using Statistical Table A (pg 704 of your textbook), what is the chi-square crit
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using Statistical Table A (pg 704 of your textbook), what is the chi-square critical value with significance level of a-0 057 (report exactly the value will you reject or fail to reject the null hypothesis? (answer either reject or fail to reject) 98 ESTION 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 links to the Math Bench web page on performing chi-squared tests here 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 h e 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 Using this information as possble to calc te he expected phen type atio of the omspn g of two organ sms, assuming the two genes are NOT linked If the genes a e? offspring will not exhibit the expected p for both the arr gene and the bee gene is crossed with an organism that is homozygous recessive for both genes 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 null hypothesis? . Enter either 1 or 2 in the answer box The table below depicts the observed genotypic abundances of the offispring Aa Bb Aa bb a Bb 47 ed test to assess if two gones arr and bee are linked Below is a video about how to perform the ca the ca ChiSauaredinExcel (05:241 Click Save and Submit to save and submit. Click Save All Answers to save allExplanation / 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 = (47+23+31+39)/4= 35
Expected frequency of each genotype = 35
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 9.1429 with 3 degree of freedom is: P = 0.02745. The result is not significant at p < 0.05.
Since P-value (0.02745) is less than the significance level (0.05), therefore Null Hypothesis is rejected. Based on the evidence, their is significant different between the observed and expected phenotypic ratio. The genes are linked.
Map Distance = Number of F2 recombinants / Total number of F2 offsprings x 100
= 100*(23+31) / (47+23+31+39) = 38.6 m.u
ab ab ab ab AB AaBb AaBb AaBb AaBb Ab Aabb Aabb Aabb Aabb aB aaBb aaBb aaBb aaBb ab aabb aabb aabb aabbRelated Questions
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