Part 1
Purpose:
The purpose of this lab is to find the most efficient way to make cheese. Efficiency is determined by the process' speed, simplicity, amount of materials, and the taste product. The amount of materials was a constant so we judged the most efficient culture, by it’s curdling speed.
Hypothesis:
My hypothesis for this lab is that buttermilk will curdle the milk the fastest. Each curdling agent is an independent variable, and the curdling time is dependent.
Procedure:
Data:
Curdling Agent
Curdling Time
Weight of Cone and Curds
Weight of Cone
Weight of Curds
Rate
Chymosin
FPC
5 minutes
4.69g
1.15g
3.54g
708 mg/m
Chymosin
NCB
1440 minutes
2.94g
1.15g
1.79g
1.24mg/m
Buttermilk
24 hours
2.8g
1.15g
1.65g
1.14mg/m
Water
24 hours
2.36g
1.15g
Jhuu 1.22g
0.84 mg/m
Analysis:
From the data, we can see that FPC chymosin curds milk the fastest. My hypothesis was that buttermilk would curd the milk the fastest. My hypothesis was incorrect, FPC was the most effective curdling agent. One error in this lab was that we were unable to check every half hour for curdling after 30 minutes. It was the end of the period, so we left it overnight. This error did not affect the results in any way, the FPC still curdled before the others. One way to improve the lab would be to repeat the curdling process twice, to check if the results were consistent. This would lead to further investigations, such as if the heat was higher, would the milk curdle faster.
Conclusion:
We discovered FPC chymosin curdles milk at a faster rate than water, buttermilk, and NCB chymosin. The experiment consisted of putting equal amounts of milk with four different curdling agents(chymosin FPC, water, buttermilk, and NCB chymosin) in our armpits, and recording the rate it took each agent to curdle the milk, from the time and amount of curds. As you can see from the table above, our data showed that chymosin FPC curdles milk at a fast rate. It had a rate of 708 mg/s while the next closest was only 1.24 mg/s. This means that FPC chymosin had a very high rate, and that it is the most effective method of making cheese. Again, FPC had the highest rate of curdling and is very effective.
Part 2
Purpose:
The purpose of this lab was to determine if a larger amount of FPC resulted in a faster curdling rate.
Hypothesis:
My hypothesis is that a 200 ul amount of FPC chymosin with have a faster curdling rate than a 100 ul amount of FPC.
Purpose:
Data:
Curdling agent
Weight of cone and curds
Weight of cone
Weight of curds
Rate
Curdling time
FPC 100ml
3.16g
.75g
2.4g
204 mg/m
10 min
FPC 200ml
3.15g
.75g
2.41
482 mg/m
4 min
Analysis:
From this data, we can determine that 200 ul of FPC curdles milk at a faster rate than 100 ul. This means that the rate of curdling depends on the amount of FPC. My hypothesis was correct, the larger amount of FPC does have a much higher curdling rate. A possible error could be that the temperature of the two armpits were different temperatures, possibly from the difference in shirt material. A way to solve this would be to use both armpits of the same person, or put both tubes together in one armpit. This would improve the accuracy of our results. This lab leads to further investigation, such as if the amount of fat in milk changes the rate of curdling.
Conclusion:
We discovered that with a larger amount of FPC, the milk curdles at a quicker rate. Our experiment included curdling milk with 100 ul of FPC and 200 ul of FPC, and recording the curdling time of both. The 100 ul amount of FPC curdled milk at a rate of 204 mg/s while the 100 ul amount of FPC curdled the milk at a rate of 482 mg/s. The amount of curds was equal in both. The time of curdling was shorter in the larger amount. From this data we can see that the 200 ul amount of FPC curdled the same amount of milk in 4 minutes versus 10 minutes from the 100 ul amount. This means that to curdle cheese effectively, you must use the largest amount of FPC possible. The 200 ul amount had a faster curdling rate than the 100 ul solution of 482 mg/s vs 204 mg/s.
Part 3
Purpose:
The purpose of this lab is to determine what macromolecules are in our cheese.
Hypothesis:
My hypothesis is that our cheese contains all four macromolecules.
Procedure:
-Monosaccharide indicator test
1.Test for glucose: In a test tube, mix 2 ml of cheese solution with 2 ml of benedict’s solution. Heat for 2 minutes in boiling hot water bath (100 ml of water in a 250-ml beaker at 100 degrees C) record all color changes and the length of time for each color change to appear.
-Polysaccharide test
2. Test for starch: In a test tube, mix 2 ml of well-mixed cheese solution with 0.25 ml of
Lugol’s Iodine. Gently swirl to mix. DO NOT HEAT. record the color change.
-Protein test
3. Test for protein: place 2 ml of cheese solution in test tube. Wearing goggles and gloves, add 0.5 ml of 10% NaOH and gently vortex to mix. Add 0.25 ml of 5% copper sulfate (CuSO4) and gently mix. The NaOH and CuSO4 mixture is called biuret reagent. Mix well. Record color change after 30 seconds.
-Lipid test
4. Test for lipids): Sudan IV. add 6 microliters of Sudan IV to 2 ml of cheese sample. Gently mix, red is negative for lipid test orange is positive test.
5. Repeat every test, substituting macromolecule for our cheese.
Data:
Standard
Indicator used
Description + control
Description - control
Glucose
Benedict's Solution
yellow-dark red
blue
Starch
Lugol's Iodine
Black
Red
Protein
Biuret Reagent
Purple
Light blue
Fat/Lipids
Paper test/ sudan IV
Orange
Red
Macromolecule
Cheese does
Contain
(Positive Result) ✅
Glucose
✅
Starch
Protein
✅
Lipid
✅
Analysis:This data means that our cheese contains glucose, protein, and lipids. It does not contain starch. My hypothesis was that our cheese contained all four macromolecules. This was incorrect, we only had three of four macromolecules. Some possible errors could be that skin cells from our hands fell onto the cheese altering the results. To improve the accuracy of the results, we could wear gloves when holding cheese. A further investigation would be to see how much of each macromolecule there is in the cheese.
Conclusion:
We discovered out cheese contains the macromolecules; glucose, protein, and lipids. We also know that there is no starch in our cheese. Our experiment consisted of testing whether our cheese had macromolecules by testing the reactions of the cheese with other substances. Our data table shows that the cheese tested positive for glucose, protein, and lipids. Our data shows cheese tested negative for starch. This means that the cheese contains glucose, protein and lipids but does not have any starch. Again, the tests concluded with the cheese having glucose, protein and lipids. There are no starches in cheese.
Purpose:
The purpose of this lab is to find the most efficient way to make cheese. Efficiency is determined by the process' speed, simplicity, amount of materials, and the taste product. The amount of materials was a constant so we judged the most efficient culture, by it’s curdling speed.
Hypothesis:
My hypothesis for this lab is that buttermilk will curdle the milk the fastest. Each curdling agent is an independent variable, and the curdling time is dependent.
Procedure:
- Label 4 6ml tubes with the type of curdling agent and group number.
- Use a large pipet to transfer 3 ml of milk to each of the 6ml tubes.
- Use a small pipet and transfer the entire contents of the tubes, 100 ul, of fermentation produced chymosin, natural bovine chymosin or buttermilk to the labeled tube containing the milk. For water, fill the small transfer pipet to the bottom of the bulb and add to the labeled tube containing the milk. Use a different pipet for each transfer to avoid cross contamination.
- Cap the tubes, invert 3 times, then transfer to armpit.
- Check for curdling every 5 minutes by examining the curds.
- Record the time when the milk begins to curdle, small lumps or solidified.
- If the milk hasn’t curdled in 30 minutes, check every half hour.
- Record in minutes when milk curdles in data table.
- Determine the amount of curds produced by each solution
- For each treatment weigh a paper one and record its weight
- Transfer the entire contents of a tube into a labeled filter paper cone over a collection vessel. Once liquid has drained, dry the curds overnight
- Weigh the cone with curds. Subtract the one weight. Record the weight of the curds in mg.
- Repeat for each treatment
- Make a data table showing the rate of curd production by each curdling agent.
- Create a data table showing the rate.
Data:
Curdling Agent
Curdling Time
Weight of Cone and Curds
Weight of Cone
Weight of Curds
Rate
Chymosin
FPC
5 minutes
4.69g
1.15g
3.54g
708 mg/m
Chymosin
NCB
1440 minutes
2.94g
1.15g
1.79g
1.24mg/m
Buttermilk
24 hours
2.8g
1.15g
1.65g
1.14mg/m
Water
24 hours
2.36g
1.15g
Jhuu 1.22g
0.84 mg/m
Analysis:
From the data, we can see that FPC chymosin curds milk the fastest. My hypothesis was that buttermilk would curd the milk the fastest. My hypothesis was incorrect, FPC was the most effective curdling agent. One error in this lab was that we were unable to check every half hour for curdling after 30 minutes. It was the end of the period, so we left it overnight. This error did not affect the results in any way, the FPC still curdled before the others. One way to improve the lab would be to repeat the curdling process twice, to check if the results were consistent. This would lead to further investigations, such as if the heat was higher, would the milk curdle faster.
Conclusion:
We discovered FPC chymosin curdles milk at a faster rate than water, buttermilk, and NCB chymosin. The experiment consisted of putting equal amounts of milk with four different curdling agents(chymosin FPC, water, buttermilk, and NCB chymosin) in our armpits, and recording the rate it took each agent to curdle the milk, from the time and amount of curds. As you can see from the table above, our data showed that chymosin FPC curdles milk at a fast rate. It had a rate of 708 mg/s while the next closest was only 1.24 mg/s. This means that FPC chymosin had a very high rate, and that it is the most effective method of making cheese. Again, FPC had the highest rate of curdling and is very effective.
Part 2
Purpose:
The purpose of this lab was to determine if a larger amount of FPC resulted in a faster curdling rate.
Hypothesis:
My hypothesis is that a 200 ul amount of FPC chymosin with have a faster curdling rate than a 100 ul amount of FPC.
Purpose:
- Label 1 6ml tube with 100 ul of FPC and another with 200 ul of FPC.
- Use a large pipet to transfer 3 ml of milk to each of the 6ml tubes.
- Use a small pipet and transfer the entire contents of the tubes, 100 ul and 200 ul of fermentation produced chymosin.
- Cap the tubes, invert 3 times, then transfer to armpit.
- Check for curdling every 5 minutes by examining the curds.
- Record the time when the milk begins to curdle, small lumps or solidified.
- If the milk hasn’t curdled in 30 minutes, check every half hour.
- Record in minutes when milk curdles in data table.
- Determine the amount of curds produced by each solution
- For each treatment weigh a paper one and record its weight
- Transfer the entire contents of a tube into a labeled filter paper cone over a collection vessel. Once liquid has drained, dry the curds overnight
- Weigh the cone with curds. Subtract the one weight. Record the weight of the curds in mg.
- Repeat for each treatment
- Make a data table showing the rate of curd production by each curdling agent.
- Create a data table showing the rate.
Data:
Curdling agent
Weight of cone and curds
Weight of cone
Weight of curds
Rate
Curdling time
FPC 100ml
3.16g
.75g
2.4g
204 mg/m
10 min
FPC 200ml
3.15g
.75g
2.41
482 mg/m
4 min
Analysis:
From this data, we can determine that 200 ul of FPC curdles milk at a faster rate than 100 ul. This means that the rate of curdling depends on the amount of FPC. My hypothesis was correct, the larger amount of FPC does have a much higher curdling rate. A possible error could be that the temperature of the two armpits were different temperatures, possibly from the difference in shirt material. A way to solve this would be to use both armpits of the same person, or put both tubes together in one armpit. This would improve the accuracy of our results. This lab leads to further investigation, such as if the amount of fat in milk changes the rate of curdling.
Conclusion:
We discovered that with a larger amount of FPC, the milk curdles at a quicker rate. Our experiment included curdling milk with 100 ul of FPC and 200 ul of FPC, and recording the curdling time of both. The 100 ul amount of FPC curdled milk at a rate of 204 mg/s while the 100 ul amount of FPC curdled the milk at a rate of 482 mg/s. The amount of curds was equal in both. The time of curdling was shorter in the larger amount. From this data we can see that the 200 ul amount of FPC curdled the same amount of milk in 4 minutes versus 10 minutes from the 100 ul amount. This means that to curdle cheese effectively, you must use the largest amount of FPC possible. The 200 ul amount had a faster curdling rate than the 100 ul solution of 482 mg/s vs 204 mg/s.
Part 3
Purpose:
The purpose of this lab is to determine what macromolecules are in our cheese.
Hypothesis:
My hypothesis is that our cheese contains all four macromolecules.
Procedure:
-Monosaccharide indicator test
1.Test for glucose: In a test tube, mix 2 ml of cheese solution with 2 ml of benedict’s solution. Heat for 2 minutes in boiling hot water bath (100 ml of water in a 250-ml beaker at 100 degrees C) record all color changes and the length of time for each color change to appear.
-Polysaccharide test
2. Test for starch: In a test tube, mix 2 ml of well-mixed cheese solution with 0.25 ml of
Lugol’s Iodine. Gently swirl to mix. DO NOT HEAT. record the color change.
-Protein test
3. Test for protein: place 2 ml of cheese solution in test tube. Wearing goggles and gloves, add 0.5 ml of 10% NaOH and gently vortex to mix. Add 0.25 ml of 5% copper sulfate (CuSO4) and gently mix. The NaOH and CuSO4 mixture is called biuret reagent. Mix well. Record color change after 30 seconds.
-Lipid test
4. Test for lipids): Sudan IV. add 6 microliters of Sudan IV to 2 ml of cheese sample. Gently mix, red is negative for lipid test orange is positive test.
5. Repeat every test, substituting macromolecule for our cheese.
Data:
Standard
Indicator used
Description + control
Description - control
Glucose
Benedict's Solution
yellow-dark red
blue
Starch
Lugol's Iodine
Black
Red
Protein
Biuret Reagent
Purple
Light blue
Fat/Lipids
Paper test/ sudan IV
Orange
Red
Macromolecule
Cheese does
Contain
(Positive Result) ✅
Glucose
✅
Starch
Protein
✅
Lipid
✅
Analysis:This data means that our cheese contains glucose, protein, and lipids. It does not contain starch. My hypothesis was that our cheese contained all four macromolecules. This was incorrect, we only had three of four macromolecules. Some possible errors could be that skin cells from our hands fell onto the cheese altering the results. To improve the accuracy of the results, we could wear gloves when holding cheese. A further investigation would be to see how much of each macromolecule there is in the cheese.
Conclusion:
We discovered out cheese contains the macromolecules; glucose, protein, and lipids. We also know that there is no starch in our cheese. Our experiment consisted of testing whether our cheese had macromolecules by testing the reactions of the cheese with other substances. Our data table shows that the cheese tested positive for glucose, protein, and lipids. Our data shows cheese tested negative for starch. This means that the cheese contains glucose, protein and lipids but does not have any starch. Again, the tests concluded with the cheese having glucose, protein and lipids. There are no starches in cheese.
Reflection
This project went well. I did well following directions, analyzing our data, and creating new experiments. I had success in finding results, and also the new experiment was conclusive. Some improvements I must make, is paying attention to the content beforehand, editing my write up, and putting in some extra research on my own. This lab was very important.