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oksian1 [2.3K]
3 years ago
7

DNA transcription-to-translation # 1 Homework Unanswered Due in 4 days Given the following sequence of the coding strand, writte

n in standard 5' to 3' orientation, locate the start codon. CGTTATGTGGACTCTCTGGTATGACTCACCTTAT Starting with and including the start codon, enter the sequence of protein that would be produced by this. Enter your answer without spaces and use single letter abbreviations for the amino acids.
Chemistry
1 answer:
uysha [10]3 years ago
3 0

Explanation:

Translation is the process by which a polypeptide is polymerized from genetic information.

Firstly we have to make a transcription from the coding DNA strand to a single RNA strand (mRNA). RNA pol reads from 5' to 3' of the template strand and nucleotides are added by complementarity ( Adenine with Uracil, Thymine with Adenine and Cytosine with Guanine, Guanine with Cytosine).

DNA:  5'-  CGTTATGTGGACTCTCTGGTATGACTCACCTTAT -3'

mRNA: 5'-GCAAUACACCUGAGAGACCAUACUGAGUGGAAUA -3'

mRNA goes to the ribosomes where translation takes place. The enzyme will read every three letters (codon) starting at the start codon sequence (TAC in DNA, AUG in mRNA). According to codons tRNA carrying the amino acids will place it (by complementary to their anticodon) and the enzyme will join it to the nascent polypeptide or protein.

In order to do this we need to look up the genetic code and assign the proper amino acids.

Unfortunately the given strand does not have a start codon TAC codifying for initial methionine.

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What is the molarity of a 17.0% by mass solution of sodium acetate, NaC2H3O2 (82.0 g/mol), in water? The density of the solution
sattari [20]

Answer:

[NaCH₃COO] = 2.26M

Explanation:

17% by mass is a sort of concentration. Gives the information about grams of solute in 100 g of solution. (In this case, 17 g of NaCH₃COO)

Let's determine the volume of solution, by density

Mass of solution / Volume of solution = Solution density

100 g / Volume of solution = 1.09 g/mL

100 g / 1.09 g/mL = 91.7 mL

17 grams of solute is contained in 91.7 mL

Molarity (M) = Mol of solute /L of solution

91.7 mL / 1000 = 0.0917L

17 g / 82 g/m = 0.207 moles

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3 years ago
A certain amount of hydrogen peroxide was dissolved in 100. mL of water and then titrated with 1.68 M KMnO4. What mass of H2O2 w
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Answer:

see explanation below

Explanation:

First to all, this is a redox reaction, and the reaction taking place is the following:

2KMnO4 + 3H2SO4 + 5H2O2 -----> 2MnSO4 + K2SO4 + 8H2O + 5O2

According to this reaction, we can see that the mole ratio between the peroxide and the permangante is 5:2. Therefore, if the titration required 21.3 mL to reach the equivalence point, then, the moles would be:

MhVh = MpVp

h would be the hydrogen peroxide, and p the permanganate.

But like it was stated before, the mole ratio is 5:2 so:

5MhVh = 2MpVp

Replacing moles:

5nh = 2MpVp

Now, we just have to replace the given data:

nh = 2MpVp/5

nh = 2 * 1.68 * 0.0213 / 5

nh = 0.0143 moles

Now to get the mass, we just need the molecular mass of the peroxide:

MM = 2*1 + 2*16 = 34 g/mol

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m = 0.0143 * 34

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2 years ago
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The half life of oxygen is 2 minutes. What fraction of a sample of 0.15 will remain after 5 half lives?​
Natasha_Volkova [10]

Answer:

3.13%.

Explanation:

The following data were obtained from the question:

Original amount (N₀) = 0.15

Half life (t½) = 2 mins

Number of half-life (n) = 5

Fraction of sample remaining =.?

Next, we shall determine the amount remaining (N) after 5 half-life. This can be obtained as follow:

Amount remaining (N) = 1/2ⁿ × original amount (N₀)

NOTE: n is the number of half-life.

N = 1/2ⁿ × N₀

N = 1/2⁵ × 0.15

N = 1/32 × 0.15

N = 0.15/32

N = 4.69×10¯³

Therefore, 4.69×10¯³ is remaining after 5 half-life.

Finally, we shall the fraction of the sample remaining after 5 half-life as follow:

Original amount (N₀) = 0.15

Amount remaining (N) = 4.69×10¯³

Fraction remaining = N/N₀ × 100

Fraction remaining = 4.69×10¯³/0.15 × 100

Fraction remaining = 3.13%

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3 years ago
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