Biological Molecules Questions Flashcards
Describe the chemical reactions involved in the conversion of polymers to monomers and monomers to polymers.
Give two named examples
A condensation reaction joins monomers together and forms a
(chemical) bond and releases water;
A hydrolysis reaction breaks a (chemical) bond between
monomers and uses water;
A suitable example of polymers and the monomers from which
they are made;
A second suitable example of polymers and the monomers
from which they are made;
What is a monomer?
(a monomer is a smaller / repeating) unit / molecule from which larger
molecules / polymers are made;
Describe a biochemical test to show that raffinose solution contains a non reducing sugar
Heat with acid and nuetralise
Heat with benedicts solution
Red precipitate colour
Suggest method to measure quanitity of a reducing sugar
Filter and dry precipitate
Find mass/weight
Give one similarity and one difference between structyures of lactulose (galactose and one molecule of fructose)
And lactose
Both contain galactose / a glycocidic bond
Lactulose contains fructose whereas lactose contains glucose
Describe two differences between the structure of a cellulose molecule and a glycogen molecule
Cellulose is made up of β-glucose (monomers) and glycogen is
made up of α-glucose (monomers);
Cellulose molecule has straight chain and glycogen is
branched;
Cellulose molecule has straight chain and glycogen is coiled;
glycogen has 1,4- and 1,6- glycosidic bonds and cellulose has
only 1,4- glycosidic bonds
Describe and explain two features of starch that make it a good storage
molecule.
Insoluble (in water), so doesn’t affect water potential;
Branched / coiled / (α-)helix, so makes molecule compact;
OR
Branched / coiled / (α-)helix so can fit many (molecules) in
small area;
Polymer of (α-)glucose so provides glucose for respiration;
Branched / more ends for fast breakdown / enzyme action;
Large (molecule), so can’t cross the cell membrane
Describe structure of glycogen
Polysaccharide of α-glucose;
OR
polymer of α-glucose;
2. (Joined by) glycosidic bonds
OR
Branched structure;
Describe how glycogen acts as a source of energy
Hydrolysed (to glucose);
2. Glucose used in respiration;
Explain difference in sturcture of starch molecule and celullose molecule shown in diagram
Starch formed from α-glucose but cellulose formed from β-glucose;
Position of hydrogen and hydroxyl groups on carbon atom 1
inverted.
Explain one way in which starch molecules are adapted for their function in plant cells.
Insoluble
Does not affect water potential
Helical;
4. Compact;
OR
5. Large molecule;
6. Cannot leave cell.
Explain how cellulose molecules are adapted for their function in plant
cells.
Long and straight chains;
2. Become linked together by many hydrogen bonds to form
fibrils;
3. Provide strength (to cell wall).
Describe how you would text for the presence of a lipid in a liquid sample of food.
Shake with) ethanol / alcohol;
1. Accept named alcohol
2. Then add (to) water;
2. Order must be correct
3. White / milky / cloudy (layer indicates oil).
Describe how a triglyceride molecule is formed.
One glycerol and three fatty acids;
2. Condensation (reactions) and removal of three molecules of water;
3. Ester bond(s) (formed)
Describe how an ester bond is formed in a phospholipid molecule.
Condensation (reaction)
2. OR
Loss of water;
Between of glycerol and fatty acid;
Describe the induced-fit model of enzyme action and how an enzyme acts
as a catalyst.
Substrate binds to the active site/enzyme
OR
2. Enzyme-substrate complex forms;
Accept for ‘binds’, fits
Active site changes shape (slightly) so it is complementary to
substrate
OR
Active site changes shape (slightly) so
distorting/breaking/forming bonds in the substrate;
3. Reduces activation energy;
A competitive inhibitor decreases the rate of an enzyme-controlled
reaction.
Inhibitor similar shape to substrate;
Reject same shape
Accept ‘complementary to active site’
2. Fits/binds to active site;
3. Prevents/reduces enzyme-substrate complex forming
Describe how the structure of a protein depends on the amino acids it
contains.
Structure is determined by (relative) position of amino acid/R group/interactions;
Primary structure is sequence/order of amino acids;
Secondary structure formed by hydrogen bonding (between amino
acids);
Tertiary structure formed by interactions (between R groups);
Creates active site in enzymes
Explain how the active site of an enzyme causes a high rate of reaction.
Lowers activation energy;
2. 3. Induced fit causes active site (of enzyme) to change shape;
(So) enzyme-substrate complex causes bonds to form/break
Describe two other ways in which all dipeptides are similar and one way in
which they might differ.
mine/NH2 (group at end);
Accept amino/NH3
+
2. Carboxyl/COOH (group at end);
Accept carboxylic / COO−
3. Two R groups;
4. All contain C and H and N and O;
Accept examples of different R groups
Difference
5. Variable/different R group(s);
3
(c) 1. Moved to negative (electrode) because positive(ly charged);
2. 3. (Spots move) different distances/rates because (amino acids)
different charge/mass;
Accept size for ma
Describe how a non-competitive inhibitor can reduce the rate of an
enzyme-controlled reaction.
Attaches to the enzyme at a site other than the active site;
Accept ‘attaches to allosteric/inhibitor site’
Changes (shape of) the active site
OR
Changes tertiary structure (of enzyme);
(So active site and substrate) no longer complementary so
less/no substrate can fit/bind;
Describe how a peptide bond is formed between two amino acids to form a
dipeptide.
Condensation (reaction) / loss of water;
Accept each marking point if shown clearly in
diagram.
2. Between amine / NH2 and carboxyl / COOH;
The secondary structure of a polypeptide is produced by bonds between
amino acids.
Describe how
Hydrogen bonds;
Accept as a diagram
Reject N - - - C / ionic / disulfide bridge / peptide
bond
2. Between NH (group of one amino acid) and C=O (group);
OR
Forming β pleated sheets / α helix;
Two proteins have the same number and type of amino acids but different
tertiary structures.
Explain why
Different sequence of amino acids
OR
Different primary structure;
If candidate assumes proteins are the same, accept
effect of different pH/ temperature
2. Forms ionic / hydrogen / disulfide bonds in different places
Formation of an enzyme-substrate complex increases the rate of reaction.
Explain how.
Reduces activation energy;
Accept ‘reduces Ea’.
2. Due to bending bonds
OR
Without enzyme, very few substrates have sufficient energy for
reaction;
Describe the structure of DNA.
Polymer of nucleotides;
Accept ‘Polynucleotide’
Accept for ‘phosphate’. phosphoric acid
2. Each nucleotide formed from deoxyribose, a phosphate (group) and
an organic/nitrogenous base;
3. Phosphodiester bonds (between nucleotides);
4. Double helix/2 strands held by hydrogen bonds;
5. (Hydrogen bonds/pairing) between adenine, thymine and
cytosine, guanine;
Describe how a phosphodiester bond is formed between two nucleotides
within a DNA molecule.
Condensation (reaction)/loss of water;
2. (Between) phosphate and deoxyribose;
3. (Catalysed by) DNA polymerase;
Describe how the separation of strands occurs.
DNA helicase;
2. Breaks hydrogen bonds between base pairs/ AT and
GC/complementary bases
OR
Breaks hydrogen bonds between polynucleotide strands;
Describe the role of DNA polymerase in the semi-conservative replication of DNA
Joins (adjacent DNA) nucleotides;
Reject suggestions that it forms hydrogen bonds or
joins complementary bases.
Reject ‘nucleotide bases’.
2. (Catalyses) condensation (reactions);
3. (Catalyses formation of) phosphodiester bonds (between adjacent
nucleotides);
Use your knowledge of semi-conservative replication of DNA to suggest:
1. the role of the single-stranded DNA fragments
2. The role of the DNA nucleotides.
Role of single-stranded DNA fragments
1. Template;
2. Determines order of nucleotides/bases;
Role of DNA nucleotides
3. Forms complementary pairs / A – T, G - C
OR
Forms complementary (DNA) strand;
Give two features of DNA and explain how each one is important in the
semi-conservative replication of DNA.
W eak / easily broken hydrogen bonds between bases allow two
strands to separate / unzip;
Two strands, so both can act as templates;
may appear in the same feature
Complementary base pairing allows accurate replication;
Describe the role of two named enzymes in the process of semi-
conservative replication of DNA.
DNA) helicase causes breaking of hydrogen/H bonds (between DNA
strands);
Reject ‘helicase hydrolyses hydrogen bonds’.
DNA polymerase joins the (DNA) nucleotides;
Forming phosphodiester bonds
Describe the function of each of these enzymes.
DNA helicase_
DNA polymerase ___
DNA helicase – (unwinding DNA and)
breaking hydrogen bonds / bonds between
chains / bases / strands;
2. DNA polymerase – joins (adjacent)
nucleotides OR forms phosphodiester bond /
sugar-phosphate backbone;
Adenosine triphosphate (ATP) is a nucleotide derivative.
Contrast the structures of ATP and a nucleotide found in DNA to give two
differences.
ATP has ribose and DNA nucleotide has
deoxyribose;
2. ATP has 3 phosphate (groups) and DNA
nucleotide has 1 phosphate (group);
3. ATP – base always adenine and in DNA
nucleotide base can be different / varies;
