IL-1ß is a protein released by cells under conditions of infection. Excessive release of IL-1ß can be very damaging and can lead to severe disease. In a typical experiment, the typical concentration of IL-1ß around stressed cells is 1100 ± 115 pg/ml. A researcher is testing drugs that can inhibit IL-1ß release. She tests two drugs, drug-1 and drug-2, for their effects on IL-1ß release. If the drugs have a z-score of - 3, she will consider them promising for future study. In her experiment, stressed cells treated with drug-1 yielded 1469 pg/ml of IL-1ß, while drug-2 yielded 777 pg/ml of IL-1ß. Is she going to continue studying either drug, and if so, which one?

A. She will continue studying drug-1 but not drug-2.

B. She will continue studying both drug-1 and drug-2.

C. She will continue studying drug-2 but not drug-1.

D. She will not study drug-2 or drug-1.

Question

Pooky

Mathematics

31 October, 17:05

IL-1ß is a protein released by cells under conditions of infection. Excessive release of IL-1ß can be very damaging and can lead to severe disease. In a typical experiment, the typical concentration of IL-1ß around stressed cells is 1100 ± 115 pg/ml. A researcher is testing drugs that can inhibit IL-1ß release. She tests two drugs, drug-1 and drug-2, for their effects on IL-1ß release. If the drugs have a z-score of - 3, she will consider them promising for future study. In her experiment, stressed cells treated with drug-1 yielded 1469 pg/ml of IL-1ß, while drug-2 yielded 777 pg/ml of IL-1ß. Is she going to continue studying either drug, and if so, which one?

A. She will continue studying drug-1 but not drug-2.

B. She will continue studying both drug-1 and drug-2.

C. She will continue studying drug-2 but not drug-1.

D. She will not study drug-2 or drug-1.

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Answers (1)

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Adrianna Le · Answer 1

D.) would be the best choice, with neither meeting the - 3 z-score requirement. C.) may also be used, if not D.), as it is fairly close to - 3. With a standard deviation of 115 and an average of 1100, a z score of - 3 would be 3*115 = (345), subtracted from the mean, 1100-345=755. Drug-2 has results fairly close to this, at 777 so if either were going to be studied it would be this. A positive 3 would be 1100+345=1445. Drug-1 clearly has a positive z score that would be greater than three.