AQA AS Bio Specimen Paper 2014

Question 1

Describe how you could use cell fractionation to isolate chloroplasts from leaf tissue. [3 marks]

Answer 1

1. How to break open cells and remove debris;
2. Solution is cold/isotonic/buffered;
3. Second pellet is chloroplast;

Question 2

Name two structures in a eukaryotic cell that cannot be identified using an optical microscope. [1 mark]

Two should be named for one mark.

Answer 2

• Mitochondrion
• Ribosome
• Endoplasmic Reticulum
• Lysosome
• Cell-surface membrane

Question 3

A technician investigated the effect of temperature on the rate of an enzyme-controlled reaction. At each temperature, he started the reaction using the same volume of substrate solution and the same volume of enzyme solution.

Give one other factor the technician would have controlled. [1 mark]

Answer 3

Concentration of substrate solution / of enzyme

solution / pH;

Question 4

Describe and explain the differences between the two curves. [5 marks]

Answer 4

1. Initial rate of reaction faster at 37 °C;
2. Because more kinetic energy;
3. So more E–S collisions/more E–S complexes
   formed;
4. Graph reaches plateau at 37 °C;
5. Because all substrate used up;

Question 5

Describe and explain the appearance of one of the chromosomes in cell X. [3 marks]

Answer 5

1. Chromosome is formed of two chromatids;
2. (Because) DNA replication (has occurred);
3. (Sister) chromatids held together by
   centromere;

Question 6

Describe what has happened during division 1 in Figure 3. [2 marks]

Answer 6

1. Chromosomes in homologous pair;
2. One of each into daughter cells / haploid
   number;

Question 7

Identify one event that occurred during division 2 but not during division 1. [1 mark]

Answer 7

Separation of (sister) chromatids / division of
centromere;

Question 8

Name two ways in which meiosis produces genetic variation. [2 marks]

Answer 8

1. Independent segregation (of homologous
   chromosomes) ;
2. Crossing over / formation of chiasmata;

Question 9

The arrows in Figure 5 show the directions in which each new DNA strand is being
produced.
Use Figure 4, Figure 5 and your knowledge of enzyme action to explain why the
arrows point in opposite directions. [4 marks]

Answer 9

1. (Figure 4 shows) DNA has antiparallel
   strands/described;
2. (Figure 4 shows) shape of the nucleotides is
   different/nucleotides aligned differently;
3. Enzymes have active sites with specific shape;
4. Only substrates with complementary
   shape/only the phosphate end (of the
   developing strand) can bind with active site

Question 10

Cholesterol increases the stability of plasma membranes. Cholesterol does this by
making membranes less flexible.

Suggest one advantage of the different percentage of cholesterol in red blood cells compared with cells lining the ileum. [1 mark]

Answer 10

Red blood cells free in blood/not supported by

other cells so cholesterol helps to maintain shape;

Question 11

E. coli has no cholesterol in its cell-surface membrane. Despite this, the cell maintains a constant shape. Explain why. [2 marks]

Answer 11

1. Cell unable to change shape;
2. (Because) cell has a cell wall;
3. (Wall is) rigid/made of peptidoglycan/murein;

Question 12

When HIV infects a human cell, the following events occur.
• A single-stranded length of HIV DNA is made.
• The human cell then makes a complementary strand to the HIV DNA.
The complementary strand is made in the same way as a new complementary
strand is made during semi-conservative replication of human DNA.
Describe how the complementary strand of HIV DNA is made. [3 marks]

Answer 12

1. (Complementary) nucleotides/bases pair
   OR A to T and C to G;
2. DNA polymerase;
3. Nucleotides join together (to form new
   strand)/phosphodiester bonds form;

Question 13

Contrast the structures of DNA and mRNA molecules to give three differences.
[3 marks]

Answer 13

1. DNA double stranded/double helix and
mRNA single-stranded;
2. DNA (very) long and RNA short;
3. Thymine/T in DNA and uracil/U in RNA;
4. Deoxyribose in DNA and ribose in RNA;
5. DNA has base pairing and mRNA doesn’t/
DNA has hydrogen bonding and mRNA
doesn’t;
6. DNA has introns/non-coding sequences
and mRNA doesn’t;

Question 14

This fat substitute cannot be digested in the gut by lipase.
Suggest why.
[2 marks]

Answer 14

1. (Fat substitute) is a different/wrong
shape/not complementary;
OR
Bond between glycerol/fatty acid
and propylene glycol different (to
that between glycerol and fatty
acid)/no ester bond;
2. Unable to fit/bind to (active site of)
lipase/no ES complex formed

Question 15

This fat substitute is a lipid. Despite being a lipid, it cannot cross the cell-surface
membranes of cells lining the gut.
Suggest why it cannot cross cell-surface membranes.
[1 mark]

Answer 15

It is hydrophilic/is polar/is too large/is

too big;