Solutions to the Practice Questions Set

Refer to the Resources website for background on the various techniques used in these problems. I saw no need to repeat them all here.

1. 1 unit volume of "stuff" + 1 unit volume of solvent = 8 total volumes. The dilution factor is therefore 1:8 or an 8-fold dilution.

2. Use VC=VC approach.

What volume of stock (V1)*100mg/ml = 30 ml * 25 mg/ml?
V1 = 30 ml * 25 mg/ml) = 7.5 ml stock; bring to 30 ml with 22.5 ml water
..............100 mg/ml

 

3. step serial dilution, 1:10 dilution at each step, 7 ml total volume per step;
unit volume = 7/10 ml = 0.7 ml

Step 1: ........0.7 ml unknown + 6.3 ml water = 1:10 dilution
Step 2: ........0.7 ml 1:10 product + 6.3 ml water = 1:100 dilution
Step 3: ........0.7 ml 1:100 product + 6.3 ml water = 1:1000 dilution

4. A molarity problem: a 1 molar solution = 1 molecular weight of stuff per liter

0.18 moles/L * 58.44 g/mole * 0.5 L = 5.25 g/ 500 ml

5. Percent solutions are mass/volume relationships. A 1% soluition is equivalent to 1 g stuff/100 ml of solvent. In the case of seawater, the stuff is a mix of salt and minerals ("salts") whose combined concentration we call salinity. If we evaporate (or distill) 100 ml of seawater that is 3.5% "salts" we will get approximately 3.5 g of residues (the salts).

6. Chi-square problem; 4 phenotypes so df = 4-1 = 3, critical value at df=3 and probability of 0.05 = 7.8. Since the test stat (4.67) < critical value (7.8), we do not reject the NULL.

A plausible results entence: "The distribution of F2 phenotypes indicated that two genes were involved and that they appear to assort independently." Note that no stat summary needs to be included since there was not a significant difference.

7. Read the error term as SD not SEM (we haven't used SEM so you wouldn't know how to deal with it - mia culpa)

a) If we adhere strictly to our stranded-on-a-desert island rubric for deciding significance, i.e., is there overlap in the range of the SD's, we must say that the difference isn't significant because the SD's overlap by about 0.3.

standard diet SD range is 55.6 - 62.8
high prot diet SD range is 62.5 - 68.9

However, because we have a large sample size (n =50 for each group), and because the overlap is very small, we, as thinking and rational biologists, are more inclined to believe that the difference in the means is really significant. If only we had a computer.....

b) Oh joy! Look what washed up on shore! A Dell PC! Still runs great though! Now we can run that unpaired t-test!