Login

Biochem Help?Moderator: BioTeam
15 posts • Page 1 of 2 • 1, 2
Biochem Help?What is the concentration in µg ml1 of a solution of glycine of 23 g litre1? (2 marks)
(Mol. wt of glycine = 75.07) Thanks ? Can I get some pointers on where to start 5. The following colony counts on nutrient agar plates were obtained from dilutions of an overnight culture of the bacterium Pseudomonas aeruginosa: 105 = >1000 106 = 622 107 = 74 108 = 14 Choose an appropriate colony count to determine how many colonyforming units (cfu) were present in 1 ml of the original culture (assuming that 100 l of the diluted culture was spread onto each agar plate)? (4 marks) 6. 50 mg of the amino acid isoleucine (molecular weight of 131.2) has been dissolved in 50 ml water to produce a stock solution. What is the molarity of the stock solution? The assay mixture requires a final concentration of 200 µM isoleucine in 10 ml buffer. What volume of stock solution needs to be made up to 10 ml? (4 marks) 7. The optical density OD (or absorbance) at 600 nm of an overnight culture of the bacterium Vibrio fischeri is 0.4. What volume of this suspension do you need to make up to 100 ml with fresh medium to give an OD of 0.1 at 600 nm? 10. From a bottle of molten nutrient agar you need to pour 20 agar plates each containing 50 g ml1 ampicillin. Each plate will contain ~ 25ml of nutrient agar, and you have a stock ampicillin antibiotic solution of 200 mg ml1. What volume of ampicillin stock solution do you need to add to the bottle of molten agar before pouring the plates? (4 marks)
Re: Biochem Help?
Pointers: 1 convert g in mol with MW and you have your answer. 5 your prof should have told you about the best numbers for counting. Basically if a plate is too dense or too sparse the value cannot be trusted (in one case because there are too many to insure correct isolation, in the other because the effect of random distributio becomes too important too be negligible) 6 same as the first one 7 You have 4$ one dollar bills, you owe me 1 dollar, how many bills do you give me? No seriously, this is as simple... 8 calculate the concentration in the bottle, the rest is just there to complicate things.
Patrick
Science has proof without any certainty. Creationists have certainty without any proof. (Ashley Montague)
23/75.07=0.31 mol.L1 and then you convert to µmol/ml
Patrick
Science has proof without any certainty. Creationists have certainty without any proof. (Ashley Montague)
Maybe I've just confused myself, but even after converting 0.31 mol.L1 to µmol/ml how do I get to the µg/ml the question wanted?
Ah wait... so mass/molar mass = 23/75.07 = 0.306mol Concentration = mol/volume in L = 0.306mol/1 = 0.306mol/L therefore 0.306mol/1000 (total ml in L) = 0.000306 (3.06 * 10^4) so 0.000306mol/mL then weight = mol * molar mass = 0.000306*75.07 = 0.023g per mL 1000000μg = 1g so 0.23g * 1000000 = 23,000μg/mL Is that looking correct now?
I think I might be responsible... I read µmol, not µg and give pointers fo this conversion. In this case going through the molarity is adding an useless layer of complexity. But the answer is correct
Patrick
Science has proof without any certainty. Creationists have certainty without any proof. (Ashley Montague)
Haha, ty,
I've gone back through my notes and completed question 5 now, and from your pointers I've had a go at question 7. If i'm right in the head i'm going to assume that I only need 25ml. I have 100ml that absorbs 0.4 units So I just divde that suspension into 4 parts (25ml) and dilute it with 75ml of fresh medium? This would make it 4 times weaker (I assume) and mean that it only absorbs 0.1 units?  And then on that note, I also had a go at question 6 here how much I did... I started off converting the 50mg to grams so 0.05g So mass divided by molar mass = 0.05g/131.2 = 0.000381 mol Now the molarity is mols per litre So I have 0.000381 moll and 50ml And this is where I lost myself last night (was too tired) But I assumed that as 50ml is 0.05 litres, Would I just multiply both sides by 20 and then divide them? So 0.000381 mol * 20 / 50ml * 20 Which is the same as just dividing them off the bat right? Yeah I got lost then... Any help on where I went wrong there? Because I'm sure I got to the mol part right but couldn't work out last night how to get to the molarity. 7.62 × 10 ^ 6 was my answer anyway..
Question 7 is correct
Question 6: your calculation are correct until the last step. You have 3.81e4 moles in 50 ml, so in 1L as you pointed out you would have 20 times more: 3.81e4 x 20=7.62e3 mol (per Liter) so the molarity is 7.62e3 M No you just calculate the dilution as in question 7 to know how much of the stock solution you need to get a 200µM solution (200µM = 200 e6 M = 2e4 M)
Patrick
Science has proof without any certainty. Creationists have certainty without any proof. (Ashley Montague)
Thankyou for the help with question 7 I had an idea it was a trick question and just looked harder but felt I was being stupid =/
As for question 6... I have 50ml of stock solution with a molarity of 7.62e3 (mol l1 (i think)) And I need a final solution to have 2e4M Soo... I realise that the one I have is too big... But the question also asks me to make it so that I have a volume that is less than 10ml (From the question I have 50ml atm) Now I think I've got the understanding for what the question wants... But I don't know where to go... I worked out the molarity of my stock if I just took 10ml of it, and the result is: 7.62e5 But this value is far far too low for the answer i'm supposed to be getting =/ I then worked out that I need 26.25 milliliters of my stock solution to have a molarity of 2e4 But the question reads What volume of stock solution needs to be made up to 10 ml So my answer can't be 26.25 ml can it?
No what you want is 10 ml of 200µmol per liter that is what µM means.
The traditional way of calculating is Ci x Vi=Cf x Vf Which basically says that you have the same number of molecule in the initial solution than in the final one (concentration x volume = #of molecules in moles) Ci: initial concentration, in mol per liter, here 7.62e3 Vi: volume of initial solution, that is what you want to calculate Cf: final concentration, in mol per liter, here 200e6 Vf: fianl volume, here 10ml You just need to solve this for Vi and you have your answer
Patrick
Science has proof without any certainty. Creationists have certainty without any proof. (Ashley Montague)
Been a little busy with other lecture work so have only just looked back at this.
Ci x Vi = Cf x Vf so Vi = Cf x Vf / Ci 2e4 x 10 / 7.62e3 = 0.264 ml? Converting the 10ml into litres would just give me the answer in litres I thought, so I left it as ml =/ Think I did this right anyway
15 posts • Page 1 of 2 • 1, 2
Who is onlineUsers browsing this forum: No registered users and 3 guests 
© BiologyOnline.org. All Rights Reserved. Register  Login  About Us  Contact Us  Link to Us  Disclaimer & Privacy  Powered by CASPION