Question

1. 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.

   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.

   1.) 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

   2.) 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.

   3.) Using an alpha value of 0.05, do you fail to reject or reject the null hypothesis?

2. 4.)  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 to FOUR decimal places or

Answer 1

Answer:

The cross seems to be Dihybrid testcross & its ratio would be 1:1:1:1

Total progeny = 140

AaBb = 140/4 = 35.000

Aabb = 140/4 = 35.000

aaBb=140/4 = 35.000

aabb=140/4 = 35.000

Phenotype	Observed(O)	Expected (E)	O-E	(O-E)2	(O-E)2/E
AaBb	47	35.0000	12.0000	144.0000	4.1143
Aabb	23	35.0000	-12.0000	144.0000	4.1143
aaBb	31	35.0000	-4.0000	16.0000	0.4571
aabb	39	35.0000	4.0000	16.0000	0.4571
Total	140	4.0000			9.1429

X^2 value = 9.1429

Degrees of freedom = Number of phenotypes – 1

Df = 4-1=3

P value value = 7.81

The chi-square value of 9.1429 is greater than the p value of 7.81. We can reject null hypothesis; hence, the two genes are not assorted rather, they are linked.

Hint: Non-recombinant progeny are more common in linkage.

Recombination frequency (RF)= (no. of recombinant progeny / total progeny)100

RF= (23+31 / 140)100 = 38.5714%

RF(%) = Distance between genes (mu)

Distance between the two genes = 38.5714 m.u.

Answer 2

To perform a chi-squared test and assess if the genes arr and bee are linked, we need to compare the observed genotypic abundances with the expected genotypic ratios under the assumption of independent assortment.

Based on the observed genotypic abundances:

Genotype Abundance Aa Bb 47 Aa bb 23 aa Bb 31 aa bb 39

To determine the expected genotypic ratios under independent assortment, we need to calculate the expected frequencies for each genotype. The expected frequency for each genotype can be calculated by multiplying the marginal frequencies of each allele. Since the genes are independent, the expected frequency for each genotype is the product of the frequencies of each allele:

Expected frequency = (frequency of arr allele) * (frequency of bee allele)

Let's calculate the expected frequencies:

Expected frequency of Aa Bb = (frequency of Aa) * (frequency of Bb) = [(47 + 23 + 31 + 39) / total] * [(47 + 31) / total] = (140 / total) * (78 / total)

Expected frequency of Aa bb = (frequency of Aa) * (frequency of bb) = [(47 + 23 + 31 + 39) / total] * [(23 + 39) / total] = (140 / total) * (62 / total)

Expected frequency of aa Bb = (frequency of aa) * (frequency of Bb) = [(47 + 23 + 31 + 39) / total] * [(47 + 31) / total] = (140 / total) * (78 / total)

Expected frequency of aa bb = (frequency of aa) * (frequency of bb) = [(47 + 23 + 31 + 39) / total] * [(23 + 39) / total] = (140 / total) * (62 / total)

To calculate the chi-squared statistic, we compare the observed and expected frequencies using the formula:

chi-squared = Σ [(observed frequency - expected frequency)^2 / expected frequency]

Calculating the chi-squared value:

chi-squared = [(47 - Expected frequency of Aa Bb)^2 / Expected frequency of Aa Bb] + [(23 - Expected frequency of Aa bb)^2 / Expected frequency of Aa bb] + [(31 - Expected frequency of aa Bb)^2 / Expected frequency of aa Bb] + [(39 - Expected frequency of aa bb)^2 / Expected frequency of aa bb]

Degrees of freedom (df) = (number of genotypes - 1) = (4 - 1) = 3

Now, to answer your questions:

1.) The calculated value of chi-squared depends on the actual frequencies and the total number of offspring, which is not provided in the question. You will need to substitute the actual values and perform the calculations using Excel or statistical software to obtain the chi-squared value.

2.) The P-value can be determined using the formula =(1-chisq.dist(x, df, TRUE)) in Excel, where x is the calculated value of chi-squared and df is the degrees of freedom. You can substitute the calculated chi-squared value to obtain the P-value.

3.) With an alpha value of 0.05, if the P-value is less than 0.05, you reject the null hypothesis. Otherwise, you fail to reject the null hypothesis.

4.) If the genes are linked, you can use the percent of offspring that experienced crossing over to estimate the distance between the two genes. However, the information regarding the percent of offspring that