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kifflom [539]
1 year ago
3

Prehensile tails are:

Biology
1 answer:
Ad libitum [116K]1 year ago
8 0
<h3>What ape species have prehensile tails?</h3>

Howler monkeys (Alouatta), spider monkeys (Ateles), woolly monkeys (Lagorix), capuchin monkeys, and woolly spider monkeys (Brachyteles) are the only primate genera (Figure 1) that have prehensile tails (Cebus).

<h3>Are the tails of spider monkeys prehensile?</h3>

Spider monkeys can swing through the trees thanks to their long, slender arms and hook-like hands. Their thumbs are not opposable. The brown-headed spider monkey may use its prehensile tail as a fifth limb to grasp trees because it can grasp objects.

<h3>What number of brown-headed spider monkeys are still present?</h3>

One of the rarest primates in the world is the brown-headed spider monkey from Ecuador, which is in grave risk of extinction. 250 people are thought to be the species' total global population.

To know more about prehensile tails visit;

brainly.com/question/15034552?

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Suppose 64 percent of people living in a remote, isolated mountain village can taste phenylthiocarbamide (PTC) and must, therefo
worty [1.4K]

Answer:

The percentage of the population that is heterozygous for this trait is 48%

Explanation:

They are two alleles, the phenylthiocarbamide tasters (PTC) and the non phenylthiocarbamide tasters (non PTC). PTC testers are dominant and non PTC tasters are recessive.

let the frequency of the dominant allele(A) be p

and the frequency of the recessive allele(a) be q

We are told that 64 percent of people living in a remote, isolated mountain village can taste phenylthiocarbamide (PTC) and must, therefore, have at least one copy of the dominant PTC taster allele (that is AA and Aa)

Frequency of AA = p², Frequency of Aa = 2pq and Frequency of aa = q²

Therefore p² + 2pq = 64% = 0.64

According to Hardy–Weinberg:

p² + 2pq + q² = 1 and

p + q = 1

Since p² + 2pq = 0.64

∴ 0.64 + q² = 1

q² = 1 - 0.64 = 0.36

q = √0.36 = 0.6

Since p + q = 1

p = 1 - q = 1 - 0.6 = 0.4

The frequency of heterozygous = 2pq = 2 × 0.4 × 0.6 = 0.48

Therefore the percentage of the population that is heterozygous for this trait is 48%

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<h3>The Prokaryotic Cell</h3>

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