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Tom [10]
3 years ago
10

While using Biuret reagent, you test not only the mystery food sample, but also gelatin and water. Which solution acts as a posi

tive control?
Biology
2 answers:
Otrada [13]3 years ago
8 0
The answer is gelatin. Correct me if im wrong.. Since gelatin consist of the elements Carbon, Hydrogen, Nitrogen and Oxygen. (same as the protein elements.) [biuret test is used to test the presence of amino acids]
babunello [35]3 years ago
5 0

Answer:

gelatin

Explanation:

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The production of orchids by cutting is an example of______.
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Answer:

The production of orchids by cutting is an example of <em><u>divine creation</u></em>

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Which area shows the climax community of secondary succession?
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Usually the place with the most stable community
(ex: Tall growing trees, plants that live a long time)
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3) Two students decided to repeat the Hershey and Chase experiment, with modifications for a science competition. They radioacti
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Answer:

Explanation:

In the Hershey and Chase experiment, they aimed to show where the phage DNA went when it exited the protein coat and entered the bacteria. In this scenario, the two student's approach won't work because amino acids (and therefore all proteins) also have nitrogen atoms which ultimately means that the radioactivity would not distinguish between DNA and proteins and the experiment will not provide the correct results.

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3 years ago
Which of the following carries the hereditary information for cells?
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Are you asking for DNA or chromosomes or something? You asked for which of the following but then dont give the following

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3 years ago
For this question, we will utilize a population of Martians that is in Hardy Weinberg Equilibrium. The dominant Martian phenotyp
Brrunno [24]

Answer:

d.0.48

Explanation:

When a population is in Hardy Weinberg equilibrium the <u>genotypic </u>frequencies are:

freq (AA) = p²

freq (Aa) = 2pq

freq (aa) = q²

<em>p</em> is the frequency of the dominant <em>A</em> allele and <em>q</em> is the frequency of the recessive <em>a</em> allele.

In this population of 100 individuals, 84 martians have the dominant phenotype and 16 have the recessive phenotype.

Therefore:

q²=16/100

q² = 0.16

q=√0.16

q = 0.4

And p+q=1, so:

p = 1 - q

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p = 0.6

The frequency of heterozygotes is:

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freq (Aa) = 0.48

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