2 Transcription, Translation

 

Session Level Objectives (SLOs): after completing the session, students will be able to:

SLO 1. Explain how cells overcome three major challenges to replicate their genomes.

SLO 2. Explain why different genes need to be expressed at different rates, in different cells, at different times.

SLO 3. Describe the functions of the three mammalian RNA polymerases.

SLO 4. Explain the structure of a mammalian gene, including regulatory and coding elements.

SLO 5. Describe how RNA polymerase II (RNAP II) is directed to gene promoters and the roles of general transcription factors.

SLO 6. Explain how transcription factors control which proteins are made in different cell types.

SLO 7. Outline the processing reactions that precede export of an mRNA from nucleus to cytoplasm.

SLO 8. Outline the major steps of protein synthesis, including the roles of mRNA, tRNA, tRNA synthetases and ribosomes.

SLO 9. Describe the basic ways in which microRNA (miRNA) molecules control gene expression.

Common Themes of Information Transfer

DNA —replication—> DNA —transcription—> RNA —translation—> polypeptide

There are common themes in these reactions.

• Biological polymerization reactions always have an intrinsic directionality (polarity): In DNA replication and RNA transcription, the template strand is always read 3´-to– 5´, and the nascent chain is always synthesized 5´-to-3´. In protein synthesis (translation), the mRNA template is always read 5´-to-3´, and the nascent polypeptide is always synthesized N-to-C.
• The polymerase must be accurately positioned at a start site on the template. In each case this process is called initiation, and in each case initiation entails several regulated steps.
• The polymerase has an elongation cycle, which is what it sounds like. This is the major biosynthetic stage.
• A signal on the template signals termination of polymerization. Termination involves disassembly of the elongation machinery and the release of templates and products.

This is a general framework. We will not focus on every stage for each process, but rather on key stages that illustrate important concepts.

SLO 1. Explain how cells overcome three major challenges to replicate their genome.

To replicate its genome, the cell must overcome several challenges:

• • There are two DNA strands to be replicated.
  • The two strands run in opposite directions (they’re antiparallel).
  • Replication must be accurate.
  • Enormous amounts of DNA are replicated: 6 Gbp/cell (= 6×109 base pairs per cell).
  • One and only one copy of each of the 46 chromosomes must be segregated into each daughter cell.

How is this done? Here we provide a cursory outline of DNA replication. Later in the course we will look more closely at replication, mitosis, and meiosis in the contexts of the cell division cycle, genetic inheritance, and cancer.

For now, there are three core concepts about DNA replication that you need to know:

• DNA replication is semiconservative (Fig. 1). This means that when a cell divides, the DNA duplexes in each daughter cell contain one of the parent cell’s original DNA strands (which is used as a template for polymerization, and one newly synthesized or nascent strand.
• DNA replication is semi-discontinuous (Fig. 2). This means that the nascent strand associated with one of the two strands is synthesized in short segments called Okazaki fragments that are then knitted together.
  

  

Discontinuous replication on one strand is necessary because the DNA polymerase can only add nucleotides to the 3´end of a nascent chain, but the template strands are antiparallel (Fig. 1).

An important difference between DNA and RNA polymerases is that DNA polymerases can only extend pre-existing nascent chains, while RNA polymerases can begin new chains from a single nucleotide. Thus, DNA polymerases invariably require a short RNA primer. The primer is made by a special RNA polymerase called primase (Fig. 2).

• DNA replication is bidirectional (Fig. 3). This means that at the DNA replication origin — the site where polymerization is initiated — two replication forks diverge from the
  

  

  origin. The replication machinery at each fork synthesizes one leading strand and one lagging strand (which is assembled from Okazaki fragments).

Because human chromosomes can be hundreds of millions of base pairs in length, replication is done in parallel starting at many replication origins on each chromosome.

Each chromosome begins as one piece of double-stranded DNA. Replicating the very ends of the chromosomes, the telomeres, presents special problems on the lagging strand. Telomere DNA is therefore maintained by a special enzyme called telomerase.

SLO 2. Explain why different genes need to be transcribed at different rates, in different cells, at different times.

Most of the mammalian genome consists of non-coding DNA. A minority (~2%) of the human genome actually serves to encode mRNAs that are used as templates for protein synthesis. A similarly small fraction of the genome encodes other varieties of biologically important RNA molecules.

The fundamental question about gene expression is this: each of us has many different cell types, but only one genome. The different cell types are different because they make different proteins: muscle cells make contractile proteins; nerve cells have the enzymes needed to manufacture neurotransmitters; osteoblasts have the protein machinery needed to manufacture bone, and so on.

Moreover, even a single cell type needs to make different proteins at different times: we synthesize and secrete insulin (a peptide hormone) when we eat. We remodel entire tissues and organs throughout growth, in response to injury, and during pregnancy. How does this happen?

DNA —transcription—> RNA —translation—> polypeptide

The level of any given protein is a function of competing processes: synthesis and destruction. Protein synthesis requires an mRNA template, and the abundance of the mRNA encoding any given protein is also regulated by a balance of synthesis and destruction.

SLO 3. Describe the functions of the three mammalian RNA polymerases.

RNA Transcription

The cell makes RNA molecules that do different things:

• mRNA is the template for protein synthesis (translation).
• tRNA and rRNA are core parts of the protein synthesis machinery.
• Diverse RNA molecules are involved in regulating gene expression and other processes. Examples include micro RNA (miRNA) and long non-coding (lncRNA). Many other examples are emerging.

The various RNAs are made by three RNA polymerase (RNAP) enzymes:

• RNAP I makes most of the ribosomal rRNA — the most important part of the ribosome, the enzyme that synthesizes polypeptides.
• RNAP II makes all of the messenger mRNA — the templates for polypeptide synthesis.
• RNAP III makes transfer tRNA — the carrier of activated amino acids for polypeptide synthesis.

We will focus on transcription by RNAPII, because its activity controls the levels of mRNA templates for protein synthesis. The underlying principles by which RNAP I and III operate are similar.

SLO 4. Explain the structure of a mammalian gene, including regulatory and coding elements.

A gene contains two kinds of sequences:

1. The transcription unit is the DNA sequence used as a template to synthesize RNA.
2. Regulatory sequences tell RNAP where to initiate and terminate transcription. They allow cells to control which genes are actively transcribed (“expressed”), and which are silent. These sequences can be further sub-divided:

• • The promoter sequence directs RNAP II and associated general transcription factors to the transcriptional start site.
  • Enhancer sequences, usually 10-30 bp in length, bind transcription factors, or activator proteins, that instruct RNA polymerase to become active at the promoter. Each gene is controlled by different enhancer elements.
  • Different cells contain specific sets of transcription factors. This is the main basis for cell-type-specific gene regulation! Enhancers can sit right next to the promoter, or tens of thousands of base pairs distant. Enhancers are usually upstream of the promoter but they can also be embedded within the transcription unit or even downstream of it.
  • The terminator tells RNAP that it has reached the end of the transcription unit.
  • There are other regulatory elements in the genome as well. For example, silencer sequences bind factors that, as the name suggests, decrease transcription. Insulator elements ensure that different genes are subject to independent regulation.
    

    

SLO 5. Describe how RNA polymerase II (RNAP II) is directed to gene promoters and the roles of general transcription factors.

Sequence of Events in Transcription:

1. Genomic DNA is packaged into chromatin. Transcription is regulated in part by how densely packaged a given gene is, and hence, how accessible its regulatory sequence elements are. Later, we will discuss how DNA packaging is controlled.
2. To initiate transcription of a gene, RNAP II must be directed to the promoter. This is done by the General Transcription Factors (Figs. 5 and 6). The GTFs recognize and bind to promoter sequences. They place RNAP II at the start site. GTFs then locally melt the DNA at the promoter, separating the two DNA strands to form a transcription bubble.
3. The GTFs are “general” transcription factors because they are always needed for initiation of transcription. However, the GTFs do not have the ability to regulate when RNAP actually initiates RNA synthesis.
4. In other words, GTFs are necessary but not sufficient for initiation of transcription.
5. Transcription factors that bind to regulatory promoter, so intervening DNA must loop out in order for transcription factors, coactivators, and the initiation complex (GTFs + RNAP II) to touch one another. enhancer sequences are needed to activate transcription by RNAP II and the GTFs (Figs. 5 and 6)
6. Activating transcription factors “talk” to RNAP II and the GTFs by binding to coactivators that touch RNAP II and the GTFs (Fig. 6). Together, these events cause the pre-initiation complex — RNAP II and the GTFs — to initiate RNA polymerization. As we will see in a later session, transcription factors can also control chromatin structure.
7. The chemistry of RNA polymerization is similar to the chemistry of DNA polymerization. The most important differences are:
   • No primer is needed for RNA synthesis.
   • NTPs are used for RNA synthesis, not 2-deoxy dNTPs.
8. As RNAP II elongates the nascent mRNA chain, it moves along the template strand of the transcription unit (Fig. 5). The replication bubble moves as RNAP II “crawls” along the DNA template strand. In other words, the DNA double helix melts in front of RNAP II and re- hybridizes (anneals) behind it.
9. When RNAP II reaches a terminator sequence (Fig. 5), the newly-synthesized RNA chain is released, RNAP is removed from the template strand, and the transcription bubble collapses.

SLO 6. Explain how transcription factors control which proteins are made in different cell types.

The reason we care so much about the mechanics of RNAP II transcription is that this process controls which mRNA transcripts are produced, and in what abundance. This in turn controls the specific repertoire of proteins that can be made by each cell.

Fig. 7 shows the regulatory sequences of genes that encode some key proteins made only in specific kinds of cells: skeletal muscle cells, heart muscle cells, and cells in the lens of the eye.

Each type of DNA enhancer element sequence is recognized and bound by specific activating transcription factors — shown here by colored shapes. Humans have about 2,000 different transcription factors.

By producing specific combinations of transcription factors, each cell specifies which subsets of genes are actively transcribed, and in what quantities.

This concept is so important that it bears repeating:

The specific array of transcription factors, present in a given cell, shapes that cell’s pattern of gene expression — and thus, that cell’s overall protein complement, the cell’s identity (muscle, fibroblast, neuron, etc.) and its functional characteristics.

Another important point is that transcription factors are proteins. Consequently, the genes that encode transcription factors are themselves subject to transcriptional regulation. By transcribing and translating specific transcription factors the cell can execute temporary or stable programs of gene expression in response to developmental cues and other signals, such as food or infection.

Fig. 8. Positive feedback loops maintain gene expression programs. Here, an “initiator” transcription factor binds an enhancer on the gene encoding a “terminal selector” transcription factor. When this gene is transcribed and the resulting mRNA is translated, the resulting protein binds to another enhancer in its own gene, and ensures that the terminal selector gene continues to be transcribed. The terminal selector also stimulates transcription of other genes needed for specific functions. Source: PNAS 110:7101

SLO 7. Outline the processing reactions that precede export of an mRNA from nucleus to cytoplasm.

mRNA Processing and Export

DNA replication and RNA transcription both occur in the cell’s nucleus. However, proteins are synthesized in the cytoplasm. To serve as templates for protein synthesis, mRNA molecules must be exported from the nucleus to the cytoplasm. This occurs at a special portal in the nuclear membrane, the nuclear pore (Fig. 9). The nuclear pore is an immense molecular assemblage that precisely controls the passage of mRNA and other macromolecules into, and out of, the nucleus. This is another theme that we will encounter again and again: biosynthetic products made in one cellular organelle are shuttled to another location — as with stations on an assembly line.

 

In the nucleus, the initial “raw” RNA transcript made by RNAP II must be processed (Fig. 10). Only then is the mature mRNA exported from the nucleus to the cytoplasm, where it will be used as a template for protein synthesis.

1. At the 5´end of the mRNA (the “beginning” of the transcript), a special nucleotide, 7– methylguanosine, is attached through a covalent bond. This is called the 5´ cap.
   • Later, in the cytoplasm, the 5´ cap will tell the protein synthesis machinery that the RNA bearing the cap is a messenger mRNA — a template for protein synthesis — and not some other type of RNA.
2. At the 3´ end of the transcript, a string of adenosine (A) nucleotides is added. This is the poly–A tail of the mRNA. The poly-A tail is constructed by a special enzyme (poly-A polymerase) through a non–templated polymerization process.
   • The poly-A tail will signal export of the mRNA from nucleus to cytoplasm. It will also control the stability (the half-life) of the mRNA once it’s in the cytoplasm.
3. The mRNA is spliced to remove introns and ligate (join) exons together. This step is somewhat involved and extremely important, so we’ll examine it in a bit more detail.

Differential mRNA splicing

The maturation of a large mRNA molecule may entail dozens of splicing reactions. In different cell types, these splicing reactions may be regulated so that not every exon ends up in each final, mature mRNA molecule. Consequently, a single transcription unit may encode more than one mRNA variant, with each derived from a different combination of exons (Fig. 11).

Differential splicing allows the ~20,000 protein-coding genes in the human genome to encode substantially more than 20,000 distinct mRNA templates and, thus, a much greater diversity of proteins.

 

To summarize:

• Transcription initiation controls how many mRNA transcripts get made.
• Different cells have different activating transcription factors.
• Different genes have different enhancers that bind different transcription factors.
• mRNA cap addition signals that the mRNA will be a template for protein synthesis.
• Differential splicing controls which exons are in the mature mRNA template, and thus the sequence of the resulting polypeptide.
• The poly-A tail (along with other features of the mRNA) controls nuclear export and the stability of the mRNA — how long it persists in the cytoplasm.

SLO 8. Outline the major steps of protein synthesis, including the roles of mRNA, tRNA, tRNA synthetases and ribosomes.

Here we summarize how polypeptide chains are synthesized and how they fold into their correct three-dimensional configurations.

The notes for this section begin with two charts: first, the assignments of RNA codons to amino acids (the genetic code); second, the chemical and physical properties of the 20 regular amino acids listed by features. This table also includes the rare amino acid selenocysteine, sometimes considered as the “21st amino acid,” although it is a modification of cysteine.

You do NOT need to memorize Tables 1 and 2! You do need to be able to apply their content.

Table 1

Table 2

Source: Wikimedia

Protein Synthesis: the Genetic Code

Like nucleic acid synthesis, protein synthesis is a template-directed process. However, in protein synthesis, the process is a bit more complicated because the template does not directly interact with the polypeptide product. How this works has never been explained more plainly than by Francis Crick, in an astonishing proposal that he made in 1955:

…Each amino acid would combine chemically, at a special enzyme, with a small molecule which, having a specific hydrogen-bonding surface, could combine specifically with the nucleic acid template. This combination would also supply the energy necessary for polymerization… there would be 20 different kinds of adaptor molecule, one for each amino acid, and 20 different enzymes to join the amino acid to their adaptors…

We now know that the “adapter” is tRNA. Crick’s proposal was correct in every detail save one: there are 61 possible combinations of 3-base mRNA codons (see Table 1), so there are more than 20 tRNA “adapters.” This system provides the biochemical basis of the genetic code — the rules through which each 3-base codon in an mRNA template specifies one specific amino acid.

tRNA and aminoacyl tRNA synthetase enzymes

1. 

   Each tRNA has 2 “business ends”:

   • The anticodon pairs with the 3-base codon on the mRNA template. Look closely at the diagram (Fig. 13). Note that as in other nucleic acid hybrids, the two strands are antiparallel.
   • The aminoacyl acceptor site is a terminal adenosine (A) nucleotide, where the carboxyl group of the amino acid is esterified to the 3´-OH of the adenine. Note that this is a relatively unstable, high–energy bond. It will make polypeptide elongation thermodynamically downhill, and hence favorable — exactly as predicted by Crick.
2. Any given tRNA can (in principle) be esterified to any amino acid. However, the fidelity of translation depends absolutely on the accurate matching of a tRNA bearing a specific anticodon to the corresponding amino acid.
3. The job of coupling each amino acid to its corresponding tRNAs is done by 20 different enzymes: the aminoacyl tRNA synthetases (also called aaRS enzymes). This is an absolutely key point: the specificity of the genetic code is controlled by the tRNA synthetases.
4. The tRNA synthetases produce aminoacylated tRNA molecules in a two-step reaction (Fig. 14).
   • First, an ATP is coupled to the amino acid to form an amino adenylate. That is, the amino acid is coupled to AMP. This is a high-energy activated intermediate.
     

     

   • In the first step, pyrophosphate (PPi) is also released. As we saw in DNA and RNA polymerization, pyrophosphate is immediately destroyed by pyrophosphatase, making the first sub-reaction an irreversible committed step.
   • Second, the aminoacyl group is transferred to the tRNA. The products are an aminoacyl tRNA, and AMP. Because we start with ATP, and end up with AMP and two inorganic phosphates, coupling of an amino acid to tRNA has an energetic cost of 2 ATP equivalents.

The Ribosome

The “polypeptide polymerase” is the ribosome, an enormous ribonucleoprotein complex. The ribosome has two subunits. Each subunit contains both RNA and many different polypeptides.

1. The small subunit (it is not small, just not as big as the large subunit!) is the “decoding center.” The small subunit’s job is to match each codon on the template to a corresponding aminoacyl-tRNA. This is no easy task. There are 61 codons that specify 20 amino acids (Table 1), so a large majority of the aminoacyl-tRNA molecules that enter the ribosome must be rejected.
2. When a correct codon-anticodon interaction is detected by the small subunit, the large subunit catalyzes the peptidyltransfer reaction — the chemistry of polypeptide elongation.

Remarkably, although each ribosome subunit contains both peptides and rRNA, both the decoding center within the small subunit, and the peptidyltransfer center in the large subunit, are made of ribosomal rRNA. The enzymatic core of the ribosome is a “ribozyme”. This is probably a relic of the ancient origin of ribosomes at the dawn of life, in the so-called RNA world.