Science Buddies: "Ask an Expert"

oh ok. I will do more research.. I just tested the uv strips for the sun exposure (only for like 30 min though) and it gave me a reading of 100 mj/cm squared. When I expose them for a whole day, I might get a different reading though. I am pretty sure that this is the amount the plants receive per day. If not, do I have to convert this reading into something else like milliwatts? Please let me know. Thanks

Yes--read current papers about photomorphogenesis until you feel confident that you understand the experiments, data, and conclusions well enough to explain it to someone who asks you a question. Don't make any statement that you cannot back up with evidence from other scientists' research or from your own experiments.

I'm glad to hear that the UV test strips are working. Let me know when you test the lamp and make the comparison between it and sunlight. If i understand the company's description of how the test works, it shows the accumulated UV dose for the time the strip was exposed to the light. So, the longer you leave the strip under the light, the darker it will get. I would try several different times [say 0.5 hr, 1 hr, 2 hr] for sunlight and several times for the lamp so you can get an accurate average hourly dose. If the response is linear then you should see a reading for 1 hr that is twice that for 0.5 hr, and the 2 hr reading should be twice the 1 hr reading. If you exposed the lettuce to 12 hr of sunlight then all you have to do is multiply 12 times the hourly dose reading to get the dose for the entire period. You would not need to leave the strips in the light for 12 hrs.

Try doubling the exposure time and see if it doubles the exposure dose. Let me know. The UV dose for the treated plants will be the dose from sunlight plus the dose from the lamp. Remember to take pictures of the strips immediately after exposure because the color may change with time. You can put the photos on your presentation board along with the actual dose readings.

I think you really have done a great project and i hope the judges think so too. I will be very interested to hear later what questions they asked and what comments they made.

Regards,

Sybee

Oh Ok. I already finished exposing the strips for 2 hours under the uv lamp and it gave me a reading of 200 mj/ cm squared. Does this mean that this is how much uv the plants in this group are receiving. I tried doing the student t test but i am unsure of what results I got. I did look up a guide on youtube and it told me that the probability that I would get would determine whether my null hypothesis is supported or not. If my null hypothesis was that the means were similar and I got a probability of about 0.2, does this mean that the means are't similar and if so what does this mean to my project?

Thanks

Did you place the test strips at about midway relative to the leaves of the lettuce plants? The closer the leaves are to the source, the greater the dose, but you can't test the UV for each leaf, so you will have to aim for a distance that looks like an average dose. Measure this distance and record it in your lab book along with measurements of the distances of the top and bottom leaves from the lamp. Are you taking pictures of the strips? If the color is not stable, then a photo will be the only record you have, so DON'T FORGET!

I am not following why you are trying to do a t-test on your UV readings. What are you comparing and why? The purpose of measuring the UV is simply to get the dose. It doesn't prove or disprove anything. What you need to compare are the average weights and the chlorophyll densities of the plants.

200 mJ/cm2 is the UV dose that the plants received from the lamp in two hours, so your 10-hour plants would have received 1000 mJ/cm2. In your previous post you said 30 min of sunshine gave 100 mJ/cm2. What time of day was that? The sun's intensity is greatest at noon, so you probably should do one reading at 9 am, one at noon and one at 3 pm and average them to get an average dose.

The number you are looking to find is how much MORE UV the treated plants received. They got the same 12 hours of sunshine each day as the non-treated plants, but they got 2 or 10 hours additional under the lamp. So, just add the UV dose from sunlight to the UV dose from the lamp and you have it.

You can divide the lamp UV dose by the sunlight UV dose x 100 to get the percent of additional UV. This number is important because you want to compare it to how much the solar UV is expected to change if the ozone layer decreases by some amount. You need to be able to say: "If the ozone decreases by one-third, the solar UV level will increase by x% and this compares to the percent increase in my experiment." I have no idea what it would take to decrease the ozone level by one-third. I just used that as an example so you can understand what I'm telling you to do.

Calculate how much additional UV your treated plants got as a percentage of the UV dose of sunlight and send me your calculations.

Sybee

I meant calculating the student t test for the chlorophyll content (a.k.a the means of the density of the leaf)

Oh, now i understand. The null hypothesis says that the two means are not different. If your t-test disproves that, then the means ARE different and your conclusion that the extra light resulted in extra chlorophyll production is correct.

You said you got higher density readings for the leaves from plants exposed to the UV lamp. Did you calculate the standard deviation of the means? When you plot the data you can include the S.D. values as error bars. Do you know how to do that in Excel? Also, you need to calculate p values for your two means/S.D. If the p value is 0.05 or less then you can be 95% certain that the means ARE statistically different. http://www.danielsoper.com/statcalc3/calc.aspx?id=8

Thank You. I got a probability of 0.2 approximately so does that mean it is proving the null hypothesis? When I searched on youtube, they said if it is under 0.5 then the null hypothesis is disproved. When showing this data, do I explain to everyone that these numbers are the densities of the leaves of each group and that these numbers prove how much chlorophyll content are within the leaves and the more chlorophyl, the more resistant the leaves are to uv.

The Image J program measures the darkness or lightness of an image. The more chlorophyll a leaf has, the darker it will appear in a photograph. The program measures the density [‘darkness’] of the IMAGE, the photos you took--not the leaf.

What are the means and the standard deviations of your density readings?

What was the result of your t-test?

Did you calculate how much more UV the plants received from the lamp?

Did you compare the weights of the plants yet?

I would not say that ONLY chlorophyll makes plants resistant to UVB. There are other chemical compounds in leaves that can absorb UVB, just like a sunscreen does that you rub on your skin. Look at those references again on photoregeneration. They talked about UV protection. And you can always google it and find out what plants use other than chlorophyll to block excess UV.

Always think carefully about your conclusions in science because you have to defend them.

Hi Sybee,
I've been dong research for whether plants such as phytoplankton undergo photomorphogenesis. I found a website that briefly talks about it and I wanted to let you see it to see what you think about it. Based on my understanding, phytoplankton do undergo photomorphogenesis but i may be wrong. Also, I found out that most organisms in the plantae kingdom undergo photomorphogenesis and my hypothesis in this is because, since most plants are sessile or immobile, they had to develop adaptations and mechanisms to defend themselves against uv radiation. That's why overtime, the plants developed a way to undergo photomorphogenesis.

http://www.photobiology.info/Hader.html

I'm not sure how assential this is anymore but I tried sending you the picture of the spectrum. Unfortunately, the picture is too big to send. Should I still put it on my board and say that the lamp emits more than just uvb and if they ask what I would have done differently, I could say that I could just use a source of light that only emits uvb. Oh also, I've conducted the student t test on excel and got a probability of 0.1 when I said that the tails was 2 while I got a probability of 0.07 when i said the tails was 1. So my question is, what tail and type do i input?

I've officially finished my project and other than typing my written report and application, I would say I am done. As soon as I get your input, I will fix and send you my application. So here is the information I got for The Effects of UV Radiation on Lettuce Growth

Data : all 3 plants received 3,600 mj/cm squared (noting that we receive 9 hours of sunlight) per day.
The 2 hour plants received an additional 200 mj/cm squared while the 12 hour plants received an additional 1200 mj/cm squared. So, the 2 hour plants received an additional 6% of uvb radiation per day and the 12 hours received an additional 34% of radiation per day. I understood that you wanted me to compare this to the amount the ozone layer would deplete. Should I search based on what people measure on how much the ozone layer will deplete in a few years and how much more uvb radiation we would receive so I could use that information to compare to the percentage of increase I got? For example, if the ozone layer depletes by 1/12 in a few years, we would receive say around an additional 10% of radiation. This approximately compares with the percentage of increase received by the 2 hour plants. Thus, our plants won't get negatively affected ( I should connect stuff like that to prove why the plants won't get negatively affected).

The Density of the Lettuce
0 hours - 103 arbitrary units
2 hours - 130 arbitrary units
12 hours - 133 arbitrary units

I tried using the t test but stumbled (as you saw earlier)

Weights of Plants
0 hour - 223.4 grams
2 hour - 262 grams
12 hour - 297.4 grams

The 2 hour plants received a 117% increase in fresh weight compared to the control and the 12 hour plants received a 133% increase compared to the control.

These were all the measurements I needed for my project and they all support the idea that a low dose of uv radiation stimulates growth. I have some questions to touch up my project though.

Should I convert the millijoules to nanometers due to the fact that the wavelengths for everything are in nanometers and the action wavelength for photomorphogenesis was also shown in nanometers? Also, if you have a solution to the uv intensity comparison to depletion, please let me know. Thank You very much for helping me accomplish this far.

You are on the right track with your statements about the ozone layer, but be sure you include citations to all the sources of information that you consulted in your paper. The standard way to do this is to put a number in the text like this: [1] then list all the references at the end in numerical order.

You should try to understand how global ozone levels are measured, how much the level has changed over the past several decades and what the future changes are predicted to be. The wiki on ozone depletion [http://en.wikipedia.org/wiki/Ozone_depletion] seems to be pretty accurate. There are two aspects of the ozone layer to think about. One is the ozone ‘hole’ over the poles and the other is the general ozone depletion over the middle latitudes where most of the farming takes place. The ozone hole over the poles would allow a greater flux of UVB there but since there is not a lot of vegetation at the poles, I would not expect this increased UVB to have any effect on plant life. The rest of the world, however, is where the depletion DOES have an effect.

What I am looking for is the relationship between DECREASE in ozone concentration and INCREASE in UVB at the land surface. It probably is not linear, by which I mean you cannot say that a 5% decrease in ozone will cause a 5% increase in UVB. Where did you get the value of 10% UV increase for a 1/12 decrease in ozone layer?

Here is the only statement I found about this in the wiki: “Ozone, while a minority constituent in Earth's atmosphere, is responsible for most of the absorption of UVB radiation. The amount of UVB radiation that penetrates through the ozone layer decreases exponentially with the slant-path thickness and density of the layer. When stratospheric ozone levels begin depleting, higher levels of UVB reaching the Earth’s surface will become more frequent. This means that the less ozone there is, the less protection there will be, hence more UVB reaches the Earth.[31] Correspondingly, a decrease in atmospheric ozone is expected to give rise to significantly increased levels of UVB near the surface. Ozone-driven phenolic formation in tree rings has dated the start of ozone depletion in northern latitudes to the late 1700s.[citation needed]”

What I still haven’t found is the exact mathematical relationship between ozone depletion and UVB increase. All I know is that the relationship is exponential which means that a small change in ozone can cause a large change in UVB levels—but exactly how much I can’t say without having the formula. Maybe you can find it. Why don’t you try calling NOAA, the National Oceanographic and Atmospheric Association? They are the ones, I think, who are involved in measuring depletion of the ozone layer and its effects on UVB, so they should be able to give you a more exact function for UV vs ozone.

Here’s a good article explaining the atmospheric ozone layer and how it affects the planet: http://www.ucar.edu/communications/gcip ... OR_GFX.pdf

Check pp. 5-6 of this monograph for information about UV absorbance and how the thickness of the ozone layer is measured using a Dobson spectrophotometer.

Your average values for image density and weight look good but what I don’t see is the standard deviation or standard error of the mean. To prove your results statistically, you must calculate these values. You can do this in Excel, by using a scientific calculator or online. You don’t need any advanced math. Just follow the directions, plug in your readings and the calculations will be done for you. Check youtube for a video that shows you how to do that, like this one: http://www.youtube.com/watch?v=sCbslydajOM

What you are comparing is the mean of the control, which is no extra light, with the mean readings from the plants that received 2 hr and 10 hr extra light. You have to do a t-test to prove that the means are not the same [the null hypothesis]. The youtube video above will show you how. Let me know how your t-tests turn out. Hopefully, they will prove that the increased chlorophyll and weight were the result of the increased light.

MINOR TECHNICAL POINTS
1. the correct unit for millijoules is mJ [not mj]
2. the correct unit for ‘cm squared’ is cm2 [the 2 should be a superscript]
3. uvb should be capitalized—UVB

FORMAT FOR REPORT
Are you using the standard journal style for a report of original research? This form is the best and is the way a paper published in a scientific journal is written: title, author and school name on the first page; a 300 [approx.] word Abstract that summarizes your work; Introduction in which you give some background information on the subject of your research; Methods in which you list everything you did and how you did it; Results; Conclusions; References.


I think you have done really well on a difficult project and I would hope the science fair judges think so too! Please let us know later how your presentation went.

All the best,

Sybee

Thank You, I tried making an example for the ozone depletion numbers, I doubt it's right. I will do the t test, standard deviation and graphs today and send them to you. Did you see my lasts post, I sent you a link regarding whether phytoplankton undergo photomorphogenesis

Thanks for the tips sybee. Would you say my data is paired or unpaired? As soon as I get a reply, I will do the t test. Also, did you see my last post, I sent a link I wanted you to take a look at about whether phytoplankton undergo photomorphogenesis

Hi again,

Sorry I missed the post with the link among all the others but I just read the article and it is very good. I knew that phytoplankton was more at risk from UV because of their exposure in the open ocean but I did not know that they were able to do PM to protect themselves. You should mention this in your report as phytoplankton are the plants from which most of the oxygen we breathe comes from.

I forgot to answer one of your questions from a previous post. You had asked about converting mJ to nanometers: “Should I convert the millijoules to nanometers due to the fact that the wavelengths for everything are in nanometers and the action wavelength for photomorphogenesis was also shown in nanometers?” You cannot convert mJ to nm because one is a measure of energy and the other is wavelength. The energy comes from light and different wavelengths have different energies. The shorter wavelengths are more energetic. Maybe I do not understand what you meant. Can you clarify what you are thinking of?

In regard to your t-test, the values you are comparing are unpaired because they come from two separate groups—one that received only sunlight and one that received sunlight plus light from the UVB lamp.

Good luck!

Sybee

Thank You. I will add the fact that even phytoplankton can undergo photomorphogenesis in my application. I will send you basically my whole written report (if you don't mind) when I am done. Regarding the t-test, mu null hypothesis was supported for I got a probability of 0.1 comparing the control to the 2 hour and a probability of 0.07 comparing the control to the 12 hour. Does this mean, there wasn't really any significant change in chlorophyll content? For the standard deviation, I got + or - 29 comparing the control to the 12 hour and a standard deviation of 31 comparing the control to the 2 hour. When plotting graphs, would you recommend me say 0 hour, 2 hour and 12 hour or the intensities the plants received in mJ

We usually set our cutoff for statistical significance at the 95% confidence interval which would give a p-value of 0.05, so the 12 hr treatment is almost there. Can you send me the density readings for each plant? There is a statistical test that will show whether one of the values is incorrect and can be discarded. One bad value can skew the mean so that it is incorrect.

When you plot the means do it as a bar chart. You should have the intensity values on the y-axis and the x-axis will be just the labels—
Control 2 hr UV 12 hr UV

You also need to have error bars on the graph and I would suggest using the standard error of the mean [S.E.M.] rather than the standard deviation. The S.E.M. is the standard deviation divided by the square root of the number of samples, in your case the number of plants, in each group.

Please also send me the raw data for the weights of all the plants and I will do the same statistical test for outliers.

I might have did it wrong, but I used different sections of the leaf and averaged that out first, then I averaged the averages of the two leaves I used per group. If it is wrong, please let me know but here is the data I got.
0 Hour - 137, 97, 86, 134,
0 hour second leaf - 130, 70, 76, 126

2 hour first leaf - 140, 89, 101, 180
2 hour second leaf - 117, 101, 160, 154

12 hour first leaf - 136, 156, 140, 116
12 hour second leaf - 179, 115, 127, 93

I'm not quite sure what you meant by raw data but here are the weights anyways. Including the soil and pot, the 0 hour weighed 553 grams, the two hour weighed, 591.6 grams and the 12 hour weighed 607 grams. Subtracting the soil and pot, the fresh weight for the 0 hour was 223.4, 2 hour was 262 and the 12 hour was 297.4 grams. When plotting in the bar graph, did you mean the densities in the y-axis for I don't really understand what the intensities would do in the y-axis. Thank You

I'm confused. How many plants were in each group? Did you read two leaves from each plant and do four readings per leaf?

What is meant by the raw data is the actual weights of EACH plant, before you average them. I need to see each individual weight so i can test whether any of the values should be discarded. You did weigh every plant,right?

Sorry. I meant average density on the y-axis.

Yes, I took 2 leaves per group but measure 4 different corners per leaf and averaged that first. That's why there are so many numbers. For the raw data of the weights of my plants, I never really averaged them so I am quite confused. The way I interpreted it was the weights of the plants before I subtracted the weight of the pot and soil from it.

I understand that you made 4 density readings per leaf on 2 leaves, but that’s not what I wanted to know.

What I mean is when you started this experiment, you had some number of plants in each of your three groups—HOW MANY?

Now, when you measured the density of the leaves, you were supposed to do it for EACH plant in the group, then average the density readings for that group. Did you not do this?

I don’t see density readings for EACH plant in EACH of the three groups.

Same for the weights. What is the weight of EACH plant?

Oh, I see. I had 2 plants per group and yes, I took one leaf per plant and measured it. Yes, I did find the average of the densities for a certain group. That's how I got my numbers. The average density for the first plant of the control group is 113.5. The average density of the leaf for the second plant of the control group is 100.5. The average density of the leaf for the first plant of the 2 hour group is 127.5 and the density for the leaf of the second plant of the 2 hour group is 133. Finally, the density of the leaf of the first plant in the 12 hour group is 137 and the density of the leaf in the second plant of the 12 hour group is 128.5.

When you do an experiment the number of subjects per group is VERY important. At the very minimum you should have had 3 plants per group--5 would have been better. You can do a t-test with only 2 subjects but it is not very accurate. I should have noticed back in December that you said you planned to use 2 plants per group and told you to use 5 per group.

I hope you have learned a lot in this project about how to conduct scientific experiments. When you are planning your next project, keep in mind all the details we have talked about in this project, especially with regard to statistics which is so important.

Best wishes,

Sybee

Thank you. So for the student t test, did it prove that there was any significant change amongst the plants? If so, what should I do? Also, should I look up a video on how to graph error bars in a bar graph?

You can say that the trend shows an increase in chlorophyll density and mass with longer light exposure, but the differences are not statistically significant.

If you are using Excel to make the bar graphs, search youtube for videos on how to add the error bars. There are lots of videos to choose from. It is much easier to learn by watching it than it is to read a tutorial--at least i think so.

Thank You for the information. I have now completed my graphs and have the result that as I increased the amount of uv radiation, plant growth increased. For the application, this may still be an assumption, but now knowing even phytoplankton can undergo photomorphogenesis as well as most plants due to the fact that they are sessile, that as the ozone layer depletes, most of our plants won't get negatively affected and some may even thrive more. This project also raises alertness for the protein uvr8 and the fact that it can stimulate plant growth. Knowing this information, genetic engineerers can incorporate this protein into plants that lack it (in order to stimulate their growth) and allow our world to have more plant life. Should I do some research on genetic engineering for the protein due to the fact that genetic engineering may not always help plants. Should I not even incorporate this fact and just say for my application that most of our plant live will thrive as the ozone layer depletes. Regarding questions the judges may ask, such as why this is the case, I would explain the idea of photomorphogenesis and how photoreceptors can repair plants and help them develop more. If they ask why I chose a low dosage instead of a high, even though a high dosage would have killed the plants , I would say that the point of my project is to observe the effects of the ozone depletion and the more radiation that our plants would be exposed to. Most likely, the ozone layer wouldn't allow that much uvb radiation to come in unless it vanishes (that's why i did a low dose) these are the two questions I am expecting to ask. If there any loop holes to my project please let me know for I need to make some touch ups to my project since the fair is in a week. Finally, this topic was brought up before but I do have the spectrum for the lamp. I probably will add it to my board and tell the judges that even though results still showed that uvb did have an impact on plant growth, the fact that the lamp emitted other light such as visible light might have affected the results. So next time, I should use just a uvb lamp to observe the effects.

If you are going to propose engineering plants with UVR8, first you need to do some research to find out which plants have it and which plants don't. Maybe they ALL have the gene or something similar. Being a scientist is something like being an attorney--no one will believe your statements unless you present evidence.

One point that you are overlooking is the dose effect. Plants can adapt to 'some' increase in UVB [I don't know how much] but at some point the UV will be too much for them--and probably us too. I don't know how much is too much. How much UVB would we be exposed to if half the ozone layer was gone? What if it was all gone? I asked you to call the National Oceanographic and Atmospheric Administration [NOAA] and ask them how much extra UVRB exposure would occur for a certain decrease in ozone. I tried to find this value but it wasn't in any of the sources i checked.

Yes--definitely show the lamp spectrum and tell the judges what you said in your post. It is good that you do have the UV dose data from the test strips to prove how much UV the plants got.

If I think of anything else that you should add or be prepared for I will let you know. Have some copies of your key references with you in case the judges want to look at them.

Congratulations,

Sybee

Ok. I'll try calling them this weekend and see what I get to add on to my project. For the error bars, though I looked up a video and know how to do it, I have a few questions on it. First, is the point of the error bar to show where/what part of the mean is inaccurate? Also, when finding standard error of mean, do I take both the means of the control and 2 hour at the same time to calculate (and similarly the means of the 0 and 12 hour) or do I find the standard error the means separately? Thank You

The S.E.M. is calculated from the standard deviation which is a measure of how much the actual measurements differ from the mean of the measurements. that is why the more readings you have per plant and the more plants you have per group, the more accurate your statistics are. When we make groups for our experiments we use a minimum of 4 subjects per group and usually 6 to 8. Also, every experiment is repeated at least once--usually twice. This is the only way to be sure your data are accurate.

Each S.E.M. is calculated for each mean separately.

So, I finished calculating the S.E.M for each group and found that the 0 hour had a standard error of mean of 19.6, the 2 hour had one of 23.3 and the 12 hour group had a s.e.m of 18.8. Would lower numbers mean that it is closer or farther to the mean? If so, what do I saw when presenting the error bars and s.e.m? Also, what you saw arbitrary units are? I am aware that they are vague units but can't really say much about it. Thank You