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. 2009 Spring;8(1):29–43. doi: 10.1187/cbe.08-07-0039

Table 5.

Postsemester interview results: summary of significant differences in tool usage and answer quality between experimental and control groups

Question Bloom's classification Tool used to ID molecule? Tool used for longer time period before answering question? Which group produced better answers?
5. Can you tell me what you've learned about protein 1°, 2°, 3°, and 4° structure? Note: 17 of the 20 students answered this question without consulting any tool. Knowledge Experimental (1 of 4 assessors; Kruskal-Wallis P value = 0.025, t test P value = 0.020)
6. What kind of biomolecule do you think this is: lipid, protein, nucleic acid, or carbohydrate? Application Experimental group used models (Kruskal-Wallis P value = 0.088, t test P value = 0.072) Experimental group used models (Kruskal-Wallis P value = 0.073, t test P value = 0.131). Experimental (1 of 4 assessors; Kruskal-Wallis P value = 0.097, t test P value = 0.069)
8. Can you find the N terminus? Application Control group used PE (Kruskal-Wallis P value = 0.091, t test P value = 0.181) Experimental (2 of 4 assessors; (Kruskal-Wallis P values = 0.059 and 0.013, t test P values = 0.034 and 0.013, respectively).
9. Identify alpha helices/beta sheets Application Experimental (2 of 4 assessors; both Kruskal-Wallis P values = 0.075, both t test P values = 0.074)
10. Does this biomolecule show any 4° structure? Application Experimental (2 of 4 assessors; Kruskal-Wallis P values = 0.075 and 0.057, t test P values = 0.066 and 0.071, respectively)
16a. Propose function for colored region in biomolecule Synthesis Experimental (1 of 4 assessors; Kruskal-Wallis P value = 0.076, t test P value = 0.071)
20 & 21. Propose mutations which constitutively activate/deactivate biomolecule Synthesis Experimental group used models (Kruskal-Wallis P value for Q20 = 0.058, t test P value = 0.046; Kruskal-Wallis P value for Q21 = 0.061, t test P value = 0.126) Experimental (Q21) (1 of 4 assessors; (Kruskal-Wallis P value = 0.085, t test P value = 0.085)

Cell entry in the final column indicates how many of the four assessors found significant differences between the two groups and statistical output.