Using the diagram above, a key characteristic of aquatic dead zones is ________ and which activities is most directly related to
the dead zones in the Gulf of Mexico just west of the Mississippi River?
A) high concentration of dissolved oxygen, and extensive shrimp fishing
B) low concentration of dissolved oxygen, and farming in the Midwest
C) high concentration of nitric oxides and burning coal to generate electricity in the Northeast.
D) low concentration of dissolved phosphates an Barge traffic on the Mississippi River
A) high concentration of dissolved oxygen, and extensive shrimp fishing
B) low concentration of dissolved oxygen, and farming in the Midwest
C) high concentration of nitric oxides and burning coal to generate electricity in the Northeast.
D) low concentration of dissolved phosphates an Barge traffic on the Mississippi River
1 answer:
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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).