FOR THIS QUESTION, EXCEL *MUST* BE USED TO CALCULATE THE P-VALUE FOR CHI SQUARED
ID: 204162 • Letter: F
Question
FOR THIS QUESTION, EXCEL *MUST* BE USED TO CALCULATE THE P-VALUE FOR CHI SQUARED!
Here is a link to the Khan Academy page on dihybrid crosses and gene mapping. Here is a link to the Math Bench web page on performing chi-squared tests.
There are two alleles for the arr gene, A and a. There are also two alleles for bee gene, B and b. The two genes are on a chromosome that is 1.5 micrometers in length, but it has yet to be determined if the two genes are linked. Below is data from a study to determine if the genes are linked, and if so, how far apart the genes are on the chromosome.
An organism that is heterozygous for both the arr gene and the bee gene is crossed with an organism that is homozygous recessive for both genes.
Using this information, it is possible to calculate the expected phenotypic ratio of the offspring of the two organisms, assuming the two genes are *NOT* linked. If the genes are linked, the offspring will not exhibit the expected phenotypic ratio.
The arr and bee genes are independent of each other.
The arr and bee genes are linked.
Above are the two hypotheses. Which of the two hypotheses is the null hypothesis?
The table below depicts the observed genotypic abundances of the offspring.
Use the predicted and observed data to perform a chi-squared test to assess if two genes arr and bee are linked.
What is your calculated value of chi squared?
Report your answer to FOUR decimal places
FURTHER CALCULATIONS SHOULD BE PERFORMED IN EXCEL ON THE UNROUNDED NUMBER
What is the P-value determined using the formula =(1-chisq.dist(x, df, TRUE)) in Excel, where x is the calculated value of chi squared, df is the degrees of freedom, and TRUE indicates a cumulative distribution function.
Report your answer to FOUR decimal places
Using an alpha value of 0.05, do you fail to reject or reject the null hypothesis?
Enter either fail to reject, or reject.
If the genes are linked, use the percent of offspring that experienced crossing over to determine the distance between the two genes in micrometers. If the genes are not linked, enter n/a.
Report your answer in micrometers
Report your answer to four decimal places.
Genotype Abundance Aa Bb 40 Aa bb 30 aa Bb 40 aa bb 35Explanation / Answer
Hypothesis : The arr and bee genes are independent of each other
we know that genotypic ratio for dyhybrid cross = 9:3:3:1
total progeny = 145
Genotype Observed(O) Expected(E) (O -E) (O -E)2 (O - E)2/E
aa bb 35 27 8 64 2.3703
Aa bb 30 9 21 441 49
total 145 145 78.7758
Hence, P calculated = 78.7758
df = 3
Ho is rejected: The arr and bee genes are linked and are not independent of each other.
Aa Bb 40 82 -42 1764 21.1463 aa Bb 40 27 13 169 6.2592aa bb 35 27 8 64 2.3703
Aa bb 30 9 21 441 49
total 145 145 78.7758
Hence, P calculated = 78.7758
df = 3
Ho is rejected: The arr and bee genes are linked and are not independent of each other.
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