Surface plasmon resonance method

Principle:

This sensitive technique provides the information of the binding interactions between an injected analyte and an immobilized biomolecule in real time without the presence of a non-invasive, label means. As the analyte binds to the immobilized particle, this forms an associated increase in mass that proportionally increases in refractive index which is observed as a shift in the resonance angle. A flow injection analysis configuration is commonly employed in which the analyte of interest, solvated in a buffer solution, is transported across the sensing surface, where it interacts with the immobilized biomolecule. When combined with appropriate surface chemistry, microfluidics and software, this technique can be used in many applications including:

• Affinity analysis
• Kinetic analysis
• Concentration assays
• Active concentration assays
• Binding stoichiometry
• Thermodynamic analysis
• Study of interaction mechanisms
• Dependence of interaction on environmental conditions
• Routine screening
• Ligand-fishing
• Epitope mapping

From RNA to translation - special features

The scanning model and the Kozak consensus sequence and how it influences initiation of translation in eukaryotes

Ribosomes enter at the 5’end of the mRNA and scan down the RNA to look for the first AUG codon that is in “good context.” The nucleotides that influence initiation are A/G at -3 and G at +4.

Diagram the features of a sequence that facilitates ribosomal frame-shifting in Eukaryotes (only)

RNA synthesis and Processing

The differences in transcription elongation of prokaryotes and eukaryotes

Eukaryotes:

-          Transcription elongation in eukaryotes is tightly coupled to RNA processing.

-          The RNA splicing removes intron sequences from newly transcribed pre-mRNA and join the exons (coding regions) together.

-          The splicing process of RNA is catalyzed by spliceosome. Sometimes, alternative splicing happens, allows differential expression of exons .

-          The mRNa is capped and polyadenylated in the nucleus and then exported to the cytoplasm for translation.

-          The nucleolus is a ribosome-producing factory.

-          Bote: although intron will be removed during the process of splicing, but it is very important because it controls many process in transcription.

Prokaryotes:

-          Both transcription and translation occur in the same compartment. There are no introns to splice out and the mRNA is not capped or polyadenylated.

The sequence required for intron removal in eukaryotic RNA and the 3 major steps in the RNA splicing reaction

-    In the mRNA strand, especially in the intron sequence, the adenine initiates its attack on 5’ splice site (the upstream of the intron sequence).

-    3’-OH of upstream exon reacts with 5’ of the downstream exon.

-    Exons are joined, and the lariat is released.

-    In the intron, the 5’ upstream has GU invariant bases while the 3’downstrem has AG invariant bases

The major steps in generating the 3’ end of eukaryotic mRNA

As RNA pol II reaches the end of a gene, the position of the 3’end of each mRNA molecule is ultimately specified by a signal encoded in the genome. These signals are transcribed into RNA as the RNA pol II moves through them, and they are then (as RNA) by a series of RNA-binding proteins and RNA-processing enzymes.  Two multisubunit proteins, CPSF and CstF are part of the RNA pol tail, and then transferred to the 3’end processing sequence on the RNA.

Once these proteins bind to the specific nucleotides, additional proteins assemble with them to create the 3’end of the mRNA. First, the RNA is cleaved. Next poly-A pol enzymes adds, one at a time, nucleotides to the 3’end produced by the cleavage without the requirement of template ( Note that: poly-A pol is not directly encoded in the genome).

As the poly-A tail is synthesized, proteins called poly-A-binding protein assemble onto the poly-A tail and remain their bound to direct mRNA to travel from the nucleus to cytosol as well we the synthesis of a protein on the ribosome.

Some of random questions for science

Assuming that there were no time constraints on replication of the genome of a human cell, what would be the minimum number of origins that would be required?  If replication had to be accomplished in an 8-hour S phase and replication forks moved at 50 nucleotides/second, what would be the minimum number or origins required replicating the human genome? (The human genome comprises a total of 6.4 X 10⁹ nucleotides on 46 chromosomes).

Number of nucleotides to fully replicate in 8 hr:

50 (nucleotide/sec ) × 8hr × 3600 sec  * 2= 2880000 nucleotides for 1 origin

Number of origins required to replicate the human genome:

            6.4×10⁹ / 2880000 = 2222 origins

Explain the following:  Spontaneous depurination and removal of a deaminated C by uracil DNA glycosylase both leave an identical intermediate, which is the substrate recognized by AP endonuclease.

Ans: both spontaneous depurination formed by spontaneous hydrolysis and removal of a deaminated C by uracil DNA glucosylase resulted an identical intermediate which is the phosphate sugar backbone, thus one strand of DNA helix has a missing base. This missing base is recognized by AP endonuclease and phhosphodiesterase that cleaves the sugar-phosphate group which is absent of a nitrogen base and fill the gap by DNA polymerase and ligation.

A particular 225-bp segment of human DNA was used to assemble nucleosomes and then digested with micrococcal nuclease. Next, the reaction mixture was extracted to remove proteins, leaving only the DNA for subsequent analysis. This resulted in a quantitative yield of DNA fragments 146-bp in length. Explain this result.

When a segment of human DNA is used to assemble nucleosomes, parts of the DNA strand wrap around the histone complex to form the nucleosome. Each nucleosome is linked together by the same DNA strand.  From the previous statement above, the treatment with micrococcal nuclease results the break of DNA linkage which connected nucleosomes together. Now, only nucleosomes are present but they do not connect to each other. Extraction to remove proteins will leave only the DNA segment which the segment that associates with histone complex. Thus, after treatment, the DNA fragment only has 146 bp in length rather than 225 bp. The missing base pairing can only be found from the other DNA fragments that belong to DNA linkage which is digested with micrococcal nuclease at the beginning.

Subsequent digestion of these fragments with a restriction enzyme that cuts once within the original 225-bp sequence produced two well-defined bands at 37 and 109 bp. Why do you suppose two well-defined fragments were generated by restriction digestion, rather than a range of fragments of different sizes?

Restrictive enzyme is the special enzyme that binds at certain site (which is known as restriction site), and cleaves the DNA ino 2 specific fragments rather than multiple fragments catalyzed by common nuclease. Thus, in this case, only 2 fragments are expected to be present.

Cre recombinase is a site-specific enzyme that catalyzes recombination between two LoxP DNA recognition sequences.   Cre recombinase pairs two PoxP sites in the same orientation, breaks both duplexes at the same point within the LoxP sites, and joins the ends with new partners so that each LoxP site is regenerated. Predict the rearrangement of sequences that will be generated by Cre-mediated site-specific recombination for each of the two DNAs shown below.(draw one picture for each DNA).

You hypothesize that in order for new histone subunits to be incorporated into chromatin after DNA replication or repair, histone H3 must be able to be phosphorylated at serine 10.  Describe a possible approach to test your hypothesis.

To test this hypothesis, first of all phosphorylated of the histone H3 and then let this histone H3 incorporate into chromatin where it associates with other histones and DNA strand. When H2A. H2B, H3 and H4 associates together with DNA strand, this complex will be called nucleosome where each nucleosome connects to each other by DNA strand. Using micrococcal nuclease is to digest the DNA linkage, that leaves the nucleosome. Extraction of proteins from the nucleosome is then used to measure the amount of proteins.

Apply the same step above with histone without phosphorylation is used as a control to compare the level of proteins present from nucleosome after phosphorylation. If phosphorylated H3 actually incorporates into the chromatin, the level of proteins present in nucleosome will be greater; otherwise, if the level of proteins present in nucleosome will low, this indicates that phosphorylated histone H3 cannot incorporate into chromatin.

You suspect that exposure of a cancer cell line to a specific chemotherapeutic drug is causing DNA double-strand breaks in these cells.  You want to determine whether the cancer cells activate the DNA repair process.  Describe one possible experiment you could do to determine if this is true. 

At the end of replication, to protect the end of DNA strands, telomerase must be induced to generate telomere sequence. Thus, in the cancer cell, the level of telomerase is highly greater comparing to the normal cell because cancer cell is immortal. Measurement of the telomerase level will indicate whether cancer cell is still active or inactive by extraction of telomerase before and after chemotherapeutic. If the level of telomerase declines, this means that most of cancer cell is killed; otherwise, if the level of telomerase is still the same, this suggests that this chemotherapeutic is failed.

Transposable elements make up over 40% of the human genome, and are inserted more or less randomly throughout the human genome, with a few exceptions.  Some of these exceptions include the Hox clusters of genes.  Shown in the figure are examples of two human genomic regions.  The top region codes for part of one of the human Hox clusters, the HOXA cluster.  Shown is the region coding for HOXA4, A5, A6, A7, A10, A11, A13 (there is no HOXA8 or A12).  The bottom region is that which codes for the MLL gene on chromosome 11.  The dark rectangles on the mRNA lines indicate exons.  The locations of repeat sequences within both gene regions are shown as vertical lines (as indicated by SINE, LINE, LTR, which are examples of types of repeat sequences). The appropriate differential expression of the Hox genes along the anteroposterior axis of the developing embryo establishes the basic body plan for humans (and for other animals).  This is a very highly regulated process, with critical timing for when the individual genes in the cluster get turned on or off, and to what level and where they are expressed.  One Hox cluster located on chromosome 2 is comprised of 9 genes and covers about 100 kb of DNA (only about 10 kb of this DNA codes for the Hox proteins. 

Why do you suppose that transposable elements are so rare in the Hox cluster as compared to other regions? HOXA4 through HOXA13

From the figures of both Hox cluster gene and MLL gene and based on the presence of LINE and SINE sequence, this indicates that Hox gene is located at the unique sequence while MLL gene is located at the repeat sequence. In nature, only transposon elements can be found in repeat sequence region on chromosome. In fact, the unique sequence region only contains protein-coding region ( a small portion of sequence found in unique sequence) for protein expression. Thus, this is the reason to explain why Hox cluster gene do not show a lot transposable elements in its DNA sequence.  

Competitive ELISA

Simple idea for extraction nucleic acids from cell

Principle methods for extraction of nucleic acids from cell

Cell is cultured and harvested
Lyse cell will detergent to disrupt membrane and inactivate nuclease
Using phenol to seperate nucleic acids from proteins ( Note: proteins will be precipitated while nucleic acids remain in aqueous phase)

Separation of DNA from RNA is based on its solubility in alcoohol
Using DNase or RNase to remove the nucleic acids that is not necessary

Blocking buffer

Before using antibodies to detect proteins that have been dotted or transferred to a membrane, the remaining binding surface must be blocked to prevent the nonspecific binding of the antibodies. Otherwise, the antibodies or other detection reagents will bind to any remaining sites that initially served to immobilize the proteins of interest. In principle, any protein that does not have binding affinity for the target or probe components in the assay can be used for blocking. In practice, however, certain proteins perform better than others because they bind to the membrane or other immobilization surface more consistently or because they somehow stabilize the function of other system components. In fact, no single protein or mixture of proteins works best for all Western blot experiments, and empirical testing is necessary to obtain the best possible results for a given combination of specific antibodies, membrane type and substrate system.

Blocking Nonspecific Sites

The membrane supports used in Western blotting have a high affinity for proteins. Therefore, after the transfer of the proteins from the gel, it is important to block the remaining surface of the membrane to prevent nonspecific binding of the detection antibodies during subsequent steps. A variety of blocking buffers ranging from milk or normal serum to highly purified proteins have been used to block free sites on a membrane. The blocking buffer should improve the sensitivity of the assay by reducing background interference and improving the signal to noise ratio. The ideal blocking buffer will bind to all potential sites of nonspecific interaction, eliminating background altogether without altering or obscuring the epitope for antibody binding.

The proper choice of blocker for a given blot depends on the antigen itself and on the type of detection label used. For example, in applications where alkaline phosphatase conjugates are used, a blocking buffer in TBS should be selected because PBS interferes with alkaline phosphatase. For true optimization of the blocking step for a particular immunoassay, empirical testing is essential. Many factors, including various protein:protein interactions unique to a given set of immunoassay reagents, can influence nonspecific binding. The most important parameter when selecting a blocker is the signal:noise ratio, measured as the signal obtained with a sample containing the target analyte, as compared to that obtained with a sample without the target analyte. Using inadequate amounts of blocker will result in excessive background staining and a reduced signal:noise ratio. Using excessive concentrations of blocker may mask antibody:antigen interactions or inhibit the marker enzyme, again causing a reduction of the signal:noise ratio. When developing any new immunoassay, it is important to test several different blockers for the highest signal:noise ratio in the assay. No single blocking agent is ideal for every occasion since each antibody-antigen pair has unique characteristics.