Answer:
Option D- frequency of the heterozygous qenotype
Explanation:
In the Hardy-Weinberg equilibrium , 2pq stands for the genotype frequency of heterozygotes (Aa) in a population at the state of equilibrium. dominant homozygotes (AA) is represented by the p-square and q- square represents the recessive homozygotes (aa) frequency. p represents the allele frequency of A Allele and q stands for the allele frequency of the allele a. Summation of the terms gives an equal to 1(100%). The sum of all of these terms is always equal to due to the fact that the 3 frequencies in the equation are the available or possible allele combinations.
Hardy-Weinberg is relevant for it helps us in comparison of a population's actual genetic structure as time goes on along with the genetic structure we are to get if really population was in Hardy-Weinberg equilibrium state that is population was not evolving.
The body contains about a trillion cells
The Griffith's experiment, the Avery-MacLeod-McCarty experiment, and the Hershey–Chase experiments were the set of experiments that established DNA as the key hereditary molecule. The Avery-MacLeod-McCarty experiment was an extension to the Griffith's experiment. The heat killed virulent S strain cells of the Griffith's experiment were lysed to form a supernatant containing a mix of RNA, DNA, proteins and lipids from the cell. The supernatent was equally divided into 3 parts after the removal of the lipids. The 3 parts were respectively treated with an RNAase to degrade the RNA, DNAase to degrade the DNA and proteinase to degrade the proteins. The treated supernatant was then added into the culture containing the non-virulent R cells. In case of the supernatant treated with the DNAse, no transformation of R cells into S cells occurred. The transformation of R cells to S cells occurred in the proteinase and the RNAse cases. This indicated that DNA was the hereditary molecule and not protein or RNA.
