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Taya2010 [7]
3 years ago
6

When researchers need to prepare many reactions for polymerase chain reaction (PCR) amplification, they often create a "master m

ix" solution. A master mix contains the reagents common to all the planned PCR amplifications, regardless of the target DNA. Making a master mix is a way to minimize the number of pipetting steps.Suppose a researcher needs to PCR amplify seven different genes of interest from different organisms. The researcher prepares a master mix and dispenses it to seven different PCR tubes, one for each gene of interest.Select the PCR components the researcher must add to each of the seven tubes of master mix to selectively amplify each gene of interest.dNTPsMg2+-Mg2+-based bufferprimersDNA polymeraseDNA template
Chemistry
1 answer:
Margaret [11]3 years ago
6 0

Answer:

The master mix contains the following reagents: dNTPs, DNA Polymerase, PCR buffer and MgCl2.

Explanation:

The DNA templates are the gene fragments to amplify by PCR, thereby they have to be added separately in each tube. Moreover, the primer pairs are specific for each gene, thereby they have to be added separately in each tube.  

Deoxynucleotide triphosphates (dNTPs) are the building blocks of the DNA molecules: dGTP, dATP, TTP, and dCTP.

The PCR buffer provides a suitable medium for the activity of the DNA polymerase, often it contains Tris-Hcl and KCl.

MgCl2 is a cofactor for the activity of the DNA Polymerase.

The DNA Polymerase is an enzyme that amplifies DNA by adding nucleotides to the 3' end.

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Can you draw the peptide structure at pH 7.0 (including peptide bonds)
ELEN [110]

Answer:

See image below

Explanation:

The image is labeled according to the sequence N'-Trp-Ser-Asg-Gly-Cys-His-COOH' which means that in the main chain of the peptide, the amino group of the Tryptophan and the carboxylic group of the Histidine are free and thus its charge depends on the pH; other groups that rely on the pH are the side groups of the Cysteine and the Histidine.

Overall, ionizable groups in this peptide are:

  • Amino Group of the Tryptophan (pKa = 9.39)
  • SH group of the Cysteine (pKa = 8.18)
  • Secondary amine of the Histidine (pKa = 6.00)
  • Carboxylic Group of the Histidine (pKa = 1.82)

Then, the amino group of Trp and SH group of Cys are protonated since the peptide is at a pH below the pKa. The secondary amine of the Histidine is deprotonated because the pH is greater than the pKa, as well with its carboxylic acid group.

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2 years ago
Which of the following describes a property of water that helps sustain marine life forms?
Mariana [72]

Answer:

Open spaces in water's solid structure makes its solid state less dense than its liquid state.

Explanation:

  • Water unlike other liquids is special. It contracts when cooled, down to a temperature of 4°C but thereafter begins to expand as it reaches 0°C and turns into ice.
  • This property is useful for the preservation of marine life in very cold temperatures. During winter, the surface water in water lakes and rivers starts cooling. Upon reaching the temperature of 4°C, the surface water descends to the bottom as it denser.
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4 0
3 years ago
Polonium- 210 , Po210 , decays to lead- 206 , Pb206 , by alpha emission according to the equation Po84210⟶Pb82206+He24 If the ha
elixir [45]

Answer:

0.269 g

Explanation:

Po_{210} ^{84}  ⟶  Pb_{206} ^{82}+  + He_{4} ^{2}

Data:

Half-life of  Po_{210} ^{84}  (T(1/2)) = 138.4 days

Mass of PoCl4 = 561 mg (0,561 g) and molecular weight of PoCl4 = 350. 79 g/mol  

Time = 338.8 days  

Isotopic masses  

Po_{210} ^{84} = 209.98 g/mol  

Pb_{206} ^{82} = 205.97 g/mol  

Concepts  

Avogadro’s number: This is the number of constituent particles that are contained in a mol of any substance. These constituted particles can be atoms, molecules or ions). Its value is 6.023*10^23.  

The radioactive decay law is  

N=Noe^(-λt)

Where:  

No = number of atoms in t=0

N = number of atoms in t=t (now) in this case t=338.8 days  

λ= radioactive decay constant  

The radioactive constant is related to the half-life by the next equation  

λ= \frac{ln 2}{t(1/2)}

so  

λ= \frac{ln2}{138.4 days}  =0,005008 days^(-1)

No (Atoms of  Po_{210} ^{84}  in t=0)

To get No we need to calculate the number of atoms of  Po_{210} ^{84}   in the initial sample. We have a sample of 0,561 g of PoCl4. If we get the number of moles of PoCl4 in the sample, this will be the number of moles of  Po_{210} ^{84}  in the initial sample.  

This is:

\frac{0,561 g of PoCl4}{350. 79 g of PoCl4 /mol} = 0,001599 mol of  PoCl4

This is the number of mol of  Po_{210} ^{84} in the initial sample.

To get the number of atoms in the initial sample we use the Avogadro’s number = 6.023*10^23  

0,001599 mol of  Po_{210} ^{84} * 6.023*10^23 atoms/ mol of  Po_{210} ^{84} = 9.632 *10^20 atoms of  Po_{210} ^{84}

Atoms after 338.8 days

We use the radioactive decay law to get this value  

N=Noe^(-λt)

N=9.632*10^20 e^(-0,005008 days^(-1) * 338.8 days) =1.765*10^20

This is the number of atoms of  Po_{210} ^{84} in the sample after 338.8 days has passed  

The number of atoms  Po_{210} ^{84} transformed is equal to the number of atoms of Pb_{206} ^{82}  produced.  

The number of atoms of Po_{210} ^{84} transformed is No - N  

9.632 *10^20 – 1.765 *10^20 = 7.866*10^20

So, 7.866*10^20 is the number of atoms of Pb_{206} ^{82} produced  

We can get the mass with the Avogadro’s number

(7.866*10^20 atoms of Pb_{206} ^{82} ) / ( 6.023*10^23 atoms of Pb_{206} ^{82} / mol of Pb_{206} ^{82} =  0.001306 moles of Pb_{206} ^{82}

This number of moles have a mass of:

(0,001306 moles of Pb_{206} ^{82} )* (205.97 g of Pb_{206} ^{82} /mol of Pb_{206} ^{82} ) = 0.269 g  

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3 years ago
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Ivanshal [37]
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