Consider two loci A and B that are on the same chromosome but very far apart suc
ID: 3206351 • Letter: C
Question
Consider two loci A and B that are on the same chromosome but very far apart such that their measured recombination frequency (RF) is 0.45. Therefore, if you were to sample 20 gametes, on average you would find that 11 were parental-type and 9 were recombinant type. [Print out your R console from parts (c) and (d) and submit it with your problem set. Each student should perform the R analysis and turn in his or her own print out. Using these numbers (11 and 9) as your observed values, set up a chi-square test to show that a sample size of 20 gametes is insufficient to confidently detect linkage between genes separated by an RF of 0.45. Clearly state (in words) the null hypothesis for this test, and list the expected values for parental-type and recombinant type gametes given this null hypothesis. Calculate the corresponding chi-square value. A widely used threshold for statistical significance is pExplanation / Answer
Solution:
a) The null hypothesis is that the two loci are unlinked (i.e., they assort independently) and, therefore, that the number of parental-type and recombinantt-type gametes will be equal.
Under the null hypothesis, the expected numbers of parental-type and recombinant-type gametes would both be 10.
Chi-sq = ((11-10)^2)/10 + ((9-10)^2)/10 =0.2
b) THe calculated chi-square value is much less than 3.84, so we would fail to reject the null hypothesis. This means that we cannot conclude that the genes are linked.
c) > phenotypes = c(11,9)
> expected = c(0.5,0.5)
> chisq.test(phenotypes, p=expected)
Chi-squared test for given probabilities
data: phenotypes
X-squared = 0.2, df = 1, p-value = 0.6547
d) > phenotypes = c(11,9)
> expected = c(0.5,0.5)
> chisq.test(phenotypes, p=expected)
Chi-squared test for given probabilities
data: phenotypes
X-squared = 0.2, df = 1, p-value = 0.6547
e) Following the doubling scheme described in part (d), the values 352 vs. 288 were the first to fall below p = 0.05 (actual p-value was 0.011), so a total smaple size in the rough vicinity of 500 would likely be sufficient to detect significant linkage.