Answer: <u>Endonuclease enzymes used in molecular biology that cut DNA at specified points.</u>
Explanation:
Enzymes are specific protein types which bind to a substrate within a reaction, to increase the rate of reaction within the solution- they speed up the rate of reaction.
Restriction enzymes are bacteria-derived enzymes; these make cuts on deoxyribonucleic acid molecules or DNA. These are also called restriction endonucleases. They are utilized in molecular biology for DNA cloning and sequencing and cut DNA into smaller pieces called fragments.
Restriction enzymes make directed cuts on DNA molecules. They precisely target sites on DNA to produce mostly identical or homogenous, discrete fragments of equal sizes, producing blunt or sticky ends. In order to do this, they recognize sequences of nucleotides that correspond with a complementary sequence on the endonuclease called restriction sites.
There are several kinds that may require cofactors (chemical or metallic compounds that aid in enzyme activity) :
- Type I: cleave far away from the recognition site; require ATP and SAMe S-Adenosyl-L-Methionine
- Type II: cleave near to the site; require Magnesium
- Type III: cleave near to the site; require ATP which is not hydrolysed but SAMe S-Adenosyl-L-Methionine is optional
- Type IV: cleavage targeted to DNA that have undergone post transcriptional modification through certain types of methylation (addition of a methyl group)
Answer:
12.02 g
Explanation:
From the question given above, the following data were obtained:
Half life (t½) = 2 days
Original amount (N₀) = 96 g
Time (t) = 6 days
Amount remaining (N) =..?
Next, we shall determine the rate of disintegration of the isotope. This can be obtained as follow:
Half life (t½) = 2 days
Decay constant (K) =?
K = 0.693 / t½
K = 0.693 / 2
K = 0.3465 /day
Therefore, the rate of disintegration of the isotope is 0.3465 /day.
Finally, we shall determine the amount of the isotope remaining after 6 days as follow:
Original amount (N₀) = 96 g
Time (t) = 6 days
Decay constant (K) = 0.3465 /day.
Amount remaining (N) =.?
Log (N₀/N) = kt / 2.303
Log (96/N) = (0.3465 × 6) / 2.303
Log (96/N) = 2.079/2.303
Log (96/N) = 0.9027
Take the anti log of 0.9027
96/N = anti log (0.9027)
96/N = 7.99
Cross multiply
96 = N × 7.99
Divide both side by 7.99
N = 96 /7.99
N = 12.02 g
Therefore, the amount of the isotope remaining after 6 days is 12.02 g
Answer: 2) 2HCl(sq) + CaCO3(s) CaCl2(sq) + CO2(g) + H2O (l) No of moles of CaCO3 = amount of the CaCO3 (g)/mw of CaCO3 (g/mole)= 0.8085 g/100 g/mole = 0.008085
Explanation:
Answer:
uhhh im not a boy ima just answer this for points
Explanation:
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whatever u r mixing together will indicated what the reaction or the product will be