Relationships Protein Comparison Revision...Chimpanzee T R P Bonobo T R P Gorilla T K P Orang utan K R Q (ii)€€€€ Hybrid DNA was made from the gene for chimpanzee haemoglobin

Relationships Protein Comparison Revision

May 2103

114 minutes

101 marks

Page 1 of 35

Q1.          This question should be written in continuous prose, where appropriate.

Quality of Written Communication will be assessed in these answers.

(a)     Use your knowledge of classification to arrange class, phylum, genus and family in order of decreasing number of species.

          largest number of                                                               smallest number of species                                                                                                 species

......................             ......................           ......................           ...................... (1)

(b)     The diagram shows an amoeba. This is a single-celled organism.

Amoeba is classified as a protoctist. Giving a different answer in each case, explain why it is not

(i)      a prokaryote;

............................................................................................................

............................................................................................................

(ii)     a fungus.

............................................................................................................

............................................................................................................. (2)

(c)     Cytochrome c is a protein involved in one of the reactions of aerobic respiration in a mitochondrion. The molecular structure of cytochrome c from different species has been analysed. More similarities are present in the structure of cytochrome c in closely related species than in distantly related species.

(i)      Explain what is meant when two species are described as being closely related.

............................................................................................................

............................................................................................................ (1)

Page 2 of 35

(ii)     A difference in the molecular structure of cytochrome c may arise in a small population that becomes geographically isolated. Explain how the difference may arise and how it may spread in the population.

............................................................................................................

............................................................................................................

............................................................................................................

............................................................................................................

............................................................................................................

............................................................................................................

............................................................................................................

............................................................................................................ (4)

(Total 8 marks)

Q2.          (a)     A fish uses its gills to absorb oxygen from water. Explain how the gills of a fish are adapted for efficient gas exchange.

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

...................................................................................................................... (6)

Page 3 of 35

Mackerel live in the surface waters of the sea. Toadfish live on the seabed in deep water.

(b)     The concentration of oxygen is higher in the surface waters than it is in water close to the seabed. Suggest why.

......................................................................................................................

......................................................................................................................

......................................................................................................................

...................................................................................................................... (2)

(c)     The graph shows oxygen dissociation curves for toadfish haemoglobin and for mackerel haemoglobin.

Explain how the shape of the curve for toadfish haemoglobin is related to where the toadfish is normally found.

......................................................................................................................

......................................................................................................................

......................................................................................................................

...................................................................................................................... (2)

Page 4 of 35

(d)     Scientists analysed the sequence of amino acids in one polypeptide chain in the haemoglobin of four different species of ape. The only difference they found affected the amino acids at three positions in the polypeptide chain. Their results are shown in the table. The letters are abbreviations for particular amino acids.

(i)      What information do the data in the table suggest about the relationships between the chimpanzee, the bonobo and the gorilla? Explain your answer.

.............................................................................................................

.............................................................................................................

.............................................................................................................

............................................................................................................. (2)

Species Position 87 Position 104 Position 125

Chimpanzee T R P

Bonobo T R P

Gorilla T K P

Orang utan K R Q

(ii)     Hybrid DNA was made from the gene for chimpanzee haemoglobin and the genes for the haemoglobin of the other three species of ape. Which of the three samples of hybrid DNA would separate into two strands at the lowest temperature?

Explain your answer.

.............................................................................................................

.............................................................................................................

.............................................................................................................

.............................................................................................................

.............................................................................................................

............................................................................................................. (3)

(Total 15 marks)

Page 5 of 35

Q3.          Read the following passage.

Soon a single drop of blood might be enough to reveal, at a very early stage, if a patient has cancer. It could also tell us what type of cancer it is and whether it is treatable. Fragments of DNA from body cells are present in blood plasma. Some of these fragments may be from cancer cells. The fragments can be detected by a new test in which a test strip containing

5     nucleic acid binds to sections of altered DNA.

Other cancer-detecting techniques involve removing a tissue sample from a patient. The tissue sample is used to obtain mRNA. By examining the mRNA, scientists can discover whether cancer is present.

Use information from the passage and your own knowledge to answer the questions.

(a)     Describe how altered DNA may lead to cancer.

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

...................................................................................................................... (6)

(b)     Explain why fragments of DNA from cancer cells may be present in blood plasma (lines 3-4).

......................................................................................................................

......................................................................................................................

......................................................................................................................

...................................................................................................................... (2)

Page 6 of 35

(c)     Explain why the nucleic acid on the test strip will only bind to altered DNA (lines 4-5).

......................................................................................................................

......................................................................................................................

......................................................................................................................

...................................................................................................................... (2)

(d)     This test strip will allow cancers to be detected at a very early stage. Explain why cancer is more likely to be treated successfully if the disease is detected at a very early stage.

......................................................................................................................

......................................................................................................................

......................................................................................................................

...................................................................................................................... (2)

(e)     Explain how examining mRNA (line 7) enables scientists to discover whether cancer is present.

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

...................................................................................................................... (3)

(Total 15 marks)

Q4.          (a)     Class, family, genus and kingdom are terms used in classifying organisms. Write the terms in the correct sequence.

Largest number                                                                          Smallest number of species                                                                                   of species

          .......................          .......................             .......................        ....................... (1)

Page 7 of 35

(b)     Cytochrome c is a protein. The table shows the sequence of the last six amino acids in cytochrome c in humans and three other animals.

•        The three other animals are a monkey, a fish and a horse. •        One of the three is in the same order as humans. •        Two are in the same class.

(i)      Complete the table to show the animal from which each sample of cytochrome c was taken.

(1)

Animal Sequence of amino acids in

cytochrome c

Human lys–ile–phe–ile–met–lys

  lys–th–rphe–va–lglu–lys

  lys–ile–phe–ile–met–lys

  lys–ile–phe–val–glu–lys

(ii)     Explain your answer.

.............................................................................................................

............................................................................................................. (1)

(c)     DNA hybridisation shows similarities between DNA samples. Explain why

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

...................................................................................................................... (3)

(Total 6 marks)

Page 8 of 35

Q5.          The diagram shows a seahorse. A seahorse is a fish. Mating in seahorses begins with courtship behaviour. After this, the female transfers her unfertilised eggs to the male’s pouch. Most male fish fertilise eggs that have been released into the sea. However, a male seahorse fertilises the eggs while they are inside his pouch. The fertilised eggs stay in the pouch where they develop into young seahorses.

(a)     Give two ways in which courtship behaviour increases the probability of successful mating.

1 ...................................................................................................................

......................................................................................................................

2 ...................................................................................................................

...................................................................................................................... (2)

(b)     Give one way in which reproduction in seahorses increases the probability of

(i)      fertilisation

.............................................................................................................

............................................................................................................. (1)

(ii)     survival of young seahorses.

.............................................................................................................

............................................................................................................. (1)

          Scientists investigated the effect of total body length on the selection of a mate in one Australian species of seahorse. The scientists used head length as a measure of total body length.

(c)     (i)      Use the diagram to suggest why the scientists measured head length rather than total body length.

.............................................................................................................

............................................................................................................. (1)

Page 9 of 35

(ii)     Suggest why the scientists were able to use head length as a measure of total body length.

.............................................................................................................

............................................................................................................. (1)

The scientists measured the head lengths of the female and male of a number of pairs. The results are shown in the graph.

(d)     The scientists concluded that total body length affects the selection of a mate. Explain how the results support this conclusion.

......................................................................................................................

...................................................................................................................... (1)

(e)     A female with a head length of 50 mm selected a mate. Explain how you could use the graph to predict the total head length of the mate selected.

......................................................................................................................

......................................................................................................................

......................................................................................................................

...................................................................................................................... (2)

Page 10 of 35

(f)      Scientists studied two species of North American seahorse. They thought that these two species are closely related. Describe how comparisons of biological molecules in these two species could be used to find out if they are closely related.

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

...................................................................................................................... (6)

(Total 15 marks)

Page 11 of 35

Q6.          Cytochrome c is a protein found in all eukaryotes. In humans it consists of 102 amino acids. Biologists have compared the amino acid sequence in some other species with that in humans. The table shows amino acids 9 to 13 in the amino acid sequences of cytochrome c from four species.

(a)     What do the results suggest about the relationship between humans and the other three species?

......................................................................................................................

......................................................................................................................

......................................................................................................................

...................................................................................................................... (2)

  Amino acid in this position in cytochrome c

Species 9 10 11 12 13

Human Ile Phe Ile Met Lys

Chicken Ile Phe Val Gln Lys

Dogfish Val Phe Val Gln Lys

Chimpanzee Ile Phe Ile Met Lys

(b)     Suggest one advantage of using cytochrome c to determine relationships between species.

......................................................................................................................

...................................................................................................................... (1)

(c)     Comparing the base sequence of a gene provides more information than comparing the amino acid sequence for which the gene codes. Explain why.

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

...................................................................................................................... (2)

(Total 5 marks)

Page 12 of 35

Q7.          Hummingbirds belong to the order Apodiformes. One genus in this order is Topaza.

(a)     (i)      Name one other taxonomic group to which all members of the Apodiformes belong.

............................................................................................................... (1)

(ii)     Name the taxonomic group between order and genus.

............................................................................................................... (1)

The crimson topaz and the fiery topaz are hummingbirds.

Biologists investigated whether the crimson topaz and the fiery topaz are different species of hummingbird, or different forms of the same species.

They caught large numbers of each type of hummingbird. For each bird they

•        recorded its sex

•        recorded its mass

•        recorded the colour of its throat feathers

•        took a sample of a blood protein.

The table shows some of their results.  

(b)     (i)      Explain how the standard deviation helps in the interpretation of these data.

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................

............................................................................................................... (2)

  Crimson topaz Fiery topaz

  Male Female Male Female

Mean mass / g (± standard deviation)

13.6 (±1.9) 10.8 (±1.3) 14.2 (±1.6) 11.6 (±0.63)

Colour of throat feathers Green Grey edges Yellowish green No grey edges

Page 13 of 35

(ii)     In hummingbirds throat colour is important in courtship. Explain the evidence in the table that shows that the crimson topaz and the fiery topaz may be different species of hummingbird.

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................

............................................................................................................... (2)

(c)     The biologists analysed the amino acid sequences of the blood protein samples from these hummingbirds.

Explain how these sequences could provide evidence as to whether the crimson topaz and the fiery topaz are different species.

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................ (2)

(Total 8 marks)

Q8.          Cranes are large birds. One of the earliest methods of classifying cranes was based on the calls they make during the breeding season.

(a)     Explain why biologists could use calls to investigate relationships between different species of crane.

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

...................................................................................................................... (2)

Page 14 of 35

(b)     More recently, biologists have used DNA hybridisation to confirm the relationships between different species of crane. They made samples of hybrid DNA from the same and from different species. They measured the percentage of hybridisation of each sample. The results are shown in the table.

Species of crane from which hybrid DNA was made Percentage DNA

hybridisation

Grus americana                and            Grus monachus 97.4

Grus monachus                and            Grus rubicunda 95.7

Grus americana                and            Grus rubicunda 95.5

Grus rubicunda                 and            Grus rubicunda 99.9

Grus americana                and            Grus americana 99.9

Grus monachus                and            Grus monachus 99.8

(i)      Which two species seem to be the most closely related? Explain your answer.

.............................................................................................................

.............................................................................................................

.............................................................................................................

.............................................................................................................

............................................................................................................. (2)

(ii)     The biologists measured the temperatures at which the samples of hybrid DNA separated into single strands. Explain why these temperatures could be used to find the percentage of DNA hybridisation.

.............................................................................................................

.............................................................................................................

.............................................................................................................

.............................................................................................................

............................................................................................................. (2)

Page 15 of 35

(c)     Biologists can also use protein structure to investigate the relationship between different species of crane. Explain why.

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

...................................................................................................................... (2)

(Total 8 marks)

Q9.          (a)     Scientists can use protein structure to investigate the evolutionary relationships between different species. Explain why.

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................ (2)

(b)     Comparing the base sequence of genes provides more evolutionary information than comparing the structure of proteins. Explain why.

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................ (2)

Page 16 of 35

(c)     The proteins of different species can be compared using immunological techniques. The protein albumin obtained from a human was injected into a rabbit. The rabbit produced antibodies against the human albumin. These antibodies were extracted from the rabbit and then added to samples of albumin obtained from four different animal species. The amount of precipitate produced in each sample was then measured. The results are shown in the table.

What do the results suggest about the evolutionary relationship between humans and the other species?

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................ (2)

(Total 6 marks)

Species from which albumin was obtained

Amount of precipitate / arbitrary units

Rat 23

Chimpanzee 96

Marmoset 65

Trout 11

Q10.          Phenylketonuria is a disease caused by mutations of the gene coding for the enzyme PAH. The table shows part of the DNA base sequence coding for PAH. It also shows a mutation of this sequence which leads to the production of non-functioning PAH.

(a)     (i)      What is the maximum number of amino acids for which this base sequence could code?  

(1)

DNA base sequence coding for PAH

C A G T T C G C T A C G

DNA base sequence coding for non-functioning PAH

C A G T T C C C T A C G

Page 17 of 35

(ii)     Explain how this mutation leads to the formation of non-functioning PAH.

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................

(Extra space) ........................................................................................

...............................................................................................................

............................................................................................................... (3)

PAH catalyses a reaction at the start of two enzyme-controlled pathways. The diagram shows these pathways.

(b)     Use the information in the diagram to give two symptoms you might expect to be visible in a person who produces non-functioning PAH.

1 .....................................................................................................................

2 ..................................................................................................................... (2)

(c)     One mutation causing phenylketonuria was originally only found in one population in central Asia. It is now found in many different populations across Asia. Suggest how the spread of this mutation may have occurred.

........................................................................................................................

........................................................................................................................

........................................................................................................................ (1)

(Total 7 marks)

Page 18 of 35

Q11.          Cranes are large birds. One of the earliest methods of classifying cranes was based on the calls they make during the breeding season.

(a)     Explain why biologists could use calls to investigate relationships between different species of crane.

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

...................................................................................................................... (2)

(b)     More recently, biologists have used DNA hybridisation to confirm the relationships between different species of crane. They made samples of hybrid DNA from the same and from different species. They measured the percentage of hybridisation of each sample. The results are shown in the table.

Species of crane from which hybrid DNA was made Percentage DNA

hybridisation

Grus americana                and            Grus monachus 97.4

Grus monachus                and            Grus rubicunda 95.7

Grus americana                and            Grus rubicunda 95.5

Grus rubicunda                 and            Grus rubicunda 99.9

Grus americana                and            Grus americana 99.9

Grus monachus                and            Grus monachus 99.8

(i)      Which two species seem to be the most closely related? Explain your answer.

.............................................................................................................

.............................................................................................................

.............................................................................................................

.............................................................................................................

............................................................................................................. (2)

Page 19 of 35

(ii)     The biologists measured the temperatures at which the samples of hybrid DNA separated into single strands. Explain why these temperatures could be used to find the percentage of DNA hybridisation.

.............................................................................................................

.............................................................................................................

.............................................................................................................

.............................................................................................................

............................................................................................................. (2)

(c)     Biologists can also use protein structure to investigate the relationship between different species of crane. Explain why.

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

...................................................................................................................... (2)

(Total 8 marks)

Page 20 of 35

M1.          (a)     phylum, class, family, genus; 1

(b)     (i)      presence of a nucleus / membrane bound organelles / named organelles only 80S ribosomes / lacks a cell wall;

1

(ii)     lacks a cell wall / no chitin / is motile / has one nucleus / no hyphae;

(do not credit it has a nucleus) (credit only one answer relating to a lack of cell wall; if more than one answer is given in (i) and / or (ii), incorrect answers negate)

1

(c)     (i)      more recent common ancestor / DNA in common; 1

(ii)     mutation; there is variation; genes (coding) for protein / cytochrome c with different structures; EITHER individuals with a modified cytochrome c have a selective advantage / are selected for; these individuals are more likely to survive to have offspring / have more offspring;

(must link a comparison of survival to reproduction)

gene / allele frequency changes over generations / time; OR changed structure does not affect protein function; these structural differences accumulate over time;

4 max [8]

M2.          (a)     1       Large surface area provided by lamellae/filaments;

Q Candidates are required to refer to lamellae or filaments. Do not penalise for confusion between two

2       Increases diffusion/makes diffusion efficient;

3       Thin epithelium/distance between water and blood;

4       Water and blood flow in opposite directions/countercurrent;

5       (Point 4) maintains concentration gradient (along gill)/equilibrium not reached;

5 Not enough to say gives steep concentration gradient

6       As water always next to blood with lower concentration of oxygen;

7       Circulation replaces blood saturated with oxygen;

8       Ventilation replaces water (as oxygen removed); 6-8 Accept answers relating to carbon dioxide

6 max

Page 21 of 35

(b)     Mixing of air and water (at surface);

Air has higher concentration of oxygen than water;

Diffusion into water;

Plants/seaweeds near surface/in light;

Produce oxygen by photosynthesis; 2 max

(c)     Not much oxygen near sea bed;

Toadfish haemoglobin (nearly) saturated/loads readily at /has higher affinity for oxygen at low partial pressure (of oxygen);

2

(d)     (i)      The chimpanzee and the bonobo are more closely related (than to the gorilla);

They have identical amino acids/one of the amino acids is different in the gorilla;

2

(ii)     (Chimpanzee) orang-utan;

Amino acids different so bases different;

Few hydrogen bonds; 3

[15]

M3.          (a)     1  (DNA altered by) mutation; 2  (mutation) changes base sequence; 3  of gene controlling cell growth / oncogene / that monitors cell division; 4  of tumour suppressor gene; 5  change protein structure / non-functional protein / protein not formed; 6  (tumour suppressor genes) produce proteins that inhibit cell division; 7  mitosis; 8  uncontrolled / rapid / abnormal (cell division); 9  malignant tumour;

max 6

(b)     cancer cells die / break open; releasing DNA;

2

(c)     normal DNA and changed DNA have different sequences; DNA only binds to complementary sequence;

2

(d)     fewer abnormal / cancerous cells / smaller tumours; less cell damage; less spread / fewer locations to treat;

max 2

Page 22 of 35

(e)     mRNA base sequence has changed; gene / DNA structure is different / has mutated; cancer gene active / tumour suppressor gene inactive;

3 [15]

M4.          (a)     Kingdom, class, family, genus; 1

(b)     (i)      (Human) Fish Rhesus monkey Horse;

1

(ii)     As aminals closely related, more amino acids in sequence; 1

(c)     The more similar the DNA, the more similar the base sequences; The greater the number of hydrogen bonds/bonds between base pairs; More energy/heat needed to separate strands;

Q Correct terminology of base, base pair and hydrogen bond must be used as specified in scheme.

3 [6]

M5.          (a)     Recognition of same species;

Stimulates release of gametes;

Recognition of mate/opposite gender;

Indication of sexual maturity/fertility; 2 max

(b)     (i)      Internal fertilisation/fertilisation occurs in pouch/limited area; Q The term fertilisation is not required in the answer but must be implied.

1

(ii)     Protection from predators (developing in pouch); 1

(c)     (i)      Less stress caused to seahorse/quicker/more accurate method/ body is curved/head is linear;

Q Do not accept “easier” unless qualified. 1

(ii)     Head length proportional to body length/or described; 1

Page 23 of 35

(d)     Positive correlation between head/body lengths of male and female/ female and male with similar head/body lengths pair together;

1

(e)     Use line of best fit;

And extrapolate/extend line as required; 2

(f)      (Compare) DNA;

Sequence of bases/nucleotides;

DNA hybridisation;

Separate DNA strands/break hydrogen bonds;

Mix DNA/strands (of different species);

Temperature/heat required to separate (hybrid) strands indicates relationship;

Compare same/named protein;

Sequence of amino acids/primary structure;

Immunological evidence – not a mark

Inject (seahorse) protein/serum into animal;

(Obtain) antibodies/serum;

Add protein/serum/plasma from other (seahorse) species;

Amount of precipitate indicates relationship; Q The marks awarded for reference to DNA and sequence of bases/nucleotides must be in a different context to DNA hybridisation.

6 max [15]

M6.          (a)     Most closely (related) to chimpanzee/most recent common ancestor; 1

Least (related) to dogfish/least recent common ancestor; Allow ‘chicken is second’ to chimpanzee as equivalent to second mark point. Allow answers which compare similarity in DNA/genetic material. Marks should not be awarded for answers which only compare amino acid sequences without any indication of relationships. Allow ‘monkey’ for chimpanzee and ‘fish’ for dogfish

1

Page 24 of 35

(b)     Is present in all eukaryotes; 1

(c)     Reference to base triplet/triplet code/more bases than amino acids/longer base sequence than amino acid sequence;

Introns/non-coding DNA;

Same amino acid may be coded for/DNA code is degenerate; Different (base) triplets code for same amino acid = 2 marks Reject different amino acids are formed/produced. Ignore reference to codon.

2 max [5]

M7.         (a)      (i)     Kingdom / phylum / class;

Accept Animalia / animal kingdom / Chordata / Chordates / Aves Allow phonetic spelling

1

(ii)     Family; 1

(b)     (i)     1.      Shows the spread of the data / how data varies; 1.   Reject range.     Accept varies from the mean

2.      Overlap = no difference / due to chance / not significant; 2.   Allow converse

3.      Low SD means results more reliable / repeatable; 3.   Ignore accurate / valid

2 max

(ii)     1.      Different colour / different feathers / different throat;

2.      Birds don’t mate / pair bond with / recognise other species; 2.   Reference to courtship alone is not sufficient

2

(c)     1.      Different species would have different amino acid sequences; Accept more closely related = more similar sequence

2.      Amino acid sequence is the result of DNA / alleles / base sequence; References to incorrect statements about coding negates second mark

2 [8]

Page 25 of 35

M8.          (a)     Is species specific / allows recognition of same species;

Greater similarity in calls the closer the relationship (between the species);

Accept: ‘Similar species have similar calls’ as first marking point. Reference to courtship on its own is not sufficient for a mark. Must refer to relationship for second marking point.

2

(b)     (i)      G. americana and G. monachus; Highest percentage (DNA hybridisation) / more bases are similar/complementary / more hydrogen bonds / more base pairings;

Second marking point can be awarded without first marking point. 2

(ii)     Higher temperature / more energy (required) the higher the percentage DNA hybridisation / more bases are similar/complementary / more base pairings;

Correct reference to breaking hydrogen bonds / more/less hydrogen bonds being present;

Accept: ‘The greater the number of hydrogen bonds the higher the temperature/more energy required to break them’ for one mark.

2

(c)     1. More closely related (species) have more similarities in amino acid sequence/primary structure;

2. In same protein / named protein e.g. albumin;

3. Amino acid sequence is related to (DNA) base/triplet sequence;

OR

4. Similar species have a similar immune response to a protein/named protein;

5. More closely related (species) produce more ‘precipitate’ / antibody-antigen     (complexes) / agglutination;

Accept: ‘Similar species have similarities in amino acid sequence’ for first marking point. Accept: Converse for marking points 1, 4 and 5. Marking point 5 is for measuring the extent of the immune response.

2 max [8]

Page 26 of 35

M9.          (a)    1.      Amino acid sequences / primary structure;

More closely related (species) have more similarities in amino acid sequence/primary structure = two marks;

2.      Closer the (amino acid) sequence the closer the relationship;

3.      (Protein structure) related to (DNA) base/triplet sequence; Amino acid sequence is related to (DNA) base/triplet sequence = two marks;

2 max

(b)     1.      Reference to base triplets/triplet code / more bases than amino acids / longer base sequence than amino acid sequence;

Different (base) triplets code for same amino acids = 2 marks; Degeneracy of triplet code = 2 marks

2.      Introns / non-coding DNA; Ignore reference to codon.

3.      Degeneracy of code / more than one code for each amino acid; 3. Allow ‘more than one base sequence can code for a protein’

2 max

(c)     1.      Most closely related to chimpanzee;

2.      Least closely related to trout; 2

[6]

M10.         (a)      (i)     4; 1

(ii)     1.      Change in amino acid / (sequence of) amino acids / primary structure; 1.  Reject = different amino acids are ‘formed’

2.      Change in hydrogen / ionic / disulphide bonds;

3.      Alters tertiary structure / active site (of enzyme); 3.  Alters 3D structure on its own is not enough for this marking point.

4.      Substrate not complementary / cannot bind (to enzyme / active site) / no enzyme- substrate complexes form;

3 max

(b)     1.      Lack of skin pigment / pale / light skin / albino;

2.      Lack of coordination / muscles action affected; 2 max

Page 27 of 35

(c)     Founder effect / colonies split off / migration / interbreeding; Allow description of interbreeding e.g. reproduction between individuals from different populations

1 [7]

M11.          (a)     Is species specific / allows recognition of same species;

Greater similarity in calls the closer the relationship (between the species);

Accept: ‘Similar species have similar calls’ as first marking point. Reference to courtship on its own is not sufficient for a mark. Must refer to relationship for second marking point.

2

(b)     (i)      G. americana and G. monachus; Highest percentage (DNA hybridisation) / more bases are similar/complementary / more hydrogen bonds / more base pairings;

Second marking point can be awarded without first marking point. 2

(ii)     Higher temperature / more energy (required) the higher the percentage DNA hybridisation / more bases are similar/complementary / more base pairings;

Correct reference to breaking hydrogen bonds / more/less hydrogen bonds being present;

Accept: ‘The greater the number of hydrogen bonds the higher the temperature/more energy required to break them’ for one mark.

2

(c)     1. More closely related (species) have more similarities in amino acid sequence/primary structure;

2. In same protein / named protein e.g. albumin;

3. Amino acid sequence is related to (DNA) base/triplet sequence;

OR

4. Similar species have a similar immune response to a protein/named protein;

5. More closely related (species) produce more ‘precipitate’ / antibody-antigen     (complexes) / agglutination;

Accept: ‘Similar species have similarities in amino acid sequence’ for first marking point. Accept: Converse for marking points 1, 4 and 5. Marking point 5 is for measuring the extent of the immune response.

2 max [8]

Page 28 of 35

E1.          (a)     The vast majority of candidates answered this question correctly.

(b)     Candidates tended to answer part (i) more successfully than they answered part (ii).

(i)      Most candidates were able to state a eukaryotic feature, such as membrane-bound organelle or a named organelle. Some failed to gain the mark by failing to make clear what organism they were discussing in their answer.

(ii)     Commonly, the lack of a cell wall was given as a correct answer, but a significant number of answers made an invalid reference to saprophytic digestion.

(c)     (i)      Many candidates answered this question correctly by stating similarities between genes or DNA. Few considered recent ancestry. A significant number referred only to phenotypic similarities or they used a definition of species to suggest that the organisms would fail to produce fertile offspring.

(ii)     Most candidates picked up marks for referring to mutations and for an understanding of variation, although a significant proportion stated that the mutation occurred in the structure of the protein, or implied that cytochrome was an organism. Unfortunately, most candidates failed to link their answer to cytochrome c and relied on generalisations about ‘natural selection’ and ‘survival’ without focusing on why differences in the protein would be the basis of evolutionary change. Also, very few candidates either considered time or multiple generations in their explanation about a change in allele frequency; in many answers, evolution occurred in a single step.

E2.          (a)     Candidates showed a good understanding of the adaptations of gills for efficient gas exchange. Although there were some who wrote in very general terms about ‘gills’, most candidates linked surface area to the possession of gill filaments or lamellae and to diffusion. The principle of counter-current flow was frequently mentioned and it was clear that most candidates had an excellent understanding of this concept. Some illustrated their answers with diagrams and these were occasionally very helpful. Candidates should be aware, however, that marks can only be awarded for diagrams that are properly labelled. There were numerous sketches on which were written figures that might have represented anything. Some points were made less frequently or less convincingly. There was relatively little mention of the roles of ventilation and circulation in maintaining the concentration gradient and many struggled to describe the short diffusion path in sufficient detail to gain credit. There were also a number of frequent misconceptions. These included references to air passing over the gills; to diffusion only being able to take place in water, and to the presence of carbon dioxide being essential for the diffusion of oxygen.

(b)     Successful responses to this part of the question usually referred to photosynthesis or to the diffusion of oxygen from the higher concentration in the air. There were many answers, however, that involved fanciful ideas about generation of oxygen at depth and this bubbling to the surface, or incorporated the concept of need, such as that there was less oxygen at depth because the toadfish did not need it.

(c)     This answer illustrated a common failing among less able candidates in answering questions that involve application of knowledge. They were often inclined to rely on recall and, while most were able to indicate that the toadfish environment was low in oxygen, they not infrequently related this to high altitude. There was also a tendency to give answers that were too brief, omitting reference to the context of low partial pressure when describing the high affinity of toadfish haemoglobin for oxygen.

Page 29 of 35

(d)     Answers to part (i) tended to fall into two categories. Either candidates gave very good answers that made the points in the mark scheme succinctly, or they wrote at length about the three organisms without ever quite answering the question. However, it was encouraging to see many excellent answers to a question set in a context which is new to the specification. Part (ii) discriminated effectively across the full mark range. Where a single mark was obtained, it usually came from the correct identification of the hybrid DNA from the chimpanzee and the orang-utan separating at the lowest temperature. Some candidates then unfortunately suggested that weaker rather than fewer hydrogen bonds were formed. It was only in the best answers that differences in amino acid sequence were successfully linked to differences in base sequence.

E3.          (a)     Many candidates gave a good account of the changes a mutation could produce and those with clear expression achieved full marks; many scored three or four marks. Uncontrolled cell division and malignant tumors were frequently referred to and some appreciated that genes which controlled cell division could have changed. References to benign tumours or cell mutations were irrelevant in the context of this question.

(b)     Very few candidates achieved marks here, mainly because they did not read the question. Whole cells in the blood were not required, but the understanding that cancer cells could burst or die and release their DNA was.

(c)     Few seemed to understand this and restated the question without reference to the changed base sequences to which the strip would bind.

(d)     This was generally well known. The main reason for failing to gain marks was a reference to an undefined ‘it’ which would be growing, dividing or spreading, causing undefined damage.

(e)     Here too some candidates who understood the problem found it hard to explain that changes in the mRNA would reflect mutations in the DNA and would show that a cancer gene was active.

E5.          (a)     Most candidates had little difficulty obtaining at least one mark often for stating that courtship behavior enables recognition of the same species. Over a third of candidates gained a second mark. These candidates often linked courtship behaviour to sexual maturity or to the release of gametes. Most candidates failing to gain two marks often provided only one suggestion or there was a lack of clarity in their answers.

(b)     (i)      Almost two thirds of candidates gained this mark. Most candidates used the information in the stem of the question to explain that fusion between gametes would be more likely within a limited area.

(ii)     Less than half the candidates obtained this mark by indicating that developing within the pouch protected young seahorses from predators.

(c)     (i)      Most candidates gained this mark often by stating that the curved tail made it difficult to obtain an accurate measurement of body length.

(ii)     This proved slightly more difficult with a number of candidates providing the same answer as in (c)(i). Nevertheless, over 60% of candidates did obtain the mark by suggesting that body length is proportional to head length.

Page 30 of 35

(d)     The vast majority of candidates obtained this mark by describing the trend of seahorses with similar head/body lengths pairing together.

(e)     This was generally well answered with most candidates obtaining the first marking point by referring to drawing a line of best fit. Over 50% of candidates gained the second mark by explaining how extrapolation of the graph could be used to predict the total head length of selected mate.

(f)      This question proved an excellent discriminator. The vast majority of candidates described how DNA hybridisation could be used to find out if the two species of seahorses are closely related. Not surprisingly, the quality of the descriptions of DNA hybridisation varied considerably and a variety of alternative methods were credited. Most candidates gained a couple of marks for naming the technique and for the principle of mixing the DNA strands of the two species. Many candidates also appreciated that a higher temperature would be required to separate hybrid strands from closely related species. A maximum of four marks was available for a full description of DNA hybridisation.

Other methods described included; comparing DNA base sequences, comparing amino acid sequences and immunological studies. There was considerable confusion between the first two methods with many candidates referring to ‘amino acid sequences of DNA’. Few candidates appreciated that the same or a named protein should be studied when comparing amino acid sequences. Descriptions of immunological investigations were relatively infrequent and apart from some notable exceptions, were generally of poor quality displaying little understanding of even the basic principles. Nevertheless, over a third of candidates obtained four or more marks for this question with many providing outstanding detailed descriptions of the various methods involved.

E6.          (a)     Most candidates obtained at least one mark for stating that humans are most closely related to chimpanzees. Approximately half the candidates also gained the second marking point by indicating that humans are least closely related to dogfish.

(b)     Many candidates used the stem of the question to state correctly that cytochrome is found in all eukaryotes. Incorrect responses usually referred to cytochrome being present in all species.

(c)     Many candidates did refer to the base triplet code and gained a single mark point. Better candidates referred to introns or to different base triplets coding for the same amino acid. However, a significant number of candidates failed to gain any marks, often describing a base triplet as a gene.

E7.         (a)      (i)      Almost all students gained this mark.

(ii)      Again, almost all gained this mark, with many writing a mnemonic of one form or another in the margin.

(b)     (i)       Most students gained the first mark for a simple definition of standard deviation in terms of the spread of the data. A few failed to gain the mark by using the word ‘range ’ as an alternative to ‘spread’. The interpretation of standard deviation in terms of overlap was less well understood, and very few students suggested that a low standard deviation was related to closely grouped and therefore reliable data.

Page 31 of 35

(ii)     This question was mostly answered successfully. However some students used up all the space describing the differences in colours, and did not link the different colours of the two types of hummingbird to successful mating or to species recognition. They referred only to courtship which, as the term appeared in the stem, did not gain credit.

(c)     The majority of low marks gained in this question resulted from students failing to respond to the question ‘ …these sequences (i.e. the amino acid sequences) could provide evidence … ’and going on to describe how different DNA base sequences would give different proteins. Although students seemed to appreciate that different species have different amino acid sequences in the same protein, few could link this to differences in the DNA base sequence. Students seemed unclear about the relationships between the DNA base sequences and the amino acid sequence, and the use of incorrect terminology made their answers even more opaque.

E8.          (a)     Most candidates obtained at least one mark for indicating that the calls made by cranes are species specific. Many candidates, however, then proceeded to discuss courtship or simply to state that biologists could compare the calls. Relatively few candidates suggested that the greater the similarity in calls, the closer the relationship between different species is likely to be.

(b)     (i)      Almost eighty percent of candidates obtained both marks. They identified G. americana and G. monachus as being the most closely related because the percentage of DNA hybridisation between these species was highest. A significant minority of candidates suggested that G. rubicanda and G. americana are most closely related because these species having the highest intraspecific percentage DNA hybridisation.

(ii)     Most candidates obtained one mark by linking the requirement of a higher temperature with a higher percentage of DNA hybridisation. Approximately fifty percent of candidates obtained a second mark for a correct reference to hydrogen bonds. A common error by weaker candidates was to refer to hydrogen bonds between amino acids.

(c)     Very few candidates obtained both marks and over fifty percent scored zero. Answers were often expressed poorly with many candidates interchanging the terms amino acids and DNA bases. The omission of the word ‘sequence’ often prevented candidates gaining credit. A small number of candidates approached this question via immunological comparisons. These candidates often obtained a single mark for suggesting that ‘more precipitate’ would be formed in closely related species.

Page 32 of 35

E9.          (a)     Over 40% of students failed to score on this question. Many of these students suggested that proteins consist of bases and the confusion between bases and amino acids pervaded their responses. Although a number of students did correctly refer to the sequence of amino acids, only better students linked the similarity of the amino acid sequence with a close evolutionary relationship between different species.

(b)     This question also proved challenging with less than 50% of students gaining any marks. However, a significant number of students did gain one mark for reference to the triplet code and students appreciating the degeneracy of this code were able to gain both marks. Some students gained credit by referring to introns or non-coding DNA.

(c)     This question caused few problems with three out of four students gaining both marks. Almost every student stated that the results showed that humans and chimpanzee have the closest evolutionary relationship. However, some students did not always clearly express the idea that the trout was least closely related to humans.

E10.         (a)      (i)       Over 90% of students correctly determined that base sequence could code for a maximum number of four amino acids.

(ii)     The vast majority of students gained at least one mark, often by mentioning a change in the sequence in amino acids. However, a significant number of students incorrectly referred to ‘different amino acids being formed’. Most students gained a second mark for explaining that the active site/ tertiary structure would be altered. Over 50% of students gained maximum marks either by linking this to enzyme-substrate complexes not being formed or to changes in hydrogen bonds.

(b)     Most students had little difficulty in using the information to give two symptoms of phenylketonuria and gained both marks.

(c)     The majority of students obtained this mark, often by referring to migration or by describing interbreeding. However, over a third of students failed to gain credit and often accounted for the spread of phenylketonuria by horizontal or vertical gene transfer.

E11.          (a)     Most candidates obtained at least one mark for indicating that the calls made by cranes are species specific. Many candidates, however, then proceeded to discuss courtship or simply to state that biologists could compare the calls. Relatively few candidates suggested that the greater the similarity in calls, the closer the relationship between different species is likely to be.

(b)     (i)      Almost eighty percent of candidates obtained both marks. They identified G. americana and G. monachus as being the most closely related because the percentage of DNA hybridisation between these species was highest. A significant minority of candidates suggested that G. rubicanda and G. americana are most closely related because these species having the highest intraspecific percentage DNA hybridisation.

(ii)     Most candidates obtained one mark by linking the requirement of a higher temperature with a higher percentage of DNA hybridisation. Approximately fifty percent of candidates obtained a second mark for a correct reference to hydrogen bonds. A common error by weaker candidates was to refer to hydrogen bonds between amino acids.

Page 33 of 35

(c)     Very few candidates obtained both marks and over fifty percent scored zero. Answers were often expressed poorly with many candidates interchanging the terms amino acids and DNA bases. The omission of the word ‘sequence’ often prevented candidates gaining credit. A small number of candidates approached this question via immunological comparisons. These candidates often obtained a single mark for suggesting that ‘more precipitate’ would be formed in closely related species.

Page 34 of 35

Page 35 of 35