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Cambridge International A Level Biology Revision Guide

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<strong>Cambridge</strong> <strong>International</strong> AS <strong>Level</strong> <strong>Biology</strong><br />

Answers to end-of-chapter questions<br />

6 dissolve easily in water;<br />

sweet;<br />

general formula (CH 2 O) n / contain the<br />

elements carbon, hydrogen and oxygen /<br />

hydrogen and oxygen are present in ratio of<br />

2 : 1;[3]<br />

7 a lactose could be a source of energy;<br />

it could be digested to, monosaccharides /<br />

glucose and galactose, which could then be<br />

used as building blocks for larger molecules;<br />

[2]<br />

b condensation;[1]<br />

c glycosidic bond;[1]<br />

d<br />

OH on C atom 1 above<br />

ring (β-galactose)<br />

6 CH 2 OH<br />

5<br />

O<br />

OH<br />

OH<br />

H<br />

4 1<br />

DH H<br />

H<br />

H<br />

3 2<br />

H OH<br />

galactose<br />

(drawn according to convention)<br />

H<br />

3<br />

H<br />

2<br />

H<br />

OH H<br />

4 1<br />

OH<br />

H<br />

5<br />

O<br />

OH<br />

6 CH 2 OH<br />

6 CH 2 OH<br />

5<br />

O<br />

H<br />

H<br />

H<br />

4 1<br />

OH H<br />

OH<br />

2 OH<br />

3<br />

H OH<br />

α-glucose<br />

OH<br />

galactose<br />

(molecule rotated 180° in order<br />

to line up with α-glucose)<br />

OH on C atom 1<br />

below ring<br />

glucose correctly drawn;<br />

galactose correctly drawn;[2]<br />

Carbon atoms need not be numbered. Note<br />

that galactose will probably be drawn<br />

‘upside down’ as in the disaccharide – the<br />

conventional way of drawing it is also shown<br />

in the diagram above. The form used to make<br />

the disaccharide is the beta form of galactose,<br />

but students will not need to know this, other<br />

than for interest.<br />

e alpha glucose / α-glucose;<br />

the –OH group on carbon atom 1 is below the<br />

ring;[2]<br />

f carry out a Benedict’s test on both solutions;<br />

lactose would give a brick-red / brown<br />

precipitate, sucrose would not;<br />

accept positive result for lactose, negative<br />

result for sucrose[2]<br />

[Total: 10]<br />

8 a i<br />

H<br />

H<br />

N<br />

CH 3 O H<br />

C<br />

H<br />

vi alternative answers<br />

A<br />

H<br />

H<br />

O<br />

v peptide bonds<br />

A<br />

H<br />

OH<br />

CH 2<br />

C N C C N C C<br />

C of COOH joined to N of NH 2 for both peptide<br />

bonds;<br />

peptide bonds shown as C=O joined to –NH<br />

(i.e. water has been eliminated);<br />

all three amino acids joined and in correct<br />

sequence; accept even if errors in bonding[3]<br />

ii primary structure;[1]<br />

iii water;[1]<br />

iv ring drawn around –OH or whole R group<br />

(–CH 2 OH) of serine;[1]<br />

v rings drawn around two peptide bonds<br />

and bonds labelled appropriately;[1]<br />

vi ring drawn around –NH group one side of a<br />

peptide bond and group labelled A;[1]<br />

b held in place by hydrogen bonding;<br />

secondary structures;<br />

all the –NH and –C=O groups of, peptide<br />

bonds / polypeptide backbone, are involved;<br />

[3]<br />

c molecule made from repeating subunits;<br />

subunits similar or identical to each other;<br />

giant molecule / macromolecule; [max. 2]<br />

d i XXX, XXY, XYY, XYX, YYY, YYX, YXX, YXY;[1]<br />

ii 2 3; [1]<br />

[Total: 15]<br />

H<br />

iv<br />

O<br />

OH<br />

<strong>Cambridge</strong> <strong>International</strong> AS and A <strong>Level</strong> <strong>Biology</strong> © <strong>Cambridge</strong> University Press 2014

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