Question

1-How many base pairs are in a molecule of phage T2 DNA 52 mm long? (2 points)

2- It has been shown that infectious agents such as viruses often exert a dramatic effect on their host cell’s genome architecture. In many cases, viruses induce methylation of host DNA sequences in order to enhance their infectivity. What specific host gene functions would you consider as strong candidates for such methylation by infecting viruses? (2 points)

3- Mammals contain a diploid genome consisting of at least 109 bp. If this amount of DNA is present as chromatin fibers, where each group of 200 bp of DNA is combined with 9 histones into a nucleosome and each group of 6 nucleosomes is combined into a solenoid, achieving a final packing ratio of 50, determine (a) the total number of nucleosomes in all fibers (2 points), (b) the total number of histone molecules combined with DNA in the diploid genome (2 points), and (c) the combined length of all fibers (2 points).

Answer 1

1. No of base pairs present in 3.4nm long DNA = 10bp

So, no of base pairs in 1nm = 10/3.4 = 2.94 bp

Lebgth of T2 phage DNA= 52nm = 5.2 x 10⁷nm

No of base pairs in this T2 phage = 2.94 x 5.2 x 10⁷= 15.28 x 10⁷= 1.53 x 10⁵bp

2. Viruses infect host cell for resources for their own replication to enhance their infectivity.Thus, the most common approach to increase the infectivity is manipulation of cell cycle to achieve a favorable cellular environment. Therefore, viruses methylate gene involved in the cell cylt to activate their DNA amplification and hence enhance their infectivity.

3. a) No of base pair in a nucleosome = 200bp

No of given base pair = 10⁹bp

No of nucleosomes in the given genome= no of given bp/no of base pair in one nucleosome

= 10⁹/200 = 5 x 10⁶nucleosomes are present

b) Now no fo histone protein in one nucelosome = 9

No of histone in the total genome = no of nucleosomes in the genome x no of histones in one nucleosome

= 5x 10⁶ x 9 = 4.5 x 10⁷

c) Total base pairs in the genome = 10⁹

Length of 1 base pair = 3.4 anstrong

packing ratio = 50

So the combined length of all the fibres  = (10⁹ x 3.4)/50 = 6.8 x 10⁷