2012 a Level Answers P1 and P2 Compiled Final

2012 A Level H2 Biology Paper 2 Suggested Answers

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1

(a) (i) Identify the stage of mitosis shown in Fig. 1.1. [1]

Metaphase; 1m

(b) (ii) Describe what happens in the next stage of mitosis. [3]

1. Anaphase;

2. Centromeres divide,

3. Sister chromatids separate and are pulled apart to opposite poles;

4. By spindle fibres attached to kinetochore at the centromere;

5. Resulting in equal distribution of chromosomes at each pole of cell; @1m

6. Poles of the cell move farther apart as the polar/non-kinetochore mitotubules/

spindle fibres slide past each other, hence elongating the cell;

(c) Outline the role of centrioles in mitosis. [2]

1. Centrioles act as microtubule organizing centres

2. They organise the synthesis of spindle fibres;

3. That pull the sister chromatids apart to opposite poles of the cell; @1m

4. Centrioles exist as a pair at each pole which is important in determining polarity

of the cell;

(d) Describe and explain what happens to the nuclear envelope during mitosis. [4]

1. Nuclear envelope disintegrates during prophase;

2. Releasing the condensed chromosomes into cytoplasm;

3. For the attachment of spindle fibres to the kinetochores at metaphase;

4. And the separation of sister chromatids at anaphase;

5. Nuclear envelope reformed around chromosomes at each pole during telophase;

6. For compartmentalize chromosomes/chromatin to allow for proper division of

cytoplasm during cytokinesis;

@1m

(e) Suggest the specific cell structures targeted by the two fluorescent dyes used in this

preparation. Their binding is shown on Fig. 1.1 in regions A and B. [2]

Structures in region A: Chromosomes;

Structures in region B: Microtubules/spindle fibres (and asters/centrioles);

[Total: 12]

2

(a) Explain why chloride ions can only cross cell surface membranes via such proteins [3]

1. (chloride) ions are charged/ hydrophilic R! polar;

2. cannot pass through hydrophobic core in phospholipid bilayer;

3. transmembrane CFTR protein provides a hydrophilic channel/pore;


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(b) Describe how the membrane holds onto these proteins. [2]

1. Hydrophobic interactions occur between hydrophobic R groups of residues of

the CFTR protein in contact with fatty acids/hydrophobic core of phospholipid

bilayer;

2. Hydrogen bonds and ionic bonds form between charged/polar R groups of

residues of the CFTR protein in contact with the phosphate heads of

phospholipids;

(c)

Suggest the role of ATP in opening the channel. [2]

1. ATP allows the phosphorylation of the CFTR channel / ATP binds to ATP-binding

site;

2. 3D conformational change in CFTR opens the channel;

(d) State the type of gene mutation that causes the commonest mutation to the CFTR gene. [1]

1. Deletion (of 3 nucleotides);

(e)

Explain how such structural changes in a protein may affect its functioning. [3]

1. Loss of phenylalanine results in a change in primary structure;

2. which leads to a change in the tertiary structure/3D conformation;

3. CFTR cannot be phosphorylated by ATP / cannot be opened;

(f) Explain why only one copy of the normal allele of the CFTR gene is required to prevent

cystic fibrosis in individuals with the most common mutation present in the other allele.

[2]

1. The normal (dominant) allele is sufficient for the cell to express sufficient

functional CFTR chloride ion channels;

2. Allowing adequate efflux of Cl- to the exterior of cell followed by osmosis of

water to prevent formation of thick mucus.

3. Dominant allele is able to mask the effect of the recessive allele;

[Total: 13]

3

(a) State the name of the structures shown in Fig. 3.1. [4]

A: Capsid

B: tail sheath

C: tail fibers

D: base pin @1m

(b) (ii) Describe the role of viral DNA in the infection process of the T4 bacteriophage. [3]

1. Codes for structural component of virus like capsid, tail sheath, tail fibres

and base pin and enzymes, lysozyme, nuclease;

2. Allows virus to bind specifically to bacteria using tail fibres

3. Phage DNA encoded nuclease degrades host cell DNA;


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4. Phage DNA encoded lysozyme breaks down peptidoglycan cell wall

5. Synthesis of phage encoded proteins using host cell RNA polymerase and

ribosome; (Reject metabolic machinery)

6. As a template for Replication of virus DNA using host cell DNA polymerase;

(c)

Suggest the advantages of the lysogenic pathway to the lambda phage virus. [2]

1. Bacteria do not die allowing phage DNA to be replicated along with bacteria

genome during binary fission;

2. Hence results in large numbers of bacteria with phage DNA;

3. Allows virus to be assembled and released when external conditions are right;

[Total: 9]

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(a) Suggest how transcription factors such as TBP bind to DNA. [2]

1. Positively charged R groups on (binding site of) TBP is able to bind to negatively-

charged DNA (with ref. to electrostatic attraction);

2. conformation of binding site on proteins is complementary in shape to that of the

TATA box;

R! ref. to complementary base pairing (which is applicable only to nucleic acids)

Please take note that the complementary base pairing is only between the bases of

the double helix.

(b) Suggest why the unwinding of the double helix of DNA promotes transcription. [2]

1. Template strand of DNA used in transcription must be single-stranded/not bound to another strand;

2. Allows RNA polymerase/ transcription factors to bind at the promoter;

3. Allows formation of hydrogen bonds by complementary base pairing with the

incoming ribonucleotide incorporated into the RNA strand being synthesised;

(c) Explain why such processing of pre-mRNA molecules is necessary. [3]

1. Intron is a non-coding region;

2. It does not code for amino acids in the polypeptide;

3. if it is not removed, the ribosome will read the intron and incorporate wrong amino acids into the polypeptide/ cause premature termination of translation / cause a frameshift mutation / removal of the intron allows the ribosome to read the exons properly and synthesise the correct amino acid sequence in the polypeptide;

[Total: 7]

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(a) Explain the term epistasis in this context. [3]


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1. Epistasis is a form of gene interaction in which a allele at one gene locus

alters the phenotypic expression of a gene at a second locus

2. a dominant allele at both loci must be present to produce purple pigment

3. this purple pigment is a product of a two step biochemical pathway which

requires the functional enzymes encoded by these two dominant alleles

(b) Use the symbols A, a and B, b to draw a genetic diagram to explain the results shown

in the F2 generation.

Parental genotype: AAbb x aaBB

Gametes: Ab aB

F1 genotype AaBb

F1 phenotype Purple

F2 x F2 AaBb x AaBb

Gametes:

AB Ab aB ab

AB

AABB Purple

AABb Purple

AaBB Purple

AaBb Purple

Ab

AABb Purple

AAbb White

AaBb Purple

Aabb White

aB

AaBB Purple

AaBb Purple

aaBB White

aaBb White

ab

AaBb Purple

Aabb White

aaBb White

Aabb White

F2 phenotypic ratio: 9 Purple : 7 White

9 : 7

(c) Explain how the chi-squared calculated value of 1.60 supports the statement that epistasis

is the correct explanation of these results. [3]

1. Calculated x2 value of 1.60 < critical x2 value of 3.841 df=1

2. P > 0.05

3. Difference between obs and exp value is not significant / expected ratio 9:7 is

valid and the difference between the obs and exp value is due to chance@1m

(d) Suggest why pea plants are good experimental organisms for carrying out such crosses. [2]

4. Peas are available easily and in many varieties with identifiable phenotype ;

5. Peas have both male and female organs, can carry out self pollination / cross


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pollination

6. Short life cycle/ generation time

@1m

[Total: 12]

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(a) Explain why there is a smaller yield of ATP per molecule of glucose from anaerobic

respiration than aerobic respiration. [4]

1. incomplete oxidation of glucose due to lack of oxygen / no oxygen to act as

final electron acceptor and there is no regeneration of NAD

2. only glycolysis occurs (in cytoplasm) to produce a net yield of 2 ATP (by

substrate-level phosphorylation)

3. Lack of O2 results in link reaction, Krebs cycle and oxidative phosphorylation

not taking place to produce more ATP

4. pyruvate is instead converted into ethanol (in yeast & plants) and lactate (in

animals), which still contains a lot of energy ;

(b) Using your knowledge of ATP yields, calculate an estimate of how many times faster

glycolysis must be under anaerobic conditions when compared to aerobic conditions in

order to produce the same number of ATP molecules in the same time.

Aerobic respiration = 38 ATP

Anaerobic respiration = 2 ATP

38/2 = 19 times faster

(c) Describe and explain the effect on PFK activity of increasing ATP concentration. [3]

1. PFK is inhibited by high ATP concentration at low substrate concentration;

2. ATP acts as an allosteric inhibitor by binding at the allosteric site of PFK/ at a site

other than the active site,

3. changes 3D conformation of active site of PFK, substrate (F6P) unable to bind to

active site of PFK and glycolysis is slowed down

4. At low ATP, PFK acitivity is higher than that at high ATP

5. PFK acts to regulate the rate of glycolysis /ref. to eg. of negative feedback

inhibition where high ATP levels inhibits rate of glycolysis and hence inhibits

further formation of ATP @1m

(d) Explain the meaning of the term facilitated diffusion. [2]

1. Diffusion of substance with the assistance of a transport protein/ channel

protein;

2. Substance transported is polar or charged;


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3. Movement of substance is passive process down the concentration gradient and

does not involve ATP;

4. The channel only allow specific molecules to diffuse through;

@1m

[Total: 12]

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(a) Explain the ways in which islands favour the formation of new species. [6]

1. ref. to allopatric speciation/ geographical isolation;

2. phenotypic variation exists in original population;

3. environment on different islands exert different selection pressures;

4. different reproductive success takes place in each of the island populations/

ref. to certain alleles passed to next generation at a higher frequency;

5. over time, allele frequencies change in each population / forming distinct

gene pools;

6. gene pool also changes due to genetic drift;

7. ref. to reproductive isolation/no gene flow between the island populations;

(b) Suggest advantages of using mitochondrial DNA. [2]

1. Mitochondrial DNA is small / completely sequenced;

2. Ref. to higher chance of preserving/extracting mitochondrial DNA as there

are multiple copies per cell;

3. Ref. to highly-conserved regions which can be used to compare distantly-

related organisms /

Ref. to high rate of mutation/high level of variability in sequence, allowing

(closely-related) organisms to be distinguished;

4. Ref. to because it is inherited maternally, it is easy to trace the inheritance;

5. It is inherited in a haploid manner and due to the lack of

recombination/crossing over in mtDNA it is sensitive to changes brought

about due to genetic drift in the sub population level, hence can compare

closely related sp or within between subpopulation of sp.

(c) Describe the advantages of using nucleotide data such as this in classifying organisms. [3]

1. Objective (quantitative) analysis of data;

2. Allows greater accuracy in measurement of the degree of relatedness between

groups of organisms as it can be compared using statistical software;

3. Ability to cross-reference phylogenetic tree derived with other phylogenetic

trees obtained based on other molecules;

4. Able to use both living and dead specimen / only a small part of specimen

required (as DNA can be easily amplified using PCR);


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5. Removes the necessity for fossil evidence which is sometimes rare or

impossible to obtain

(d) Outline two of the functions of these protein complexes. [2]

1. Election carriers pass electrons through a series of redox reactions of

decreasing energy levels during oxidative phosphorylation;

2. Proton pumps which pump protons into the intermembrane space to form a

proton gradient/pool / for chemiosmotic synthesis of ATP;

3. ATP synthase synthesise ATP using the proton motive force/flow of protons;

[Total: 13]

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(a) Outline the molecular structure of phospholipids in relation to their function in cell

membranes. [7]

1. Phospholipid = glycerol + 2 fatty acids + phosphate group (joined by ester linkages);

2. Amphipathic phosphate head is hydrophilic and hydrocarbon tail is hydrophobic;

3. Phospholipid arranges themselves in aqueous envt / to form bilayer; 4. From hydrophobic barrier to regulate movement of ions/polar and large

molecules in and out of the cell; 5. To maintain a different cellular environment from the external. 6. Enable proteins to be embedded via hydrophobic interactions between fatty

acid tail and proteins (which in turn serves functions of transport, recognition and reception) ;

7. Kink structure of phospholipid hydrocarbon tail coupled with weak hydrophobic interactions maintains fluidity of cell membrane;

8. Fluidity of membrane allows cell uptake/secretion via endocytosis and exocytosis

(b) Explain the role of Golgi body and its link to the rough endoplasmic reticulum [7]

1) Golgi body is a stack of flattened cisternae;

Role of Golgi body

2) Enzymes present in the cisterna of golgi body;

3) That carries out post translational modification of proteins (eg hydroxylation,

glycosylation);

4) Sort protein into their final destination / secretion out of cell;

5) By packaging them into secretory vesicles, lysosome;

Link to RER

6) RER and Golgi body plays a role in protein synthesis;

7) Golgi body receives proteins synthesized by ribosome on RER in transport

vesicles;

8) Transport vesicles fuses with golgi body at the cis face and move towards the


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trans face

9) Vesicles from RER coalesces to form new cis Golgi cisternae;

(c) Describe the structure of ribosomes. [6]

1) Ribosome consist of 2 subunits, the large subunits and the small subunit;

2) Made up of ribosomal proteins and ribosomal ribonucleic acid;

3) Assembly of 2 subunits to form ribosome only occur during translation

4) Large subunit consist of ribozyme form by rRNA/peptidyl transferase which

catalyses the formation of peptide bonds between amino acids;

5) Large subunit consist of E site, P site and A site;

6) Small subunit consist mRNA binding site;

7) Ref. to 2 types of ribosomes, 70s found in prokaryotes, mitochondria &

chloroplasts and 80s found in cytoplasm of eukaryotic cells

9 (a) Outline the light dependent reactions of photosynthesis. [7]

1 Light energy is converted to chemical energy in the form of ATP and reduced NADP during the light-dependent reactions;

2 Light (of particular wavelengths ) is absorbed by chlorophylls (and accessory pigments)/photosystems on the thylakoid membrane, and excites the electron at the reaction centre;

3 Excited electron is emitted and transported through the electron transport chain; 4 In cyclic photophosphorylation, the electron is emitted from PS I and is accepted

by PS I; 5 In non-cyclic photophosphorylation, the electron emitted from PS II is accepted

by PS I while the electron emitted by PS I is accepted by NADP+ and reduced NADP is formed;

6 Energy released from electrons passing through the ETC is used to pump protons into the thylakoid space;

7 Hence forming a proton gradient / proton pool / proton motive force; 8 As protons move from the thylakoid space to the stroma through ATP synthase,

ATP synthase uses this energy to synthesise ATP by chemiosmosis (from ADP and Pi)

9 Water undergoes photolysis (at PS II), resulting in electrons which fill the electron gap in (reaction centre of) PS II, H+ and O2 which is released by a by-product;

(b) Explain the role of membranes in the chloroplast. [7]

1 Many thylakoids are stacked to form grana (and grana are connected by intergrana);

2 This increases the surface area for the attachment of chlorophyll/photosynthetic

pigments/photosystems, electron carriers and ATP synthase;

3 Thylakoid membrane compartmentalises the chloroplast, separating the stroma

from the thylakoid space;

4 Thylakoid membrane is impermeable to protons, hence allowing formation of the

proton pool/gradient/proton motive force ;

5 Ref. to these are important for the light-dependent stage of photosynthesis /

photophosphorylation

6 Chlorophyll/photosynthetic pigments/photosystems absorb light energy and

become photoactivated/emit an excited electron;


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7 (PS II embedded in) thylakoid membrane is also where photolysis of water takes

place;

8 The electron transport chain uses energy from the movement of electrons to pump

protons into thylakoid space;

9 Protons can only move from the thylakoid space to the stroma via ATP synthase

10 ATP synthase uses proton motive force to form ATP by chemiosmosis;

11 (not as important as the earlier points) Double membrane of chloroplast controls

the substances that enter and leave the chloroplast/ named e.g. of substance such

as GALP, enzymes involved in photosynthesis;

(c) Describe the effect of increasing light intensity on the rate of photosynthesis. [6]

1 At low light intensity, the rate of photosynthesis increases linearly with increasing light intensity ;

2 as more photosystems are photoactivated/ more electrons at the photosystems are excited;

3 leading to a higher rate of photophosphorylation; 4 Which results in greater formation of ATP and reduced NADP 5 giving rise to a higher rate of Calvin cycle and formation of GALP/sugars; 6 As light intensity increases, the light compensation point is reached where

CO2 is neither evolved nor absorbed / CO2 taken in by photosynthesis is equal to CO2 given out by respiration;

7 When light intensity reaches the light saturation point, rate of photosynthesis plateaus/reaches a maximum;

8 light intensity is no longer a limiting factor / some other factor eg. [CO2], temperature is a limiting factor;

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2012 a Level Answers P1 and P2 Compiled Final