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Contact [7]
3 years ago
13

Several new materials have been developed for use in the microbiology lab. These materials must be sterilized before use, but ca

nnot withstand a temperature over 60oC and can’t be exposed to water. Which method of sterilization would be applicable?
Biology
1 answer:
asambeis [7]3 years ago
4 0

Answer:

Ionizing radiation

Explanation:

Ionizing radiation is a method of sterilization that can be used to sterilze material that can not withstand some higher temperature and cannot be exposed to water for sterilization.

It is form of energy that is released when atoms is released to move in the form of electromagnetic waves or particles. such as; X-rays or Gamma rays particles,neutrons, alpha or beta.

It could be dangerous if atoms inliving cells is exposed to it because the cell will be ionized, and it can lead to death of cell , or the mutatation of cell and can lead to cancer. Although not all cells are affected by ionizing radiation.

1)X-Rays, X-rays move in the form of photons energy .

2)Alpha Particles. Alpha particles have less penetration than gamma ray they don't penetrate skin as x-rays

3)Beta Particles, they are particles of electrons which are faster than alpha particles.

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3 years ago
Hydrosphere Part 1 / 34 of 46
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3 years ago
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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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4 years ago
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The correct answer is: exonuclease activity.

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DNA Polymerase has an exonuclease activity in which mismatched nucleotides are removed from the newly replicated DNA strand in order to prevent mutations that can lead to malfunctioning or harmful proteins.

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4 years ago
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The DNA sequences that have been transcribed into RNA (and will later become proteins) are called _____________.
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rodons

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