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otez555 [7]
2 years ago
12

Many small poisonous animals are brightly colored, including the poison dart frog and some species of fish, octopi, and insects.

What is common in their environments that would lead to these different animals having a similar trait?
Biology
2 answers:
Anvisha [2.4K]2 years ago
5 0
The common factor between these traits are in fact the predators and prey defense mechanisms.

For example, a poison Dart frog’s colorful appearance gives off aposematic signals. “Aposematism, also called aposematic mechanism, biological means by which a dangerous, or noxious, organism advertises its dangerous nature to a potential predator. The predator, having recognized the dangerous organism as an unfavourable prey, thereupon desists from attacking it.”
Fudgin [204]2 years ago
3 0

Answer:

trees brah

Explanation:

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If the frequency of homozygous dominant is 60%, the frequency of heterozygous is 20%, and the frequency of homozygous recessive
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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.
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