Life Sciences lab classes explained

I expect a lot of you have been struggling with the table on page 41 of your 109 workbook.  Let me try to work through it for you (this is going to be a long post!). This table looked something like this: If you read the previous question, you will realise that this table wants values for alanine.  In the previous question you were asked to calculate the  relative molecular mass  of alanine - remember, this is equivalent to the  mass of 1 mole of alanine .  To calculate this, you have to add up the atomic masses of all of the elements that make up alanine, and if you do this you get a relative molecular mass of  89  (which means, 1 mole of alanine weighs 89g). ******************************************************* A quick re-cap on moles and relative molecular mass.... Remember from my previous post,  moles are coming , that a mole is simply a quantity, it is  6.02 x 10 23  of something.  And relative molecular mass ...

Struggling with the Cytochrome C question in your 109 workshop 2.... Then this post is for you! Let me show you how I would do this question: The Question: A lot of information followed by... The Biochemist prepared 2.5 ml of cytochrome c (Mr = 12000) in water (solution A). It was then diluted by addition of 1.0 ml solution A to 9.0 ml water to give solution B. Finally 2.0 ml solution B was diluted 20-fold (i.e. the ratio of the final volume to the initial volume is 20) with water to give a final solution C which was determined to have a concentration of 0.15 mg/ml. Before I even look at the questions, I would draw out the information provided in a kind of flow diagram: Ok, now I have a clear idea of what I have been told, I will look at the questions: 1)  What volume of water was added to give the 20-fold dilution of cytochrome C in the preparation of solution C? So, we know that we are talking about the second dilution step, where 2.0 ml of solution B was taken and dilu...