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Lina20 [59]
2 years ago
15

which of the following lacks sufficient penetrating power for bulk sterilization? a. ultraviolet (uv) radiation at 260 nm.b. x r

ays.c. gamma radiation.d. alpha and beta particles.
Biology
1 answer:
harina [27]2 years ago
6 0

Alpha and beta particles lacks sufficient penetrating power for bulk sterilization

<h3>What is Bulk sterilization?</h3>

Bulk sterilization is a means of reducing contaminants using a vessels at a very high temperature.

Among the particles we have, alpha particles is known to have a low penetrating power compared to gamma radiation. Hence we can conclude that  alpha and beta particles lacks sufficient penetrating power for bulk sterilization.

Learn more on bulk sterilization here: brainly.com/question/14642957

#SPJ11

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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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