When a genetic population follows Hardy-Weinberg Equilibrium (HW), it states that certain biological tenets or requirements must be met. Given so, then HW states that the total frequency of all homozygous dominant alleles (p) and the total frequency of all homozygous recessive alleles (q) for a gene, account for the total # of alleles for that gene in that HW population, which is 100% or 1.00 as a decimal. So in short: p + q = 1, and additionally (p+q)^2 = 1^2, or 1
So (p+q)(p+q) algebraically works out to p^2 + 2pq + q^2 = 1, where p^2 = frequency of homozygous dominant individuals, 2pq = frequency of heterozygous individuals, and q^2 = frequency of homozygous recessive individuals.
So the problem states that homozygous dominant individuals (p^2) account for 60%, or 0.60. Thus the square root (sr) of p^2 = p or the dominant allele frequency in the population. So sr(p^2) = sr(0.60) -->
p = 0.775 or 77.5%
Homozygous recessive individuals (q^2) account for 20%, or 0.20. Thus sr(q^2) = q or the recessive allele frequency in the population. So sr(q^2) = sr(0.20) --> q = 0.447 or 44.7%
But since 44.7% + 77.5% = 122.2%, which is not equal to 1, we have a situation in which the allele frequencies do not match up, therefore this population cannot be determined using the Hardy-Weinberg Equation.
The optical microscope, also referred to as a light microscope, is a type of microscope that commonly uses visible light and a system of lenses to generate magnified images of small objects
The correct answer would be A
Answer:
The experiment conducted in 1952 by hershey and chase consisted in the use of sulfur to track proteins and radioactive phosphorus to track DNA from T2 phages that infect bacterial cells, the key result was that phage proteins were outside the cell during the infection but the DNA was inside the cells because new phages were obtained with radioactive phosphorus. This allows us to conclude that DNA acts as a genetic material of phage T2
