Chapter 1 Review Section Answers
1 Making connections � The key words listed above can be called concepts. Concepts can be related to each other by the
use of linking words or phrases to form propositions (see the example at left). An arrow shows the sense of the relationship.
When several concepts are related in a meaningful way, a concept map is formed. Because concepts can be related in
many ways, there is no single correct concept map. Figure 1.35 shows one concept map containing some of the key words
from this chapter. Add to this map or prepare another by using key words from the list above and other words of your choice.
Checklist
� Check the meaning of each of the key words of this chapter.
Comments
There are many correct answers for this question. The students’ word choice will govern their outcome. One concept map
has been shown on page 29. Another map given below contains some of the key words of the chapter, and presents
another equally correct map.
2 Applying your understanding
You wish to carry out some research on molecular structure. What advantage might it be to use a synchrotron (see page 3),
compared with the light and electron microscopes currently in use?
Checklist
� Read the Biotech box on page 3 in Nature of Biology Book 2, Third Edition.
Comments
The advantage of the synchrotron is that the intensity of the complex radiation allows detection of the structure of molecules.
A light microscope transmits an image and can only magnify material up to 1000 if oil immersion is used. This means some
organelles in a cell cannot be seen (e.g. mitochondria) and certainly the molecular structure of a molecule is not visible. An
electron microscope uses a beam of electrons instead of light and can resolve an image with more detail than a light
microscope. Hence fine structure of organelles within a cell can be determined.
3 Communicating your understanding _ Answer the following questions.
a Explain why water is solid by the time it reaches 0°C.
b What is the chemical formula of glucose?
c What is the basic difference between a monosaccharide molecule and a
disaccharide molecule?
d How many glycerols are there in a triglyceride fat molecule?
e Amino acids have the same basic structure and yet they are all different. Explain
which part of their structure confers this difference.
f If an amino acid is involved in the formation of a sulfur bond, what can you
conclude about the molecular formula of that amino acid?
g Explain what is meant by the tertiary structure of a protein.
h How many different nitrogenous bases are in DNA?
i How many chains of nucleotides are in a DNA molecule?
j How many chains of nucleotides are in a molecule of messenger RNA?
k In what form is glucose stored in the human body?
Checklist
� Review chapter 1.
Comments
a As the temperature drops there is not sufficient movement to break the hydrogen bonds so a lattice structure forms.
b C6H12O6
c A monosaccharide consists of one carbon ring (a sugar) and a disaccharide consists of two sugar carbon rings and has
the formula C12H22O11.
d One
e The side chain or R group
f The molecular formula of that amino acid will contain sulfur
g The irregular folding of polypeptides held together by ionic or hydrogen bonds
h Four
i Two
j One
k Glycogen
4 Applying your understanding � What is a characteristic of fats that makes them energy dense?
Checklist
_ Read chapter 1 (pp. 213).
Comments
Fats store twice as much energy as the same weight of polysaccharide. The lack of water in the fat molecules means that
they carry more energy per molecule.
5
Applying understanding and drawing conclusions � During an experiment, you analyse a compound that has a
chemical composition of carbon, hydrogen and oxygen in the ratio of 1:2:1 and a six-sided molecular shape. What
predictions would you make with regard to classification of the compound?
Checklist
_ Read chapter 1 (pp. 910).
Comments
This compound is likely to be a monosaccharide (C6H12O6) and since it is a hexagonal molecule it is most likely glucose.
6 Applying your knowledge and understanding �
a When two monomers such as amino acids join together, a molecule of water is produced. Use two amino acid molecules
to explain where this water comes from.
b How many water molecules would be required to completely hydrolyse a carbohydrate polymer that contained 100
monomers?
Checklist
_ Read chapter 1 (pp. 1011 and pp. 1415).
Comments
a Refer to figure 1.17.
b 50, one water per pair of monsaccharides
7
Applying understanding and drawing conclusions � Before the introduction of genetically engineered insulin for use
by people with diabetes, the protein hormone was extracted from beef or pig pancreas. Explain how you would expect the
sequence of amino acids in the beef and pig insulin to compare with that of humans.
Checklist
_ Read chapter 1 (p. 20).
Comments
You would expect the amino acid sequence from beef or pig insulin to be similar to that of humans. The active and inactive
form of the molecule relies on specific disulfide bonds between the amino acids.
8 Analysing data and drawing conclusions
�
A particular small polypeptide contains nine amino acids. The polypeptide has been fragmented in various experiments by
breaking particular peptide bonds. The fragments obtained were:
ser – cys – his – pro – arg – cys
pro – arg – cys
X – gly – met – cys
his – pro – arg – cys
X – gly – met – cys – ser – cys
X is known to be the first amino acid in the polypeptide. What is the primary structure of the polypeptide?
Checklist
_ Read chapter 1 (p. 15).
Comments
X – gly – met – cys – ser – cys his – pro – arg – cys
9 Analysing data and drawing conclusions
� Assume
that the nitrogen base sequence in one of the strands in a DNA molecule is:
A-T-C-G-A-C-A-T-G-G-A-A-T-A-C-C-T-C.
a What is the base sequence in the complementary strand?
b How many amino acids does this piece of DNA code for?
Checklist
_ Read chapter 1 (pp. 23–5).
Comments
a T-A-G-C-T-G-T-A-C-C-T-T-A-T-G-G-A-G
b6
10 Analysing data and drawing conclusions
�A
piece of RNA has the base sequence:
U-A-C-G-A-U-U-C-G-A-A-C-A-U-G
a Are you able to determine the kind of RNA this molecule is without any further information?
b If you were told this piece of RNA carries the code for five amino acids, would you change your answer to part a? If so,
what would your explanation be for change?
Checklist
_ Read chapter 1 (p. 26).
Comments
a No
b You now know it is messenger RNA because it is the only RNA which is
translated into a polypeptide.
11
Applying your knowledge � From the table below match each statement on the left with the most appropriate term on
the right.
Statements
Term
Some organic molecules store and process
hydrophilic
1
information at the molecular level.
Some substances can dissolve other substances.
proteins
2
Some molecules dissolve readily in water.
nucleic acids
3
Our fingernails and hair are made of a particular kind of molecule.
carbohydrates
4
5
One group of organic molecules contains polysaccharides.
solvent
Checklist
� Review chapter 1.
Comments
1. Nucleic acids
2. Solvent
3. Hydrophilic
4. Proteins
5. Carbohydrates
12
Using the web _ Go to www.jaconline.com.au/natureofbiology/natbiol2-3e and click on the ‘Protein’ weblink for this
chapter.
a Click on ‘Download’ to obtain a pdf file of the protein illustrations and print a copy.
b Draw two columns, one at each side of the illustrations. Give one column the heading ‘Example’ and the other ‘Function’.
c Carry out web research or use the textbook to enter relevant information in the
columns for each protein structure.
Checklist
� Read chapter 1 (pp. 1422).
Comments
The figure shows the four levels of organisation of proteins.
After the web search students may come up with many different examples.

primary protein structure e.g. insulin, lysozyme (enzyme)

secondary protein structure e.g. alpha helix, alpha keratin;

beta sheets e.g. spider silk or fibroin

tertiary protein structure e.g. hexokinase (an enzyme involved in glycolysis)

quaternary protein structure e.g. haemoglobin (carry oxygen)
BIOCHALLENGE (p. 28)
1.
a amino acids
b secondary
c The polypeptide chains of the silk are extended to their maximum.
2.
a and b
Fat consists of a long chain of fatty acids attached to a glycerol. A phospholipid has two fatty acid chains attached
to a glycerol and also has a phosphate group.
The steroid lipid molecules shown do not consist of a chain of fatty acids, they are composed of four interconnected rings of
carbon and differ from each other in the functional groups attached to these rings.
3.
a A = a polymer, monomer is a nucleotide
b nucleic acid, since single-stranded RNA
c plants and animals in the nucleus and at the ribosome
a B = a polymer, monomer is a dipeptide
b composed of glycine and alanine
c plants and animals as part of a polypeptide or protein molecule
a C = a polymer, monomer is a phospholipid
b bilipid layer made up of phospholipids
c plant and animal cell membranes
a D = a monomer,
b glucose, a carbon ring
c plant and animal cells in the mitochondria, blood and extracellular fluid in animals, and in the cells after
photosynthesis and in the phloem in plants
a E = a polymer, monomer is an amino acid
b Haemoglobin composed of two globin molecules (alpha and beta) which in turn are composed of amino acids.
The molecule has been folded and shows quaternary structure (combination of two different polypeptide chains).
c animal cells, in particular red blood cells
a F = a polymer, monomer is glucose
b starch
c plants as a storage molecule, in plastids