What difference in traits is caused by chromosomes?
1 answer:
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Answer: The blanks can be correctly filled up with prophase I and metaphase II.
Explanation:
In females, finite number of oocytes are present since birth. The process of oogenesis initiate in embryonic stage only. However, oogonia divide and mature to form primary oocytes. These primary oocytes start meiotic division but arrest in prophase I stage before birth.
After a girl attains puberty, one primary oocyte resumes its division each month during a menstrual cycle. The primary oocyte divides to form secondary oocyte and polar body.
Polar bodies degenerate after completing meiosis II. However, secondary oocyte halts its division again at metaphase II until fertilization takes place. Once fertilization takes place it completes its meiosis II and results into mature ovum and polar body.
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).