Mark Botirius

1 Two of the three key parts of evolution are: 1. Changes in the genetic material; 2. Natural selection. Discuss how each of process acts on each of the following:a. DNA (generally) b. rRNA genes c. mRNA genes d. tRNA genes

1a,1. General changes in DNA happen in a variety of ways. Overall, the most common types of changes are base substitutions, indels, and inversions. (Rogers 173). Base substitutions are further categorized as either transitions (a purine or pyrimidine is exchanged for a base of the same type) or transversions (a purine or pyrimidine is exchanged for one of the opposite type) with transitions being more common than transversions. Base substitutions are point mutations that, due to the redundant nature of the genetic code, can either result in a synonymous or nonsynonymous mutation. Synonymous mutations, of course, will have little or no effect (codon bias notwithstanding), while the effect of nonsynonymous mutations will depend on the properties of the replacement amino acid and can range from no effect to an extremely deleterious one. Next, there are indels. As its name indicates, this is a change in the DNA that results in the addition or deletion of genetic material. The effect of these mutations can vary enormously depending on what is being inserted or deleted, and where it is inserted or deleted. For example, if the indel is a point mutation, it often causes a frameshift that changes the reading frame of the gene that alters the entire amino acid sequence. Usually this produces a nonworking protein. Of course, indels are not limited to single bases. For example, replicative transposons (Pierce 304) can insert long stretches of DNA, often disrupting host genes in the process. Also, Indels can have no effect as well. Much of eukaryotic DNA is non-coding. If the insertion or deletion occurs in an area of non-coding DNA that does not also have some other function (such as a promoter), or an area of an intron not involved in some other activity (such as splicing), then the indel can have no effect. Indels can also come from a prophage left behind by viruses during their lysogenic cycle that has since mutated. (Rogers 192-196)

I would be negligent if I didn’t pause here to discuss what could be the most important indel of all, that is also largely unique to this course. That is, the changes in DNA that is the direct result of endosymbiotic evolutionary pathways. I would give a specific citation here, except this source of genetic variation is so important and fundamental to this course I would have to cite the first nine chapters of Integrated Molecular Evolution. Examples of this type of indel are the presence of mitochondrial and chromosomal genes in the host chromosome. These genes were once a part of the endosymbiont genome that have since moved (inserted) into the host chromosome and have been subsequently lost (deleted) from the original endosymbiont. It should be noted that examples of this type of indel are not limited to bacterial endosymbionts within eukaryotes. A case in point is when a eukaryote has another eukaryote as an endosymbiont.

Lastly, there are inversions. In this genetic change, the DNA forms a loop during recombination when one of the strands breaks in two places. When the strand is reintegrated it is in a reverse orientation causing the sequence to be subsequently inverted. The effects from this mutation can be either minimal, apoptotic, or carcinogenic. (Rogers 173) (Iwasa 478)

1a,2. Having discussed the effects that genetic changes have on DNA (generally), I now turn my attention to the effects of natural selection on these changes. In other words, how does natural selection influence changes at the level of DNA? One example is the proofreading function of DNA polymerase. This is one instance where the first change in the genetic material discussed (i.e. base substitutions) encounters a natural selection process. There are many DNA repair pathways designed to correct base substitutions. For example, DNA polymerase has proofreading

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Mark Botirius

capabilities during replication and its corrective capabilities can vary depending on the type of base substitution. For instance, if the base substitution is due to deamination, DNA polymerase easily corrects the deamination of cytosine to uracil but does not correct the deamination of 5-methylcytosine to thymine. The reason is because uracil does not belong in DNA, while thymine does. Base substitutions of cytosine to uracil are selected against, while base substitutions of 5-methylcytosine to thymine are not. (Pierce 494) Likewise, transversions cause the DNA helix to distort more than translations because purines consist of two rings and pyrimidines consist of only one ring. As a result, base substitutions that are transversions are also selected against, so that transitions are more common. Indels can also exhibit selection in that some types are more common than others. For example, genome wide repeats alone comprise over 43% of the human genome. (Watson 206) Genome wide repeats are caused almost exclusively by transposons, meaning that in terms of different types of indels, transposons are highly selected for. Now that I have covered natural selection as it affects general DNA changes, I can address the more specific question of what it means with respect to a particular gene, namely rRNA, mRNA, and tRNA genes.

1b, c, d; 1. All genes have a few basic and fundamental parts in common that relate to the structure and function of the molecules they produce. These parts are the promoter, 5’ UTR (or the first exon), exons (e.g. euchromatin), introns (e.g. heterochromatin in eukaryotes), 3’ UTR (the last exon), and finally a termination site. All of the genetic changes I have discussed so far have the ability to occur in any of these fundamental gene segments and result in a range consisting of no effect to a deleterious effect for reasons I have already described. Therefore, it is much more efficient to simply state what can happen when these regions are mutated, since I have already covered the types of mutations that can occur, and then apply that to rRNA mRNA, and tRNA genes.

The promoter is the location on the gene where transcription is initiated. Changes to the genetic material of the promoter can result in the gene being up regulated, down regulated, or silenced altogether. This would correspondingly result in a positive selection (if the genes are up regulated) to a negative selection (if the genes are down regulated or silenced). If these genes are the ones currently being examined, the results are a corresponding change in the expression of rRNA, mRNA, or tRNA genes. Next is the 5’ UTR.1 This region codes for the 5’ UTR region of mRNA, which is where the 5’ cap is added. This cap is important because it is where cap binding proteins attach. CBP’s are needed for proper attachment of the mRNA to the ribosome. A deleterious mutation in this part of the gene could result in improper binding of mRNA to the ribosome and therefore negative selection. (Rogers 163) Following the 5’ UTR region on the list are the exons. This is the region where the amino acids are coded. Genetic changes here affect the proteins that the gene produces. The range of outcomes proceeds from none (i.e. a synonymous or other type of silent mutation) to beneficial, (e.g. an amino acid change that produces a beneficial protein) to lethal (e.g. a frameshift mutation that eliminates a critically needed protein). The resulting selection possibilities correspondingly range from strongly selected for to strongly selected against.

Usually, exons refer to genes that code for proteins. With regards to rRNA, and tRNA however, the coding regions analogous to exons are the SSU, LSU, 5s, 5.8s, ITS1, and ITS2. Mutations in these genes can have a have a variety of effects. The function of the SSU, for example, is to transport the mRNA transcript to the LSU, where it associates with the LSU and properly positions the transcript in the ribosome duplex for translation. If this gene has a deleterious

1In DNA, this is not always stated as a separate region like it is for mRNA, however it is always located within the first exon downstream from the promoter (in eukaryotes). In other words, the DNA code for the 5’ UTR of mRNA is located in the first exon. Likewise, the 3’ UTR is located in the last exon.

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Mark Botirius

mutation, then it is possible that the SSU will not be able to transport the mRNA, or will not be able to properly interact with the LSU resulting in an inability to translate mRNA to proteins.

The large subunit is where the enzymatic function is located. It actually catalyzes the reaction that joins the amino acid to the growing peptide chain. Mutations in this gene could result in a loss of enzymatic function, an inability to properly associate with the SSU, or an inability to properly associate with the mRNA and/or the growing peptide chain. All of this could lead to a loss of translation of the mRNA. Furthermore, mutations in either the SSU or LSU genes could distort the molecules, making it impossible to exit the nuclear pores to travel to the cytoplasm, where they need to be.

The function of the tRNA’s is to bring the amino acids to the ribosome and they also set the genetic code. Mutations in these genes will likewise affect these functions. Changes in the anticodon region of the tRNA could therefore change the amino acid code for that tRNA or cause it to lose the ability to bind to the tRNA altogether. In addition, the tRNA itself may lose the ability to interact properly with the ribosome. All of these possibilities would result in a negative selection, which is why these genes are so strongly conserved

Next on my list are introns. Surprisingly, introns can be found in all three types of RNA genes. Changes in these genes can affect how the introns are spliced or even if they are spliced at all. With regards to mRNA, this could, of course, cause a failure to produce a needed protein or produce a mutated protein. In the case of rRNAs and tRNAs, this would cause a distortion in the shape of the molecules which would certainly result in a non-functional molecule.

Lastly, a mutation in the 3’ UTR region of the gene that codes for mRNA could affect the 3’ tail that is added at the end of the molecule. The 3’ poly A tail functions to protect the mRNA from degradation. Without this tail, the life of the mRNA is greatly reduced, thereby destroying its functionality.

Finally, it would be irresponsible if I didn’t mention one last facet of selection on genes in general, and rRNA, mRNA, and tRNA genes in particular. From the standpoint of evolution, it is probably one of the most important and again, unique to this course. It is the selection that occurs following an endosymbiotic event. After an endosymbiotic event, there exists two sets of genes for

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Figure 1. This is an illustration from a power point presentation from Dr. Rogers class depicting the rRNA, and to a lesser extent, tRNA genes.

Mark Botirius

many cellular processes. Modern analysis shows, however, that the genes that finally wind up in the host chromosome are a mixture of host and symbiont genes. Therefore, although it could be due to purely random events, it is also likely that the genes that ultimately wound up in the chromosome were the result of selection. Those rRNA, mRNA, and tRNA genes that best served the organism were positively selected.

2. Ribosomal genes are needed in multiple copies. Explain why. Outline some of the mechanisms for creating or maintaining copy numbers sufficient for cells and organelles. Explain how these processes might have developed and been selected for during evolutionary processes.

The reason why ribosomal genes are needed in multiple copies can be demonstrated with a simple mathematical argument. There are an estimated 106 proteins in prokaryotes and 109 for each eukaryote (Rogers 76) and the ribosome has a clock speed of only 20 amino acids per second. (Watson 521) Therefore, in order to produce the needed number of proteins, prokaryotes have approximately 20,000 ribosomes, and eukaryotes contain approximately ten million or more. The speed at which RNA polymerase can synthesize a ribosome is between 50 to 150 nucleotides per second. (Rogers 76) Utilizing Dr. Rogers’ figure of 100 nucleotides per second gives us a total of one day and three hours for a single gene to produce enough ribosomes for a prokaryote, and more than 30 years for a eukaryote. (Rogers 76-77) Clearly, one gene is not sufficient. In fact, the gene copy number per cell is typically in the hundreds.

So how do cells create enough ribosomal genes? Several methods have been discussed in class. The two I find the most interesting are the method used by the protozoan Tetrahymena and the frog Xenopus laevis. On either side of the Tetrahymena ribosomal gene are special sequences (called A’ and M repeats) that are recognized by an endonuclease. A section of DNA containing the ribosomal gene is cut out, and telomeres are added to all of the ends except the upstream end of the excerpt. This end circularizes and a replication bubble forms that travels towards the ribosomal gene. The replication bubble travels all the way to the open end of the excerpt, replicating the ribosomal gene in the process, and creating a DNA segment that is symmetrical about its center.

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Figure 2. An illustration from Dr. Rogers’ PowerPoint. The top picture shows how one end of the excerpt circularizes. The other end is protected from degradation by a telomere. The replication bubble travels downstream, replicating the ribosomal gene in the process. The end result is a DNA segment that contains two copies of the ribosomal gene and is symmetrical about its center.

Mark Botirius

The new, linear DNA segment contains A’-M sites and it undergoes further rounds of replication to produce the needed copies of ribosomal genes.

The mechanism for the frog Xenopus laevis is similar. Like Tetrahymena, cuts are made at signal sequences flanking the ribosomal gene. In the case of Xenopus laevis, however, the cuts are made at one or more ribosomal gene repeats producing several excerpts. These excerpts individually circularize and then undergo rolling circle replication to increase their number to produce the needed gene copy number.

The reason I find these two mechanisms interesting is that the specificity of the splicing enzymes makes them very similar to restriction endonucleases found in bacteria. In addition, rolling circle replication is also found in bacterial conjugation. This strongly suggests to me that these mechanisms may have been obtained by these two eukaryotes from their bacterial endosymbionts in the course of their evolution. This answers the question of how these processes might have developed in the course of the evolution of these two particular organisms. These bacterial mechanisms were conducive to

creating needed copies of ribosomal genes and were therefore subjected to positive selection and retained.

Another mechanism for increasing ribosomal gene copy number is provided by amphibian oocytes. The amphibian oogonium divides asymmetrically, with virtually all of the mRNAs and ribosomes going to the larger oocyte. This effectively doubles the ribosomal number for the primary oocyte thereby providing the needed number of ribosomes. One way this mechanism could have evolved is by selective pressures at the organismal level. Initially, the division of the oogonium was symmetrical, producing zygotes of equivalent fitness and ribosomal numbers. Eventually, some of the oogonium produced daughter oocytes with slightly different proportions of the parental ribosomes. Those oocytes that contained a slightly larger proportion of ribosomes were a little more fit, and so were positively selected to a slight degree. Eventually, time and natural selection resulted in the mechanism that produced the fittest zygote, and that mechanism was one where the oogonium divides asymmetrically.

Lastly, one of the most common mechanisms to produce the needed ribosomal number is present in plants all around us. Many plants are polyploid through a process called endomitosis, where the genome is copied without cytokinesis. In other words, instead of increasing only the ribosomal genes, the entire genome is duplicated leading to a very high C value and therefore a very high number of ribosomal genes as well. (Rogers 121) How could have polyploidy evolved in plants? Personally, I see a few intriguing clues in the characteristics of the life cycles of many plants. In animals (like humans) our multicellular stage is diplontic. That is, when we exist as a multicellular organism, our genome is diploid. Plants, on the other hand, are not limited to this diploidy. During

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Figure 3. Another Dr. Rogers’ PowerPoint slide showing the rolling circle replication mechanism used by Xenopus laevis increase ribosomal gene copy number.

Mark Botirius

part of their multicellular life cycle, they are diploid, like us. During a different part of their multicellular life cycle, they are haploid. This flexibility in ploidy I think is what eventually led to their ability to be polypoid, and it also explains why animals, such as humans, do not tolerate polyploidy (it is lethal to us). I think a possible evolutionary route to polyploidy is that some ancient plant ancestor whose life cycle involved “alternation of generations” failed to reduce its chromosome number (N) during meiosis. This is possible because we know that polyploidy is normally due to a failed reduction division in meiosis. (Futuyma 505) This increase in ploidy led to an increase in beneficial and needed genes, such as ribosomal genes, which led to an increase in fitness and positive selection.

Works CitedFutuyma, Douglas J. Evolution. 3rd. Sunderland: Sinauer Associates, 2013. Hardback.

Iwasa, J and Marshall, W. Karp's Cell and Molecular Biology. 8th. Hoboken: Wiley, 2016. Book.

Pierce, Benjamin A. Genetics, A Conceptual Approach. New York: W.H. Freeman and Company, 2012.

Rogers, Scott Orland. Integrated Molecular Evolution. 2nd. Boca Raton: CRC Press, 2017. Hardback.

Watson, et al. Molecular Biology of the Gene. 7th. Boston: Pearson, 2014.

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