Chemicals of LifeData Based Questions

Elemental Composition of Living Organisms - page 44

1. In the bar chart for Canis (dog) the three most occurring elements are N (nitrogen), P (phosphorus) and Ca (calcium). In Chara (pondweed) the most occurring elements are N (nitrogen), K (potassium) and Ca (calcium). Overall, the three most occurring ele-ments in both are K (potassium), Ca (calcium) and N (nitrogen).

2. A) In Canis the the fourth and fifth most frequently occurring elements are K (potas-sium) and Na (sodium).

B) In Chara the fourth and fifth most frequently occurring elements are Na (sodium) and Mg (magnesium).

3. A) Chara = 21 mg g-1 , Canis = 83 mg g-1. 83-21 = 62 mg g-1. There is a difference of

62 mg g-1 of nitrogen between Canis and Chara.B) There may be a difference in the nitrogen content because of the type of organism

they are. Canis is a dog while Chara is a pondweed. Dogs require more nitrogen for the metabolic processes that they carry out such as the digestion of food and dogs also have an abundance of nitrogen in their bodies which is why their urine tends to ruin lawns. While the pondweed does require nitrogen for processes such as photosynthesis or chemosynthesis, far less is required to carry out the processes which causes the dif-ferences in nitrogen levels.

4. From the bar graphs of the elemental composition of Canis and Chara one is able to tell that they both have an abundance of both nitrogen and calcium although Canis has a much higher amount of both. One can also see that both have approximately the same amount of sodium standing at around 8 mg g-1 which applies to sulphur as well. Both organisms require phosphorus but Chara requires a significant amount more which is the same for magnesium. Lastly, there is a difference of one element in the composi-tion of these organisms. Chara has a small amount of manganese while Canis has some chlorine.

Emperor Penguins - page 50

1. A) wild before = (11.8+7.7+0.4+17.3= 37.2), wild after = (6.9+0.4+14.4+2.2= 23.9) 37.2-23.9 = 13.3. There was a mass loss of 13.3 kilograms in the wild birds.

B) captive before = (0.4+18.2+12.0+8.0= 38.6), captive after = (14.3+0.5+6.8+0.8= 22.4) 38.6-22.4 = 16.2 . There was a mass loss of 16.2 kilograms in the captive birds.

2. In the captive birds before there were 12.0 kg of lipids while in the wild there was 11.8 kg. This is a difference of 0.2 kg higher for the captive birds. After the 14 weeks, the captive birds had a lipid content of 2.2 kg, 9.6 kg less than before and 1.4 kg more than the captive birds after who dropped 11.2 kg after the 14 weeks. The captive birds relied more on their stored lipids than those who were in the wild causing them to lose a significant amount more.

3. The lipids also provide insulation for the penguins to keep them warm in the harsh winters. This would also provide an explanation for why the lipid numbers were different between the captive and wild birds, less lipids are used for insulation when they are huddled in packs.

Working with data: using the Ti-83 graphing calculator to perform the t-test - page 56

1. Standard deviation is the measure of the variation and indicates how closely the set of data is to the mean. It is used to show how closely the data is related to the mean and if any variations occurred by chance or if they are different for a reason.

2. In this, the differences in standard deviation are that the non blow-dried hair has a higher standard deviation than the blow-dried hair. Through the t-test one is able to see that the differences in variation from the mean is not by chance in the experiment but is instead a significant difference that was caused by a variable.