There are two alleles for the arr gene, A and a. There are also two alleles for
ID: 204198 • Letter: T
Question
There are two alleles for the arr gene, A and a. There are also two alleles for beegene, 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.
39
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 in the box below
micrometers
Report your answer in micrometers
Report your answer to four decimal places
Or enter n/a if the genes are not linked
Genotype Abundance Aa Bb 47 Aa bb 23 aa Bb 31 aa bb39
Explanation / 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 aabb