QUESTION 3 FOR THIS QUESTION, EXCEL \"MUST BE USED TO CALCULATE THE P-VALUE FOR

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QUESTION 3 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 aro two alleles for the arr gene, A and a·There 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 gone and the bee gone 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 1. The arr and bee genes are independent of each other 2. The arr and bee genes are linked. Above are the two hypotheses, which of the two hypotheses is the hypothesis? 1 Enter either 1 or 2 in the answer box · The table below depicts the observed genotypic abundances of the offspring Genotype Abundance Aa Bb Aa bb aa Bb 40 30 40 35 Use the predicted and observed data to perform a chi-squared test to assess if two genes arr and bee are linked. Below is a video about how to perform the calculations in Excel Name: ChiSquaredinExcel (05:24) Duration: 05:24 Added: 20 Feb 2018 05:04 PM Added By: Joseph Battistelli tests chisquare Tags: chi square, dihybrid cross What is your calculated value of ch squared? Watch Media Click Save and Submit to sqve and submit. Click Save All Ansuwers to save all ansivers Save All Anwers

Explanation / Answer

Answer:

Based on the given information:

1) Which of the above hypothesis is the Null Hypothesis:

2) The given genetic cross is: AaBb x aabb. The parental gametes are: AB, Ab, aB, ab and ab, ab, ab, ab

It can be represented in the form of punnett square as below:

As per the punnett square representation, there are 4 different genotypes (AaBb, Aabb, aaBb, aabb) in 1:1:1:1 ratio.

The expected genotype frequency/abundance if the genes are not linked or independent of each other can be calculted as below:

Expected genotype abundance = Total number of individuals with different genotypes / 4 = (40+30+40+35)/4= 36.25

Expected frequency of each genotype = 36.25

Chi Square analysis:

Degree of freedom = Number of categories - 1 = 4 - 1 = 3

P-value can be calculated using an online tool or Excel:

Probabilty value corresponding to a Chi square value of 1.8966 with 3 degree of freedom is: P = 0.594143. The result is not significant at p < 0.05.

Since P-value is more than the significance level (0.05), therefore Null Hypothesis is not rejected. Based on the evidence, their is no significant different between the observed and expected phenotypic ratio. The genes are not linked and are indepedent of each other.

ab ab ab ab AB AaBb AaBb AaBb AaBb Ab Aabb Aabb Aabb Aabb aB aaBb aaBb aaBb aaBb ab aabb aabb aabb aabb

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