(i) AAU - UUA
(ii) CGA - GCU
(iii) UAU - AUA
(iv) GCA - CGU

tRNA on one hand binds to specific amino acids and on the other hand reads the codon of the amino acid bound to it through its anticodon, it is called as adapter molecule.

There are 1.4 million locations where single base DNA differences (Called SNPs - Single Nucleotide Polymorphism - pronounce as snips) occurs in human's genome.
(i) By tracing human history
(ii) By finding chromosomal locations for disease associated sequences

When lactose is absent "i" gene regulates and produces repressor mRNA in the absence of lactose, which blocks the transcription. The repressor protein binds to the operator region of the operon and as a result prevents RNA polymerase from binding to the operon. The operon will be switched off (stop) in this situation.

(i) Identify all the genes (approximately 30000) in human DNA.
(ii) To store this information in databases.
(iii) Improve tools for data analysis.  

(i) In eukaryotes, chromatin is formed by a series of repeating units called nucleosomes.
(ii) Korenberg proposed a model for the nucleosome, in which 2 molecules of the four histone proteins H₂A, H₂B, H3 AND H4 are organized to form a unit of eight molecules called histone octamere.
(iii) The negatively charged DNA is wrapped around the positively charged histone octamer to form a structure called nucleosome. A typical nucleosome contains 200 bp of DNA helix.
(iv) The histone octameres are in close contact and DNA is coiled on the outside of nucleosome.
(v) Neighbouring nucleosomes are connected by linker DNA (Hl) that is exposed to enzymes. The DNA makes two complete turns around the histone octameres and the two turns are sealed off by an HI molecule
(vi) Chromatin lacking HI has beads on a string appearance in which DNA enters and leaves the nucleosomes at random places.
(vii) HI of one nucleosome can interact with HI of the neighbouring nucleosomes resulting in the further folding of the fibre.

1. They wanted to know whether the proteins coat or the DNA core of the virus entered , into the infected cells. This would help them to determine whether genes were made up protein or of DNA.
2. They used radioactive isotopes of phosphorous - 32 and sulfur - 35 as markers. So they could trace the proteins and DNA to see what they were doing to the cell.
3. They were perfect because proteins contain almost no phosphorous and DNA contains no sulfur, so the radioactive isotopes would show up in the cells if the P-32 or S-35 appeared. If S-35 shown up in the bacteria. It meant the viruses protein had been injected into the bacteria, if the P-32 was found in the bacteria the DNA had been injected into the bacteria.
4. By that they concluded the genetic material of the bacteriophage was DNA, not protein.
5. If in Hershey and Chase's experiments, the phages were labeled with radioactive carbon, the label would be found in all components of the phages and of the phase ghosts (empty capsids) and possible in all components of the bacteria because carbon is found in all organic material.
6. If Hershey and Chase used radiolabeled nitrogen they would see that the cells would be labelled in the nucleus, in the membrane and possibly throughout the cell. Additionally, numerous components of the phage ghosts and the progeny phage would be labelled.
7. Nitrogen is found in proteins and in DNA so they were not able to get actual result that the genetic material of the phage was DNA, if they use radio isotopes labelled with carbon and nitrogen.

Structural difference between DNA and RNA.