SHOW ALL YOUR WORK! (3 points) You have 2.0 mL of an extract of protein from rat

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SHOW ALL YOUR WORK! (3 points) You have 2.0 mL of an extract of protein from rat skeletal muscle. You have done a assay for total protein on a set of standards and on several dilutions of your unkn According to your standard curve, the own sample. protein concentration of a 120 dilution of the unknown sample is 86 microgramsmhL (86 g/mL) (See the above discussion of standard curves.) a. What is the concentration of protein in the undiluted anknown sample (report in mg/ml rt in mg/mL)? b. What is the total protein yield from your rat muscls extract (report in mg)? c. If one gram of rat skeletal muscle typically contains about0.07 mgof protein per gram of tissue, approximately how much rat skeletal muscle must you have extracted in this experiment (in grams)? 2. (2 points) Fill in the blanks: ("a" is an example) 10 110', or , or or -fold a. Add 0.5 mL sample to 4.5 mL ofdiluent. Dilution- b. Add 2.0 mL sample to 8.0 mL of diluent. Dilution c. Add 0.1 mL sample to 0.9 mL of diluent.Dilution -fold d. Add 0.3 mL sample to 29.7 mL of diluent. Dilution , or_-_fold Dilution- , or -fold e. Add 100 ul. sample to 300 L of diluent. 3. (1 point) Fill in the missing information. The volumes indicated in the tubes are starting volumes, and the volumes indicated by the arrows are the volumes being added. a. 2 ml 1 ml Final dilution or, -told Dilutions-

Explanation / Answer

1) A) 1 to 20 fold dilution concentration is 86 ug/ml. So, 86 X 20 = 1720 ug/ml. 1720 ug = 1.72 mg

So, protein concentration in undiluted unknown sample is 1.72 mg/ml.

B) Total protein extract is 2 ml. Therefore, total protein extract is 1.72 X 2 = 3.44 mg

C) 1 gram tissue contains 0.07 mg of protein. Therefore, 3.44/0.07 = 49.14

So, we need 49.14 grams of tissue extracts.

2) B) 10/2 = 5. Therefore 1 to 5 or 5 fold

C) 1/0.1 = 10. Therefore 1 to 10 or 10 fold

D) 30/0.3 = 100. Therefore 1 to 100 or 100 fold

E) 400/100 = 4. Therefore 1 to 4 or 4 fold

3) 4/2 = 2 fold. 5/1 = 5 fold. So, Final dilution = 1 to 7 or 7 fold

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