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.