Retroviruses do no follow central dogma. Comment.
Retroviruses do not follow central dogma of biology (DNA $\rightarrow$ RNA $\rightarrow$ Protein) because their genetic material is not DNA. Instead they have RNA that is converted to DNA by the enzyme reverse transcriptase.
In an experiment, DNA is treated with the compound which tends to place itself amongst the stacks of nitrogenous base pairs. As a result of this, the distance between two consecutive base increases. From $0.34-0.44 \mathrm{~nm}$ calculate the length of DNA double helix (which has $2 \times 10^9 \mathrm{bp}$ ) in the presence of saturating of this compound.
$$ \text { The length of DNA double helix }=2 \times 10^9 \times 0.44 \times 10^{-9} / \mathrm{bp} . $$
What would happen if histones were to be mutated and made rich in acidic amino acids such as aspartic acid and glutamic acid in place of basic amino acids such as lysine and arginine?
If histones were mutated and made rich in acidic amino acids. They will not be able to serve the purpose of keeping the DNA coiled around them. This is because DNA is negatively charged molecule and histones are positively charged because of basic amino acids.
So, they are attracted to each other. If histones become negatively charged, instead of binding, they will rather repel DNA. The packaging of DNA in eukaryotes would not happen. Consequently, the chromatin fibre would not be formed.
Recall the experiments done by Frederick Griffith, Avery, MacLeod and McCarty, where DNA was speculated to be the genetic material. If RNA, instead of DNA was the genetic material, would the heat killed strain of Pneumococcus have transformed the R-strain into virulent strain? Explain.
RNA is more liable and prone to degradation (owing to the presence of $2^{\prime} \mathrm{OH}$ group in its ribose). Hence, heat-killed S-stain may not have retained its ability to transform the R-strain into virulent form if RNA was its genetic material.
You are repeating the Hershey-Chase experiment and are provided with two isotopes ${ }^{32} \mathrm{P}$ and ${ }^{15} \mathrm{~N}$ (in place of ${ }^{35} \mathrm{~S}$ in the original experiment). How does you expect your results to be different?
Use of ${ }^{15} \mathrm{~N}$ will be inappropriate because method of detection of ${ }^{32} \mathrm{P}$ and ${ }^{15} \mathrm{~N}$ different ( ${ }^{32} \mathrm{P}$ being a radioactive isotope while ${ }^{15} \mathrm{~N}$ is non-radioactive but is the heavier isotope of nitrogen).
Even if ${ }^{15} \mathrm{~N}$ was radioactive then its presence would have been detected, both inside the cell ( ${ }^{15} \mathrm{~N}$ incorporated as introgenous base in DNA) as well as in the supernatant, because ${ }^{15} \mathrm{~N}$ would also get incorporated in amino group of amino acids in proteins. Hence, the use of ${ }^{15} \mathrm{~N}$ would not give any conclusive results.