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Chapter 9: Nutrition and the Digestive System
Mini Investigation: Digestion Can Be a Mouthful
(Page 395)
A. The evidence of starch was that the cracker turned black when the Lugol’s was added.
B. It didn’t taste sweet so I can infer that the cracker didn’t contain sugar.
C. As I chewed the cracker, it became sweeter. This indicated that there was a chemical change and
sugar was then present.
D. The saliva in my mouth might be responsible for the chemical change.
E. The purpose of the stain was to indicate the presence or absence of starch.
F. Amylase is an enzyme that breaks down starch into sugars. It is produced in the salivary glands in
your mouth where digestion begins. It is also produced in the pancreas and released into the small
intestine where the digestion of starch continues.
Section 9.1: Why We Need to Eat
Mini Investigation: How Much Energy Do You Need?
(Page 398)
1. Answers will vary.
Sample Answers:
Given:
Student A is female, 16 years old, 165 cm tall, and has a mass of 50 kg
Required: BMR (Student A)
Analysis: BMR (females)
= [655 + (9.6 × mass (kg)) + (1.8 × height (cm))
– (4.7 ×age (years))] × 4.18
Solution: BMR (Student A) = [655 + (9.6 × 50) + (1.8 × 165) – (4.7 × 16)] × 4.18
= [655 + (480) + (297) - (75.2)] × 4.18
= [1356.8] × 4.18
= 5671.424 kJ, rounded to 5700 kJ
Statement: Student A’s BMR is 5700 kJ.
Given:
Required:
Analysis:
Student B is male, 16 years old, 165 cm tall, and has a mass of 50 kg
BMR (Student B)
BMR (males) = [66 + (13.7 × mass (kg)) + (5.0 × height (cm))
– (6.8 × age (years))] × 4.18
Solution: BMR (Student B) = [655 + (9.6 × 50) + (1.8 × 165) – (4.7 × 16)] × 4.18
= [66 + 822 + 875 – 108.8] × 4.18
= [1654.2] × 4.18
= 6 914.556 kJ, rounded to 6900
Statement: Student B’s BMR is 6 900 kJ.
A. Answers will vary.
Sample answers: Using the examples from above, Student A exercises vigorously or plays sports
6–7 days/week. Student B exercises lightly or plays sports 1–3 days/ week.
Student A
Given: Student A’s BMR is 5700 kJ. She exercises vigorously or plays sports 6–7 days/week.
Required: Average daily energy requirement
Analysis: Average daily energy requirement = BMR × 1.725
Solution: Average daily energy requirement = 5700 kJ × 1.725
= 9832.5 kJ, rounded to 9800 kJ
Copyright © 2011 Nelson Education Ltd.
Chapter 9: Nutrition and the Digestive System
9.1-1
Statement: Student A’s average daily energy requirement is 9800 kJ.
Student B
Given: Student B’s BMR is 6900 kJ. He exercises lightly or plays sports 1–3 days/ week.
Required: Average daily energy requirement
Analysis: Average daily energy requirement = BMR × 1.375
Solution: Average daily energy requirement = 6 900 kJ × 1.375
= 9497.5 kJ, rounded to 9500 kJ
Statement: Student B’s average daily energy requirement was 9500 kJ.
Section 9.1 Questions
(Page 399)
1. Two factors that determine how much and how often an animal eats are its size and whether it is
endothermic (warm-blooded) or ectothermic (cold-blooded).
2. Physical activity requires more energy because muscles require energy to produce movement. The
larger the body size, the more energy required to perform basic life functions. Males are larger than
females, on average, and have a greater muscle mass so their metabolic rates tend to be higher and
they therefore require more energy.
3. Answers will vary. Sample answer for a 60 kg person.
Activity
Time (h)
Energy requirement Total energy used
(kJ/kg/h)
(kJ)
Sleeping
7.5
4.1
1845
Cycling (15.3 km/h)
1.5
25.8
2322
Sitting
6.0
5.2
1872
Walking (4.8 km/h)
1.0
16.2
972
Using a computer
3.0
9.0
1620
keyboard
Writing
1.0
6.0
360
Running (8.8 km/h)
0.5
37.5
1125
Cooking
1.0
10.5
630
Standing
1.5
6.3
567
Playing piano
1.0
11.2
672
Total
24.0
------11 985
4. (a) Playing a computer game requires 9.0 kJ/kg/h. A 70 kg person would require
9.0 kJ/kg/h × 70 kg or 630 kJ/h.
Time of Activity = Total amount of energy available ÷ Amount of energy used per hour
= 2500 kJ ÷ 630 kJ/h
= 3.968 h
To use 2500 kJ of energy, Paul would have to play a computer game for
3.968 h or 3 hours 58 minutes (approximately 4 hours).
(b) From Table 2 (p. 399), running at 12.9 km/h requires 62.0 kJ/kg/h.
A 70 kg person would require 62.0 kJ/kg/h × 70 kg or 4 340 kJ/h.
Time of Activity = Total amount of energy available ÷ Amount of energy used per hour
= 2 500 kJ ÷ 4 340 kJ/h
= 0.576 h
To use 2 500 kJ, Paul would have to run at 12.9 km/h for 0.576 h, or 34.6 minutes (approximately
35 minutes).
(c) From Table 1 (p. 399), 2alking at 6.4 km/h requires 20.6 kJ/kg/h.
A 70 kg person would require 20.6 kJ/kg/h × 70 kg, or 1442 kJ/h.
In 2 h, Paul would therefore require 1442 kJ/h × 2 h, or 2884 kJ.
If Paul walked at 6.4 km/h for 2 h, he would use 2884 kJ of energy. This amount of activity would
use up all the energy from his meal and an additional 384 kJ.
Copyright © 2011 Nelson Education Ltd.
Chapter 9: Nutrition and the Digestive System
9.1-2
5. Since the workday activities of an average office worker (e.g., sitting, standing, using a
computer), do not require as much energy as those of an average letter carrier (walking), the letter
carrier would require more energy.
6. (a) The daily energy requirement for an active 5 year old girl is 6270 kJ.
(b) The daily energy requirement for an inactive 65 year old man is 8987 kJ.
(c) The daily energy requirement for a very active 16 year old male is 12 540 kJ.
(d) The daily energy requirement for a very active 16 year old female is9823 kJ.
Copyright © 2011 Nelson Education Ltd.
Chapter 9: Nutrition and the Digestive System
9.1-3