When amylase break down starch what sugars are produced??
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The given question says that b species diverged around 60,000 year ago from species a. Since the divergence, they have a nucleotide difference of 15 nucleotide.
We need to find out the nucleotide difference in another 240,000 year, which means a total of 300,000 year since divergence.
The nucleotide difference in 60,000 years = 15 nucleotide.
The nucleotide difference in 1 year = nucleotide.
The nucleotide difference in 300,000 years = nucleotide.
The nucleotide difference in 300,000 years = 75 nucleotide.
Hence, after another 240,000 years or 300,000 years since divergence, the species and species b would show a difference of 75 nucleotide.
In another 240,000 years, there would be 60 nucleotide difference, as 15 of the nucleotide are already different in initial 60,000 years from the total of 75 nucleotide difference in 300,000 years. So, 60 (75-15) nucleotide difference would be observed in next 240,000 years.
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).