Question

This sequence represents the non-template strand of the entire transcribed region (i.e., the first G is +1) of the ILTCD gene (which is the “I Love The Central Dogma” gene):

5’GAGATTCGATGGTAAGTCTCATTGCGTCCTGAGTCCTAATTTAAATAAAGCCTTTGTAATACAGGGCAATAAAGGCCTACGC 3’

1. What are the sequences of each of the three possible introns in this provided sequence.
2. If the first hexanucleotide is used for 3’ end processing, how many exons will the final mRNA contain?
3. If a regulatory protein sat down on the first hexanucleotide and blocked it, describe what will now happen to make the new RNA, beginning with the skipping of this hexanucleotide. Be sure to indicate how many exons would be in this new final mRNA.
4. Write out the coding sequence (from the initiator codon to the stop codon) for the shorter of the two mRNAs, and then write out the amino acid sequence it codes for.
5. How many tRNAs will be needed to translate this protein?
6. If this gene is a housekeeping gene, but the regulatory protein mentioned in ## (let’s call it CPSF-blocker) was only active in neurons, which mRNA would you find in neurons, and which one would you find in liver cells?
7. The expression of CPSF-blocker is regulated by a regulatory transcription factor – why would you expect this regulatory transcription factor to interact with a HAT protein in neurons?

Answer 1

Ans:- The intron has been a big biological mystery since it was first discovered in several aspects. First, all of the completely sequenced eukaryotes harbor introns in the genomic structure, whereas no prokaryotes identified so far carry introns. Second, the amount of total introns varies in different species. Third, the length and number of introns vary in different genes, even within the same species genome. Fourth, all introns are copied into RNAs by transcription and DNAs by replication processes, but intron sequences do not participate in protein-coding sequences. The existence of introns in the genome should be a burden to some cells, because cells have to consume a great deal of energy to copy and excise them exactly at the correct positions with the help of complicated spliceosomal machineries. The existence throughout the long evolutionary history is explained, only if selective advantages of carrying introns are assumed to be given to cells to overcome the negative effect of introns. In that regard, we summarize previous research about the functional roles or benefits of introns. Additionally, several other studies strongly suggesting that introns should not be junk will be introduced.

Introns are a common eukaryotic event. Several features of interrupted genes are:

• The sequence order is the same as in the mRNA
• The structure of an interrupted gene is identical in all tissues.
• Introns of nuclear genes have termination codons in all three reading frames.

Exon - RNA sequences in the primary transcript that are found in the mRNA

Intron - RNA sequences between exons that are removed by splicing

Introns have been found in eukaryotic mRNA, tRNA and rRNA, as well as chloroplast, mitochondrial and a phage of E. coli. Eubacteria are the only species in which introns have not been found. In general, genes that are related by evolution have related organizations with conservation of the position at least some introns. Furthermore, conservation of introns is also detected between genes in related species.

The amount and size of introns varies greatly. The mammalian DHFR has 6 exons that total about 2000 bases, yet the gene is 31,000 bases. Likewise, the alpha-collagen has 50 exons that range from 45-249 bases and the gene is about 40,000 bases. Clearly two genes of the same size can have different number of introns, and introns that vary in size.

Some species will have an intron in a gene, but another species may not have an intron in the same gene. An example is the cytochrome oxidase subunit II gene of plant mitochondria where some plant species have an intron in this gene and others do not.

Features of Nuclear Splicing Junctions

1. No extensive homology exists between the ends of an intron.
2. The intron/exon junctions, though, do have well-conserved short sequences.

  EXON | INTRON | EXON | | \_/ \_/ A G G T A........C A G 64 73 100 100 62 65 100 100 Percent occurrence 

Splicing of hnRNAs

1. The left exon is cleaved to produce a linear molecule and a right intron/exon molecule.
2. The left end of the right intron/exon molecule forms a 5'-2' linkage to the adenosine in the sequence 5'-CUGAC-3'. This sequence is about 30 bases upstream of the right exon junction. This produces a lariat structure.
3. The right junction is cut, the lariat becomes single-stranded and is degraded, and the exons are spliced.

Splicing appears to involve a complex called the spliceosome. This complex consists of RNA and protein, and appears to be composed of a group of small nuclear ribonucleoprotein particles or snRNPs. These snRNPs each seem to have a role in the splicing process. We will talk about just one snRNP, U1.

U1 binds to the 5' splice site. The RNA is complementary to 4-6 nucleotides of the 5' end, but RNA cannot bind alone, it requires the proteins constituent of the particle. An important question is whether U1 is required. If the sequence of the splice site is mutated, binding to the left junction will not occur. But if the U1 RNA is altered to be complimentary to the mutation in the left junction, binding is restored.

The genes in DNA encode protein molecules, which are the "workhorses" of the cell, carrying out all the functions necessary for life. For example, enzymes, including those that metabolize nutrients and synthesize new cellular constituents, as well as DNA polymerases and other enzymes that make copies of DNA during cell division, are all proteins.

In the simplest sense, expressing a gene means manufacturing its corresponding protein, and this multilayered process has two major steps. In the first step, the information in DNA is transferred to a messenger RNA (mRNA) molecule by way of a process called transcription. During transcription, the DNA of a gene serves as a template for complementary base-pairing, and an enzyme called RNA polymerase II catalyzes the formation of a pre-mRNA molecule, which is then processed to form mature mRNA (Figure 1). The resulting mRNA is a single-stranded copy of the gene, which next must be translated into a protein molecule.

Figure 1: A gene is expressed through the processes of transcription and translation.

During transcription, the enzyme RNA polymerase (green) uses DNA as a template to produce a pre-mRNA transcript (pink). The pre-mRNA is processed to form a mature mRNA molecule that can be translated to build the protein molecule (polypeptide) encoded by the original gene.

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Figure Detail

During translation, which is the second major step in gene expression, the mRNA is "read" according to the genetic code, which relates the DNA sequence to the amino acid sequence in proteins (Figure 2). Each group of three bases in mRNA constitutes a codon, and each codon specifies a particular amino acid (hence, it is a triplet code). The mRNA sequence is thus used as a template to assemble—in order—the chain of amino acids that form a protein.