Answer:
Population size of elk increased while the population of aspen tree declined.
Explanation:
As per question, removal of wolves from the system would result in increased elk population due to absence of its predator (wolves). The increased elk population would extensively feed on shrubs and trees like aspen and cottonwood trees. This would cause decline in population size of these shrubs and trees.
Answer:
Centrioles
Explanation:
If we look at ths other options Chloroplast and Cell Wall are both needed for a plant to survive and function. So we automatically know that it's going to be there so those aren't an option. We know that both animal and plant cells have a vacuole but plant cells have larger ones so we know that's not an answer. Leaving us with the only other possible answer left, Centrioles.
In DNA<span>, the code letters are A, T, G, and C, which stand </span>for<span> the chemicals </span>adenine<span>, thymine, guanine, and cytosine, respectively. In base pairing, </span>adenine always pairs<span> with thymine, and guanine </span>always pairs<span> with cytosine.</span>
<span> The answer is <span>centriole</span>
The majority of cells have mitochondrion as they need them for their respiration. Chloroplast is the structure plants use to produce their organic matter, the vacuole is very developed in plant cells and is central where as you usually have many vacuoles in animal cells, and the centriole is usually present in animal cells only as well as the wall of cellulose that wraps up the plant cell.</span>
Answer:
1. Allele frequency of b = 0.09 (or 9%)
2. Allele frequency of B = 0.91 (0.91%)
3. Genotype frequency of BB = 0.8281 (or 82.81%)
4. Genotype frequency of Bb = 0.1638 (or 16.38%)
Explanation:
Given that:
p = the frequency of the dominant allele (represented here by B) = 0.91
q = the frequency of the recessive allele (represented here by b) = 0.09
For a population in genetic equilibrium:
p + q = 1.0 (The sum of the frequencies of both alleles is 100%.)
(p + q)^2 = 1
Therefore:
p^2 + 2pq + q^2 = 1
in which:
p^2 = frequency of BB (homozygous dominant)
2pq = frequency of Bb (heterozygous)
q^2 = frequency of bb (homozygous recessive)
p^2 = 0.91^2 = 0.8281
2pq = 2(0.91)(0.9) = 0.1638