Of the following locations you would most likely find mosses growing in?
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fine details such as the wings, legs and other fragile structures are preserved effectively giving us a window into how the preserved organism would have appeared in life. without the presence of amber, we wouldn't be able to examine a majority of small prehistoric organisms.
Explanation: amber is able to preserve a snapshot of the ancient world in ways that other forms of fossilization simple fails to do. while an imprint of a feather could be preserved in rock the actual feather could be captured within amber. Amber also preserves small organisms that otherwise would have been lost to time such as the insects seen in the photo above. And much like the gecko seen below it can show us how these organisms truly appeared in life.
Two traits that we are examining are the color (purple/white) and the smoothness.
In our sample, we have 135 individuals ( 75 +28+24+8= 135).
Since we know that expected ratio of a dihybrid cross, is <span>9:3:3:1 we can calculate what is the expected values of each phenotype is.
So, for the white wrinkled phenotype, we expect that there will be one-sixteenth of the whole sample ( there are 16 parts of the whole sample 9+3+3+1=16).
So, we multiply the whole sample- 135 with one-sixteenth (or </span>0.0625) and get 8,4375.
When you calculate the values for all phenotypes you get results shown in the attached excel table.
When you have your expected and experimental values you compare them with a chi-square test. (The test determines if the difference between the expected and experimental results is statistically significant).
In our sample, we have 135 individuals ( 75 +28+24+8= 135).
Since we know that expected ratio of a dihybrid cross, is <span>9:3:3:1 we can calculate what is the expected values of each phenotype is.
So, for the white wrinkled phenotype, we expect that there will be one-sixteenth of the whole sample ( there are 16 parts of the whole sample 9+3+3+1=16).
So, we multiply the whole sample- 135 with one-sixteenth (or </span>0.0625) and get 8,4375.
When you calculate the values for all phenotypes you get results shown in the attached excel table.
When you have your expected and experimental values you compare them with a chi-square test. (The test determines if the difference between the expected and experimental results is statistically significant).