Q. I need bacteria only (no other types of organisms) and the specific strains of bacteria and description of them or any sources where I can find the information. Thanks.
A. Wine is primarily made using yeasts such as Saccharomyces cerevisiae, via alcoholic fermentation. The pyruvate decarboxylase enzyme used by these yeasts to produce ethanol is rare in bacteria, however there are a few bacterial organisms than can carry out alcoholic fermentation.
Zymomonas mobilis is one of the rare species that does possess the pyruvate decarboxylase, and would be able to produce ethanol. It can be isoalted from the Mexcian alcoholic beverage "pulque".
Sarcina ventriculi and Ervinia amylovora are other representative organisms that can produce ethanol from fermentation.
Many lactic acid bacteria, enterobacteria, and clostridia can produce considerable ethanol without the pyruvate decarboxylase. Instead, they convert:
pyruate -> acetyl-CoA -> acetaldehyde -> ethanol, alcohol dehydrogenase.
It has been of high interest to the alcohol industry to find organisms that can produce ethanol under high temperature, to prevent contamination of unwanted organisms. Some bacterial species capable of such alcoholic fermentation at high temperatures include: Clostridium thermohydrosulfuricum, Thermoanaerobium brockii, and Thermoanaerobacter ethanolicus. Note that these organisms lack pyruvate decarboxylase and produce considerable acetate and lactate in addition to ethanol.
Hope this helps!
Zymomonas mobilis is one of the rare species that does possess the pyruvate decarboxylase, and would be able to produce ethanol. It can be isoalted from the Mexcian alcoholic beverage "pulque".
Sarcina ventriculi and Ervinia amylovora are other representative organisms that can produce ethanol from fermentation.
Many lactic acid bacteria, enterobacteria, and clostridia can produce considerable ethanol without the pyruvate decarboxylase. Instead, they convert:
pyruate -> acetyl-CoA -> acetaldehyde -> ethanol, alcohol dehydrogenase.
It has been of high interest to the alcohol industry to find organisms that can produce ethanol under high temperature, to prevent contamination of unwanted organisms. Some bacterial species capable of such alcoholic fermentation at high temperatures include: Clostridium thermohydrosulfuricum, Thermoanaerobium brockii, and Thermoanaerobacter ethanolicus. Note that these organisms lack pyruvate decarboxylase and produce considerable acetate and lactate in addition to ethanol.
Hope this helps!
What is the best method to know when bacteria is dying, or the best method to count bacteria in a petri dish?
Q. To clarify, if bacteria is in two petri dishes, and I add two different chemicals detrimental to the bacteria, what is the best way to tell which chemical is killing the bacteria fastest? Thanks!
A. I would suggest a method similar to that used in testing antimicrobial susceptibility. (1) Impregnate an known similar concetration of your chemical each on its own paper disk. (2) spreas your bacteria evenly on appropriate medium. (3) stamp the chemical disks on top of medium and incubate at 37 degrees for a day. (4) measure the zone of inhibition (area surrounding chemical disk with no bacterial growth). The larger the area, the more effective the chemical.
N.B. there may also be issues of chemical diffusion etc.
N.B. there may also be issues of chemical diffusion etc.
What happens to the bacteria after extracting the insulin or hormone?
Q. I know that we use bacteria for DNA recombinant. After we extract whatever we need from the bacteria, what happens to the bacteria? I really want to know.
A. The bacteria are usually killed in the process.
After all, it's not as if you could milk them.
In some cases, the desired end product can be sseparated from the bacteria without killing them, such as skimming it off. In that case a continuous production is possible. Otherwise, you make it in batches.
After all, it's not as if you could milk them.
In some cases, the desired end product can be sseparated from the bacteria without killing them, such as skimming it off. In that case a continuous production is possible. Otherwise, you make it in batches.
The count in a bacteria culture was 400 after 15 minutes and 1400 after 30 minutes. What was the initial siz?
Q. The count in a bacteria culture was 400 after 15 minutes and 1400 after 30 minutes. What was the initial siz?
Find the doubling period in minutes
Find the population after 115 minutes
When (in minutes) will thle population reach 15000?
4 hours ago - 4 days left to answer.
Find the doubling period in minutes
Find the population after 115 minutes
When (in minutes) will thle population reach 15000?
4 hours ago - 4 days left to answer.
A. Long answer...
A = Pe^rt
A -represents the amount after a certain amount of time
P- represents the the amount at the start
r -represents the rate or speed and is always represented as a decimal
t- represents the amount of time in years, months, days, hours , minutes
e- is natural log = 2.718281828
see: http://www.algebralab.org/Word/Word.aspx?file=Algebra_InterestIII.xml
1st we find the rate or speed of growth represented by letter r
A = xe^rt
400= xe^r15
400/x =e^r15
As rule: ln e^x =x
Ln(400/x) = ln e^r15= r15
Rule: ln(a/b) = ln a - lnb
Ln 400 �ln x =15r
Ln 400 -15r =lnx
5.99146-15r =lnx
Have to do the same for the 1400 bacteria.
1400= xe^r30
1400/x =e^r30
Ln(1400/x) = ln e^r30= r30
Ln 1400 �ln x =30r
Ln 1400 -30r =lnx
7.242275 -30r =lnx
Set the ln x�s to equal
7.242275 -30r = 5.99146-15r
1.252763=15r
1.252763/15 = r =0.083517531
Faster way to find x:
400/x e^r15 =1400/xe^r30
e^r30/e^r15 =1400/400 (the x's cancel)
e^r15 = 3.5
ln e^r15 = ln 3.5
r15 =1.252762968
r =1.252762968/15 =0.083517531
Now we can find initial amount x.
400= xe^(0.083517531)(15)
400 =x e^1.252763
e- is natural log = 2.718281828 , plug # in calculator and hit inverse ln works faster
e^1.252763 =3.5
400=x(3.5)
400/3.5 = x = 114.286 =P
The values for P and R never change.
What was the initial size? P =114.286 is initial amount
Find the doubling period in minutes. For doubling
Let left side (A/P) equal 2 and then find time.
2 = e^(0.083517531)t
Ln 2 =(0.083517531)t
0.69314718/0.083517531 = t = 8.3 min
Find the population after 115 minutes. Let t =115 and find A = 1,695,041
When (in minutes) will the population reach 15000? Let A =15000 and find t = 58.4 minutes
A T e^rt
400153.5
228.57148.292
1,00025.978.75
1,4003012.25
15,00058.39131.25
210,087.5901838.266
1,695,04111514831.61
5,932,64413051910.64
31,526,254150275854.7
A = Pe^rt
A -represents the amount after a certain amount of time
P- represents the the amount at the start
r -represents the rate or speed and is always represented as a decimal
t- represents the amount of time in years, months, days, hours , minutes
e- is natural log = 2.718281828
see: http://www.algebralab.org/Word/Word.aspx?file=Algebra_InterestIII.xml
1st we find the rate or speed of growth represented by letter r
A = xe^rt
400= xe^r15
400/x =e^r15
As rule: ln e^x =x
Ln(400/x) = ln e^r15= r15
Rule: ln(a/b) = ln a - lnb
Ln 400 �ln x =15r
Ln 400 -15r =lnx
5.99146-15r =lnx
Have to do the same for the 1400 bacteria.
1400= xe^r30
1400/x =e^r30
Ln(1400/x) = ln e^r30= r30
Ln 1400 �ln x =30r
Ln 1400 -30r =lnx
7.242275 -30r =lnx
Set the ln x�s to equal
7.242275 -30r = 5.99146-15r
1.252763=15r
1.252763/15 = r =0.083517531
Faster way to find x:
400/x e^r15 =1400/xe^r30
e^r30/e^r15 =1400/400 (the x's cancel)
e^r15 = 3.5
ln e^r15 = ln 3.5
r15 =1.252762968
r =1.252762968/15 =0.083517531
Now we can find initial amount x.
400= xe^(0.083517531)(15)
400 =x e^1.252763
e- is natural log = 2.718281828 , plug # in calculator and hit inverse ln works faster
e^1.252763 =3.5
400=x(3.5)
400/3.5 = x = 114.286 =P
The values for P and R never change.
What was the initial size? P =114.286 is initial amount
Find the doubling period in minutes. For doubling
Let left side (A/P) equal 2 and then find time.
2 = e^(0.083517531)t
Ln 2 =(0.083517531)t
0.69314718/0.083517531 = t = 8.3 min
Find the population after 115 minutes. Let t =115 and find A = 1,695,041
When (in minutes) will the population reach 15000? Let A =15000 and find t = 58.4 minutes
A T e^rt
400153.5
228.57148.292
1,00025.978.75
1,4003012.25
15,00058.39131.25
210,087.5901838.266
1,695,04111514831.61
5,932,64413051910.64
31,526,254150275854.7
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