based on the data in the graph, which of the following should be used to calculate the difference in ld50 for the two different species of mice?

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Answer 1

Based on the data in the graph 575mg−490mg should be used to calculate the difference in ld50 for the two different species of mice.

ToxicologyThe median lethal dose, also known as LD50, LC50, or LCt50, is a hazardous unit used in toxicology to assess the deadly dose of a toxin, radiation, or disease. The dose necessary to cause the death of 50% of a population under test after a predetermined test period is known as the LD50 value for a drug.Values for LD50 and LC50 are used to determine acute toxicity. The insecticide is more hazardous the lower the LD50. An illustration would be that a pesticide with an LD50 of 5 mg/kg is 100 times more toxic than one with an LD50 of 500 mg/kg. The following two values are given in milligrams per kilogram of the animal's body weight (mg/kg body wt.).

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Based On The Data In The Graph, Which Of The Following Should Be Used To Calculate The Difference In

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as characteristized in lecture, an oil spill is an example of: question 12 options: competition antagonism ammensalism commensalism

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Antagonism can take one of two forms: competitive and non-competitive. Ketamine, on the other hand, is a non-competitive foe of the NMDA-glutamate receptor, whereas naloxone is a competitive foe of all opioid receptors. The correct answer is Antagonism.

Competitive antagonism is based on the idea that an agonist or antagonist can bind to the same recognition site(s) on the receptor and compete for those sites when they are present simultaneously.

A competitive antagonist blocks the agonist's action by binding to the same site as the agonist but not activating it. To prevent the receptor from being activated, a non-competitive antagonist binds to an allosteric (non-agonist) site on the receptor.

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You are crossing two pea plants. One is heterozygous for yellow. The second pea plant is homozygous for green. Use "G/g" as the letter to represent the gene for this problem.

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Answer:

Parent 1 genotype: Gg

Parent 2 genotype: gg


gg's bro

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