jake_f

I got it finally. I took the number 37 and 159 and multiply to their dilution factor and average it. And did the same for plate 2, 45 and 220. Then I average the answer of the two plates. Thanks for you help.

My lecturer told me that the answer for that question is 3.00 X 10 ^8. But I still could not get it. I use all the various number that is between 25-250 and calculated back based on dilution but I still couldn't get the answer.

I don't know how to start, what does it mean by counting the number of CFU between 25 – 250 colonies? Do I take the number, 37 from plate one and 45 from plate 2, divide it by 0.1ml and multiply by 10^6?

How do you calculate that? The following results were obtained when a sample of microbes was serially diluted and the diluted samples used for spread plate experiment (0.1 ml per plate). Determine the CFU/ml in the original sample (express your answer to 2 decimal places). Dilution | Plate 1 | plate 2| 10^-4 | 450 | 360 | 10^-5 | 159 | 220 | 10^-6 | 37 | 45 | 10^-7 | 7 | 10 | The steps my lecturer taught me is: -Count the number of CFU between 25 – 250 colonies -Multiply the number by the reciprocal of the dilution -Gives an estimate of the number of bacteria/ml What does it means?

Please point out where I got wrong. The question: At a sand gravel plant, sand is falling off a conveyor belt and onto a conical pile at the rate of 10m^3 / min. The diameter of the base of the cone is approximately 3 times the altitude. At what rate is the height of the pile changing when it is 15 meters high? My solution: Let V be the volume of the conical pile, and the height be h, and the radius, r. Given, dv/dt = 10. Since the diameter of the base is approximately 3 times the altitude, h= 6r, and r = 2.5 m h=6r dh/dr = 6 v=1/3 πr^2h v=1/3 πr^2(6r) v=2 πr^3 dv/dr = 6πr^2 dv/dt= dr/dt X dv/dr dr/dt= 10 / 6π(2.5)^2 = approximately 0.0848826 m/min dh/dt = dr/dt X dh/dr =0.0848826 X 6 =0.5092956 m/min (my wrong Ans) But the answer should be 0.0063m/min. Please explain to me where I got wrong.

Thanks to dttom and CharonY, I guess I am finally on the right track. So the reason for the white blood cell to have RNA, is to synthesis proteins, but the functions of the proteins synthesized are related to the different functions of the various types of white blood cells. Thanks, I think I can try to attempt that question now.

I think it is as compared to red blood cell. So is the high concentration of RNA due to the protein synthesis that is needed in the leukocytes?

I am Jake, just eighteen years old. I am just a student and definitely not the brightest crayons in the box. I hoped that you guys wouldn't mind my 'poor science' , I am still working hard. Pleased to meet you all.

Leukocytes is divided into two groups, Granulocytes and Agranulocytes. I know that the DNA is used for amino acid sequences of the proteins and enzymes. The question I would like to ask: Why do Granulocytes/ Agranulocytes have a high concentration of RNA? I also know that there are the messenger RNA (mRNA), ribosomal RNA (rRNA) and transfer RNA (tRNA). I know their functions too. But the reason is I don't really know why the Leukocytes must have a high concentration of RNA. Is it because of the white blood cell function? Please enlighten me.

Actually, I just have some doubts with my experiment result. The equation for the reaction, C7H6O3 + C4H6O3 -> C9H8O4 +CH3COOH I started off with 2.42g of salicylic acid (C7H6O3) which is about 0.175 mole of it. By right, based on the equation, one mole of salicylic acid should yield one mole of aspirin (C9H8O4). So 0.175 mole of Aspirin is expected at the end of the experiment. And the expected mass of aspirin should be 3.15g. But at the end of the experiment, I only got 1.1g of aspirin. Which when calculated in percentage yield, is only 34.92%. So should I presume that I made some mistake during the course of the experiment? When I asked my lecturer about it, he just give me a smile which make me even more suspicious about my results.