• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

An investigation into the effect of pectinase on fruit juice production from an apple.

Extracts from this document...


An investigation into the effect of pectinase on fruit juice production from an apple Aim: To investigate how the affect how adding pectinase to chopped apples varies the amount of juice that can be taken out from it. Background Pectinases are used in the processing of non-citrus fruits to maximise the production of clear juice. Nearly all fruits and berries contain pectins and other polysaccharides such as starch. Pectins hold the fruit cells together like a "glue" and result in less of juice being released during crushing. The presence of soluble pectins in the juice also causes hazing. The addition of pectinases such as pectin methyl esterase, polygalacturonase and pectin lyase, at the crushing (or pulping) stage increases the yield of juice and helps in the clarification. Pectinases are particularly important in the production of fruit juice concentrates (like Vimto) as pectins can form very viscous gels which stop filtration and that leads to a high concentration of dissolved solids. Prediction I predict that the amount of juice collected with pectinase will be 20% more of that collected without pectinase. ...read more.


My second graph of rate of juice collection against time shows that the pectinase increased the rate at which fruit juice was collected. At first the rate is very high for both lines but then they start to lower until the two lines converge. This shows that for both with and without pectinase, the rate of collection will decrease. The rate at which juice is collected is higher for pectinase for a certain amount of time then after 4 minutes it returns to a slower rate which is the same as the rate without pectinase. My third graph shows that pectinase increased the yield of fruit juice between 200% and 300% which is far more than the 20% that was my prediction. Conclusion In conclusion I can see that pectinase increased the fruit juice yield and the rate at which fruit juice is collected. Pectinase does this because it breaks down a protein called pectin which holds together the cellulose found in the cell walls of the fruit. ...read more.


This is because the amount of juice collected changes quickly over the first few minutes and having readings every minute does not monitor this change effectively. I could have read the amount of juice every 30 seconds and that would have given me more I didn't encounter any problems while doing this experiment. Ways I could improve this experiment are taking more readings to improve reliability, this would increase my reliability a lot. I could monitor the surface area of the apple chinks which would improve my reliability slightly. I could have repeated the experiment to improve my reliability greatly. This experiment is biologically significant because pectinase is used in the fruit juice industry. They need to maximise the amount of fruit juice from each apple and to do that, pectinase is needed as just demonstrated from my experiment. Ideas for further study are investigating the ratio of apple to pectinase and seeing if that is proportional to the amount of juice collected. Also I could see if pectinase works with other fruits such as oranges and compare to see in which fruit pectinase works best. ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our University Degree Botany section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related University Degree Botany essays

  1. Investigation Into the effect of pectinase on apple juice extraction.

    Conclusion Although there are a number of anomalies and inaccuracies within the results there is a trend and this trend proves my hypothesis. The results do show that as the amount of enzyme is increased so to is the volume of apple juice.

  2. Factors affecting the rate of photosynthesis

    This would have increased the rate of reaction of the plant's enzymes which would have increased the photosynthesis rate. 0.1M NaHCO3 The 0.1M NaHCO3 produced the steepest line. Near the end of the line it looks as if the rate of reaction is hit by another limiting factor.

  1. An investigation to find the effect of bile salts on the digestion of fats.

    If too much bile is added, then the solution will be too alkaline, and the optimum pH will be lost and the breakdown of fat will not be as effective (see fig 2) The pH of the pancreatic swine lipase we are using is 6.6 and it changes the initial

  2. An Investigation into Microclimate on a Sand Dune System

    I also doubt the consistency of the equipment itself. I did predict that the humidity would be higher at the beginning and end of the data, closer to the sea and then in the woodland area. This has almost worked out on my graph as the highest points on my graph are at the first and last few recordings.

  1. A review of the development, production and post harvest requirements of Gerberas

    3.1 Propagation: * Seed: Gerberas have large seeds, which germinate at about 70% under suitable environmental conditions (Rogers and Tjia, 1990). Plants are genetically varied so Gerbera produced from open pollinated seed are unlikely to be similar, which is no good for commercial production.

  2. Investigating the inhibitory effect of reserpine on locomotor activity in mice, and its reversal ...

    Statistical Analysis * To determine whether there was a significant difference between the locomotor activity of the control mouse and the mouse treated with reserpine (2hrs), a 2 sample unpaired t-test was conducted with Mouse A and Mouse C.

  1. This investigation aims to determine what effect an increase in the surrounding temperature has ...

    I will now work out the ?. To do this I will need to add the values from the (x-x)�: 0.45 + 0.11 + 0.11 = 0.67 I will now divide the above value by 2 0.67 / 2 = 0.335 To finish the equation I need to now square route 0.335 V0.335 = 0.56 The standard deviation for 53 C is 0.56.

  2. Comparing the yields of fruit juice produced from different types of fruits using Pectinase.

    Enzymes have catalytic properties; in other words, they alter the rate of reaction without themselves undergoing a permanent change. Most chemical reactions require an initial input of energy, called activation energy, to enable them to occur. Enzymes reduce the need for activation energy and so allow reactions to take place

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work