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nekit [7.7K]
4 years ago
13

You've collected several samples of DNA at a crime scene, but they are all short strands. What technique would you rely on to as

sist with analysis?
A) RFLP

B) STR

C) PCR

D) APA

The answer is C) PCR.
Chemistry
1 answer:
Nezavi [6.7K]4 years ago
3 0

Option C

PCR technique would you rely on to assist with analysis

<u>Explanation:</u>

Experts elaborate this tiny specimen by a means acknowledged as a polymerase chain reaction, or PCR. PCR produces replicas of the DNA significant like DNA duplicates itself in a cell, offering virtually any coveted measure of the ancestral element.

Following the accumulation of biological matter from a crime picture, the DNA is beginning derived from its biological source element and then estimated to assess the number of DNA rescued. Subsequent detaching the DNA from its cells, precise areas are imitated with a procedure acknowledged as the polymerase chain reaction, or PCR.

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9. Given the following reaction:CO (g) + 2 H2(g) CH3OH (g)In an experiment, 0.45 mol of CO and 0.57 mol of H2 were placed in a 1
WITCHER [35]

Answer:

Keq=11.5

Explanation:

Hello,

In this case, for the given reaction at equilibrium:

CO (g) + 2 H_2(g) \rightleftharpoons CH_3OH (g)

We can write the law of mass action as:

Keq=\frac{[CH_3OH]}{[CO][H_2]^2}

That in terms of the change x due to the reaction extent we can write:

Keq=\frac{x}{([CO]_0-x)([H_2]_0-2x)^2}

Nevertheless, for the carbon monoxide, we can directly compute x as shown below:

[CO]_0=\frac{0.45mol}{1.00L}=0.45M\\

[H_2]_0=\frac{0.57mol}{1.00L}=0.57M\\

[CO]_{eq}=\frac{0.28mol}{1.00L}=0.28M\\

x=[CO]_0-[CO]_{eq}=0.45M-0.28M=0.17M

Finally, we can compute the equilibrium constant:

Keq=\frac{0.17M}{(0.45M-0.17M)(0.57M-2*0.17M)^2}\\\\Keq=11.5

Best regards.

3 0
4 years ago
What of the following could be classified as matter
Sholpan [36]

Answer:

Desk

Water

Cloud

Helium

Explanation:

7 0
3 years ago
Which statements are true of heterogeneous mixtures?
LiRa [457]

Answer:

1 and 2

Explanation:

Message me for explanation.

6 0
4 years ago
Compare the shape of a chemical signalling molecule and a receptor.
Andru [333]

Cell signalling is a process by which cells communicate with each other and transfer messages.

<h3>What is a receptor?</h3>

A molecule inside or on the surface of a cell binds to a specific substance and causes a specific effect in the cell.

Cell signalling is a process by which cells communicate with each other and transfer messages. When a signal molecule binds to a receptor protein, changes occur in the receptor proteins. Two changes are listed below:

1. The receptor protein undergoes conformational changes and acts as an enzyme.

2. The receptor protein that makes a second messenger molecule that leads the reaction further to other cells or molecules.

Learn more about the receptor here:

brainly.com/question/6438216

#SPJ1

6 0
2 years ago
Imagine that you are given the mass spectra of these two compounds, but the spectra are missing the compound names.
12345 [234]

The structures of the isomers and the m/z values of their peaks are not given in the question. The complete question is provided in the attachment

Answer:

Compound 2 (2,5-dimethylhexane) will not have the peaks at 29 and 85 m/z

Explanation:

The fragmentation of molecules by electron ionization of mass spectrometer occurs according to Stevenson's Rule, which states that "The most probable fragmentation is the one that leaves the positive charge on the fragment with the lowest ionization energy". This is much like the Markovnikov's Rule in organic chemistry which has predicted the formation of most stable carbocation and the addition of hydrogen halide to it.

The mass spectra of compound 1 (2,4-dimethylhexane) will contain all the m/z values mentioned in the question. Each peak indicate towards homologous series of fragmentation product of the compound 1. The first peak can be attributed to ethyl carbocation (m/z = 29), with the increase of 14 units the next peak indicates towards propyl carbocation (m/z = 43) and onwards until molecular ion peak of 114 m/z.

Compound 2 (2,5-dimethylhexane) structure shows that the cleavage  of C-C bond will not yield a stable ethyl and hexyl carbocation. Hence, no peaks will be observed at 29 and 85 m/z. The absence of these two peaks can be used to distinguish one isomer from the other.

5 0
4 years ago
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