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.

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 offspring.

Genotype	Abundance
Aa Bb	47
Aa bb	23
aa Bb	31
aa bb	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

Answer 1

(a)

Option :1 (The arr gene and bee gene are independent of each other)

Explanation:

According to the question , the two variables are

Arr gene alleles : A and a    and bee gene alleles: B and b .

Therefore to test wheteher there are any relationship between this two variable, we can define the null hypothesis as:

The arr gene and bee gene are independent of each other

against the alternative hypothesis as :

The arr gene and bee gene are not independent of each other.

(b)

According to the question there are 4 different genotypes (AaBb, Aabb, aaBb, aabb) in 1:1:1:1 ratio.

Therefore the expected genotype frequency, whether they are not linked or independent of each other can be calculated as below:

	Oi	Ei	(Oi-Ei)^2/Ei
0.25	47	35	4.1143
0.25	23	35	4.1143
0.25	31	35	0.4571
0.25	39	35	0.4571
1	140	140	9.1429
		TS	9.1429
		critical value	7.8147
		p-value	0.0275

Formulas

	Oi	Ei	(Oi-Ei)^2/Ei
0.25	47	=$B$7*A2	=(B2-C2)^2/C2
0.25	23	=$B$7*A3	=(B3-C3)^2/C3
0.25	31	=$B$7*A4	=(B4-C4)^2/C4
0.25	39	=$B$7*A5	=(B5-C5)^2/C5
=SUM(A2:A5)	=SUM(B2:B5)	=SUM(C2:C5)	=SUM(D2:D5)
		TS	=D7
		critical value	=CHISQ.INV(0.95,3)
		p-value	=CHISQ.DIST.RT(D7,3)

Calculated value of chi-squared = 9.1429
p-value = 0.0275
since p-value < alpha
we reject the null hypothesis,