaches and starts transcription. At the end of the open reading frame, the regulatory sequence contains a terminator (not shown.)The middle panel shows a pre mRNA which is modified by excising introns and keeping exons. This is called post transcription modification. A mature mRNA contains a 5' cap and poly-A tail. The bottom panel shows a synthesis of protein via translation.

Types of RNA (Ribonucleic Acid): 4 Types

It is the most abundant RNA (70-80% of total) which has 3-4 types. Some of its types (23S, 28S) are the longest of all RNAs. As the name indicates, rRNA is a constituent of ribosomes.

Here it lies coiled in between and over the protein molecules. Depending upon their sedimentation coefficient, RNAs of eukaryotes are of four types— 28S, 18S, 5.8S and 5S.

Procaryotic ribosomes have three types of RNAs— 23S, 16S and 5S. 28S, 5.8S and 5S (23S and 5S in prokaryotes) occur in larger subunit of ribosome while 18S (16 S in prokaryotes) is found in smaller subunit of ribosome. rRNA is transcribed in the form of a longer chain of 45S in eukaryotes and 30S in prokaryotes.

In eukaryotic transcript the arrangement in 5′ → 3′ direction is 18S — 5.8S — 28S. Several methylations occur prior to removal of spacer RNA. Removal of spacer RNA breaks the transcript into 2-3 parts. 5S is often transcribed separately.

(i) rRNAs bind protein molecules and give rise to ribosomes,

(ii) У end of 18S rRNA (16S in prokaryotes) has nucleotides complementary to those of cap region of mRNA.

(iii) 5S rRNA and surrounding protein complex provide binding site for tRNA.

rRNAs get associated with specific proteins to form ribosome subunits. 50S subunit of prokaryotic ribosome contains 23S rRNA, 5S rRNA and some 32 protein molecules. 30S subunit of prokaryotic ribosome has 16S rRNA and about 21 protein molecules.

60S subunit of eukaryotic ribosome contains 28S rRNA, 5S rRNA, 5.8S rRNA and about 50 protein molecules. 40S subunit of eukaryotic ribosome consists of 18S rRNA and some 33 protein molecules.

Type # 2. Transfer RNA (tRNA):

It is also called soluble or sRNA. There are over 100 types of tRNAs. Transfer RNA constitutes about 15% of the total RNA. tRNA is the smallest RNA with 70-85 nucleotides and sedimentation coefficient of 4S. The nitrogen bases of several of its nucleotides get modified e.g., pseudouridine (ψ), dihydrouridine (DHU), inosine (I).

This causes coiling of the otherwise single-stranded tRNA into L-shaped form (three dimen­sional, Klug, 1974) or clover-like form (two dimensional, Holley, 1965). About half of the nucleotides are base paired to produce paired stems. Five regions are unpaired or single stranded— AA-binding site, T ψ С loop, DHU loop, extra arm and anticodon loop.

(i) Anticodon:

It is made up of three nitrogen bases for recognising and attaching to the codon of mRNA.

(ii) AA-Binding Site:

It lies at the 3′ end opposite the anticodon and has CCA— OH group. This CCA group is added after the transcription (5′ end bears G). Amino acid or AA-binding site and anticodon are the two recognition sites of tRNA.

(iii) T ψ C Loop:

It contains pseudouridine. The loop is the site for attaching to ribosomes,

(iv) DHU Loop:

The loop contains dihydrouridine. It is binding site for aminoacyl synthetase enzyme,

(v) Extra Arm:

It is a variable site arm or loop which lies between T ψ C loop and anticodon. The exact role of extra arm is not known.

(i) tRNA is adapter molecule which is meant for transferring amino acids to ribosomes for synthesis of polypeptides. There are different tRNAs for different amino acids. Some amino acids can be picked up by 2-6 tRNAs. tRNAs carry specific amino acids at particular points during polypeptide synthesis as per cidons of mRNA.

Codons are recognised by anticodons of tRNAs. Specific amino acids are recognised by particular activating or aminoacyl synthetase enzymes,

(ii) They hold peptidyl chains over the mRNAs.

Type # 3. Messenger RNA (mRNA):

It is a long RNA which constitutes 2-5% of the total RNA content. It brings instructions from the DNA for the formation of particular type of polypep­tide. The instructions are present in the base sequence of its nucleotides. Ii is called genetic code. Three adjacent nitrogen bases specify a particular amino acid.

Formation of polypep­tide occurs over the ribosome. mRNA gets attached to ribosome. tRNAs are induced to bring amino acids in a particular sequence according to the sequence of codons present over mRNA. mRNA has methylated region at the 5′ terminus.

It functions as a cap for attachment with ribosome. Cap is followed by an initiation codon (AUG) either immediately or after a small noncoding region. Then there is coding region followed by termination codon (UAA, UAG or UGA). There is then a small noncoding region and poly A area at the 3’ terminus (Fig. 9.24). An mRNA may specify only a single polypeptide or a number of them.

The former is called monocistronic while the latter is known as polycistronic. Polycistronic mRNA is more common in prokaryotes. Eukaryotic mRNA is usually monocistronic.

The life time of mRNA is also variable. In some lower forms it is from a few minutes to a few hours. On the other hand the mRNAs of higher forms seem to have a long life. It is several days in case of young red blood corpuscles which continue to form haemoglobin even when nucleus has degenerated.

(i) mRNA carries coded information for translation into polypeptide forma­tion.

(ii) Through reverse transcription it can form compact genes which are used in genetic engineering. The phenomenon also occurs in nature and has added certain genes in the genomes,

Model the process of polypeptide synthesis

The whole process of transcription occurs in 3 stages. They are:

  • Initiation:
    • Similar to replication, the DNA double helix unwinds forming a transcription bubble.
    • The hydrogen bonds between the nucleotide base pairs of the two anti-parallel strands are broken creating two separated strands, a template strand (3’→5’) and a non-template strand (5’→3’).
    • There is a particular sequence in the uncoiled DNA which indicates the initiation site of transcription known as promoter.
    • RNA Polymerase enzyme binds to the promoter and begins to add complementary bases to the template strand, the process being similar to replication. The only difference here is that in case of mRNA, the nitrogenous base Thymine is replaced by Uracil.
    • RNA Polymerase enzyme moves along the 3’→5’ template strand and adds new nucleotide base pairs complementary to that of the template strand.
    • With the addition of new nucleotide base pairs, the RNA strand keeps elongating.
    • RNA Polymerase encounters a definite DNA sequence known as the terminator which acts as the stop signal for termination. Once it encounters this sequence, transcription stops.
    • The freshly synthesized mRNA is known as nascent mRNA or pre-mRNA (direction is 5’→3’) and will undergo some changes before it enters the translation phase which are known ads post-translational modifications. Some of which are:
      • Some sections of the newly created mRNA contain sequences that do not code for proteins. In a process called splicing, these non-coding sequences (also termed as introns) are removed and the coding regions (known as exons) are joined together.
      • When the mRNA is exposed to the cytoplasmic environment during its journey to the ribosome, there are chances of it being destroyed by certain cytoplasmic enzymes called ribonucleases. To avoid this, at the 5’ end a 7-methylguanosine cap and at the 3’ end, an extension of around 250 Adenine residues known as Poly (A) tail are added.


      • Process by which mRNA is transcribed to protein.
      • Involves two types of RNA tRNA and mRNA.
      • Takes place in ribosome.
      • Before translation begins, nucleotide base pairs are clustered in groups of 3 known as codons. Each codon contains 3 nucleotide base pairs formed by the combination of A, U, G and C.
      • Out of possible 64 codons, 61 codons code for different amino acids which are the backbones of the polypeptide/protein synthesized after translation. The other 3 codons act as terminators of the translation process and are termed as stop codons.
      • The amino acids are brought by the tRNA molecules.

      The whole process of translations is also divided into 3 steps:

      • Initiation:
        • As soon as the start codon AUG is identified, two subunits of ribosome, the large subunit and the small subunit associate together.
        • The tRNA molecule carries the amino acid for the start codon AUG, Methionine. The tRNA molecule contains an anti-codon region complementary to the codons in the mRNA which allows them to join.
        • The ribosome moves along the mRNA and keeps adding amino acids based on the codon in the mRNA. The amino acid of the second codon forms peptide bond with the amino acid of the first codon and in the same manner as the ribosome moves, each new amino acid forms bond with the previous one forming a polypeptide chain.
        • As the peptide chain elongates, the ribosome is divided into three sites. The A site is where new tRNA molecules enter the ribosome, the P site is where peptide bonds are formed and E site is the exit site from which empty tRNA molecules whose amino acids have already bonded with existing amino acids exit the ribosome.
        • The polypeptide chain keeps elongating until any of the three stop codons UAA, UAG or UGA are detected. Once the ribosome encounters the stop codon, it triggers a series of events releasing a polypeptide chain.
        • The ribosomal subunits dissociate as soon as the polypeptide chain is released.
        • The polypeptide chain undergoes different structural modifications to form a functional protein.

        Importance of mRNA and tRNA:

        • mRNA is considered as the first expression of genes.
        • It holds information for precise synthesis of protein and each type of protein to be made from an mRNA strand is decided by the codon arrangements in the strand.
        • tRNA molecules are carriers of amino acids which are the backbones of protein molecules.

        Importance of polypeptide synthesis:

        • For creating proteins that carry out different functions in our bodies. For example, proteins actin and myosin build up our muscles.
        • For creating enzymes that control different biochemical pathways happening inside the cells. For example, cellular respiration takes in multiple steps which are controlled by different enzymes.
        • Proteins control different characteristics of every living organism. Thus, what we are overall is because of different types of protein expressions. Without polypeptide synthesis, life would have been small and our existence wouldn’t have been very different from what it is now.
        • Polypeptide synthesis forms products that are necessary to carry out replication, transcription and translation as well.

        Effects of environment in phenotypic expression:

        • Expression of phenotype is often controlled by environmental factors.
        • For example, natural factors such as light, temperature, nutrient availability, water etc. can affect phenotypic expression of plants.
        • An example of how environment affects phenotype is hydrangeas. Hydrangeas are plants that have different flower colours (pink and blue) depending on the PH of the soil they are in (environment). Soils with pH less than 5 (acid) they are blue and soils with pH more than 7 (alkaline) they are pink.

        Effects of genes in phenotypic expression:

        Interaction between genes can affect phenotypic expressions. Some examples are as follows:

        Causes of Mutations

        Mutations can be spontaneous or induced. They may also occur during the process of DNA replication. Here are some of the important causes of mutations.

        Spontaneous Mutations

        Some mutations can arise spontaneously in any type of cell. Most of such mutations are seen in highly proliferating cells such as cells of the intestines, skin cells, etc. They occur at the frequency of 10 -4 to 10 -7 per cell in one generation. The following different changes can occur as a result of spontaneous mutations at the molecular level.

        • Tautomerization, in which the position of a hydrogen atom is changed altering the base pairing due to different pattern of hydrogen bonding.
        • Depurination, a purine base is lost resulting in an empty site.
        • Deamination, hydrol7ysis of the amino group changes base from cytosine to uracil and adenine to hypoxanthine.

        Such mutations can arise in normal healthy cells with even zero-probability of being affected.

        Mutations during DNA Replication

        Despite the most effective proof-reading, some mutations may arise during the process of DNA replication. These include the addition of a nucleotide, mismatched pairing, and deletion of nucleotides. Such mutations are normally removed by the DNA repair system. They only pass to the next generation if the DNA repair mechanism fails.

        Induced Mutations

        Most of the mutations are caused by external environmental factors. Such factors that damage the DNA and cause mutations are called carcinogens.

        Carcinogens have two major types chemical carcinogens and radiations.

        Chemical Carcinogens

        They include five major classes of chemical compounds.

        • Polycyclic aromatic hydrocarbons such as benzopyrene present in cigarettes
        • Aromatic amine such as 4-acetylaminofluorene
        • Nitrosamines such as diethylnitrosamine and dimethylnitrosamine
        • Drugs like cyclophosphamide and diethylstilbestrol
        • Naturally occurring compounds like aflatoxin made by some fungi

        These chemical compounds can cause mutations via different mechanisms.


        The exposure of cells to different types of radiations also cause gene mutations. The most harmful radiations are UV-light and ionized radiations like X-rays.

        They can cause the following types of damage to the DNA.

        • Formation of pyrimidine dimers within the DNA strand
        • Removal of bases from the gene resulting in apurinic or apyrimidinic sites
        • Single or double-strand breaks within the DNA
        • Cross-linking of DNA strands

        Protein Translation in Parkinson’s Disease

        J.W. Kim , . V.L. Dawson , in Parkinson's Disease , 2017

        1.1 Translation in Eukaryotic Cells

        In eukaryotes, mature mRNAs are exported from the nucleus to the cytosol after splicing. The cap-binding eukaryotic initiation factor 4F (eIF4F), a complex that consists of eIF4E, eIF4G, eIF4A, binds to the mRNAs and activates them. With the help of initiation factors such as eIF1, eIF1A, and eIF3, the small ribosomal subunit (40S in eukaryotes) and ternary complex, comprising initiator methionyl-tRNA ( Met-tRN A i Met ), eIF2 and GTP, form the 43S preinitiation complex (PIC). The eIF4F cap complex recruits the PIC to the mRNA and forms the 48S initiation complex. Once the 48S complex is formed, it scans the mRNA until it finds a start codon. Upon recognition of the start codon, the GTP in the ternary complex is hydrolyzed with the aid of eIF5, an eIF2-specific GTPase-activating protein (GAP). eIF5 and eIF5B facilitate the dissociation of the initiation factors and the joining of the large ribosomal subunit (60S in eukaryotes). The large subunit joins to make an 80S ribosome, which has peptidyltransferase activity and thereby synthesizes polypeptide by forming peptide bonds between amino acids. With the help of eukaryotic elongation factor 2 (eEF2), an aminoacyl-tRNA is recruited to the acceptor site (A site) of the ribosome. A new peptide bond is formed between the amino acids in the peptidyl (P)- and A-sites. eEF2 also promotes ribosomal translocation, and the whole cycle repeats until the ribosome encounters a stop codon. At the stop codon, instead of a charged tRNA, release factors approach the P-site and mediate termination of protein synthesis ( Fig. 9.1 ). 4

        Figure 9.1 . Translation initiation in eukaryotic cells.

        The canonical pathway of eukaryotic translation initiation is divided into seven different stages. (1) mRNA activation: mRNA is activated by eIF4F complex formation, poly-A binding protein (PABP) binding to the poly-A tail, and subsequent circularization of the mRNA. m 7 G: 7-methylguanosine “cap” eIF4F cap-binding complex consists of eIF4E, eIF4G, and eIF4A. eIF4E binds directly to the m7G “cap,” eIF4G works as a scaffold protein and eIF4A is an RNA helicase, whose function is assisted by eIF4B. IRES: internal ribosome entry site there are several types of IRESs with different initiation factor requirements the illustration depicts a certain type of IRES (eg, encephalomyocarditis virus) with eIF4G/A/B requirements. (2) Attachment to mRNA: a ternary complex consists of eIF2, Met-tRN A i Met initiator tRNA, and GTP. A 43S PIC comprises a 40S small subunit, a ternary complex, eIF1, eIF1A, eIF3, and eIF5. A 43S PIC is recruited to the activated mRNA by interaction between initiation factors, primarily by eIF4G-eIF3 interaction. (3) 5′ to 3′ scanning: A 43S PIC joins to the activated mRNA to make a 48S initiation complex, and starts scanning. It is noteworthy that there is a type of IRES (cricket paralysis virus) that doesn’t require a scanning process for initiation. (4) Start codon recognition (5) Initiation factor dissociation (6) 60S joining: after start codon recognition, eIF5 and eIF5B are known to mediate the eIF2-bound GTP hydrolysis, the initiation factor dissociation, and the 60S subunit joining. (7) 80S complex formation: after formation of an 80S complex, elongation step succeeds.

        eIF2 is one of the major regulatory targets in translational initiation. eIF2 is a component of the ternary complex and it binds to GTP and Met-tRN A i Met . More specifically, it delivers the Met-tRN A i Met to the 40S small ribosomal subunit. With the help of other initiation factors, the ternary complex forms the 48S initiation complex and enables the 40S subunit to scan, find the start codon, and initiate protein synthesis. Once the 48S complex finds the start codon, GTP is hydrolyzed to GDP, which is an essential step for initiation factor dissociation and 60S large subunit joining. eIF2 is a heterotrimeric protein consisting of subunits alpha, beta, and gamma. Serine 51 on eIF2α can be phosphorylated by various kinases in response to stress signals, and phosphorylated eIF2α is resistant to the guanine nucleotide exchange factor (GEF) activity of eIF2B, therefore it remains in an inactive eIF2-GDP-tRNAi state. 1,2 Hence, it decreases active ternary complex availability and thereby reduces global protein synthesis, for example, under conditions of stress. 5 It is also known that the phosphorylation of eIF2α increases the expression of a group of genes with stress-responsive functions, such as activating transcription factor 4 (ATF4), through upstream open reading frame (uORF)-mediated translational regulation. 5,6 uORF-mediated regulation is involved in start codon selection upon arrangement of the different start codons, different ORFs can be chosen, resulting in the production of different proteins. Briefly, when ternary complex availability is low, start codon recognition by the 48S complex is hampered and the balance of start codon selection is skewed toward the downstream start codons. Several stress-responsive genes have multiple uORFs encoding nonfunctional gene products. Under normal conditions, the uORFs prevent the major ORF from being translated. However, when the ternary complex is depleted, the downstream start codons are preferred, leading to the synthesis of the desired functional protein. 1,2,5 It is noteworthy that eIF2α-mediated translational regulation has been shown to play important roles in processes relevant to neurons, like synaptic plasticity or neurodegeneration. 7–9

        Module 5 / Inquiry Question 3

        So far, in the previous weeks, we have touched on how genetic variation supports the continuity of a species. We have also explored how genetic variation can be created via different processes occurring during meiosis such as crossing over, independent assortment and random segregation. Also, during fertilisation of gametes.

        Earlier in Week 1 notes, we briefly mentioned that an organism’s gene contribute towards an organism’s characteristics – structure, physiological and behavioural. In this week, we will learn how an organism’s gene code for a specific polypeptide chain which is used to produce proteins. The proteins are what determine an organism’s phenotype as well as physiological and behavioural traits traits . For example, a protein that codes for an organism’s eye colour (phenotype – physical/structural trait)

        It is important to note that the protein synthesis process in living organisms produce protein that are determined by the organisms’ genes. The definition of a gene is a segment of a DNA. So, before we dive into the process of protein synthesis, we need to understand how DNA exists in prokaryotes and eukaryotes living organisms.

        Learning Objective #1 - Construct appropriate representations to model the forms in which DNA exists in prokaryotes and eukaryotes

        There are two types of DNA namely Chromosomal DNA and Extrachromosomal DNA.

        The difference between these two are that the former is located inside a cell’s nucleus whereas the latter is located outside a cell’s nucleus. Remember that there are many cells in the body

        Plasmids are circular DNA molecules that are found outside the nucleus in prokaryotes’ cells. Hence, they are a type of extrachromosomal DNA.

        Mitochondrial DNA and chloroplast DNA are also classified as extrachromosomal DNA as they are not found within the nucleus of a cell. These two types of circular DNA are found in eukaryotes.

        Chromosomes are classified as linear chromosomal DNA as they contain linear DNA and are located within the nucleus of a cell. Chromosomes are found in eukaryotes’ cells.

        The table below highlights more differences between the DNA found in the cells of prokaryotes and eukaryotes.

        Now, that we have talked about the different forms in which DNA can exist in eukaryotes and prokaryotes, let’s have a look at these different forms of DNA through some visual diagrams.

        Prokaryotes - Circular DNA in cytoplasm with Plasmids Model

        Prokaryotes have circular DNA as shown in the diagram. In the case of bacteria, it has circular bacterial DNA.

        Notice that bacterial DNA is double stranded but it is circular in shape.

        Prokaryotes also contain plasmids which are circular DNA molecules.

        Eukaryotes - Nuclear DNA Model

        Eukaryotes - Mitochondrial DNA Model

        Eukaryotes - Chloroplast DNA Model

        Now, let’s explore how DNA is involved in the protein synthesis process where proteins produced determine the traits (physical, physiological and behaviour traits) of living organisms.

        We will first be focusing on protein synthesis in eukaryotes which consists of two separate stages. i.e. transcription and translation. Then after that, we will look into protein synthesis in prokaryotes.

        Learning Objective #2 :

        Model the process of polypeptide synthesis - Transcription

        *Typo in diagram: Ribonucleotide Triphosphate should read Ribonucleoside triphosphate. These are RNA nucleotides.

        RNA nucleotides are different to DNA nucleotides as they do not contain thymine nitrogenous base. They have Uracil (U) nitrogenous base instead. Furthermore, they different in their sugar molecule. More specifically,

        A RNA nucleotide can have – C, G, A or U nitrogenous bases. Each nucleotide has a ribose sugar and phosphate group.

        A DNA nucleotide can have – G, C, A or T nitrogenous base. Each nucleotide has a deoxyribose sugar and phosphate group

        In eukaryotes, transcription is the first stage of the two-stage protein synthesis process.

        Transcription (‘transcribing’) is the process whereby the genetic information of a gene, situated in DNA strand, is copied to mRNA molecule that is synthesised during the process.

        Step 1: RNA polymerase (enzyme) attaches itself to the DNA at the promoter sequence region, breaking the hydrogen bonds that is bonded the nitrogenous bases. This results in the unwinding of a section of the DNA double helix. This process is known as initiation.

        Note that RNA polymerase does not unwind the entire DNA molecule, only a section consisting of a gene that is required to code for a particular polypeptide chain.

        Step 2: The section of DNA is unwound to allow free ribonucleoside triphosphate molecules to perform complementary base pairing with the template strand of the DNA. That is, Adenine will pair with Uracil and Cytosine will pair with Guanine.

        Note that there is NO thymine base in ribonucleoside triphosphate molecules, they are replaced with Uracil. This is because the mRNA molecule consists of Uracil, Adenine, Cytosine and Guanine whereas the DNA strand consists of Thymine, Adenine, Cytosine and Guanine.

        NOTE: Only one strand (the template strand) of DNA gets transcribed.

        Fun point about enzymes: Most enzymes have their names ending with ‘ase’. For example, DNA polymerase, RNA polymerase and helicase.

        Step 3: RNA polymerase moves downstream of the DNA (from 3′ to 5′ as shown in diagram) as more nitrogenous bases on the template strand are being paired with ribonucleoside triphosphate. These free RNA nucleotides are found in the nuclear sap, this is the same place where the free nucleotides are found during DNA replication. This process is known as elongation or sometimes also referred to as propagation because more RNA nucleotides are being paired with the nitrogenous bases on the template strand. As RNA polymerase moves downstream of the DNA, the enzyme rewinds the DNA behind it to reform the double helix.

        Step 4: The propagation stage of transcription stops when the RNA polymerase arrives at a termination sequence. Here, the enzyme releases the chain of ribonucleoside triphosphate from the complex, creating an mRNA strand that has identical genetic information as the coding strand of the DNA (only difference is that uracil is present rather than thymine) as it is formed via complementary base pairing using the template DNA strand.

        Learning Objective #3 :

        Model the process of polypeptide synthesis - Translation

        Translation is the process whereby mRNA information is used to create polypeptide chain and specifying its amino acid sequence.

        A polypeptide chain is consist of a chain of amino acids. There are proteins that are only made up of one polypeptide chain but there are also proteins that are made up of more than one polypeptide chain.

        Step 1: The mRNA migrates out of the cell nucleus and into the cell’s cytoplasm via the nuclear membrane pore.

        Step 2: A small ribosomal unit attaches to the mRNA

        Step 3: Following the small ribosomal unit, the large ribosomal unit attaches to the mRNA

        With the mRNA enclosed by the ribosome, it means that translation occurs within ribosomes. The rough endoplasmic reticulum (organelle) contains many ribosomes on its surface. This was from the Year 11 Preliminary HSC Biology Syllabus.

        Step 4: There are tRNA molecules found in the cytoplasm that have an anticodon. One anticodon is made up of three RNA nucleotides, each RNA nucleotide has a nitrogenous base – A, U, C or G. Each of these tRNA molecules can bind with a specific type of amino acid that is specific to its anticodon. This binding process requires the assistance of an enzyme (enzyme name not necessary for HSC Biology).

        Step 5: The mRNA codon specifies the tRNA, carrying an amino acid, with the complementary anticodon to bind with itself. The ribosome reads the mRNA codons so that the tRNA molecules with the correct anticodon bind with the correct mRNA codon.

        Note: Each codon (a sequence of three RNA nucleotide) on the mRNA strand will specify the tRNA anticodon for successful binding. For example, an mRNA with the codon AUC will specify and only allow an tRNA molecule with the UAG anticodon to bind with it.

        In effect, each mRNA codon specifies a tRNA molecule (based on complementary anticodon) and, thus, specifies the amino acid. Hence, the amino acid is specific to the mRNA codon (as well as specific to tRNA anticodon, of course).

        In the HSC Exam: You will be expected to know that each mRNA codon specifies an amino acid. This is why the translation table shows the mRNA codon that corresponds to an amino acid.

        Step 6: As the next mRNA codon specifies the another tRNA to bind with it, the prior tRNA molecule will detach from the mRNA and ‘transfer’ its amino acid to the new tRNA molecule that entered the ribosome complex. The amino acids undergo a condensation chemical reaction to bond with each other via a peptide bond. As this elongation process of building the amino acid chain progresses, the ribosome unit moves along the mRNA to continue reading subsequent mRNA codons.

        Step 7: The elongation process of building the amino acid chain stops when the ribosome complex reads the mRNA stop codon (a sequence of three RNA nucleotides). At this stage, a release factor comes into ribosome and binds with stop codon. This release factor causes the amino acid chain to separate from the tRNA molecule, resulting a polypeptide chain.

        Step 8: The ribosome complex separates into its small and large ribosomal subunits and mRNA separates into its individual nucleotides, i.e. ribonucleoside triphosphate.

        Step 9: The polypeptide chain coils up as the amino acids forms hydrogen bonding with each other. This single polypeptide can be a protein. (In diagram, I added extra amino acids, solely to depict coiling of polypeptide).

        NOTE for Step 9: Whether or not this polypeptide is a protein will depend on the type of protein that the original organism’s gene was expressing at the start of protein-synthesis. However, if the expressed protein has only required one polypeptide chain then it will be called a protein. If the protein to be expressed requires more than one polypeptide, then the polypeptide chain form will interact and bond with other polypeptides which together will fold or coil to form the protein.

        Protein Synthesis in Prokaryotes

        What we have just explored is protein-synthesis in eukaryotes. But how does the process of protein-synthesis differ in prokaryotes? Let’s find out

        In eukaryotes’ protein-synthesis there are regions called introns and exons. Exons are areas responsible gene expression and introns are non-coding regions of DNA that does not specify for an amino acid. Both of these DNA sequences get copied to the mRNA strand.

        Before translation, splicing occurs where introns DNA segments are removed from the mRNA strand.

        For prokaryotes, they have minimal introns and thus splicing is rarely occurs.

        In prokaryotes, transcription and translation occur simultaneously rather as separate steps because prokaryotes’ DNA are not separated from the cytoplasm via nuclear membrane.

        Lastly, each mRNA in prokaryotes contains genetic information from multiple genes . Comparatively, each mRNA in eukaryotes contain genetic information for only a single gene .

        So, in general, this means one mRNA in a prokaryote can code for more than one polypeptide compared to one mRNA in an eukaryote

        Learning Objective #4 - The importance of mRNA

        Part of the transcription involves the creation of a mRNA molecule that contains nitrogenous bases that are complementary to those nitrogenous bases found in the template strand or identical to nitrogenous base to those bases found in the DNA coding strand (except that uracil is present rather than thymine).

        From this, we can see that mRNA is important in ensuring that the organisms’ genes code for the correct mRNA codons. This allows the correct tRNA molecule with matching anticodons that correct the amino acid that corresponds to the mRNA codon to form the correct amino acid sequence for the polypeptide chain. Thus, the polypeptide chain(s) can fold correctly resulting in a correct protein structure and function.

        I know the word ‘correct’ was used many times there but that was for you to understand how the amino acid attached to the mRNA is SPECIFIC to the mRNA codon!

        So, how can i word this in the exam without being repetitive you say? Well, In the exam, you can write the following: The correct gene will allow the correct mRNA, formed from complementary base pair, to specify the correct tRNA carrying a specific amino acid to bind with the matching mRNA codon. This ensures that the right amino acids sequence of the resulting polypeptide chain and, hence, the correct protein to be created.

        Learning Objective #4 - The importance of tRNA

        The tRNA’s role is important in ensuring that it’s anticodon specifies and binds to the correct amino acid. This will ensure that the resulting polypeptide chain will have the right amino acid sequence that allow the protein-folding process to occur correctly. If not, the protein will not have the correct shape (primary structure)! The shape of the protein is critical in determining its function!

        This can be seen in the example of enzymes, a type of protein.

        Without enzymes, many metabolic processes such as cellular respiration simply will not occur as the reactants will not form chemical bonds with each other to create the products.

        The enzymes involved in catalysing mammals’ metabolic processes are proteins. Enzymes allows reactions to occur with lower energy at faster rates. Enzymes work by binding reactants together at the enzymes’ active site to weaken their chemical bonds and create products.

        The enzyme’s active site is critical in ensuring that the enzyme is able to attach the specific reactants by their specific shape. The enzyme’s active site and the specific reactants’ shape matches specifically! So, if the enzymes’ (protein) shape is not correct due to incorrect amino acid sequence in protein synthesis, the enzymes’ (protein) cannot correctly perform its function in catalysing the required metabolic process such as cellular respiration.

        Without cellular respiration, the cells of the organism will not be able to able ATP (Energy). Without specific energy, the organism cannot perform daily activities such as hunting for food, cell growth, cell repair, cell division, maintain its core temperature, or even walk.

        This means that the organism will die.

        Apart from enzymes being made up of proteins, it is important to note that proteins also specify an organism’s characteristics (structural/physical, physiological and behavioural traits). A structural (or physical) characteristic may be hair colour. The details of how this works is beyond the HSC Biology Syllabus.

        In the final learning objective of this week’s notes, we will explore more examples of proteins, apart from enzymes which we just talked about, in terms of how their structure is related to their functions.

        So, to wrap up, we should now realise that proteins are important. Protein shape is important to their function. Therefore, the right amino acid sequence is important as well as the right mRNA codon sequence and, hence, DNA is important.

        Previously, we have touched on the importance of DNA in terms of how genetic variation (allele combinations) are important in specifying the right characteristics of an organism for tolerating against ambient environment’s selective pressures.

        Now, we have touched on how DNA is important coding for the right protein which specifics for physical traits as well as catalysing necessary metabolic processes. We have come a long way already :)

        Learning Objective #5 - Analyse the importance and function of polypeptide synthesis

        This learning objective would be related to the purpose and importance in creating polypeptide chains and therefore proteins.

        This has already been discussed towards the end of the the previous learning objective. So, scroll up a little to refresh your mind (if you are returning to this set of notes for exam revision).

        However, to add on to what has already been said, here are some more pointers that is specific to this learning objective.

        As already mentioned, gene expression refers to process whereby polypeptides are produced by the coding of genes. Hence, gene expression is part of the polypeptide synthesis and, thus, protein synthesis process. Polypeptide synthesis is a highly regulated process.

        For example, our white blood cells only produce proteins known as antibodies to immobilise and defend against foreign matter such as bacteria when necessary. After a successful defence, our white blood cells will stop producing antibodies. That is, gene expression will stop (we will learn more about regulatory DNA sequences that control gene expression in Module 6).

        While antibodies are only produced when the first and second line of defence has failed to defend against foreign matter such as bacteria.

        Haemoglobin is a protein molecule that is inside our red blood cells are continuously produced in the bone marrow. Haemoglobin is a protein that is made up of four polypeptide chain and helps increase the amount of oxygen that our red blood cells can carry to our cells for cellular respiration.

        Learning Objective #6 - Assess how genes and environment affect phenotypic expression

        The relationship between genes and phenotypic expression

        This has already been discussed. The flow chart below summarises the process of how genes are responsible for an organism’s traits via protein synthesis.

        Gene —> mRNA —> tRNA attached to amino acid —> Polypeptide chain —> Protein —> Codes for a characteristic which can be structural, physiological or behavioural.

        For example, a structural or physical trait can be eye colour.

        In reality, most traits are specified by more than one gene.

        The relationship between environment and phenotypic expression

        So far, we have touched on how gene expression in protein synthesis is responsible for an organism’s phenotype.

        However, what studies have found is that there are in fact many environmental factors that will affect an organism’s gene expression and thus phenotype

        Some examples are outlined below.

        The organism’s diet or availability of food/water: Pea plants that have limited availability of water would be shorter than pea plants that have access to abundant volumes of water, provided their pea plants are genetically identical.

        pH of soil: Hydrangeas exhibit pink and blue colours depending on the pH of the soil. If the soil pH is less than 6, the hydrangeas will be blue. If the pH is greater than 7, they are pink. Hydrangeas colour are dependent on the concentration of aluminium ions in the soil where ion availability is affected by pH.

        Temperature of ambient environment: Himalayan rabbits are found to have different fur colour based on temperature affecting their gene expression in producing fur pigments. Above thirty five degrees celsius, they have white fur (better at reflecting heat). Below thirty degrees celsius, they have black fur (better at absorbing and trapping heat).

        You will learn about mutation in future modules. However, just to touch on it a little bit, mutations can also affect an organism’s phenotype by altering an organism’s DNA sequence. For example, UV radiation will modify the DNA sequence of a gene. If the mutation affects a gene that is a oncogene which controls cell growth, then the mutation may lead to uncontrolled growth of unspecialised cells (cancer cells). These unspecialised cells take the nutrients from surrounding specialised (useful) cells, leading to the death of specialised cells and increasing numbers of unspecialised cells (cancer cells). Such abnormal cell growth may appear as ‘bumps’ on an organism’s phenotype.

        In short, you can think that Genotype + Environmental Factors = Phenotype.

        Learning Objective #7 - Investigate the structure and function of proteins in living things

        We have already briefly touched on the relationship between protein structure and function in living organisms. R

        ecall that in learning objective #4, we used the example of enzymes (example of a type of protein) and their unique active site’s shape (protein structure) in catalysing specific chemical reactions by binding to specific reactants (protein function as catalyst for metabolic processes).

        To add what we have already explored in learning objective #4, for this learning objective, we will explore couple of more examples of how proteins can perform various functions and how their structure help support the protein’s success in performing such activities.

        Keratin is a protein that has the function in providing the structure for an organism. Specifically, keratin provides the flexibility of an organism’s skin (protein’s function). The highly-coiled, rope-like protein structure of keratin allows the protein to stretch by without snapping (i.e. chemical bonds does not break), thus, allows the skin to withstand stretching forces.

        Another example is involves protein that adopt the function as part of an organism’s immune response. In short, when a foreign substance such as a bacteria enters the organism’s blood stream, plasma B cells gets activated leading to the release of proteins called antibodies. These antibodies are produced (via gene expression) have the same shape as the antigen (proteins) on the bacteria. This allowed antibody to bind with the antigen and neutralise the bacteria, preventing the bacteria to produce any toxic chemicals.

        Note that this is an investigation task. So, if you wish you can do further research other types of proteins that exist in living organisms. During your research, you can then document how the protein’s structure is related to their specific function in supporting the survival of the living organism.

        More on protein structure specifically

        In terms of protein structure specifically, there can be four levels of structure for a protein:

        Primary structure, secondary structure, tertiary structure and quaternary structure.

        Primary structure – is related to the amino acid sequence of the polypeptide chain(s). The way amino acids are ordered will determine how chemical bonds can and cannot be formed between amino acids. The primary structure is critical in determining how the secondary, tertiary and quaternary structures (shape) would be like! Remember, protein shape is related to its function.

        Secondary structure – is related to the way that each polypeptide chain will coil up into helixes by forming more hydrogen bonds between carboxyl and amine groups. Secondary structure is critical in providing chemical stability to polypeptide chains.

        Tertiary structure – is related to the way that the now-coiled-up polypeptide chains further coils up to form an irregular three-dimensional structure. This is critical in providing the shape of the eventual protein which is critical for the protein’s function. Up until this point, the main forms of chemical bonds are hydrogen bonding. However, to create this tertiary structure, there are other forms of bonds including ionic bonds. These new bonds increases the protein’s ability to tolerate variations in pH and temperature. However, at the end of the day, the range of temperatures and pH in which proteins can operate efficiently are generally quite narrow.

        Quaternary structure – is related to the way different polypeptides interact with each other via hydrogen bonding, forming a functional protein such as an enzyme that is able to catalyse a chemical reaction. If you zoom into any section of the quaternary structure of the protein using an electron microscope, you will be able to see first see the tertiary structure then secondary structure and eventually the primary structure (amino acid sequence) of the protein.

        Week 3 Homework Questions

        Week 3 Homework Question #1: What is difference between genome, genotype and phenotype? (4 marks)

        Week 3 Homework Question #2: Explain why there are two stages of protein-synthesis in eukaryotes and only one in prokaryotes (3 marks)

        Week 3 Homework Question #3: Describe the process of transcription in protein-synthesis (4 marks)

        Week 3 Homework Question #4: Describe the process of translation in protein-synthesis (4 marks)

        Week 3 Homework Question 5: Explain how environmental factors are contribute towards an organism’s phenotype (4 marks)

        Week 3 Homework Question 6: Distinguish between eukaryotes and prokaryotes in terms of how DNA exists in their cell(s). (4 marks)

        Week 3 Homework Question 7: Explain how genes contribute towards an organism’s phenotype (4 marks)

        Week 3 Homework Question 8: Describe the importance of mRNA (3 marks)

        Week 3 Homework Question 9: Describe the importance of tRNA (3 marks)

        Week 3 Homework Question 10: Explain how the structure of protein is related to its function, provide two examples in your answer (4 marks)

        Week 3 Curveball Questions

        Curveball Question 1: Explain the importance of polypeptide synthesis (8 marks)

        Curveball Question 2: Explain the specific relationship between mRNA, tRNA and amino acids (5 marks)

        Curveball Question 3: Assess whether errors in DNA replication (e.g. an adenine base pairs with a cytosine base instead of thymine), may affect an organism’s phenotype and thus the continuity of species. Include the mechanisms of transcription and translation in your answer (8 marks)

        Curveball Question 4: Describe the importance of mRNA and tRNA for the survival of a living organism and in supporting the continuity of a species’s population (4 marks)

        Curveball Question 5: Describe the four different levels structures of a protein may have (4 marks)

        True or False: Errors during interphase I of meiosis may change the phenotype of the organism (parent)?

        True or False: Errors during Interphase I of meiosis may change the phenotype of the organism’s offspring?

        True or False: Errors during Interphase I of mitosis may change the phenotype of the organism (parent)?

        RNA Types and Structure

        Ribonucleic acid (RNA), like deoxyribonucleic acid (DNA), is a polymer of nucleotides that is essential to cellular protein synthesis. Unlike DNA, RNA is a single-stranded structure containing the sugar moiety ribose (instead of deoxyribose) and the base uracil (instead of thymine). While DNA stores the genetic information, RNA generally carries out the instructions encoded in the DNA but RNA also executes diverse non-coding functions. There are 3 major types of RNA that perform different but collaborative roles in protein synthesis: messenger RNA (mRNA), transfer RNA (tRNA), and ribosomal RNA (rRNA). During transcription, RNA is synthesized from DNA through a series of steps catalyzed by the enzyme RNA polymerase. The mRNA formed will serve as an amino acid template for protein synthesis. Translation proceeds with the tRNA transporting the corresponding amino acid based on the deciphered nucleotide sequence (codon) in the mRNA. The ribosomes, which are composed of rRNA, then facilitate the assembly of amino acids into a polypeptide. These components work together to convert the mRNA template obtained from DNA into the desired protein.


        During transcription, a section of the genetic code is transcribed into a single strand of RNA. Such sections are termed coding regions. This happens under the action of the enzyme RNA polymerase.

        The coding DNA strand serves as a matrix for the construction of an RNA strand. The synthesized mRNA encodes a protein in a process of translation.

        Prokaryotes have no nucleus and exhibit transcription in the cytoplasm. While in the eukaryotes, the nuclear genome is transcribed in the karyoplasm of the cell nucleus.

        A simplified flow of messenger RNA (mRNA) formation. Image Source: Dovelike, Wikimedia Commons CC-BY-SA 3.0

        In prokaryotes, ribosomes can already attach to the not yet completely synthesized mRNA sequence. And then begin translation. Hence, the synthesis of proteins can start at the same time as transcription, which enables special forms of gene regulation.

        In eukaryotes, the primary RNA transcript is at first subjected to various processes in the cell nucleus. Only then it is exported from the nucleus as mRNA into the cytoplasm where the ribosomes are located.

        Prokaryotes possess only one type of RNA polymerase for the synthesis of RNA. In contrast, eukaryotes possess different types of RNA polymerases. And primarily the RNA polymerase II catalyzes the synthesis of pre-mRNA.

        A major difference between prokaryotic and eukaryotic messenger RNA is that prokaryotic mRNA is usually polycistronic, while eukaryotic messenger RNA is usually monocistronic. This enables prokaryotes to have the information of several genes on only one single mRNA transcript. So synthesis of the encoded proteins and mRNA synthesis occurs simultaneously. One such jointly transcribed region of functionally related genes on the DNA is called an operon.

        Eukaryotic pre-messenger RNA (pre-mRNA) processing

        In eukaryotic cells, a mature messenger RNA is produced by processing its precursor. The precursor is termed as the hnRNA (heterogeneous nuclear RNA) or pre-mRNA (precursor messenger RNA, pre-mRNA).

        These steps take place in the cell nucleus. Then the mRNA enters the cytoplasm through nuclear pores. And eventually, protein biosynthesis takes place via ribosomes.

        • Capping: The 5′ end of the RNA molecule gets a 5′ cap structure. This cap consists of a modified form of guanosine, 7-methylguanosine (m7G). The cap protects the RNA from degradation by nucleases and allows the cap-binding complex. This is important for nuclear export, among other things. After transport into the cytosol, the cap aids in the recognition of the mRNA. It does so with the help of a small ribosomal subunit. This helps in initiating translation.
        • Polyadenylation: The RNA undergoes polyadenylation at the 3′ end. During this process, a poly-A tail consisting of 30 to 200 adenine nucleotides is attached. This also protects the messenger RNA from enzymatic degradation. In addition, it facilitates both nuclear export and the translation of the mRNA.
        • Splicing: Splicing removes certain RNA segments from the original transcript known as introns. Introns usually do not contribute towards the coding information. The remaining segments are joined together as exons. This process takes place in the spliceosome.

        The spliceosome is a complex of the hnRNA and the so-called snRNPs (small nuclear ribonucleoproteins). Spliceosome consists of the snRNAs U1, U2, U4, U5, and U6 and about 50 proteins. By alternative splicing, different mRNAs can thus be produced from the same hnRNA. These results when translated can also lead to different proteins.

        Spliceosome complex that helps produce messenger RNA. Image source: Agathman, Wikimedia Commons, CC-BY-SA 3.0

        This is also where various regulatory processes of the cell intervene. Antisense RNA and RNA interference can be used to degrade mRNA. Thus this prevents translation.

        Furthermore, nucleotides in a messenger RNA are changed sometimes by the RNA editing process. An example is the mRNA of apolipoprotein B. For example, in some tissues editing in mRNA of apolipoprotein B creates a second stop codon upstream. This codes for a shorter protein with a different function.

        Untranslated regions on mRNA are also responsible for regulating transcription as well as translation.

        Examiners report

        Many candidates gained full marks for their diagrams of joined DNA nucleotides. As mentioned earlier, the problem for some candidates was their misinterpretation of &ldquoa single strand of DNA.&rdquo Though appropriate shapes were given, the bonding was improper.

        In their outlines of gene transfer, candidates (as a group) eventually included each of the ten marking points. A number of candidates thoroughly understood the topic, while others wrote about meiosis and crossing over! The nature of the topic allowed candidates to express their ideas in a logical sequence.

        The process of translation has been examined frequently on past papers. Though the topic involves many different molecular structures and events, some candidates seemed to correctly grasp much of the detail and overall result. Some excellent answers appeared. However, as in previous years, there were candidates who confused translation with transcription (perhaps a reading error after glancing at the question?) and those who mixed accurate with inaccurate information.

        Watch the video: SARS-CoV-2: Corona-Impfstoffe im Vergleich mRNA und Vektor (January 2022).