Answer:
The correct answer would be - Characteristics can be lost in evolution.
Explanation:
The new evidence helps in developing a new hypothesis. In this case, new evidence proved that the Chondrichthyes diverged after the evolution of bone had started instead of before the evolution started. This process called atavism where an ancestral genetic trait reappears after having lost. This leads to loss of the traits in the evolution
This can take place by knocking the mutation out to overriding the gene by the old gene or overriding the new trait by the old trait during the evolution period.
Answer:
In a particular case of secondary succession, three species of wild grass all invaded a field. By the second season, a single species dominated the field and the other two species had a lower relative abundance. A possible factor contributing to the abundances of these species in this example of secondary succession is <u>inhibition</u>.
Explanation:
Trees are great examples of allelopathy in plants. Some use their allelochemicals to inhibit germination or impede development of nearby plant life. Most allelopathic trees release these chemicals through their leaves, which are toxic once absorbed by other plants. Black walnut is a prime example of this.
To be able to compare the results to a sample without any of the independent variable added.
Hardy-Weinberg Equation (HW) states that following certain biological tenets or requirements, 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 decimel. 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 = genotype frequency of homozygous dominant individuals, 2pq = genotype frequency of heterozygous individuals, and q^2 = genotype frequency of homozygous recessive individuals.
The problem states that Ptotal = 150 individuals, H frequency (p) = 0.2, and h frequency (q) = 0.8.
So homozygous dominant individuals (HH) = p^2 = (0.2)^2 = 0.04 or 4% of 150 --> 6 people
Heterozygous individuals (Hh) = 2pq = 2(0.2)(0.8) = 0.32 or 32% of 150
--> 48 people
And homozygous recessive individuals (hh) = q^2 = (0.8)^2 = 0.64 = 64% of 150 --> 96 people
Hope that helps you to understand how to solve these types of population genetics problems!
