answer

watching

views

25 Jan 2019

if the purpose of a RT-qPCR is to normalize mRNA in order to have a proper comparison/quantification of mRNA levels, why is the ending result DNA or more so, how does that tie into the normalized mRNA levels that are being expressed?

answer

watching

202

views

For unlimited access to Homework Help, a Homework+ subscription is required.

You have 0 free answers left.
Get unlimited access to 3.8 million step-by-step answers.

Get unlimited access

Already have an account? Log in

Lelia LubowitzLv2

27 Jan 2019

Unlock all answers

Get 1 free homework help answer.

Already have an account? Log in

Related questions

You have discovered that the RNA transcribed from Gene X in maize is required for the regulation of Gene Y. The RNA of Gene X is not translated into proteins. However, in its absence, the levels of protein Y (product of Gene Y) are significantly increased. Propose one likely mechanism for how Gene X may be controlling the levels of Protein Y.

This is the current answer I have:

mRNA of Gene X in maize is a non-coding RNA

It is functional but it does not get translated into a protein

It binds to the complementary sequence that is present on Gene Y mRNA

Therefore, Gene Y mRNA undergoes cleavage or degradation

This results in the blocking of translation of Gene Y mRNA

Therefore, levels of protein Y are absent in presence of Gene X mRNA

In the absence of Gene X mRNA, Gene Y mRNA is not cleaved or degraded

As a result, Gene Y mRNA is translated and this results in significantly increased levels of Gene Y protein in the absence of Gene X mRNA.

Is this correct?

Unit 2: Genetics Assignment

In this assignment you will convert a DNA sequence into proteins through several steps. If you have no idea where to begin I recommend that you watch the âHow to read the Genetic Codeâ and âProtein Synthesisâ videos

Here is the DNA sequence:

TACCACGCCAAGCGAAGCGGGATT

Part 1: Transcribe the sequence into mRNA and separate the mRNA into codons. Name the enzyme responsible and the start and stop codons used.

Part 2: Give the tRNA anticodons that match the mRNA.

Part 3: Translate the mRNA into amino acids.

Part 4: Give an example of how a mutation to this sequence could cause a silent mutation.

Hereâs an example that you can complete:

Normal DNA = CAC Mutated DNA = CAG

Normal mRNA = Mutated mRNA=

Normal amino acid = Mutated amino acid=

*What do you notice happened above?

*What does this tell you about the effect of a silent mutation?

Part 5: Give an example of how a mutation to this sequence could cause a missense mutation.

Hereâs an example that you can complete:

Normal DNA = CAC Mutated DNA = GAC

Normal mRNA = Mutated mRNA=

Normal amino acid = Mutated amino acid=

*What do you notice happened above?

*What does this tell you about the effect of a missense mutation?

Part 6: Besides mutations how else do bacteria acquire new genes? Define each of these in your own words:

*transformation

*transduction

*conjugation

salmonrabbit236

How can a cell remember which gene it needs to express (turn on)and which gene does from one generation to the next?

1.Methylation of cytosine bases

2.heritable chromatin structure

3.both of the above

4. none of the above

If you were to make 2 liters of a 2 molar solution of SodiumChloride (MW 58.4) How many grams of Sodium Chloride would youdissolve?

/1.58.4 /2.29.2 /3.116.8 /4.233.6

After transcription, nuclear RNA is capped, polyadenylated,spliced. After successful splicing, proteins are added to themature mRNA at the exon junction complex and then transported tothe cytoplasm. Which of the following is a NOT a reason why thesemodifications are added to

mRNA?

1.to identify the RNA molecule as mRNA

2.to stabilize mRNA molecule

3.to aid in transport from ribosome to cytoplasm throughpore

4.to signal to the cell that mRNA is ready for translation

A cell is being held in G1 phasedue to DNA damage caused by UV radiation. When the DNA is damaged,p53 is phorphorylated and precents entry into whatphase?

1.M phase 2. s-phase 3.G0 phase 4. none of the above