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241Salters Advanced Chemistry, Pearson Education Ltd 2009.
University of York. This document may have been altered from the
original.
TLTHE THREAD OF LIFE
Concept mapThis concept map shows how the major chemical ideas
in this teaching module develop throughout the course.
Concept First introduced in module
Developed in module(s)
Assumed in module(s)
Relative atomic mass and relative formula mass EL all
Amount of substance EL DF allChemical formulae and inorganic
nomenclature EL ES allBalanced chemical equations EL DF allAtomic
structure EL allCovalent bonding EL DF, ES, PR allShapes of
molecules EL DF allRelationship between properties, and bonding and
structure
EL MR, TL, AI, CD, O DF, ES, PR, SS
Catalysis DF A, TL, SS, AI severalIsomerism DF PR, TL
MDHomologous series DF PR severalNomenclature of organic compounds
DF ES, PR, WM, MR,
TL, CDMD
Structural formulae (full, shortened and skeletal) DF ES, PR all
organic modules
Organic functional groups DF ES, PR, WM, MR, TL, CD
MD
Properties of alkanes DF severalStructure of benzene DF CD WM,
MR, TL, MDIonic equations ES SS severalIonic substances ES O
severalElectronegativity and bond polarity ES PR, O
severalIntermolecular bonds ES PR, MR, TL, O AI, CD, MDRedox ES SS,
AI severalIndustrial applications ES WM, MR, TL, SS,
AIMD
Atom economy ES WM TL, AI, MDRates of reactions A TL, AI Bond
ssion A PR severalElimination reactions PR severalCarboxylic acids
PR WM, MR TL, AI, CD, MDPolymers and polymerisation PR MR, TL
MDChromatography WM TL, CD AI, MDAcids and bases WM O TL, SS,
MDCondensation reactions WM MR, TL MDDelocalisation of electrons WM
CD severalAmines MR TL CD, MDAmides MR TL CD, MDGreen chemistry MR
TL, AI CD, MDAmino acids TL MDProteins and enzymes TL MDDNA and
protein synthesis TL Molecular recognition TL MD
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242Salters Advanced Chemistry, Pearson Education Ltd 2009.
University of York.
This document may have been altered from the original.
TL Advance warning
Advance warningThe following items needed for activities in this
module may not be in your school currently, and might take a little
time to obtain.
Activity Item(s) Essential/optional Typical quantity per
activity
TL2.1 glycineethanoyl chloride
EssentialEssential
2 g2 cm3
TL2.3 aspartame (e.g. Canderel) tabletsaspartic
acidphenylalanine
EssentialEssentialEssential
13 tabletsSmall amounts for chromatographySmall amounts for
chromatography
TL2.4 mirror Essential access to 1
TL3 small sweets e.g. Micro Mix or Jelly Bearsstrawberry sweet
lacescocktail sticks
EssentialEssentialEssential
1016
TL4.1 glucose test strips (such as Clinistix or Diastix)*
Essential 510
TL4.2 potassium peroxodisulfate(VI) (K2S2O8) Essential 0.0400
mol dm3 (10 cm3)
TL4.5 set of graphs and description boxes Essential 1 set per
group
TL5 plastic-coated wire (e.g. Radio Spares 30 0.25 mm strand:
code RS360239 to RS360295 depending on colour)*Minit molecular
model peg type a: Ref 7a293 (white) to 7a300 (green) depending on
colour*21 cm plastic tubes: Ref 72289 (red)*RASMOL or other
electronic molecular structure les
Essential
Essential
EssentialOptional
1 m
30
81
* Current suppliers are listed on the Salters Advanced Chemistry
website.
Storyline: answers to assignments 1 a i Amino acids with
non-polar side chains are: Gly Ala Val Leu Ile Phe Pro (Trp Met) ii
Amino acids with polar side chains are: Ser Thr Cys Asp Glu Asn Gln
Tyr His Lys Arg iii Amino acids with ionisable groups on their
side
chains are: Asp Glu His Lys Arg Tyr b The side chains in Leu and
Ile make them structural
isomers. c i Ser ii Thr iii Tyr iv Either Asp or Glu2 a
CH COOHCO NH CHH2N
HCH3 b i Ser ii Ala3 a Butenedioic acid. b
Z E c No, since butanedioic acid will bind to the active site
in
only one way, so only one E/Z isomer will be formed.
d HOOCCH2CH2CH2COOH It might if the relevant parts can t on to
the active
site. This is unlikely, however, so it probably would not react
and might be an inhibitor. If a product were formed it would be
pentenedioic acid.
4 a Franklin was close. The DNA structure is helical, with
phosphate groups on the outside. She did not postulate a double
helix, though.
b
O
O
O
OMg2
H
H
H H
H
H
H
H
Mg2+ already hydrated so unable to bond to phosphate groups.
c
O
OH
PHO
OH All the OH groups.
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Activities: notes and answers to questions TL
243Salters Advanced Chemistry, Pearson Education Ltd 2009.
University of York. This document may have been altered from the
original.
TL2.1 Investigating an amino acid
Safety note Information about hazardous chemicals is given on
the activity sheet.1 Glycine is soluble in water. It is largely
present as
zwitterions, which are solvated by water resulting in iondipole
interactions.
2 Glycine solution is close to neutral (slightly acid). The pH
should remain fairly constant throughout the additions of acid and
alkali. The theory is covered in the Chemical Ideas section on
amino acids.
3 Only butylamine reacts with ethanoyl chloride. Both dissolve
in water. Butylamine is basic but glycine is neutral. The
zwitterion form of glycine is responsible for its different
behaviour. There is no NH2 group with its lone pair of electrons to
react with the acid chloride. The zwitterion can be either a proton
donor or a proton acceptor in solution, which is why the amino acid
is neutral. Butylamine dissolves in water as a result of hydrogen
bonding, glycine dissolves as a result of iondipole bonds.
4 Only relatively weak intermolecular bonds (dipoledipole bonds,
dipoleinduced dipole bonds and hydrogen bonds) between molecules of
butylamine. Glycine exists as zwitterions with strong ionic bonds
between the particles.
TL2.2 The structures of peptides
This activity provides an opportunity for students to practise
using molecular models and molecular drawing software to reinforce
their ideas about molecular structures.
1 The structural formulae of the three amino acids are:
CH
H
COOH
CH
CH3
COOH
C
COOH
H2N CH
CH3
CH3H
H2N H2N
5 One possible dipeptide is:
Gly Ala
COOH
CH3
C
H
H
CH2N
H
C
H
N
O
The six dipeptides are: Gly Ala, Gly Val, Ala Gly, Ala Val, Ala
Val, Val Ala6 One possible tripeptide is:
C
C
H H O
C CC COOHCNN
H CH
CH3H3C
H2N
H H HCH3Gly Ala Val
The six tripeptides are: Gly Ala Val, Gly Val Ala, Ala Gly Val,
Ala Val Gly, Val Gly Ala, Val Ala Gly
Activities: notes and answers to questions
5 a Both condensation. b
H
H
C
H
H
H
H
HH
CO
C
C
HH
C
O
O
6 a Only two bases are important in coding for: Ser Leu Pro Arg
Thr Val Ala Gly b Three bases are important for: Phe Tyr Cys Trp
His Glu Ile Met Asn Lys Asp Gln7 a i Lys Lys ii Arg Ala Arg Ala iii
Tyr Leu Thr b i ACC ii CUA or CUG8 a GUCA b GTCA9 There are a few
correct answers here. The object of the
assignment is to start the students thinking about these
matters. A few suggested answers are given to the right to open
discussion.
a i The individual him/herself. Their doctors, for diagnosing
genetic diseases but
see below. The police, for solving crimes this is of benet
to
the whole community, although many might disagree.
ii Anyone else not mentioned above this is condential personal
information.
b Should parents have the right to have their children
tested?
Yes could help them to make decisions about how their children
could be treated for genetic diseases.
No children may object to it later on. Should testing be
performed for genetic diseases for
which there is no cure? Yes know areas where care is needed;
cure may be
discovered soon. No just makes an individual anxious and they
may
not suffer as a result of the condition. Should an individual be
given his/her personal genetic
information? Yes its his/her right, the information relates
to
themselves. No it may make the individual anxious (possibly
without cause). Should insurance companies make use of the data
to
x premiums? Yes its only an extension to different premiums
on
grounds of age and gender, for example; those with healthy genes
will benet.
No they should distribute the risk equally for all.
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TL Activities: notes and answers to questions
244Salters Advanced Chemistry, Pearson Education Ltd 2009.
University of York.
This document may have been altered from the original.
TL2.3 Whats in aspartame?
Safety note Information about hazardous chemicals is given on
the activity sheet. Check that the articial sweetener you use
contains aspartame and not saccharin. Also, avoid aspartame tablets
such as Hermesetas, which also contain leucine.Peptide hydrolysis
normally requires reux for several hours with moderately
concentrated acid. However it is possible to get adequate results
after 30 minutes reux with 4 mol dm3 HCl.1
NH
Acidhydrolysis
aspartame
aspartic acid phenylalanine methanol
O
OH2N OCH3
HO
O
H2N
HO
OH2N
OH
O
O
OHCH3OH+
2 It is important that the hydrolysed aspartame is checked
against both amino acids. Aspartame itself, being only a dipeptide,
will travel up the paper, and could be confused with an amino acid.
Also, some tablets contain phenylalanine along with the aspartame.
Both phenylalanine and aspartic acid must be detected to be sure
hydrolysis has occurred.
TL2.4 The shapes of A-amino acidsThe model building in step 3
can be shared round members of a group to save time. If you have
molecular modelling software this can be used to good effect
alongside the model kits, and students can be given print-outs of,
say, D-alanine and L-alanine at the end of the session.1
109109
109 120
N CC
H HO
OH
H
H
2 a
C H All angles 109H CC
H H
H
H
H
b
All angles 109C HO
H
H
c
H All angles 109C S
H
H
C
H
H
C
H
H
d
109
109 120
CC
H HO
OH
5 a
COOH
C(H3C)2HC
HH2N
HOOC
CCH(CH3)2
HNH2
b
7 Molecules that are non-superimposable on their mirror images
are chiral. A carbon atom that is surrounded by four different
groups is called a chiral centre.
TL3 Modelling protein structures
This activity aims to reinforce students ideas about the
structures of proteins in a more unusual way. If students are
allowed to eat the sweets they have used after completing the
activity, it is recommended that this activity does not take place
in a laboratory unless careful measures have been taken to ensure
that the sweets cannot become contaminated with chemicals.1 The
small sweets represent amino acids.2 Peptide links join the amino
acids together (the lace).3 Condensation reactions are involved in
creating peptide
links.4 The sequence of amino acids is the primary structure
of
the protein.5 The helix or pleated sheet is the secondary
structure of
the protein.6 The structure created when sections of the helix
or
pleated sheet is folded upon itself is the tertiary structure of
the protein.
7 The cocktail sticks could represent instantaneous
dipoleinduced dipole, permanent dipolepermanent dipole, hydrogen,
ionic and sulfursulfur covalent bonds.
8 Phenylalanine and leucine are likely to form instantaneous
dipoleinduced dipole bonds.
9 Serine and asparagine are likely to form hydrogen bonds.
H
H
CN
CH2 H2C
OH
O10 Aspartic acid and lysine are likely to form ionic bonds.
H2C
H2C CH2H2C
H3N CH2
O
OC
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Activities: notes and answers to questions TL
245Salters Advanced Chemistry, Pearson Education Ltd 2009.
University of York. This document may have been altered from the
original.
TL4.1 Testing for glucose
Safety note Information about hazardous chemicals is given on
the activity sheet. In a simple experiment, the test strips could
be placed in dilute acid or alkali, or in boiling water, before
being used to prove that these conditions deactivate the enzyme.
More careful investigation might make use of solutions of different
pH or water at different temperatures, and students might try to
time how long a particular depth of colour takes to develop. The pH
of urine is 4.87.5. The enzyme has no effect on other sugars. The
use of test strips makes this activity quicker and more convenient
than some other methods of studying enzyme catalysis. They can,
however, be expensive for large groups. It helps to cut the test
strips into thinner pieces.
TL4.2 Using the iodine clock method to nd the order of a
reaction
Safety note Information about hazardous chemicals is given in
the activity sheet.2 Iodide concentrations vary from 0.5 mol dm3
to
0.1 mol dm3. The corresponding times should be between about 80
s and 400 s.
3 I is always in excess.4 8 s 105 mol I2 can be produced in each
case.5 a 2 s 105 mol S2O32 b 1 s 105 mol I2 c 12.5% of the total
reaction is studied.7 The reaction is rst order with respect to
iodide.8 a Rate = k[I] [S2O82] b Second order
c k = Rate _________ [I][S2O82]
units of k are dm3 mol1 s1
Rate ____ [I] can be found from the gradient of the graph.
d The value of k depends on the temperature.
TL4.3 Methods of following reactions
Safety note Information about hazardous chemicals is given in
the activity sheet.
Two methods are described for following the course of this
reaction; by titration and by using a colorimeter. Teachers may
arrange for groups of students to use different methods, so that
they can compare the methods and the results they obtain. The graph
of concentration of iodine against time is a straight line. This
indicates that the rate of reaction is the same at different iodine
concentrations. The results show that the reaction is zero order
with respect to the iodine and the students should conclude that
iodine is not involved in the rate-determining step.
TL4.4 Enzyme kinetics
1 At high substrate concentration the process is almost zero
order with respect to the substrate.
2 If the rst step in the mechanism were the rate-determining
step, the rate of the reaction would depend on [S]. (It would be
rst order with respect to the substrate.)
3 At saturation, [ES] will be constant, and so the rate of
conversion of bound substrate to unbound product (ESmP + E) remains
constant. It is independent of [S], i.e. the reaction is zero order
with respect to substrate.
4 At lower substrate concentrations, the rate at which the
substrate binds to the enzyme decreases as the [S] falls. Enzyme
active sites will no longer be full. The stage E + SmES eventually
becomes rate-determining and the reaction becomes rst order with
respect to substrate.
5 The reaction is always rst order with respect to enzyme.6 The
enzyme concentration is always low compared with
the substrate concentration. The concentration of ES formed
depends on [E]. So the rate always depends on the enzyme
concentration, no matter which step is rate-determining.
TL4.5 Kinetics graphs
In this activity students choose from a set of eight graphs the
one which best matches what they expect from a number of different
situations related to rates of reaction (see below).
Description Graph Reason
[reactant] v time for a zero-order reaction (with respect to
this reactant)
C rate of change of concentration with time is constant
[reactant] v time for a rst-order reaction (with respect to this
reactant)
E graph illustrates a change of concentration with a constant
half-life
[reaction product] v time F concentration increases from zero
and reaches a constant value as the reaction is completed
rate of reaction v [reactant] for a zero-order reaction (with
respect to this reactant))
A the rate is constant
rate of reaction v [reactant] for a rst-order reaction (with
respect to this reactant)
B straight line, because rate is directly proportional to
[reactant]
rate of reaction v [reactant] for a second-order reaction (with
respect to this reactant)
D an exponentially increasing line
rate of reaction v [reactant]2 for a second-order reaction (with
respect to this reactant)
B a straight line, because rate is directly proportional to
[reactant]2
rate of reaction v [substrate] for an enzyme-catalysed
decomposition reaction
F the reaction is rst order at low substrate concentrations and
changes to zero order at higher substrate concentrations
number of collisions with Kinetic Energy E v Kinetic Energy (E)
at temperature T K
G same prole as graph H, but is more narrow and has a higher
peak
number of collisions with Kinetic Energy E v Kinetic Energy (E)
at temperature (T + 10) K
H same prole as graph G, but is wider and has a lower peak
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TL Activities: notes and answers to questions
246Salters Advanced Chemistry, Pearson Education Ltd 2009.
University of York.
This document may have been altered from the original.
TL5 Modelling DNA
Students gain a much better understanding of the structure of
DNA if they draw and build models of the bases and the double
helix. By doing this activity they should also realise that
hydrogen bonding and instantaneous dipoleinduced dipole bonding are
important for holding the structure together.
O
OH
OH
O
O
O
O
O
P
OH
OH
sugarphosphate backbone
O
OH
OH
O
O
O
O
O
POH
N
NO
HO
CH3
sugarphosphate backbone with base thymine
H
N
NN
NN
O
N
CH3
O
N
HH
base pair thymine and adenine
N
NN
NN
N
O
O
NH
N
CH3
H
H
H
H
base pair cytosine and guanine
1 The centre of the double helix is full; there is no empty
space. This contrasts with the traditional ladder representation of
DNA, which gives the mistaken impression that there are large gaps
between the bases.
2 The helix could not be twisted more tightly because the bases
are already at their closest.
3 Instantaneous dipoleinduced dipole bonding.4 Less twisting
would take the bases further apart and the
structure would lose the instantaneous dipoleinduced dipole
bonding. This bonding may be weak between any pair of bases, but it
is signicant over the whole polymer.
TL7 Check your knowledge and understanding
This activity ensures that students are aware of the learning
outcomes (specication statements) that their assessment will be
based on, and provides an opportunity for them to reect on how well
they understand the ideas that they have covered in this module.
Crucially, it enables teachers to identify areas where individual
students are less condent, and to provide appropriate additional
support to improve their understanding. This activity could be used
as part of the preparation for an end of module test.