Question

1. (4 points) Given that the average molecular weight of an amino acid is 110 daltons (Da), determine the following. Remember to convert between daltons and kilodaltons (kDa) where necessary. Each of your answers must be numerically correct and expressed in proper units (1 pt) and show how the value was arrived at using the amino acid molecular weight above (1 pt) for you to receive full credit. A. A molecule of human a-tubulin has a molecular weight of 50.1 kDa. Approximately how many amino acids does it contain? Round to the nearest whole number. B. Protein C contains 105 amino acids. What is its molecular weight in daltons (Da)? 2. (6 points) In a brief paragraph, explain how soluble proteins were obtained from the fish muscle samples. Your explanation should include: 1. the purpose of mincing (cutting) and vortexing the tissue (2 pts), 2. the purpose of the homogenization buffer (2 pts), and 3. the purpose of the centrifugation step (2 pts). You must clearly describe what role each of these things played in extracting and solubilizing protein from the samples to receive full credit. When considering the homogenization buffer, revisit the “Micropipetting and Preparing Solutions" exercise to remind yourself of what the buffer ingredients were.

Answer 1

Answer 1:

Answer 1: • Given, average molecular weight of an amino acid is 110 Da, Aw) • Formula to be used: Molecular weight of protein = Number of amino acids X Aw => M = NX AW -(1) (A) Given, molecular weight of x- tubulin protein = 50.1 kDa = 50100 Da We need to find no. of amino acids in it. Using the above mentioned formula, we can follow that, 50100 = N X 110 => N = 50100 / 110 I N = 455,45 amino acids. I * Thus, number about 455 of amino acids in 50.1 kDa protein will be ( mearest whole number). B) Given, a protein C contains 105 amino acids. We need to find its molecular weight, M. Using (i), we get, M= NX AW = M = 105 X 110 = 11,550 Daltons. * I hus molecular weight of protein C is 11,550 Da

(A) 455 amino acids

(B) 11,550 Da