Transcription - Biochemistry

Transcription is the second process involved in the production of proteins from a gene. The three processes are DNA replication, transcription, and translation. DNA replication involves replication of DNA from a parent strand, transcription involves the synthesis of a RNA molecule from a DNA molecule, and translation involves the conversion of the mRNA molecule to a polypeptide.

As mentioned, transcription produces an RNA molecule from a DNA molecule (parent strand). Recall that RNA molecules have uracil, whereas DNA molecules have thymine; therefore, the product will contain more uracil.

Amino acids are found in proteins. Since the products of transcription are nucleic acids (RNA molecules) they won’t contain any amino acids. Recall that a nucleic acid consists of pentose sugar molecules (ribose in RNA and deoxyribose in DNA), nitrogenous bases (adenine, guanine, cytosine, thymine (in DNA), and uracil (in RNA)), and phosphate groups.

RNA polymerase is an important enzyme involved in transcription. Its function is to add nucleotides to the growing mRNA chain. Although it adds complementary nucleotides to the DNA, RNA polymerase itself doesn’t bind to complementary DNA sequences, rather it binds at promoters.

There are three types of RNA molecules. First, mRNA molecules are the main products of transcription that undergo translation to produce most of the proteins found in a cell. Second, tRNA molecules are special RNA molecules that facilitate the addition of amino acids to a growing polypeptide chain during translation. Third, rRNA molecules are components of ribosomes and are synthesized in the nucleolus (location of assembly of ribosomes). The enzyme in this question is involved in the production of rRNA molecules. RNA polymerase I is used in production of rRNA molecules. RNA polymerase II is used for mRNA molecules and RNA polymerase III is used for tRNA molecules.

Transcription is the process of producing RNA molecules from a parent strand of DNA. The first step in transcription involves the binding of a sigma factor to inactive RNA polymerase, which in turn binds to the promoter region on the DNA molecule. After the binding of RNA polymerase to the DNA promoter, RNA polymerase gets activated and removes the hydrogen bonds between nucleotides of the double stranded DNA. Finally, RNA polymerase adds complementary nucleotides to the growing RNA molecule.

After completion of transcription, RNA molecules undergo post-transcriptional modifications and exit the nucleus and enter cytoplasm, where it can undergo translation.

Prokaryotic and eukaryotic transcription are similar in many ways; however, they are also different from one another. One main difference is that in eukaryotic transcription there are several events that occur after the completion of transcription. These events are called post-transcriptional modifications.

There are three main post-transcriptional modifications: polyadenylation, capping, and splicing. Polyadenylation involves the addition of multiple adenine molecules at the 3’ end of the newly synthesized RNA molecule. This segment of RNA with multiple adenine molecules is called the poly A tail. Capping involves the addition of a methyl cap to the 5’ end of the RNA molecule. Splicing involves the excision of segments of the mRNA molecule that aren’t used in translation. These segments are called introns and the usable segments are called exons. The exons are ligated back together and the end product is released into the cytoplasm where it can undergo translation.

Products of prokaryotic transcription don’t undergo any of these post-transcriptional modifications and are immediately translated into proteins.

If the coding DNA reads 5’ AATGACGTC 3’, then the template strand would read 3’ UUACTGCAG 5’. The mRNA transcription would read 5’ AAUGACGUC 3’. The corresponding tRNA anti-codons would be 3’ UUA 5’ 3’ CUG 5’; 3’ CAG 5’, which produce the amino acids Asn-Asp-Val. To determine the amino acid sequence, you find the portion of the genetic code table corresponding to the DNA or mRNA, not the tRNA, nucleotides. (That's why ”Leu-Leu-Gln" is incorrect.)

The "central dogma" of biology says that information goes from DNA via transcription to RNA via translation to proteins. Reverse transcriptases, however, employed by retroviruses, synthesize DNA from RNA. As for the other enzymes: one function of helicases (among others) is to pull apart double helix strands. Catalase breaks down hydrogen peroxide. Carboxylase adds a carboxyl group to a substrate, and a ligase creates a bond between two molecules, for example, via a phosphodiester bond.

The spliceosome first releases the 5' exon by forming a lariat structure (2'-5' phosphodiester) bond between two introns in a transesterification reaction. Then, the exons are spliced together with another transesterifaction reaction, and the intron lariat is released.

RNase H is involved in reverse transcriptase, but it is not a ribozyme. RNase P is a ribonuclease that cleaves/processes rRNA and generates 5' ends. The RNA component of RNase P is its catalytic subunit.

In eukaryotes, transcription and splicing could occur simultaneously. Both of these processes take place in the nucleus of the cell, while translation takes place in the cytoplasm. Therefore, translation could not happen at the same time as either transcription or splicing. Replication occurs totally independently from all of the other processes listed.

While some of these answer choices are true, others are false. But even for the choices that are true, only one of them directly answers the question. Let's go ahead and look at each choice.

• Some amino acids can be coded for by more than just one codon

This is a true statement, and is also the correct answer. The degeneracy of the code is due to the fact that, for some amino acids, a number of different codons can result in the same amino acid. For example, the amino acid tyrosine can be coded for by either UAU or by UAC.

• A change in the reading frame alters expression of all subsequent codons

While this is a true statement, it does not answer the question. Nonetheless, it's important to know that mutations which either insert or delete a nucleotide will change the entire rest of the reading frame. Consequently, there are likely to be many errors and the resultant polypeptide will likely not be functional.

• Some codons are able to code for more than just one amino acid

This is a false statement. But be careful. This answer choice looks a lot like the correct one. The difference is that, for this choice, we're talking about a single codon being able to code for more than one amino acid. This is not the case.

• The sequence of one type of molecule is able to code for the sequence of another type of molecule

This is a true statement with regards to gene expression. For transcription, the DNA (sequence of deoxyribonucleotides) serves as a template for the formation of mRNA (sequence of ribonucleotides). And for translation, that same mRNA serves as a precursor for the formation of a polypeptide (sequence of amino acids).

• The same genetic code is not shared by all species

From what we know, this is a false statement. The genetic code is universal, meaning that all living things have been found to have the same codon-amino acid relationship.

The saying that genes can run in either direction is referring to how they can be expressed. Genes can be found on either strand of the DNA molecule. Since these two strands run antiparallel to one another, the genes being read on one strand would be read in the opposite direction on the other strand.

Genes can also be located on various loci within a given chromosome. Also, it is true that genes assort independently during cell division, but this doesn't answer the question. Also, genes can only be read in the direction. They can never be read in the direction. This is because DNA polymerase, the enzyme responsible for elongating the chain, can only add nucleotides to a chain that has a hydroxyl group at the position.

For biochemical purposes, it is important to have an understanding of the "Central Dogma" of molecular biology. DNA multiplies via replication, is turned into RNA via transcription, and finally to proteins via translation. Going back to DNA from RNA is known as reverse transcription, and is the correct answer. The term "reverse translation" can refer to an aspect of cloning, but does not naturally occur.

After transcription, mRNA is modified so that it can be preserved for a longer time in the cell. A nucleotide cap structure is attached to the 5' end of the mRNA and a poly A tail is attached to the 3' end of the mRNA in order to accomplish this goal.

The reverse transcriptase activity of retroviruses like HIV is used to replicate their RNA genome in the host cell. This activity is not needed or present in DNA viruses which can use the host's enzymes to replicate. The reverse transcriptase activity of telomerases is used to prevent telomere ends shortening after multiple replications in somatic cells. Telomeres are short sequences at the end of chromosomes that prevent deterioration of the chromosomes. Retrotransposones are elements that amplify repetitive sequences in the DNA of eukaryotes.

The RNA transcript contains nucleotide bases at each position, which are complementary to the DNA. RNA is synthesized in the 5' to 3' direction from a DNA template strand with antiparallel direction (3' to 5').The coding DNA strand is identical to the RNA transcript with the exception that thymine is replaced with uracil in RNA.

Eukaryotes have three types of RNA polymerase, I, II, and III. RNA polymerase II recognizes the promoter and binds to the promoter forming a preinitiation complex. The polymerase is composed of 10-12 subunits. Transcription factors also bind the promoter (the region of DNA upstream of the start or origin of transcription).In eukaryotes sigma protein factor is not required for transcription to occur.

Termination of transcription in prokaryotes (intrinsic termination) is mediated by special DNA secondary structures (stem and loop structures). Stem and loop structures have nucleotides that are complementary with the adjacent nucleotides. Along with a poly uracil sequence, these structures do not allow transcription to go further.In rho-dependent termination, rho binds to RNA until it reaches a RNA–DNA helical region, where it acts as a helicase and unwinds the complex. This in turn stops transcription. Sigma and TFIID are important in transcription in eukaryotes, not prokaryotes.

We're given the type of enzyme contained within a virus, and we're asked to make a determination of the virus' genetic makeup.

To begin with, we're told that the enzyme is an RNA-dependent RNA polymerase. The name of the enzyme gives us insight into what it does. It requires RNA as a template to produce more RNA.

So if this enzyme can convert RNA into RNA, where does the original RNA come from? The answer is that it must come from the virus. This means that we must be dealing with single-stranded RNA.

Now, the question is to determine the sense of the RNA genome of the virus. That is to say, it can be minus or plus. A minus-sense RNA is one whose complementary sequence can be translated into protein. A plus-sense RNA is one that doesn't need any processing to be translated. Rather, plus-sense RNA can be translated right away. Since we know that the enzyme present is going to produce RNA from RNA, we can then reason that the viral genome is likely minus-sense. When the minus-sense RNA is enacted on by this enzyme, the result is a new strand of RNA that can be translated into protein to serve the needs of the virus.

The correct answer is "six general transcription factors." RNA polymerase I is used to transcribe rRNAs, not mRNAs. The rho factor is used to initiate transcription in prokaryotes, not eukaryotes. Elongator is a factor used by eukaryotes to elongate transcripts and speed up transcription, but is not required to initiate transcription.