Relationships DNA Comparison Revision

Transcription

1 Relationships DNA Comparison Revision May minutes 57 marks Page 1 of 19

2 Q1. 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. DNA base sequence coding for PAH DNA base sequence coding for non-functioning PAH C A G T T C G C T A C G C A G T T C C C T A C G (a) (i) What is the maximum number of amino acids for which this base sequence could code? (1) Explain how this mutation leads to the formation of non-functioning PAH (Extra space) (3) Page 2 of 19

3 PAH catalyses a reaction at the start of two enzyme-controlled pathways. The diagram shows these pathways. Use the information in the diagram to give two symptoms you might expect to be visible in a person who produces non-functioning PAH 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 3 of 19

4 Q2. 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) Explain why fragments of DNA from cancer cells may be present in blood plasma (lines 3-4). Page 4 of 19

5 Explain why the nucleic acid on the test strip will only bind to altered DNA (lines 4-5). (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. (e) Explain how examining mrna (line 7) enables scientists to discover whether cancer is present. (3) (Total 15 marks) Page 5 of 19

6 Q3. (a) The cheetah, Acinonyx jubatus, and other cat species belong to the family Felidae. Complete the table to show the classification of the cheetah. Kingdom Animalia Chordata Mammalia Carnivora Family Felidae Genus This system of classification is described as hierarchical. Explain what is meant by a hierarchical classification. (1) Despite differences in form, leopards, tigers and lions are classified as different species of the same genus. Cheetahs, although similar in form to leopards, are classified in a different genus. (i) Describe one way by which different species may be distinguished. (1) Suggest two other sources of evidence which scientists may have used to classify cheetahs and leopards in different genera (Total 6 marks) Page 6 of 19

7 Q4. (a) Class, family, genus and kingdom are terms used in classifying organisms. Write the terms in the correct sequence. Largest number of species Smallest number of species (1) 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. Animal Human Sequence of amino acids in cytochrome c lys ile phe ile met lys lys th rphe va lglu lys lys ile phe ile met lys lys ile phe val glu lys 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. Explain your answer. (1) (1) DNA hybridisation shows similarities between DNA samples. Explain why (3) (Total 6 marks) Page 7 of 19

8 Q5. 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. 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. Page 8 of 19

9 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. Biologists can also use protein structure to investigate the relationship between different species of crane. Explain why. (Total 8 marks) Q6. (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 9 of 19

10 Mackerel live in the surface waters of the sea. Toadfish live on the seabed in deep water. The concentration of oxygen is higher in the surface waters than it is in water close to the seabed. Suggest why. 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. Page 10 of 19

11 (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. Species Position 87 Position 104 Position 125 Chimpanzee T R P Bonobo T R P Gorilla T K P Orang utan K R Q (i) What information do the data in the table suggest about the relationships between the chimpanzee, the bonobo and the gorilla? Explain your answer. 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 11 of 19

12 M1. (a) (i) 4; 1 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 1. Lack of skin pigment / pale / light skin / albino; 2. Lack of coordination / muscles action affected; 2 max Founder effect / colonies split off / migration / interbreeding; Allow description of interbreeding e.g. reproduction between individuals from different populations 1 [7] M2. (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 (d) cancer cells die / break open; releasing DNA; normal DNA and changed DNA have different sequences; DNA only binds to complementary sequence; fewer abnormal / cancerous cells / smaller tumours; less cell damage; less spread / fewer locations to treat; 2 2 max 2 Page 12 of 19

13 (e) mrna base sequence has changed; gene / DNA structure is different / has mutated; cancer gene active / tumour suppressor gene inactive; 3 [15] M3. (a) phylum, class, order; species, Acinonyx jubatus; 2 larger groups containing smaller groups; 1 (i) do not interbreed to produce fertile offspring / different DNA / different niches; 1 fossil record; evolutionary history/phylogeny; biochemical differences e.g. DNA/proteins/cytochromes; homologous features / named feature; karyotype / number and form of chromosomes; (discount any example credited in (i)) 2 [6] M4. (a) Kingdom, class, family, genus; (i) (Human) Fish Rhesus monkey Horse; 1 1 As aminals closely related, more amino acids in sequence; 1 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] Page 13 of 19

14 M5. (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. (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 2 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 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 14 of 19

15 M6. (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 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 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 (Chimpanzee) orang-utan; Amino acids different so bases different; Few hydrogen bonds; 3 [15] Page 15 of 19

16 E1. (a) (i) Over 90% of students correctly determined that base sequence could code for a maximum number of four amino acids. 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 enzymesubstrate complexes not being formed or to changes in hydrogen bonds. Most students had little difficulty in using the information to give two symptoms of phenylketonuria and gained both marks. 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. E2. (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. (d) (e) 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. Few seemed to understand this and restated the question without reference to the changed base sequences to which the strip would bind. 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. 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. E3. Although only a few candidates obtained maximum marks on this question, most candidates were able to gain between two and four marks. (a) Most candidates correctly filled in phylum, class and order, however, weaker candidates often put common names such as cat or cheetah in the species box. Approximately half the candidates gained this mark. Most candidates who did not, either provided definitions which were too vague or referred to different sizes of organisms. Page 16 of 19

17 (i) Most candidates obtained this mark by referring to the breeding of organisms to produce fertile offspring. Very few candidates gained two marks with the majority simply referring to different diets, colours or behaviour. Correct answers usually referred to fossils, evolutionary history or biochemical differences. E5. (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. (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. 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. 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. E6. (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. Page 17 of 19

18 (d) 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. 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. 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 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. Page 18 of 19

19 Page 19 of 19