http://www.umanitoba.ca/afs/plant_science/courses/PLNT3140/as4/as4.html

PLNT3140 INTRODUCTORY CYTOGENETICS    

Assignment 4


This assignment is worth 5% of the course grade.

Due Wed. Dec 6, 2023.




Figure 3.

Several inversions were found in polytene chromosomes of the dipteran midge Axarus varvestris.

Inversion heterozygotes are shown for four chromosomes exhibiting inversions. Arms are named using the letters A - G. Centromeres are indicated by arrows.


For inversions on arms A, C, F and G, Table 2 shows the frequencies of larvae homozygous for the standard chromosome (AA), heterozygous for the inversion (Aa), or homozygous for the inversion (aa) at one location on two different sampling dates. If the population was in Hardy-Weinberg equilibrium for each inversion, we would expect to see AA, Aa and aa progeny in the proportions of p2 : 2pq : q2, where q is the frequency of the inverted chromosome and frequency of the standard chromosome  p is equal to 1-q. For example, from a sample of 90 larvae, given an allele frequency q = 0.44 of the C inversion, the expected number of  aa larvae would be 0.442 x 90 = 17.4*.   (Note: "p" in the table is NOT the same p as in the equation above, but rather refers to the probability of seeing these ratios by chance if the population was in HW equilibrium).


* - The expected numbers of progeny in the table differ slightly from those predicted by the given value of q. The differences are small enough as to be attributable to rounding errors, for our purposes just ignore those differences.

1. (1 point) Based on the Hardy-Weinberg calculation, some of the expected number of aa larvae are less than 1. You can't have a fraction of a larva. What is the meaning of a number less than 1?

2. (1 point) The results for the A, C and F inversions are in strong agreement with HW frequencies. The authors state that the observed genotype counts G-arm inversion do not agree with HW equilibrium. Using q and n from the table, re-calculate the HW frequencies and expected number of progeny for the G inversion, and at least one of A, C or F. For your convenience, a spreadsheet template is provided in as4.spreadsheet_template.xlsx. When finished, paste your spreadsheet results into the report. Explain how the results tell us that G is out of equilibrium in this population.

3. (1 point) On both sampling dates, the observed number of aa larvae for G is 0. What is a possible explanation for this result?

4. (1 point) Figure 6 shows mirror histograms of how many larvae were found with the standard chromosome (above the 0 line), and how many were found with the inverted chromosome (below the line). The numbers of Standard/Inversion larvae are roughly symmetrical for G and C, while for A and F the Standard counts above the 0 line are much greater than the Inversion counts below the 0 line. Based on the data, what is the basis for the lack of symmetry for the A and F data? In other words, why is the area under the curve about the same above and below the line for G and C, whereas most of the area under the curve very little for the A and C inversions?

Figure 6


5. (1 point) The G inversion data in Fig. 6 seems skewed, compared to A, C and F. Based on the data given below, what does the data tell us about G larvae, compared to A, C or F larvae?



Submitting your assignment

Note on grading: In assigning a grade, some consideration may be given to how the answer communicates your ideas. Keep in mind the following:




Notes: If you have questions, it may help to send me a messge at frist@cc.umanitoba.ca.