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Law Incorporation [45]
3 years ago
8

30 POINTS AND BRIANLIEST ANSWER

Biology
1 answer:
lianna [129]3 years ago
5 0

Mice (AND RATS) And Men.

Mice and rats have been following humans since at least the origins of agriculture more than 10,000 years ago. It is easy to imagine we have probably been trying to kill them for nearly as long. More recently, however, we've been poisoning these pests, offering them tempting treats laced with deadly chemicals. Rats living in forests and other wild places are attracted to new foods in particular and so feed readily from such baits. Rats living with humans are not, at least not anymore. Present them with a new food and they will wait. Several authors have suggested that this "neophobia" in urban rats has evolved in response to the threat posed to rats and mice by our new "foods." For now, the little we know about the evolution of neophobia fits with this idea. The clearest evolutionary change in rats and mice as a result of our interference has been the evolution of resistance to the rat poison warfarin. We then created superwarfarin to target these resistant populations, but resistance to this poison has recently evolved (Mayumi et al., 2008). Once again our garden of neglect is seemingly growing out of our control.

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What animals have been dominant in the Cenozoic era
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Explanation:

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3 years ago
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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