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1. EcoR1 is a restriction enzyme that will cut DNA at a certain base sequence to make DNA fragments suitable for gel electrophoresis. (a) Below is a diagram showing part of a DNA molecule that has been cut with EcoR1 into two sections called P and Q. The four bases at the cut (sticky) end of section P have been labelled. (i) Name a component of the DNA molecule found in the part labelled X. .................................................... ..................................................... ....................... (1) (ii) State the letter of the four complementary bases for the sticky end of section P.

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1. EcoR1 is a restriction enzyme that will cut DNA at a certain base sequence to make DNA fragments suitable for gel electrophoresis. (a) Below is a diagram showing part of a DNA molecule that has been cut with EcoR1 into two sections called P and Q. The four bases at the cut (sticky) end of section P have been labelled. (i) Name a component of the DNA molecule found in the part labelled X. ................................................................................................................................ (1) (ii) State the letter of the four complementary bases for the sticky end of section P. ................................................................................................................................ ................................................................................................................................(2) (iii) Name the base T. ................................................................................................................................(1)(b) EcoR1 cut a piece of DNA which is shown below.

(i) The letters bp stand for base pairs when referring to DNA. Name the type of bond that joins two bases together to form a base pair. ................................................................................................................................(1) (ii) A gel electrophoresis study was undertaken with 5 samples of DNA. Each sample was made up of a 5500bp section of DNA that had been mixed with two restriction enzymes. These enzymes cut the DNA section into smaller fragments. The restriction enzymes were different in each of the five samples. The results of the study are shown in the diagram of the gel electrophoresis plate below.

In one sample the restriction enzymes used were Hind and Bam HI. These enzymes cut the DNA section as shown below.

Use both pieces of information to choose the sample that correctly represents the DNA mixed with Hind and Bam HI. Sample number: ................................

(1) (Total 6 marks)Jun 2008 6131Ans6(a)(i) {phosphate / phosphoric acid } / {deoxyribose / pentose / eq} (sugar) ; (1) 6(a)(ii) AA ; TT ; (2) 6(a)(iii) thymine ; (1) 6(b)(i) hydrogen / H (bond) ; (1) 6(b)(ii) (sample) 4 ; (1)

2. 8 Huntingtons disease is a genetic condition that leads to a loss in brain function. The gene involved contains a section of DNA with many repeats of the base sequence CAG. The number of these repeats determines whether or not an allele of this gene will cause Huntingtons disease.l An allele with 40 or more CAG repeats will cause Huntingtons disease.l An allele with 36 39 CAG repeats may cause Huntingtons disease.l An allele with fewer than 36 CAG repeats will not cause Huntingtons disease.The graph shows the age at which a sample of patients with Huntingtons disease first developed symptoms and the number of CAG repeats in the allele causing Huntingtons disease in each patient.

8 (a) (i) People can be tested to see whether they have an allele for this gene with more than 36 CAG repeats. Some doctors suggest that the results can be used to predict the age at which someone will develop Huntingtons disease.Use information in the graph to evaluate this suggestion.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3 marks)(Extra space) .....................................................................................................................................................................................................................................................................................................................................................................................8 (a) (ii) Huntingtons disease is always fatal. Despite this, the allele is passed on in human populations. Use information in the graph to suggest why.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2 marks)

8 (b) Scientists took DNA samples from three people, J, K and L. They used the polymerase chain reaction (PCR) to produce many copies of the piece of DNA containing the CAG repeats obtained from each person. They separated the DNA fragments by gel electrophoresis. A radioactively labelled probe was then used to detect the fragments. The diagram shows the appearance of part of the gel after an X-ray was taken. The bands show the DNA fragments that contain the CAG repeats.

8 (b) (i) Only one of these people tested positive for Huntingtons disease. Which person was this? Explain your answer.Person ...............................................................................................................................Explanation ........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2 marks)8 (b) (ii) The diagram only shows part of the gel. Suggest how the scientists found the number of CAG repeats in the bands shown on the gel.................................................................................................................................................................................................................................................................................................................................................................................................(1 mark)8 (b) (iii) Two bands are usually seen for each person tested. Suggest why only one band was seen for Person L.................................................................................................................................................................................................................................................................................................................................................................................................(1 mark)Total 9 marks Aqa Biology BIOL5Unit 5 Control in cells and in organismsFriday 22 June 2012 9.00 am to 11.15 am

3. The black mamba is a poisonous snake. Its poison contains a toxin.The table shows the base sequence of mRNA that codes for the first two amino acids of this toxin.

Complete the table to show1 (a) (i) the base sequence of the anticodon on the first tRNA molecule that would bind to this mRNA sequence (1 mark)1 (a) (ii) the base sequence of the DNA from which this mRNA was transcribed. (1 mark)1 (b) The length of the section of DNA that codes for the complete toxin is longer than the mRNA used for translation. Explain why.................................................................................................................................................................................................................................................................................................................................................................................................(1 mark)1 (c) A mutation in the base sequence of the DNA that codes for the toxin would change the base sequence of the mRNA.Explain how a change in the base sequence of the mRNA could lead to a change in the tertiary structure of the toxin.................................................................................................................................................................................................................................................................................................................................................................................................(1 mark)1 (d) The black mambas toxin kills prey by preventing their breathing. It does this by inhibiting the enzyme acetylcholinesterase at neuromuscular junctions. Explain how this prevents breathing.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3 marks)(Extra space) .....................................................................................................................................................................................................................................................................................................................................................................................Answer

4. Only one strand of the DNA of a gene (the sense strand) is normally transcribed tomRNA. The complementary strand of DNA is the antisense strand, which is notnormally transcribed. By inserting a promoter at the end of the antisense sequence, this forming an antisense gene, RNA transcription can occur from it. This anti-mRNA bonds to normal mRNA to form double-stranded RNA (duplex RNA), which cannot be translated by ribosomes. This technique can be used to suppress specific genes. The sequence of events is shown in Fig. 2.1.

(a) Explain(i) The bonding of anti-mRNA with normal mRNA to form duplex RNA..........................................................................................................................................................................................................................................................................................................................................................................................[3](ii) Why the resulting duplex RNA cannot be translated..........................................................................................................................................................................................................................................................................................................................................................................................[3]A promoter sequence was inserted into the antisense strand of DNA.(b) Outline how the promoter sequence may have been inserted into the antisense strand of DNA.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................[5]A recently identified gene (pTOM13) in tomato plants is expressed during the ripening of tomato fruits.Tomato plants were transformed by adding an antisense gene to pTOM13 to the normal genome. Antisense genes, once introduced, are inherited. When transformed plants that each contain one antisense gene are crossed, the offspring include plants with no antisense gene, one antisense gene and two antisense genes in a 1:2:1 ratio.Ethene production during fruit ripening was measured in plants with one and with two antisense genes, and in normal plants. The results are shown in Fig. 2.2.

(c) Suggest:(i) a function of the protein encoded by pTOM13........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................[2](ii) a reason for the difference in ethene production between transformed plantswith one antisense gene and those with two antisense genes.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................[2](d) Explain how two plants, each with an antisense gene added to the normal genome, can produce offspring with no antisense gene, one antisense gene and two antisense genes in a 1:2:1 ratio.......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................[3](e) Suggest one benefit and one hazard of the genetic engineering of plants.benefit ............................................................................................................................................................................................................................................................hazard..................................................................................................................................................................................................................................................[2][Total : 20]5. 5 All the living affected individuals in a family were found to have a mutation in the gene locus coding for a kinase enzyme. DNA profiles of the same part of the normal and mutant alleles of the gene are shown in Fig. 5.1. Such profiles could form the basis of genetic screening for the condition.

(a) Explain how such DNA profiles are produced.(In this question, 1 mark is available for the quality of written communication.).................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................. [9](a) (i) two mRNAs are complementary / shown in diagram;bases H-bond;base pairing;A-U and C-G [3 marks max](ii) no exposed bases / binding sites blocked;cannot bind to, ribosomes/rRNA;cannot bind tRNA;no start signal;so no protein produced [3 marks max]www.XtremePapers.netSpecimen Materials OCR 2000Biology Oxford, Cambridge and RSA Examinations145(b) restriction enzyme / named restriction enzyme;both DNAs cut with same enzyme;sticky ends;DNAs join by complementary sticky ends;nucleotides added to make sticky ends;phosphate-sugar backbones sealed;ligase [5 marks max](c) (i) enzyme;involved in ethene production / other sensible suggestion;[2 marks max](ii) antisense gene blocks pTOM13;antisense gene not expressed as rapidly as normal gene;with 1 copy not all mRNA duplexed;greater chance that with 2 copies of antisense gene;if even a small quantity of enzyme produced it will have a detectableeffect;ref partial/incomplete/co-dominance [2 marks max](d) gene added to one (strand of DNA of) one chromosome;plant (effectively) heterozygote;chromosomes segregate in meiosis;two types of gamete;shown in diagram [3 marks max](e) one sensible benefit e.g. slows ripening / increases keeping time;one sensible hazard e.g. inherited so passed to other tomatoes and ruinripening / very difficult to contain [2 marks max][Total: 20]5 (a) Quality of written communication assessed in this answer.source of DNA sample;gene/chromosome/DNA, cut up;by restriction enzyme/endonuclease;separate fragments of different sizes;by electrophoresis; samples in wells at, one / cathode end;fragments move different distances;migrate to anode;shortest, furthest distance; ref to chargetransfer via Southern Blotting;description;bands invisible;(incubate) with radioactive probe;or use stain;autoradiograph/expose to photographic plate;final banding pattern / ref to bands in Fig. 5.1 [8 marks max]Q clear, well organised answer using specialist terms [1 mark][9 marks max]

6. The main protein in yoghurt is casein. The synthesis of casein in cells is shown in the diagram below.

(i) At each stage, one or more nucleic acids is involved. Give the nucleic acid or nucleic acids involved at each stage on the dotted line next to the box. The first one has been done for you.

(4) (ii) Name the site within a cell where stage 3 occurs. ................................................................................................................................(1)7. (a) Scientists can use restriction enzymes to cut a sample of DNA into shorter pieces.The lengths of the DNA pieces can be measured in units called base pairs.Explain why a base pair is a suitable unit for measuring the length of a piece of DNA......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... (2 marks)8 (b) Sickle cell anaemia is caused by a point mutation in the haemoglobin gene. Doctors have developed a test for the allele causing sickle cell anaemia using a restriction enzyme called Ddel. The allele for normal haemoglobin, HA, can be cut by Ddel.The allele for sickle-cell haemoglobin, HS, cannot be cut by Ddel.The method used by the doctors involves the following steps.1. Take samples of cells from a person and extract the DNA.2. Treat the DNA with restriction enzymes other than Ddel to cut the DNA into many fragments.3. Carry out PCR (the polymerase chain reaction) using primers that are specific for part of the haemoglobin gene.4. Add Ddel enzyme to the DNA produced by the PCR.5. Find the lengths of the resulting DNA pieces.8 (b) (i) What is a primer?................................................................................................................................................................................................................................................................(1 mark)8 (b) (ii) How can a primer be specific to part of the haemoglobin gene (step 3)?................................................................................................................................................................................................................................................................(1 mark)8 (b) (iii) Why is it necessary to carry out PCR on the extracted DNA (step 3)?................................................................................................................................................................................................................................................................(1 mark)8 (c) Mustapha and Shahira are both carriers of sickle cell anaemia. Shahira is pregnant and they want to know if their baby will have sickle cell anaemia.The doctors tested samples of DNA from Mustapha, from Shahira and from the fetus.They used the method described in part (b) to test for the allele responsible for sickle cell anaemia. The primers used in this method were specific for the base sequences that occur 55 base pairs before the point of mutation and 55 base pairs after the point of mutation in the haemoglobin gene.The table shows the results.

8 (c) (i) Explain the results of the tests on Mustaphas and Shahiras DNA............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................. (3 marks)Total 10 marks

AqaHuman Biology HBIO4Unit 4 Bodies and cells in and out of controlFriday 15 June 2012 9.00 am to 11.00 am7.The polymerase chain reaction (PCR) is a means of generating large numbers of identical copies of a sample of DNA. PCR technology allows scientists to remove tiny samples of DNA from a single hair follicle or white blood cells in a drop of blood and then make numerous copies of it. This process can be very useful in forensic science. The steps involved in PCR technology are shown in the diagram below.

(a) Suggest how heating of the original DNA sample to 94C (step 1) would bring about separation of the two strands of DNA. .......................................................................................................................................................................................................................................................................................................................................................................................... [1] (b) In step 2, small lengths of single stranded DNA known as primers are added to the sample to stop the two sides of the sample DNA molecule from rebinding with each other. Why is it important that the sample DNA now remains as two separate strands?............................................................................................................................................................................................................................................................[1] (c) In step 3, describe how the enzyme DNA polymerase would produce new double stranded copies using the single DNA strands and the pool of nucleotides provided. ........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................ [2] (d) Suggest why it is important that some of the DNA copies manufactured are recycled in step 4. ............................................................................................................................................................................................................................................................(1)In criminal cases, DNA found on the victim can undergo PCR and eventually allow the production of a DNA fingerprint. The figure below illustrates the DNA fingerprint from a victim, the DNA specimen found on the victim and the DNA fingerprints of three potential suspects.

(e) (i) Use the DNA fingerprints to identify the guilty suspect. ..............................................................................................................................(1)(ii) Why is it important to include the victims DNA fingerprint for analysis? ..............................................................................................................................(1)12. 2 The diagram below shows a stage in the synthesis of a part of a polypeptide. (a) Name the organelle illustrated in the diagram. [1] (b) Using the information in the diagram above determine: (i) the m-RNA codon at 1 (ii) the t-RNA anticodon at 2 (iii) the type of bond formed at 4 [3]

[7] (a) Heating breaks down hydrogen bonds/bonds holding the two strands together/bonds between the bases; (b) Each strand acts as a template against which free nucleotides combine (as in semi-conservative model); (c) Any two from nucleotides are arranged in place opposite the exposed bases on each strand according to the complementary base pairings (A-T, C-G, T-A, G-C) the complementary bases are held together by hydrogen bonds the sugar of one nucleotide is joined to the phosphate group of the next nucleotide a condensation reaction forms the bond (d) This ensures that multiple copies of the DNA are manufactured; (e) (i) Suspect one; (ii) To ensure that the DNA specimen is not that of the victim; 12 answer(a) ribosome; (b) (i) CCU; (ii) UUC; (iii) peptide bond;