Answer:- The wavelength is
and the right option is D.
Solution:- Wavelength is inversely proportional to the frequency and the equation is:
![\lambda =\frac{c}{\nu }](https://tex.z-dn.net/?f=%5Clambda%20%3D%5Cfrac%7Bc%7D%7B%5Cnu%20%7D)
where,
is the wavelength, c is the speed of light and
is frequency.
Frequency is
and the speed of light is
.
Since, ![1Hz=1s^-^1](https://tex.z-dn.net/?f=1Hz%3D1s%5E-%5E1)
So, ![2.20*10^-^4Hz=2.20*10^-^4s^-^1](https://tex.z-dn.net/?f=2.20%2A10%5E-%5E4Hz%3D2.20%2A10%5E-%5E4s%5E-%5E1)
Let's plug in the values in the equation:
![\lambda =\frac{3.00*10^8m.s^-^1}{2.20*10^-^4s^-^1}](https://tex.z-dn.net/?f=%5Clambda%20%3D%5Cfrac%7B3.00%2A10%5E8m.s%5E-%5E1%7D%7B2.20%2A10%5E-%5E4s%5E-%5E1%7D)
![\lambda =1.36*10^1^2m](https://tex.z-dn.net/?f=%5Clambda%20%3D1.36%2A10%5E1%5E2m)
Hence, the right option is D)
.
Answer:
Number of peptide fragments resulting from cleaving with cyanogen bromide? A: Three peptide fragments
Number of peptide fragments resulting from cleaving with trypsin? A: Four peptide fragments
Which of these reagents gives the smallest single fragment (in number of amino acid residues)? A: CnBr, a dipeptide fragment consisting of AL (Alanine-Leucine)
Explanation:
Cyanogen bromide cleaves the methionine C-terminus, then we have a first fragment of 8 amino acids: DSRLSKTM, a second fragment of 15 aas YSIEAPAKLDWEQNM, and a last fragment of only 2 aas is produced, AL
Trypsin cuts the C-terminus of Arginine and Lysine, then we'll have a first fragment of 3 aas DSR, a second fragment consisting of also 3 aas LSK, a third fragment of 10 aas TMYSIEAPAK, and a last fragment of 9 aas LDWEQNMAL. All produced in three cut sites.