1. (a) The inheritance of the ability to produce hydrogen cyanide is controlled by two genes which are located on different chromosomes. The dominant allele of one gene, G, controls the production of enzyme G which converts a precursor to linamarin. The dominant allele of the other gene, E, controls the production of enzyme E which converts linamarin to hydrogen cyanide. This is summarised in the diagram. A llele G E nzym e G A llele A E nzym e E P re c u rso r L inam arin H y d ro g en cy a n id e Explain why plants homozygous for the allele g will not produce hydrogen cyanide when their tissues are damaged. (2) Recently a strain of genetically engineered clover has been developed which has a high concentration of proteins rich in sulphur-containing amino-acids. A piece of DNA was prepared which contained the three different genes. This was inserted into a clover plant. G ene o b tain ed fro m su n flo w e r se e d s. T h is g en e co d e s fo r a p ro tein rich in su lp h u r-co n tain in g am in o a c id s. 1 G ene en su re s th at th e p ro tein rich in su lp h u r-co n tain in g am in o a cid s is p ro d u ced in lea f cells. 2 G ene p re v en ts th is p ro tein b ein g d ig e ste d in th e ru m en o f sh eep. 3 (b) The copy of Gene 1 used in this experiment was obtained from the mRNA of the sunflower seeds. (i) Explain how enzymes could be used to obtain the gene from the mRNA. (3) (ii) Explain why it would be an advantage to obtain the gene from the mRNA rather than from the DNA of the sunflower seeds. (2) (Total 7 marks) 2. One technique used to determine the sequence of nucleotides in a Woodford High School 1
Transcript
1. (a) The inheritance of the ability to produce hydrogen cyanide is controlled by two genes which are located on different chromosomes. The dominant allele of one gene, G, controls the production of enzyme G which converts a precursor to linamarin. The dominant allele of the other gene, E, controls the production of enzyme E which converts linamarin to hydrogen cyanide. This is summarised in the diagram.
A lle le G E n zy m e G
A lle le A E n zy m e E
P recu rso r
L in am arin
H y d ro g e n cy a n id e
Explain why plants homozygous for the allele g will not produce hydrogen cyanide when their tissues are damaged.
(2)
Recently a strain of genetically engineered clover has been developed which has a high concentration of proteins rich in sulphur-containing amino-acids. A piece of DNA was prepared which contained the three different genes. This was inserted into a clover plant.
G e n e o b ta ine d f ro m su n flo w er seed s . T h is g e ne c o d es fo r a p ro te in r ic h in su lp h u r-c o n ta in in g a m in o acid s .
1
G e n e en su re s th a t th e p ro te in rich in su lp h u r-co n ta in in g a m in o a c id s is p ro d u ce d in lea f ce lls .
2
G e n e p rev e n ts th is p ro te in b e in g d ig es ted in th e ru m en o f sh e ep .
3
(b) The copy of Gene 1 used in this experiment was obtained from the mRNA of the sunflower seeds.
(i) Explain how enzymes could be used to obtain the gene from the mRNA.(3)
(ii) Explain why it would be an advantage to obtain the gene from the mRNA rather than from the DNA of the sunflower seeds. (2)
(Total 7 marks)
2. One technique used to determine the sequence of nucleotides in a sample of DNA is the Sanger procedure. This requires four sequencing reactions to be carried out at the same time.The sequencing reactions occur in four separate tubes. Each tube contains
a large quantity of the sample DNA
a large quantity of the four nucleotides containing thymine, cytosine, guanine and adenine
DNA polymerase
radioactive primers
Woodford High School 1
A modified nucleotide is also added to each tube, as shown in Figure 1.
Tu b e 1A sm all qu a n tity o f
n u c leo tid es co n ta in in gm o d ified th y m in e
Tu b e 2A sm all qu a n tity o f
n u c leo tid es co n ta in in gm o d ified cy to s in e
Tu b e 3A sm all q u a n tity o f
n u c leo tid es co n ta in in gm o d ifie d g u an in e
Tu b e 4A sm all q u a n tity o f
n u c leo tid es co n ta in in gm o d ifie d ad en in e
F ig u re 1
(a) A large quantity of the DNA sample is required for this procedure. Name the reaction used to amplify small amounts of DNA into quantities large enough for this procedure.
(c) (i) When a modified nucleotide is used to form a complementary DNA strand, the sequencing reaction is terminated. Suggest how this sequencing reaction is terminated.
(d) A different sample of DNA was then analysed. The DNA fragments from the four tubes were separated in a gel by electrophoresis and analysed by autoradiography. Figure 2 shows the banding pattern produced.
T C G A
D irec tio n o fm o v em en t o ffra g m en ts
– v e
+ v e
(i) Explain why the DNA fragments move different distances in the gel.
One aim of cancer therapy is to find a magic bullet that seeks out and kills tumour cells butleaves normal cells unharmed. For this to work, the bullet needs to be able to recognise adifference between the two types of cell.
Some tumours grow so fast that they outgrow their blood supply and the oxygen concentration5 in their cells falls. Drugs are being developed that are only effective once they reach the low
oxygen conditions inside a tumour cell. Here enzymes called reductase enzymes activate thedrug which then kills the cell.
Professor Stratford and his colleagues at Manchester are taking advantage of the fact that theP450 reductase gene is only switched on in an environment which is low in oxygen. His team
10 have constructed the piece of DNA which is shown in the diagram.
Region of DNA which switches P450 reductase gene Gene coding for protein whichgene on in low oxygen acts as a marker on plasmaconcentrations membrane
This piece of DNA was injected into breast cancer cells and the cells were grown in the laboratory. The marker protein was used to identify cells with the injected gene. When the oxygen concentration was reduced, the concentration of P450 reductase increased.
Use information from the passage and your own knowledge to answer the following questions.
(a) Apart from the rates at which they grow, give one way in which tumour cells differ from normal cells.
(e) (i) The investigators added the gene coding for the protein which acts as a marker on the plasma membrane to their specially constructed piece of DNA. Explain why.
(ii) Some antibodies fluoresce when illuminated with ultraviolet light. Suggest why these antibodies could be used to identify the cells which had the marker protein on their plasma membranes.
(b) There are many different restriction endonucleases. Each type cuts the DNA of a plasmid at a specific base sequence called a restriction site. The diagram shows the position of four restriction sites, J, K, L and M, for four different enzymes on a single plasmid. The distances between these sites is measured in kilobases of DNA.
5 0 k b
4 k b1 0 k b
36 kbM
LKJ
1 kb = 1 kilobase
The plasmid was cut using only two restriction endonucleases. The resulting fragments were separated by gel electrophoresis. The positions of the fragments are shown in the chart below.
(iii) The table shows a sequence of bases from the mRNA coding for insulin.Complete the table to show the sequence of bases you would expect in the single-stranded DNA copy.
mRNA base sequence U C A A C C
DNA base sequence(1)
(b) What is the role of DNA ligase in producing genetically-modified bacteria?
7. Bacteria can be genetically modified to produce insulin for human use. To achieve this, human insulin genes are transferred into bacteria. Plasmids containing two antibiotic resistance genes, one coding for resistance to tetracycline and one for resistance to ampicillin, are used to carry out this transfer.
A restriction enzyme was used to cut up the human DNA and plasmids. Figure 1 shows the different fragments of human DNA and the type of cut plasmid that was produced.
The fragments of human DNA and the cut plasmids were mixed together with DNA ligase. Several types of plasmid were formed. Some contained human DNA in the centre of the gene coding for resistance to tetracycline. The different types of plasmid are shown in Figure 2.
Woodford High School 9
Tetra cy clin eres is tan c e g e n e
In su ling en e
H u m a nD N A
A m p ic illinr es ista n ceg e n e
A m p icillinre s istan c eg e n e
A m p ic illinre s is tan c eg e ne
F ig u re 2
P lasm id co n ta in in gn o ex tra D N A
P lasm id co n ta in in gth e in su lin g e n e
P lasm id co n ta in in g h u m a n D N Afrag m en t w ith o u t in sulin g e ne
(c) Explain what causes several types of plasmid to be formed.
(d) The plasmids are mixed with the bacteria. Some bacteria take up the plasmids.
(i) Explain how it is possible to distinguish between bacteria which have taken up a plasmid with human DNA and those which have taken up a plasmid without any extra DNA.
Large numbers of possums in New Zealand are eating crops and spreading disease betweencattle. The use of shotguns and poisons by farmers has not greatly reduced possum numbers.A better solution to the possum problem may have been found. A crop of carrots has beengenetically modified to produce a ‘sterility protein’. This sterility protein prevents possums
5 producing offspring.
First, scientists identified the gene that codes for this sterility protein. Several copies of thesterility gene were cut out from long sections of DNA using a special enzyme. The sameenzyme was also used to cut open plasmids which had been removed from bacterial cells. Adifferent enzyme joined together the ‘sticky ends’ of a plasmid and of a sterility gene to
10 produce a recombinant plasmid.
The scientists then tried to put these plasmids back into the bacteria. Each plasmid alsocontained a gene, giving resistance to an antibiotic which normally kills bacteria. Because ofthis resistance gene, the scientists could identify bacteria containing the sterility gene andisolate them from bacteria which had not taken up this gene. Finally, carrot seedlings were
15 sprayed with bacterial cells. The plasmids entered the carrot seedlings and carried copies ofthe sterility gene into the DNA of carrot cells.
The genetically modified crop will be harvested and the carrots scattered across land populated by possums.
Use information from the passage and your own knowledge to answer the following questions.
(d) Explain how the presence of the antibiotic resistance gene allowed scientists to identify and isolate the bacteria which contain the sterility gene (lines 12 - 14)
The table shows the number of base pairs present in the fragments.
Fragment Number of base pairs (× 103)
1 4.65
2 5.72
3 10.71
4 2.39
5 5.35
6 7.53
The diagram shows the electrophoresis gel used. The mixture of fragments was placed at the start point marked S and the process started. The boxes indicate the positions reached by the different fragments.
Woodford High School 12
S
D irec tio n o f m o v em en t o f fra g m en ts
(b) Explain why base pairs are a suitable way of measuring the length of a piece of DNA.
10. -thalassaemia is a genetic condition in which abnormal haemoglobin is produced. In one form, the recessive allele for -thalassaemia, t, differs from the normal allele, T, by a single base-pair. A radioactive DNA probe was used to investigate the genotypes of four members of one family. The flowchart summarises the technique involved.
D N A sam p le s ex tra c ted an d cu t in to frag m en ts u s in g are s tric tio n en zy m e
F rag m en ts sep a ra ted fro m each o th e r b y e lec tro p h o res is
S u rp lu s p ro b e w ash ed o ff
O n e reg io n o f th e re su ltin g g e l w as b lo tted w ith tw op ieces o f filte r p ap er. T h e firs t w as so ak ed in a so lu tio n
co n ta in in g a rad io ac tiv e D N A p ro b e fo r th e n o rm a l a lle le .T h e seco n d w as so ak ed in a so lu tio n co n ta in in g a
rad io ac tiv e D N A p ro b e fo r th e -th a la ssaem ia alle le .
The diagram below shows the appearance of the two pieces of filter paper which resulted from the investigation.
F a th e r M o th e r F irs tch ild
S ec o n dc h ild
F ilte r p ap e r so ak edw ith p ro b e fo rn o rm a l a lle le
F ilte r p ap e r so ak edw ith p ro b e fo r
-th a la ssa em iaa lle le
(a) What is the probability that the next child that this couple have is a girl who has-thalassaemia? Explain your answer.
(b) (i) The fragment of DNA containing the normal allele and the fragment with the -thalassaemia allele moved the same distance on the gel. Explain why.
(ii) The allele for -thalassaemia differs from the normal allele by only one base-pair.Explain why the probe used to identify these alleles consists of a piece of DNA twenty bases in length and not just one base.
11. Genetic engineering has made it possible to transfer genes from one species to another. For example, a gene that gives resistance to herbicide and another gene which gives resistance to insect attack have been transferred into maize. Some people think that there will be great advantages in growing maize with these genes. Others are equally convinced that there are long-term dangers in growing crops of this maize.
12. DNA sequencing is a technique used to find the sequence of nucleotides in a sample of DNA. Enzymes are used to cut the DNA sample into fragments of different lengths.The ends of these fragments are then labelled using radioactive probes.Four different probes attach to the end of DNA fragments in which the terminal nucleotide is adenine, cytosine, thymine and guanine respectively. The labelled fragments are then separated.
The diagram shows apparatus used to separate the DNA fragments and the end result of the process.
T C G A
N e g a tiv ete rm in a l
F o ur sam p leso f rad io ac tiv e lylab e lled D N Afrag m en ts a rep lace d in th efo u r w e lls
G el
P o s itiv ete rm in a l
D N A frag m en tsv isib le
(a) (i) Use the information from the diagram and your knowledge of DNA sequencing techniques to draw a flow chart for the process. The first box has been completed for you.
Woodford High School 16
F o u r sam p le s o f rad io ac tiv e ly lab e lle d D N Afra g m e n ts a re p lace d in th e fo u r w ells
(4)
Woodford High School 17
(ii) Why do the DNA fragments move different distances in the gel?
(b) On the diagram, draw the DNA molecules that would be present at the end of Cycle 2. Use the same method of shading to distinguish between original DNA strands and the new DNA strands.
(2)
(c) Describe one example of the use of the polymerase chain reaction.
Q1. (a) Do not produce enzyme G/produce non-functional enzyme G;No linamarin formed; 2
(b) (i) Reverse transcriptase;makes single strand of DNA/cDNA from RNA;Double strand then formed;DNA polymerase; max 3
(ii) If from DNA all genes are present in cell;mRNA from activated genes only/codes for one protein;DNA has introns/non-coding /junk DNA;Have been edited out in mRNA; max 2
(ii) enables replication / sequencing to start / keeps strands separate; 1
(c) (i) (modified nucleotide) does not form bonds/react with other nucleotides; 1does not “fit” DNA polymerase/enzyme/active site;
(ii) AC; 1 max(accept reading from right hand side i.e. TC)
(d) (i) different lengths / sizes / mass; 1
(ii) radioactive primer; 1
(iii) GAAGTCTCAG; 1(accept reading from autoradiogram i.e. CTTCAGAGTC)
8]
Q3(a) Lower oxygen concentration/more reductase/relatively large nucleus/difference in metabolism/continue to divide when not in contact with another surface; 1
(b) High respiratory rate associated with rapid growth andinsufficient blood supply to supply enough oxygen; 1
(c) They have enough oxygen present;therefore reductase enzymes not produced;inactive drug is harmless; Max 2
(d) (i) Restriction enzyme/endonuclease; 1
(ii) Ligase; 1
(e) (i) The marker could be identified easily;so it would be known which cells contain the piece of DNA; 2
(ii) Antibodies are specific;proteins;will only attach to marker protein;correspond in shape; Max 2
(f) mRNA makes a copy of the DNA code;during transcription;chains of DNA separate and mRNA nucleotides;line up with complementary bases;
Woodford High School 22
tRNA carries specific amino acid;to correct position on mRNA;anticodon and codon match; Max 5
[15]
Q4. (a) (i) transfer/carry genes from one organism to another/into bacteria/cells; 1
(ii) cut open plasmid;cut donor DNA, to remove gene/length of DNA;cut donor DNA and plasmid with the same enzyme/enzymethat cuts at the same base sequence;sticky ends/(overhanging) ends with, single strand/bases exposed;association/attachment/pairing of complementary strand; 2 max
Q5. (a) (i) Difficulty of finding one gene among all the genes in the nucleus / largeamounts of mRNA coding for insulin will be present in insulin producingcells / idea that mRNA will be ‘edited’ 1
(ii) Reverse transcriptase; 1
(iii) AGTTGG; 1
(b) Joins the gene for insulin into the plasmid; 1
(c) Allows transformed bacteria to be separated from non-transformed;Further detail e.g. transformed bacteria survivewhen antibiotic applied to medium; 2
[6]
Q6 (a) Locate with the use of a gene probe;Use restriction enzymes / endonucleases;To cut at specific base sequence;By hydrolysing / breaking sugar-phosphate bonds; max 2
(b) Same restriction enzymes;Cut at same base sequence in bacterial DNA;Leaving sticky ends / hydrogen bonds break;Join / splice with ligase;Use of plasmid; max 4
[6]
Q7. (a) circular DNA;separate from main bacteria] DNA;contains only a few genes; 2 max
(b) enzymes only cut DNA at specific base sequence/recognition site/specificpoint;sequence of bases/recognition site/specific point (on which enzyme acts)
Woodford High School 23
occurs once in plasmid and many times in human DNA;(max 1 if no reference to base sequence or recognition site) 2
(c) all cut DNA have same/complementary base sequence at ends orsame/complementary sticky ends;random process by which sticky ends join; 2
(d) (i) replica plating;use of pad/velvet surface to transfer bacteria;use of agar plate containing ampicillin/no tetracycline and agarplate containing tetracycline;in bacteria with human DNA tetracycline gene no longer functional/not resistant to tetracycline;bacteria with human DNA grow on plate with ampicillin/notetracycline but are killed by tetracycline;bacteria with no extra DNA in plasmid not killed, 4 max
(ii) use of gene probes;bacteria with insulin gene produce insulin; 1 max
[11]
Q8. Quality of written communication should be considered in crediting points in the marking scheme. In order to gain credit, answers must be expressed logically in clear, scientific terms.
(a) (i) Restriction; I endonuclease only 1
(ii) Ligase / ligating; 1
(b) Cut ends of DNA;one strand longer than the other / staggered cut / unpaired bases;Can attach to complementary DNA / bases; max 2
(d) Acts as a marker;Both genes on same plasmid / bacteria take up both genes;Reference to antibiotic;Kills bacteria without (either) gene / plasmid /only those with gene/plasmid will grow / survive; max 3
(e) ForMore effective than other methods;More humane method than shooting;Poisons may harm other animals;Prevent spread of disease / destruction of crops;Economic benefit to farmer / consumer; max 3
AgainstPlasmid / (either) gene may enter another species;May eradicate / greatly reduce possums;May sterilise other species;Disruption of food chain / change in numbers of a species;Not immediate / only affects next generation;Reference to antibiotic resistance in bacteria;Possums resistant to sterility protein; max 3
[15]
q.9(a) Endonuclease / restriction enzyme; 1
Woodford High School 24
(b) DNA made of base pairs;Each base pair is same length / occupies same distance along backbone; 2
(c) (i) Second blank box from left labelled 6; 1
(ii) Distance moved depends on length / number of base pairs /second longest fragment / second shortest distance identified; 1
(d) 5; 1[6]
Q10(a) Mother and father both heterozygotes / Tt / carriers;Probability of thalassaemia 1/4 and female 1/2;Probability of both 1/8; 3
(b) (i) Cut at same base sequence as same enzyme used;Fragments are same length / size / have same charge;Only differs by a single base; max 2
(ii) Single base occurs many times;Sequence of 20 unlikely to occur elsewhere; 2Allow one mark for establishing the principle where neither markingpoint clearly made.
[7]
Q11 positive:
less crops lost to insect damage/ diseases spread by insects;
can spray herbicide with no loss to crop/reduce competition from weeds;
more saleable product;
less use of insecticide;
possibly cheaper food; max 3
negative:
gene transfer to non-crop species;
consumer resistance to “un-natural” products;
transfer of genes into food chains/effect of food chains/examples;
creation of “plague” weeds/uneconomic plants;
excessive use of herbicides;
Reject disadvantages of selective breeding max 3[6]
Q12(a) (i) current switched on / fragments move due to electrical attraction;several hours to run;DNA transferred to nylon membrane / ‘southern blot’;(wrapped) photographic film placed on gel;film developed / radioactivity darkens film max 4
(ii) different lengths / mass; 1
(iii) e.g. progress towards cure for genetic diseases 1
(b) Quality of written communication.The answer to this part of the question requires continuous prose.
Woodford High School 25
To gain one mark for Quality of Written Communication these answers should be presented in clear, scientific English. Technical terminology should have been used effectively and should usually be accurate.
maximum 4 marks for generic genetic engineering techniques:human gene identifiedremoved from human DNA using endonuclease;detail e.g. sticky ends;same endonuclease;used to cut plasmid;role of ligase; max 4
combined with promoter sequence;gene / DNA (+ promoter sequence) injected into nucleus of fertilisedsheep egg;detail e.g. micropipette;embryo inserted into sheep uterus;enzyme obtained from sheep milk. Overall max 6
[12]
Q13(a) replace defective genes/treat genetic diseases with (healthy) genes; 1
(b) one amino acid missing/different/changed; 1
(c) (i) gene is expressed;healthy genes replicated with cells so not lost; 1 max
(ii) gamete cells are not affected/do not take up the healthy gene;still able to pass on the defective gene; 2
[5]
Q14(a) (i) heat (to about 90C) 1
(ii) primers / short nucleotide chains / RNA addedindividual (DNA) nucleotides then added;by (DNA) polymerase 3
(b) 2 double stranded molecules with original (white) and new (black) DNAstrands;2 double stranded molecules with new (black) DNA strands 2
(c) provides multiple copies of a DNA fragment;e.g. to analyse in forensic detection 2
