mv^2 =2* e*40k J/Cv= sqrt(2* e*40k/m)
mv^2/r = F centripetal = evB2* e*40k/(210*sqrt(2* e*40k/m)*e) =BB=√e∗80,000∗m−−−−−−−−−−−/210
Answer:
After this treatment, the investigators should expect to get a mixture of the desired enzyme, plus fragments of the peptide used to desorb the enzyme in question.
This would be the result of using a peptide as a desorption solution when the desired protein is a protease,
Assuming that the protease retains its activity in the medium in question, and that the peptide can act as a substrate (which would make sense), as the peptide solution is added, it will interact with and bind to the antibody, but some molecules will also interact with the active site of the enzyme as it desorbs and passes through, culminating on the elution of the hydrolized part of the peptide along with the enzyme.
Answer:
the answer is C
Explanation:
As you move from left to right across a period, the number of protons in the nucleus increases. The electrons are thus attracted to the nucleus more strongly, and the atomic radius is smaller (this attraction is much stronger than the relatively weak repulsion between electrons).
1.true
2.true
3.fasle
4.true
5.false
6.true